How to store user input in an int array - c++

Question

I need to ask the user to input 8 zip codes and then store them in an array of integers and then to output them one by one , each being in a new line.These two things should be done in separate functions. But when it runs the code first time it shows only menu, then second time in loop when i enter L and input 8 zipcodes it shows this error
Enter your choice: libc++abi.dylib: terminating with uncaught exception of type std::out_of_range: basic_string(in here 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

Now I have not had a hand on C++ for a long time, but this will work.
You should look on the variable declarations and some other things though.
#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;
}
}
You can add remaining of your functions with case 3 and 4.
Try to debug the code and see how it work before you post :)

Answer 2

You need to look closer at the statements
return ZipCodes[i];
and
ZipCodes[SIZE]= LoadCityZipCodes(ZipCodes, SIZE);
In both cases you use ZipCodes[8] which is definitely out of range.
To solve both problems, don't return anything from LoadCityZipCodes (i.e. make it return void), because it already sets the values in the array.

Answer 3

You need to remeber that when programming we start counting at 0. That means for an array that has the size 8 all the valid indexes are: 0, 1, 2, 3, 4, 5, 6, 7. When you try to access your array with ZipCodes[8] you are asking for the 9th element which is out of range.