In C programming, using the `%p` format specifier to print pointers is the standard practice for outputting the address of a pointer variable. According to the C language standard (e.g., ISO/IEC 9899), when using the `printf` function with the `%p` format specifier to print a pointer, a `void*` pointer must be passed.

Regarding null pointers (typically defined as `NULL`), the standard specifies that a `void*` pointer should be passed when using `%p` for printing. Although `NULL` represents an invalid address, using `%p` to print it is well-defined behavior. Typically, printing a `NULL` pointer yields results such as `(nil)` or `0x0`, depending on the specific implementation and platform.

Example: The following code snippet demonstrates how to safely print a null pointer in a C program:
```c
#include <stdio.h>
int main() {
    void *ptr = NULL;
    printf("Pointer address: %p\n", ptr);
    return 0;
}
```

In this example, `ptr` is a null pointer initialized to `NULL`. When printed using `%p`, the output is expected to be either `(nil)` or `0x0`, depending entirely on the compiler and runtime platform implementation.

To summarize, using `%p` to print a null pointer in C is well-defined and legal behavior, not causing undefined behavior. However, developers should ensure that the correct type (`void*`) is passed during printing to avoid potential type mismatch issues.

Printing null pointers with %p is undefined behavior?

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