C Programming - Control Instructions - Discussion

Hey you are right I do not understand how the value is assigned to I. After calculating expression it gives 1 that is fine but there is no assignment is taking place. Please explain@shilpa.

Thanks shilpa!

If you compile the code in gcc, the output will be different.
After 32767, it is printing from 4294934528 to 4294967295.

This option is wrong, actually there is no right option. The code will print first "1" to "32767" after it will increase as usual and print "-32768" to "-1" and now if increase in i++ the value will be "0" so the condition becomes false. Exit the loop.

I've tested this piece of software on Linux with GCC and I agree with Prakhar.

Sai, you can see the output Prakhar is suggesting if you change the output format option from %u to %d. However you're also right about those long numbers representation. I think it happens because the compiler uses a 4-bit equivalent representation for the negative values when you use the %u option and I overflows, but the number of iterations are exactly the same, and when it reaches 0, as Prakhar said, the loop finishes.

Does this answer mean the i increases from 1 to 32767(0x7fff) and increases again to become -32768(0x1000), and again it becomes -32767(0x1001). By this incremental rule, the last two value is -1(0xffff) and 0(overflow?)

And because it printf as an unsigned int, it shows 1..65535?

I'm confused, can anyone please help me?

At first understand the basic for loop structure how is it works.

for(i=0;i<=5;i++)
{
printf("%d",i); //statement//
}
for(initialization;check condition; updating)
{
statement;
}


It works as follows --1->at first I is assign as 0 // only once //.

2->check the condition.
3-> statement is executed.
4-> increment the value //In updating section//.
5-> check the condition.
6-> the again statement. //Do not go to initialization block//.

Now concentrate about given prob--.

for(i<=5 && i>=-1; ++i; i>0)
printf ("%u, ", i);

So here is works as follows:

1-> when i=0 0<=5 && 0>=-1 //True means 1//.
2-> ++i mean = 2; //Means non zero means true//.
3-> statement;.
4-> 2>0 //True//.
5-> ++i means 3.
6-> statement.

In this procedure it will prints a infinite loops.

%u returns the address of specified variable here that variable is i.

Can anyone explain me why output starts from 1, not from 2 as initialization expression evaluates to 1 and after it, ++i increment i to 2?

Here is a syntax error, as the place of loop condition and increment have been interchanged.