最近我遇到了一个问题,不知何故(但只是某种程度上)对我有意义.它基于解释临时的构造作为单个(!)构造函数参数的声明.请看下面的最小例子.
#include <iostream>
class Foo0{
public:
Foo0(int a){};
void doStuff() {std::cout<<"maap"<<std::endl;};
};
class Foo1{
public:
Foo1(int a){};
void doStuff() {std::cout<<"maap"<<std::endl;};
};
class Foo2{
public:
Foo2(int a){};
void doStuff() {std::cout<<"maap"<<std::endl;};
};
class Bar{
public:
Bar(Foo0 foo0, Foo1 foo1, Foo2 foo2){};
};
int main () {
int x = 1;
Bar bar0(Foo0(x), Foo1(x), Foo2(x)); // Does not work: conflicting declaration ‘Foo1 x’ previous declaration as ‘Foo0 x’; conflicting declaration ‘Foo2 x’ previous declaration as ‘Foo0 x’
Bar bar1(Foo0{x}, Foo1(x), Foo2(x)); // Works …Run Code Online (Sandbox Code Playgroud) 我有这段代码可以用clang(甚至是-Weverything)编译好,但gcc会发出错误.
#include <iostream>
#include <vector>
#include <fstream>
using namespace std;
class PhonebookWriter
{
public:
PhonebookWriter(const string& fname):
fname_(fname), names_(), numbers_() {}
PhonebookWriter& operator()(const string& name,
const string& number)
{
names_.push_back(name);
numbers_.push_back(number);
return *this;
}
~PhonebookWriter(void)
{
ofstream f(fname_.c_str());
for(size_t i=0;i<names_.size();++i)
f << names_[i] << " " << numbers_[i] << "\n";
f.close();
}
private:
const string fname_;
vector<string> names_;
vector<string> numbers_;
};
namespace {
void write_guests_data(const string& fname)
{
PhonebookWriter(fname)("Mr Foo Bar","12345")("Mrs Bar Foo","54321");
}
}
int main(void)
{
write_guests_data("phone_book.txt");
return …Run Code Online (Sandbox Code Playgroud)