CODEWITHSUNDEEP
c++stlvector
murrekatt
murrekatt

Reputation: 6129

What's this unexpected std::vector behavior?

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.

Upvotes: 1

Views: 243

Answers (2)

Mike Seymour
Mike Seymour

Reputation: 254461

In the first case, you have undefined behaviour. reserve sets the capacity, but leaves the size as zero. Your loop then writes to invalid locations beyond the end of the vector. Printing using the (invalid) pointer appears to work (although there is no guarantee of that), since you've written the string to the memory that it points at; printing using the iterator range prints nothing, because the vector is still empty.

The second loop correctly increases the size each time, so that the vector actually contains the expected contents.

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

[] is intended to be as fast as possible, so it does no range checking. If you want a range check, use at(), which will throw an exception on an out-of-range access. If you want to resize the array, you have to do it yourself.

Upvotes: 5

Matthieu M.
Matthieu M.

Reputation: 299820

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.

Upvotes: 11

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