Segmentation fault while copying char * to Char *[ ] array

Asked Jan 05 '18 at 08:00

Active Jan 05 '18 at 08:00

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I am trying to copy char * element to char *[] array . I am taking each token values and copying to another array . Here i am using strcpy function , This is the C code i have written :

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

int main()
{
    char input[20];
    char *output[20];
    char *token = NULL;
    int count = 0,i=0;
    printf("Enter the input string\n");
    scanf("%s",input);
    token = strtok(input,",");
    while(NULL != token)
    {
        strcpy(*(output+i),token);
        i++;
        count++;
        token = strtok(NULL,",");
    }
    return 0;
}

I am getting segmentation fault on strcpy(*(output+i),token); this line . What exactly i am missing here ? Anybody explain a little bit would be more helpful to understand .

asked Jan 05 '18 at 08:00

user2986042

1,098
2
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37

You have defined `output` as an array of pointers – Inian Jan 05 '18 at 08:02
2

Don't copy to `output[i]`, ***assign*** to it instead: `output[i] = token;` – Some programmer dude Jan 05 '18 at 08:04
... and make sure tokenisation does not result in more then 20 tokens. – alk Jan 05 '18 at 08:06
any explanation ? what will happen when i copy with strcpy ? – user2986042 Jan 05 '18 at 08:07
2

What does e.g. `output[0]` point to? It's *uninitialized*, and therefore *indeterminate*. It will point to some seemingly random location. What happens when you copy some data to that location? [*Undefined behavior*](https://en.wikipedia.org/wiki/Undefined_behavior). – Some programmer dude Jan 05 '18 at 08:09
1

Please consider reading the answers to the question, your question had been close for as being a duplicate question to. – alk Jan 05 '18 at 08:20

Segmentation fault while copying char * to Char *[ ] array

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