Question

Consider the following nested if-statements:

int x = N;
int y = 1;
if (++x == 1) y = 2;
else if (++x == 3) y = 3;
else if (++x == 5) y = 4;
else if (++x == 7) y = 5;
N is an integer to which x is initialized. Depending on the value of N, y will have some value at the end. Consider the cases where N is 0, 1,..., 10. Then, identify, in the list below, the possible combination of values for N and y.


a) N = 2; y = 1.
b) N = 5; y = 4.
c) N = 2; y = 4.
d) N = 4; y = 5.

Answer 1

no alternative

Answer 2

u mean no right answer?

Answer 3

yes, no right answer

Answer 4

Thanks a lot. My professor said every problem has a correct answer, damn I am depressed.

Answer 5

sorry, I was wrong, the correct answer is the letter c) N = 2; y = 4.

Answer 6

Could you explain to me why it is c? I don't understand. Coz I think when N=2 then x+1 = 3 so y = 3.But c) is y=4. How can I understand this?

Answer 7

é meio complicado, numa sequencia de N inicializando em 0 ate 10, trying to show the code:
int x = N; N=2
int y = 1; ----
if (++x == 1) y = 2; + + X, increment N +1, then N=3
else if (++x == 3) y = 3; + + X, increment N +1, then N=4
else if (++x == 5) y = 4; + + X, increment N +1, then N=5,,enters the condition,
else if (++x == 7) y = 5;

ok?

Answer 8

good question!!

Answer 9

Yes c is true.