Do you want us to explain Lambda functions or the code?

Both, if you may. The reason why I want to understand it is because someday I might have to use it. Gotta know how it work and how to use.

Thanks!

By the way, I found a page introducing Lambda expression.

http://en.cppreference.com/w/cpp/language/lambda

Even the very first syntax part confuses my head. Oh my...

How familiar are you with C++?

Basically I am more familiar with C like Linked lists, stacks, pointers, etc.

C++ part is not that good, especially polymorphism and inheritance.

Learn c++ before doing Cocos2d-x

Check this out http://www.sonarlearning.co.uk/coursepage.php?topic=desktop&course=cplusplus

I have finished most of the lessons. I just picked those I am not so familiar with.

ok thats cool

Holy crap I have just closed the page without submitting. What a dumb I am.

I have just discovered a page that explains lambda expression. Try to explain it with my words.

Lambda Expression is a new function in C++11, which brings programmers flexibility and conciseness.

It is an expression of anonymoius function, which allows the programmers create a function without declaring it. 

[=] (int x) mutable throw() -> int
{
  //compound-statement
  int n = x + y;
  return n;
}

[=]: lambda-introducer, aka capture clause.

It is necessary to every lambda expression. Not only it is the keyword to lambda expression, but also it captures variables.

• [ ]: Do not capture the external variables.
• [=]: Pass every variables by value.
• [&]: Pass every variables by reference.
• [x, &y]: Pass x by value and pass y by reference.
• [=, &y]: Pass y by reference and pass the rest by value.
• [&, x]: Pass x by value and pass hte rest by reference.

And we should put the capture-default(=,&) in the first.

(int x): lambda declarator, aka parameter list.

There are some restrictions about it.

• No default assignment of the parameter.
• No parameter list with dynamic size.
• The parameter must include named parameter.

Lambda declarator is not necessary. If we do not need to pass any parameter we could just get rid of it.

mutable: mutable specification, which allows us to modify the variables passed by value. It is not necessary either.

throw(): exception specification.

(I do not know what the function exception is, but it says this could handle the exception part.)

→ int: return type. What lambda expression is returning. If we do not return anything, we could skip it.

--

Well, I think I get the most of it.

Now let’s see the code and discuss it, shall we?

void FloatWord::floatIn(const float delay, std::function callback)
{
	Size visibleSize = Director::getInstance()->getVisibleSize();
	//float in to the center of the screen. Therefore, the height is unchanged.
	MoveTo* move = MoveTo::create(delay, Point(visibleSize.width / 2, _begin.y));

	//This
	CallFunc* call = CallFunc::create(callback);
	//This

	Sequence* action = Sequence::create(move, call, NULL);
	_label->runAction(action);
}

void GameLayer::floatLevelWord()
{
	Size visibleSize = Director::getInstance()->getVisibleSize();
	_levelMsg = FloatWord::create("Level" + cocos2d::String::createWithFormat(": %d", GAMEDATA::getInstance()->getNextLevel())->_string, /*display level string*/
		visibleSize.width * FONT_SIZE, /*set the font size*/
		Point(visibleSize.width, visibleSize.height / 3 * 2)); /*set the position*/

	this->addChild(_levelMsg,1);
	_levelMsg->floatIn(FLOAT_EFFECT_DELAY_SEC, CC_CALLBACK_0(GameLayer::floatTargetScoreWord, this));
}

So is he passing the CC_CALLBACK_0 into the lambda function? Or did I just misunderstand?

Thank you.

Is this lambda expression?

If it’s not bound to an identifier then it’s a lambda function.