c function definition calls for pointer but example does not use pointers

Question

I relatively new to low level programming such as c. I am reviewing the strstr() function here. When reviewing the function definition char *strstr(const char *str1, const char *str2); I understand that function will return a pointer or a NULL depending if str2 was found in str1.

What I can't understand though, is if the funciton requires the two inputs to be pointers, when does the example not use pointers?

#include <string.h>
int main ()
{
  char string[55] ="This is a test string for testing";
  char *p;
  p = strstr (string,"test");
  if(p)
  {
    printf("string found\n" );
    printf ("First occurrence of string \"test\" in \"%s\" is"\
           " \"%s\"",string, p);
  }
  else printf("string not found\n" );
   return 0;
}

Strings are arrays. And arrays become (decay to) pointers when passed to functions. — Eugene Sh., Feb 04 '21 at 14:57
Why do you think this example does not use pointers? `string` is a `char *`, as is `"test"`. (Rather `string` is a char array, which decays to a `char *`) — William Pursell, Feb 04 '21 at 14:59

score 0 · Accepted Answer · answered Feb 04 '21 at 15:09

In strstr(string,"test");, string is an array of 55 char. What strstr needs here is a pointer to the first element of string, which we can write as &string[0]. However, as a convenience, C automatically converts string to a pointer to its first element. So the desired value is passed to strstr, and it is a pointer.

This automatic conversion happens whenever an array is used in an expression and is not the operand of sizeof, is not the operand of unary &, and is not a string literal used to initialize an array.

"test" is a string literal. It causes the creation of an array of char initialized with the characters in the string, followed by a terminating null character. The string literal in source code represents that array. Since it is an array used in an expression, it too is converted to a pointer to its first element. So, again, the desired pointer is passed to strstr.

You could instead write &"test"[0], but that would confuse people who are not used to it.

c function definition calls for pointer but example does not use pointers

1 Answers1