Debuggergdb.cpp redundant code

[AndrewCot]

In src\plugins\contrib-wip\debuggerGDB_MI\debuggergdb.cpp the OnTimer function is:

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void DebuggerGDB::OnTimer(cb_unused wxTimerEvent& event)
{
    // send any buffered (previous) output
    ParseOutput(wxEmptyString);

    CheckIfConsoleIsClosed();

    wxWakeUpIdle();
}
The ParseOutput function is:

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void DebuggerGDB::ParseOutput(const wxString& output)
{
    if (!output.IsEmpty() && m_State.HasDriver())
    {
        m_State.GetDriver()->ParseOutput(output);
    }
}

So my conclusion is that the OnTimer line

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ParseOutput(wxEmptyString); is redundant. Have I missed something?

[ollydbg]

From my point of view, you are correct.

[Pecan]

That's just one of those laughable moments that make me think: "I hope I didn't do that!"

[AndrewCot]

Another code snippet to check out, but this one will not slow things down as much as the previous one:

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        buffer.Remove(idx);
        // remove the '>>>>>>' part of the prompt (or whats left of it)
        int cnt = 6; // max 6 '>'
        while (!buffer.empty() && buffer.Last() == _T('>') && cnt--)
            buffer.RemoveLast();
        if (!buffer.empty() && buffer.Last() == _T('\n'))
            buffer.RemoveLast();
        cmd->ParseOutput(buffer.Left(idx));

Check the first line and then ask yourself why the last line is not

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cmd->ParseOutput(buffer);

[ollydbg]

Another code snippet to check out, but this one will not slow things down as much as the previous one:

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        buffer.Remove(idx);
        // remove the '>>>>>>' part of the prompt (or whats left of it)
        int cnt = 6; // max 6 '>'
        while (!buffer.empty() && buffer.Last() == _T('>') && cnt--)
            buffer.RemoveLast();
        if (!buffer.empty() && buffer.Last() == _T('\n'))
            buffer.RemoveLast();
        cmd->ParseOutput(buffer.Left(idx));

Check the first line and then ask yourself why the last line is not

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cmd->ParseOutput(buffer);

The first line is: Removes all characters from the string starting at idx, but it just returned the modified buffer, so the buffer is not changed. Right?

The last line is: Removes all the characters after the idx, and pass to the ParseOutput function.

So, the first line did nothing?

[AndrewCot]

The

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buffer.Remove(idx); removes the characters in the buffer, so it removed from the idx offset to the end.
The RemoveLast calls remove the '>' if they are there.

So the last line calling with only the Left idx characters is redundant as the first line removed the characters after idx.

[ollydbg]

The

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buffer.Remove(idx); removes the characters in the buffer, so it removed from the idx offset to the end.
The RemoveLast calls remove the '>' if they are there.

So the last line calling with only the Left idx characters is redundant as the first line removed the characters after idx.

OK, you are correct!