Lambdas

Exceptions

#include <iostream>
#include <stdexcept>

using namespace std ;

int main()
{

  auto divide = [](int a, int b) -> int
    {
        if (b == 0) {
            throw runtime_error("Division by zero!");
        }
        return a / b;
    };

    try
    {
        int result = divide(10, 0);
        cout << "Result: " << result << endl;
    }
    catch (const runtime_error& e)
    {
        cerr << "Error: " << e.what() << endl;
    }


  auto divide1 = [](int a, int b) noexcept  -> int
    {
        if (b == 0) {
            cout << "Divisor is zero. Can't divide." << endl ;
            return 0;
        }
        return a / b;
    };

  auto divide2 = [](int a, int b) noexcept  -> int
    {
        if (b == 0) {
            cout << "Divisor is zero. Can't divide." << endl ;
             throw runtime_error("Division by zero!");
        }
        return a / b;
    };


}

Output:
$ rm a.exe ; g++ except1.cpp ; ./a.exe
except1.cpp: In lambda function:
except1.cpp:41:14: warning: ‘throw’ will always call ‘terminate’ [-Wterminate]
   41 |              throw runtime_error("Division by zero!");
      |              ^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Error: Division by zero!


We can have a lambda expression throw an exception. The below snippet
shows how this can be done.

  auto divide = [](int a, int b) -> int
    {
        if (b == 0) {
            throw runtime_error("Division by zero!");
        }
        return a / b;
    };

    try
    {
        int result = divide(10, 0);
        cout << "Result: " << result << endl;
    }
    catch (const runtime_error& e)
    {
        cerr << "Error: " << e.what() << endl;
    }

 The lambda expression checks if the divisor is 0 and if so
 throws a "runbtime_error" exception which is caught later on in
 a try block upon the execution of the lambda object "divide" .

  auto divide1 = [](int a, int b) noexcept  -> int
    {
        if (b == 0) {
            cout << "Divisor is zero. Can't divide." << endl ;
            return 0;
        }
        return a / b;
    };

 We can also have exception specification as shown above. The
 "noexcept" is stating that this lambda expression does not
 throw an exception.


  auto divide2 = [](int a, int b) noexcept  -> int
    {
        if (b == 0) {
            cout << "Divisor is zero. Can't divide." << endl ;
             throw runtime_error("Division by zero!");
        }
        return a / b;
    };

 In the above we have the "noexcept" specification but the
 lambda expression throws an exception. In this case we
 get a compiler warning.

Exercise

 #include <iostream>
 #include <stdexcept>
 #include <vector>

 using namespace std ;

 int main()
 {
   vector<int> v1 = { 10,20 } ;

   auto f1 = [=]()
     {
         cout << "Lambda Step Begin" << endl ;
         cout << v1.at( 2 ) << endl ;
         cout << "Lambda Step End" << endl ;
     };

     try
     {
         f1();
         cout << "After f1(): " <<  endl;
     }
     catch (const out_of_range& e)
     {
         cerr << "Error: Out of range " << endl;
     }


   return 0 ;
}

Solutions