Thread: Another Simple Array Question

Originally Posted by guyfromfl

++i means you start i as i + what i was, i++ means you start i as what it was then start adding...correct?

++i would mean you start the loop as 1 if i=0?

Not in the slightest. You have to remember what for means:

Code:

for(statement1; statement2; statement3)
{
   statement4;
   statement5;
}

means

Code:

statement1;
while (statement2) {
    statement4;
    statement5;
    statement3;
}

Pre-incrementing is powerful, but it can't get past two whole other statements. ETA: I mean, the pre-incrementing in statement3 still can't happen all the way before statements 4 and 5.

In one of my texts it says the basic code for looping through an array is

Code:

int i;
const int NUM = 10;
int my_array[NUM]:
for (i = 0; i < NUM; i++) {
    do something with my_array[i];
}

which I thought could be shortened to

Code:

int i, my_array[10]:
for (i = 0; i < 10; i++) {
    do something with my_array[i];
}

which is very similar to the code I posted but I can see that even using the "basic code" it doesn't work.

Originally Posted by deadhippo

In one of my texts it says the basic code for looping through an array is

Code:

int i;
const int NUM = 10;
int my_array[NUM]:
for (i = 0; i < NUM; i++) {
    do something with my_array[i];
}

which I thought could be shortened to

Code:

int i, my_array[10]:
for (i = 0; i < 10; i++) {
    do something with my_array[i];
}

which is very similar to the code I posted but I can see that even using the "basic code" it doesn't work.

You don't see very well. That code works perfectly, assuming you do something specific with "do something". The problem with your printing is that you don't have anything in your array to print. Arrays are not initialized for you.

::ho hum:: i asked where the assigns were earlier. (the 'giving' part)
but back to a good point...

and yes your while explanation is quite eye opening.

im gonna screw with that for a while now. thanks for the heads up

Originally Posted by tabstop

You don't see very well. Arrays are not initialized for you.

Thank you very much. You are right. I didn't realize that the printf wasn't actually doing anything but printing the result of something.

Here is what I came up with from that piece of advice:

Code:

    int a[5], i;

    for (i = 0; i < 5; ++i){
        a[i] = i + 1;
        printf("%d\t", a[i]);
    }

I initialized it with i + 1 to give an array of 1 to 5. Added the tab to make it purty.
Thanks guys. You got me over another hurdle.

so wouldnt that mean he just assigned i[1] the first var?

please tell me how i am wrong with this.

and as a monday night quarterback my third recommendation was to split the outputs up ("\t or \n")

Originally Posted by guyfromfl

so wouldnt that mean he just assigned i[1] the first var?

That doesn't even mean anything.

a[0] = 1
a[1] = 2
a[2] = 3
a[3] = 4
a[4] = 5
That's what happens. Remember that assignments take the value on the right and put them in the variable on the left.

yes i am wrong. and i see whats going on now, i was at the bar too long i guess...i am right about refering arrays but missed the WHOLE part about how the loop made its own numbers as it went.

im going to bed to sleep the poison out.

sorry...
mark

Someone already posted an easier example of this, but when you declare an array you are simply reserving memory. Anything can be in that memory. You need to initialize each of the elements in that array.

tzuch

Code:

int array[5], i;

for(i=0;i<5;i++)
{
array[i]=0; //or any other input, usually from the user
}

Hope this helps.

Originally Posted by tzuch

Hope this helps.

Thanks for the input.