I have a following program in c++ with an output

Question

#include <exception>
#include <iostream>

using namespace std;
class A
{
public:
    A(int n) : i(n) { cout << i; }

    ~A() { cout << i; }

private:
    int i;
};

int f(int n)
{
    if (n == 1) {
        throw std::logic_error("0");
        cout << "7" << endl;
    }
    A a(n);
    return f(n - 1);
}

int main() //main function
{
    try {
        int ret = f(3);
        A a(ret);
    }
    catch (const std::exception &e) {
        cout << e.what() << endl;
    }
    return 0;
}

Why the output of this c++ in exceptional handling is 32230 ? I am little bit confused.

I am using ATOM for programming Thanks

The simplest thing for you to do is to get out your line by line debugger and step through. — Bathsheba, Jun 28 '18 at 12:29
https://stackoverflow.com/questions/2331316/what-is-stack-unwinding I think that may help you undestand what is going on — Ch1v1, Jun 28 '18 at 12:33
If the cause of your confusion is seeing one large integer then I suggest you get into the habit of outputting a newline "\n" after outputting a number. Then it'll be more apparent that you're actually outputting 5 individual numbers. — acraig5075, Jun 28 '18 at 12:36
`//main function`... A good habit is to avoid such comments. — StoryTeller - Unslander Monica, Jun 28 '18 at 12:36
@L_Church - If you are genuinely curious, and don't mind I took a peek in the OP's profile to be sure, [this is probably why](https://www.quora.com/Why-do-Indians-use-the-honorific-sir-so-freely-What-are-its-strange-origins). — StoryTeller - Unslander Monica, Jun 28 '18 at 12:39
Why was this closed? It's clear what the question is, and the code snippet compiles and demonstrates the question. Nothing unclear or broad about it. — acraig5075, Jun 28 '18 at 12:42
@acraig5075: Remember that closing is done by voting which is a proxy for the opinion of the community. For what it's worth, I pointed out to the OP that the reason becomes obvious if you debug line by line, then voted to close the question. — Bathsheba, Jun 28 '18 at 12:45

score 1 · Answer 1 · answered Jun 28 '18 at 12:29

This is due to the order of your operations.

1) First, creation of A with n = 3 is called, so it prints 3.

2) Then, trhough f(n-1), creation of A with n = n-1 = 2 is called, it prints 2.

3) Then through f(n-1=1), throw 0 is called. It leaves the function f.

4) when leaving step 3), the programs leaves then the function f(2) called in step 2). So it destroy the object a created in step 2), it prints 2

5) When leaving step 4), the programs leaves then f(3) called in step 1). So it destroy the a created in step 1), it prints 3

6) It prints the error that was thrown: 0

You can see how all the steps are done either with a debugger and going step by step, or even following the algorithm on a paper..

score 0 · Answer 2 · answered Jun 28 '18 at 12:34

You are calling function f recursive. This means first you create an object with 3, the constructor outputs 3, next recursion with 2, constructor outputs 2, next recursion 1, now you will run in the exception handler.

Exception raises back through the recursion so you get 2 from destructor in second recursion level and 3 from destructor 1 level. Finally, you output 0 with e.what().

Result 32230.

I have a following program in c++ with an output

2 Answers2