How could "printf (cout)" change uninitialized local variable?

Question

I made a mistake and used uninitialized local variable in a function and happened to find a single line of "cout" could change the uninitialized value. To simplify, I show the problem as follow.

#include <stdio.h>

void foo(void){
  int i;
  printf("%d\n", i);
  i = 777;
}
int main(void){
  foo();
  //cout << "hello!" << endl;
  foo();
  return 0;
}

The line I comment out will change the output from (unknown#;777) to (unknown#;hello!;0). Could someone help explain? Thanks.

You are printing out an uninitialized variable? What value do you think it could be - anything perhaps? — Ed Heal, Feb 21 '16 at 08:01
undefined behaviour cannot be defined - ie. looking at an uninitialised variable is looking at arbitrary memory - what is in that memory may depend on what code has executed previously but may or may not be repeatable depending on your environment — Matt Coubrough, Feb 21 '16 at 08:02
`i` is (at least by most compilers) placed on the stack, and you read a value that happens to be there. Function calls (for example output to cout) use the stack too, so it just happens cout changes the stack in that way. — Joachim Isaksson, Feb 21 '16 at 08:02
I get `777 777` and `777 hello 777` with Clang and -O3. It goes to show you that you can't even rely on the stack explanation. — chris, Feb 21 '16 at 08:13
@EdHeal he does say if he uncomments that line it changes the output of the program. He is learning. I probably made similar mistakes in 1995. — graham.reeds, Feb 21 '16 at 08:26

frogatto · Accepted Answer · 2016-02-21T08:18:41.477

This is because of stack behavior.

The first case:

int main(void){
  foo();
  foo();
  return 0;
}

The first foo() builds the stack and initialize the location of i with 777 and leave it as is. Then, the second call to foo() lays on the same stack structure where location of i is still 777. So, in this case printf prints 777.

The second case:

cout << "hello!" << endl is actually calling two functions:

cout.operator<<("hello!");
cout.operator<<(endl);

Which overwrites our previously created stack structure.

How could "printf (cout)" change uninitialized local variable?

1 Answers1