What does the C++ compiler do to ensure that different but adjacent memory locations are safe to be used on...
up vote
34
down vote
favorite
Lets say I have a struct:
struct Foo {
char a; // read and written to by thread 1 only
char b; // read and written to by thread 2 only
};
Now from what I understand, the C++ standard guarantees the safety of the above when two threads operate on the two different memory locations.
I would think though that, since char a and char b, fall within the same cache line, that the compiler has to do extra syncing.
What exactly happens here?
c++ multithreading thread-safety
|
show 1 more comment
up vote
34
down vote
favorite
Lets say I have a struct:
struct Foo {
char a; // read and written to by thread 1 only
char b; // read and written to by thread 2 only
};
Now from what I understand, the C++ standard guarantees the safety of the above when two threads operate on the two different memory locations.
I would think though that, since char a and char b, fall within the same cache line, that the compiler has to do extra syncing.
What exactly happens here?
c++ multithreading thread-safety
10
On a lot of platforms (for example, x86), it doesn't have to do anything. It just works (it means that the HW does the necessary extra stuff).
– geza
2 days ago
6
Yes. But the exact hit could vary on different generations/vendors of CPU. Do a search on "false sharing".
– geza
2 days ago
4
This is handled by the hardware, not the compiler, as far as I am aware. This is called false sharing
– NathanOliver
2 days ago
1
I think the only CPUs that C++ has actually been implemented on that the compiler would have to do anything special to support the C++ memory model are early Alpha CPUs which lacked instructions that could atomically set a single byte (or 16-bit) memory location. See Peter Cordes answer to a related question for details: stackoverflow.com/a/46818162/3826372 As far as know there's no compiler implementations that have been updated to support the C++11 memory model on these long obsolete Alpha CPUs.
– Ross Ridge
2 days ago
1
@RossRidge - well, there's the even more obsolete TMS9900 that has the same issue (only with 8-bit values, as it assumes 16-bit alignment) -- there's a port of gcc 3.4 to the architecture, but I don't know whether any particular C++ variant is supported. I'm also not aware of any extant dual-processor TMS9900 machines that would ever actually see this issue. The TMS9900 has instructions that operate on single bytes, but the bus implementation always fetches both bytes in a 16-bit word and rewrites the unchanged one.
– Jules
yesterday
|
show 1 more comment
up vote
34
down vote
favorite
up vote
34
down vote
favorite
Lets say I have a struct:
struct Foo {
char a; // read and written to by thread 1 only
char b; // read and written to by thread 2 only
};
Now from what I understand, the C++ standard guarantees the safety of the above when two threads operate on the two different memory locations.
I would think though that, since char a and char b, fall within the same cache line, that the compiler has to do extra syncing.
What exactly happens here?
c++ multithreading thread-safety
Lets say I have a struct:
struct Foo {
char a; // read and written to by thread 1 only
char b; // read and written to by thread 2 only
};
Now from what I understand, the C++ standard guarantees the safety of the above when two threads operate on the two different memory locations.
I would think though that, since char a and char b, fall within the same cache line, that the compiler has to do extra syncing.
What exactly happens here?
c++ multithreading thread-safety
c++ multithreading thread-safety
asked 2 days ago
Nathan Doromal
1,49111520
1,49111520
10
On a lot of platforms (for example, x86), it doesn't have to do anything. It just works (it means that the HW does the necessary extra stuff).
– geza
2 days ago
6
Yes. But the exact hit could vary on different generations/vendors of CPU. Do a search on "false sharing".
– geza
2 days ago
4
This is handled by the hardware, not the compiler, as far as I am aware. This is called false sharing
– NathanOliver
2 days ago
1
I think the only CPUs that C++ has actually been implemented on that the compiler would have to do anything special to support the C++ memory model are early Alpha CPUs which lacked instructions that could atomically set a single byte (or 16-bit) memory location. See Peter Cordes answer to a related question for details: stackoverflow.com/a/46818162/3826372 As far as know there's no compiler implementations that have been updated to support the C++11 memory model on these long obsolete Alpha CPUs.
