Arduino: incorrect calculation of long integer
I'm doing a simple calculation with integers (on Arduino with ESP8266 12E), but I can't get the expected result and can't find the error. Can someone guide me?
#define A 200
#define B A * 62
#define C 500
void setup() {
Serial.begin(9600);
Serial.println("");
unsigned long aux = 0;
aux = (B * 500) / C; // (12400 * 500) / 500 = 12400
Serial.printf("aux = %dn", aux);
aux = aux * C; // 12400 * 500 = 6200000
Serial.printf("aux = %dn", aux);
// ERROR: Should result in "500", but is resulting in "1922000"
aux = aux / B; // 6200000 / 12400 = 500
Serial.printf("aux = %dn", aux); // It's printing "1922000"
}
esp8266
asked 2 days ago
wBB
1134
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This question came from our site for electronics and electrical engineering professionals, students, and enthusiasts.
add a comment |
I'm doing a simple calculation with integers (on Arduino with ESP8266 12E), but I can't get the expected result and can't find the error. Can someone guide me?
#define A 200
#define B A * 62
#define C 500
void setup() {
Serial.begin(9600);
Serial.println("");
unsigned long aux = 0;
aux = (B * 500) / C; // (12400 * 500) / 500 = 12400
Serial.printf("aux = %dn", aux);
aux = aux * C; // 12400 * 500 = 6200000
Serial.printf("aux = %dn", aux);
// ERROR: Should result in "500", but is resulting in "1922000"
aux = aux / B; // 6200000 / 12400 = 500
Serial.printf("aux = %dn", aux); // It's printing "1922000"
}
esp8266
asked 2 days ago
wBB
1134
migrated from electronics.stackexchange.com 2 days ago
This question came from our site for electronics and electrical engineering professionals, students, and enthusiasts.
add a comment |
I'm doing a simple calculation with integers (on Arduino with ESP8266 12E), but I can't get the expected result and can't find the error. Can someone guide me?
#define A 200
#define B A * 62
#define C 500
void setup() {
Serial.begin(9600);
Serial.println("");
unsigned long aux = 0;
aux = (B * 500) / C; // (12400 * 500) / 500 = 12400
Serial.printf("aux = %dn", aux);
aux = aux * C; // 12400 * 500 = 6200000
Serial.printf("aux = %dn", aux);
// ERROR: Should result in "500", but is resulting in "1922000"
aux = aux / B; // 6200000 / 12400 = 500
Serial.printf("aux = %dn", aux); // It's printing "1922000"
}
esp8266
asked 2 days ago
wBB
1134
I'm doing a simple calculation with integers (on Arduino with ESP8266 12E), but I can't get the expected result and can't find the error. Can someone guide me?
#define A 200
#define B A * 62
#define C 500
void setup() {
Serial.begin(9600);
Serial.println("");
unsigned long aux = 0;
aux = (B * 500) / C; // (12400 * 500) / 500 = 12400
Serial.printf("aux = %dn", aux);
aux = aux * C; // 12400 * 500 = 6200000
Serial.printf("aux = %dn", aux);
// ERROR: Should result in "500", but is resulting in "1922000"
aux = aux / B; // 6200000 / 12400 = 500
Serial.printf("aux = %dn", aux); // It's printing "1922000"
}
esp8266
esp8266
asked 2 days ago
wBB
1134
asked 2 days ago
wBB
1134
asked 2 days ago
wBB
1134
asked 2 days ago
wBB
1134
asked 2 days ago
wBB
1134
1134
migrated from electronics.stackexchange.com 2 days ago
This question came from our site for electronics and electrical engineering professionals, students, and enthusiasts.
migrated from electronics.stackexchange.com 2 days ago
This question came from our site for electronics and electrical engineering professionals, students, and enthusiasts.
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
In your #define of B you missed parenthesis (). Change your definition to:
#define B (A * 62)
Without parenthesis you first divide 6200000 by 200 and then multiply result by 62, which is not what you intend.
