#include <stdio.h>
#define sqr(x) (x*x)
int main()
{
int x=2;
printf("value of x=%d",sqr(x+1));
}
What is the value of x?
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Answer Posted / divakar
ouput :value of x=5
bcoz sqr(x+1)=(x+1*x+1) if u substitute x=2 u will get 5
since '*' is having more priority than '+'
at the of prog if u add prinf("%d",x); u will get 2
bcoz x value is not changed
| Is This Answer Correct ? | 32 Yes | 4 No |
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