Array of array of string in c with malloc

Question

Can you tell me what am I doing wrong here ? I don't get the text when I print printf("%s\n",text[0]);

I created char **text; and malloced all pointers.

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


char **text;

int main()
{
    int i; 
    text = malloc(20000);


    for(i=0; i < 20000; i++) {
        text[i] = malloc(4);

        memcpy(text[i], "test",4);
    }
    printf("%s\n",text[0]);
    printf("%s\n",text[1]);
}

Also, malloc(20000) allocates 20000 bytes, and you're then trying to fill it with 20000 character pointers (which take several bytes each, and are overwriting each other). — Lee Daniel Crocker
– Lee Daniel Crocker, Commented May 5, 2013 at 23:27
Yeah, I should have done this text = malloc(20000 * sizeof(char *)); — user_iv
– user_iv, Commented May 5, 2013 at 23:29
And nick is also right...you need to allocate 5 bytes for the string "text" to accommodate the null terminator. — Lee Daniel Crocker
– Lee Daniel Crocker, Commented May 5, 2013 at 23:30

nickb · Accepted Answer · 2013-05-05 23:31:26Z

2

I believe you're looking for something like this:

int numElements = 20000, i;
// Allocate an array of char pointers
text = malloc(numElements * sizeof( char *)); 

for( i = 0; i < numElements; i++) {
    // 4 for the length of the string "test", plus one additional for \0 (NULL byte)
    text[i] = malloc( (4 + 1) * sizeof( char));
    memcpy( text[i], "test\0", 5);
}

Here is an online demo showing it working, as it produces the output:

test
test

answered May 5, 2013 at 23:31

nickb

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Comments

BLUEPIXY · Accepted Answer · 2013-05-05 23:25:51Z

1

for(i=0; i < 20000/sizeof(char*); i++) {
    text[i] = malloc(5);
    memcpy(text[i], "test",5);

}

answered May 5, 2013 at 23:25

BLUEPIXY

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1 Comment

Lee Daniel Crocker Over a year ago

Or, of course, change the first malloc(20000) to malloc(20000 * sizeof(char *)).

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