How to dynamically allocate a two dimensional array of pointers ? (C)

Question

I have an assignment to make a dictionary. It will contain an x amount of words and their definitions (input by user). Instructions say that the dictionary should be of type char*** (2D array of pointers=arrays=strings), but I've got absolutely no idea of how to dynamically allocate the size of the array. it should have 2 lines, 1 for words and another 1 for their definitions, and the number of columns is decided by how many words are in the dictionary. While looking for help online i came up with this:

char** AllocateArray(int line, int column)
{
    char** pArray=(char**)malloc(line*sizeof(char*));
    int i;
    for(i=0;i<2;i++) 
        pArray[i]=(char*)malloc(column*sizeof(char));

    return pArray;
}

What changes should i make in the code for it to work with my char*** ? Using Visual studio 2012

Edit:

I have a problem with this right now:

void inputString(char* p1)
{
    char buffer[80];
    printf("\nEnter a word:");
    scanf("%s",buffer);
    p1=(char*)malloc(strlen(buffer)+1);
    if(p1!=NULL)
    {
        strcpy(p1,buffer);
        free(buffer);
    }
}

it crashes right after i input a word. the char* that the function receives is dictionary[i][j]. –

Being a "three star programmer" is not usually considered a good thing. — Tim Seguine, Dec 15 '13 at 20:01

score 2 · Answer 1 · answered Dec 15 '13 at 22:03

2

Don't free() anything allocated on the stack (i.e. buffer). Also, your function inputString() will not tell its client what memory it had allocated, since p1 is local to it.

answered Dec 15 '13 at 22:03

Fredrik Lundström

74
5

score 0 · Answer 2 · answered Dec 15 '13 at 19:45

0

Here is an example.

char*** dictionary;
int i = 0;
int j = 0;
int lines = 10;
dictionary = (char***)malloc(sizeof(char**)*lines);

for(i=0;i<lines;i++)
{
    dictionary[i] = (char**)malloc(sizeof(char*)*4);
    for(j=0;j<4;j++)
        dictionary[i][j] = (char*)malloc(sizeof(char)*25);
}

You have to modify the malloc's parameters in order to adapt to your problem/ or modify them when you need more memory for your strings. Also it might be a good idea to try and free memory when you do not need it

answered Dec 15 '13 at 19:45

Silent Control

614
10
22

[You Shouldn't cast the results of malloc](http://stackoverflow.com/questions/605845/do-i-cast-the-result-of-malloc/605858#605858) – Leeor Dec 15 '13 at 19:47
I agree that in C (and not C++) it is not necessary, but writing the casts at least while he's a beginner in dynamic memory allocation is a good practice, because it helps him understand better pointers (which are his main problem). – Silent Control Dec 15 '13 at 19:51

score 0 · Answer 3 · edited Dec 15 '13 at 22:01

0

Don't forget to malloc like this...

dictionary[i][j] = (char*)malloc(sizeof(char)*strlen(word_to_insert)+1);

...because each word end with a supplementary byte filled with 0 "null terminate string".

edited Dec 15 '13 at 22:01

Fiddling Bits

8,712
3
28
46

answered Dec 15 '13 at 19:58

SRedouane

498
5
10

score 0 · Answer 4 · answered Dec 15 '13 at 23:00

a sample

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

int main(){
    char ***dictionary;
    const char *words[] = { "ASEAN", "United Nations", "OPEC" };
    size_t howManyWords = sizeof(words)/sizeof(*words);
    int i;

    dictionary = malloc(howManyWords*sizeof(char**));
    printf("Please enter the definition of this word\n");
    for(i = 0; i < howManyWords; ++i){
        char buff[80];
        char **keyValue;

        printf("%s : ", words[i]);
        fgets(buff, sizeof(buff), stdin);
        keyValue = malloc(2*sizeof(char*));
        keyValue[0] = (char*)words[i];
        keyValue[1] = malloc(strlen(buff)+1);
        strcpy(keyValue[1], buff);
        dictionary[i] = keyValue;
    }

    //print
    for(i=0;i<howManyWords;++i){
        printf("%s : %s", dictionary[i][0], dictionary[i][1]);
    }
    //release
    for(i=0;i<howManyWords;++i){
        free(dictionary[i][1]);
        free(dictionary[i]);
    }
    free(dictionary);
    return 0;
}

How to dynamically allocate a two dimensional array of pointers ? (C)

4 Answers4