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#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 / jugal
Sorry guys,
my bad,
i thought it was
#define sqr(x) ((x)*(x))
the output wud be 5
but still the value of will be 2 only
| Is This Answer Correct ? | 2 Yes | 0 No |
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