std :: setw，std :: setfill等的实际返回类型是什么？

Question

I'm curious of the real return type of std::setw and std::setfill

As we see references, their datatype of return value is "undefined". However, Is it possible to declare a function without return type?

In a situation that I have to develop a method supplying extended features for 'cout' or 'cin', and this method should be called like

cout << foo(32, 'A', 0.00f) << "Hello world!";

How should I declare the method?

Answer 1

The return type of std::setw et al is unspecified because each C++ implementation may decide to do it differently, so there's no one answer - you have to survey the compilers/versions you're interested in.

Looking at the libstdc++ used with GCC, we see:

00214   struct _Setw { int _M_n; };
00215 
00216   /**
00217    *  @brief  Manipulator for @c width.
00218    *  @param  n  The new width.
00219    *
00220    *  Sent to a stream object, this manipulator calls @c width(n) for
00221    *  that object.
00222   */
00223   inline _Setw 
00224   setw(int __n)
00225   { return { __n }; } 

_Setw is a small struct to capture the width parameter, and which std::ostream& operator<<(std::ostream&, _Setw) and ...>>... can then handle to set the width in the stream:

00227   template<typename _CharT, typename _Traits>
00228     inline basic_istream<_CharT, _Traits>& 
00229     operator>>(basic_istream<_CharT, _Traits>& __is, _Setw __f)
00230     {
00231       __is.width(__f._M_n);
00232       return __is; 
00233     }
00234 
00235   template<typename _CharT, typename _Traits>
00236     inline basic_ostream<_CharT, _Traits>& 
00237     operator<<(basic_ostream<_CharT, _Traits>& __os, _Setw __f)
00238     {
00239       __os.width(__f._M_n);
00240       return __os; 
00241     }

As we see references, their datatype of return value is "undefined".

It's unspecified , not undefined.

However, Is it possible to declare a function without return type?

No - every function must have a return type, even if only void .

In a situation that I have to develop a method supplying extended features for 'cout' or 'cin', and this method should be called like

cout << foo(32, 'A', 0.00f) << "Hello world!";

How should I declare the method?

You can do something similar and have function foo return an object that you write streaming operators for. Those streaming functions should then manipulate the stream: you'll need to use iword and xalloc to give the streams extra state to track the potentially modified behaviours you're going to add - see this answer .

Answer 2

It is unspecified, or in other words, not specified in the C++ standard and depends on your library implementation. More on that here .

You can't declare your own function to manipulate the insertion/extraction operators unless you overload them to accept a new type.