using boolean expressions instead of if/else or similar condition - c

Question

I need to use boolen expressions instead of using terms such as: c: b? A, switch, else ... if. and using for / while-do / while as if is also not allowed.
My question is when receiving a name, based on its value we decide what to print. For example, if the name is shorter than 6 letters we print "player A is sent", if the letters sum (in ascii values) is more than 20 we send player B and stuff like that. In addition, if we can send more than one player, we will choose to send the largest lexicographic player (playerC < playerB < playerA.)
what I tried to do is something like this
bool sum = AsciiSum(name)>=20;
bool len = NameLength(name)<6;
...
so the rest should be that
if(sum)
printf("%c\n" ,'playerA');
else if (len)
printf("%c\n" ,'playerB');
else
printf("%c\n" ,'no player was sent');
But as I said , I cant use if and else or similar conditions. Will be grateful for any help in that last part.

Answer 1

One way of doing this would be to reroute your input through a custom print function which return a boolean value. That was, you could then just use the AND (&&) operator to call it.
Example code:
bool printMessage(char* string)
{
printf("%s", string);
return true;
}
bool sum = AsciiSum(name)>=20 && printMessage("playerA");
bool len = NameLength(name)<6 && printMessage("playerB");
// ... etc
Why does this work? The AND operator will only allow the next condition to be checked if the first one was true. Since your condition is that sum must be true in order to print the message, this is a way to call the method to print without directly using any if/else statements
Edit: An even better way, as suggested by Johnny Mopp is to use stdio puts(), which returns an integer (that's what a boolean is functionally).
bool sum = AsciiSum(name)>=20 && puts("playerA");
bool len = NameLength(name)<6 && puts("playerB");
With this, you don't even have to create a printMessage function, though you should stay with whichever feels more comfortable to work with.

Answer 2

Given an output function that always returns true :
bool print( const char* msg )
{
puts( msg ) ;
return true ;
}
You can then exploit short-circuit evaluation thus:
bool sum = AsciiSum(name)>=20;
bool len = NameLength(name)<6;
(sum && print("playerA")) ||
(len && print("playerB")) ||
print("no player was sent") ;
Testing all combinations thus:
bool sum = false, len = false ;
(sum && print("playerA")) || (len && print("playerB")) || print("no player was sent") ;
sum = false, len = true ;
(sum && print("playerA")) || (len && print("playerB")) || print("no player was sent") ;
sum = true, len = false ;
(sum && print("playerA")) || (len && print("playerB")) || print("no player was sent") ;
sum = true, len = true ;
(sum && print("playerA")) || (len && print("playerB")) || print("no player was sent") ;
correctly outputs:
no player was sent
playerB
playerA
playerA
Your compiler may issue a diagnostic regarding the result of the expression being unused; that can be supressed by a dummy assignment:
bool dummy = (sum && print("playerA")) ||
(len && print("playerB")) ||
print("no player was sent") ;
or by casting the whole expression to void:
(void)((sum && print("playerA")) ||
(len && print("playerB")) ||
print("no player was sent")) ;
The only requirement of the output function is that it returns a non-zero value to short-circuit the ||, to that end printf() will work if the message is not an empty string, and avoid the need to define a specific output function.:
(void)((sum && printf("playerA\n")) ||
(len && printf("playerB\n")) ||
printf("no player was sent\n")) ;
puts() however only guarantees to return a non-negative value on success, which includes zero, so should not be used.