Control Instructions

When 1 is entered, The output of the code below is?

#include 
    void main()
    {
        int ch;
        printf("enter a value btw 1 to 2:");
        scanf("%d", &ch);
        switch (ch)
        {
        case 1:
            printf("1\n");
        default:
            printf("2\n");
        }
    }

Predict the output of the below program:

#include 
#define EVEN 0
#define ODD 1
int main()
{
    int i = 3;
    switch (i & 1)
    {
        case EVEN: printf("Even");
                break;
        case ODD: printf("Odd");
                break;
        default: printf("Default");
    }
    return 0;
}

What will be the output of the program, if a short int is 2 bytes wide?

#include
int main()
{
    short int i = 0;
    for(i<=5 && i>=-1; ++i; i>0)
        printf("%u,", i);
    return 0;
}

Point out the error, if any in the while loop.

#include
int main()
{
    void fun();
    int i = 1;
    while(i <= 5)
    {
        printf("%d\n", i);
        if(i>2)
            goto here;
    }
return 0;
}
void fun()
{
    here:
    printf("It works");
}

The output of the code below is

#include 
    void main()
    {
        int x = 0;
        if (x == 0)
            printf("hi");
        else
            printf("how are u");
            printf("hello");
    }

The modulus operator cannot be used with a long double.

The way the break is used to take control out of switch and continue to take control of the beginning of the switch?

What is the output?

#include
int main()
{
   int n;
   for (n = 9; n!=0; n--)
     printf("n = %d", n--);
   return 0;
}

Point out the error, if any in the program.

#include
int main()
{
    int P = 10;
    switch(P)
    {
       case 10:
       printf("Case 1");

       case 20:
       printf("Case 2");
       break;

       case P:
       printf("Case 2");
       break;
    }
    return 0;
}

In _______, the bodies of the two loops are merged together to form a single loop provided that they do not make any references to each other.