Question

I found something surprising with std::vector that I thought I’d ask about here to hopefully get some interesting answers.

The code below simply copies a string into a char vector and prints the contents of the vector in two ways.

#include <vector>
#include <string>
#include <iostream>

int main()
{
    std::string s("some string");
    std::vector<char> v;
    v.reserve(s.size()+1);

    // copy using index operator
    for (std::size_t i=0; i<=s.size(); ++i)
        v[i] = s[i];

    std::cout << "&v[0]:     " << &v[0] << "\n";
    std::cout << "begin/end: " << std::string(v.begin(), v.end()) << "\n";

    // copy using push_back
    for (std::size_t i=0; i<=s.size(); ++i)
        v.push_back(s[i]);

    std::cout << "&v[0]:     " << &v[0] << "\n";
    std::cout << "begin/end: " << std::string(v.begin(), v.end()) << "\n";

    return 0;
}

Building and running this yields:

$ g++ main.cpp -o v && ./v
&v[0]:     some string
begin/end: 
&v[0]:     some string
begin/end: some string

My expectation was that it would print the string correctly in both cases, but assigning character by character using the index operator doesn’t print anything when later using begin() and end() iterators.

Why isn’t end() updated when when using []? If this is intentional, what’s the reason it’s working like this?

Is there a reasonable explanation for this behaviour? 🙂

I’ve only tried this with gcc 4.6.1 so far.

Answer 1

Typical example of Undefined Behavior.

You are only ever allowed to access elements by index (using operator[]) between 0 and v.size()-1 (included).

Using reserve does not modify the size, only the capacity. Would you have used resize instead, it would work as expected.