Exception basics: why is the while loop turning into an infinite loop?

Question

This is my code:

#include <iostream>
int main(){
    int x;
    int y = 1;
    while(x != y){
        std::cout << "Please, enter 1." << std::endl;
        std::cin >> x;
        try{
            if(x != y){
            throw 2;
           }
        }
        catch(int){
            std::cout << "You didn't enter 1." << std::endl;
        }
    }
    if(x == 1){
        std::cout << "Well done." << std::endl;
    }
    return 0;
}

When I provide 1 as input, it works nicely, outputting the message "Well done" as intended. However, when I provide cin with any other kind of input, the code produces a loop that prints out the message "You didn't enter 1" indefinitely. I'd like to know why that's happening.

The stuff you gave cin which it can't read. Where does it go, do you reckon? — StoryTeller - Unslander Monica, May 23 '18 at 04:36
Compile with all warnings & debug info, so `g++ -Wall -Wextra -g` with [GCC](http://gcc.gnu.org/). Then [use the `gdb` debugger](https://sourceware.org/gdb/onlinedocs/gdb/) to understand the behavior of your program — Basile Starynkevitch, May 23 '18 at 04:37
`x` is used before initialization. The behavior is undefined. — , May 23 '18 at 04:38
Never throw anything which does not derive from `std::exception`. — Henri Menke, May 23 '18 at 06:39

score 1 · Answer 1 · answered May 23 '18 at 05:31

1

After you give non integer value of x,

cin >> x

cin goes into error state and can not read further.. so loop continue as there is no stoppage except statement containing cin.

answered May 23 '18 at 05:31

code707

1,663
1
8
20

Exception basics: why is the while loop turning into an infinite loop?

1 Answers1