6)swap(int x,y) { int temp; temp=x; x=y; y=temp; } main() { int x=2;y=3

6)swap(int x,y)
{
int temp;
temp=x;
x=y;
y=temp;
}
main()
{
int x=2;y=3;
swap(x,y);
}
after calling swap ,what are yhe values x&y?

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6)swap(int x,y) { int temp; temp=x; x=y; y=temp; } main() { int x=2;y=3; swap(x,y); } af..

Answer / selloorhari

After calling the function swap(), the values of x,y will be
the same.

i.e. x = 2, y = 3.

The scope of the variables x,y,temp in the swap() function
lies inside the function swap() itself. So there will not be
any change in the values of x,y in the main() function..

Is This Answer Correct ?

17 Yes

0 No

6)swap(int x,y) { int temp; temp=x; x=y; y=temp; } main() { int x=2;y=3; swap(x,y); } af..

Answer / shruti

the values will be x =2 and y = 3.

the variables x and y declared in main() are local to main.

whereas

the variables x and y declared in swap() are local to swap..

the change in the value of the variables in either
function will have zero effect on the other function.

Hence the value remains teh same.

Is This Answer Correct ?

16 Yes

1 No

6)swap(int x,y) { int temp; temp=x; x=y; y=temp; } main() { int x=2;y=3; swap(x,y); } af..

Answer / shankar

Value remain same hare....I.E :- X=2,y=3.Because here rules of call by value is applied. If we use call by reference the only value is changed .

Is This Answer Correct ?

1 Yes

0 No

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