Need help with mock exam question

Hi,

I have an exam coming up and I need to complete this mock exam. If i post the code, can anyone tell me if my understanding is correct and where to look at next:

What will this code output:

Code:

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int i=1, j=1
for (i=0; i>0; i--)
{
      j += i*(i-1);
};
printf("%d",j+1);

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What I understand is this. First i and j are defined as intergers, 1 value for both. Then there is a for loop, with the starting condition as i=0, theloop continues while i is more than zero and 1 is taken away from i each loop.

The problem I have up to this point is initially i is set as 1, but then in the loop x is set at 0? Also what does += mean?

This is a trick question; the loop won't iterate so the answer is 2.

Code:

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i=0; i>0;  // hmmm

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Hi,

Thank you for the help. Im going to need to know how to answer these questions though so i still need a little help getting the answer.

It seems like i is defined twice with two different values (0 and 1). Is this correct? If so, what does this mean? Which value is followed?

Also does += mean more than or equal to? Much like >=.

x+=y means x = x + y

x += y is the same as x = x + y, except that x is evaluated only once. This matters if x is an operation with side effects, like "++ptr".

I went completely through one of these the other day with transgalactic2 and at the very end, after I manage to demonstrate that the 60-70 of obfuscationism by his prof didn't do anything at all, s/he tells me the real question now is "what is the return value of what2"?

So in the Original Post,

Quote:

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What will this code output:

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This code will output "2". Now:

Quote:

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It seems like i is defined twice with two different values (0 and 1). Is this correct? If so, what does this mean? Which value is followed?

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This code does nothing. This code does not mean anything. You will never, ever, ever, ever, ever see this code in a real program. Ever. So I don't know what you are suppose to do with it, analysis wise, but there is no point in asking "what does this mean".

But that is correct, i is defined twice [no, I take that back, see below]. The first definition is superfluous, but part of the declaration, which is necessary.

Quote:

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Originally Posted by spadez

It seems like i is defined twice with two different values (0 and 1). Is this correct? If so, what does this mean? Which value is followed?

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No, look again. i is define once. The second time it is simply assigned a new value. Redefining i would require "int i=0". This is valid only in C99.

I assent to King Mir on this one. Unless you want to distinguish between redefinition (i=) and redeclaration (int i=).

Hi guys, thank you so much for the help. So that I understand this, i is being declared at the top and assigned the value 1. Below in the code it is then reassigned (but not redeclared) a different value, in this case 0.

So it would make no difference if the code at the top was instead:

Code:

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int i
int j=1

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Therefore not giving i a value, since it will be set a value later on in the loop.

To check my understanding of the next part, if x=1 and y=2, would the code:

Code:

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x += y

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Mean that the new value of X would be 3?
(3[x] = 1[x] + 2[y])

Finally printf("%d",j+1); would just simply add one to the value of j and then print its value?

Quote:

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Originally Posted by spadez

Mean that the new value of X would be 3?

Finally printf("%d",j+1); would just simply add one to the value of j and then print its value?

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1) Yes, "+=" adds the right side to the left side.
2) Yes.