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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?
- 16 Answers
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Answer / 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 |
Answer / gg
Ans : 5
I agree with Divakar ans & similar answers.
sqr(x+1)=x+1*x+1=2+1*2+1=5, but not 2*2+1
| Is This Answer Correct ? | 17 Yes | 0 No |
Answer / pravin
sqrt(x+1)(x+1*x+1)
as x=2;result will be 2+1*2+1=5;
thank u
| Is This Answer Correct ? | 6 Yes | 1 No |
Answer / 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 |
Answer / fazlur
Macro will blindly substitutes, it doesn't calculate. So
remember whenever you encounter the macro, you first
substitute the value then calculate later. Ofcourse the
answer would be 5 in this case.
| Is This Answer Correct ? | 2 Yes | 0 No |
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