'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);

Isolative

You need the malloc, otherwise msg is just a dangling pointer. – Tola

Use malloc, but remove the cast and sizeof(char). The correct usage is char *msg = malloc(15); – Hung

Also malloc() is declared in <stdlib.h> not <malloc.h> – Protractile

And the return value from malloc() should ALWAYS be checked: char *msg = malloc(15); if (msg == NULL) /* not ok to proceed */; – Protractile

@MateuszPiotrowski: if you don't check you have no way of knowing whether it "worked". Returning NULL is malloc's way of telling you something went wrong. – Protractile

For people (like me) curious why omitting sizeof(char) is OK in this instance (as in @R..'s comment): This question explains that sizeof(char)==1 per the C99 standard. – Genro

Answer 1

Your original code does not assign msg. Attempting to strcpy to it would be bad. You need to allocate some space before you strcpy into it. You could use malloc as you suggest or allocate space on the stack like this:

char msg[15];

If you malloc the memory you should remember to free it at some point. If you allocate on the stack the memory will be automatically returned to the stack when it goes out of scope (e.g. the function exits). In both cases you need to be careful to allocate enough to be able to copy the longest string into it. You might want to take a look at strncpy to avoid overflowing the array.

Phonogram

if the size of msg is less than the string's length, say "char msg[3]; strcpy(msg, "abcdefg");", would be ok? If now, I cout msg, will the output be "abcdefg"? – Halflength

No it will not. char msg[3] allocates space for 3 characters. You cannot copy 8 characters into that space (7 letters plus a null terminator). – Phonogram

Answer 2

strdup does the malloc and strcpy for you

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

Transcendent

strdup is not C standard, not C89 and not C99 – Doorman

POSIX is a standard. There are a lot of things that we use that aren't C89 or C99. That isn't reason not to use something so simple. Please use strdup before writing a dense macro like MYSTRDUP(). strdup takes 1 line to implement as a function, and frankly, should be in the standard. – Numerary

Certainly, but it is worth mentioning that this is in POSIX. – Hyksos

just had weird issues with strdup but only on Win 10, malloc+strcpy instead then it works?! – Indecorous

I would be very suspicious of a statement like 'just had weird issues with strdup'. – Transcendent

Don't quote me on this but I think the reason it isn't in the C standard is because it allocates memory and only the malloc/calloc set of operators are supposed to do that. Otherwise, it causes a side effect. This is why it has never been included. – Topflight

strdup is included in C2x. en.cppreference.com/w/c/string/byte/strdup – Buggs

Answer 3

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!!!");

Fulbert

don't cast the return of malloc and don't use sizof(char). – Psychodiagnostics

why not sizeof(char)? – Transcendent

char is guaranteed to be 1 byte in c – Coatbridge

@Coatbridge so sizeof(char) is both correct and clear but redundant. No need to say 'dont use' – Transcendent

Answer 4

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");

Doorman

except in the case that malloc failed – Transcendent

missing a (char*) cast #define MYSTRDUP(str,lit) strcpy(str = (char*)malloc(strlen(lit)+1), lit) – Boarder

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!!!");

Sanitation

(UB == undefined behavior) – Moreville

Answer 6

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

Pons