Why exception caught without throwing in C++?

Question

I am running this program. As per my understanding, there should be a bad allocation exception raised within the try block and should cause a runtime failure. However, the program executes without any problem. I have the following questions:

Why is there no run-time error?

How does the control reach catch block even when no exceptions were thrown?

#include <iostream>
using namespace std;

int main() {
    int b =4, *c = NULL, i=-1;
    try{
        c = new int [i]; 
        b--;
    }catch (exception& e){
        cout << "coming here" << endl;
        c = new int[1];
        b++;
    }
    cout << b << endl;
    return 0;
}

For me, it comes in the catch block. Where did you run this? — starboy_jb, Dec 09 '20 at 08:09
How this is supposed to behave depends on the language version. Until C++11, it has undefined behaviour and shouldn't do anything in particular. — molbdnilo, Dec 09 '20 at 08:10
The exception **is** the run-time error you are looking for. — n. m. could be an AI, Dec 09 '20 at 08:20

Bathsheba · Accepted Answer · 2020-12-09T08:42:45.457

3

-1 will be converted to the std::size_t value std::numeric_limits<std::size_t>::max() due to the implicit conversion to an unsigned type. It is indeed unlikely that you have that much memory available.

However a std::bad_alloc might not be thrown if your operating system doesn't actually allocate the memory until you consume it (common on linux platforms).

See lazy allocation for c++ object arrays

edited Dec 09 '20 at 08:42

answered Dec 09 '20 at 08:16

Bathsheba

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score 1 · Answer 2 · answered Dec 09 '20 at 08:09

1

Your c = new int [i] within the try block throws the exception because it cannot allocate the memory (i=-1), so the catch block will be executed.

answered Dec 09 '20 at 08:09

trenki

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Why exception caught without throwing in C++?

2 Answers2