Change string by index

Question

I am a beginner in C++ and I am currently working with strings.

My question is why when compiling the code I\'m providing below, I can get the string\'s characters w

Answer 1

You have not resized your altered string to fit the length of the original string before the loop, thus your code exhibits undefined behavior:

altered[i] = original[i] + 5; // UB -  altered is empty

To fix this, resize altered before the loop:

altered.resize(original.size());

Or use std::string::operator+= or similar to append to altered:

altered += original[i] + 5;

This way, it can be empty before the loop, it will automatically resize itself to contain appended characters.

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Explanation

The way UB is happening here, is that you're succeeding in writing the data in the static array, which std::string uses for short string optimization (std::string::operator[] does no checks if you're accessing this array past the std::string::size()), but std::string::size() remains 0, as well as std::string::begin() == std::string::end().

That's why you can access the data individually (again, with UB):

cout << altered[0] << " " << altered[1] << " " << altered[2] << endl;

but cout << aligned does not print anything, considering simplified operator<< definition for std::string looks functionally like this:

std::ostream &operator<<(std::ostream &os, std::string const& str)
{
    for(auto it = str.begin(); it != str.end(); ++it) // this loop does not run
        os << *it;

    return os;
}

In one sentence, std::string is not aware of what you did to its underlying array and that you meant the string to grow in length.

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To conclude, way of doing this transformation:

std::transform(original.begin(), original.end(),
    std::back_inserter(altered), // or altered.begin() if altered was resized to original's length
    [](char c)
    {
        return c + 5;
    }

(required headers: , )