How to create an uninitialised array of strings in C?

Question

I'm trying to create an array of strings using calloc method and I'm getting quite a lot many errors.

int main() {
    int T,i;
    char *w;
    char **s;

    w=(char*)calloc(100,sizeof(char));
    scanf("%d",&T);
    s=(char**)calloc(T,sizeof(char));
    s=(char*)calloc(100,sizeof(char));
    for(i=0;i<T;i++)
    { 
      scanf("%s",w);
      s[i]=w;
    }   
}

In the above code T is the number of strings and w is the maximum size of the string. Please shed some light on as to how I should be declaring string arrays dynamically and statically.

calloc(T,sizeof(char)) ==> calloc(T,sizeof(char*)). But anyway, you immediately overwrite that pointer. — Weather Vane
– Weather Vane, Commented Oct 17, 2020 at 16:52
1) Declare and array of strings statically: char[NSTRINGS][STRINGLEN]. 2) Allocate an array of strings dynamically: calloc(NSTRINGS,sizeof(char*)). 3) Allocate one string dynamically: calloc(STRINGLEN,sizeof(char)) — paulsm4
– paulsm4, Commented Oct 17, 2020 at 17:01

dspr · Accepted Answer · 2020-10-17 17:17:10Z

1

If your array stores string pointers, a new string must be allocated for each of them :

int main() {
    int arraySize,i;
    char *str;
    char **arrayOfPtr;
    scanf("%d", &arraySize);
    arrayOfPtr = (char**)calloc(arraySize,sizeof(char*));
    
    for(i=0;i<arraySize;i++)
    {  
      str =(char*)calloc(100,sizeof(char)); //<== string allocation
      scanf("%s", str);
      arrayOfPtr[i]= str;
    }

And you must free strings and array separately because the string memory is not that of the array :

    for (i=0;i<arraySize;i++) free(arrayOfPtr[i]);
    free(arrayOfPtr);
}

answered Oct 17, 2020 at 17:17

dspr

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