带有while语句范围catch的if语句中的C ++错误
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在大多数情况下,我有这个代码工作。
我遇到的唯一问题是当我输入兼容的答案后输入不兼容的答案时,第一次,程序终止而不是重置。
代码采用整数1..4
作为答案,或-1
作为标志终止序列,然后cout
计数的结果。
如果没有输入这五个答案中的一个,则应该再次请求兼容的输入。
如果输入一次超出范围的输入,则在代码开始时,它将捕获并正常运行。
但是,在输入正确答案一次后,如果收到的答案超出范围或不是-1
,则序列停止而不是要求正确的输入。
#include <iostream>
using namespace std;
int main()
{
int coffee = 0;
int tea = 0;
int coke = 0;
int orangeJuice = 0;
int person = 1;
int choice;
cout << "Please input the favorite beverage of person #" << person << ": "
<< endl
<< "Choose 1, 2, 3, or 4 from the above menu or -1 to exit the program"
<< endl;
cin >> choice;
while (choice < 1 || choice > 4) {
cout << "Invalid choice" << endl;
cout << "Please input the favorite beverage of person #" << person << ": " << endl
<< "Choose 1, 2, 3, or 4 from the above menu or -1 to exit the program" << endl;
cin >> choice;
}
while (choice != -1 && choice >= 1 && choice <= 4) {
if (choice == 1) {
person++;
coffee++;
cout << "Please input the favorite beverage of person #" << person <<
": " << endl
<< "Choose 1, 2, 3, or 4 from the above menu or -1 to exit the program" << endl;
cin >> choice;
}
else if (choice == 2) {
person++;
tea++;
cout << "Please input the favorite beverage of person #" << person <<
": " << endl
<< "Choose 1, 2, 3, or 4 from the above menu or -1 to exit the program" << endl;
cin >> choice;
}
else if (choice == 3) {
person++;
coke++;
cout << "Please input the favorite beverage of person #" << person <<
": " << endl
<< "Choose 1, 2, 3, or 4 from the above menu or -1 to exit the program" << endl;
cin >> choice;
}
else if (choice == 4) {
person++;
orangeJuice++;
cout << "Please input the favorite beverage of person #" << person <<
": " << endl
<< "Choose 1, 2, 3, or 4 from the above menu or -1 to exit the program" << endl;
cin >> choice;
}
}
cout << "The total number of people surveyed is " << person << ". The results are as followed." << endl;
cout << "Beverage Number of Votes" << endl;
cout << "**********************************" << endl;
cout << "Coffee " << coffee << endl;
cout << "Tea " << tea << endl;
cout << "Coke " << coke << endl;
cout << "Orange Juice " << orangeJuice << endl;
return 0;
}
答案
您没有进行任何错误处理以确保cin >> choice
成功。如果用户输入的内容不能转换为int
(比如字母而不是数字),operator>>
会失败并在流上设置错误标志,choice
要么是不确定的要么是0(取决于实现),并且输入中的错误输入会保留在输入中缓冲。在清除错误状态和输入缓冲区之前,operator>>
将继续失败。
此外,您的第一个while
循环不允许用户输入-1
作为答案。它满足choice < 1
条件,使循环提示用户输入。
此外,您在处理输入和输出的方式上只有很多冗余。
尝试更像这样的东西:
#include <iostream>
#include <limits>
using namespace std;
int main()
{
int coffee = 0;
int tea = 0;
int coke = 0;
int orangeJuice = 0;
int person = 0;
int choice;
do
{
cout << "Please input the favorite beverage of person #" << person+1 << ": " << endl;
do
{
cout << "Choose 1, 2, 3, or 4 from the above menu, or -1 to exit the program" << endl;
if (!(cin >> choice))
{
cout << "Invalid input" << endl;
cin.ignore(numeric_limits<streamsize>::max(), '
');
cin.clear();
continue;
}
if (((choice >= 1) && (choice <= 4)) || (choice == -1))
break;
cout << "Invalid choice" << endl;
}
while (true);
if (choice == -1)
break;
switch (choice)
{
case 1:
++coffee;
break;
case 2:
++tea;
break;
case 3:
++coke;
break;
case 4:
++orangeJuice;
break;
}
++person;
}
while (true);
cout << "The total number of people surveyed is " << person << ". The results are as followed." << endl;
cout << "Beverage Number of Votes" << endl;
cout << "**********************************" << endl;
cout << "Coffee " << coffee << endl;
cout << "Tea " << tea << endl;
cout << "Coke " << coke << endl;
cout << "Orange Juice " << orangeJuice << endl;
return 0;
}
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