# [C] Play Again Loop Logic

Discussion in 'C' started by anubizs, Jun 11, 2015.

1. ### anubizsNew Member

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I made a basic calculator using this codes. What i want to learn is how to make a loop after the program ask "try again y/n"? it will return to choose an operation.

Code:
```#include<stdio.h>
#include<conio.h>

int main(){
int num1, num2, choice;

printf("Choose an operation\n\n");
printf("[1] Add\n[2] Subtract\n[3] Multiply\n[4] Divide\n[5] Exit\n");
scanf("%d", &choice);

switch(choice){

case 1:
printf("Enter 1st number:\n");
scanf("%d", &num1);
printf("Enter 2nd number:\n");
scanf("%d", &num2);
printf("\n%d", (num1+num2));
break;
case 2:
printf("Enter 1st number:\n");
scanf("%d", &num1);
printf("Enter 2nd number:\n");
scanf("%d", &num2);
printf("\n%d ", (num1-num2));
break;
case 3:
printf("Enter 1st number:\n");
scanf("%d", &num1);
printf("Enter 2nd number:\n");
scanf("%d", &num2);
printf("\n%d", (num1*num2));
break;
case 4:
printf("Enter 1st number:\n");
scanf("\n%d", &num1);
printf("Enter 2nd number:\n");
scanf("%d", &num2);
printf("\n%d", (num1/num2));
break;
case 5:
return 0;

default:
printf("That is not a valid choice.");
break;
}
getch();
}

```

2. ### xpi0t0sMentor

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I would just enclose the lot in a for loop:
Code:
```for (int quit=0; !quit; )
{
printf("Choose an operation\n\n");
//...
case 5: quit=1; break;
//...
}
```

shabbir likes this.