Exit case in switch in C - c

Question

I am trying this code with exit case, the exit case doesn't work it still asks me for the two numbers and then it exits. I have provided the output. What is the alternative to the switch case that I can use for menu driven programs?
Sample input/output:
::::::Menu:::::
1. Addition:
2. Subtraction
3. Multiplication
4. Division
5. Mod
6. Exit
Enter your choice:6
Enter the values of a and b:3 4
#include<stdio.h>
#include<stdlib.h>
int main(){
int a,b,c;
int no;
do{
printf("\n::::::Menu:::::\n");
printf(" 1. Addition:\n 2. Subtraction \n 3. Multiplication \n 4. Division \n 5. Mod \n 6. Exit");
printf("\nEnter your choice:");
scanf("%d",&no);
printf("\nEnter the values of a and b:");
scanf("%d%d",&a,&b);
switch(no){
case 1:
c = a + b;
printf("\nAddition is:%d",c);
break;
case 2:
c = a - b;
printf("\nSubtraction is:%d",c);
break;
case 3:
c = a * b;
printf("\nMultiplication is:%d",c);
break;
case 4:
c = a / b;
printf("\nDivision is:%f",c);
break;
case 5:
c = a % b;
printf("\nMod is:%d",c);
break;
case 6:
exit(1);
default:
printf("\nInvalid choice\n");
break;
}
}while(no!=6);
return 0;
}

Answer 1

Your program asks for two numbers because you are checking for the exit code after the second scanf statement. If you wish the program to exit when 6 is entered, then you have to add an if statement in between the first and second scanf. You should additionally removed the exit case from your switch statement. Below is an example:
printf("\nEnter your choice:");
scanf("%d",&no);
if (no == 6)
exit(0);
printf("\nEnter the values of a and b:");
scanf("%d%d",&a,&b);

Answer 2

You're reading the user's menu option, then asking for the next two numbers before you get to the switch statement, so of course it will always ask for those two numbers.
You need to either check specifically for 6 right after reading the menu input, or you need to move the second prompts to only where they're needed, i.e. inside each case.

Answer 3

It keeps asking for the two numbers because the switch statement is positioned right AFTER you ask for the user input. I restructured your code, this should work:
#include<stdio.h>
#include<stdlib.h>
int main(){
int a,b,c;
int no;
while(1){
printf("\n::::::Menu:::::\n");
printf(" 1. Addition:\n 2. Subtraction \n 3. Multiplication \n 4. Division \n 5. Mod \n 6. Exit");
printf("\nEnter your choice:");
scanf("%d",&no);
if (no == 6)
break;
printf("\nEnter the values of a and b:");
scanf("%d%d",&a,&b);
switch(no){
case 1:
c = a + b;
printf("\nAddition is:%d",c);
break;
case 2:
c = a - b;
printf("\nSubtraction is:%d",c);
break;
case 3:
c = a * b;
printf("\nMultiplication is:%d",c);
break;
case 4:
c = a / b;
printf("\nDivision is:%f",c);
break;
case 5:
c = a % b;
printf("\nMod is:%d",c);
break;
default:
printf("\nInvalid choice\n");
break;
}
}
return 0;
}
Note how I changed the do-while to a while-true and explicitly check for no == 6 PRIOR to asking for user input.

Answer 4

The statements in C are meant to be executed sequentially unless you disrupt the normal flow of action using jumps. In your case
printf("\nEnter your choice:");
scanf("%d",&no);
printf("\nEnter the values of a and b:");
scanf("%d%d",&a,&b);
// Only now does the switch start.
You ask for the choice first and then for the two numbers. That is why you always end up in entering the two values. One way is to take the exit out of the switch-case which perhaps is the simplest solution I guess. Something like
printf("\nEnter your choice:");
scanf("%d",&no);
if ( 6 == no )
exit(1);
What is the alternative to the switch case that I can use for menu
driven programs
Well, switch-case is exactly made for that. Why think of an alternative?

Answer 5

#include<stdlib.h>
#include<conio.h>
#include<stdio.h>
int main()
{
char ch;
while(1)
{
printf("l.print l c.print c q. exit \n");
printf("enter choice ");
scanf("%c",&ch);
switch(ch)
{
case 'l':
printf("You have typed l \n");
break;
case 'c':
printf("yoh have typed c \n");
break;
case 'q':
exit(0);
}
}
return 0;
}