Passing two dimensional array of strings to function?

Question

I am trying to send this array to the function sortByLength.

The function prototype is

sortByLength(char *words[],int * lengths, const int size);

Here is the rest of my code up to where I try to pass it to the function:

#include <stdio.h>
#include "task2.h"
#define SIZE 3

int main(void)
{
    int i;
    int lengths[SIZE] = {3, 4, 6};
    char words[][20] = {"Sun", "Rain", "Cloudy"};

    sortByLength(words[][20],lengths,SIZE);

    return 0;
}

@kiks73 It will not work. Because he declared that as a 2-D array. But he getting that as pointer to pointers. — Karthikeyan.R.S
– Karthikeyan.R.S, Commented Mar 24, 2015 at 5:06
@Karthikeyan.R.S; Its not array of pointers but pointer to pointer. — haccks
– haccks, Commented Mar 24, 2015 at 5:07
@haccks so 'char words[][20]' is a pointer that points to the pointer that points to the first element "Sun"? — MilkyCode
– MilkyCode, Commented Mar 24, 2015 at 14:34

haccks · Accepted Answer · 2015-03-24 05:11:09Z

2

The prototype

sortByLength(char *words[],int * lengths, const int size);

is equivalent to

sortByLength(char **words,int * lengths, const int size);

words[][20] will converted to pointer to array of 20 chars when passed to a function. char ** and char (*)[20] are incompatible types. Change function prototype to

sortByLength(char (*words)[20], int *lengths, const int size);

answered Mar 24, 2015 at 5:11

haccks

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1 Comment

Spikatrix Over a year ago

or use sortByLength(words,lengths,SIZE); instead of sortByLength(words[][20] ,lengths,SIZE);.

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