Thread: Help Working with a char*

Originally Posted by joatmon

I have a function that returns a char*. No problem. But I need to concatenate another array with the results of this function. I'm getting a segmentation error.

What am I doing wrong? Thanks.

Code:

char* temp;
//but this next line causes a segmentation fault
strncpy(temp, get_filename(selection), sizeof(get_filename));

In the first line, you create a variable called temp which stores an address (of a character / of the first of a sequence of characters). Temp starts out, like most other variables, with random contents - it has a random memory address.

Now, your next line of code tries to write to that random area of memory, without knowing what is there or even if that address is valid.

You need another line where you actually allocate memory to hold the result of strncpy, and store the address of that memory in temp. Since you allocate this memory, at some point when you're done with it, you need to free it.

I don't think you really don't need to copy the string to another location. You could just do:

Code:

char *filename = get_filename(selection);

This depends on other details of your program, but it might be what you're looking for.

Oh, and the last parameter of strncpy is wrong. You're passing in the size of a function ponter, not the size of the string!

To do it safely, say with dynamic allocation (in case that's what you need) :

Code:

char *temp = NULL;
char *filename = get_filename(selection);
size_t len = strlen(filename);
temp = malloc(len + 1);
if (temp == NULL) {
    /* malloc failed */
}
strncpy(temp, filename, len);

That won't work because get_filename has already removed the brackets from your string, replacing them with ascii nuls. You can only call it once on selection unless selection has been reloaded with the string. That's why I only called it once in the code displayed.

You've gone back to not allocating any space for temp.
You have to do the malloc, and you'll have to malloc (at least) enough to hold the final string size.

Also, I'm not sure why you're subtracting 1 from strlen(temp) ???

No problem! In some ways it can be more confusing coming from Java to C than learning C as your first language. You have to "unlearn" some things and learn the C way.

Java is far simpler for memory handling.
Even C++ is easier (in some ways) than C for memory handling.
C is a fairly low-level language that doesn't really do anything you don't ask for.
That's its strength.
And its weakness.

So here's how I'd do it:

Code:

char *dir = "client/";
char *filename = get_filename(selection);
if (filename == NULL) {
    /* get_filename failed */
}
char *path = malloc(strlen(dir) + strlen(filename) + 1);
if (path == NULL) {
    /* malloc failed */
}
strcpy(path, dir);
strcat(path, filename);

Let's analyse this part:

Code:

char* temp = get_filename(selection);
size_t len = strlen(temp);
 
temp = malloc(len + 1);
fprintf(stderr, "temp: %s\n", temp);  //EMPTY!!!!!!!!!!!!!!

First, you call get_filename and store the result in temp. Then you overwrite temp with the return value of malloc. Then, you print temp, but temp now points to uninitialised storage, hence you risk undefined behaviour.

I would get rid of these two lines:

Code:

temp = malloc(len + 1);
fprintf(stderr, "temp: %s\n", temp);  //EMPTY!!!!!!!!!!!!!!

Originally Posted by joatmon

I guess that I'm confused as to when I have to call malloc and when I don't.

let's say in java you have classes named foo and bar, and bar contains a function called get_foo() that returns a foo object. you wouldn't do something like this:

Code:

bar b = new bar();
foo f = b.get_foo();
f = new foo();

the same thing applies to malloc. you don't call it if you've already assigned something that want to use.