Answer:
[tex]a = 1[/tex]
[tex]b = 1[/tex]
[tex]c = 0[/tex]
Explanation:
Given
The above code segment
Required
The final values of [tex]a,\ b\ \&\ c[/tex]
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:
[tex]a = 1; b =0; c = -1[/tex]
Next, the following if condition is tested
if ((b + 1) == a)
[tex]b + 1 =0 +1[/tex]
[tex]b + 1 = 1[/tex]
And:
[tex]a = 1[/tex]
So,
[tex]b + 1 = a = 1[/tex]
Since the condition is true, the statements in the curly brace { } will be executed;
[tex]b++ \to b = b+1 \to b = 0 + 1 \to b = 1[/tex]
So:
[tex]b = 1[/tex]
[tex]c += b \to c = c+b \to c = -1 + 1 \to c = 0[/tex]
So:
[tex]c = 0[/tex]
Next, the following if condition is tested
c == a
c = 0 and a =1
So:
[tex]c \ne a[/tex]
That means, the statements in its curly brace will not be executed.
So, the final values of a, b and c are:
[tex]a = 1[/tex]
[tex]b = 1[/tex]
[tex]c = 0[/tex]
