可能重複:
Why can templates only be implemented in the header file?未定義的參考`無效排序::交換<int>(INT *,INT,INT)」
這是我的make文件:
#!/usr/bin/make -f
compiler = g++
compiler_flags = -Wall -I /usr/include/c++/4.5
debug_flags = -D DEBUG -g
binary_filename = sort_testing.bin
all: clean release
release:
$(compiler) $(compiler_flags) main.cpp sort.o -o $(binary_filename)
debug: sort.o
$(compiler) $(debug_flags) $(compiler_flags) main.cpp sort.o -o $(binary_filename)
run:
./$(binary_filename)
clean:
rm -f *.o $(binary_filename)
sort.o:
$(compiler) $(debug_flags) $(compiler_flags) -c sort.cpp
這裏是我的C++文件:
// sort.hpp
#ifndef SORT_H
#define SORT_H
namespace sort{
template<class T> void swap(T*,int,int);
}
#endif
// sort.cpp
#include "sort.hpp"
namespace sort{
template<class T>
void swap(T* items, int index_a, int index_b){
T t = items[index_a];
items[index_a] = items[index_b];
items[index_b] = t;
}
}
// main.cpp
#include <iostream>
#include <exception>
#include <time.h>
#include <stdlib.h>
#include <stdio.h>
using namespace std;
#include "sort.hpp"
using namespace sort;
#define NUM_INTS 5
int main(int argc, char** argv){
try{
cout << "\n\n\n";
srand(time(NULL));
int * int_coll = new int[NUM_INTS];
for (int x = 0; x < NUM_INTS; x++)
int_coll[x] = rand() % 100 + 1;
cout << "Before swap" << endl;
for (int x = 0; x < NUM_INTS; x++)
cout << "int " << x << " == " << int_coll[x] << endl;
cout << "\n\n\n";
cout << "Swapping ints" << endl;
swap<int>(int_coll, 0, 1);
cout << "AFter swap" << endl;
for (int x = 0; x < NUM_INTS; x++)
cout << "int " << x << " == " << int_coll[x] << endl;
}catch(exception& e){
cout << "Exception: " << e.what() << endl;
}
return 0;
}
而且,這裏是我的問題:
./make clean debug
rm -f *.o sort_testing.bin
g++ -D DEBUG -g -Wall -I /usr/include/c++/4.5 -c sort.cpp
g++ -D DEBUG -g -Wall -I /usr/include/c++/4.5 main.cpp sort.o -o sort_testing.bin
/tmp/ccRl2ZvH.o: In function `main':
/home/dev/c++/sorting/main.cpp:33: undefined reference to `void sort::swap<int>;(int*, int, int)'
collect2: ld returned 1 exit status
make: *** [debug] Error 1
不知道如何解決這個問題?
你不需要'swap'。該類型可以從'int_coll'推導出來。 –
Xeo
2011-05-16 06:10:47