Looping Procedural Programming
Selamat sore sobat blogger kali ini saya akan share program procedural programming c++ ynag telah saya buat, walaupun programnya sangat sederhana tetapi semoga programnya bisa bermanfaat bagi sobat semua. langsung aja lihat programnya dibawah ini.
#include <iostream>
#include <conio.h>
#include <stdlib.h>
using namespace std;
int main()
{
a:
int pilih;
char kembali;
cout<<" 1. Looping "<<endl;
cout<<" 2. Array "<<endl;
cout<<" 3. Keluar "<<endl;
cout<<"MasuKkan Pilihan Anda (1/2/3) :";
cin>> pilih;
switch(pilih)
{
case 1:
char kembali;
int a;
for(int i=1,a=5;i<=5;i++)
{ cout<<"\nNilai ke "<<i<<" = "<<a;
a+=5; }
cout<<endl;
cout<<"ingin menghitung lagi? [Y/N] "<<endl;
cin>>kembali;
if (kembali == 'y')
{
cout<<"selesai"<<endl;
system("cls");
goto a;
}
else (kembali == 'n');
{
cout<<"selesai";
exit(0);
}
break;
case 2:
int x[6],limit;
char nama[6][20];
char nik[6][20];
cout<<"masukkan limit = ";
cin>>limit;
for(int a=0;a<=limit-1;a++)
{
cout<<"nama"<<a+1<<"=";
cin>>nama[a];
cout<<"nik"<<a+1<<"=";
cin>>nik[a];
}
cout<<endl;
cout<<"NO"<<"\t"<<"NAMA"<<"\t"<<"NIK"<<endl;
for(int a=0;a<=limit-1;a++)
{
cout<<a+1<<"\t"<<nama[a]<<"\t"<<nik[a];
cout<<endl;
}
cout<<"ingin menghitung lagi? [Y/N] "<<endl;
cin>>kembali;
if (kembali == 'y')
{
cout<<"selesai"<<endl;
system("cls");
goto a;
}
else (kembali == 'n');
{
cout<<"selesai";
exit(0);
}
break;
case 3:
cout<<"ingin menghitung lagi? [Y/N] "<<endl;
cin>>kembali;
if (kembali == 'y')
{
cout<<"selesai"<<endl;
system("cls");
goto a;
}
else (kembali == 'n');
{
cout<<"selesai";
exit(0);
}
}
} 
