Блог пользователя bhikkhu

here is a puzzle
Автор bhikkhu, 8 лет назад, По-английски 
#include <vector>
#include <utility>
#include <cassert>
int main() {
    std::vector<int> vec;
    vec.push_back(0);
    int & r = vec[0];
    if (r == 0) {
        r = 1;
        vec.push_back(0);
    }
    assert(r > 0);
    return 0;
}

Find out why this code is dangerous :D

I became a victim of this problem while solving a problem using logic similar to the code shown.

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8 лет назад, # |
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By inserting another element, the vector will double its capacity and thus the memory address may change. Is it correct?

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8 лет назад, # |
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It's dangerous because in the second push_back, the vector reallocates new array of double length, and deletes the old one. (If you don't know how vector works, see this.)

Hence, r becomes pointing to an invalid memory position, so its value will be just garbage. One way to make sure of this, is to add the following line before doing any push_backs.

vec.reserve(2);
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