Dude, you're 100% right. I've spent several hours trying to figure out what was wrong ... Thank you so much!
– wBB
2 days ago
2
@wBB remember that in C, macros are replaced in the code, exactly as you wrote them, in the preprocessor step before the code gets compiled. So it helps as a sanity-check in these cases to expand the macros yourself in your code to see if you're getting what you intended.
– brhans
2 days ago
It really is was my lack of exeperience in C that led me to the problem. Thanks!
– wBB
2 days ago
add a comment |
Fully-parenthesizing macros (as noted in answer by dmz) solves one class of problem.
Another thing you should do is, in any arithmetic expression which involves literal constants, use the L suffix on at least one of the constants involved if there's any chance the result will exceed 32767 (the maximum guaranteed-representable value for int). The type of an arithmetic operation in C is based on the types of the operands of that operation only; the type of the variable to which the result is assigned is irrelevant.
answered 2 days ago
mlp
1311
2
...and the format specifier for anunsigned longis%lunot%d- the compiler for the asker's esp8266 uses a 32-bitintso they get away with some things they would not on an ATmega-based Arduino where anintis the minimum 16 bit size allowed by the specification.
– Chris Stratton
2 days ago
@mlp usually I use a typecast on almost everything. Example:int J = -1, typecast(unsigned char) J //prints 255. When you talk about the suffixL, what do you mean? Can you give an example?
– wBB
2 days ago
@ChrisStratton, thanks for explanation.
– wBB
2 days ago
add a comment |
As explained in previous answers, fully parenthesizing the macros is the
standard solution to this problem in C. However, on Arduino you are
programming in C++, and in C++ it is considered good practice to replace
this usage of #define by explicit constants:
const int A = 200;
const int B = A * 62;
const int C = 500;
Not only this makes the initial problem go away, it also provides some
type safety: you can choose to give these constants other types (e.g.
long) if appropriate.
answered 2 days ago
Edgar Bonet
24k22344
Some compilers will interpret "const" as a variable going into program memory which then in turn results in issues with memory spaces. Consts offer type protection, which can also be (partially) achieved by casting inside the define "#define A ((int)200)" for instance.
– le_top
yesterday
@le_top: Do you have a specific example of the kind of “issues with memory spaces” you can get? I doubt you could find an example that does not invoke undefined behavior.
– Edgar Bonet
yesterday
Something along these lines for example: "const int a=100; int b=200; void setB(const int *c) {b=c;} void ex1() {setB(&a);}" . But there are other cases. When "const" puts "a" in ROM, some embedded compilers can not cope with this kind of assignment and do not report all violating cases. So I tend to put "CONST" if there is a future risk for this. (with a "#define CONST const" if possible).
– le_top
17 hours ago
@le_top: I guess you meanb=*c. This sounds like a compiler bug to me. What compiler had issues with this? I tried your code on avr-gcc, and it had no issues, even when I replacedconstby__flash const, which has the effect of puttingain flash.
– Edgar Bonet
14 hours ago
Yes, b=*c. It will depend on the compiler and the uC. I file reports when I find bugs - when the compiler does not warn about it it is a bug, but otherwise it is a documented limitation.On Arduino you should use PROGMEM rather than __flash. I do not want to put a specific compilrer forward, my comment was mainly about warning that there are compilers for embedded systems that interpret const in a way that breaks code compatibility. That's also why you're required to add "__flash" to get the variable in FLASH - keeping it in RAM eases code generation (and speed) for those processors.
– le_top
12 mins ago
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
In your #define of B you missed parenthesis (). Change your definition to:
#define B (A * 62)
Without parenthesis you first divide 6200000 by 200 and then multiply result by 62, which is not what you intend.
Dude, you're 100% right. I've spent several hours trying to figure out what was wrong ... Thank you so much!
– wBB
2 days ago
2
@wBB remember that in C, macros are replaced in the code, exactly as you wrote them, in the preprocessor step before the code gets compiled. So it helps as a sanity-check in these cases to expand the macros yourself in your code to see if you're getting what you intended.
