问题描述
有一个原因,为什么传递引用STL映射作为const导致[]运算符破裂?我使用const时得到这个编译器错误(gcc 4.2):
这是函数原型:
void func(const char ch,std :: string& str,const std :: map< std :: string,std :: string>& map);
而且,我应该提到当我删除 const 。
如果我被正确指示,[]操作符将实际插入一个新对如果它没有找到钥匙,这当然会解释为什么会发生,但我不能想象这将是可以接受的行为。
如果有一个更好的方法,如使用 find 而不是[],我会很感激。我似乎无法找到工作,不过...我收到 const 不匹配的迭代器错误。
谢谢。
是的,您不能使用 operator []
。使用 find
,但注意返回 const_iterator
而不是 iterator
:
std :: map< std :: string,std :: string> :: const_iterator it;
it = map.find(name);
if(it!= map.end()){
std :: string const& data = it-> second;
// ...
}
就像指针。您不能将 int const *
分配给 int *
。同样,您不能将 const_iterator
分配给迭代器
。
Is there a reason why passing a reference to a STL map as const causes the [] operator to break? I get this compiler error (gcc 4.2) when I use const:
Here's the function prototype:
void func(const char ch, std::string &str, const std::map<std::string, std::string> &map);
And, I should mention that there is no problem when I remove the const keyword in front of std::map.
If I've been instructed correctly, the [] operator will actually insert a new pair into the map if it doesn't find the key, which would of course explain why this happens, but I can't imagine that this would ever be acceptable behavior.
If there is a better method, like using find instead of [], I'd appreciate it. I can't seem to get find to work either though... I receive const mismatched iterator errors.
Thanks.
Yes you can't use operator[]
. Use find
, but note it returns const_iterator
instead of iterator
:
std::map<std::string, std::string>::const_iterator it;
it = map.find(name);
if(it != map.end()) {
std::string const& data = it->second;
// ...
}
It's like with pointers. You can't assign int const*
to int*
. Likewise, you can't assign const_iterator
to iterator
.
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