strcpy Crashes C Program

Question

Please tell me why strcpy fails. Im trying to copy the value str1 into one of the element of str2.

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

int main(int argc, char *argv[]) {

    char str1[]={"cat"}; //string
    char *str2[]={"mouse","dog"}; //array
    strcpy(str2[0],str1);       

    return 0;
}

You are trying to write into a read-only hard coded memory. Don't do it. A good book about pointers will be a good idea. Any way those strings are literal strings Google is your best friend. — Michi, Jul 14 '16 at 03:22
By the way *str2[] is not an Array, is an Array of Pointers. — Michi, Jul 14 '16 at 03:25
See [this post](http://stackoverflow.com/questions/1088622/how-do-i-create-an-array-of-strings-in-c) for better understanding how to create an array of strings. — iRove, Jul 14 '16 at 03:35

score 1 · Accepted Answer · edited May 23 '17 at 12:14

1

char str1[]={"cat"}; //string

is wrong in this context

"" is a replacement for tedious {'a','b'...'\0'}.

Do either

char str1[]="cat";

or

char str1[]={'c','a','t','\0'};

Even then your code wouldn't work

strcpy(str2[0],str1);

because you are trying to write into a read-only hard coded memory

as [ @michi ] mentioned in his comment.

But below would work

str2[0]=malloc(sizeof(str1)); // You allocate memory for str2[0];
strcpy(str2[0],str1); 
printf("str2[0] : %s\n",str2[0])

Also remember to free the allocated memory after use

free(str2[0]);

edited May 23 '17 at 12:14

Community

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answered Jul 14 '16 at 03:34

sjsam

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This helps the OPs problem? – Michi Jul 14 '16 at 03:37
@Michi : Updated. Guess the op need to be good with the terminology and sure they need a good book – sjsam Jul 14 '16 at 03:39
1

thank you. Changing the declaration to str2[2][20] makes it editable. I'm just wondering how to make str2 as parameter of a function and reference each element by index; – Gerie Tanabe Jul 14 '16 at 04:04

strcpy Crashes C Program

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