Inserting addresses into sets, the size of the sets is smaller than expected

Question

I am getting a size of 1. Shouldn't it be 4? I am inserting the addresses of integers into the sets.

void func(set<int*>& s1, set<int*>& s2, int a)
{
    s1.insert(&a);
    s2.insert(&a);
}

int main()
{
    set<int*> s1, s2;

    int a = 1, b = 2, c = 3, d = 4;
    func(s1, s2, a);
    func(s1, s2, b);
    func(s1, s2, c);
    func(s1, s2, d);

    cout<<"  s1.size = "<<s1.size()<<"  s2.size = "<<s2.size()<<endl;
}

Really? 3 people think this question should be closed because it "does not appear to be about programming"? — user200783, Jul 05 '17 at 09:40
Surprisingly, GCC (`g++ -W -Wall -Wextra -Werror`) and Clang compile this code without any warnings. — pts, Jul 05 '17 at 11:29
@user200783 _"... within the scope defined in the [help center](https://stackoverflow.com/help/on-topic)."_ — YSC, Jul 05 '17 at 11:59
This is neither "a problem that can no longer be reproduced" nor "a simple typographical error". Voted to reopen. — user200783, Jul 05 '17 at 14:26

Holt · Answer 1 · 2017-07-05T06:24:33.517

44

&a inside func is the address of the local parameter a, not the address of the original variable (a, b, c or d), which:

Can take various kind of values between different calls of func;
Becomes an invalid pointer once you reach the end of the scope in which a is declared (the end of func here).

In your case, since you do nothing but calling func 4 times in a row, the address of this parameter happens to not change, which is why you get a size of 1 (you insert the same pointer 4 times).

This behavior is implementation defined (thanks @Rakete1111) since sets will need to compare invalid pointers.

Accessing the pointed element through s1 or s2 (e.g. **s1.begin()) is undefined behavior though.

edited Jul 05 '17 at 06:24

answered Jul 05 '17 at 06:05

Holt

36,600
7
92
139

3

@Holt It looks like it is [implementation defined behavior](https://stackoverflow.com/a/30694084/3980929). – Rakete1111 Jul 05 '17 at 06:20
2

@Rakete1111 Thanks, I did not know that. I have updated the answer. – Holt Jul 05 '17 at 06:24

songyuanyao · Answer 2 · 2017-07-05T06:54:36.087

The parameter a is passed by value, that means what you're inserting is the address of the local parameter copied from the arguments. a will be constructed everytime when func is called, their addresses might be same or different; nothing is guaranteed.

And it's worth noting that the parameter a will be destroyed when get out of the function, then the pointers inserted will become dangled.

You should make it pass-by-reference, then the addresses of the arguments (i.e. the variable a, b, c, d in main()) will be inserted. These addresses are different, and won't become dangled inside the body of main().

void func(set<int*>& s1, set<int*>& s2, int& a)
{
    s1.insert(&a);
    s2.insert(&a);
}

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Inserting addresses into sets, the size of the sets is smaller than expected

2 Answers2