我有三个文件.main.cpp 的内容是
I have three files . The contents of main.cpp are
#include<iostream>
#include<QString>
#include "util.h"
int main()
{
using Util::convert2QString;
using namespace std;
int n =22;
QString tmp = convert2QString<int>(n);
return 0;
}
util.h
namespace Util
{
template<class T>
QString convert2QString(T type , int digits=0);
}
util.cpp
namespace Util
{
template<class T>
QString convert2QString(T type, int digits=0)
{
using std::string;
string temp = (boost::format("%1%") % type).str();
return QString::fromStdString(temp);
}
}
当我尝试使用以下命令编译这些文件时,出现未定义的引用错误
When I try to compile these files with following command I get undefined reference error
vickey@tb:~/work/trash/template$ g++ main.cpp util.cpp -lQtGui -lQtCore -I. -I/usr/local/Trolltech/Qt-4.8.0/include/QtCore -I/usr/local/Trolltech/Qt-4.8.0/include/QtGui -I/usr/local/Trolltech/Qt-4.8.0/include
/tmp/cca9oU6Q.o: In function `main':
main.cpp:(.text+0x22): undefined reference to `QString Util::convert2QString<int>(int, int)'
collect2: ld returned 1 exit status
模板声明或实现有问题吗?为什么我会收到这些链接错误:?
Is there something wrong with the template declaration or implementation ? why M I getting these linking errors :?
非专用模板的实现必须对使用它的翻译单元可见.
The implementation of a non-specialized template must be visible to a translation unit that uses it.
编译器必须能够看到实现,才能为代码中的所有特化生成代码.
The compiler must be able to see the implementation in order to generate code for all specializations in your code.
这可以通过两种方式实现:
This can be achieved in two ways:
1) 将实现移到标题内.
1) Move the implementation inside the header.
2) 如果您想将其分开,请将其移动到包含在原始标题中的不同标题中:
2) If you want to keep it separate, move it into a different header which you include in your original header:
util.h
namespace Util
{
template<class T>
QString convert2QString(T type , int digits=0);
}
#include "util_impl.h"
util_impl.h
namespace Util
{
template<class T>
QString convert2QString(T type, int digits=0)
{
using std::string;
string temp = (boost::format("%1") % type).str();
return QString::fromStdString(temp);
}
}
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