Working with "multidimensional" "dynamic arrays" in C

Question

I have been trying to execute the following code. Though the task is trivial, I am getting a segmentation fault.

The aim of the code snippet betlow is to create a multi dimensional array of maximum row size 4 and column size 33. Then after creation, it should set the contents of all the rows as 0, followed by a '\0' character. Then in the end, it should display the output on the stdout.

Although I am not new to programming, I keep on getting similar errors, so if possible please explain me how can I avoid such mistakes in the future.

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

int main(int argc, char* argv[])
{
        int i,j,k,x,y;
        char** arr;
        arr = (char**) malloc(4 * sizeof(char*));
        for ( i = 0; i < 4; i++) {
                arr[i] = (char*) malloc(9 * sizeof(char));
                memset(arr,0,8);
                arr[i][8] = '\0';
        }
        for ( j = 0; j<4; j++) {
                puts(arr[j]);
        }
        return 0;
}

Casting `malloc` is generally wrong, as it can potentially hide errors. ([Relevant reading](http://stackoverflow.com/questions/605845/do-i-cast-the-result-of-malloc)) — Cody Gray - on strike, Jan 10 '12 at 00:27
The thing you are building here is often called a "ragged array" or "jagged array" and it is different from a vanilla two dimensional array such as you can declare with `char arr[33][4]`; Ragged arrays are only one of several solutions to the problem of dynamic multidimensional arrays in c. — dmckee --- ex-moderator kitten, Jan 10 '12 at 00:29
You also need to free() the memory blocks allocated with malloc(). Add for(j=0;j<4;++j)free(arr[j]);free(arr); before return 0; — al01, Jan 10 '12 at 00:34

Mysticial · Accepted Answer · 2012-01-10T00:38:44.977

5

You're memseting the wrong pointer.

Instead of:

memset(arr,0,8);

you want:

memset(arr[i],0,8);

So you're off by one level of indirection.

As pointed out in the comments, here are some optimizations:

for ( i = 0; i < 4; i++) {
    arr[i] = (char*) malloc(9 * sizeof(char));
    memset(arr[i],0,9);
}

or

for ( i = 0; i < 4; i++) {
    arr[i] = (char*) calloc(9, sizeof(char));
}

Note that the cast to char* isn't necessary in C.

If you wanted to the character '0' instead of the null-character, then you should go with this:

for ( i = 0; i < 4; i++) {
    arr[i] = (char*) malloc(8 * sizeof(char));
    memset(arr[i],'0',8);
    arr[i][8] = '\0';
}

edited Jan 10 '12 at 00:38

answered Jan 10 '12 at 00:26

Mysticial

464,885
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+1: and the code should be using `memset(arr[i], '\0', 9);` so that the separate assignment of the extra 0 byte is not needed. – Jonathan Leffler Jan 10 '12 at 00:29
Or just use `calloc()` instead of `malloc()` and forget about it :) – Brian Roach Jan 10 '12 at 00:29
1

+1 for addressing the seg-fault; but from the description, I think the OP *might* actually need `memset(arr[i],'0',8)` (setting each element to a length-eight null-terminated string of zeroes). – ruakh Jan 10 '12 at 00:30
@ruakh It could be, it's not entirely clear if the OP wants '\0' or '0'. – Mysticial Jan 10 '12 at 00:31
The OP wants to fill the string with '0', printing '\0' would not produce any output. – al01 Jan 10 '12 at 00:39

score 1 · Answer 2 · answered Jan 10 '12 at 00:28

1

memset(arr,0,8);

is the issue, you are clearing the array of pointers

try:

memset(arr[i], '0', 8);

assuming you want to fill the string with zeros

answered Jan 10 '12 at 00:28

al01

122
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v01pe · Answer 3 · 2012-01-10T00:34:27.730

0

in the for loop I think you have an logical error:

memset(arr[i],0,8);

you also don't need to set this:

arr[i][8] = '\0';

instead you could do this:

memset(arr[i],0,9);

the '\0' character is nothing different then 0.

edited Jan 10 '12 at 00:34

answered Jan 10 '12 at 00:29

v01pe

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Working with "multidimensional" "dynamic arrays" in C

3 Answers3