The method signature for getenv char *getenv(const char *name) The return value is a pointer to char .

Now if we look at the below example :

#include <stdio.h>
#include <stdlib.h>
int main() 
{
  
char *p;
p = getenv("PATH");
if(p != NULL)
 printf("Current path is %s", p);

return 0;


}

The only confusion here is the "p" printed with format specifier "%s" as String but what about the Pointer Dereference If I put "*p" instead, I end up with segmentation fault while if I want to get the pointer address I can do this printf("%p", p); confusion about getting the value by deferencing the point "p" in our case. Please explain

CodePudding user response：

Looks like you're confused about some fundamental basics in C.

In case of

char *p;

• p is a pointer to a series of char, which is in fact a \0 terminated string. This string can be printed with printf("%s", p);
• Since p is a pointer you can print the memory address it points to printf("%p", p);, which is exactly the same as if p was defined as void *p;
• *p dereferences the first char in the string pointed to by p, which is essentially the same as p[0]. This string can be printed with printf("%c", *p); or printf("%c", p[0]);