问题描述
为什么下面程序的输出是什么？
Why is the output of the below program what it is?
#include <iostream>
using namespace std;
int main(){
cout << "2+3 = " <<
cout << 2 + 3 << endl;
}
产生
2+3 = 15
而不是预期的
2+3 = 5
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推荐答案
无论是有意还是无意，您在末尾都有<< 第一条输出行，您可能的意思是; 。因此，您基本上拥有
Whether intentionally or by accident, you have << at the end of the first output line, where you probably meant ;. So you essentially have
cout << "2+3 = "; // this, of course, prints "2+3 = "
cout << cout; // this prints "1"
cout << 2 + 3; // this prints "5"
cout << endl; // this finishes the line
所以问题归结为：为什么 cout<< cout; 打印 1 ？
So the question boils down to this: why does cout << cout; print "1"?
这可能是令人惊讶的，微妙。 std :: cout 通过其基类 std :: basic_ios 提供，旨在在布尔上下文中使用，例如
This turns out to be, perhaps surprisingly, subtle. std::cout, via its base class std::basic_ios, provides a certain type conversion operator that is intended to be used in boolean context, as in
while (cout) { PrintSomething(cout); }
这是一个非常糟糕的示例，因为很难使输出失败-但 std :: basic_ios 实际上是输入流和输出流的基类，对于输入来说，它更有意义：
This is a pretty poor example, as it's difficult to get output to fail - but std::basic_ios is actually a base class for both input and output streams, and for input it makes much more sense:
int value;
while (cin >> value) { DoSomethingWith(value); }
（在流结束时退出，或者当流字符不构成有效字符时退出循环
(gets out of the loop at end of stream, or when stream characters do not form a valid integer).
现在，此转换运算符的确切定义在标准的C ++ 03和C ++ 11版本之间已更改。在旧版本中，它是 operator void *（）const; （通常实现为 return fail（）？NULL：this; ），而在较新的版本中，它是显式运算符bool（）const; （通常简单地实现为 return！fail（）; ）。这两个声明在布尔上下文中都可以正常工作，但是在（错误）在此类上下文之外使用时的行为会有所不同。
Now, the exact definition of this conversion operator has changed between C++03 and C++11 versions of the standard. In older versions, it was operator void*() const; (typically implemented as return fail() ? NULL : this;), while in newer it's explicit operator bool() const; (typically implemented simply as return !fail();). Both declarations work fine in a boolean context, but behave differently when (mis)used outside of such context.
尤其是在C ++ 03规则下， cout<< cout 将解释为 cout<< cout.operator void *（）并打印一些地址。根据C ++ 11规则， cout<< cout 根本不应该编译，因为运算符被声明为 explicit ，因此不能参与隐式转换。实际上，这是进行此更改的主要动机-防止不必要的代码编译。符合这两个标准的编译器都不会生成打印 1 的程序。
In particular, under C++03 rules, cout << cout would be interpreted as cout << cout.operator void*() and print some address. Under C++11 rules, cout << cout should not compile at all, as the operator is declared explicit and thus cannot participate in implicit conversions. That was in fact the primary motivation for the change - preventing nonsensical code from compiling. A compiler that conforms to either standard would not produce a program that prints "1".
显然，某些C ++实现允许混合和匹配编译器和库，从而产生不一致的结果（引用@StephanLechner：我在xcode中找到了一个设置，该设置会产生1，而另一个设置会产生一个地址：语言方言c ++ 98结合使用标准库libc ++（具有c ++ 11支持的LLVM标准库）可生成1，而将c ++ 98与libstdc（gnu c ++标准库）结合可生成地址；）。您可以使用不理解 explicit 转换运算符（C ++ 11中的新功能）的C ++ 03风格的编译器，并结合使用C ++ 11-样式库，将转换定义为运算符bool（）。通过这种混合， cout<< cout 解释为 cout<< cout.operator bool（），而这只是 cout<< true 并显示 1 。
Apparently, certain C++ implementations allow mixing and matching the compiler and the library in such a way that produces non-conforming outcome (quoting @StephanLechner: "I found a setting in xcode which produces 1, and another setting that yields an address: Language dialect c++98 combined with "Standard library libc++ (LLVM standard library with c++11 support)" yields 1, whereas c++98 combined with libstdc (gnu c++ standard library) yields an address;"). You can have a C++03-style compiler that doesn't understand explicit conversion operators (which are new in C++11) combined with a C++11-style library that defines the conversion as operator bool(). With such a mix, it becomes possible for cout << cout to be interpreted as cout << cout.operator bool(), which in turn is simply cout << true and prints "1".
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