如何修复c++;不输出输入 如何修复C++不输出输入,我不知道!所以请帮助我,我是新到C++!
如何修复c++;不输出输入 如何修复C++不输出输入,我不知道!所以请帮助我,我是新到C++!,c++,C++,#包括 使用名称空间std; int main() { std::cout输入; } 如果(输入=“交付”) { 难道你不是在征求意见吗 std::cout << "Please type if you want pickup or delivery: "; 接下来,您需要交换getline(cin,address);和cin.ignore();around..当前您正在请求输入,然后清除缓冲区 如果(地址==地址) 你想在这里实现什么 您的代码可以这样编写 #
#包括
使用名称空间std;
int main()
{
std::cout输入;
}
如果(输入=“交付”)
{
难道你不是在征求意见吗
std::cout << "Please type if you want pickup or delivery: ";
接下来,您需要交换getline(cin,address);
和cin.ignore();
around..当前您正在请求输入,然后清除缓冲区
如果(地址==地址)
你想在这里实现什么
您的代码可以这样编写
#include <iostream>
using namespace std;
int main()
{
std::cout << "Please type if you want pickup or delivery: ";
string input;
string pickup;
string delivery;
string address;
cin>>input;
if (input == "pickup")
{
cout << "What pizza do you want(press r for a random pizza and c for cheese pizza!): ";
cin >> input;
}
if (input == "delivery")
{
cout << "Tell me your adress and i will deliver a random pizza to ur house: ";
cin.ignore();
getline(cin,address);
cout << "Thx we will delivery the pizza to " << address << " tommorow at 3PM!" << endl;
}
return 0;
}
#包括
使用名称空间std;
int main()
{
std::cout>输入;
如果(输入=“拾取”)
{
cout>输入;
}
如果(输入=“交付”)
{
coutstd::cin.ignore()
之前的std::getline()
。我想说,如果我输入地址,它会打印单词和输入的人的地址(但thx你解决了我的问题),我的代码会这样做?因为你有cin.ignore地址没有打印,但从一开始你就没有输入收货或送货
#include <iostream>
using namespace std;
int main()
{
std::cout << "Please type if you want pickup or delivery: ";
string input;
string pickup;
string delivery;
string address;
cin>>input;
if (input == "pickup")
{
cout << "What pizza do you want(press r for a random pizza and c for cheese pizza!): ";
cin >> input;
}
if (input == "delivery")
{
cout << "Tell me your adress and i will deliver a random pizza to ur house: ";
cin.ignore();
getline(cin,address);
cout << "Thx we will delivery the pizza to " << address << " tommorow at 3PM!" << endl;
}
return 0;
}