#include <stdio.h>
#define SQRT(X)X*X
int main(){
int x=16/SQRT(4);
printf("%d",x);
return 0;
}
this output is 16 why? firstly i defined macro and then try do calculation
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Damien
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Ashan Tharaka
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1@AlexF, what `SQRT(2+2)` would be? – tstanisl Nov 23 '21 at 09:59
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may I ask why the word "square" is abbreviated as "SQRT"? There is no "t" there. – tstanisl Nov 23 '21 at 10:03
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1@tstanisl - Always keep your enemies guessing. Up is down, left is right, and squaring is SQRT. – StoryTeller - Unslander Monica Nov 23 '21 at 10:09
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You need to study how to write function-like macros in your beginner-level C book before using them. This here is a super-common FAQ that any half-decent book will address. – Lundin Nov 23 '21 at 10:09
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SQRT usually does something right opposite `sqrt(4) = 2`, `sqrt(9) = 3` This name is **taken**!!. Do not use it unless you want confuse the code readers.# – 0___________ Nov 23 '21 at 10:11
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Why do you want to use macro for that? – 0___________ Nov 23 '21 at 10:13
1 Answers
4
Currently int x=16/SQRT(4); will expand to int x=16/4*4; which is clearly 16. Use brackets to ensure the macro expansion is as intended:
#define SQRT(X) ((X)*(X))
Aside: SQRT is an odd name for the macro as that usually means "square root". SQR or SQUARE would be a more appropriate name.
kaylum
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I don't know C exactly, but in C++ this may still produce incorrect results (undefined behavior) if you use it like this: `int x = 2; int result = SQRT(x++);` Use a `constexpr` function instead and no macro at all. – Thomas Weller Jul 19 '23 at 07:42
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