Question

Int a = 1; int b = 0; int c = -1; if ((b + 1) == a) { b++; c += b; } if (c == a) { a--; b = 4; }

Answer 1

Answer:

$a = 1$

$b = 1$

$c = 0$

Explanation:

Given

The above code segment

Required

The final values of $a,\ b\ \&\ c$

The following line declares and initializes the values of a, b and c

int a = 1;     int b = 0;     int c = -1;

So, we have:

$a = 1; b =0; c = -1$

Next, the following if condition is tested

if ((b + 1) == a)

$b + 1 =0 +1$

$b + 1 = 1$

And:

$a = 1$

So,

$b + 1 = a = 1$

Since the condition is true, the statements in the curly brace { } will be executed;

$b++ \to b = b+1 \to b = 0 + 1 \to b = 1$

So:

$b = 1$

$c += b \to c = c+b \to c = -1 + 1 \to c = 0$

So:

$c = 0$

Next, the following if condition is tested

c == a

c = 0 and a =1

So:

$c \ne a$

That means, the statements in its curly brace will not be executed.

So, the final values of a, b and c are:

$a = 1$

$b = 1$

$c = 0$