What is the trick behind strcpy()/uninitialized char pointer this code?

问题

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

void main ()
{
  char *imsi;
  unsigned int i;
  int val;
  char *dest;

  imsi = "405750111";

  strncpy(dest,imsi,5);

  printf("%s",dest);

  /*  i = 10; */
}

In the above code, with the i = 10 assignment is commented as above, the code works fine without error. When assignment is included for compilation, the error (segmentation fault) occurs at strncpy(dest,imsi,5);.

By avoiding optimization to variable i (i.e., volatile int i;), the error is cleared even with the assignment (i = 10) included.

回答1:

In your code, by saying

 strncpy(dest,imsi,5);

you're trying to write into an unitialized pointer dest. It can (and most possibly, it will) point to some memory which is not accessible from your program (invalid memory). It invokes undefined behavior.

There is nothing that can be guaranteed about a program having UB. It can work as expected (depends on what you're expecting, actually) or it may crash or open your bank account and transfer all money to some potential terrorist organization.

N.B - I hope by reading last line you got scared, so the bottom line is

Don't try to write into any uninitialized pointer (memory area). Period.

回答2:

The behaviour of this code is unpredictable because the pointer dest is used before it is initialised. The difference in observed behaviour is only indirectly related to the root cause bug, which is the uninitialised variable. In C it is the programmers responsibility to allocate storage for the output of the strncpy() function and you haven't done that.

The simplest fix is to define an output buffer like this: char dest[10];

来源：https://stackoverflow.com/questions/36477421/what-is-the-trick-behind-strcpy-uninitialized-char-pointer-this-code

标签

pointers

memory-management

strncpy