#define swap1(a,b) a=a+b;b=a-b;a=a-b; main() { int x=5,y=10; swap1(x,y);

#define swap1(a,b) a=a+b;b=a-b;a=a-b;
main()
{
int x=5,y=10;
swap1(x,y);
printf("%d %d\n",x,y);
swap2(x,y);
printf("%d %d\n",x,y);
}

int swap2(int a,int b)
{
int temp;
temp=a;
b=a;
a=temp;
return;
}

what are the outputs?

Question Posted / naveed

4 Answers
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#define swap1(a,b) a=a+b;b=a-b;a=a-b; main() { int x=5,y=10; swap1(x,y); printf("%d %d\..

Answer / goodhunter

10 5
10 5

Is This Answer Correct ?

19 Yes

2 No

#define swap1(a,b) a=a+b;b=a-b;a=a-b; main() { int x=5,y=10; swap1(x,y); printf("%d %d\..

Answer / jaroosh

Result of the above program will probably be sth like :
compile error `return' with no value, in function returning
non-void
or
function swap2(...) should return a non-void value.
thats because from the erroneous code YOU CANT PREDICT what:
return;
in swap2 function was about to return.
It may sound that Im picking, but as an interviewer myself,
I have to say it is CRUCIAL on an interview to pinpoint
errors in the code, NEVER assume that its just a
misspelling, some of those errors are, some of them aren't
and are there to check if you read code thoroughly, its
always better to point such things.

Assuming the code was right and the swap2 signature was
void swap2(int a, int b)
code result will be :
10 5
10 5
switching values of a and b in swap2 doesnt affect x and y
values in program because they are being passed BY VALUE to
swap2.

Is This Answer Correct ?

7 Yes

2 No

#define swap1(a,b) a=a+b;b=a-b;a=a-b; main() { int x=5,y=10; swap1(x,y); printf("%d %d\..

Answer / mannucse

10 5
5 10

Is This Answer Correct ?

9 Yes

10 No

#define swap1(a,b) a=a+b;b=a-b;a=a-b; main() { int x=5,y=10; swap1(x,y); printf("%d %d\..

Answer / chitra

10 5
5 5

Is This Answer Correct ?

3 Yes

9 No

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