我不明白为什么这并不编译:模板模板类
struct A
{};
template<class T>
struct B
{};
template<template<class> class T1, class T2>
struct C
{};
int
main (int ac, char **av)
{
typedef B<double> b; //compiles
typedef B<const double> b_const; //compiles
typedef B<A> ba; //compiles
typedef B<const A> ba_const; //compiles
typedef C<B,double> c1; //compiles
typedef C<B,const double> c2; //compiles
typedef C<const B,double> c3; //ISO C++ forbids declaration of ‘type name’ with no type
}
(我找到参照标准有点神秘)
我有什么改变,使其编译?
编辑:
编译器的细节(这好像是培训相关):
Using built-in specs.
Target: x86_64-linux-gnu
Configured with: ../src/configure -v --with-pkgversion='Ubuntu/Linaro 4.4.4-14ubuntu5' --with-bugurl=file:///usr/share/doc/gcc-4.4/README.Bugs --enable-languages=c,c++,fortran,objc,obj-c++ --prefix=/usr --program-suffix=-4.4 --enable-shared --enable-multiarch --enable-linker-build-id --with-system-zlib --libexecdir=/usr/lib --without-included-gettext --enable-threads=posix --with-gxx-include-dir=/usr/include/c++/4.4 --libdir=/usr/lib --enable-nls --with-sysroot=/ --enable-clocale=gnu --enable-libstdcxx-debug --enable-objc-gc --disable-werror --with-arch-32=i686 --with-tune=generic --enable-checking=release --build=x86_64-linux-gnu --host=x86_64-linux-gnu --target=x86_64-linux-gnu
Thread model: posix
gcc version 4.4.5 (Ubuntu/Linaro 4.4.4-14ubuntu5)
EDIT2:
通过交代的手段,我试图做这样的事情:
template<template<class> class TheContainer, class T>
struct Iterator
template<class T>
struct Container
typedef Iterator<Container, double> iterator;
typedef Iterator<const Container, double> const_iterator;
非模板容器的技术可在此升压文档的末尾找到:http://www.boost.org/doc/libs/1_46_1/libs/iterator/doc/iterator_facade.html
我猜测解决方案并不是为了模板。回想起来,这似乎很明显。
这篇文章有一个很棒的主题:) – Laserallan 2011-05-13 23:03:48
@Laserallan它使我成为Perl的口号 – Tom 2011-05-13 23:09:35
“我需要改变什么才能编译它?”显然,你需要删除最后一个typedef。更严重的是,你想通过这样做完成什么? – 2011-05-14 00:24:54