Question

我需要让用户输入8个邮政编码，然后将它们存储在一个整数数组中，然后逐个输出，每个都在一个新的行中。这两件事应该在不同的函数中完成。但是当它第一次运行代码时它只显示菜单，然后第二次循环，当我输入L并输入8个zipcodes时显示此错误输入您的选择：libc ++ abi.dylib：以类型为std :: out_of_range的未捕获异常终止：basic_string（此处为optionInChar = optionInString.at（0）;）

using namespace std;

void DisplayCityZipCodes();
int LoadCityZipCodes(int ZipCodes[],int SIZE);
void DisplayCityZipCodes(int ZipCodes[],int SIZE);
void DisplayMenu();
char GetOption();

int main(){

int const SIZE=8;
int ZipCodes[SIZE]={0};

bool moreWork=true;



option=GetOption();

 while(moreWork) {

  DisplayMenu();

 option=GetOption();

  switch(option){

     case 'L':
    ZipCodes[SIZE]= LoadCityZipCodes(ZipCodes,  SIZE);

     break;

        cout<<"D";
     case 'D': DisplayCityZipCodes(ZipCodes,  SIZE);
     break;

   }
 }
}

void DisplayMenu(){
cout<<"       **********************\n\n";

cout<<"       San Jose City Zip codes\n\n";

cout<<"       **********************\n\n";

cout<<"1. Load City zip codes\n";
cout<<"2. Display all City zip codes\n";
cout<<"3. Search a City zip code\n";
cout<<"4. Reverse the City zip code List\n";
cout<<"5. Quit\n";

}

char GetOption(){
string optionInString="";
char optionInChar='a';
cout<<"\n\nEnter your choice: ";
getline(cin,optionInString);
optionInChar=optionInString.at(0);
cout<<"\n";
return optionInChar;

  }



int LoadCityZipCodes(int  ZipCodes[],int  SIZE){

cout<<"PLease enter 8 city Zip Codes ";
int i=0;
for(;i<8;i=i+1){
cin >>ZipCodes[i];
}
return ZipCodes[i];


}

void DisplayCityZipCodes(int ZipCodes[],int SIZE){
int i=0;
for(;i<8;i=i+1){
cout<<ZipCodes[i]<<endl;
}


}

Answer 1

现在我已经很长时间没有使用过C ++了，但是这样可行。您应该查看变量声明和其他一些内容。

#include<iostream>
#include<conio.h>

using namespace std;

void DisplayCityZipCodes();
void LoadCityZipCodes();
void DisplayMenu();
int const SIZE = 8;
int ZipCodes[SIZE];

int main()
{
    DisplayMenu();
    return 0;
}

void DisplayMenu()
{
    int ch;
    cout << "       **********************\n\n";
    cout << "       San Jose City Zip codes\n\n";
    cout << "       **********************\n\n";
    do
    {

        cout << "1. Load City zip codes\n";
        cout << "2. Display all City zip codes\n";
        cout << "3. Search a City zip code\n";
        cout << "4. Reverse the City zip code List\n";
        cout << "5. Quit\n";

        cout << "\nPlease enter your choice:";
        cin >> ch;
        switch (ch)
        {
        case 1:
            LoadCityZipCodes();
            break;
        case 2:
            DisplayCityZipCodes();
            break;
        case 5:
            exit(0);   //You will need to include math.h for this.
        default:
            cout << "please enter a proper choice!";
            break;
        }
    } while (1);

}

void LoadCityZipCodes()
{
    cout << "PLease enter 8 city Zip Codes ";
    for (int i=0; i<8; i++)
    {
        cin >> ZipCodes[i];
    }

}

void DisplayCityZipCodes()
{
    for (int i=0; i<8; i++)
    {
        cout << ZipCodes[i] << endl;
    }
}

您可以使用案例3和4添加剩余的功能。在发布之前尝试调试代码并查看它是如何工作的：）

Answer 2

您需要仔细查看陈述

return ZipCodes[i];

和

ZipCodes[SIZE]= LoadCityZipCodes(ZipCodes,  SIZE);

在这两种情况下，您使用的ZipCodes[8]肯定超出了范围。

要解决这两个问题，请不要从LoadCityZipCodes返回任何内容（即返回void），因为它已经设置了数组中的值。

Answer 3

你需要记住，在编程时我们从0开始计数。这意味着对于大小为8的数组，所有有效索引都是：0,1,2,3,4,5,6,7。当你尝试使用ZipCodes [8]访问你的数组，你要求的第9个元素超出范围。

如何在int数组中存储用户输入

3 个答案: