In C++14 is it valid to use a double in the dimension of a new expression?
In C++14 given the following code:
void foo() {
double d = 5.0;
auto p1 = new int[d];
}
clang compiles this without diagnostic while gcc on the other hand produces the following diagnostic (see it live in godbolt):
error: expression in new-declarator must have integral or enumeration type
7 | auto p1 = new int[d];
| ^
I specifically labeled this C++14 because in C++11 mode clang treats this as ill-formed and produces the following diagnostic (see it live in godbolt):
error: array size expression must have integral or unscoped enumeration type, not 'double'
auto p1 = new int[d];
^ ~
Is clang correct? If so what changed in C++14 to allow this?
c++ c++14 language-lawyer new-expression
asked Dec 12 '18 at 14:25
Shafik Yaghmour
125k23322532
|
show 4 more comments
In C++14 given the following code:
void foo() {
double d = 5.0;
auto p1 = new int[d];
}
clang compiles this without diagnostic while gcc on the other hand produces the following diagnostic (see it live in godbolt):
error: expression in new-declarator must have integral or enumeration type
7 | auto p1 = new int[d];
| ^
I specifically labeled this C++14 because in C++11 mode clang treats this as ill-formed and produces the following diagnostic (see it live in godbolt):
error: array size expression must have integral or unscoped enumeration type, not 'double'
auto p1 = new int[d];
^ ~
Is clang correct? If so what changed in C++14 to allow this?
c++ c++14 language-lawyer new-expression
asked Dec 12 '18 at 14:25
Shafik Yaghmour
125k23322532
2
Out of curiosity, a double allows for fractional quantities, so how would you allocate 0.75 of an integer, such asint * p_array = new int [0.75];? Or take something like 0.33333333, which is kind of difficult to allocate.
– Thomas Matthews
Dec 12 '18 at 15:13
5
@ThomasMatthews no this would end up being a float to integral conversion and would truncate the float
– Shafik Yaghmour
Dec 12 '18 at 15:59
1
@tomerzeitune Well, that (0.75 -> 0) is not what you wrote (0.75 -> 16360) ;)
– Bob__
Dec 13 '18 at 9:50
1
@tomerzeitune Whyint? Why notunsignedorlong?
– curiousguy
Dec 13 '18 at 22:35
1
@KeithThompson I think the commenters are joking around
– Shafik Yaghmour
Dec 14 '18 at 18:11
|
show 4 more comments
In C++14 given the following code:
void foo() {
double d = 5.0;
auto p1 = new int[d];
}
clang compiles this without diagnostic while gcc on the other hand produces the following diagnostic (see it live in godbolt):
error: expression in new-declarator must have integral or enumeration type
7 | auto p1 = new int[d];
| ^
I specifically labeled this C++14 because in C++11 mode clang treats this as ill-formed and produces the following diagnostic (see it live in godbolt):
error: array size expression must have integral or unscoped enumeration type, not 'double'
auto p1 = new int[d];
^ ~
Is clang correct? If so what changed in C++14 to allow this?
c++ c++14 language-lawyer new-expression
asked Dec 12 '18 at 14:25
Shafik Yaghmour
125k23322532
In C++14 given the following code:
void foo() {
double d = 5.0;
auto p1 = new int[d];
}
clang compiles this without diagnostic while gcc on the other hand produces the following diagnostic (see it live in godbolt):
error: expression in new-declarator must have integral or enumeration type
7 | auto p1 = new int[d];
| ^
I specifically labeled this C++14 because in C++11 mode clang treats this as ill-formed and produces the following diagnostic (see it live in godbolt):
error: array size expression must have integral or unscoped enumeration type, not 'double'
auto p1 = new int[d];
^ ~
Is clang correct? If so what changed in C++14 to allow this?
c++ c++14 language-lawyer new-expression
c++ c++14 language-lawyer new-expression
asked Dec 12 '18 at 14:25
Shafik Yaghmour
125k23322532
asked Dec 12 '18 at 14:25
Shafik Yaghmour
125k23322532
edited Dec 13 '18 at 21:10
asked Dec 12 '18 at 14:25
Shafik Yaghmour
125k23322532
asked Dec 12 '18 at 14:25
Shafik Yaghmour
125k23322532
asked Dec 12 '18 at 14:25
Shafik Yaghmour
125k23322532
125k23322532
2
Out of curiosity, a double allows for fractional quantities, so how would you allocate 0.75 of an integer, such asint * p_array = new int [0.75];? Or take something like 0.33333333, which is kind of difficult to allocate.