– Ross Ridge
2 days ago
1
@RossRidge - well, there's the even more obsolete TMS9900 that has the same issue (only with 8-bit values, as it assumes 16-bit alignment) -- there's a port of gcc 3.4 to the architecture, but I don't know whether any particular C++ variant is supported. I'm also not aware of any extant dual-processor TMS9900 machines that would ever actually see this issue. The TMS9900 has instructions that operate on single bytes, but the bus implementation always fetches both bytes in a 16-bit word and rewrites the unchanged one.
– Jules
yesterday
|
show 1 more comment
10
On a lot of platforms (for example, x86), it doesn't have to do anything. It just works (it means that the HW does the necessary extra stuff).
– geza
2 days ago
6
Yes. But the exact hit could vary on different generations/vendors of CPU. Do a search on "false sharing".
– geza
2 days ago
4
This is handled by the hardware, not the compiler, as far as I am aware. This is called false sharing
– NathanOliver
2 days ago
1
I think the only CPUs that C++ has actually been implemented on that the compiler would have to do anything special to support the C++ memory model are early Alpha CPUs which lacked instructions that could atomically set a single byte (or 16-bit) memory location. See Peter Cordes answer to a related question for details: stackoverflow.com/a/46818162/3826372 As far as know there's no compiler implementations that have been updated to support the C++11 memory model on these long obsolete Alpha CPUs.
– Ross Ridge
2 days ago
1
@RossRidge - well, there's the even more obsolete TMS9900 that has the same issue (only with 8-bit values, as it assumes 16-bit alignment) -- there's a port of gcc 3.4 to the architecture, but I don't know whether any particular C++ variant is supported. I'm also not aware of any extant dual-processor TMS9900 machines that would ever actually see this issue. The TMS9900 has instructions that operate on single bytes, but the bus implementation always fetches both bytes in a 16-bit word and rewrites the unchanged one.
– Jules
yesterday
10
10
On a lot of platforms (for example, x86), it doesn't have to do anything. It just works (it means that the HW does the necessary extra stuff).
– geza
2 days ago
On a lot of platforms (for example, x86), it doesn't have to do anything. It just works (it means that the HW does the necessary extra stuff).
– geza
2 days ago
6
6
Yes. But the exact hit could vary on different generations/vendors of CPU. Do a search on "false sharing".
– geza
2 days ago
Yes. But the exact hit could vary on different generations/vendors of CPU. Do a search on "false sharing".
– geza
2 days ago
4
4
This is handled by the hardware, not the compiler, as far as I am aware. This is called false sharing
– NathanOliver
2 days ago
This is handled by the hardware, not the compiler, as far as I am aware. This is called false sharing
– NathanOliver
2 days ago
1
1
I think the only CPUs that C++ has actually been implemented on that the compiler would have to do anything special to support the C++ memory model are early Alpha CPUs which lacked instructions that could atomically set a single byte (or 16-bit) memory location. See Peter Cordes answer to a related question for details: stackoverflow.com/a/46818162/3826372 As far as know there's no compiler implementations that have been updated to support the C++11 memory model on these long obsolete Alpha CPUs.
– Ross Ridge
2 days ago
I think the only CPUs that C++ has actually been implemented on that the compiler would have to do anything special to support the C++ memory model are early Alpha CPUs which lacked instructions that could atomically set a single byte (or 16-bit) memory location. See Peter Cordes answer to a related question for details: stackoverflow.com/a/46818162/3826372 As far as know there's no compiler implementations that have been updated to support the C++11 memory model on these long obsolete Alpha CPUs.
– Ross Ridge
2 days ago
1
1
@RossRidge - well, there's the even more obsolete TMS9900 that has the same issue (only with 8-bit values, as it assumes 16-bit alignment) -- there's a port of gcc 3.4 to the architecture, but I don't know whether any particular C++ variant is supported. I'm also not aware of any extant dual-processor TMS9900 machines that would ever actually see this issue. The TMS9900 has instructions that operate on single bytes, but the bus implementation always fetches both bytes in a 16-bit word and rewrites the unchanged one.
