Using C-string gives Warning: "Address of stack memory associated with local variable returned"

Question

I am not a C programmer, so I am not that familiar with C-string but now I have to use a C library so here is a shortened version of my code to demonstrate my problem:

char** ReadLineImpl::my_completion () {
    char* matches[1];
    matches[0] = "add";

    return matches;
}

I am getting this warning:

Warning - address of stack memory associated with local variable 'matches' returned

And my program does not seem to work properly (might be because of the above mentioned warning).

What does the warning imply? and will it cause any problems?

You're returning the address to the first character pointer which is allocated on stack and which ceases to exist after you exit the function. — Uchia Itachi, Aug 04 '13 at 08:27

Nemanja Boric · Accepted Answer · 2020-03-23T09:17:23.457

73

Variable char* matches[1]; is declared on stack, and it will be automatically released when current block goes out of the scope.

This means when you return matches, memory reserved for matches will be freed, and your pointer will point to something that you don't want to.

You can solve this in many ways, and some of them are:

Declare matches[1] as static: static char* matches[1]; - this will allocate space for matches in the static space and not on the stack (this may bite you if you use it unappropriately, as all instances of my_completion function will share the same matches variable).

Allocate space in the caller function and pass it to my_completion function: my_completion(matches):

char* matches[1];
matches = my_completion(matches);

// ...

char** ReadLineImpl::my_completion (char** matches) {
     matches[0] = "add";

     return matches;
}

Allocate space in the called function on heap (using malloc, calloc, and friends) and pass the ownership to the caller function, which will have to deallocate this space when not needed anymore (using free).

edited Mar 23 '20 at 09:17

answered Aug 04 '13 at 08:27

Nemanja Boric

21,057
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1

Thanks I got it. The worst thing is that this is the second time that get this kind of problem:) anyways, thank you very much – khajvah Aug 04 '13 at 08:29
Assuming that it's the `GNU readline`, this will cause a crash, since `readline` will free the memory returned by the completion function. – Mats Petersson Aug 04 '13 at 08:49
@MatsPetersson This does seems like readline - and in that case you are right. I will upvote your answer! – Nemanja Boric Aug 04 '13 at 09:02
5

declaring `matches` as `static` will allocate space on static space, not on heap. – MAnyKey Aug 04 '13 at 09:32
2

Third option is allocate memory in heap by calling calloc – Mike Glukhov Aug 29 '16 at 11:03

Mats Petersson · Answer 2 · 2013-08-04T08:56:43.967

When you return the matches array, what you are returning is the address of the first element. This is stored on the stack inside my_completion. Once you return from my_completion that memory is reclaimed and will (most likely) eventually be reused for something else, overwriting the values stored in matches - and yes, that may well be why your application doesn't work - if it isn't right now, it probably will be once you have fixed some other problems, or changed it a bit, or something else, because this is not one of those little warnings that you can safely ignore.

You can fix this in a few different ways. The most obvious is to simply use std::vector<char *> [or better yet std::vector<std::string>] instead:

std::vector<std::string> ReadLineImpl::my_completion ()
{
    std::vector<std::string> strings;
    strings.push_back("add");
    return strings;
}

Edit: So, if the library requires a char ** as per the readline interface,then use this:

char** ReadLineImpl::my_completion ()
{
    char **matches = static_cast<char **>malloc(1 * sizeof(char *));
    matches[1] = "add";
    return matches;
}

Problem solved!

I would use a vector as a c++ programmer but the library demands char**, so then I have to convert to char**. — khajvah, Aug 04 '13 at 08:32
So, assuming you are using the `readline` interface, then you MUST allocate the memory with malloc, as `readline` frees it later. I will edit. — Mats Petersson, Aug 04 '13 at 08:46
Good, just one thing, I had to use (char**)malloc(1 * sizeof(char *)) but thanks, your answer is better, though I am not gonna change — khajvah, Aug 04 '13 at 08:54

score 1 · Answer 3 · edited Dec 06 '21 at 22:09

1

Use heap instead of stack

It's better to allocate memory in heap for this case by using:

int* someDataForParams(void *_params) {
    
    ...
    int* charCounts = (int*) calloc(96, sizeof(char*));
    ...
    
    return charCounts;
}

96 is just a string length(Just a magic number)

edited Dec 06 '21 at 22:09

Alec von Barnekow

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answered Aug 29 '16 at 11:05

Mike Glukhov

1,656
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score 0 · Answer 4 · answered Jan 24 '22 at 16:29

0

There are 2 solutions: the first one is to declare the variable as a static variablr with the keyword static and the second solution is to use dynamic allocation with malloc or calloc

answered Jan 24 '22 at 16:29

thiere18

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score -2 · Answer 5 · answered Apr 26 '15 at 00:59

-2

change

char* matches[1];

to

char *matches = new matches[1];

answered Apr 26 '15 at 00:59

Shammel Lee

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10

Why? Maybe, You should describe the reason to do so. – WY Hsu Nov 29 '17 at 14:26

Using C-string gives Warning: "Address of stack memory associated with local variable returned"

5 Answers5

Use heap instead of stack

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