如果抛出会发生什么;语句在catch块外部执行? [英] What happens if a throw; statement is executed outside of catch block?

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问题描述

在C ++ throw; 当在catch块内执行时,会重新抛出当前捕获的异常。



此回答中,提出了异常分派器的概念当使用复杂的异常处理时,通常会减少代码重复:

  try {
CodeThatMightThrow
} catch(...){
ExceptionHandler();
}

无效ExceptionHandler()
{
try {
throw;
} catch(FileException * e){
//处理一些复杂的逻辑
delete e;
} catch(GenericException * e){
//处理其他复杂逻辑
delete e;
}
}

抛出指针或值不会



如果ExceptionHandler()不是从catch块调用,会发生什么?



<我尝试这个代码与VC7:

  int main(int,char **)
{
try {
throw;
} catch(...){
MessageBox(0,,,0);
}
return 0;
}



首先它使调试器指示第一次机会异常,未处理的异常。如果我在调试器外运行这个代码,程序崩溃的方式与调用abort()时一样。



这种情况下的预期行为是什么?

解决方案

从标准中,15.1 / 8


如果目前没有处理异常,则执行不带操作数调用 std :: terminate ()的 throw-expression



In C++ throw; when executed inside a catch block rethrows the currently caught exception outside the block.

In this answer an idea of exception dispatcher is brought up as a solution to reducing code duplication when using complex exception handling often:

try {
    CodeThatMightThrow();
} catch(...) {
    ExceptionHandler();
}

void ExceptionHandler()
{
    try {
        throw;
    } catch( FileException* e ) {
        //do handling with some complex logic
        delete e;
    } catch( GenericException* e ) {
        //do handling with other complex logic
        delete e;
    }
}

Throwing a pointer or a value doesn't make any difference so it's out of the question.

What happens if ExceptionHandler() is called not from a catch block?

I tried this code with VC7:

int main( int, char** )
{   
    try {
        throw;
    } catch( ... ) {
        MessageBox( 0, "", "", 0 );
    }
    return 0;
 }

First it causes the debugger to indicate a first-chance exception, then immediately an unhandled exception. If I run this code outside the debugger the program crashes the same way as if abort() has been called.

What is the expected behaviour for such situations?

解决方案

From the Standard, 15.1/8

If no exception is presently being handled, executing a throw-expression with no operand calls std::terminate().

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