How to properly copy char array to char array?

Question

I am quite a newbie on C and am trying to get a C routine working.

const char *filenames[entry_count];
for (i = 0; i < entry_count; i++) {
    char filename_inzip[256];
    unz_file_info file_info;
    err = unzGetCurrentFileInfo(uf, &file_info, filename_inzip,
            sizeof(filename_inzip), NULL, 0, NULL, 0); 

    //This doesn't work. My array gets filled with the last entry over and over.
    filenames[i] = filename_inzip; //Option 1

    //Here I get SIGSEV errors, or gibberish in output when debugging values
    strcpy(filenames[i], filename_inzip); //Option 2
}

I guess the first assignment doesn't work because it just points to the same temporary memory address of the char[256] in the loop.

The second option doesn't work I think because I haven't use malloc or similar.

How would I get a properly filled array?

filenames[i] is not a char array. It's a pointer.

Fred Foo
– Fred Foo

2013-09-05 16:13:34 +00:00
Commented Sep 5, 2013 at 16:13 — Fred Foo
– Fred Foo, Commented Sep 5, 2013 at 16:13

simonc · Accepted Answer · 2013-09-05 16:13:05Z

2

You're correct that you need to allocate memory in the second option.

filenames[i] = strdup(filename_inzip);

is the easiest way of doing this.

strdup is a Posix function rather than standard C. If it isn't available for you

filenames[i] = malloc(strlen(filename_inzip)+1);
if (filenames[i] != NULL) {
    strcpy(filenames[i], filename_inzip);
}

does an equivalent job

Note that you'll also have to call free for each array element you allocate later in your program.

answered Sep 5, 2013 at 16:13

simonc

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4 Comments

Peterdk Over a year ago

Ok, will I need to use a loop and then call free on each?

simonc Over a year ago

Yes. Take care not to call free on any element whose value is unassigned however. Behaviour of free(NULL) is well defined so setting all elements to NULL at the start of your code would allow you to just loop through indexes [0..entry_count] later on without worrying about how many elements were actually allocated.

Peterdk Over a year ago

I guess there's not a way to free the complete array with a single call and not in a loop?

simonc Over a year ago

No. As a general rule, every call to malloc must have exactly one later matching call to free. If you call malloc in a loop you should therefore expect to later call free in a loop also.

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