如果我抛出一个字符串字面量我应该捕获什么类型？ [英] What type should I catch if I throw a string literal?

问题描述

我在Linux下使用g ++在C ++中编写一个非常简单的应用程序，我试图抛出一些原始字符串作为异常（是的，我知道，它不是一个好的做法）。

I am writing a pretty simple application in C++ using g++ under Linux and I am trying to throw some raw strings as exceptions (yes, I know, its not a good practise).

我有以下代码（简化）：

I have the following code (simplified):

int main()
{
  try
  {
    throw "not implemented";

  }
  catch(std::string &error)
  {
    cerr<<"Error: "<<error<<endl;
  }
  catch(char* error)
  {
    cerr<<"Error: "<<error<<endl;
  }
  catch(...)
  {
    cerr<<"Unknown error"<<endl;
  }
}

我得到 Unknow错误。但是如果我静态转换文本字符串到std :: string或char *它打印错误：未实现如预期。我的问题是：如果我不想使用静态转型，那么我应该捕获什么类型？

And I get Unknow error on the console. But if I static cast the literal string to either std::string or char* it prints Error: not implemented as expected. My question is: so what is the type I should catch if I don't want to use static casts?

推荐答案

用 char const * 而不是 char * 来捕获它。 std :: string 或 char * 都不会捕获它。

You need to catch it with char const* instead of char*. Neither anything like std::string nor char* will catch it.

捕获对于匹配的类型有限制的规则。规范说明（其中cv意味着const / volatile组合或两者都不）。

Catching has restricted rules with regard to what types it match. The spec says (where "cv" means "const/volatile combination" or neither of them).

类型E的异常对象if

A handler is a match for an exception object of type E if

• 处理程序类型为cv T或cv T&和E和T是相同类型（忽略顶级cv限定符），或


• 处理程序类型为cv T或cv T&并且T是E的无歧义的公共基类，或者



• 处理程序类型为cv1 T * cv2，E是可以转换为类型的指针类型处理程序

• The handler is of type cv T or cv T& and E and T are the same type (ignoring the top-level cv-qualifiers), or
• the handler is of type cv T or cv T& and T is an unambiguous public base class of E, or

• the handler is of type cv1 T* cv2 and E is a pointer type that can be converted to the type of the handler by either or both of

• 一个标准指针转换（4.10），不涉及到指向私有或受保护或模糊类的指针转换


• 资格转换

$ b b

字符串文字具有类型 char const [N] ，但是抛出一个数组将会衰减数组，并实际抛出一个指向它的第一个元素的指针。所以你不能通过 char * 捕获一个抛出的字符串文字，因为当它匹配时，它需要匹配 char * 到 char const * ，这将抛弃一个const（限定符转换只允许添加 const）。仅当需要专门转换字符串字面量时，才考虑将字符串文字特殊转换为 char * 。

A string literal has type char const[N], but throwing an array will decay the array and actually throws a pointer to its first element. So you cannot catch a thrown string literal by a char*, because at the time it matches, it needs to match the char* to a char const*, which would throw away a const (a qualification conversion is only allowed to add const). The special conversion of a string literal to char* is only considered when you need to convert a string literal specifically.

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