– brhans
2 days ago
It really is was my lack of exeperience in C that led me to the problem. Thanks!
– wBB
2 days ago
add a comment |
In your #define of B you missed parenthesis (). Change your definition to:
#define B (A * 62)
Without parenthesis you first divide 6200000 by 200 and then multiply result by 62, which is not what you intend.
Dude, you're 100% right. I've spent several hours trying to figure out what was wrong ... Thank you so much!
– wBB
2 days ago
2
@wBB remember that in C, macros are replaced in the code, exactly as you wrote them, in the preprocessor step before the code gets compiled. So it helps as a sanity-check in these cases to expand the macros yourself in your code to see if you're getting what you intended.
– brhans
2 days ago
It really is was my lack of exeperience in C that led me to the problem. Thanks!
– wBB
2 days ago
add a comment |
In your #define of B you missed parenthesis (). Change your definition to:
#define B (A * 62)
Without parenthesis you first divide 6200000 by 200 and then multiply result by 62, which is not what you intend.
In your #define of B you missed parenthesis (). Change your definition to:
#define B (A * 62)
Without parenthesis you first divide 6200000 by 200 and then multiply result by 62, which is not what you intend.
answered 2 days ago
dmz
Dude, you're 100% right. I've spent several hours trying to figure out what was wrong ... Thank you so much!
– wBB
2 days ago
2
@wBB remember that in C, macros are replaced in the code, exactly as you wrote them, in the preprocessor step before the code gets compiled. So it helps as a sanity-check in these cases to expand the macros yourself in your code to see if you're getting what you intended.
– brhans
2 days ago
It really is was my lack of exeperience in C that led me to the problem. Thanks!
– wBB
2 days ago
add a comment |
Dude, you're 100% right. I've spent several hours trying to figure out what was wrong ... Thank you so much!
– wBB
2 days ago
2
@wBB remember that in C, macros are replaced in the code, exactly as you wrote them, in the preprocessor step before the code gets compiled. So it helps as a sanity-check in these cases to expand the macros yourself in your code to see if you're getting what you intended.
– brhans
2 days ago
It really is was my lack of exeperience in C that led me to the problem. Thanks!
– wBB
2 days ago
Dude, you're 100% right. I've spent several hours trying to figure out what was wrong ... Thank you so much!
– wBB
2 days ago
Dude, you're 100% right. I've spent several hours trying to figure out what was wrong ... Thank you so much!
– wBB
2 days ago
2
2
@wBB remember that in C, macros are replaced in the code, exactly as you wrote them, in the preprocessor step before the code gets compiled. So it helps as a sanity-check in these cases to expand the macros yourself in your code to see if you're getting what you intended.
– brhans
2 days ago
@wBB remember that in C, macros are replaced in the code, exactly as you wrote them, in the preprocessor step before the code gets compiled. So it helps as a sanity-check in these cases to expand the macros yourself in your code to see if you're getting what you intended.
– brhans
2 days ago
It really is was my lack of exeperience in C that led me to the problem. Thanks!
– wBB
2 days ago
It really is was my lack of exeperience in C that led me to the problem. Thanks!
– wBB
2 days ago
add a comment |
Fully-parenthesizing macros (as noted in answer by dmz) solves one class of problem.
Another thing you should do is, in any arithmetic expression which involves literal constants, use the L suffix on at least one of the constants involved if there's any chance the result will exceed 32767 (the maximum guaranteed-representable value for int). The type of an arithmetic operation in C is based on the types of the operands of that operation only; the type of the variable to which the result is assigned is irrelevant.
answered 2 days ago
mlp
1311
2
...and the format specifier for anunsigned longis%lunot%d- the compiler for the asker's esp8266 uses a 32-bitintso they get away with some things they would not on an ATmega-based Arduino where anintis the minimum 16 bit size allowed by the specification.
– Chris Stratton
2 days ago
@mlp usually I use a typecast on almost everything. Example:int J = -1, typecast(unsigned char) J //prints 255. When you talk about the suffixL, what do you mean? Can you give an example?
– wBB
2 days ago
@ChrisStratton, thanks for explanation.
– wBB
2 days ago
add a comment |
Fully-parenthesizing macros (as noted in answer by dmz) solves one class of problem.
Another thing you should do is, in any arithmetic expression which involves literal constants, use the L suffix on at least one of the constants involved if there's any chance the result will exceed 32767 (the maximum guaranteed-representable value for int). The type of an arithmetic operation in C is based on the types of the operands of that operation only; the type of the variable to which the result is assigned is irrelevant.
answered 2 days ago
mlp
1311
2
...and the format specifier for anunsigned longis%lunot%d- the compiler for the asker's esp8266 uses a 32-bitintso they get away with some things they would not on an ATmega-based Arduino where anintis the minimum 16 bit size allowed by the specification.
– Chris Stratton
2 days ago
@mlp usually I use a typecast on almost everything. Example:int J = -1, typecast(unsigned char) J //prints 255. When you talk about the suffixL, what do you mean? Can you give an example?
– wBB
2 days ago
@ChrisStratton, thanks for explanation.
– wBB
2 days ago
add a comment |
Fully-parenthesizing macros (as noted in answer by dmz) solves one class of problem.
Another thing you should do is, in any arithmetic expression which involves literal constants, use the L suffix on at least one of the constants involved if there's any chance the result will exceed 32767 (the maximum guaranteed-representable value for int). The type of an arithmetic operation in C is based on the types of the operands of that operation only; the type of the variable to which the result is assigned is irrelevant.
answered 2 days ago
mlp
1311
Fully-parenthesizing macros (as noted in answer by dmz) solves one class of problem.
Another thing you should do is, in any arithmetic expression which involves literal constants, use the L suffix on at least one of the constants involved if there's any chance the result will exceed 32767 (the maximum guaranteed-representable value for int). The type of an arithmetic operation in C is based on the types of the operands of that operation only; the type of the variable to which the result is assigned is irrelevant.
answered 2 days ago
mlp
1311
answered 2 days ago
mlp
1311
answered 2 days ago
mlp
1311
answered 2 days ago
mlp
1311
1311
2
...and the format specifier for anunsigned longis%lunot%d- the compiler for the asker's esp8266 uses a 32-bitintso they get away with some things they would not on an ATmega-based Arduino where anintis the minimum 16 bit size allowed by the specification.
– Chris Stratton
2 days ago
@mlp usually I use a typecast on almost everything. Example:int J = -1, typecast(unsigned char) J //prints 255. When you talk about the suffixL, what do you mean? Can you give an example?
– wBB
2 days ago
@ChrisStratton, thanks for explanation.
– wBB
2 days ago
add a comment |
2
...and the format specifier for anunsigned longis%lunot%d- the compiler for the asker's esp8266 uses a 32-bitintso they get away with some things they would not on an ATmega-based Arduino where anintis the minimum 16 bit size allowed by the specification.
– Chris Stratton
2 days ago
@mlp usually I use a typecast on almost everything. Example:int J = -1, typecast(unsigned char) J //prints 255. When you talk about the suffixL, what do you mean? Can you give an example?
– wBB
2 days ago
@ChrisStratton, thanks for explanation.
– wBB
2 days ago
2
2
...and the format specifier for an
unsigned long is %lu not %d - the compiler for the asker's esp8266 uses a 32-bit int so they get away with some things they would not on an ATmega-based Arduino where an int is the minimum 16 bit size allowed by the specification.– Chris Stratton
2 days ago
...and the format specifier for an
unsigned long is %lu not %d - the compiler for the asker's esp8266 uses a 32-bit int so they get away with some things they would not on an ATmega-based Arduino where an int is the minimum 16 bit size allowed by the specification.– Chris Stratton
2 days ago
@mlp usually I use a typecast on almost everything. Example:
int J = -1, typecast (unsigned char) J //prints 255. When you talk about the suffix L, what do you mean? Can you give an example?– wBB
2 days ago
@mlp usually I use a typecast on almost everything. Example:
int J = -1, typecast (unsigned char) J //prints 255. When you talk about the suffix L, what do you mean? Can you give an example?– wBB
2 days ago
@ChrisStratton, thanks for explanation.
