Predict the output of the below program
#define square(x) x*x int main() { int x; x = 36/square(6); printf("%d",x); getchar(); return 0; } |
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Output: 36
Explanation:
Preprocessor replaces square(6) by 6*6 and the expression becomes x = 36/6*6 and value of x is calculated as 36. If we want correct behavior from macro square(x), we should declare it as
#define square(x) ((x)*(x)) /* Note that the expression
(x*x) will also fail for square(6-2) */
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