35. Find the ouput.

#include<stdio.h>
int i;
int fun();

int main()
{
    while(i)
    {
        fun();
        main();
    }
    printf("Hello\n");
    return 0;
}
int fun()
{
    printf("Hi");
}

Answer: Hello
Explanation:
Step 1: int i; The variable i is declared as an integer type.
Step 1: int fun(); This prototype tells the compiler that the function fun() does not accept any arguments and it returns an integer value.
Step 1: while(i) The value of i is not initialized so this while condition is failed. So, it does not execute the while block.
Step 1: printf("Hello\n"); It prints "Hello".
Hence the output of the program is "Hello". 

28. Find the output

Program:

main()
{
    int a[5] = {5, 1, 15, 20, 25};
    int i, j, m;
    i = ++a[1];
    j = a[1]++;
    m = a[i++];
    printf("%d, %d, %d", i, j, m);
}

Ans : 3,2,15

Explanation :

Step 1: int a[5] = {5, 1, 15, 20, 25}; The variable arr is declared as an integer array with a size of 5 and it is initialized to

a[0] = 5, a[1] = 1, a[2] = 15, a[3] = 20, a[4] = 25 .

Step 2: int i, j, m; The variable i,j,m are declared as an integer type.

Step 3: i = ++a[1]; becomes i = ++1; Hence i = 2 and a[1] = 2

Step 4: j = a[1]++; becomes j = 2++; Hence j = 2 and a[1] = 3.

Step 5: m = a[i++]; becomes m = a[2]; Hence m = 15 and i is incremented by 1(i++ means 2++ so i=3)

Step 6: printf("%d, %d, %d", i, j, m); It prints the value of the variables i, j, m

Hence the output of the program is 3, 2, 15

24. Find the output.

void main()
{
int i=3, *j, k;
j = &i;
printf("%d\n", i**j*i+*j);
}

Answer: 
30

16. Find the output.

int main()
{
int ***r, **q, *p, i=8;
p = &i;
q = &p;
r = &q;
printf("%d, %d, %d\n", *p, **q, ***r);
return 0;
}

Answer:
8,8,8

6. Predict the output

main()
{
static int var = 5;
printf("%d ",var--);
if(var)
main();
}

Answer:
5 4 3 2 1
Explanation:
When static storage class is given, it is initialized once. The change in the value of a static variable is retained even between the function calls. Main is also treated like any other ordinary function, which can be called recursively.

Find the output.

void main()
{
static int i=5;
if(--i)
{
main();
printf("%d ",i);
}
}

Output: 
0 0 0 0


Explanation:
The variable 'i' is declared as static, hence memory for 'i' will be allocated
for only once, as it encounters the statement. The function main() will be called
recursively unless 'i' becomes equal to 0, and since main() is recursively called, so
the value of static 'i' ie., 0 will be printed every time the control is returned.

Find The Output

main()
{
i=5;
printf("%d",i++ + ++i);
printf("\n%d",i++);
printf("\n%d",++i);
}

Answer:
12
7
9

Find the output.

main()
{
int i=0;
while(+(+i--)!=0)
i-=i++;
printf("%d",i);
}

Answer:
-1

Explanation:

Unary + is the only dummy operator in C. So it has no effect on the expression and now the while loop is, while(i--!=0) which is false and so breaks out of while loop. The value –1 is printed due to the post-decrement operator.

Find the output.

Try this..
main()
{
printf("\nab");
printf("\bsi");
printf("\rha");
}
What is the output of above program??

Answer:
hai

Explanation:
The first printf statement prints 'ab'
The second printf statement prints 'asi'(bcoz \b backspace cmd prints next to 'a' and overwrites 'b')
The third printf statement prints 'hai' (bcoz \r return cmd overwrites on 'asi')

Find the output.

main()
{
int i=-1;
+i;
printf("i = %d, +i = %d \n",i,+i);
}
what's the output??

Answer: 
i = -1, +i = -1

Explanation: 
   Unary + is the only dummy operator in C. Where-ever it comes you can
just ignore it just because it has no effect in the expressions (hence the
name dummy operator).