Initialize string array in C

Question

I was wondering that if someone gave me an input file of many words:

Such as: "AB1CDE1 FG2HIJ2 KL3MNO3"

and I wanted to put each word into a char array in which the resulting outputs would be:

printing this ->

array[0] = "AB1CDE1"
array[1] = "FG2HIJ2"
array[2] = "KL3MNO3"

I would need to allocate memory for this is what I understand.

So if I use char *array = (char*)malloc(sizeof(*array) * num_arrays) wouldn't this only give me the first values of each input given to me?

How would I initialize this array to store all the values in each array?

(I'm new to C coding so might be a really easy question.)

Ammar Faizi · Accepted Answer · 2021-03-15 08:01:42Z

1

You can use pointer to array of chars, and then allocate enough space with malloc for that.

Code Example

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

int main() {
    char (*array)[8] = malloc(3u * sizeof(char [8]));

    strcpy(array[0], "AB1CDE1");
    strcpy(array[1], "FG2HIJ2");
    strcpy(array[2], "KL3MNO3");

    printf("array[0] = %s\n", array[0]);
    printf("array[1] = %s\n", array[1]);
    printf("array[2] = %s\n", array[2]);

    free(array);
    return 0;
}

Run

ammarfaizi2@integral:/tmp$ gcc test.c -o test
ammarfaizi2@integral:/tmp$ ./test
array[0] = AB1CDE1
array[1] = FG2HIJ2
array[2] = KL3MNO3

answered Mar 15, 2021 at 8:01

Ammar Faizi

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4 Comments

2AM Coffee Over a year ago

What does the 3u mean when using malloc? Is it specifically the numbers that I am using? I understand the 3 but not the u

Jabberwocky Over a year ago

The u means "unsigned int", you don't need it. malloc(3 * ... is enough.

Ammar Faizi Over a year ago

Yeah, it is unsigned int. Not strictly required, it is just my habit to emphasize that the argument is unsigned type.

pmg Over a year ago

I prefer char (*array)[8] = malloc(3u * sizeof *array); so that there's only one place to change when you need longer strings.

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