– Thomas Matthews
Dec 12 '18 at 15:13
5
@ThomasMatthews no this would end up being a float to integral conversion and would truncate the float
– Shafik Yaghmour
Dec 12 '18 at 15:59
1
@tomerzeitune Well, that (0.75 -> 0) is not what you wrote (0.75 -> 16360) ;)
– Bob__
Dec 13 '18 at 9:50
1
@tomerzeitune Whyint? Why notunsignedorlong?
– curiousguy
Dec 13 '18 at 22:35
1
@KeithThompson I think the commenters are joking around
– Shafik Yaghmour
Dec 14 '18 at 18:11
|
show 4 more comments
2
Out of curiosity, a double allows for fractional quantities, so how would you allocate 0.75 of an integer, such asint * p_array = new int [0.75];? Or take something like 0.33333333, which is kind of difficult to allocate.
– Thomas Matthews
Dec 12 '18 at 15:13
5
@ThomasMatthews no this would end up being a float to integral conversion and would truncate the float
– Shafik Yaghmour
Dec 12 '18 at 15:59
1
@tomerzeitune Well, that (0.75 -> 0) is not what you wrote (0.75 -> 16360) ;)
– Bob__
Dec 13 '18 at 9:50
1
@tomerzeitune Whyint? Why notunsignedorlong?
– curiousguy
Dec 13 '18 at 22:35
1
@KeithThompson I think the commenters are joking around
– Shafik Yaghmour
Dec 14 '18 at 18:11
2
2
Out of curiosity, a double allows for fractional quantities, so how would you allocate 0.75 of an integer, such as
int * p_array = new int [0.75];? Or take something like 0.33333333, which is kind of difficult to allocate.– Thomas Matthews
Dec 12 '18 at 15:13
Out of curiosity, a double allows for fractional quantities, so how would you allocate 0.75 of an integer, such as
int * p_array = new int [0.75];? Or take something like 0.33333333, which is kind of difficult to allocate.– Thomas Matthews
Dec 12 '18 at 15:13
5
5
@ThomasMatthews no this would end up being a float to integral conversion and would truncate the float
– Shafik Yaghmour
Dec 12 '18 at 15:59
@ThomasMatthews no this would end up being a float to integral conversion and would truncate the float
– Shafik Yaghmour
Dec 12 '18 at 15:59
1
1
@tomerzeitune Well, that (0.75 -> 0) is not what you wrote (0.75 -> 16360) ;)
– Bob__
Dec 13 '18 at 9:50
@tomerzeitune Well, that (0.75 -> 0) is not what you wrote (0.75 -> 16360) ;)
– Bob__
Dec 13 '18 at 9:50
1
1
@tomerzeitune Why
int? Why not unsigned or long?– curiousguy
Dec 13 '18 at 22:35
@tomerzeitune Why
int? Why not unsigned or long?– curiousguy
Dec 13 '18 at 22:35
1
1
@KeithThompson I think the commenters are joking around
– Shafik Yaghmour
Dec 14 '18 at 18:11
@KeithThompson I think the commenters are joking around
– Shafik Yaghmour
Dec 14 '18 at 18:11
|
show 4 more comments
2 Answers
2
active
oldest
votes
Clang is correct, the key wording in [expr.new]p6 changes from the following in the C++11 draft:
Every constant-expression in a noptr-new-declarator shall be an integral constant expression ([expr.const]) and evaluate to a strictly positive value. The expression in a noptr-new-declarator shall be of integral type, unscoped enumeration type, or a class type for which a single non-explicit conversion function to integral or unscoped enumeration type exists ([class.conv]). If the expression is of class type, the expression is converted by calling that conversion function, and the result of the conversion is used in place of the original expression. …
to this in the C++14 draft:
Every constant-expression in a noptr-new-declarator shall be a converted constant expression ([expr.const]) of type
std::size_tand shall evaluate to a strictly positive value. The expression in a noptr-new-declarator is implicitly converted tostd::size_t. …
In C++14 the requirement for the expression in a noptr-new-declarator was weakened to not require an integral, unscoped enumeration or a class with a
single non-explicit conversion function to one of those types but just allow implicit conversions to size_t.
