Mixing
How does the Haskell compiler emit code for `(==) (fromInteger 0) (fromInteger 0)` when the intermediary...
My question is related to type class instance deduction in the presence of ambiguously typed intermediary values
Prelude> :t fromInteger 0
fromInteger 0 :: Num a => a
Prelude> :t (==)
(==) :: Eq a => a -> a -> Bool
Prelude> :t (==) (fromInteger 0)
(==) (fromInteger 0) :: (Eq a, Num a) => a -> Bool
Prelude> :t (==) (fromInteger 0) (fromInteger 1)
(==) (fromInteger 0) (fromInteger 1) :: Bool
Prelude> (==) (fromInteger 0) (fromInteger 1)
False
Magic! It's unclear how or whether a was made concrete, yet the code ran successfully!
According to the type inference rules, the type variables denoted by a above unify successfully with each other because they have compatible Num a constraints across the different terms. However, a never binds to a concrete type. My question is, at runtime, which instance dictionary (or specialization, whatever) is used for the (==) function?
Is this a case where Haskell relies on a simple binary memcmp style comparison? Or perhaps it just picks the first instance in its list of Num instances since in theory it shouldn't matter (as long as the algebraic properties of that instance are implemented correctly...)
haskell instance
asked Jan 2 at 2:52
Will BradleyWill Bradley
1,3561125
add a comment |
My question is related to type class instance deduction in the presence of ambiguously typed intermediary values
Prelude> :t fromInteger 0
fromInteger 0 :: Num a => a
Prelude> :t (==)
(==) :: Eq a => a -> a -> Bool
Prelude> :t (==) (fromInteger 0)
(==) (fromInteger 0) :: (Eq a, Num a) => a -> Bool
Prelude> :t (==) (fromInteger 0) (fromInteger 1)
(==) (fromInteger 0) (fromInteger 1) :: Bool
Prelude> (==) (fromInteger 0) (fromInteger 1)
False
Magic! It's unclear how or whether a was made concrete, yet the code ran successfully!
According to the type inference rules, the type variables denoted by a above unify successfully with each other because they have compatible Num a constraints across the different terms. However, a never binds to a concrete type. My question is, at runtime, which instance dictionary (or specialization, whatever) is used for the (==) function?
Is this a case where Haskell relies on a simple binary memcmp style comparison? Or perhaps it just picks the first instance in its list of Num instances since in theory it shouldn't matter (as long as the algebraic properties of that instance are implemented correctly...)
haskell instance
asked Jan 2 at 2:52
Will BradleyWill Bradley
1,3561125
add a comment |
My question is related to type class instance deduction in the presence of ambiguously typed intermediary values
Prelude> :t fromInteger 0
fromInteger 0 :: Num a => a
Prelude> :t (==)
(==) :: Eq a => a -> a -> Bool
Prelude> :t (==) (fromInteger 0)
(==) (fromInteger 0) :: (Eq a, Num a) => a -> Bool
Prelude> :t (==) (fromInteger 0) (fromInteger 1)
(==) (fromInteger 0) (fromInteger 1) :: Bool
Prelude> (==) (fromInteger 0) (fromInteger 1)
False
Magic! It's unclear how or whether a was made concrete, yet the code ran successfully!
According to the type inference rules, the type variables denoted by a above unify successfully with each other because they have compatible Num a constraints across the different terms. However, a never binds to a concrete type. My question is, at runtime, which instance dictionary (or specialization, whatever) is used for the (==) function?
Is this a case where Haskell relies on a simple binary memcmp style comparison? Or perhaps it just picks the first instance in its list of Num instances since in theory it shouldn't matter (as long as the algebraic properties of that instance are implemented correctly...)
haskell instance
asked Jan 2 at 2:52
Will BradleyWill Bradley
1,3561125
My question is related to type class instance deduction in the presence of ambiguously typed intermediary values
Prelude> :t fromInteger 0
fromInteger 0 :: Num a => a
Prelude> :t (==)
(==) :: Eq a => a -> a -> Bool
Prelude> :t (==) (fromInteger 0)
(==) (fromInteger 0) :: (Eq a, Num a) => a -> Bool
Prelude> :t (==) (fromInteger 0) (fromInteger 1)
(==) (fromInteger 0) (fromInteger 1) :: Bool
Prelude> (==) (fromInteger 0) (fromInteger 1)
False
Magic! It's unclear how or whether a was made concrete, yet the code ran successfully!
According to the type inference rules, the type variables denoted by a above unify successfully with each other because they have compatible Num a constraints across the different terms. However, a never binds to a concrete type. My question is, at runtime, which instance dictionary (or specialization, whatever) is used for the (==) function?