– Jules
yesterday
@RossRidge - well, there's the even more obsolete TMS9900 that has the same issue (only with 8-bit values, as it assumes 16-bit alignment) -- there's a port of gcc 3.4 to the architecture, but I don't know whether any particular C++ variant is supported. I'm also not aware of any extant dual-processor TMS9900 machines that would ever actually see this issue. The TMS9900 has instructions that operate on single bytes, but the bus implementation always fetches both bytes in a 16-bit word and rewrites the unchanged one.
– Jules
yesterday
|
show 1 more comment
2 Answers
2
active
oldest
votes
up vote
30
down vote
accepted
This is hardware-dependent. On hardware I am familiar with, C++ doesn't have to do anything special, because from hardware perspective accessing different bytes even on a cached line is handled 'transparently'. From the hardware, this situation is not really different from
char a[2];
// or
char a, b;
In the cases above, we are talking about two adjacent objects, which are guaranteed to be independently accessible.
However, I've put 'transparently' in quotes for a reason. When you really have a case like that, you could be suffering (performance-wise) from a 'false sharing' - which happens when two (or more) threads access adjacent memory simultaneously and it ends up being cached in several CPU's caches. This leads to constant cache invalidation. In the real life, care should be taken to prevent this from happening when possible.
2
care should be taken to prevent this from happening when possible.
How would you suggest one going about doing that?
– ArtB
2 days ago
3
@ArtB there is no hard and fast rule. Designing program correctly from the scratch is always the best approach. You can also try profiling tools, such as valgrind and analyze the number of cache misses.
– SergeyA
2 days ago
1
@ArtB - can be done by adding padding members to structs to separate fields into different cache lines. There are plenty of papers/blog posts/etc out there discussing how to measure to see if you've got a problem and then how to ameliorate it.
– davidbak
2 days ago
@ArtB: C++17 provides interference sizes to help guide such design.
– Davis Herring
2 days ago
add a comment |
up vote
18
down vote
As others have explained, nothing in particular on common hardware. However, there is a catch: The compiler must refrain from performing certain optimizations, unless it can prove that other threads don't access the memory locations in question, e.g.:
std::array<std::uint8_t, 8u> c;
void f()
{
c[0] ^= 0xfa;
c[3] ^= 0x10;
c[6] ^= 0x8b;
c[7] ^= 0x92;
}
Here, in a single-threaded memory model, the compiler could emit code like the following (pseudo-assembly; assumes little-endian hardware):
load r0, *(std::uint64_t *) &c[0]
xor r0, 0x928b0000100000fa
store r0, *(std::uint64_t *) &c[0]
This is likely to be faster on common hardware than xor'ing the individual bytes. However, it reads and writes the unaffected (and unmentioned) elements of c
at indices 1, 2, 4 and 5. If other threads are writing to these memory locations concurrently, these changes could be overwritten.
For this reason, optimizations like these are often unusable in a multi-threaded memory model. As long as the compiler performs only loads and stores of matching length, or merges accesses only when there is no gap (e.g. the accesses to c[6]
and c[7]
can still be merged), the hardware commonly already provides the necessary guarantees for correct execution.
(That said, there are/have been some architectures with weak and counterintuitive memory order guarantees, e.g. DEC Alpha does not track pointers as a data dependency in the way that other architectures do, so it is necessary to introduce an explicit memory barrier in some cases, in low level code. There is a somewhat well-known little rant by Linus Torvalds on this issue. However, a conforming C++ implementation is expected to shield you from such issues.)
It's not the optimization that are unsafe in a multithread model, but the code itself. Unless the whole thing is protected by a mutex, code that reads and writes those locations from different threads is already invalid and whatever happens happens. This means that the compiler is indeed free to optimize this code with a single read-write operation because no valid code would be able to tell the difference.
– 6502
22 hours ago
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
30
down vote
accepted
This is hardware-dependent. On hardware I am familiar with, C++ doesn't have to do anything special, because from hardware perspective accessing different bytes even on a cached line is handled 'transparently'. From the hardware, this situation is not really different from
char a[2];
// or
char a, b;
In the cases above, we are talking about two adjacent objects, which are guaranteed to be independently accessible.
