如果我抛出一个字符串文字，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

我正在使用 Linux 下的 g++ 在 C++ 中编写一个非常简单的应用程序，并且我试图抛出一些原始字符串作为异常(是的，我知道，这不是一个好习惯).

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;
  }
}

我在控制台上得到 Unknow error.但是，如果我将文字字符串静态转换为 std::string 或 char * 它会按预期打印 Error: not implemented .我的问题是:如果我不想使用静态强制转换，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

你需要用 char const* 而不是 char* 来捕捉它.std::string 和 char* 之类的东西都不会捕捉到它.

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

Catching 对匹配的类型有限制性的规则.规范说(其中cv"表示const/volatile 组合"或两者都不是).

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

如果

• 处理程序的类型为 cv T 或 cv T&并且 E 和 T 是同一类型(忽略顶级 cv 限定符)，或者
• 处理程序的类型为 cv T 或 cv T&并且 T 是 E 的明确公共基类，或者

• 处理程序的类型为 cv1 T* cv2，E 是指针类型，可以通过其中一个或两个转换为处理程序的类型

• 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

• 标准指针转换 (4.10) 不涉及到指向私有或受保护或不明确类的指针的转换
• 资格转换

字符串字面量的类型为 char const[N]，但抛出数组会使数组衰减，实际上会抛出指向其第一个元素的指针.因此，您无法通过 char* 捕获抛出的字符串文字，因为在匹配时，它需要将 char* 与 char const*，这将丢弃一个 const(限定转换只允许 add const).仅在需要专门转换字符串文字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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