Thread: Using pointer to array of chars in order to find memory adress of '\0' character

"p" is a char*, "s1" is a char* (arrays are essentially pointers). your trying to assign a "char**" (&s1) to a "char*" (p). Basically, change it to "p = s1".

EDIT: This illustrates that "*(p + i)" is equivalent to "p[i]".
Also, you have:

Code:

printf("found \0 character\n");

You should escape this character and use "\\0".

Originally Posted by nadroj

"p" is a char*, "s1" is a char* (arrays are essentially pointers). your trying to assign a "char**" (&s1) to a "char*" (p). Basically, change it to "p = s1".

EDIT: This illustrates that "*(p + i)" is equivalent to "p[i]".
Also, you have:

Code:

printf("found \0 character\n");

You should escape this character and use "\\0".

That is incorrect. The actual type is char (*)[5], as illustrated in this error message:
Error 1 error C2440: '=' : cannot convert from 'char (*)[5]' to 'char *'
It is actually a pointer to an array of 5 elements, not a pointer to pointer.

I also suggest you indent more correctly, something along the lines of:

Code:

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

int main(void) {

	char s1[] = "name";   
	int i;
	char *p;

	p = &s1;

	for (i=0;i<=strlen(s1);i++){

		if (*(p + i) != '\0') {
			printf("not yet\n");
		}
		else if (*(p + i) == '\0')
		{
			printf("found \0 character\n");

			printf("memory adress: %d\n",p+i);
			printf("character: %c\n",*(p+i));
		}
	}

	system("PAUSE");

	return 0;

}

Originally Posted by nadroj

"s1" is a char* (arrays are essentially pointers).

To make Elysia's point more explicit: arrays are not essentially pointers. Rather, an array is converted to a pointer to its first element in many contexts.