C++ lambda函数，效率也许不如想象的好

C++ lambda函数，当函数参数用复制的方式捕获时，参数存放在哪里呢？

开始以为在栈里，好像不对。

带着这个疑问，写一段递归函数，将参数的地址打出来看看。

参考代码如下：

#include <stdio.h> 
#include <functional> 

void fun2( std::function<void()> callback ) { 
    (callback)(); 
} 

void fun1(int n) { 
    if(n <= 0) return; 
    printf("stack address = %p, ", &n); 

    fun2([n]() { 
        printf("capture address = %p\n", &n); 
        fun1(n - 1); 
    }); 
} 

int main() { 
    fun1(200); 
    return 0; 
}

(编译环境：mingw64， 版本x64-4.8.0-release-posix-seh-rev2）

结果令人吃惊。以下为输出结果，注意捕获参数的地址，并不在程序的stack区域，

并且地址还不总是连续的。

很明显，lambda表达的复制捕获的参数，是动态分配出空间存放的。

特别关注效率或动态分配的场合，还是小心为妙。现实很骨感。

stack address = 000000000022F1E0, capture address = 00000000002F6D20
stack address = 000000000022F0C0, capture address = 00000000002F6D40
stack address = 000000000022EFA0, capture address = 00000000002F6D60
stack address = 000000000022EE80, capture address = 00000000002F6D80
stack address = 000000000022ED60, capture address = 00000000002F6DA0
stack address = 000000000022EC40, capture address = 00000000002F6DC0
stack address = 000000000022EB20, capture address = 00000000007A7810
stack address = 000000000022EA00, capture address = 00000000007A7820
stack address = 000000000022E8E0, capture address = 00000000007A7830
stack address = 000000000022E7C0, capture address = 00000000007A7840

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