Proper way to copy C strings

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You could use strdup() to return a copy of a C-string, as in:

#include <string.h>

const char *stringA = "foo";
char *stringB = NULL;

stringB = strdup(stringA);
/* ... */
free(stringB);

You could also use strcpy(), but you need to allocate space first, which isn’t hard to do but can lead to an overflow error, if not done correctly:

#include <string.h>

const char *stringA = "foo";
char *stringB = NULL;

/* you must add one to cover the byte needed for the terminating null character */
stringB = (char *) malloc( strlen(stringA) + 1 ); 
strcpy( stringB, stringA );
/* ... */
free(stringB);

If you cannot use strdup(), I would recommend the use of strncpy() instead of strcpy(). The strncpy() function copies up to — and only up to — n bytes, which helps avoid overflow errors. If strlen(stringA) + 1 > n, however, you would need to terminate stringB, yourself. But, generally, you’ll know what sizes you need for things:

#include <string.h>

const char *stringA = "foo";
char *stringB = NULL;

/* you must add one to cover the byte needed for the terminating null character */
stringB = (char *) malloc( strlen(stringA) + 1 ); 
strncpy( stringB, stringA, strlen(stringA) + 1 );
/* ... */
free(stringB);

I think strdup() is cleaner, myself, so I try to use it where working with strings exclusively. I don’t know if there are serious downsides to the POSIX/non-POSIX approach, performance-wise, but I am not a C or C++ expert.

Note that I cast the result of malloc() to char *. This is because your question is tagged as a c++ question. In C++, it is required to cast the result from malloc(). In C, however, you would not cast this.

EDIT

There you go, there’s one complication: strdup() is not in C or C++. So use strcpy() or strncp() with a pre-sized array or a malloc-ed pointer. It’s a good habit to use strncp() instead of strcpy(), wherever you might use that function. It will help reduce the potential for errors.

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