Is this a case where Haskell relies on a simple binary memcmp style comparison? Or perhaps it just picks the first instance in its list of Num instances since in theory it shouldn't matter (as long as the algebraic properties of that instance are implemented correctly...)
haskell instance
haskell instance
asked Jan 2 at 2:52
Will BradleyWill Bradley
1,3561125
asked Jan 2 at 2:52
Will BradleyWill Bradley
1,3561125
edited Jan 2 at 3:01
Will Bradley
asked Jan 2 at 2:52
Will BradleyWill Bradley
1,3561125
asked Jan 2 at 2:52
Will BradleyWill Bradley
1,3561125
asked Jan 2 at 2:52
Will BradleyWill Bradley
1,3561125
1,3561125
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
Yes it is made concrete by defaulting, which is pretty much your second theory.
The expression [..] is ambiguous because the literal 4 is of Num a => a type in Haskell. 4 can be an Int, a Float or any other type that is an instance of Num, so the compiler can’t choose any particular type for the same reason above. But the Haskell Committee thought that this is too much restriction. After much debates, they compromised and added an ad-hoc rule for choosing a particular default type.
(Although it does matter which instance is chosen. For example 2^64 == 0 is True if Int is chosen, but False if Integer is chosen. Integer is first in the default list for good reason.)
edited Jan 2 at 3:14
user2864740
44.1k671150
answered Jan 2 at 3:10
luquiluqui
48.3k6114168
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
StackExchange.using("externalEditor", function () {
StackExchange.using("snippets", function () {
StackExchange.snippets.init();
});
});
}, "code-snippets");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "1"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f54000708%2fhow-does-the-haskell-compiler-emit-code-for-frominteger-0-frominteger-0%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Yes it is made concrete by defaulting, which is pretty much your second theory.
The expression [..] is ambiguous because the literal 4 is of Num a => a type in Haskell. 4 can be an Int, a Float or any other type that is an instance of Num, so the compiler can’t choose any particular type for the same reason above. But the Haskell Committee thought that this is too much restriction. After much debates, they compromised and added an ad-hoc rule for choosing a particular default type.
(Although it does matter which instance is chosen. For example 2^64 == 0 is True if Int is chosen, but False if Integer is chosen. Integer is first in the default list for good reason.)
edited Jan 2 at 3:14
user2864740
44.1k671150
answered Jan 2 at 3:10
luquiluqui
48.3k6114168
add a comment |
Yes it is made concrete by defaulting, which is pretty much your second theory.
The expression [..] is ambiguous because the literal 4 is of Num a => a type in Haskell. 4 can be an Int, a Float or any other type that is an instance of Num, so the compiler can’t choose any particular type for the same reason above. But the Haskell Committee thought that this is too much restriction. After much debates, they compromised and added an ad-hoc rule for choosing a particular default type.
(Although it does matter which instance is chosen. For example 2^64 == 0 is True if Int is chosen, but False if Integer is chosen. Integer is first in the default list for good reason.)
edited Jan 2 at 3:14
user2864740
44.1k671150
answered Jan 2 at 3:10
luquiluqui
48.3k6114168
add a comment |
Yes it is made concrete by defaulting, which is pretty much your second theory.
The expression [..] is ambiguous because the literal 4 is of Num a => a type in Haskell. 4 can be an Int, a Float or any other type that is an instance of Num, so the compiler can’t choose any particular type for the same reason above. But the Haskell Committee thought that this is too much restriction. After much debates, they compromised and added an ad-hoc rule for choosing a particular default type.
(Although it does matter which instance is chosen. For example 2^64 == 0 is True if Int is chosen, but False if Integer is chosen. Integer is first in the default list for good reason.)
edited Jan 2 at 3:14
user2864740
44.1k671150
answered Jan 2 at 3:10
luquiluqui
48.3k6114168
Yes it is made concrete by defaulting, which is pretty much your second theory.
The expression [..] is ambiguous because the literal 4 is of Num a => a type in Haskell. 4 can be an Int, a Float or any other type that is an instance of Num, so the compiler can’t choose any particular type for the same reason above. But the Haskell Committee thought that this is too much restriction. After much debates, they compromised and added an ad-hoc rule for choosing a particular default type.
(Although it does matter which instance is chosen. For example 2^64 == 0 is True if Int is chosen, but False if Integer is chosen. Integer is first in the default list for good reason.)
edited Jan 2 at 3:14
user2864740
44.1k671150
answered Jan 2 at 3:10
luquiluqui
48.3k6114168
edited Jan 2 at 3:14
user2864740
44.1k671150
edited Jan 2 at 3:14
user2864740
44.1k671150
edited Jan 2 at 3:14
user2864740
44.1k671150
44.1k671150
answered Jan 2 at 3:10
luquiluqui
48.3k6114168
answered Jan 2 at 3:10
luquiluqui
48.3k6114168
answered Jan 2 at 3:10
luquiluqui
48.3k6114168
48.3k6114168
add a comment |
add a comment |
Thanks for contributing an answer to Stack Overflow!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f54000708%2fhow-does-the-haskell-compiler-emit-code-for-frominteger-0-frominteger-0%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
