计划1：它的工作原理
#include <iostream>
int main(){
auto accumulator = [](int x) {
return [=](int y) -> int {
return x+y;
};
};
auto ac=accumulator(1);
std::cout << ac(1) << " " << ac(1) << " " << ac(1) << " " << std::endl;
std::cout << ac(1) << " " << ac(1) << " " << ac(1) << " " << std::endl;
std::cout << ac(1) << " " << ac(1) << " " << ac(1) << " " << std::endl;
}
产量符合预期：
2 2 2
2 2 2
2 2 2
2.第2号方案：关闭,工作正常
#include <iostream>
int main(){
auto accumulator = [](int x) {
return [&](int y) -> int {
return x+=y;
};
};
auto ac=accumulator(1);
std::cout << ac(1) << " " << ac(1) << " " << ac(1) << " " << std::endl;
std::cout << ac(1) << " " << ac(1) << " " << ac(1) << " " << std::endl;
std::cout << ac(1) << " " << ac(1) << " " << ac(1) << " " << std::endl;
}
输出是：
4 3 2
7 6 5
10 9 8
程序3：带有std :: function的程序1,Works Fine
#include <iostream>
#include <functional> // std::function
int main(){
typedef std::function<int(int)> fint2int_type;
typedef std::function<fint2int_type(int)> parent_lambda_type;
parent_lambda_type accumulator = [](int x) -> fint2int_type{
return [=](int y) -> int {
return x+y;
};
};
fint2int_type ac=accumulator(1);
std::cout << ac(1) << " " << ac(1) << " " << ac(1) << " " << std::endl;
std::cout << ac(1) << " " << ac(1) << " " << ac(1) << " " << std::endl;
std::cout << ac(1) << " " << ac(1) << " " << ac(1) << " " << std::endl;
}
输出是：
2 2 2
2 2 2
2 2 2
程序4：带有std :: function,Undefined Behavior的程序2
#include <iostream>
#include <functional> // std::function
int main(){
typedef std::function<int(int)> fint2int_type;
typedef std::function<fint2int_type(int)> parent_lambda_type;
parent_lambda_type accumulator = [](int x) -> fint2int_type{
return [&](int y) -> int {
return x+=y;
};
};
fint2int_type ac=accumulator(1);
std::cout << ac(1) << " " << ac(1) << " " << ac(1) << " " << std::endl;
std::cout << ac(1) << " " << ac(1) << " " << ac(1) << " " << std::endl;
std::cout << ac(1) << " " << ac(1) << " " << ac(1) << " " << std::endl;
}
该计划的第一次运行给出：
4 3 2
4 3 2
12364812 12364811 12364810
第二轮同一程序：
4 3 2
4 3 2
1666060 1666059 1666058
第三个：
4 3 2
4 3 2
2182156 2182155 2182154
我如何使用std :: function打破代码？为什么程序1 – 3运行良好,而程序4正确调用ac(1)三次(！)？为什么4号程序卡在接下来的三个案例中,好像变量x已被值捕获,而不是参考.并且ac(1)的最后三个调用完全不可预测,好像对x的任何引用都将丢失.
解决方法
I hope to find out why there is an
undefined behavior in the code
每次我处理复杂而复杂的lambda时,我觉得首先将其转换为函数 – 对象形式会更容易.因为lambdas只是函数对象的语法糖,并且对于每个lambda,都存在与对应的函数对象的一对一映射.本文解释了如何进行翻译：
http://blogs.msdn.com/b/vcblog/archive/2008/10/28/lambdas-auto-and-static-assert-c-0x-features-in-vc10-part-1.aspx
例如,你的程序没有2：
#include <iostream>
int main(){
auto accumulator = [](int x) {
return [&](int y) -> int {
return x+=y;
};
};
auto ac=accumulator(1);
std::cout << ac(1) << " " << ac(1) << " " << ac(1) << " " << std::endl;
std::cout << ac(1) << " " << ac(1) << " " << ac(1) << " " << std::endl;
std::cout << ac(1) << " " << ac(1) << " " << ac(1) << " " << std::endl;
}
将由编译器大致翻译成这一个：
#include <iostream>
struct InnerAccumulator
{
int& x;
InnerAccumulator(int& x):x(x)
{
}
int operator()(int y) const
{
return x+=y;
}
};
struct Accumulator
{
InnerAccumulator operator()(int x) const
{
return InnerAccumulator(x); // constructor
}
};
int main()
{
Accumulator accumulator;
InnerAccumulator ac = accumulator(1);
std::cout << ac(1) << " " << ac(1) << " " << ac(1) << " " << std::endl;
std::cout << ac(1) << " " << ac(1) << " " << ac(1) << " " << std::endl;
std::cout << ac(1) << " " << ac(1) << " " << ac(1) << " " << std::endl;
}
现在,问题变得非常明显：
InnerAccumulator operator()(int x) const
{
return InnerAccumulator(x); // constructor
}
这里InnerAccumulator的构造函数将引用x,这是一个局部变量,一旦退出operator()范围就会死亡.所以,是的,你只是得到了一个明确的旧的未定义的行为,因为你怀疑.
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