Allocating a char pointer array

Question

I've got a number of files that I am reading the contents into a char array, I would like to store the contents I read from each file in an array of characters so that I could access the separate character arrays (file contents) by index, I am not sure of the syntax on how to do this.

My understanding is that I need to set each of the memory addresses in my patterns array to one memory address per char array; to the char array stored in patternData but I do not know how to set this. What is the syntax that I am missing that I need to get this to happen?

What I've tried

I would have thought that If I was storing a type of char* then I would need a type char** to store the separate arrays of char arrays.

I would access a char* by using the following notation to set the memory address of the pattern index to

&patterns[INDEX] = &pData;

However this does not work. There is a plethora of "char pointer array" questions but I'm not sure of the correct way to do this simple assignment of pData to an index of patterns.

char *tData;
int tLength;
char *pData;
int pLength;

char **patterns;

void ReadFromFile(FILE *f, char **data, int *length) //This is what is passed into function


    int main(int argc, char **argv)
    {
        for (int i = 1; i <= 5; i++)
        {
            FILE *f;
            char fileName[1000];
            sprintf(fileName, "file%d.txt", i);
            f = fopen(fileName, "r");
            if (f == NULL)
                return 0;

            ReadFromFile(f, &pData, &pLength); //works to here, contents of file assigned to pData
            fclose(f);

            &patterns[i - 1] = &pData;

        }

        return 0;

    }

Shouldn't it read patterns[i - 1] = pData;?

nlu
– nlu

2015-03-18 15:39:54 +00:00
Commented Mar 18, 2015 at 15:39 — nlu
– nlu, Commented Mar 18, 2015 at 15:39

Sergey Kalinichenko · Accepted Answer · 2015-03-18 15:41:36Z

2

This line is incorrect:

&patterns[i - 1] = &pData;

You are trying to assign to the result of the "take an address" operator, which is not possible because it's not an lvalue.

The assignment should be as follows:

patterns[i - 1] = pData;

but you need to allocate patterns before you do this. You can use

patterns = malloc(sizeof(char*)*5);

or simply declare patterns to be an array of five:

char *patterns[5];

This assumes that your ReadFromFile function allocates a char*, and assigns it to the address pointed to by pData. Note that you need to free all pointers that were obtained through malloc/calloc/realloc.

answered Mar 18, 2015 at 15:41

Sergey Kalinichenko

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Comments

interestedparty333 · Accepted Answer · 2015-03-18 15:38:53Z

0

I haven't read particularly carefully, but it strikes me as odd that you're taking the reference of patterns when it's a **. This is in the line:

&patterns[i-1] = &pData;

answered Mar 18, 2015 at 15:38

interestedparty333

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1 Comment

Oliver Over a year ago

Sorry I think thats a line written with inherent confusion, I basically need to assign the memory address at that patterns index to the address of pData

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