警告级别4中的代码不可达警告

Question

Can someone explain me why this gives unreachable code warning for *a=9; line. I'm using VS2015 preview and warning level 4

int main(){
int* a = foo();
try{
 *a = 5;
}catch(int)
{
 *a=9;
}
  return 0;
}

Answer 1

This gives unreachable code warning for *a=9 , because your compiler knows that the code in try block will never throw any kind of exception. So your catch block will never execute.

Use try/catch when code is prone to throw an exception.

Answer 2

The reason it is unreachable is because your code fragment *a = 5 cannot throw an exception. It might give an access violation (for example if foo returns nullptr), but it will not be handled as a C++ exception.

If that is something you want, you CAN catch access violations using __try/__except, but I think you should only use this under exceptional circumstances.

https://msdn.microsoft.com/en-us/library/s58ftw19%28v=vs.80%29.aspx?f=255&MSPPError=-2147217396