How do I get strcpy destination string just big enough? - c

Question

I've resumed C coding for fun after a several year absence.
I've given myself an exercise to safely copy text from standard input to strings using fgets(), and copy to a string just big enough, i.e. only with enough capacity to hold the no. of chars I've actually typed, ultimately to make lists, stacks etc. from scratch, in other words, playing with pointers.
The only way I've managed this smells of kludge to me, as I'm defining the destination string variable for strcpy() late in the control flow. Is there a more elegant/dynamic way to do this?
#inlcude <stdio.h>
#include <string.h>
#define MAXLENGTH 20
int main(int argc, char *argv[]) {
char message[MAXLENGTH];
printf("Enter a string: \n");
fgets(message, MAXLENGTH, stdin);
/* various tests here, omitted for brevity */
char destinationString[strlen(message)];
/*
* Just testing to prove that
* the strlen() of the destination
* string is LESS than MAXLENGTH
*/
printf("Here's the strlen() of destinationString: %lu\n", strlen(destinationString));
printf("Here's the sizeof() destinationString: %lu,\n" sizeof(destinationString));
printf("Here's the contents of the copy: %s", destinationString);
return 0;
}

Answer 1

You can certainly do this dynamically by using malloc.
Consider something like this:
int main(int argc, char *argv[]) {
char *destinationString;
/* ... */
/* Don't forget to allocate one extra byte for the termination character */
destinationString = malloc(strlen(message) + 1);
if (!destinationString)
return -1;
strcpy(destinationString, message);
/* Note: Normally, you should probably use strncpy to avoid overflow
but here, we're sure that there's enough space so strcpy is acceptable */
/* ... */
free(destinationString); /* When you're done using it */
/* ... */
}
I also pointed this out in the comments but to re-iterate, you actually need to allocate strlen(message) + 1 bytes in your destination string buffer or else it will overflow. The extra character is to store the null termination character at the end of C strings.

Answer 2

Code has a number of choices. Here are 2:
malloc() and later free() right sized memory similarly answered by #tangrs. Note that sizeof() destinationString will be the size of a pointer.
size_t size = strlen(message) + 1;
char *destinationString = malloc(size);
memcpy(destinationString, message, size);
Use variable length array, VLA, available in C99 and optionally in C11.
VLA approach with code clean-up
#include <string.h>
#define MAXLENGTH 20
int main(int argc, char *argv[]) {
char message[MAXLENGTH];
printf("Enter a string: \n");
if (fgets(message, sizeof message, stdin) == NULL) {
return -1;
}
// Use type `size_t`
size_t size = strlen(message) + 1;
char destinationString[size];
memcpy(destinationString, message, size);
// Notice "%zu"
// `sizeof destinationString` is the size of an array
printf("Here's the strlen() of destinationString: %zu\n", strlen(destinationString));
printf("Here's the sizeof() destinationString: %zu,\n" sizeof destinationString);
printf("Here's the contents of the copy: \"%s\"", destinationString);
return 0;
}
Input "Hello!" Enter
Here's the strlen() of destinationString: 8
Here's the sizeof() destinationString: 9,
Here's the contents of the copy: "Hello!
"
On my system the inputs ended with a "\r\n". To rid the buffer of those potential pesky characters, use:
fgets(message, sizeof message, stdin);
buffer[strcspn(message, "\r\n")] = '\0';
size_t size = strlen(message) + 1;
...