Regarding throw statement followed by semicolon mark

Asked Jun 08 '21 at 21:28

Active Jun 08 '21 at 21:54

Viewed 64 times

In the following program:

#include <iostream>
using namespace std;
void CleanCity(){
    try{
        throw "Garbage";
    }
    catch(const char *){
        cout << "Throw in Dustbin";
        throw;
    }
}
int main(){
    cout << "CleanCity mission starts";
    try{
        CleanCity();
    }
    catch(const char *){
        cout << "City Cleaned";
    }
    cout << "Well Done";
    return 0;
}

Can someone tell what the line throw ; does and how does this code print "City Cleaned". I also observed that if I remove throw ; then the program does not print "City Cleaned", but I can't get the reasoning behind it.

asked Jun 08 '21 at 21:28

loveofprogramming

@oveofprogramming It retranslates the exception generated in the try block of the function. – Vlad from Moscow Jun 08 '21 at 21:36
1

@GBlodgett that link title seems right, but that question is about `throw;` outside of a `catch`. – Drew Dormann Jun 08 '21 at 21:38
I'm glad I read that one instead of reflexively dupe-hammering because of an excellent title. – user4581301 Jun 08 '21 at 21:39
2

Or this [Does throw inside a catch ellipsis (…) rethrow the original error in C++?](https://stackoverflow.com/questions/2474429/does-throw-inside-a-catch-ellipsis-rethrow-the-original-error-in-c) – 1201ProgramAlarm Jun 08 '21 at 21:41

Regarding throw statement followed by semicolon mark

0 Answers0