我需要在C++中编写一个模板函数
但是当我在test.cc中取消注释
该程序链接并生成预期输出:
为什么链接失败,我该如何使其正常工作?
replace_all
,它将接受字符串、宽字符串、glibmm::ustring等,并将subject
中所有出现的search
替换为replace
。
replace_all.cc
template < class T >
T replace_all(
T const &search,
T const &replace,
T const &subject
) {
T result;
typename T::size_type done = 0;
typename T::size_type pos;
while ((pos = subject.find(search, done)) != T::npos) {
result.append (subject, done, pos - done);
result.append (replace);
done = pos + search.size ();
}
result.append(subject, done, subject.max_size());
return result;
}
test.cc
#include <iostream>
template < class T >
T replace_all(
T const &search,
T const &replace,
T const &subject
);
// #include "replace_all.cc"
using namespace std;
int main()
{
string const a = "foo bar fee boor foo barfoo b";
cout << replace_all<string>("foo", "damn", a) << endl;
return 0;
}
当我尝试使用gcc 4.1.2编译这个时
g++ -W -Wall -c replace_all.cc
g++ -W -Wall -c test.cc
g++ test.o replace_all.o
我理解为:
test.o: In function `main':
test.cc:(.text+0x13b): undefined reference to `
std::basic_string<char, std::char_traits<char>, std::allocator<char> >
replace_all< std::basic_string<char, std::char_traits<char>, std::allocator<char> > >(
std::basic_string<char, std::char_traits<char>, std::allocator<char> > const&,
std::basic_string<char, std::char_traits<char>, std::allocator<char> > const&,
std::basic_string<char, std::char_traits<char>, std::allocator<char> > const&
)
'
collect2: ld returned 1 exit status
但是当我在test.cc中取消注释
#include "replace_all.cc"
并以这种方式编译时:g++ -W -Wall test.cc
该程序链接并生成预期输出:
damn bar fee boor damn bardamn b
为什么链接失败,我该如何使其正常工作?