编译器警告：无法推断lambda返回类型 [英] Compiler warning: lambda return type cannot be deduced

问题描述

请考虑以下示例：

#include <algorithm>
#include <iostream>

int main()
{
    std::string str = "abcde4fghijk4l5mnopqrs6t8uvwxyz";
    std::string str2;

    std::remove_copy_if(str.begin(), str.end(),
        std::back_inserter(str2),
        [](char& c) {
            if (std::isdigit(c))
                return true;      // <----- warning here
            else
                return false;
        }
    );

    std::cout << str2 << '\n';
}

使用GCC 4.6.1，可以很好地进行编译并打印出预期的输出（字母），但我得到一个警告，说只有当return语句是函数体中的唯一语句时，才能推断出lambda返回类型。

With GCC 4.6.1, this compiles fine and prints expected output (the alphabet) but I get a warning saying "lambda return type can only be deduced when the return statement is the only statement in the function body".

现在，我知道如何摆脱警告（使用尾随返回类型或简单地说 return isdigit（c）; ），但是我很好奇，因为编译器不会不要警告（或应该警告）：这样的代码可能会出什么问题？标准是否对此有任何说明？

Now, I know how to get rid of the warning (using trailing return type or simply saying return isdigit(c);), but I'm curious, since compiler doesn't warn for nothing (or so it should be): what could possibly go wrong in code like this? Does standard say anything about it?

推荐答案

正如@ildjarn在其评论中所说，根据代码，您的代码只是格式不正确

As @ildjarn says in his comment, your code is simply ill-formed according to the standard.

§5.1.2[expr.prim.lambda] p4

[...]如果 lambda表达式不包含 trailing-return-type ，则为就像 trailing-return-type 表示以下类型一样：

[...] If a lambda-expression does not include a trailing-return-type, it is as if the trailing-return-type denotes the following type:

• 如果 compound-statement 的格式为
  
  { attribute-specifier-seq _opt return expression ； }
  
  左值到右值转换（4.1），数组到指针转换
  （4.2 ），以及函数到指针的转换（4.3）;


• 否则， void 。

• if the compound-statement is of the form
  { attribute-specifier-seq_opt return expression ; }
  the type of the returned expression after lvalue-to-rvalue conversion (4.1), array-to-pointer conversion (4.2), and function-to-pointer conversion (4.3);
• otherwise, void.

[...]

就是这样，基本上如果里面的代码大括号（在标准中称为 compund-statement ）除了 return some_expr; 外，标准表示返回类型不可推导，您得到 void 返回类型。

That's it, basically if the code inside the curly brackets (called compund-statement in the standard) is anything but return some_expr;, the standard says the return type is undeducible and you get a void return type.

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