The change in wording came from the proposal A Proposal to Tweak Certain C++ Contextual Conversions, v3.
answered Dec 12 '18 at 14:25
Shafik Yaghmour
125k23322532
19
I am dubious about the usefulness of allowingdoubleto be used as the size of an array... it seems more likely to let bugs pass silently than anything else :(
– Matthieu M.
Dec 12 '18 at 15:48
12
@MatthieuM. I agree, I believe it is a defect and that the intent was really to saycontextually implicitly converted. I am filing a defect report on this hopefully today but who knows maybe I am wrong :-(
– Shafik Yaghmour
Dec 12 '18 at 15:57
2
@MatthieuM. FYI I filed a defect report, processing takes a while so I don't expect to have an update anytime soon though.
– Shafik Yaghmour
Dec 14 '18 at 17:56
add a comment |
From C++14 to c++17 (for the ones that wonder like me), the phrasing remains practically the same (unlike from C++11 to C++14 as @ShafikYaghmour answered), as stated in this C++17 draft:
Every constant-expression in a noptr-new-declarator shall be a converted constant expression of type
std::size_tand shall evaluate to a strictly positive value. The expression in a noptr-new-declarator is implicitly converted tostd::size_t. [..]
with only this part ([expr.const]) missing from the C++17 draft.
answered Dec 14 '18 at 14:34
gsamaras
50.6k2399186
add a comment |
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Clang is correct, the key wording in [expr.new]p6 changes from the following in the C++11 draft:
Every constant-expression in a noptr-new-declarator shall be an integral constant expression ([expr.const]) and evaluate to a strictly positive value. The expression in a noptr-new-declarator shall be of integral type, unscoped enumeration type, or a class type for which a single non-explicit conversion function to integral or unscoped enumeration type exists ([class.conv]). If the expression is of class type, the expression is converted by calling that conversion function, and the result of the conversion is used in place of the original expression. …
to this in the C++14 draft:
Every constant-expression in a noptr-new-declarator shall be a converted constant expression ([expr.const]) of type
std::size_tand shall evaluate to a strictly positive value. The expression in a noptr-new-declarator is implicitly converted tostd::size_t. …
In C++14 the requirement for the expression in a noptr-new-declarator was weakened to not require an integral, unscoped enumeration or a class with a
single non-explicit conversion function to one of those types but just allow implicit conversions to size_t.
The change in wording came from the proposal A Proposal to Tweak Certain C++ Contextual Conversions, v3.
answered Dec 12 '18 at 14:25
Shafik Yaghmour
125k23322532
19
I am dubious about the usefulness of allowingdoubleto be used as the size of an array... it seems more likely to let bugs pass silently than anything else :(
– Matthieu M.
Dec 12 '18 at 15:48
12
@MatthieuM. I agree, I believe it is a defect and that the intent was really to saycontextually implicitly converted. I am filing a defect report on this hopefully today but who knows maybe I am wrong :-(
– Shafik Yaghmour
Dec 12 '18 at 15:57
2
@MatthieuM. FYI I filed a defect report, processing takes a while so I don't expect to have an update anytime soon though.
– Shafik Yaghmour
Dec 14 '18 at 17:56
add a comment |
Clang is correct, the key wording in [expr.new]p6 changes from the following in the C++11 draft:
Every constant-expression in a noptr-new-declarator shall be an integral constant expression ([expr.const]) and evaluate to a strictly positive value. The expression in a noptr-new-declarator shall be of integral type, unscoped enumeration type, or a class type for which a single non-explicit conversion function to integral or unscoped enumeration type exists ([class.conv]). If the expression is of class type, the expression is converted by calling that conversion function, and the result of the conversion is used in place of the original expression. …
to this in the C++14 draft:
Every constant-expression in a noptr-new-declarator shall be a converted constant expression ([expr.const]) of type
std::size_tand shall evaluate to a strictly positive value. The expression in a noptr-new-declarator is implicitly converted tostd::size_t. …
In C++14 the requirement for the expression in a noptr-new-declarator was weakened to not require an integral, unscoped enumeration or a class with a
single non-explicit conversion function to one of those types but just allow implicit conversions to size_t.
The change in wording came from the proposal A Proposal to Tweak Certain C++ Contextual Conversions, v3.
answered Dec 12 '18 at 14:25
Shafik Yaghmour
125k23322532
19
I am dubious about the usefulness of allowingdoubleto be used as the size of an array... it seems more likely to let bugs pass silently than anything else :(
– Matthieu M.
