C Returning char[] Warning "returns address of local variable"

Question

This is a portion of a homework assignment.

I am trying to read and return a single line of a file in my method getLine.

char *getLine(FILE *input) {
    char line[30];
    if(fgets(line, sizeof(line), input)!=NULL)
    {
        return line;
    }else{
        return NULL;
    }
}

This would seem to work from what I have been taught regarding pointers, however I am unable to remove the warning message warning: function returns address of local variable [enabled by default]. This warning is referring to the line return line;. My assignment requires that I have no warnings or errors when I compile. I don't see what I am doing wrong.

Most of the help I found suggested malloc-ing space for the line of text, but we haven't covered that in class yet even though I have done some in another class. Is that really the best way to do this? If so, would I be able to free anywhere in the program?

The local variable ceases to exist when the function returns, then you return a dangling pointer. Let me find a dupe. — Daniel Fischer, Feb 14 '13 at 10:21
Use `malloc` if you want to allocate memory on heap. And this question has already been asked countless times here. — KBart, Feb 14 '13 at 10:22
This is quite common mistake so you will definitely find dozens of posts about encountering this problem :) One of the reasons for this is that most of compilers don't even warn you about what you're doing: http://ideone.com/S5Se71 — LihO, Feb 14 '13 at 10:32
Which implies that you should know exactly what you're doing :) — LihO, Feb 14 '13 at 10:34
Marked duplicates don't actually contain answer (code) to this specific problem. — hyde, Dec 13 '18 at 14:49

score 14 · Accepted Answer · answered Feb 14 '13 at 10:23

char line[30]; is an array with automatic storage duration. Memory where it resides is deallocated once the execution goes out of the scope of your function, thus pointer to this memory that you return becomes invalid (dangling pointer).

Trying to access memory, that has already been deallocated, results in undefined behaviour.

You can allocate your array dynamically and let the caller explicitly deallocate it:

char *getLine() {
    char* line = malloc(30);
    ...
    return line;
}

// somewhere:
char* line = getLine();
...
free(line);

I should make a template of this answer..;) – KBart Feb 14 '13 at 10:24 — KBart, Feb 14 '13 at 10:24

hyde · Answer 2 · 2013-02-14T10:49:45.500

Alternative without using malloc, not covered by the earlier questions as far as I could find:

/* Warning, this function is not re-entrant. It overwrites result of previous call */
char *getLine(FILE *input) {
    static char line[30];
    return fgets(line, sizeof(line), input);
    /* fgets returns NULL on failure, line on success, no need to test it */
}

Explanation: static variables in function scope retain their values even after function returns. There's just one instance of such a variable, same one is used in every call to that function (so not re-entrant/thread-safe). Memory for static variables is allocated when program starts, it's neither heap nor stack but it's own reserved area in running program's memory space. Value of static variable is initialized just once, before it's first used (above does not have specific initialization, so it's filled with 0, but it could have an initializer too, and that would only be it's value when function is called first time).

While using static variable in this manner may seem a bit hacky (and I agree it is), it's still a valid pattern also employed by standard C for example with strtok() function. It also demonstrates the need for a version which is re-entrant, as C standard also has strtok_r(), which is more complex to use as it has one extra parameter instead of having local static variable in it.

C Returning char[] Warning "returns address of local variable"

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