#define d 10+10 main() { printf("%d",d*d); }

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#define d 10+10
main()
{
printf("%d",d*d);
}

Question Posted / guest

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#define d 10+10 main() { printf("%d",d*d); }..

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

#define d 10+10 main() { printf("%d",d*d); }..

Answer / hussain reddy

120

Is This Answer Correct ?

8 Yes

2 No

#define d 10+10 main() { printf("%d",d*d); }..

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

#define d 10+10 main() { printf("%d",d*d); }..

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

#define d 10+10 main() { printf("%d",d*d); }..

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

#define d 10+10 main() { printf("%d",d*d); }..

Answer / kumaran

400

Is This Answer Correct ?

3 Yes

46 No

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