how to use dynamic stirng array in c

Question

i wish to store n number of words to an array. I used pointer in the following code to do that .But the problem is that once all words are read the words are not able to print.

please provide any solution.

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

int main() {

    char buff[10];
    int i,T;
    printf("Enter the  no.of words:");
    scanf("%d",&T);

    char **word= malloc(10*T);  
    printf("Enter the words:\n");
    for(i=0;i<T ;i++){
        scanf("%s",buff);
        word[i]= malloc(sizeof(char)*10);
        word[i]=buff;
        printf("u entered %s\n",word[i]);
        if(i>0)
            printf("u entered %s\n",word[i-1]);
    }

    printf("Entered words are:\n");
    for(i=0;i<T ;i++)
        printf("%s\n",word[i]);


}

The problem is you are storing the same buff pointer to each array element. That obviously means they all contain the same value and also causes a memory leak as the malloc buffer is lost. Use strcpy or strdup. — kaylum
– kaylum, Commented Apr 24, 2017 at 3:43

ameyCU · Accepted Answer · 2017-04-24 03:45:15Z

4

You need to allocate size of char * -

char **word= malloc(sizeof(char *)*T);

And to copy string you need to use strcpy. So instead of this -

word[i]=buff;          // 1

use this -

strcpy(word[i],buff);         // include string.h header

By 1. you point to buff using a pointer but the value stored in buff changes. So value to which all pointers point is same and you get all same output i.e the most recent value stored in buff.

answered Apr 24, 2017 at 3:45

ameyCU

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1 Comment

ameyCU Over a year ago

@kilowatt glad it hlped.

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