Shouldn't a character pointer array, print a pointer? Why does it print the string it is pointed to?

Question

I have the code below:

    static char *name[] = {
            "January",
            "February",
            "March",
    };

    printf("%s", name[0]);

When I passed printf with name[0], it prints January. But shouldn't it print the address of January, since the array above stores a pointer to January?

printf's %s format wants a pointer. Your name array is an array of pointers. So name[0] is a pointer, and everyone's happy. (And, more specifically, %s wants, and name[0] is, a pointer to char, that points to the beginning of a valid, null-terminated string.) — Steve Summit
– Steve Summit, Commented Oct 18, 2021 at 16:17
See this recent question for some possibly-relevant information. — Steve Summit
– Steve Summit, Commented Oct 18, 2021 at 16:19
name[0] stores a pointer to the first character of a non-modifiable, anonymous array of char that is the string literal "January". — Ian Abbott
– Ian Abbott, Commented Oct 18, 2021 at 16:26

Vlad from Moscow · Accepted Answer · 2021-10-18 16:34:33Z

2

The conversion specifier %s interprets the corresponding argument as a pointer to first character of a string that is outputted until the terminating zero character '\0' is encountered.

If you want to output the pointer itself then you should write

printf( "%p", ( void * )name[0] );

Pay attention to that in this declaration

static char *name[] = {
        "January",
        "February",
        "March",
};

the string literals used as initializers are implicitly converted to pointers to their first characters and these pointers are stored in the array name.

edited Oct 18, 2021 at 16:34

answered Oct 18, 2021 at 16:25

Vlad from Moscow

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