$a = some object;
$b = another object;
function do_some_stuff() {
$c = $a;
$b->a_property = $c;
}
do_some_stuff();
echo $b->a_property; # Undefined, becuase $c was deleted when the func
#include <iostream>
using namespace std;
template <class T>
void swap1(T& a, T& b) {
T c = a;
a = b;
b = c;
}
int main() {
int n1 = 5;
int n2 = 7;
swap1(n1, n2);
一般STL容器不能容纳非CopyAssignable类型,如引用。如果我以不复制副本的方式构建容器,那么代码是否有效。它编译与std=c++11和c++14与某些版本的gcc-7.2,但以下有效,或我可以指望它与图书馆升级打破?在这种情况下,我应该使用reference_wrapper吗? #include <unordered_map>
struct S {};
void use (S&