C's Undefined Behaviour

Question

I have this code:

#include <stdio.h>
#include <math.h>
int *fun();
int main()
{
    int *t;
    t=fun();
    printf("%d\n",*t);
    printf("%d\n",*t);
}
int *fun()
{
    int r=95;
    return(&r);
}

The output shown by this code in codeblocks is

95
-2

I don't get why the second printf() is printing a garbage value. Can someone explain that, please?

By editing `int fun()` --> `int *fun()` after answers have pointed that deficiency out, negates that value of the answers. Better to 1) revert the edit. 2) if needed, append a note, acknowledging that issue and need for continued help. — chux - Reinstate Monica, Nov 09 '15 at 14:51
Please tell us what warning/s you got when you compiled this. — Martin James, Nov 09 '15 at 17:13

Sourav Ghosh · Answer 1 · 2015-11-09T14:15:05.883

1

This is UB, because r is a local variable inside fun(), and if you return the address of a local variable and try to use it in the caller, you're ending up using it after is has expired it's lifetime. In C, it is defined as UB.

FWIW, the address of a variable is not always equivalent to an int or can be converted to a pointer safely, thus it should be int *, at least. In your code,

t=fun();

and

return(&r);

should have given you warnings!!

edited Nov 09 '15 at 14:15

answered Nov 09 '15 at 14:07

Sourav Ghosh

133,132
16
183
261

Also, returning a pointer as `int` is bad. – MikeCAT Nov 09 '15 at 14:08
@MikeCAT yes, right. – Sourav Ghosh Nov 09 '15 at 14:09
3

It actually shall be. http://port70.net/~nsz/c/c11/n1570.html#6.3.2.3p6 . Only two safe integers to take a pointer are `(u)intptr_t`. – too honest for this site Nov 09 '15 at 14:11
@Olaf added some clarifications to that. :) – Sourav Ghosh Nov 09 '15 at 14:19

score 0 · Answer 2 · edited May 23 '17 at 12:29

You return an int* from fun, even though it's declared to return int. Change the declaration to int* fun().
An empty list allows a variable number of arguments. Use int* fun(void); instead.
Undefined behavior is undefined behavior. You try to access memory of a local variable from outside the function, so anything might happen. From the level of the C standard, it's plainly undefined whether the first, second, or no statement at all behaves weirdly (like "printing a garbage value").

C's Undefined Behaviour

2 Answers2