C strcpy does not work when copying to a pointer in a simple program?

Question

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

int main(void) {
    char *w;
    strcpy(w, "Hello Word");
    printf("%s\n", w);
    return 0;
}

What is wrong with the way the char pointer is used in the above code?

Not quite sure why you want to do that, but if a situation in which you need to do that comes up, you may find strdup() of use. — Corbin, Mar 13 '12 at 02:39
http://stackoverflow.com/questions/3584926/char-pointers-segmentation-fault , http://stackoverflow.com/questions/2868884/whats-wrong-with-strcpy — , Mar 13 '12 at 02:58

score 4 · Answer 1 · answered Mar 13 '12 at 02:38

4

You allocate no space for the string. w is just a pointer to some memory (garbage value since it's not initialized).

char w[32];

or

char *w = malloc(32);

You need to allocated the space for the characters.

answered Mar 13 '12 at 02:38

Jesus Ramos

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score 3 · Answer 2 · answered Mar 13 '12 at 02:37

3

No memory allocated.

Add

w = (char *)malloc(42);

answered Mar 13 '12 at 02:37

Java42

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2

( and don't forget to free() when you are finished with the memory ) – Java42 Mar 13 '12 at 02:41

score 3 · Answer 3 · answered Mar 13 '12 at 02:37

3

It's an uninitialized pointer. The strcpy will write to some unknown location in memory.

answered Mar 13 '12 at 02:37

Mark Tolonen

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score 2 · Accepted Answer · answered Mar 13 '12 at 02:44

Ok, you did not ask to the system for memory, to use it with the string. This code will work

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

int main() {
   char w[11];
   strcpy(w, "Hello Word");
   printf("%s\n", w);
   return 0;
}

That code declare w as an array of char, reserving the memory space for it. Other alternative is to use malloc or calloc for the char pointer. Read about that.

C strcpy does not work when copying to a pointer in a simple program?

4 Answers4

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