strcpy' with 'malloc'?

Question

Is it safe to do something like the following?

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

int main(void)
{
    char* msg;

    strcpy(msg, "Hello World!!!");  //<---------

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

    return 0;
}

Or should the following be used?

char* msg = (char*)malloc(sizeof(char) * 15);

Answer 1

strdup does the malloc and strcpy for you

char *msg = strdup("hello world");

Answer 2

The first version is not safe. And, msg should be pointing to valid memory location for "Hello World!!!" to get copied.

char* msg = (char*)malloc(sizeof(char) * 15);
strcpy(msg, "Hello World!!!");

Answer 3

Use:

#define MYSTRDUP(str,lit) strcpy(str = malloc(strlen(lit)+1), lit)

And now it's easy and standard conforming:

char *s;
MYSTRDUP(s, "foo bar");

Answer 4

You need to allocate the space. Use malloc before the strcpy.

Answer 5

 char* msg;
 strcpy(msg, "Hello World!!!");  //<---------Ewwwww
 printf("%s\n", msg); 

This is UB. No second thoughts. msg is a wild pointer and trying to dereference it might cause segfault on your implementation.

msg to be pointing to a valid memory location large enough to hold "Hello World".

Try

char* msg = malloc(15);
strcpy(msg, "Hello World!!!");

or

char msg[20]; 
strcpy(msg, "Hello World!!!");