错误:将"const std :: map< int，int>"作为“此"参数传递会丢弃限定符[-fpermissive] [英] error: passing ‘const std::map<int, int>’ as ‘this’ argument discards qualifiers [-fpermissive]

问题描述

从const C ++ std :: map获取条目无法在gcc 5.4.0上编译.

Fetching an entry from a const C++ std::map fails to compile on gcc 5.4.0.

test_map.cpp: In function ‘int main()’:
test_map.cpp:9:24: error: passing ‘const std::map<int, int>’ as ‘this’ argument discards qualifiers [-fpermissive]
                 foo[key];

最小测试用例

// Compile with
// g++ test_map.cpp -o test_map

#include <map>

int main() {
    const std::map<int, int> foo;
    foo[0]; // compiles if "const" above is suppressed
}

发布前先看

这看起来类似于通过'const，此参数将丢弃限定词[-fpermissive]是关于 Cache 的，而不是 std :: map 的.原因是:用户调用 write()方法.该方法未声明为 const ，这很有意义，因为编写大概是修改了该对象.

Look before you post

This looks similar to passing ‘const this argument discards qualifiers [-fpermissive] which is about a Cache, not a std::map. The cause there: the user calls a write() method. That method is not declared const which makes sense since writing presumably modified the object.

但是在这里，从地图中获取元素不会修改地图，是吗?

But here, fetching an element from a map does not modify the map, does it?

我真实用例中的实际地图确实是 const.它已在源代码中完全初始化.修改它没有任何意义.声明为非常量实际上可以解决该问题，但没有任何意义.

The actual map in my real use case is indeed const. It's fully initialized in the source code. It does not make sense to modify it. Declaring it non-const practically solves the problem but does not make sense.

推荐答案

operator [] 在 std :: map 中没有 const 限定词代码>，如您从文档中看到的，例如 std :: map :: operator []-cppreference.com :

operator[] hasn't a const qualifier in std::map, as you can see from the documentation, e.g. std::map::operator[] - cppreference.com:

返回对映射到与键等效的键的值的引用，如果该键尚不存在，则执行插入.

因此，您不能直接在 const 实例上使用它.请改用 at (请参考 std :: map :: at-cppreference.com )，如果您可以负担得起C ++ 11的功能.

Therefore you cannot use it directly on a const instance. Use at instead (ref std::map::at - cppreference.com) if you can afford C++11 features.

这些成员函数的声明如下:

Declarations for those member functions follow:

T& operator[](const key_type& x);
T& operator[](key_type&& x);
T&       at(const key_type& x);
const T& at(const key_type& x) const;

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