At 04:12 AM 2/24/2004, David Abrahams wrote:
Actually, if that's the way tokenizer works, you can do it, I'm pretty sure:
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I changed main() slightly (see below) and tried with several compilers: g++ 3.3.1: lvalue.cpp: In function `int main()': lvalue.cpp:32: error: syntax error before `)' token bcc32: error E2188 lvalue.cpp 32: Expression syntax in function main() error E2293 lvalue.cpp 32: ) expected in function main() Codewarrior 8.3: lvalue rvalue lvalue rvalue Intel 8.0 and VC++ 7.1: lvalue rvalue lvalue Note 4th output line missing for Intel 8.0 and VC++ 7.1 What's going on here? Mystified-in-Virginia, --Beman #include <vector> #include <iostream> template <class TokenizerFunc> struct tokenizer { struct rvalue { template <class T> rvalue(T const&) {} }; // handle lvalues template <typename Container> explicit tokenizer(Container& c, const TokenizerFunc& f = TokenizerFunc()) { std::cout << "lvalue\n"; } // handle rvalues explicit tokenizer(rvalue const& rval, const TokenizerFunc& f = TokenizerFunc()) { std::cout << "rvalue\n"; } }; int main() { std::vector<int> const x; tokenizer<int> z( (x) ); tokenizer<int> z2( (std::vector<int>()) ); tokenizer<int> z3( x ); tokenizer<int> z4( std::vector<int>() ); return 0; }