为什么在C ++ 11中按值传递的lambda参数是只读的？ [英] Why are lambda arguments passed by value read-only in C++11?

问题描述

当函数按值接受参数时，通常可以对其进行修改。但是，lambda似乎并非如此。为什么？

When a function takes an argument by value, it can usually modify it. However, this does not seem to be the case with lambdas. Why?

int main()
{
  int x = 0;
  auto lambda = [x] { x = 1; }; // error: assignment of read-only variable ‘x’
  return 0;
}

推荐答案

Herb Sutter回答了这个问题< a href = http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2012/n3424.pdf rel = noreferrer>此处如下；

Herb Sutter answered the question here as follow;

以这个稻草人为例，程序员通过值捕获本地
变量并尝试修改捕获的值（即
lambda对象的成员变量）：

Consider this straw man example, where the programmer captures a local variable by value and tries to modify the captured value (which is a member variable of the lambda object):

int val = 0;
auto x = [=](item e)            // look ma, [=] means explicit copy
            { use(e,++val); };  // error: count is const, need ‘mutable’
auto y = [val](item e)          // darnit, I really can’t get more explicit
            { use(e,++val); };  // same error: count is const, need ‘mutable’

该功能似乎是出于担心用户
可能不会意识到自己已获得副本的考虑而添加的，尤其是由于Lambda
是可复制的，因此他可能正在更改其他Lambda的副本。

This feature appears to have been added out of a concern that the user might not realize he got a copy, and in particular that since lambdas are copyable he might be changing a different lambda’s copy.

注意：这是更改功能的建议书。

Note: That is a proposal paper to change the feature.

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