Why is the loop not stopping? - c

Question

#include <stdio.h>
#include <stdlib.h>
#define NEWGEAR 15.0
#define USEDGEAR 5.0
int main() {
int type;
int num_items;
float total;
printf("Welcome to the market\n");
printf("What would you like to do\n");
printf("\t1-Buy New Gear\n");
printf("\t2-Buy used gear\n");
printf("\t3-Quit\n");
scanf("%d", &type);
while (type != 3) {
switch (type) {
case 1:
printf("How many new items would you like?\n");
scanf("%d", &num_items);
total = num_items * 15.0;
break;
This is where the code keeps on asking how many new items would you like?
case 2:
printf("How many old items would you like?\n");
scanf("%d", &num_items);
total = num_items * USEDGEAR;
break;
This is where the code keeps on asking how many old items would you like?
case 3:
break;
default:
printf("Not a valid option\n");
break;
}
}
printf("Your total cost is %f\n",total);
return 0;
}
Both my choices are looping

Answer 1

You never update the type variable to 3, so the while loop never terminates. Although you do have a break statement, it is affecting the switch and not the while loop that surrounds it.

Answer 2

I think this will handle what you want to achieve.
#include <stdio.h>
#include <stdlib.h>
#define NEWGEAR 15.0
#define USEDGEAR 5.0
int main() {
int type = 0;
int num_items;
float total;
printf("Welcome to the market\n");
printf("What would you like to do\n");
printf("\t1-Buy New Gear\n");
printf("\t2-Buy used gear\n");
printf("\t3-Quit\n");
while (type != 3) {
printf("Enter an option: ");
scanf("%d", &type);
if(type == 3)
break;
switch (type) {
case 1:
printf("How many new items would you like?\n");
scanf("%d", &num_items);
total = num_items * 15.0;
break;
case 2:
printf("How many old items would you like?\n");
scanf("%d", &num_items);
total = num_items * USEDGEAR;
break;
default:
printf("Not a valid option\n");
break;
}
}
printf("Your total cost is %f\n",total);
return 0;
}
Basically, after your user finished doing cases 1-2 or default, your program would prompt your user for an option again if he wants to quit or perform another case 1 or 2 operation.

Answer 3

Your loop logic is flawed:
you should move the prompt code inside the loop.
you should update total for each answer.
you should test if scanf() is successful at converting user input.
Here is a modified version:
#include <stdio.h>
#define NEWGEAR 15.0
#define USEDGEAR 5.0
int main() {
int type;
int num_items;
double total = 0;
printf("Welcome to the market\n");
for (;;) {
printf("What would you like to do\n");
printf("\t1-Buy New Gear\n");
printf("\t2-Buy used gear\n");
printf("\t3-Quit\n");
if (scanf("%d", &type) != 1 || type == 3)
break;
switch (type) {
case 1:
printf("How many new items would you like?\n");
if (scanf("%d", &num_items) == 1)
total += num_items * 15.0;
break;
case 2:
printf("How many old items would you like?\n");
if (scanf("%d", &num_items) == 1)
total += num_items * USEDGEAR;
break;
default:
printf("Not a valid option\n");
break;
}
}
printf("Your total cost is %f\n", total);
return 0;
}