c++ - const XX 丢弃限定符 [- fpermissive]

Question

在下面的代码片段 1 中，mKnownSRList 定义如下:

std::vector<EndPointAddr*> mKnownSRList;

我收到代码片段 2 中显示的编译错误。你能告诉我这段代码有什么问题吗？ getTipcAddress() 和 compareTo 函数的内容显示在下面的代码片段 3 和 4 中。

CODE SNIPPET 1(标出编译错误)

void 
ServiceRegistrarAPI::removeKnownSR(EndPointAddr & srEndPointAddr)
{
   auto last = 
   std::remove_if(mKnownSRList.begin(),
                  mKnownSRList.end(),
                 [srEndPointAddr]( EndPointAddr* o )
                 { 
                    //LINE 355 is the following
            EndPointTipcAddr myTipcAddress = srEndPointAddr.getTipcAddress();
                EndPointTipcAddr otherTipcAddress = o->getTipcAddress();

            return (myTipcAddress.compareTo(otherTipcAddress));
         });

    if(*last != nullptr)
    {
     delete *last;
    }

    mKnownSRList.erase(last, mKnownSRList.end());    
}

SNIPPET 2(编译错误)

  ServiceRegistrarAPI.cpp:355:72: error: passing ‘const EndPointAddr’ as ‘this’   argument of ‘EndPointTipcAddr& EndPointAddr::getTipcAddress()’ discards qualifiers [-  fpermissive]

代码片段 3(getTipcAddress 函数)

EndPointTipcAddr & getTipcAddress() { return mTipcAddress; }

CODE NIPPET 4(compareTo 函数)

  bool

  EndPointTipcAddr::compareTo(EndPointTipcAddr &rhs) 
  {     
      if( (mType == rhs.getType()) && (mInstanceNo == rhs.getInstanceNo()) )
      {
        return true;
      } 

      return false;
  }

Answer 1

参见 S5.1.2.5:

The closure type for a lambda-expression has a public inline function call operator (13.5.4) whose parameters and return type are described by the lambda-expression’s parameter-declaration-clause and trailingreturn- type respectively. This function call operator is declared const (9.3.1) if and only if the lambdaexpression’s parameter-declaration-clause is not followed by mutable. It is neither virtual nor declared volatile. Default arguments (8.3.6) shall not be specified in the parameter-declaration-clause of a lambdadeclarator. Any exception-specification specified on a lambda-expression applies to the corresponding function call operator. An attribute-specifier-seq in a lambda-declarator appertains to the type of the corresponding function call operator. [ Note: Names referenced in the lambda-declarator are looked up in the context in which the lambda-expression appears. —end note ]

基本上，这意味着生成的仿函数的 operator() 默认是 const，并且您已经按值捕获，并且这个捕获的变量是生成的仿函数的成员。

所以，您有两个选择:

1. 通过引用而非值进行捕获。

2. 将您的 lambda 更改为以下内容(注意参数声明子句后面的 mutable):

   [srEndPointAddr](EndPointAddr* o) 可变 { ... }