– wBB
2 days ago
@ChrisStratton, thanks for explanation.
– wBB
2 days ago
add a comment |
As explained in previous answers, fully parenthesizing the macros is the
standard solution to this problem in C. However, on Arduino you are
programming in C++, and in C++ it is considered good practice to replace
this usage of #define by explicit constants:
const int A = 200;
const int B = A * 62;
const int C = 500;
Not only this makes the initial problem go away, it also provides some
type safety: you can choose to give these constants other types (e.g.
long) if appropriate.
answered 2 days ago
Edgar Bonet
24k22344
Some compilers will interpret "const" as a variable going into program memory which then in turn results in issues with memory spaces. Consts offer type protection, which can also be (partially) achieved by casting inside the define "#define A ((int)200)" for instance.
– le_top
yesterday
@le_top: Do you have a specific example of the kind of “issues with memory spaces” you can get? I doubt you could find an example that does not invoke undefined behavior.
– Edgar Bonet
yesterday
Something along these lines for example: "const int a=100; int b=200; void setB(const int *c) {b=c;} void ex1() {setB(&a);}" . But there are other cases. When "const" puts "a" in ROM, some embedded compilers can not cope with this kind of assignment and do not report all violating cases. So I tend to put "CONST" if there is a future risk for this. (with a "#define CONST const" if possible).
– le_top
17 hours ago
@le_top: I guess you meanb=*c. This sounds like a compiler bug to me. What compiler had issues with this? I tried your code on avr-gcc, and it had no issues, even when I replacedconstby__flash const, which has the effect of puttingain flash.
– Edgar Bonet
14 hours ago
Yes, b=*c. It will depend on the compiler and the uC. I file reports when I find bugs - when the compiler does not warn about it it is a bug, but otherwise it is a documented limitation.On Arduino you should use PROGMEM rather than __flash. I do not want to put a specific compilrer forward, my comment was mainly about warning that there are compilers for embedded systems that interpret const in a way that breaks code compatibility. That's also why you're required to add "__flash" to get the variable in FLASH - keeping it in RAM eases code generation (and speed) for those processors.
– le_top
12 mins ago
add a comment |
As explained in previous answers, fully parenthesizing the macros is the
standard solution to this problem in C. However, on Arduino you are
programming in C++, and in C++ it is considered good practice to replace
this usage of #define by explicit constants:
const int A = 200;
const int B = A * 62;
const int C = 500;
Not only this makes the initial problem go away, it also provides some
type safety: you can choose to give these constants other types (e.g.
long) if appropriate.
answered 2 days ago
Edgar Bonet
24k22344
Some compilers will interpret "const" as a variable going into program memory which then in turn results in issues with memory spaces. Consts offer type protection, which can also be (partially) achieved by casting inside the define "#define A ((int)200)" for instance.
– le_top
yesterday
@le_top: Do you have a specific example of the kind of “issues with memory spaces” you can get? I doubt you could find an example that does not invoke undefined behavior.
– Edgar Bonet
yesterday
Something along these lines for example: "const int a=100; int b=200; void setB(const int *c) {b=c;} void ex1() {setB(&a);}" . But there are other cases. When "const" puts "a" in ROM, some embedded compilers can not cope with this kind of assignment and do not report all violating cases. So I tend to put "CONST" if there is a future risk for this. (with a "#define CONST const" if possible).
– le_top
17 hours ago
@le_top: I guess you meanb=*c. This sounds like a compiler bug to me. What compiler had issues with this? I tried your code on avr-gcc, and it had no issues, even when I replacedconstby__flash const, which has the effect of puttingain flash.
