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gcc parser bug? (using "unsigned long")

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Started by Kylotan January 29, 2007 04:32 PM

6 comments, last by Kylotan 17 years, 2 months ago

Kylotan

10,510

Author

January 29, 2007 04:32 PM

This works in MSVC 2003, but not in g++ 4.1.2:

return unsigned long();

So I have to use the following:

{
    typedef unsigned long ulong;
    return ulong();
}

The errors are: whatever.cpp:295: error: expected primary-expression before 'unsigned' whatever.cpp:295: error: expected ';' before 'unsigned' whatever.cpp:295: error: expected primary-expression before 'unsigned' whatever.cpp:295: error: expected `;' before 'unsigned' Both compilers are happy with returning int(), std::string(), etc. (For the curious, I am forcing the default constructor because it's one of a set of specialised template functions, each of which returns the default value for a given type.) Is this a bug in g++, or perhaps just a corner case that I was unaware of (and which MSVC handles 'better')?

Anonymous

January 29, 2007 06:12 PM

I don't have access to g++ right now, but I wonder if perhaps the () is binding tighter than the "unsigned", so that g++ is parsing it as "unsigned (long())", which wouldn't make sense. If that was the case one might expect "(unsigned long)()" to work correctly, so you might try that and see what happens. Just a shot in the dark.

rip-off

11,000

January 29, 2007 06:28 PM

Quote:Original post by Anonymous Poster
I don't have access to g++ right now, but I wonder if perhaps the () is binding tighter than the "unsigned", so that g++ is parsing it as "unsigned (long())", which wouldn't make sense. If that was the case one might expect "(unsigned long)()" to work correctly, so you might try that and see what happens. Just a shot in the dark.

I had already tried that, with no luck. But I'm using a mingw g++ which is version 3.something, so maybe it'll work on a newer gcc.

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hydroo

295

January 30, 2007 01:06 AM

I think it has to be like "long unsigned", because the printf format parameter is "%lu", too.

Sharlin

864

January 30, 2007 02:46 AM

gcc is conforming; the Standard requires that the type name used in a function-style cast be a simple-type-specifier, that is, either a (possibly namespace-qualified) user-defined type or typedef, or exactly one of the fundamental types char, wchar_t, bool, short, int, long, signed, unsigned, float, double, or void.

Adding parentheses doesn't work, because that would make the expression a C-style cast which doesn't allow the value-initializing "()" syntax. Yet another of the lovely quirks of C++.

Kylotan

10,510

Author

January 30, 2007 04:05 AM

Intriguing. Several 'informal' pieces of documentation include unsigned longs as part of the 'simple types', including the docs for IBM's compiler, but it would appear that the standard does not. MSVC's documentation notably doesn't include unsigned longs in there, but compiles the code anyway.

Is there any reason why unsigned longs don't fit into this type list? If not, I would think that MS's implementation is better, despite being non-standard.

Anonymous

January 30, 2007 04:20 AM

Why is it then that above code will not compile but the following does?

template<typename T>T foo(){	return T();}int main(int argc, char *argv[]){	foo<unsigned long>();	return 0;}

Kylotan

10,510

Author

January 30, 2007 04:37 AM

Presumably because it's a lexical issue, not a semantic issue. 'T' is a valid token in the context of the return statement, but 'unsigned long' isn't. That's also why my code works when you use the typedef.

gcc parser bug? (using "unsigned long")

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