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I had to print the value of a pointer: 

int *p = 0;
printf("%d", *p);

The code above throws an exception.

So I tried printf("%d", p) and that worked.

Why did it work only without the *?

Ardent Coder
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G0rdo1
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  • Does this answer your question? [How do pointer to pointers work in C?](https://stackoverflow.com/questions/897366/how-do-pointer-to-pointers-work-in-c) – Gerhard Mar 25 '20 at 09:19

1 Answers1

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When you derefererence the pointer p (as *p) you dereference a null pointer (you try to get the value where p is pointing, but it's not actually pointing anywhere). This leads to undefined behavior and very often a crash.

When you use plain p you print the contents of the pointer variable itself, not the value of where it's pointing. But that also leads to undefined behavior, because the %d format is to print an int value, not a int * value. Mismatching format specifier and argument type is UB.

Some programmer dude
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