2 ways of accessing value of an array of pointers in C

Question

The first block

#include <stdio.h>

const int MAX = 3;

int main() {

    int  var[] = { 10, 100, 200 };
    int i, *ptr[MAX];

    for (i = 0; i < MAX; i++) {
        ptr[i] = &var[i]; /* assign the address of integer. */
    }

    for (i = 0; i < MAX; i++) {
        printf("Value of var[%d] = %d\n", i, *ptr[i]);
    }

    return 0;
}

Easy to understand, since ptr is an array of int pointers. So when you need to access the i-th element, you need to dereference its value as *ptr[i].

Now the second block, just the same, but now it points to an array of char pointer:

#include <stdio.h>

const int MAX = 4;

int main() {

    char *names[] = {
        "Zara Ali",
        "Hina Ali",
        "Nuha Ali",
        "Sara Ali",
    };

    int i = 0;

    for (i = 0; i < MAX; i++) {
        printf("Value of names[%d] = %s\n", i, names[i]);
    }

    return 0;
}

This time, when we need to access its element, why don't we add a * first?

I tried to form a correct statement to print this value, seems if you dereference, it will be a single char. Why?

printf("%c", *names[1]) // Z

I know there is no strings in C, and it is a char array. And I know pointers, but still don't under the syntax here.

Because the element that we need to access is an address. The %s tells printf to print all the characters starting at the given address, and until encountering 0. In your example, the given address is the value of names[i]. The %c tells printf to print the given character.In your example, the given character is the value of *names[i], which is equivalent to names[i][0]. — barak manos
– barak manos, Commented Aug 1, 2016 at 14:17

Sourav Ghosh · Accepted Answer · 2016-08-01 14:09:35Z

4

For printf() with %s format specifier, quoting C11, chapter §7.21.6.1

s
If no l length modifier is present, the argument shall be a pointer to the initial element of an array of character type. [...]

In your case,

 printf("Value of names[%d] = %s\n", i, names[i] );

names[i] is the pointer, as required by %s. That is why, you don't dereference the pointer.

FWIW,

%c expects an int type argument (converted to an unsigned char,), so you need to dereference the pointer to get the value.
%d also expects an int argument, so you have to go ahead and dereference the pointer, as mentioned in the question.

answered Aug 1, 2016 at 14:09

Sourav Ghosh

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Comments

John Bode · Accepted Answer · 2016-08-01 14:11:18Z

1

The %d conversion specifier expects its corresponding argument to have type int; ptr[i] has type int *, so the dereference is necessary.

The %s conversion specifier expects its corresponding argument to have type char *; that is, a pointer to char. So names[i] is already of the correct type. The %s specifier tells printf to print the sequence of characters starting at the given location until it sees the string terminator.

answered Aug 1, 2016 at 14:11

John Bode

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chqrlie · Accepted Answer · 2016-08-01 14:21:01Z

0

You would use the same syntax in the second example if you were printing the first character of each array of char:

    printf("Value of names[%d] = %c\n", i, *names[i]);

But since you want to print the C strings, you pass the values of the elements of the names array, which are of the proper type, char *.

answered Aug 1, 2016 at 14:21

chqrlie

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machine_1 · Accepted Answer · 2016-08-01 14:30:05Z

0

printf() has it's own way of dealing with format type specifiers. In case of strings, that is, array of characters, it requires the %s specifier and a pointer of type char * as a corresponding argument. When printf() receives this pointer, it dereference it discreetly to print the characters that are of type char. So, that is why you pass a pointer and don't dereference when you want to print strings.

answered Aug 1, 2016 at 14:30

machine_1

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