Strange behaviour when printing pointers

Question

I have the following code:

#include 

typedef struct {
  int* arg1;
  int  arg2;
} data;

int main(int argc, char** argv) {
  data d;
  printf         


        

Answer 1

You are accessing uninitialized values. This causes undefined behaviour (meaning that anything can happen, including the program crashing).

To fix, initialize values before using them, e.g.:

 data d = { 0 };

Answer 2

d is uninitialized, so it can contain anything.

Answer 3

You are experiencing this behavior because your program invokes undefined behavior. d is uinitialized in your program and hence its value is indeterminate and this invokes undefined behavior.

C11 6.7.9 Initialization:

If an object that has automatic storage duration is not initialized explicitly, its value is indeterminate.

Quoting from this answer:

An indeterminate value even more unspecified than unspecified. It's either an unspecified value or a trap representation. A trap representation is standards-speak for some magic value which, if you try to assign it to anything, results in undefined behavior. This wouldn't have to be an actual value; probably the best way to think about it is "if C had exceptions, a trap representation would be an exception". For example, if you declare int i; in a block, without an initialization, the initial value of the variable i is indeterminate, meaning that if you try to assign this to something else before initializing it, the behavior is undefined, and the compiler is entitled to try the said demons-out-of-nose trick. Of course, in most cases, the compiler will do something less dramatic/fun, like initialize it to 0 or some other random valid value, but no matter what it does, you're not entitled to object.

Answer 4

You should initialize d to get deterministic results:

data d = {0, 42};

or:

data d;
d.arg1 = 0;
d.arg2 = 42;

Without initialization, using the values leads to undefined behavior.