Why is the output 1?

Question

This is a code snippet from a C challenge.
Should this not be an infinite loop?
It runs only once. If i print out the variabel i it have the value 0.
What make the loop to end?

unsigned int i;

int count =0;

for (i=0; i<10; i--)
count++;
printf ("%d", count );

Answer 1

Adam Aksu and A͢J - S͟o͟l͟o͟H͟e͟l͟p͟e͟r͟ An unsigned value in C can never be negative. In two's complement representation subtracting one from zero results in a very large number. See
https://en.m.wikipedia.org/wiki/Two's_complement

Answer 2

A͢J - S͟o͟l͟o͟H͟e͟l͟p͟e͟r͟ Sorry, I missed where you said that.

But I did want to make it clear that the loop ends because i becomes a large number (much greater than 10).

Answer 3

Quantum and Ong'ondi Eugene When printing an unsigned value, use the %u format. Using %d converts it to signed.

Answer 4

Steve

That's what I said "unsigned int holds only 0 and positive numbers'.

Answer 5

Steve, yes that make sense. But still when printing out the value outside the loop it has value -1.
That's what puzzles me.

Answer 6

A͢J - S͟o͟l͟o͟H͟e͟l͟p͟e͟r͟
But if i is unsigned then how come it get a negative value?
Plus it is still less than 10.

Answer 7

Or you could have incremented it to make count to reach ten because it will incrementing unsigned values

unsigned int i;

int count =0;

for (i=0; i<10; i++)
count++;
printf ("%d", count );

Answer 8

Type of variable unsigned int: represents a positive integer.
Depending on the processor architecture, it can take 2 bytes, (16 bits), or 4 bytes, (32 bits), and because of this, the range of limit values of is 0 to 65 535, (for 2 bytes), or 0 to 4 294 967 295, (for 4 bytes).

For example:
unsigned int unInt = 2;
printf("%u", unInt-=3);

Output: 4 294 967 295 (for 4 bytes) since the range of values goes in a circle ...4 294 967 295, 0, 1, 2...

https://code.sololearn.com/c2y5Ph1cyuYZ/?ref=app

Answer 9

Thanks A͢J - S͟o͟l͟o͟H͟e͟l͟p͟e͟r͟ .
But if unsigned int can only hold positive numbers and zero how can it get -1?
Should it not flip over to the highest int?

Answer 10

It actually is not negative. I should use the unsigned int format specifier, ud, in printf.
Thanks!

Answer 11

Oops, sorry. Didn't see your second response, and that you figured that out for yourself.

Answer 12

Because the first loop will be through and count becomes 1, the next round makes it false because you declared I to be unsigned so the loop ends and the final results of I becomes only one

Answer 13

Steve Yes, that's right. I forgot that.

Answer 14

Because of i-- and i is unsigned int so after first iteration i will be -1 and count will be 1

If there is no unsigned int then there would be no output because of infinite loop.

Answer 15

Adam Aksu That's because you're converting it back to a signed value with the format option, d, in the printf(). Use u if you want to see the unsigned value.

Answer 16

unsigned seems to have no effect here as it prints out -2

unsigned int i = 0;
i--;
i--;

printf("%d", i);

Answer 17

If u want an infinite loop, this one's better and small👇

int i;
for(i=1;i>0;i++)
printf("%d\n",i-1);

It will basically run infinitely printing number from 0 to infinity

Answer 18

Магистр техники и технологии. 111Стюра099 I'm not sure what is happening, but you're comment and code display all mangled up (unreadable)

Answer 19

Adam Aksu

unsigned int can only hold 0 and positive numbers so after first iteration count is 1 and i is -1

Since unsigned int can only hold positive numbers so loop will stop there for further execution.

Answer 20

Adam Aksu

unsigned int holds only 0 or positive numbers it doesn't mean i will not be -1

I would be -1 but unsigned int can't hold it so loop be break there.