Dec 12 '18 at 15:48
12
@MatthieuM. I agree, I believe it is a defect and that the intent was really to saycontextually implicitly converted. I am filing a defect report on this hopefully today but who knows maybe I am wrong :-(
– Shafik Yaghmour
Dec 12 '18 at 15:57
2
@MatthieuM. FYI I filed a defect report, processing takes a while so I don't expect to have an update anytime soon though.
– Shafik Yaghmour
Dec 14 '18 at 17:56
add a comment |
Clang is correct, the key wording in [expr.new]p6 changes from the following in the C++11 draft:
Every constant-expression in a noptr-new-declarator shall be an integral constant expression ([expr.const]) and evaluate to a strictly positive value. The expression in a noptr-new-declarator shall be of integral type, unscoped enumeration type, or a class type for which a single non-explicit conversion function to integral or unscoped enumeration type exists ([class.conv]). If the expression is of class type, the expression is converted by calling that conversion function, and the result of the conversion is used in place of the original expression. …
to this in the C++14 draft:
Every constant-expression in a noptr-new-declarator shall be a converted constant expression ([expr.const]) of type
std::size_tand shall evaluate to a strictly positive value. The expression in a noptr-new-declarator is implicitly converted tostd::size_t. …
In C++14 the requirement for the expression in a noptr-new-declarator was weakened to not require an integral, unscoped enumeration or a class with a
single non-explicit conversion function to one of those types but just allow implicit conversions to size_t.
The change in wording came from the proposal A Proposal to Tweak Certain C++ Contextual Conversions, v3.
answered Dec 12 '18 at 14:25
Shafik Yaghmour
125k23322532
Clang is correct, the key wording in [expr.new]p6 changes from the following in the C++11 draft:
Every constant-expression in a noptr-new-declarator shall be an integral constant expression ([expr.const]) and evaluate to a strictly positive value. The expression in a noptr-new-declarator shall be of integral type, unscoped enumeration type, or a class type for which a single non-explicit conversion function to integral or unscoped enumeration type exists ([class.conv]). If the expression is of class type, the expression is converted by calling that conversion function, and the result of the conversion is used in place of the original expression. …
to this in the C++14 draft:
Every constant-expression in a noptr-new-declarator shall be a converted constant expression ([expr.const]) of type
std::size_tand shall evaluate to a strictly positive value. The expression in a noptr-new-declarator is implicitly converted tostd::size_t. …
In C++14 the requirement for the expression in a noptr-new-declarator was weakened to not require an integral, unscoped enumeration or a class with a
single non-explicit conversion function to one of those types but just allow implicit conversions to size_t.
The change in wording came from the proposal A Proposal to Tweak Certain C++ Contextual Conversions, v3.
answered Dec 12 '18 at 14:25
Shafik Yaghmour
125k23322532
edited Dec 15 '18 at 15:00
answered Dec 12 '18 at 14:25
Shafik Yaghmour
125k23322532
answered Dec 12 '18 at 14:25
Shafik Yaghmour
125k23322532
answered Dec 12 '18 at 14:25
Shafik Yaghmour
125k23322532
125k23322532
19
I am dubious about the usefulness of allowingdoubleto be used as the size of an array... it seems more likely to let bugs pass silently than anything else :(
– Matthieu M.
Dec 12 '18 at 15:48
12
@MatthieuM. I agree, I believe it is a defect and that the intent was really to saycontextually implicitly converted. I am filing a defect report on this hopefully today but who knows maybe I am wrong :-(
– Shafik Yaghmour
Dec 12 '18 at 15:57
2
@MatthieuM. FYI I filed a defect report, processing takes a while so I don't expect to have an update anytime soon though.
– Shafik Yaghmour
Dec 14 '18 at 17:56
add a comment |
19
I am dubious about the usefulness of allowingdoubleto be used as the size of an array... it seems more likely to let bugs pass silently than anything else :(
– Matthieu M.
Dec 12 '18 at 15:48
12
@MatthieuM. I agree, I believe it is a defect and that the intent was really to saycontextually implicitly converted. I am filing a defect report on this hopefully today but who knows maybe I am wrong :-(
– Shafik Yaghmour
Dec 12 '18 at 15:57
2
@MatthieuM. FYI I filed a defect report, processing takes a while so I don't expect to have an update anytime soon though.
– Shafik Yaghmour
Dec 14 '18 at 17:56
19
19
I am dubious about the usefulness of allowing
double to be used as the size of an array... it seems more likely to let bugs pass silently than anything else :(– Matthieu M.
