char * variable declaration

Question

I just want to verify I got this right. The copy from sr to ds2 gives an error. Is this because ds2 is considered "const"?? Thanks and hope this isn't a bore.

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

int main(void)
{
    char *sr = "Hello World";
    char *ds1 = (char*)malloc(100 * sizeof(char));
    char *ds2 = "12345678901234567890";

    // This statement works just fine
    printf("%s\n", strcpy(ds1, sr));

    // This gives error
    strcpy(ds2, sr);


    printf("%s\n", ds2);

    return 0;
}

score 0 · Answer 1 · edited May 23 '17 at 11:50

0

Here is a similar post

difference between char* and char[] with strcpy()

When you do this

char *ds2 = "12345678901234567890";

the compiler leaves the pointer pointing to a non-writable memory region.

With this line

// This gives error
   strcpy(ds2, sr);

You are trying to do an strcpy into the non-writable memory.

You should also have a free for each malloc as you are allocating memory but not de-allocating it.

edited May 23 '17 at 11:50

Community

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answered Mar 11 '15 at 22:20

kmcnamee

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Excellent - thought something like that. Yes, I know I should have a free(). (Trying to get back on the horse after 25yrs off).I appreciate your time and assistance. Best. – Panchdara Mar 11 '15 at 23:52

char * variable declaration

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