# define x=1+4; main() { int x; printf("%d%d",x/2,x/4); }

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# define x=1+4;
main()
{
int x;
printf("%d%d",x/2,x/4);
}

Question Posted / 123ghouse@gmail.com

5 Answers
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# define x=1+4; main() { int x; printf("%d%d",x/2,x/4); }..

Answer / karthik

the preprocessor directive is not written correctly

so u r going to get CE
#define x 1+4
void main()
{
printf("%d%d",x/2,x/4);

}
will work fine and give output as 32

Is This Answer Correct ?

4 Yes

2 No

# define x=1+4; main() { int x; printf("%d%d",x/2,x/4); }..

Answer / rama krishna sidhartha

The preprocessor syntax is wrongly written. It should be as
follows :

#define x 1+4

void main()
{
printf("%d%d",x/2,x/4);
}

There is no need of declaring the variable 'x' in
between 'main()' function since it is already declared
in '#define' directive.

The output will be : 3 and 2

Is This Answer Correct ?

2 Yes

0 No

# define x=1+4; main() { int x; printf("%d%d",x/2,x/4); }..

Answer / banavathvishnu

x/2 will become 1+4/2=3
x/4 will become 1+4/4 = 2

Is This Answer Correct ?

3 Yes

2 No

# define x=1+4; main() { int x; printf("%d%d",x/2,x/4); }..

Answer / 123ghouse@gmail.com

3,2

Is This Answer Correct ?

2 Yes

2 No

# define x=1+4; main() { int x; printf("%d%d",x/2,x/4); }..

Answer / sumalatha

ans is 4 4

Is This Answer Correct ?

0 Yes

1 No

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