how to make my program run again C++ - c++

Question

I have given the following question.
Write a menu driven program with following options:
Add new value
Search value
Modify value
Print value
Print sum of all values
Quit / Terminate
You have to create 5 options as five function. Add another function to show menu options.
and here is my code:
#include<iostream>
using namespace std;
float f[100]={0};
//1st option
void AddNewValue(){int input;
cout<<"Enter a value\n";
cin>>f[input];
}
//2nd option
void SearchValue(){int i, search;
cout<<"Enter a value to search\n";
cin>>search;
int match=0;
for (int i=1;i<=100;i++)
{if (f[i]==search)
{match=1;
break;}
}
if (match==1){cout<<"Matched value found\n";
}
else {cout<<"No match found\n";}
}
//3rd option
void ModifyValue(){int input1;
cout<<"Enter the position at which you want to modify value\n";
cin>>input1;
cout<<"Enter a value\n";
cin>>f[input1-1];
}
//4th option
void PrintValue(){int i;
for (i=1;i<=100;i++)
{cout<<f[i]<<' ';}
}
//5th option
void PrintSum(){int i,sum;
for(i=1;i<=100;i++)
{sum=f[i]+f[i+1];}
cout<<"Sum is : "<<sum;
}
//starting Function
void menu(){int x;
cout<<"Enter an option: \n";
cout<<"1- Add new value\n2- Search Value\n3- Modify value\n4- Print Value\n5- Print sum of all values\n6- Quit/terminate\n";
cin>>x;
if(x==1){
AddNewValue();
}
else if (x==2){
SearchValue();
}
else if(x==3){
ModifyValue();
}
else if(x==4){
PrintValue();
}
else if(x==5){
PrintSum();
}
else{
}
}
int main(){
menu();
}
And I want to make my full program run again and again until the user enter a wrong option.

Answer 1

One option available to you is to create a while loop encapsulating the body of your menu() function, and using break; if the user enters 6.
void menu(){
int x = 0;
while(1){
cin >> x;
//your code here
if(x==6)
break;
}
}
This example will cause your menu to repeat until the user enters 6.
The purpose of break; is to 'break' from the while loop. In this case, breaking will end your menu()and return to main().
Further, it is good practice to add return 0; to the end of your main function.

Answer 2

You can use a loop and make it run untill user give a specific input. In below code once the program will run and then it will ask user to input 5 to exit or any other key to run again, and if user gives 5 as input, then it will stop or if user gives any other input then program will run again.
Hope it is clear and help you.
#include<iostream>
using namespace std;
float f[100]={0};
//1st option
void AddNewValue(){int input;
cout<<"Enter a value\n";
cin>>f[input];
}
//2nd option
void SearchValue(){int i, search;
cout<<"Enter a value to search\n";
cin>>search;
int match=0;
for (int i=1;i<=100;i++)
{if (f[i]==search)
{match=1;
break;}
}
if (match==1){cout<<"Matched value found\n";
}
else {cout<<"No match found\n";}
}
//3rd option
void ModifyValue(){int input1;
cout<<"Enter the position at which you want to modify value\n";
cin>>input1;
cout<<"Enter a value\n";
cin>>f[input1-1];
}
//4th option
void PrintValue(){int i;
for (i=1;i<=100;i++)
{cout<<f[i]<<' ';}
}
//5th option
void PrintSum(){int i,sum;
for(i=1;i<=100;i++)
{sum=f[i]+f[i+1];}
cout<<"Sum is : "<<sum;
}
//starting Function
void menu(){int x;
cout<<"Enter an option: \n";
cout<<"1- Add new value\n2- Search Value\n3- Modify value\n4- Print Value\n5- Print sum of all values\n6- Quit/terminate\n";
cin>>x;
if(x==1){
AddNewValue();
}
else if (x==2){
SearchValue();
}
else if(x==3){
ModifyValue();
}
else if(x==4){
PrintValue();
}
else if(x==5){
PrintSum();
}
else{
}
}
int main(){
int repeater;
do{
menu();
cout<<"Enter 5 to exit or any other key to run the program again :";
cin>>repeater;
}while(repeater != 5);
}

Answer 3

Keep it simple. The function in charge of displaying the menu should only display the menu and return the user's choice.
enum choice
{
ADD_NEW_VALUE,
SEARCH_VALUE,
// ...
QUIT
};
choice menu()
{
choice result = QUIT;
// display menu
// input choice
// (the more I think about it, the more I'd split it in two separate functions)
return result;
}
int main()
{
while (true) {
switch (menu())
{
case ADD_NEW_VALUE: AddNewValue(); break;
// ...
case QUIT: return 0;
}
}
}
This helps menu()'s testing and make it easier to add new options and new features.

Answer 4

If the user selects a correct option, call the menu function again. Otherwise, return control to main. I think this may work.
int menu(){int x;
cout<<"Enter an option: \n";
cout<<"1- Add new value\n2- Search Value\n3- Modify value\n4- Print Value\n5- Print sum of all values\n6- Quit/terminate\n";
cin>>x;
if(x!=1||x!=2||x!=3||x!=4||x!=5) return 0;
if(x==1){
AddNewValue();
}
else if (x==2){
SearchValue();
}
else if(x==3){
ModifyValue();
}
else if(x==4){
PrintValue();
}
else if(x==5){
PrintSum();
}
menu();
}
You could also wrap your code in a do-while loop.
void menu(){
do{
int x;
cout<<"Enter an option: \n";
cout<<"1- Add new value\n2- Search Value\n3- Modify value\n4- Print Value\n5- Print sum of all values\n6- Quit/terminate\n";
cin>>x;
if(x==1){
AddNewValue();
}
else if (x==2){
SearchValue();
}
else if(x==3){
ModifyValue();
}
else if(x==4){
PrintValue();
}
else if(x==5){
PrintSum();
}
}while(x==1||x==2||x==3||x==4||x==5);
}