Edit: Although I apparently missed the bullseye on the actual question, I think my answer is pretty good, so I'm keeping it :-) (see below).
I guess a more concise way to phrase the question might be: can a purely functional language compute anything an imperative one can?
First of all, suppose you took an imperative language like C and made it so you can't alter variables after defining them. E.g.:
int i;
for (i = 0; // okay, that's one assignment
i < 10; // just looking, that's all
i++) // BUZZZ! Sorry, can't do that!
Well, there goes your for loop. Do we get to keep our while loop?
while (i < 10)
Sure, but it's not very useful. i can't change, so it's either going to run forever or not run at all.
How about recursion? Yes, you get to keep recursion, and it's still plenty useful:
int sum(int *items, unsigned int count)
{
if (count) {
// count the first item and sum the rest
return *items + sum(items + 1, count - 1);
} else {
// no items
return 0;
}
}
Now, with functions, we don't alter state, but variables can, well, vary. Once a variable passes into our function, it's locked in. However, we can call the function again (recursion), and it's like getting a brand new set of variables (the old ones stay the same). Although there are multiple instances of items and count, sum((int[]){1,2,3}, 3) will always evaluate to 6, so you can replace that expression with 6 if you like.
Can we still do anything we want? I'm not 100% sure, but I think the answer is "yes". You certainly can if you have closures, though.
You have it right. The idea is, once a variable is defined, it can't be redefined. A referentially transparent expression, given the same variables, always yields the same result value.
I recommend looking into Haskell, a purely functional language. Haskell doesn't have an "assignment" operator, strictly speaking. For instance:
my_sum numbers = ??? where
i = 0
total = 0
Here, you can't write a "for loop" that increments i and total as it goes along. All is not lost, though. Just use recursion to keep getting new is and totals:
my_sum numbers = f 0 0 where
f i total =
if i < length numbers
then f i' total'
else total
where
i' = i+1
total' = total + (numbers !! i)
(Note that this is a stupid way to sum a list in Haskell, but it demonstrates a method of coping with single assignment.)
Now, consider this highly imperative-looking code:
main = do
a <- readLn
b <- readLn
print (a + b)
It's actually syntactic sugar for:
main =
readLn >>= (\a ->
readLn >>= (\b ->
print (a + b)))
The idea is, instead of main being a function consisting of a list of statements, main is an IO action that Haskell executes, and actions are defined and chained together with bind operations. Also, an action that does nothing, yielding an arbitrary value, can be defined with the return function.
Note that bind and return aren't specific to actions. They can be used with any type that calls itself a Monad to do all sorts of funky things.
To clarify, consider readLn. readLn is an action that, if executed, would read a line from standard input and yield its parsed value. To do something with that value, we can't store it in a variable because that would violate referential transparency:
a = readLn
If this were allowed, a's value would depend on the world and would be different every time we called readLn, meaning readLn wouldn't be referentially transparent.
Instead, we bind the readLn action to a function that deals with the action, yielding a new action, like so:
readLn >>= (\x -> print (x + 1))
The result of this expression is an action value. If Haskell got off the couch and performed this action, it would read an integer, increment it, and print it. By binding the result of an action to a function that does something with the result, we get to keep referential transparency while playing around in the world of state.