If you have a for loop like this:
for(j = 0; j<=90; j++){}
It works fine. But when you have a for loop like this:
for(j
Since nobody else has actually tackled Could someone please explain this to me?
I believe I will:
j++
is shorthand, it's not an actual operation (ok it really IS, but bear with me for the explanation)
j++
is really equal to the operation j = j + 1;
except it's not a macro or something that does inline replacement. There are a lot of discussions on here about the operations of i+++++i
and what that means (because it could be intepreted as i++ + ++i
OR (i++)++ + i
Which brings us to: i++
versus ++i
. They are called the post-increment
and pre-increment
operators. Can you guess why they are so named? The important part is how they're used in assignments. For instance, you could do: j=i++;
or j=++i;
We shall now do an example experiment:
// declare them all with the same value, for clarity and debug flow purposes ;)
int i = 0;
int j = 0;
int k = 0;
// yes we could have already set the value to 5 before, but I chose not to.
i = 5;
j = i++;
k = ++i;
print(i, j, k);
//pretend this command prints them out nicely
//to the console screen or something, it's an example
What are the values of i, j, and k?
I'll give you the answers and let you work it out ;)
i = 7, j = 5, k = 7;
That's the power of the pre and post increment operators, and the hazards of using them wrong. But here's the alternate way of writing that same order of operations:
// declare them all with the same value, for clarity and debug flow purposes ;)
int i = 0;
int j = 0;
int k = 0;
// yes we could have already set the value to 5 before, but I chose not to.
i = 5;
j = i;
i = i + 1; //post-increment
i = i + 1; //pre-increment
k = i;
print(i, j, k);
//pretend this command prints them out nicely
//to the console screen or something, it's an example
Ok, now that I've shown you how the ++
operator works, let's examine why it doesn't work for j+3
... Remember how I called it a "shorthand" earlier? That's just it, see the second example, because that's effectively what the compiler does before using the command (it's more complicated than that, but that's not for first explanations). So you'll see that the "expanded shorthand" has i =
AND i + 1
which is all that your request has.
This goes back to math. A function is defined where f(x) = mx + b
or an equation y = mx + b
so what do we call mx + b
... it's certainly not a function or equation. At most it is an expression. Which is all j+3
is, an expression. An expression without assignment does us no good, but it does take up CPU time (assuming the compiler doesn't optimize it out).
I hope that clarifies things for you and gives you some room to ask new questions. Cheers!