D-Summer-School

Function overloading quiz

What is the output of the following code?

import std.uni : toLower;

int fun(string s)
in (s == s.toLower())
{
    return -42;
}

string fun(int x)
out (; x >= 0)
{
    x *= -42;
    return "Hello World!";
}

void main()
{
    import std.stdio : writeln;

    writeln(fun("hello world!"));
    writeln(fun(-42));
}

Question

What is the output for the snippet above?

	Runtime error at 1st call of fun()	The call `fun('hello world!')` will trigger our first function(that takes as input a string). It respects the `in` contract so it will just return -42. The second call `fun(-42)` will trigger the second function. At the end of the function the value of `x` will be *-42 -42**, it respects the `out` contract, so it will return ‘Hello World!’
	Runtime error at 2nd call of fun()	The call `fun('hello world!')` will trigger our first function(that takes as input a string). It respects the `in` contract so it will just return -42. The second call `fun(-42)` will trigger the second function. At the end of the function the value of `x` will be *-42 -42**, it respects the `out` contract, so it will return ‘Hello World!’
	hello world!	The call `fun('hello world!')` will trigger our first function(that takes as input a string). It respects the `in` contract so it will just return -42. The second call `fun(-42)` will trigger the second function. At the end of the function the value of `x` will be *-42 -42**, it respects the `out` contract, so it will return ‘Hello World!’
	-42	The call `fun('hello world!')` will trigger our first function(that takes as input a string). It respects the `in` contract so it will just return -42. The second call `fun(-42)` will trigger the second function. At the end of the function the value of `x` will be *-42 -42**, it respects the `out` contract, so it will return ‘Hello World!’
	-42 Hello World!	Correct!