However, I've put 'transparently' in quotes for a reason. When you really have a case like that, you could be suffering (performance-wise) from a 'false sharing' - which happens when two (or more) threads access adjacent memory simultaneously and it ends up being cached in several CPU's caches. This leads to constant cache invalidation. In the real life, care should be taken to prevent this from happening when possible.
2
care should be taken to prevent this from happening when possible.
How would you suggest one going about doing that?
– ArtB
2 days ago
3
@ArtB there is no hard and fast rule. Designing program correctly from the scratch is always the best approach. You can also try profiling tools, such as valgrind and analyze the number of cache misses.
– SergeyA
2 days ago
1
@ArtB - can be done by adding padding members to structs to separate fields into different cache lines. There are plenty of papers/blog posts/etc out there discussing how to measure to see if you've got a problem and then how to ameliorate it.
– davidbak
2 days ago
@ArtB: C++17 provides interference sizes to help guide such design.
– Davis Herring
2 days ago
add a comment |
up vote
30
down vote
accepted
This is hardware-dependent. On hardware I am familiar with, C++ doesn't have to do anything special, because from hardware perspective accessing different bytes even on a cached line is handled 'transparently'. From the hardware, this situation is not really different from
char a[2];
// or
char a, b;
In the cases above, we are talking about two adjacent objects, which are guaranteed to be independently accessible.
However, I've put 'transparently' in quotes for a reason. When you really have a case like that, you could be suffering (performance-wise) from a 'false sharing' - which happens when two (or more) threads access adjacent memory simultaneously and it ends up being cached in several CPU's caches. This leads to constant cache invalidation. In the real life, care should be taken to prevent this from happening when possible.
2
care should be taken to prevent this from happening when possible.
How would you suggest one going about doing that?
– ArtB
2 days ago
3
@ArtB there is no hard and fast rule. Designing program correctly from the scratch is always the best approach. You can also try profiling tools, such as valgrind and analyze the number of cache misses.
– SergeyA
2 days ago
1
@ArtB - can be done by adding padding members to structs to separate fields into different cache lines. There are plenty of papers/blog posts/etc out there discussing how to measure to see if you've got a problem and then how to ameliorate it.
– davidbak
2 days ago
@ArtB: C++17 provides interference sizes to help guide such design.
– Davis Herring
2 days ago
add a comment |
up vote
30
down vote
accepted
up vote
30
down vote
accepted
This is hardware-dependent. On hardware I am familiar with, C++ doesn't have to do anything special, because from hardware perspective accessing different bytes even on a cached line is handled 'transparently'. From the hardware, this situation is not really different from
char a[2];
// or
char a, b;
In the cases above, we are talking about two adjacent objects, which are guaranteed to be independently accessible.
However, I've put 'transparently' in quotes for a reason. When you really have a case like that, you could be suffering (performance-wise) from a 'false sharing' - which happens when two (or more) threads access adjacent memory simultaneously and it ends up being cached in several CPU's caches. This leads to constant cache invalidation. In the real life, care should be taken to prevent this from happening when possible.
This is hardware-dependent. On hardware I am familiar with, C++ doesn't have to do anything special, because from hardware perspective accessing different bytes even on a cached line is handled 'transparently'. From the hardware, this situation is not really different from
char a[2];
// or
char a, b;
In the cases above, we are talking about two adjacent objects, which are guaranteed to be independently accessible.
However, I've put 'transparently' in quotes for a reason. When you really have a case like that, you could be suffering (performance-wise) from a 'false sharing' - which happens when two (or more) threads access adjacent memory simultaneously and it ends up being cached in several CPU's caches. This leads to constant cache invalidation. In the real life, care should be taken to prevent this from happening when possible.
edited 2 days ago
answered 2 days ago
SergeyA
40.2k53781
40.2k53781
2
care should be taken to prevent this from happening when possible.
How would you suggest one going about doing that?
