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#2
rjsp2023-04-06 13:50
ave(la, lb); 是不是将 template <typename type1> type1 swap(type1 a, type1 b) 实例成 swap<int>( int, int ) ？
所以  程序代码：#include <iostream> using namespace std; template<typename type1> void ave( type1 a, type1 b ) { a = b; } int main() { int a = 0; int b = 1; int& la = a; int& lb = b; ave(la, lb); cout << a << " " << b; } 等同于 程序代码：#include <iostream> using namespace std; void ave( int a, int b ) { a = b; } int main() { int a = 0; int b = 1; int& la = a; int& lb = b; ave(la, lb); cout << a << " " << b; } 展开函数调用，就等同于 程序代码：#include <iostream> using namespace std; int main() { int a = 0; int b = 1; int& la = a; int& lb = b; int ave_a = la; int ave_b = lb; ave_a = ave_b; cout << a << " " << b; } 里面没有一句在修改a |
程序代码：#include<iostream>
using namespace std;
template <typename type1>
type1 swap(type1 a, type1 b)
{
a = b;
cout << "函数模板";
return 0;
}
int main()
{
int a = 0;
int b = 1;
int& la = a;
int& lb = b;
ave(la, lb);
cout << a << " " << b;
system("pause");
return 0;
}
按照我的理解，传入a和b的引用，相当于传入了它们的地址，在函数中通过改变它们的引用，来改变它们。
