C++ exception not handling as expected

Question

I'm trying to throw an exception in my code if a vector that is created from user input is not sorted in either descending or ascending order.

using namespace std;
#include <iostream>
#include <vector>
#include <algorithm>

int main () {
vector <int> vec;

//Let user fill a vector with 12 integers.
//cout << "Please note that input data should be either increasing or decreasing." << endl;
int n = 0;
int size = 0;
while(size < 12) {
    cout << "Type integer to add to the vector." << endl;
    cin >> n;
    vec.push_back(n);
    ++size;
}

//throw exception if unsorted
try {
    if (!((is_sorted(vec.begin(), vec.end())) || (is_sorted(vec.end(), vec.begin())))) {
        throw "Input was not sorted.";
    }
}
catch(exception &error){
    cerr << "Error: " << error.what() << endl;
}

}

I have not included the rest of the code, which searches for a particular number, because I am pretty sure that it is irrelevant to this issue. When the data filled into the vector is ascending or descending, everything is fine, but when I test the exception, I get, "terminate called after throwing an instance of 'char const*' Aborted" instead of my desired error message. I don't understand what is going on here. Is it something wrong with the way I'm handling exceptions or am I using the sort() function incorrectly?

If you swap `begin()` and `end()`, you don't get an interval that goes backwards, you get an empty interval. You want to use `rbegin()` and `rend()` instead. — n. m. could be an AI, Jul 12 '17 at 04:40
There is no sense to throw and catch an exception in the same block. You should use a simple 'if' instead. — Mattia72, Jul 12 '17 at 04:44
Wow, so many parentheses! You can remove the ones around the calls to `is_sorted` -- they're redundant. And you can get rid of the outermost pair in the expression inside the `if` statement by applying De Morgan's theorem: `!(A || B)` is equivalent to `!A && !B`. — Pete Becker, Jul 12 '17 at 12:25
@n.m. -- it's worse than that. Trying to use the sequence `[end(), begin())` produces undefined behavior because the second iterator is not reachable from the first. — Pete Becker, Jul 12 '17 at 12:27
@Mattia72 -- maybe, but this is **supposed to be** a minimal example **that shows the problem**. And it does exactly that. — Pete Becker, Jul 12 '17 at 12:29

score 2 · Accepted Answer · answered Jul 12 '17 at 05:02

In C++, all types are throwable and catchable, but you are only catching subclasses of std::exception.

The best fix to your code would be changing your throw statement to:

throw std::runtime_error("Input was not sorted.");

A.S.H · Answer 2 · 2017-07-12T23:02:30.243

0

You're throwing a const char* not an std::exception. So catch it as a const char*:

  catch(const char* error) {
    std::cout << "Error: " << error << "\n";
  }

Or throw an std::exception.

Remember that you can throw many types and have many catch blocks, the one that will be invoked is the one that matches the type of the thrown exception.

edited Jul 12 '17 at 23:02

answered Jul 12 '17 at 05:01

A.S.H

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Minor niggle. I don't think `std::exception` has a constructor that eats a string or similar. – user4581301 Jul 12 '17 at 06:04

score 0 · Answer 3 · answered Jul 12 '17 at 05:05

0

If you want to catch an exception, you should throw an exception, not a const char*. See this answer: c++ exception : throwing std::string

answered Jul 12 '17 at 05:05

Mattia72

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C++ exception not handling as expected

3 Answers3