– ArtB
2 days ago
3
@ArtB there is no hard and fast rule. Designing program correctly from the scratch is always the best approach. You can also try profiling tools, such as valgrind and analyze the number of cache misses.
– SergeyA
2 days ago
1
@ArtB - can be done by adding padding members to structs to separate fields into different cache lines. There are plenty of papers/blog posts/etc out there discussing how to measure to see if you've got a problem and then how to ameliorate it.
– davidbak
2 days ago
@ArtB: C++17 provides interference sizes to help guide such design.
– Davis Herring
2 days ago
add a comment |
2
care should be taken to prevent this from happening when possible.
How would you suggest one going about doing that?
– ArtB
2 days ago
3
@ArtB there is no hard and fast rule. Designing program correctly from the scratch is always the best approach. You can also try profiling tools, such as valgrind and analyze the number of cache misses.
– SergeyA
2 days ago
1
@ArtB - can be done by adding padding members to structs to separate fields into different cache lines. There are plenty of papers/blog posts/etc out there discussing how to measure to see if you've got a problem and then how to ameliorate it.
– davidbak
2 days ago
@ArtB: C++17 provides interference sizes to help guide such design.
– Davis Herring
2 days ago
2
2
care should be taken to prevent this from happening when possible.
How would you suggest one going about doing that?– ArtB
2 days ago
care should be taken to prevent this from happening when possible.
How would you suggest one going about doing that?– ArtB
2 days ago
3
3
@ArtB there is no hard and fast rule. Designing program correctly from the scratch is always the best approach. You can also try profiling tools, such as valgrind and analyze the number of cache misses.
– SergeyA
2 days ago
@ArtB there is no hard and fast rule. Designing program correctly from the scratch is always the best approach. You can also try profiling tools, such as valgrind and analyze the number of cache misses.
– SergeyA
2 days ago
1
1
@ArtB - can be done by adding padding members to structs to separate fields into different cache lines. There are plenty of papers/blog posts/etc out there discussing how to measure to see if you've got a problem and then how to ameliorate it.
– davidbak
2 days ago
@ArtB - can be done by adding padding members to structs to separate fields into different cache lines. There are plenty of papers/blog posts/etc out there discussing how to measure to see if you've got a problem and then how to ameliorate it.
– davidbak
2 days ago
@ArtB: C++17 provides interference sizes to help guide such design.
– Davis Herring
2 days ago
@ArtB: C++17 provides interference sizes to help guide such design.
– Davis Herring
2 days ago
add a comment |
up vote
18
down vote
As others have explained, nothing in particular on common hardware. However, there is a catch: The compiler must refrain from performing certain optimizations, unless it can prove that other threads don't access the memory locations in question, e.g.:
std::array<std::uint8_t, 8u> c;
void f()
{
c[0] ^= 0xfa;
c[3] ^= 0x10;
c[6] ^= 0x8b;
c[7] ^= 0x92;
}
Here, in a single-threaded memory model, the compiler could emit code like the following (pseudo-assembly; assumes little-endian hardware):
load r0, *(std::uint64_t *) &c[0]
xor r0, 0x928b0000100000fa
store r0, *(std::uint64_t *) &c[0]
This is likely to be faster on common hardware than xor'ing the individual bytes. However, it reads and writes the unaffected (and unmentioned) elements of c
at indices 1, 2, 4 and 5. If other threads are writing to these memory locations concurrently, these changes could be overwritten.
For this reason, optimizations like these are often unusable in a multi-threaded memory model. As long as the compiler performs only loads and stores of matching length, or merges accesses only when there is no gap (e.g. the accesses to c[6]
and c[7]
can still be merged), the hardware commonly already provides the necessary guarantees for correct execution.
(That said, there are/have been some architectures with weak and counterintuitive memory order guarantees, e.g. DEC Alpha does not track pointers as a data dependency in the way that other architectures do, so it is necessary to introduce an explicit memory barrier in some cases, in low level code. There is a somewhat well-known little rant by Linus Torvalds on this issue. However, a conforming C++ implementation is expected to shield you from such issues.)