– Edgar Bonet
14 hours ago
Yes, b=*c. It will depend on the compiler and the uC. I file reports when I find bugs - when the compiler does not warn about it it is a bug, but otherwise it is a documented limitation.On Arduino you should use PROGMEM rather than __flash. I do not want to put a specific compilrer forward, my comment was mainly about warning that there are compilers for embedded systems that interpret const in a way that breaks code compatibility. That's also why you're required to add "__flash" to get the variable in FLASH - keeping it in RAM eases code generation (and speed) for those processors.
– le_top
12 mins ago
add a comment |
As explained in previous answers, fully parenthesizing the macros is the
standard solution to this problem in C. However, on Arduino you are
programming in C++, and in C++ it is considered good practice to replace
this usage of #define by explicit constants:
const int A = 200;
const int B = A * 62;
const int C = 500;
Not only this makes the initial problem go away, it also provides some
type safety: you can choose to give these constants other types (e.g.
long) if appropriate.
answered 2 days ago
Edgar Bonet
24k22344
As explained in previous answers, fully parenthesizing the macros is the
standard solution to this problem in C. However, on Arduino you are
programming in C++, and in C++ it is considered good practice to replace
this usage of #define by explicit constants:
const int A = 200;
const int B = A * 62;
const int C = 500;
Not only this makes the initial problem go away, it also provides some
type safety: you can choose to give these constants other types (e.g.
long) if appropriate.
answered 2 days ago
Edgar Bonet
24k22344
answered 2 days ago
Edgar Bonet
24k22344
answered 2 days ago
Edgar Bonet
24k22344
answered 2 days ago
Edgar Bonet
24k22344
24k22344
Some compilers will interpret "const" as a variable going into program memory which then in turn results in issues with memory spaces. Consts offer type protection, which can also be (partially) achieved by casting inside the define "#define A ((int)200)" for instance.
– le_top
yesterday
@le_top: Do you have a specific example of the kind of “issues with memory spaces” you can get? I doubt you could find an example that does not invoke undefined behavior.
– Edgar Bonet
yesterday
Something along these lines for example: "const int a=100; int b=200; void setB(const int *c) {b=c;} void ex1() {setB(&a);}" . But there are other cases. When "const" puts "a" in ROM, some embedded compilers can not cope with this kind of assignment and do not report all violating cases. So I tend to put "CONST" if there is a future risk for this. (with a "#define CONST const" if possible).
– le_top
17 hours ago
@le_top: I guess you meanb=*c. This sounds like a compiler bug to me. What compiler had issues with this? I tried your code on avr-gcc, and it had no issues, even when I replacedconstby__flash const, which has the effect of puttingain flash.
– Edgar Bonet
14 hours ago
Yes, b=*c. It will depend on the compiler and the uC. I file reports when I find bugs - when the compiler does not warn about it it is a bug, but otherwise it is a documented limitation.On Arduino you should use PROGMEM rather than __flash. I do not want to put a specific compilrer forward, my comment was mainly about warning that there are compilers for embedded systems that interpret const in a way that breaks code compatibility. That's also why you're required to add "__flash" to get the variable in FLASH - keeping it in RAM eases code generation (and speed) for those processors.
– le_top
12 mins ago
add a comment |
Some compilers will interpret "const" as a variable going into program memory which then in turn results in issues with memory spaces. Consts offer type protection, which can also be (partially) achieved by casting inside the define "#define A ((int)200)" for instance.
– le_top
yesterday
@le_top: Do you have a specific example of the kind of “issues with memory spaces” you can get? I doubt you could find an example that does not invoke undefined behavior.
– Edgar Bonet
yesterday
Something along these lines for example: "const int a=100; int b=200; void setB(const int *c) {b=c;} void ex1() {setB(&a);}" . But there are other cases. When "const" puts "a" in ROM, some embedded compilers can not cope with this kind of assignment and do not report all violating cases. So I tend to put "CONST" if there is a future risk for this. (with a "#define CONST const" if possible).
