What are Strings in C if not Simply Char Arrays?-CodePudding

I'm new to the C programming language, and I was under the impression that strings are just arrays of characters. However, when I tried the following code below (among some other tests):

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

int main(int argc, char **argv)
{
    char apple1[] = { 'a', 'p', 'p', 'l', 'e', '\0' };
    char *apple2 = "apple";
    char apple3[] = "apple";

    printf("%i\n", apple1 == apple2); // 0
    printf("%i\n", apple2 == apple3); // 0
    printf("%i\n", apple3 == apple1); // 0
    printf("%i\n", "apple" == apple1); // 0
    printf("%i\n", "apple" == apple2); // 1
    printf("%i\n", "apple" == apple3); // 0
    printf("%i\n", !strcmp(apple1, apple2)); // 1

    for (size_t i = 0; i < strlen(apple)   1; i  )
    {
        printf("%i", apple1[i] == apple2[i]);
    } // 111111

    return 0;
}

I got some unexpected results. Is there any reason for these, at least for me, counterintuitive results? Thank you very much.

CodePudding user response：

In C when you declare arrays

char apple1[] = { 'a', 'p', 'p', 'l', 'e', '\0' };
char *apple2 = "apple";
char apple3[] = "apple";

apple1,apple2 and apple3 are the starting addresses of these arrays in memory

So you cannot compare them

In C, you need to compare contents of the array with each other using functions from the string library (see string.h). E.g strcmp(apple1,apple2)

The compare function strcmp will compare each character with those two starting addresses and if all characters are the same, return 0.

CodePudding user response：

In these declarations

char apple1[] = { 'a', 'p', 'p', 'l', 'e', '\0' };
char *apple2 = "apple";
char apple3[] = "apple";

there are declared two arrays apple1 and apple3 that contain the string "apple".

In this declaration

char *apple2 = "apple";

there is declared a pointer to the string literal "apple".

In these calls of printf

printf("%i\n", apple1 == apple2); // 0
printf("%i\n", apple2 == apple3); // 0
printf("%i\n", apple3 == apple1); // 0
printf("%i\n", "apple" == apple1); // 0
printf("%i\n", "apple" == apple2); // 1
printf("%i\n", "apple" == apple3); // 0

there are compared addresses of first characters of different arrays that occupy different extents of memory. Arrays used in expressions with rare exceptions are converted to pointers to their first elements. So the result of the expressions is the integer value 0 except this call

printf("%i\n", "apple" == apple2); // 1

because in this case there are compared pointers to the same string literal (its first character) because it seems the compiler allocated one character array to store the string literal "apple" used in this call and in this declaration of a pointer

char *apple2 = "apple";

You can represent the above declaration and the call of printf the following way

char *apple2 = &"apple"[0];
//...
printf("%i\n", &"apple"[0] == apple2); // 1

However in general even if you will write for example

printf("%i\n", "apple" == "apple");

then the output can be either 0 or 1 depending on how the compiler stores identical string-literals: either as different character arrays or as one character array (it depends on compiler options).

To compare character arrays that contain strings you need to use standard C string function strcmp.