Dec 12 '18 at 15:48
I am dubious about the usefulness of allowing
double to be used as the size of an array... it seems more likely to let bugs pass silently than anything else :(– Matthieu M.
Dec 12 '18 at 15:48
12
12
@MatthieuM. I agree, I believe it is a defect and that the intent was really to say
contextually implicitly converted. I am filing a defect report on this hopefully today but who knows maybe I am wrong :-(– Shafik Yaghmour
Dec 12 '18 at 15:57
@MatthieuM. I agree, I believe it is a defect and that the intent was really to say
contextually implicitly converted. I am filing a defect report on this hopefully today but who knows maybe I am wrong :-(– Shafik Yaghmour
Dec 12 '18 at 15:57
2
2
@MatthieuM. FYI I filed a defect report, processing takes a while so I don't expect to have an update anytime soon though.
– Shafik Yaghmour
Dec 14 '18 at 17:56
@MatthieuM. FYI I filed a defect report, processing takes a while so I don't expect to have an update anytime soon though.
– Shafik Yaghmour
Dec 14 '18 at 17:56
add a comment |
From C++14 to c++17 (for the ones that wonder like me), the phrasing remains practically the same (unlike from C++11 to C++14 as @ShafikYaghmour answered), as stated in this C++17 draft:
Every constant-expression in a noptr-new-declarator shall be a converted constant expression of type
std::size_tand shall evaluate to a strictly positive value. The expression in a noptr-new-declarator is implicitly converted tostd::size_t. [..]
with only this part ([expr.const]) missing from the C++17 draft.
answered Dec 14 '18 at 14:34
gsamaras
50.6k2399186
add a comment |
From C++14 to c++17 (for the ones that wonder like me), the phrasing remains practically the same (unlike from C++11 to C++14 as @ShafikYaghmour answered), as stated in this C++17 draft:
Every constant-expression in a noptr-new-declarator shall be a converted constant expression of type
std::size_tand shall evaluate to a strictly positive value. The expression in a noptr-new-declarator is implicitly converted tostd::size_t. [..]
with only this part ([expr.const]) missing from the C++17 draft.
answered Dec 14 '18 at 14:34
gsamaras
50.6k2399186
add a comment |
From C++14 to c++17 (for the ones that wonder like me), the phrasing remains practically the same (unlike from C++11 to C++14 as @ShafikYaghmour answered), as stated in this C++17 draft:
Every constant-expression in a noptr-new-declarator shall be a converted constant expression of type
std::size_tand shall evaluate to a strictly positive value. The expression in a noptr-new-declarator is implicitly converted tostd::size_t. [..]
with only this part ([expr.const]) missing from the C++17 draft.
answered Dec 14 '18 at 14:34
gsamaras
50.6k2399186
From C++14 to c++17 (for the ones that wonder like me), the phrasing remains practically the same (unlike from C++11 to C++14 as @ShafikYaghmour answered), as stated in this C++17 draft:
Every constant-expression in a noptr-new-declarator shall be a converted constant expression of type
std::size_tand shall evaluate to a strictly positive value. The expression in a noptr-new-declarator is implicitly converted tostd::size_t. [..]
with only this part ([expr.const]) missing from the C++17 draft.
answered Dec 14 '18 at 14:34
gsamaras
50.6k2399186
answered Dec 14 '18 at 14:34
gsamaras
50.6k2399186
answered Dec 14 '18 at 14:34
gsamaras
50.6k2399186
answered Dec 14 '18 at 14:34
gsamaras
50.6k2399186
50.6k2399186
add a comment |
add a comment |
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2
Out of curiosity, a double allows for fractional quantities, so how would you allocate 0.75 of an integer, such as
int * p_array = new int [0.75];? Or take something like 0.33333333, which is kind of difficult to allocate.– Thomas Matthews
Dec 12 '18 at 15:13
5
@ThomasMatthews no this would end up being a float to integral conversion and would truncate the float
– Shafik Yaghmour
Dec 12 '18 at 15:59
1
@tomerzeitune Well, that (0.75 -> 0) is not what you wrote (0.75 -> 16360) ;)
– Bob__
Dec 13 '18 at 9:50
1
@tomerzeitune Why
int? Why notunsignedorlong?– curiousguy
Dec 13 '18 at 22:35
1
@KeithThompson I think the commenters are joking around
– Shafik Yaghmour
Dec 14 '18 at 18:11