It's not the optimization that are unsafe in a multithread model, but the code itself. Unless the whole thing is protected by a mutex, code that reads and writes those locations from different threads is already invalid and whatever happens happens. This means that the compiler is indeed free to optimize this code with a single read-write operation because no valid code would be able to tell the difference.
– 6502
22 hours ago
add a comment |
up vote
18
down vote
As others have explained, nothing in particular on common hardware. However, there is a catch: The compiler must refrain from performing certain optimizations, unless it can prove that other threads don't access the memory locations in question, e.g.:
std::array<std::uint8_t, 8u> c;
void f()
{
c[0] ^= 0xfa;
c[3] ^= 0x10;
c[6] ^= 0x8b;
c[7] ^= 0x92;
}
Here, in a single-threaded memory model, the compiler could emit code like the following (pseudo-assembly; assumes little-endian hardware):
load r0, *(std::uint64_t *) &c[0]
xor r0, 0x928b0000100000fa
store r0, *(std::uint64_t *) &c[0]
This is likely to be faster on common hardware than xor'ing the individual bytes. However, it reads and writes the unaffected (and unmentioned) elements of c
at indices 1, 2, 4 and 5. If other threads are writing to these memory locations concurrently, these changes could be overwritten.
For this reason, optimizations like these are often unusable in a multi-threaded memory model. As long as the compiler performs only loads and stores of matching length, or merges accesses only when there is no gap (e.g. the accesses to c[6]
and c[7]
can still be merged), the hardware commonly already provides the necessary guarantees for correct execution.
(That said, there are/have been some architectures with weak and counterintuitive memory order guarantees, e.g. DEC Alpha does not track pointers as a data dependency in the way that other architectures do, so it is necessary to introduce an explicit memory barrier in some cases, in low level code. There is a somewhat well-known little rant by Linus Torvalds on this issue. However, a conforming C++ implementation is expected to shield you from such issues.)
It's not the optimization that are unsafe in a multithread model, but the code itself. Unless the whole thing is protected by a mutex, code that reads and writes those locations from different threads is already invalid and whatever happens happens. This means that the compiler is indeed free to optimize this code with a single read-write operation because no valid code would be able to tell the difference.
– 6502
22 hours ago
add a comment |
up vote
18
down vote
up vote
18
down vote
As others have explained, nothing in particular on common hardware. However, there is a catch: The compiler must refrain from performing certain optimizations, unless it can prove that other threads don't access the memory locations in question, e.g.:
std::array<std::uint8_t, 8u> c;
void f()
{
c[0] ^= 0xfa;
c[3] ^= 0x10;
c[6] ^= 0x8b;
c[7] ^= 0x92;
}
Here, in a single-threaded memory model, the compiler could emit code like the following (pseudo-assembly; assumes little-endian hardware):
load r0, *(std::uint64_t *) &c[0]
xor r0, 0x928b0000100000fa
store r0, *(std::uint64_t *) &c[0]
This is likely to be faster on common hardware than xor'ing the individual bytes. However, it reads and writes the unaffected (and unmentioned) elements of c
at indices 1, 2, 4 and 5. If other threads are writing to these memory locations concurrently, these changes could be overwritten.
For this reason, optimizations like these are often unusable in a multi-threaded memory model. As long as the compiler performs only loads and stores of matching length, or merges accesses only when there is no gap (e.g. the accesses to c[6]
and c[7]
can still be merged), the hardware commonly already provides the necessary guarantees for correct execution.
(That said, there are/have been some architectures with weak and counterintuitive memory order guarantees, e.g. DEC Alpha does not track pointers as a data dependency in the way that other architectures do, so it is necessary to introduce an explicit memory barrier in some cases, in low level code. There is a somewhat well-known little rant by Linus Torvalds on this issue. However, a conforming C++ implementation is expected to shield you from such issues.)
