f() { int a=2; f1(a++); } f1(int c) { printf("%d", c); } c=?

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f()
{
int a=2;
f1(a++);
}
f1(int c)
{
printf("%d", c);
}
c=?

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f() { int a=2; f1(a++); } f1(int c) { printf("%d", c); } c=? ..

Answer / vignesh1988i

c=2

Is This Answer Correct ?

16 Yes

0 No

f() { int a=2; f1(a++); } f1(int c) { printf("%d", c); } c=? ..

Answer / jack

in line 4, a's value is pushed on to stack and then
increaments.........In line 6 the function f1 pops the a's
value and assigned to c so .......c is value is 2.......

Is This Answer Correct ?

7 Yes

2 No

f() { int a=2; f1(a++); } f1(int c) { printf("%d", c); } c=? ..

Answer / viji

a++ is postfix operator so it first assigns its values and
then incremented. In above statement the value of a is first
assigned to c and then increment. so the output of c is 2;

Is This Answer Correct ?

1 Yes

0 No

f() { int a=2; f1(a++); } f1(int c) { printf("%d", c); } c=? ..

Answer / raj

f1 should have a prototype

Is This Answer Correct ?

1 Yes

0 No

f() { int a=2; f1(a++); } f1(int c) { printf("%d", c); } c=? ..

Answer / xx

no output

Is This Answer Correct ?

1 Yes

0 No

f() { int a=2; f1(a++); } f1(int c) { printf("%d", c); } c=? ..

Answer / jasna.c

no output

Is This Answer Correct ?

2 Yes

2 No

f() { int a=2; f1(a++); } f1(int c) { printf("%d", c); } c=? ..

Answer / banavathvishnu

a value will be first assigned to C in the Function and
latter it will be incremented in f() function.
so the value C remains 2

Is This Answer Correct ?

0 Yes

0 No

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