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#define d 10+10
main()
{
printf("%d",d*d);
}
- 6 Answers
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Answer / raj
ans.
d*d will be replaced by 10+10*10+10
during runtime.
so answer is 10+100+10 = 120
| Is This Answer Correct ? | 89 Yes | 0 No |
Answer / vrushali
This boils down to (10 +10 * 10 + 10)
so answer is 120 ... but if the same macro was rewritten as
#define d (10 + 10)
then d * d = (10 + 10 ) * (10 + 10)
= 20 * 20
= 400....
Pure macro concept....
| Is This Answer Correct ? | 15 Yes | 13 No |
Answer / jyothikrishna
Ex:1
#define A 10+10
void main()
{
int a;
a = A*A;
cout<<a<<endl;
}
Ans : 120
Explanation:
When you compile this program. A*A will be converted into 10+10*10+10. Here it follows BODMAS rule first it will perform multiplication and next it will perform addition.
First 10+10*10+10 = 10 + 100 + 10
next 10 + 100 + 10 = 120 (Answer)
Ex:2
#define A (10+10)
void main()
{
int a;
a = A*A;
cout<<a<<endl;
}
Ans : 400
Explanation:
When you compile this program. A*A will be converted into (10+10)*(10+10). Here it follows BODMAS rule first it will perform Bracket values and next it will perform multiplication.
First (10+10)*(10+10) = 20 * 20
next 20 * 20 = 400 (Answer)
| Is This Answer Correct ? | 2 Yes | 1 No |
Answer / mangal bhaldare
the value of d is 10+10
so the result is
(10+10)*(10+10)
=120
| Is This Answer Correct ? | 4 Yes | 4 No |
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