Switch Argument Outside While Loop Execution Explained

[Crystal037]

TL;DR Summary

switch in c programming inside a while but argument of switch is outside the while

#include<stdio.h>
#define MAX_SIZE 6
int pop(int a[]);
void push(int a[],int e);
int pal(int a[]);
int *top;
int stack[MAX_SIZE];
int main(){
    int choice,done=0,ele;
    printf("1.Push\t2.Pop\t3.Pal\t4.Exit\n");
    printf("Enter your choice:\t");
    scanf("%d",&choice);
    while(!done){
        switch(choice){
            case 1:printf("\nEnter the element to be pushed\n");
            scanf("%d",&ele);
            push(stack,ele);
           
            case 2: if(pop(stack)==-1)
            printf("\nStack is empty\n");
            else{
             printf("\nDeleted Element=%d",pop(stack));
                }
           
            case 3:done=1;
          
            default:printf("\nInvalid Choice\n");
        }
    }
}
void push(int a[],int e){
    if(*top==MAX_SIZE-1){
        printf("\nStack Overflow\n");
    }
    else{
        a[++*top]=e;
    }
}
int pop(int a[]){
    if(*top==-1){
        printf("\nStack Underflow\n");
        return -1;
    }
    else{
        return a[*top--];
    }
}

Here switch argument is taken outside while loop so the argument of switch should remain constant and and that case should be in a loop then why is it executing only 1 time

 

[Arjan82]

because the case statements should end with a break:

switch (choice) {
    case 1:
        // do something
        break;
    case 2:
        // do something else
        break;
    default:
        // do something defaulty
        break; // <-- not sure if this last break is necessary
}

 

[Mark44]

I see several problems with this code, aside from the missing break statements for each case in the switch.

1. The input to select a choice is done outside the while loop, so whatever the first choice is, that's the one that gets performed inside the loop. You need to prompt the user to enter the choice inside the while loop.
2. The top variable is an uninitialized pointer. If the first choice is to push a value, the code attempts to dereference top, which is pointing to some garbage location.

There might be more problems, but these need to be addressed first.

 

[Crystal037]

Ev

because the case statements should end with a break:

switch (choice) {
    case 1:
        // do something
        break;
    case 2:
        // do something else
        break;
    default:
        // do something defaulty
        break; // <-- not sure if this last break is necessary
}

Even if the break isn't the switch will execute all the cases. Doesn't explain why is it executing only one time.

 

[pbuk]

Ev
Even if the break isn't the switch will execute all the cases. Doesn't explain why is it executing only one time.

Yes it does: the switch will execute all the cases including the code after case 3.

 

[Mark44]

Even if the break isn't the switch will execute all the cases. Doesn't explain why is it executing only one time.

Your post is unclear. Without the break statements on each case, all four cases will execute.

Yes it does: the switch will execute all the cases including the code after case 3.

To be clear, the snippet posted by Arjan82 that Crystal037 quoted doesn't have this problem.

@Crystal037, see my suggestions in post #3.

 

[Crystal037]

Yes it does: the switch will execute all the cases including the code after case 3.

But when executing its only executing one case given like if I give option 1 it just executes 1 and doesn't go to 2 why and even if I put break in every statement bcz of while it should execute the choice entered before again and again bcz of the while loop

 

[Crystal037]

Your post is unclear. Without the break statements on each case, all four cases will execute.

To be clear, the snippet posted by Arjan82 that Crystal037 quoted doesn't have this problem.

@Crystal037, see my suggestions in post #3.

Thats what it should do but all the cases are not getting executed only the choice entered that's getting executed.
Also can u please explain the uninitialised pointer and how does it work.

 

[Mark44]

But when executing its only executing one case given like if I give option 1 it just executes 1 and doesn't go to 2 why and even if I put break in every statement bcz of while it should execute the choice entered before again and again bcz of the while loop

Is that code that you're running exactly the same as what is shown in post #1? If the code doesn't have break statements in each case, and you enter 1 as your choice, it will execute all three cases.

Also can u please explain the uninitialised pointer and how does it work.

In line 6 you declare top

int *top;

This means that top is a pointer, but it is uninitialized, so it doesn't contain the address of any known variable.
In lines 32 and 40 (in the bodies of the push() and pop() functions) you have

void push(int a[],int e){
    if(*top==MAX_SIZE-1){                // line 32
.
.
.
int pop(int a[]){
    if(*top==-1){                             // line 40

In both lines you are dereferencing top, but since it was never initialized, you are getting access to garbage values.

 

[Mark44]

@Crystal037, you posted this question on Tuesday and it's now Sunday evening. Are you making any progress on your program?