Thread: Idiot guide

Hi,
I am just beginning to learn C and I have come across a program that I just can't understand the result of. I was hoping someone would be so kind as to explain it to me and I am sure it is pretty simple. I understand the concept of the the for statement but this program has stumped me.
*edited

Code:

int i = 9;
for (i--; i--; i--)
    printf("%d", i);

The output is : 7 5 3 1

I am confused from the beginning here. i is initially 9 and in the for statement it is initialized as 1++ which I thought would mean it is 8.

Thanks.

That is not the output I get. Are you sure it isn't this:

Code:

int i = 9;
for (i--; i--; i--)
    printf("%d", i);

Code:

int i = 9;
for (i++; i++; i++)
    printf("%d", i);

It is an infinite loop.

I think you are trying to make this:

Code:

#include <stdio.h>

int main()
{
	int i = 9;
    for (i++; i>0; i--)
        printf("\n%d", i);
	
	return 0;
}

It is an infinite loop.

Not so infinite... Just till i makes the whole loop and gets back to zero

Or maybe

Code:

int i = 9;

i--;
while(i)
{
	i--;
	printf("%d ", i);
	i--;
}

Originally Posted by Elysia

Or maybe

Code:

int i = 9;

i--;
while(i)
{
	i--;
	printf("%d ", i);
	i--;
}

That's not quite right - i will be zero at the end of that loop, whilst i will be -1 at the end of the loops posted above. Since the i-- is part of the condition of the while-loop, it's going to be executed even when i is zero, whilst your i-- being inside the loop is not executed when i is not true (i == 0).

--
Mats

I'm not sure how the assembly does it, though.
Basically,

Code:

mov esi, i;
dec esi;
jne /*jump*/;

Does jne check esi or some such?

Originally Posted by Elysia

I'm not sure how the assembly does it, though.
Basically,

Code:

mov esi, i;
dec esi;
jne /*jump*/;

Does jne check esi or some such?

It sets the CPU flags, specifically the zero flag, if esi went to 0. The CPU flags are modified with every Arithmetic - Logic operation, and some other operations also.
test eax,edx;
jz XXX;
for example will test for 0, without altering eax, while the alternative would be :
sub eax,edx;
jnz EEE;
or whatever...
Get a copy of ollydbg and run a simple executable through it, noting the operations and their effect on CPU flags will be much more enlightening to you probably given your experience.

Originally Posted by deadhippo

In the for loop the i is immediately decremented to 8, then the condition i-- is checked and as i-- is not 0 that means that the statement must be true so the concluding expression is executed decrementing it further to 7, which is printed. This continues down to 5, 3 and 1 and the one is checked again and is decremented to 0 but the condition is checked and it is also 0 and therefor false so the concluding expression is not checked and the loop ends.

Yes, that's right. The condition is checked before the loop body is entered.

Originally Posted by xuftugulus

It sets the CPU flags, specifically the zero flag, if esi went to 0. The CPU flags are modified with every Arithmetic - Logic operation, and some other operations also.
test eax,edx;
jz XXX;
for example will test for 0, without altering eax, while the alternative would be :
sub eax,edx;
jnz EEE;
or whatever...
Get a copy of ollydbg and run a simple executable through it, noting the operations and their effect on CPU flags will be much more enlightening to you probably given your experience.

You're right...
I would normally think you would have to do:

Code:

dec esi;
cmp esi, 0;
je /* test */;

Assembly is so complicated whenever all operations modify the flags. It's part of what I don't understand.
Maybe I'll do that someday.

your code is equivalent to

Code:

int i = 5;
++i;
while(i)
{
   printf("%d ", i);
   i -= 3;
   ++i;
}

do you see now the reason for the output you receive?

Stop writing such retarded loops is my advice.
Generally, I'd avoid using ++ or -- as loop condition. Do that in the next field that allows you to execute code after each loop. This applies to for loops.