local pointer variables

Question

What will be the output of the following program?

int *call();

void main() {
  int *ptr = call();
  printf("%d : %u",*ptr,ptr);
  clrscr();
  printf("%d",*ptr);
}

int *call() {
  int x = 25;
  ++x;
  //printf("%d : %u",x,&x);
  return &x;
}

Expected Output: Garbage value
Actual Output: 26 #someaddr

Since x is a local variable it's scope ends within the function call. I found this code as an example for dangling pointer.

The program exhibits undefined behavior, for the reason you mention. What is your question? You expected anything and you got something. — Pascal Cuoq
– Pascal Cuoq, Commented Oct 17, 2011 at 7:00

Jeegar Patel · Accepted Answer · 2011-10-17 07:02:00Z

2

its Undefined behaviour

since at x scope is dead after returning from call() so the pointer to that variable you can not use ahaed

BY COMPILING YOUR program you will get following error

warning: function returns address of local variable

if your program since give output 26 since its undefined behaviour. You should no do this at all.

answered Oct 17, 2011 at 7:02

Jeegar Patel

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Comments

codencandy · Accepted Answer · 2011-10-17 07:03:40Z

1

the output of this function is undefined. As you already pointed out the scope of x ends with the function. But the memory where 26 has been written is not used agian. So printing this value will give 26. If this memory is used again, it could be anything.

answered Oct 17, 2011 at 7:03

codencandy

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Comments

user379888 · Accepted Answer · 2011-10-17 10:10:41Z

0

Welcome, you have entered the Undefined Behavior valley. You can't predict what would be any value. Even if the value makes some sense, ignore it.

answered Oct 17, 2011 at 10:10

user379888

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