As others have explained, nothing in particular on common hardware. However, there is a catch: The compiler must refrain from performing certain optimizations, unless it can prove that other threads don't access the memory locations in question, e.g.:
std::array<std::uint8_t, 8u> c;
void f()
{
c[0] ^= 0xfa;
c[3] ^= 0x10;
c[6] ^= 0x8b;
c[7] ^= 0x92;
}
Here, in a single-threaded memory model, the compiler could emit code like the following (pseudo-assembly; assumes little-endian hardware):
load r0, *(std::uint64_t *) &c[0]
xor r0, 0x928b0000100000fa
store r0, *(std::uint64_t *) &c[0]
This is likely to be faster on common hardware than xor'ing the individual bytes. However, it reads and writes the unaffected (and unmentioned) elements of c
at indices 1, 2, 4 and 5. If other threads are writing to these memory locations concurrently, these changes could be overwritten.
For this reason, optimizations like these are often unusable in a multi-threaded memory model. As long as the compiler performs only loads and stores of matching length, or merges accesses only when there is no gap (e.g. the accesses to c[6]
and c[7]
can still be merged), the hardware commonly already provides the necessary guarantees for correct execution.
(That said, there are/have been some architectures with weak and counterintuitive memory order guarantees, e.g. DEC Alpha does not track pointers as a data dependency in the way that other architectures do, so it is necessary to introduce an explicit memory barrier in some cases, in low level code. There is a somewhat well-known little rant by Linus Torvalds on this issue. However, a conforming C++ implementation is expected to shield you from such issues.)
answered 2 days ago
Arne Vogel
3,74011125
3,74011125
It's not the optimization that are unsafe in a multithread model, but the code itself. Unless the whole thing is protected by a mutex, code that reads and writes those locations from different threads is already invalid and whatever happens happens. This means that the compiler is indeed free to optimize this code with a single read-write operation because no valid code would be able to tell the difference.
– 6502
22 hours ago
add a comment |
It's not the optimization that are unsafe in a multithread model, but the code itself. Unless the whole thing is protected by a mutex, code that reads and writes those locations from different threads is already invalid and whatever happens happens. This means that the compiler is indeed free to optimize this code with a single read-write operation because no valid code would be able to tell the difference.
– 6502
22 hours ago
It's not the optimization that are unsafe in a multithread model, but the code itself. Unless the whole thing is protected by a mutex, code that reads and writes those locations from different threads is already invalid and whatever happens happens. This means that the compiler is indeed free to optimize this code with a single read-write operation because no valid code would be able to tell the difference.
– 6502
22 hours ago
It's not the optimization that are unsafe in a multithread model, but the code itself. Unless the whole thing is protected by a mutex, code that reads and writes those locations from different threads is already invalid and whatever happens happens. This means that the compiler is indeed free to optimize this code with a single read-write operation because no valid code would be able to tell the difference.
– 6502
22 hours ago
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10
On a lot of platforms (for example, x86), it doesn't have to do anything. It just works (it means that the HW does the necessary extra stuff).
– geza
2 days ago
6
Yes. But the exact hit could vary on different generations/vendors of CPU. Do a search on "false sharing".
– geza
2 days ago
4
This is handled by the hardware, not the compiler, as far as I am aware. This is called false sharing
– NathanOliver
2 days ago
1
I think the only CPUs that C++ has actually been implemented on that the compiler would have to do anything special to support the C++ memory model are early Alpha CPUs which lacked instructions that could atomically set a single byte (or 16-bit) memory location. See Peter Cordes answer to a related question for details: stackoverflow.com/a/46818162/3826372 As far as know there's no compiler implementations that have been updated to support the C++11 memory model on these long obsolete Alpha CPUs.
– Ross Ridge
2 days ago
1
@RossRidge - well, there's the even more obsolete TMS9900 that has the same issue (only with 8-bit values, as it assumes 16-bit alignment) -- there's a port of gcc 3.4 to the architecture, but I don't know whether any particular C++ variant is supported. I'm also not aware of any extant dual-processor TMS9900 machines that would ever actually see this issue. The TMS9900 has instructions that operate on single bytes, but the bus implementation always fetches both bytes in a 16-bit word and rewrites the unchanged one.
– Jules
yesterday