Question

     i

What values would be stored in the given variables in each case?

a.          int n = 12 % 5;

b.          double x = 15 % 11 + 5.3 - 5 / (2.5 - 0.3);

c.   float y = 2 / (3.5 + static_cast<int>(3.5));

d.   bool z = (6 – 7 <= 2 * 1) && (5 + 4 >= 3) || (6 + 2 != 17 – 3 * 10);

Answer 1

a.          

int n = 12 % 5;

Here n contain the reminder of 12 / 5 operation.

Since reminder is 2, n=2

Therefore the value of n is 2.

b.

double x = 15 % 11 + 5.3 - 5 / (2.5 - 0.3);

It is evaluated based on the priority order.

() has highest priority so it is evaluated first.

x = 15 % 11 + 5.3 - 5 / 2.2;

% and / has next priority so evaluate it.

x = 4 + 5.3 - 2.27273;

x = 7.02727

Therefore the value of x is 7.02727

c.

float y = 2 / (3.5 + static_cast<int>(3.5));

static_cast<int>(3.5) // It change the value to integer so the value of 3.5 is changed to 3

float y = 2 / (3.5 + static_cast<int>(3.5));

float y = 2 / (3.5 + 3);

float y = 2 / (6.5);

float y = 0.307692

Therefore the value of y is 0.307692

d.

bool z = (6 - 7 <= 2 * 1) && (5 + 4 >= 3) || (6 + 2 != 17 - 3 * 10);

z = ( - 1 <= 2) && (9>= 3) || (8 != 17 - 30); // evaluate 3*10 first because multiplication has high priority

z = ( - 1 <= 2 * 1) && (9>= 3) || (8 != 13);​​

z = ( true) && (true) || (true);​​

z = true

Therefore the value of z is true.