– le_top
17 hours ago
@le_top: I guess you meanb=*c. This sounds like a compiler bug to me. What compiler had issues with this? I tried your code on avr-gcc, and it had no issues, even when I replacedconstby__flash const, which has the effect of puttingain flash.
– Edgar Bonet
14 hours ago
Yes, b=*c. It will depend on the compiler and the uC. I file reports when I find bugs - when the compiler does not warn about it it is a bug, but otherwise it is a documented limitation.On Arduino you should use PROGMEM rather than __flash. I do not want to put a specific compilrer forward, my comment was mainly about warning that there are compilers for embedded systems that interpret const in a way that breaks code compatibility. That's also why you're required to add "__flash" to get the variable in FLASH - keeping it in RAM eases code generation (and speed) for those processors.
– le_top
12 mins ago
Some compilers will interpret "const" as a variable going into program memory which then in turn results in issues with memory spaces. Consts offer type protection, which can also be (partially) achieved by casting inside the define "#define A ((int)200)" for instance.
– le_top
yesterday
Some compilers will interpret "const" as a variable going into program memory which then in turn results in issues with memory spaces. Consts offer type protection, which can also be (partially) achieved by casting inside the define "#define A ((int)200)" for instance.
– le_top
yesterday
@le_top: Do you have a specific example of the kind of “issues with memory spaces” you can get? I doubt you could find an example that does not invoke undefined behavior.
– Edgar Bonet
yesterday
@le_top: Do you have a specific example of the kind of “issues with memory spaces” you can get? I doubt you could find an example that does not invoke undefined behavior.
– Edgar Bonet
yesterday
Something along these lines for example: "const int a=100; int b=200; void setB(const int *c) {b=c;} void ex1() {setB(&a);}" . But there are other cases. When "const" puts "a" in ROM, some embedded compilers can not cope with this kind of assignment and do not report all violating cases. So I tend to put "CONST" if there is a future risk for this. (with a "#define CONST const" if possible).
– le_top
17 hours ago
Something along these lines for example: "const int a=100; int b=200; void setB(const int *c) {b=c;} void ex1() {setB(&a);}" . But there are other cases. When "const" puts "a" in ROM, some embedded compilers can not cope with this kind of assignment and do not report all violating cases. So I tend to put "CONST" if there is a future risk for this. (with a "#define CONST const" if possible).
– le_top
17 hours ago
@le_top: I guess you mean
b=*c. This sounds like a compiler bug to me. What compiler had issues with this? I tried your code on avr-gcc, and it had no issues, even when I replaced const by __flash const, which has the effect of putting a in flash.– Edgar Bonet
14 hours ago
@le_top: I guess you mean
b=*c. This sounds like a compiler bug to me. What compiler had issues with this? I tried your code on avr-gcc, and it had no issues, even when I replaced const by __flash const, which has the effect of putting a in flash.– Edgar Bonet
14 hours ago
Yes, b=*c. It will depend on the compiler and the uC. I file reports when I find bugs - when the compiler does not warn about it it is a bug, but otherwise it is a documented limitation.On Arduino you should use PROGMEM rather than __flash. I do not want to put a specific compilrer forward, my comment was mainly about warning that there are compilers for embedded systems that interpret const in a way that breaks code compatibility. That's also why you're required to add "__flash" to get the variable in FLASH - keeping it in RAM eases code generation (and speed) for those processors.
– le_top
12 mins ago
Yes, b=*c. It will depend on the compiler and the uC. I file reports when I find bugs - when the compiler does not warn about it it is a bug, but otherwise it is a documented limitation.On Arduino you should use PROGMEM rather than __flash. I do not want to put a specific compilrer forward, my comment was mainly about warning that there are compilers for embedded systems that interpret const in a way that breaks code compatibility. That's also why you're required to add "__flash" to get the variable in FLASH - keeping it in RAM eases code generation (and speed) for those processors.
– le_top
12 mins ago
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