Why apply a non-exist key to map::find would return a iterator with first value map size in C++?

Question

I have a use case like the code snippet below, using map::find in a getter returned map to find a non-exist key would actually find a iterator which first value is the size of the map (likely), thus would not behave as expect, equals with map::end

this is probably because of my map was a getter returned map. And consumed the map without assign it to a variable. So that the getter returned value may have been destructed immediately.

So if my guess is correct?
Why it returned the size of the map instead of its end iterator?


    #include <iostream>
    #include <map>


    class B {
        long long id_;

        public:
        B() = default;
        explicit B(long long);
        ~B() = default;
    };

    B::B(long long int id) : id_(id) {}


    class A {
        std::string id_;
        std::map<long long, std::shared_ptr<B>> b_;

        public:
        A() = default;
        explicit A(std::string id);
        ~A() = default;

        const std::string &id() const;

        std::map<long long, std::shared_ptr<B>> b();

    };

    A::A(std::string id): id_(id) {
        b_[1] = std::make_shared<B>(1);
        b_[2] = std::make_shared<B>(2);
    }

    const std::string &A::id() const {
        return id_;
    }

    std::map<long long, std::shared_ptr<B>> A::b() {
        return b_;
    }


    int main() {
        std::shared_ptr<A> a = std::make_shared<A>("arst");
        if (a->b().find(3) != a->b().end()) {
            std::cout << a->b().find(3)->first << std::endl;
            std::cout << a->b().at(3) << std::endl;
        }
    }

run as below:

clang --version

Apple LLVM version 10.0.0 (clang-1000.10.44.4)
Target: x86_64-apple-darwin18.2.0
Thread model: posix
InstalledDir: /Library/Developer/CommandLineTools/usr/bin

with output:

clang++ test.cc -std=c++11
./a.out

2
libc++abi.dylib: terminating with uncaught exception of type std::out_of_range: map::at:  key not found
[1]    64348 abort      ./a.out

I think it's safe to say, that given `A::id()` that returns by reference, you probably just made a typo. — StoryTeller - Unslander Monica, Jan 10 '19 at 12:43
@StoryTeller it was not a typo, but a mistake. Now I learned the right way to do it- - — Thrimbda, Jan 10 '19 at 13:03

Holt · Accepted Answer · 2019-01-10T12:42:44.317

std::map<long long, std::shared_ptr<B>> A::b();

You are returning the map by value, so each time you call a->b() you create a new copy of the map b_ which is why this kind of comparison:

a->b().find(3) != a->b().end()

...is undefined behavior since each call to b() returns a different map and comparing iterators from different container is undefined behavior.

Change your declaration (and definition) to return a (const-)reference:

const std::map<long long, std::shared_ptr<B>>& A::b();

score 9 · Answer 2 · answered Jan 10 '19 at 12:37

9

Your function A::b() returns the map by value. That means a copy is returned, a unique copy each time it is called, and whose iterators are not compatible with the iterators from any other copy.

Return by reference instead.

answered Jan 10 '19 at 12:37

Some programmer dude

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Why apply a non-exist key to map::find would return a iterator with first value map size in C++?

2 Answers2