Weird function call like syntax

Question

I have come across this toplevel function in a C++ file. So my questions are:

What exactly is the smt2::parser type declaration doing there?
What is this feature called?
How is different from a regular call to p(ctx, is, interactive);?

Here's the code:

bool parse_smt2_commands(cmd_context & ctx, std::istream & is, bool interactive) {
    smt2::parser p(ctx, is, interactive);
    return p();
}

Unfortunately, I haven't been able to find the definition(s) of p() with grep so far. I'll update the post when I find the definitions (tracking down the headers includes manually can take a while).

look for the function call operator in smt2::parser. See http://stackoverflow.com/questions/2349867/how-can-it-be-useful-to-overload-the-function-call-operator — Kate Gregory, Dec 09 '13 at 21:52
That's an object instantiation followed by a call to `smt2::parser::operator()`. — David G, Dec 09 '13 at 21:53
smt2::parser is a type. p is an instance. The (ctx, is, interactive) are the args to construct p. — Anon Mail, Dec 09 '13 at 21:54
Also, OP should have read a beginner tutorial. This is too basic. — , Dec 09 '13 at 21:54
possible duplicate of [C++ Instance Initialization Syntax](http://stackoverflow.com/questions/372665/c-instance-initialization-syntax) — , Dec 09 '13 at 21:54
Yes gang, `p` is an object that is instantiated in the middle line of the code snippet. What is happening in the last line of the code snippet? — Kate Gregory, Dec 09 '13 at 21:54
I agree with @Kate Gregory. It appears to me that smt2::parser 'class' requires an explicit constructor then a default constructor. The object may have some sort of internal state. Bad class, but that's the interface. — KeithSmith, Dec 09 '13 at 21:56
@KeithSmith There's no default constructor involved in this code whatsoever. Kate was only right in that `p()` invokes `stm2::parser::operator()`, but that is **not** the line that OP is curious of, nor does it call its default constructor. — , Dec 09 '13 at 21:57
OP hasn't asked about the last line because "I haven't been able to find the definition(s) of p() with grep so far" but I am sure it was included for a reason. — Kate Gregory, Dec 09 '13 at 21:59
Yes. I pulled the trigger too soon. @H2CO3 (are you a chemist?) and @Kate Gregory are correct, `p()` is an overloaded operator, not a constructor. Need more speed, less haste. — KeithSmith, Dec 09 '13 at 22:01
@KeithSmith No problem. No, I'm not a chemist (I *will* be a bionics engineer, though. My name is related to something else ;) — , Dec 09 '13 at 22:02

score 6 · Accepted Answer · edited Dec 09 '13 at 21:58

6

It's a little difficult to tell directly. But what I imagine is going on is smt2::parser is the name of a class or struct type. The line

stm2::parser p(ctx,is,interactive);

calls the constructor of that class/struct. The class defined by stm2::parser likely overloaded the () operator. I found a tutorial with more information about operator overloading for function calls.

edited Dec 09 '13 at 21:58

Kate Gregory

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answered Dec 09 '13 at 21:54

Behram Mistree

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@Kate Gregory, Thanks for the edit. That made the answer much more readable. – Behram Mistree Dec 09 '13 at 22:03

score 3 · Answer 2 · answered Dec 09 '13 at 21:59

In the code provided in the OP, p is an object of type smt2::parser. It calls a constructor that takes three arguments. p() calls an overloaded operator (probably operator()()) that returns a bool.

Here's an example:

class Test {
public:
Test(int i, char j, bool k) {
}
~Test() { }

bool operator()() const {
return true;
}
};

In this case, we simply call the operator on a temporary to demonstrate what is happening.

int main() {
std::cout << Test(1, 'a', true)(); // 1
return 0;
}

Weird function call like syntax

2 Answers2