为什么我可以在std :: map中使用const char *作为键<std::string, int>

Question

I have define a data structure

std::map<std::string, int> a;

I found I can pass const char* as key, like this:

a["abc"] = 1;

Which function provides automatic type conversion from const char* to std::string?

Answer 1

std::string has a constructor that allows the implicit conversion from const char* .

basic_string( const CharT* s,
              const Allocator& alloc = Allocator() );

means that an implicit conversion such as

std::string s = "Hello";

is allowed.

It is the equivalent of doing something like

struct Foo
{
  Foo() {}
  Foo(int) {} // implicit converting constructor.
};

Foo f1 = 42;
Foo f2;
f2 = 33 + 9;

If you wanted to disallow the implicit conversion construction, you mark the constructor as explicit :

struct Foo 
{
  explicit Foo(int) {}
};

Foo f = 33+9; // error
Foo f(33+9); // OK
f = Foo(33+9); // OK

Answer 2

There is a constructor for std::string which takes const char* as a parameter.

string::string(const char*);

Unless the constructor is declared explicit then the compiler will apply one use defined conversion if needed to call any function.

Answer 3

See string constructor . The constructor provides the conversion for the key in your map. It's equivalent to

a[std::string("abc")] = 1;

Answer 4

In C++ if you make a class constructor that only takes one parameter, then (unless you tell it otherwise with explicit ), that parameter's type will be implicitly convertable to your class.

std::string has such a constructor for char *

Yes, this can cause some unexpected behavior on occasion. This is why you generally should put explicit on single-parameter constructors, unless you really want these silent conversions.