Why is there a compile error in this code?

Question

I was trying to understand the difference between closures and function pointers, and I came across this answer in SO

What I don't understand is this code

BOOL (*lessThanTest)(int);
int lessThan = 100;

lessThanTest = &LessThan;

BOOL LessThan(int i) {
   return i < lessThan; // compile error - lessThan is not in scope
}

Why there is a compile error consideringn that lessThan is a global variable, it can be accessed from within LessThan function, did I miss something?

EDIT

This is not my code, it's taken from an answer in SO Function pointers, Closures, and Lambda

can you post minimum compilable code? (comment out the part that gives you the error) — Luchian Grigore, Nov 11 '11 at 15:11
That wouldn't even compile, since `LessThan` isn't yet declared before `lessThanTest = &LessThan;` — , Nov 11 '11 at 15:11
This is invalid code in C. You can achieve same result as in c# using c++ and functors. — joy, Nov 11 '11 at 15:20
@neagoegab -- it's not supposed to be valid C and it's not his example. Read the linked question for more context. — Matt Fenwick, Nov 11 '11 at 15:21
But it is still a bad example with incomplete specifications... Closure can have state same as a functor while a function pointer does not have a state and the state is maintained in a global variable or passed as an argument.... — joy, Nov 11 '11 at 15:34

score 2 · Answer 1 · answered Nov 11 '11 at 15:12

Closures take all the variables in their lexical scope along for the ride, possibly extending their lifetimes. Function pointers don't -- if the variables referenced inside their code disappear, they're hosed.

The code example you've given is a little bit confusing. I believe that it's meant to be inside of a function, meaning that lessThan is a local variable. If that scope is exited, but the function pointer still exists, then its code would have a reference to a non-existent variable -- lessThan.

score 1 · Accepted Answer · answered Nov 11 '11 at 15:13

1

You missed a paragraph in that answer:

But, now I have to pass the 2 arguments when I evaluate it. If I wished to pass this function pointer to another function where lessThan was not in scope, I would either have to manually keep it alive by passing it to each function in the chain, or by promoting it to a global.

In what you posted, int lessThan is not meant at global scope, it should be assumed to be in a function somewhere.

answered Nov 11 '11 at 15:13

Mat

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I think that paragraph is talking about the second code snippet – Mansuro Nov 11 '11 at 15:15
(I think) It talks about both. Neither are strictly valid `C` at global scope anyway. – Mat Nov 11 '11 at 15:18
"I could define a function pointer " ... "If I wished to pass **this** function pointer " – Mansuro Nov 11 '11 at 15:23
I don't see your point. If `lessThan` was at global scope, there would be no compiler error (assuming the rest of the code was correct). Both examples manipulate a function pointer, the second one has different object lifetime/visibility issues than the first, but both have the same fundamental problem - C doesn't have closures (or lambdas, or nested functions for that matter), you need to implement workarounds yourself. – Mat Nov 11 '11 at 15:32
my point is, what can be done in closures can be done using functions pointers with some workarounds like you said – Mansuro Nov 11 '11 at 15:34
Then what is your question? The code you quoted isn't valid at global scope, and you understand stuff needs to be done to simulate closures. – Mat Nov 11 '11 at 15:39
My question is about that snippet of code, for what reason did the author say it's a compile error, while it can work if lessThan variable is a global variable – Mansuro Nov 11 '11 at 15:43
It's a compiler error if `lessThan` is not in scope in the `LessThan` function. Since C doesn't have nested functions (or classes), the only ways for `lessThan` to be in scope are 1. global, 2. argument. – Mat Nov 11 '11 at 15:46

score 0 · Answer 3 · answered Nov 11 '11 at 15:14

0

Well, no, if lessThan is global, it shouldn't produce compile error, though you can hardly tell by this fragment what is meant to be where. lessThanTest=&LessThan; is definitely from some local scope, for instance.

answered Nov 11 '11 at 15:14

Michael Krelin - hacker

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score 0 · Answer 4 · edited Nov 11 '11 at 15:17

0

This code has some problems:

You declare lessThanTest as an uninitialized function pointer
You can't assign something to it later under global scope
You're using LessThan before declaring it.

edited Nov 11 '11 at 15:17

Blagovest Buyukliev

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answered Nov 11 '11 at 15:14

sidyll

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Why is there a compile error in this code?

4 Answers4