在C ++中，如果throw是一个表达式，它的类型是什么？ [英] In C++, if throw is an expression, what is its type?

问题描述

我在我的一个简短报告中选择了这项功能：

http://www.smallshire.org.uk/sufficientlysmall/2009/07/31/in-c-throw-is-an-表达式/

基本上，作者指出在C ++中：

 code> throwerror
  

是一个表达式。这在C ++标准中实际上是相当清楚的，无论是在正文还是语法。但是，什么是不清楚（至少我）是什么类型的表达式？我想 void ，但是有点试验g ++ 4.4.0和Comeau产生了这个代码：

  void f（）{
} 
 
 struct S {}; 
 
 int main（）{
 int x = 1; 
 const char * p1 = x == 1？ foo：throw S（）; // 1 
 const char * p2 = x == 1？ foo：f（）; // 2 
} 
  

编译器没有问题// 1但barfed on / / 2，因为条件运算符中的类型不同。因此， throw 表达式的类型似乎不是无效的。

那么是什么呢？ >

如果您回答，请使用标准的报价备份您的报表。

---

throw表达式如何条件运算符处理throw表达式 - 我当然没有
知道在今天之前。

解决方案

根据标准，5.16第2段点，第二个或第三个操作数（但不是两个）是一个throw-expression（15.1）;结果是另一个的类型，是一个右值。因此，条件运算符不关心throw-expression是什么类型，而只使用其他类型。

事实上，15.1，第1段明确地说 throw-expression的类型为void。

I picked this up in one of my brief forays to reddit:

http://www.smallshire.org.uk/sufficientlysmall/2009/07/31/in-c-throw-is-an-expression/

Basically, the author points out that in C++:

throw "error"

is an expression. This is actually fairly clearly spelt out in the C++ Standard, both in the main text and the grammar. However, what is not clear (to me at least) is what is the type of the expression? I guessed "void", but a bit of experimenting with g++ 4.4.0 and Comeau yielded this code:

    void f() {
    }

    struct S {};

    int main() {
        int x = 1;
        const char * p1 = x == 1 ? "foo" : throw S();  // 1
        const char * p2 = x == 1 ? "foo" : f();        // 2
    }

The compilers had no problem with //1 but barfed on //2 because the the types in the conditional operator are different. So the type of a throw expression does not seem to be void.

So what is it?

If you answer, please back up your statements with quotes from the Standard.

---

This turned out not to be so much about the type of a throw expression as how the conditional operator deals with throw expressions - something I certainly didn't know about before today. Thanks to all who replied, but particularly to David Thornley.

解决方案

According to the standard, 5.16 paragraph 2 first point, "The second or the third operand (but not both) is a throw-expression (15.1); the result is of the type of the other and is an rvalue." Therefore, the conditional operator doesn't care what type a throw-expression is, but will just use the other type.

In fact, 15.1, paragraph 1 says explicitly "A throw-expression is of type void."

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