Catch clauses not catching right? - c++

Question

I'm just learning how to use exceptions in C++ and have come across weird behavior in my "test" code. (excuse overly stupid questions like this one, please...it's NOT lack of research/effort, just lack of experience!) If I'm catching just the exception DivideByZero it works fine.
But introducing the second exception StupidQuestion makes the code not work exactly how I expected. How I wrote it below I thought it should take care of the DivideByZero exception if it needs to, and if not then check if StupidQuestion occurs, and if not just go back to the try clause and print the normal result. But if I input, say, a=3 and b=1, the program redirects to the DivideByZero try clause instead of the StupidQuestion one. The weird thing is, though, divide does seem to be throwing StupidQuestion (see via cout statement), but it's not catching right, as also seen by the absense of the cout statement.
#include <iostream>
#include <cstdlib>
using namespace std;
const int DivideByZero = 42;
const int StupidQuestion=1337;
float divide (int,int);
main(){
int a,b;
float c;
cout << "Enter numerator: ";
cin >> a;
cout << "Enter denominator: ";
cin >> b;
try{
c = divide(a,b);
cout << "The answer is " << c << endl;
}
catch(int DivideByZero){
cout << "ERROR: Divide by zero!" << endl;
}
catch(int StupidQuestion){
cout << "But doesn't come over here...?" << endl;
cout << "ERROR: You are an idiot for asking a stupid question like that!" << endl;
}
system("PAUSE");
}
float divide(int a, int b){
if(b==0){
throw DivideByZero;
}
else if(b==1){
cout << "It goes correctly here...?" << endl;
throw StupidQuestion;
}
else return (float)a/b;
}
I was wondering if it had something to do with the fact that DivideByZero and StupidQuestion were both of type int, so I changed the code to make StupidQuestion be of type char instead of int. (So: const char StupidQuestion='F'; and catch(char StupidQuestion) were really the only things changed from above) And it worked fine.
Why isn't the above code working when the two exceptions have the same type (int)?

Answer 1

Instead of this
catch(int DivideByZero) {
cout << "ERROR: Divide by zero!" << endl;
}
catch(int StupidQuestion) {
cout << "But doesn't come over here...?" << endl;
cout << "ERROR: You are an idiot for asking a stupid question like that!" << endl;
}
you are looking for
catch (int errval) {
if (errval == DivideByZero) {
cout << "ERROR: Divide by zero!" << endl;
}
else if (errval == StupidQuestion) {
cout << "ERROR: You are an idiot for asking a stupid question like that!" << endl;
}
else {
throw; // for other errors, keep searching for a handler
}
}
The variable name inside the catch clause is creating a new local variable, which has no relation to a global constant with the same name.
Also note that there will be no way to catch just one error number... but you can rethrow unknown errors as I show.

Answer 2

catch(int DivideByZero) { }
catch(int StupidQuestion) { }
Both catch blocks catch ints, they're just named differently. Only the first one can ever be entered, the second one is dead code.

Answer 3

When choosing a handler for an exception only type is taken into account, and neither values nor addresses (addresses of variables are not applicable here at all because of how exceptions work), also names of variables do not exist after compilation.
The first appropriate handler for the exception is always chosen.
Please look my answer to another question for details: https://stackoverflow.com/a/45436594/1790694