C program: segmentation fault when trying to populate dynamically created array

Question

I am working on a homework assignment that requires the creation of a dynamically allocated array which is to be populated with strings from a text file. I then need to print the array to standard output, shuffle the array, and then print it again.

My current issue is that I cannot seem to populate the array with anything without getting a segmentation fault. I tested the program with a static array and everything worked, so I know there isn't an issue with any of the other code.

Here is a section of my program.

void alloc2dArray(char ***source, int dim1, int dim2)
{
    int i = 0;

    source = malloc(sizeof(char *) * dim1);

    if(source == NULL) 
    { 
        printf("Memory full!");
        exit(EXIT_FAILURE);
    }
    for(i = 0; i < dim1; i++)
    {
            source[i] = malloc(sizeof(char) * (dim2 + 1));
            if(source[i] == NULL) 
        { 
            printf("Memory full!"); 
            exit(EXIT_FAILURE);
        }
    }
}

EDIT:

In an effort to avoid being a Three Star Programmer I changed my code to the below snippet. As luck would have it, that fixed my issue. So thanks to Kniggug for posting the link to something I had no idea about before.

char** alloc2dArray(int dim1, int dim2)
{
        int i = 0;

        char **twoDArray = malloc(sizeof(char *) * dim1);

        if(twoDArray == NULL)
        {
                printf("Memory full!");
                exit(EXIT_FAILURE);
        }
        for(i = 0; i < dim1; i++)
        {
                (twoDArray[i]) = malloc(sizeof(char) * (dim2 + 1));
                if(twoDArray[i] == NULL)
                {
                        printf("Memory full!");
                        exit(EXIT_FAILURE);
                }
        }

        return twoDArray;
}

Thank you.

The answers posted look right, but you should understand why they're correct. The above code in the alloc2dArray function is modifying a copy of the address of **source. What you need to do is modify the value pointed to by ***source, which is accomplished through *source as mentioned twice below. — JustinBlaber
– JustinBlaber, Commented Oct 28, 2013 at 21:42
jucestain, thanks for the clarification. I am still trying to wrap my head around pointers. This is only an intro course and my first time trying my hand at C. — Arioth
– Arioth, Commented Oct 28, 2013 at 21:51

perreal · Accepted Answer · 2013-10-28 21:28:27Z

4

Void alloc2dArray(char ***source, int dim1, int dim2)
{
    int i = 0;

    source = malloc(sizeof(char *) * dim1);

The assignment above has no effect outside this function, apart from leaking memory. You mean to do:

    *source = malloc(sizeof(char *) * dim1);

similarly:

(*source)[i] = malloc(sizeof(char) * (dim2 + 1));

answered Oct 28, 2013 at 21:28

perreal

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4 Comments

Arioth Over a year ago

I appreciate the quick response. I tried the changes that you suggested, but that caused a segmentation fault even if I comment out the problem strcpy line. I have included more of the program to perhaps help figure out if I have a larger issue.

perreal Over a year ago

Do you have valgrind installed? If so run under valgrind, if not use a debugger to see where it crashes.

perreal Over a year ago

@Chad, also set the maxChars to something larger than 0

Arioth Over a year ago

Thanks again for the help perreal. I tried Valgrind and found that I had several errors as well as leaks. This prompted me to try to rework the code and I was able to fix it.

digital_revenant · Accepted Answer · 2013-10-28 21:29:04Z

1

Change source to (*source) in alloc2dArray

Void alloc2dArray(char ***source, int dim1, int dim2)
{
    int i = 0;

    *source = malloc(sizeof(char *) * dim1);

    if(*source == NULL)
    {
        printf("Memory full!");
        exit(EXIT_FAILURE);
    }
    for(i = 0; i < dim1; i++)
    {
        (*source)[i] = malloc(sizeof(char) * (dim2 + 1));
        if((*source)[i] == NULL)
        {
                printf("Memory full!");
                exit(EXIT_FAILURE);
        }
    }
}

answered Oct 28, 2013 at 21:29

digital_revenant

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