String memory address location

Question

I am looking at the sequential ordering of memory address location in string arrays. I wanted to also pipe out the values at each index in the array.

I have referenced the memory address location. The issue is the output. I cannot understand why the whole string would be printed from address index 0 when I am only referring to the one address index.

#include <stdio.h> #include <string.h> int main() { char strings[] = "hello"; int i; for (i = 0; i <= strlen(strings) - 1; i++) { printf("Memory location of strings[%d] : %x - value is : %s\n", i, &strings[i], &strings[i]); } return 0; }

which returns

Memory location of strings[0] : 61fe0a - value is : hello Memory location of strings[1] : 61fe0b - value is : ello Memory location of strings[2] : 61fe0c - value is : llo Memory location of strings[3] : 61fe0d - value is : lo Memory location of strings[4] : 61fe0e - value is : o

Can you please help me to understand why each index is being printed sequentially when the expectation is something like -

Memory location of strings[0] : 61fe0a - value is : h Memory location of strings[1] : 61fe0b - value is : e Memory location of strings[2] : 61fe0c - value is : l Memory location of strings[3] : 61fe0d - value is : l Memory location of strings[4] : 61fe0e - value is : o

Aside: the practise of i <= strlen(strings)-1; is not only hard to read, but is an error waiting to happen if you do this with unsigned variables. Please use the idiomatic i < strlen(strings); — Weather Vane
– Weather Vane, Commented Aug 1, 2022 at 12:42
Because you use %s, everything until '\0' will be printed. Try %c instead — infinitezero
– infinitezero, Commented Aug 1, 2022 at 12:43
The correct format specifier to print a void * pointer is %p. And the pointer must be cast to void *. Mismatching format specifier and argument type (like using the unsigned int format %x to print a pointer) leads to undefined behavior. — Some programmer dude
– Some programmer dude, Commented Aug 1, 2022 at 12:43
you print string so its prints from adders to end for string print %c you get the char — alon
– alon, Commented Aug 1, 2022 at 12:47
As @infinitezero: instead of %s and &strings[i] please use %c and strings[i]. — Weather Vane
– Weather Vane, Commented Aug 1, 2022 at 12:47

ikegami · Accepted Answer · 2022-08-01 13:20:39Z

The program behaves as expected: you pass &strings[i] for the %s format, which expected a pointer to a null terminated C string. printf outputs all the characters in strings starting from the index i.

If you mean to only output the character at index i, you should use the %c format specifier.

Also note that you should use %p to output a pointer value and cast the argument as `(void *).

Here is a modified version:

#include <stdio.h> int main() { char string[] = "hello"; int i; for (i = 0; string[i] != '\0'; i++) { printf("Memory location of string[%d]: %p - value is: %c\n", i, (void *)&string[i], string[i]); } return 0; }

ikegami · Accepted Answer · 2022-08-01 13:58:20Z

You always pass the address to one of the characters.

In the following, string is a pointer to the location of a.

const char *string = "abc"; printf( "%s\n", string );

In the following, string is an array. But printf needs a pointer. Fortunately, an array automatically "degenerates" into a pointer to its first element when necessary.

char string[] = "abc"; printf( "%s\n", string ); printf( "%s\n", &( string[0] ) ); // Same thing!

Normally, you pass a pointer point to the first character. But there's nothing wrong with passing a pointer to the second character. printf %s will print the character at that location and characters at subsequent locations until a NUL is encountered.

If you want to print a single character, use %c.

char string[] = "abc"; printf( "%c\n", string[1] ); // b

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