传递'const这个参数丢弃限定符[-fpermissive] [英] passing ‘const this argument discards qualifiers [-fpermissive]

问题描述

我有一个类 Cache ，它有一个函数write指定为

  bool write（const MemoryAccess& memory_access，CacheLine& cl）; 
  

我这样调用这个函数。

  const Cache * this_cache; 
 c =（a == b）？my_cache：not_cache; 
 c-> write（memory_access，cl）; 
  

上面的代码给了我下面的错误：

< blockquote>

传递'const Cache'作为'bool Cache :: write（const
MemoryAccess&，CacheLine&）'的'this'参数丢弃限定符[-fpermissive]。 p>

这个参数是编译器特定的，它有助于代码修改和打破局部命名空间变量优先级。但是这样一个变量并没有被传递到这里。

解决方案

由于 c 是 const常量缓存* 类型的，你只能调用 const 成员函数。

您有两种选择：从申报中删除 const
$ b c ;

<2>更改 Cache :: write（）

  bool write（const MemoryAccess& memory_access，CacheLine& cl）const; 
  

（请注意最后添加的 const 。）

I have a class Cache which has a function write specified as

bool write(const MemoryAccess &memory_access, CacheLine &cl);

I am calling this function like this.

const Cache *this_cache;
c = (a==b)?my_cache:not_cache;
c->write(memory_access,cl);

The above line is giving me following error

"passing ‘const Cache’ as ‘this’ argument of ‘bool Cache::write(const MemoryAccess&, CacheLine&)’ discards qualifiers [-fpermissive]."

the this argument is compiler specific which helps in code-mangling and breaking local namespace variable priority. But such a variable is not being passed here.

解决方案

Since c is of type const Cache *, you can only call const member functions on it.

You have two options:

(1) remove const from the declaration of c;

(2) change Cache::write() like so:

 bool write(const MemoryAccess &memory_access, CacheLine &cl) const;

(Note the added const at the end.)

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