Lambda表达式可在if语句中返回布尔值 [英] Lambda expression to return bool in if statement

问题描述

仅此而已，我想使用<内部的lambda表达式返回 true 或 false code> if（）语句。我看到了与我的问题类似的问题： LINK 但我找不到答案。

Just to get to the point, I want to return true or false using the lambda expression inside the if() statement. I saw this question which has similar question to mine: LINK but I could not find the answer.

所以这是我的示例代码：

So here is my example code:

if([&rel_pose](Eigen::VectorXd pose)
    {
        return (sqrt(pose(0) * pose(0) + pose(1) * pose(1)) < 2) ? true : false;
    }) // do smth

当我尝试编译时，出现此错误：

When I try to compile I get this error:

 error: could not convert ‘<lambda closure object>graphslam::GraphSLAM::main_pose_callback(const ConstPtr&)::<lambda(Eigen::VectorXd)>{rel_pose}’ from ‘graphslam::GraphSLAM::main_pose_callback(const ConstPtr&)::<lambda(Eigen::VectorXd)>’ to ‘bool’
  })

好，读取错误我以为我没有调用该函数，因为编译器不将该表达式视为bool。因此，我尝试使用此代码：

Ok, reading the error I thought that I did not call the function as the compiler does not treat the expression as bool. So I tried to use this code:

if(([&rel_pose](Eigen::VectorXd pose)
    {
        return (sqrt(pose(0) * pose(0) + pose(1) * pose(1)) < 2) ? true : false;
    };)) // do smth

错误：

expected ‘)’ before ‘;’ token
  };)) return;

这似乎是一个明显的错误，但对我来说，我可能不正确地理解语法并以为询问发生了什么。

That might look like an obvious error but for me, I probably do not understand the syntax correctly and thought to ask what is happening.

编辑：
请注意，我已经简化了代码，因此您可以轻松地复制错误。我知道在这种特殊情况下lambda表达式没有任何意义。

Please note that I have simplified the code so you can replicate the error easily. I know that lambda expression in this particular case does not make any sense.

推荐答案

您忘记了调用lambda。现在，您要说的是 if（function_pointer），因此编译器无法将其转换为布尔表达式。

you forgot to call your lambda. Right now you're saying if(function_pointer) hence the compiler failing to convert that into a boolean expression.

带有布尔lambda的简单 if 子句的写法是：

A simple if clause with boolean lambda is therefore written like:

if ([]() {
    return true;
}()) {
    //do sth
}

---

同时将变量作为参数也会导致错误捕获它。您必须决定，所以要么：

---

You also have an error by having a variable be a parameter while simultaneously capturing it. You have to decide, so either:

if([](Eigen::VectorXd pose)
    {
        return (sqrt(pose(0) * pose(0) + pose(1) * pose(1)) < 2) ? true : false;
    }(rel_pose)){
    //do sth
}

或

if([&rel_pose]()
    {
        return (sqrt(rel_pose(0) * rel_pose(0) + rel_pose(1) * rel_pose(1)) < 2) ? true : false;
    }()){
    //do sth
}

---

在这种情况下是否需要lambda是有问题的，您可以摆脱lamdba，将布尔表达式保留在if子句。在谈论它时-无需在此处使用三元运算符。 返回sqrt（rel_pose（0）* rel_pose（0）+ rel_pose（1）* rel_pose（1））< 2; 就足够了，而且更具可读性。

---

The need of a lambda is in this case questionable, you could just get rid of the lamdba leaving the boolean expression in the if clause. When talking about it - no need to use a ternary operator here. return sqrt(rel_pose(0) * rel_pose(0) + rel_pose(1) * rel_pose(1)) < 2; is sufficient and more readable.

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