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C++0x Lambda 본문

Dev/C++

C++0x Lambda

디클 2009.05.20 13:43
아래 링크의 글 중 Herb Sutter 의 Lambda 예제 및 설명에 관한 리플..
http://herbsutter.wordpress.com/2008/03/29/trip-report-februarymarch-2008-iso-c-standards-meeting/

[]와 []안에 들어가는 내용에 대해서 명쾌하게 설명을..




Herb Sutter

Re binders: Okay, I give! I’ll use a better example next time.

(no name) asked: "How are local variables captured?" You have to specify whether it’s by copy or by reference. So this example is illegal because it tries to use a local variable:

    int numWidgets = 0;
    for_each( v.begin(), v.end(), []( Widget& w )
    {
        ++numWidgets;  // error, numWidgets is not in scope
    } );

If you want to update numWidgets directly, capture it by reference:

    for_each( v.begin(), v.end(), [&numWidgets]( Widget& w )
    {
        ++numWidgets;  // increments original numWidgets
    } );
    // numWidgets == v.size() here

Or use the shorthand [&] to take all captured variables implicitly by reference:

    for_each( v.begin(), v.end(), [&]( Widget& w )
    {
        ++numWidgets;  // increments original numWidgets
    } );
    // numWidgets == v.size() here

What if you want a local copy? You say to pass it by value, but for safety reasons the current proposal says you get a read-only copy that you can’t modify:

    for_each( v.begin(), v.end(), [numWidgets]( Widget& w )
    {
        int i = numWidgets; // ok
        ++i;
        // "++numWidgets;" would be an error
    } );
    // numWidgets == 0 here

Or use the shorthand [=] to take all captured variables implicitly by copy:

    for_each( v.begin(), v.end(), [=]( Widget& w )
    {
        int i = numWidgets; // ok
        ++i;
        // "++numWidgets;" would be an error
    } );
    // numWidgets == 0 here

Similarly, for the question: "What will happen in the following case:"

    int flag = 0;
    mypool.run( [] { flag = 1; } );
    cout << flag << endl;

The answer is that the code is illegal, you have to say whether you capture flag by value or by reference, which can be as simple as replacing [] with [=] or [&]. And if you capture by value, you get a read-only copy so you couldn’t assign to it.


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