Question

1. Show the output produced by each of the following program fragments. Assume that I, j, k are int variables
   1. i = 7; j = 2;

printf(“%d %d ”, i / j, i % j);

2. i = 4; j = 3;

printf(“%d”, (i + 10 ) % j);

3. i = 7; j = 8; k = 9;

printf("%d”, (i + 10) % k / j);

2. Show the output produced by each of the following program fragments. Assume that I, j, k are int variables
   1. i = 7; j = 8;

i *= j + 3;

printf(“%d %d”, i, j);

2. i = j = k = 1;

i *= j *= k;

printf(“%d %d %d”, i, j, k);

3. Show the output produced by each of the following program fragments. Assume the I, j, k are int variables.
   1. i = 4;

j = ++i * 3 – 2;

printf(“%d %d” i, j)

2. i = 3;

j = 3 – 2 * i++;

printf(“%d %d”, i, j);

3. i = 2; j = 5;

printf(“%d ”, i++ - ++j);

printf(“%d %d”, i, j);

Answer 1

For the first part:

a. 3 1

(7/2=3(gives the rounded value to the lowest whole number, so 3.5 becomes 3)).

(7%2 dividing 7 with 2 will give you a remainder of 1. So % checks the remainder also known as modulo).

b. 2

(14 divided by 3 will give us a remainder of 2) Therefore 2. to check 3*4=12, So 14-12=2.(Just another way).

c. 1

(17 divided by 9 will give a remainder of 8. So 8 divided by 8 will give a quotient of 1. Therefore answer is 1.)

For the second part:

a. 77 8

This expression is also equal to i=i*(j+3). Therefore 7*(8+3)=7*11=77; and j gets printed the same so 8.

b. 1 1 1

This expression will start from the end. so k remains the same=1 j=j*k=1 and i=i*j=1;

Suppose, instead of 1 we took 4, answer would be k=4 j=j*k, so j=4*4=16 and i=i*j, so i=4*16=64.

For the Third part:

a. 5 13

Here, i is a pre-incrementation operation. So, first it will increase the value, then it will operate. So i becomes 4+1=5, and j=(5*3)-2=13.

b. 4 -3

Here, i is a post-incrementation operation. So, it will increase its value after the operation is over. So, the expression for j=3-2*3=-3 and then i will increase its value, therefore i=4;

c. -4

i is post incrementation and j is pre-incrementation. So, j will increase its value while the arithmetic operation is taking place, whereas i will increase after the arithmentic operation is over. So, 2-6=-4

3 6

Now, since the operation is over i will increase its value by 1, j has already done it during the expression so it remains the same

i=(2+1) and j=6.