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:

1. Why is there no run-time error?

2. 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; }

Answer 1

-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

Answer 2

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.