从char指针减去char数组会产生一个int吗？

Question

In the strchr reference at http://www.cplusplus.com/reference/cstring/strchr/ , this example is provided.

/* strchr example */
#include <stdio.h>
#include <string.h>

int main ()
{
  char str[] = "This is a sample string";
  char * pch;
  printf ("Looking for the 's' character in \"%s\"...\n",str);
  pch=strchr(str,'s');
  while (pch!=NULL)
  {
    printf ("found at %d\n",pch-str+1);
    pch=strchr(pch+1,'s');
  }
  return 0;
}

Why does subtracting the char array str from the char pointer pch plus one give an int ? (as denoted by the %d format type) If I remove " -str ", the program wouldn't execute until I change %d to %s .

Answer 1

The short of it is, that's a bug :
That expression might be of type int , or you might have Undefined Behavior in the call to printf .

Let's take it step by step:

1. You are not actually subtracting an array from a pointer, but a pointer from a pointer:
   In nearly all contexts, an array decays to a pointer to its first element .

2. What type is the difference of two pointers (which is only defined if they point at or directly behind elements from the same array)?

   ptrdiff_t (That's what that typedef in <stddef.h> is for.)

3. ptrdiff_t might happen to be int , but don't depend on it.
   Instead, use the proper format-specifier: %ti .

Answer 2

Array names decays to pointers (in most cases) to its first element. Subtraction of two pointers yields an data of ptrdiff_t type, provided both points to elements of the same array.

Answer 3

You are looking at one of the oddities of C/C++ and pointers. Pointer and array syntax and semantics are simply nuts.

str is a pointer to a char (as well as an array). pch is a pointer to a char as well.

If you subtract two pointers to the same type you get the number of elements between them.

You can even do

    3 [str] ;

which is equivalent to

   str[3] ;

and is equivalent to

   *(str+3) ;

Your expression:

  pch - 1

has the type pointer to char. Your expression

  pch - str

has the type int.