[英]Over looping (Cout) in C++
#include <iostream>
#include <string>
#include <cctype>
using namespace std;
char hold;
string name;
char num1;
char num2;
int main() {
cout << "Hello!\n";
cout << "Tell me your name?: ";
cin >> name;
cout << "Well well well, if it isn't "<< name << "!\n";
cout << "Enter a NUMBER " << name << ": ";
cin >> num1;
while(!isdigit(num1)) {
cout << "Enter a NUMBER " << name << ": ";
cin >> num1;
}
cin >> hold;
system("pause");
return 0;
}
問題是,這使偵查團陷入了循環。 我如何解決它?
謝謝。
更好的方法是使用std::stringstream
(注意:include sstream
)
int getNumber()
{
std::string line;
int i;
while (std::getline(std::cin, line))
{
std::stringstream ss(line);
if (ss >> i)
{
if (ss.eof())
{
break;
}
}
std::cout << "Please re-enter your input as a number" << std::endl;
}
return i;
}
這將取代您的while循環,並且您在詢問了號碼后就進行了呼叫,因為您已經知道該怎么做。
以下是原始嘗試的簡化版本。 但是,與原始字符一樣,它僅檢查單個字符。
如果我將num1更改為int,則需要檢查輸入是否有效,如@Dieter Lucking所述。
#include <iostream>
using namespace std;
int main() {
char num1;
do {
cout << "\nEnter a number: ";
cin >> num1
} while(!isdigit(num1));
}
staticx解決方案的一些變化,它將傳遞DieterLücking的echo "" | test
echo "" | test
線。
我使用istringstream並獲得輸入,直到沒有更多標准輸入或得到有效輸入為止。 我將其全部推送到可用於任何類型的模板化Get
函數中。 您只需要向用戶提示:
template<typename T>
void Get(T& toSet, std::string prompt) // read from cin
{
std::string nextIn;
cout << prompt;
getline(cin >> std::ws, nextIn);
istringstream inStream(nextIn);
while(cin && !(inStream >> toSet))
{
inStream.clear();
cout << "Invalid Input. Try again.\n" << prompt;
getline(cin >> std::ws, nextIn);
inStream.str(nextIn);
}
if (!cin)
{
cerr << "Failed to get proper input. Exiting";
exit(1);
}
}
您將像這樣使用它:
int myNumber = 0;
Get(myNumber, "Please input a number:");
#include <iostream>
#include <string>
#include <sstream>
using namespace std;
template<typename T>
void Get(T& toSet, std::string prompt) // read from cin
{
std::string nextIn;
cout << prompt;
getline(cin >> std::ws, nextIn);
istringstream inStream(nextIn);
while(cin && !(inStream >> toSet))
{
inStream.clear();
cout << "Invalid Input. Try again.\n" << prompt;
getline(cin >> std::ws, nextIn);
inStream.str(nextIn);
}
if (!cin)
{
cerr << "\nFailed to get proper input. Exiting\n";
exit(1);
}
}
int main()
{
string name;
int num1 = -1;
cout << "\nHello!\n";
Get(name, "\nTell me your name?:");
cout << "\nWell well well, if it isn't "<< name << "!\n";
Get(num1, std::string("\nEnter a NUMBER, ") + name + ": ");
cout << "\nYou entered number: " << num1 << std::endl;
return 0;
}
聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.