Why the output of the below program is 125 and not 5?
#include<stdio.h> #define square(x) x*x int main() { int var; var = 125/square(5); printf("%d",var); return 0; } 
- The answers give are correct but maybe you would want to understand that macros is C are only text substitutions not actual code (they are writing new code for you). Macros run before the compiler tries to compile your code and are part of what is called the pre-processor. As pointed out, avoid them for math or function things and use lots of parenthesis when in doubt.bd2357– bd23572020-05-14 00:56:32 +00:00Commented May 14, 2020 at 0:56
Add a comment |
2 Answers
This line:
var = 125/square(5); is expanded into:
var = 125/5*5; which evaluates from left to right to 25*5 and then to 125;
To fix it, parenthesize the argument in the definition of square, like this:
#define square(x) ((x)*(x)) Note also the extra parenthesis around x in order to achieve expected behavior when e.g. 1+2 is passed into square.
Comments
Note that var = 125/square(5); becomes var = 125/5*5 when you compile the code.
So the compiler calculates 125/5 before 5*5. The result becomes (125/5)*5 = 125.
Instead of #define square(x) x*x, put #define square(x) (x*x).
Here is your code:
#include<stdio.h> #define square(x) (x*x) int main() { int var; var = 125/square(5); printf("%d",var); return 0; } 
1 Comment
pm100
which is why you dont put functions in defines if you can help it