Question

I have a function that accepts a char* as one of its parameters. I need to manipulate it, but leave the original char* intact. Essentially, I want to create a working copy of this char*. It seems like this should be easy, but I am really struggling.

My first (naive) attempt was to create another char* and set it equal to the original:

char* linkCopy = link; 

This doesn’t work, of course, because all I did was cause them to point to the same place.

Should I use strncpy to accomplish this?

I have tried the following, but it causes a crash:

char linkCopy[sizeof(link)] = strncpy(linkCopy, link, sizeof(link)); 

Am I missing something obvious…?

EDIT: My apologies, I was trying to simplify the examples, but I left some of the longer variable names in the second example. Fixed.

Answer 1

The sizeof will give you the size of the pointer. Which is often 4 or 8 depending on your processor/compiler, but not the size of the string pointed to. You can use strlen and strcpy:

// +1 because of '\0' at the end char * copy = malloc(strlen(original) + 1);  strcpy(copy, original); ... free(copy); // at the end, free it again. 

I’ve seen some answers propose use of strdup, but that’s a posix function, and not part of C.