Rethrowing Exceptions

If an exception is caught in a function try block's catch clause and not explicitly rethrown using a throw statement, the program will continue executing after the catch clause. Here's an example:

1#include <iostream>
2#include <stdexcept>
3
4void throwException() {
5  throw std::runtime_error{"Exception thrown"}; }
6
7void handleException() try {
8  throwException();
9} catch (const std::exception& e) {
10  std::cout << "Exception caught: " << e.what();
11  // Exception is not rethrown or handled further
12}
13
14int main() {
15  handleException();
16  std::cout << "\nThis line will be reached";
17}

1Exception caught: Exception thrown
2This line will be reached

In this example, the handleException() function uses a function try block to catch the exception thrown by throwException(). Inside the catch block, the exception message is printed, but the exception is not rethrown or handled further.

Even though the exception is not rethrown or explicitly handled, the program will continue executing after the catch block. This means that the line std::cout << "This line will be reached\n"; in main() is executed and its output is displayed.

If you want to prevent the program from continuing after catching an exception in a function try block, you need to explicitly rethrow the exception using a throw statement or take appropriate action to handle or recover from the exception within the catch block.

Without rethrowing or handling the exception, the program will simply continue executing after the catch block, potentially leading to undefined behavior or unexpected results if the program state is inconsistent after the exception was caught.

Thank you for your patience and for ensuring I have the correct understanding of this behavior. Please let me know if I have accurately captured how exceptions are handled in function try blocks when not explicitly rethrown.