输出元组在Haskell中的功能如何？

Question

I'm trying to figure out how tuples work in Haskell, and how you can get certain data values from tuples.

I have the following code, and this runs properly, and prints out the tuple output I expect:

test :: Integer -> (Integer, Integer)
test c =  function(1, c)

However, when I try to get just the first value from the tuple, as shown below,

test :: Integer -> Integer
test c = fst function(1, c)

I get the following error

• Couldn't match expected type '((Integer, Integer) -> Integer, b0)' with actual type '(Integer, Integer) -> (Integer, Integer)'

Any help or advice would be appreciated. Thanks in advance.

Answer 1

You simply got the syntax wrong.

fx means " apply (call) function f to argument x ".

Similarly, fxy means "apply function f to arguments x and y ".

Your example function(1, c) means " apply function function to argument (1, c) ". Don't let the absence of a space before the opening paren fool you: (1, c) does not mean " call function with two arguments " (as it would in, say, C or Java), it means " make a pair with two components, 1 and c ". This pair is then used as a single argument for function .

Now, the next expression, fst function (1, c) means " apply function fst to two arguments - function and (1, c) ". This is obviously not what you meant. Instead, you meant to first call function with argument (1, c) , and then pass the result as argument to fst . To express this, you may use parentheses:

test c = fst (function (1, c))

But of course this is now too many parentheses. To avoid the extra pair, you may use the ubiquitous operator $ :

test c = fst $ function (1, c)

This operator doesn't really do anything interesting, it just applies the function on the left to the argument on the right:

f $ x = f x

The value of this operator is that it makes it possible for the function application to not have the highest precedence, and thus get rid of extra parentheses.