I was looking at std::unordered_map and saw that if I wanted to use a string as the key, I'd have to create a class containing a functor.
我正在查看std :: unordered_map并看到如果我想使用字符串作为键，我必须创建一个包含仿函数的类。
Out of curiosity, I was wondering if a lambda could be used in place of this.
出于好奇，我想知道是否可以使用lambda来代替这个。
Here's the working original:
这是工作原创：
struct hf
{
size_t operator()(string const& key) const
{
return key[0]; // some bogus simplistic hash. :)
}
}
std::unordered_map<string const, int, hf> m = {{ "a", 1 }};
Here's my attempt:
这是我的尝试：
std::unordered_map<string const, int, [](string const& key) ->size_t {return key[0];}> m = {{ "a", 1 }};
That failed with the following errors:
失败时出现以下错误：
exec.cpp: In lambda function:
exec.cpp:44:77: error: ‘key’ cannot appear in a constant-expression
exec.cpp:44:82: error: an array reference cannot appear in a constant-expression
exec.cpp: At global scope:
exec.cpp:44:86: error: template argument 3 is invalid
exec.cpp:44:90: error: invalid type in declaration before ‘=’ token
exec.cpp:44:102: error: braces around scalar initializer for type ‘int’
Given the errors, it would seem that the lamba is different enough from a functor that it makes it not a constant expression. Is that correct?
鉴于错误，lamba似乎与仿函数不同，它使得它不是一个常量表达式。那是对的吗？
1 个解决方案
#1
12
The way to pass the lambda function is:
传递lambda函数的方法是：
auto hf = [](string const& key)->size_t { return key[0]; };
unordered_map<string const, int, decltype(hf)> m (1, hf);
^^^^^^^^^^^^ ^^
passing type object
The output of decltype(hf) is a class type which doesn't have default constructor (it's deleted by =delete). So, you need pass the object by constructor of unordered_map to let it construct the lambda object.
decltype（hf）的输出是一个没有默认构造函数的类类型（它被= delete删除）。因此，您需要通过unordered_map的构造函数传递对象，以使其构造lambda对象。
#1
12
The way to pass the lambda function is:
传递lambda函数的方法是：
auto hf = [](string const& key)->size_t { return key[0]; };
unordered_map<string const, int, decltype(hf)> m (1, hf);
^^^^^^^^^^^^ ^^
passing type object
The output of decltype(hf) is a class type which doesn't have default constructor (it's deleted by =delete). So, you need pass the object by constructor of unordered_map to let it construct the lambda object.
decltype（hf）的输出是一个没有默认构造函数的类类型（它被= delete删除）。因此，您需要通过unordered_map的构造函数传递对象，以使其构造lambda对象。