for loop with unsigned char gives unexpected behavior

Question

Im practicing interview questions but am having a hard time with this basic question:

How many times will this loop execute?

  unsigned char half_limit = 150;

  for (unsigned char i = 0; i < 2 * half_limit; ++i)
  {
      std::cout << i;
  }

My thought is that since an unsigned int only reaches 255 it will execute forever since when I increment an unsigned char when it is at 255 it will revert back to 0? However this thinking is wrong and what is even more strange is that this is the output cout is giving me:

 !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_`abcdefghijklmnopqrstuvwxyz{|}~���������������������������������������������������������������������������������������������������������������������������������

And when I try to limit the look with something like the following:

if (i <= 255)
  std::cout << i;
else
  break;

The loop is still infinite.

My questions is what is the expected result of the code and why is this the case?

If I'm not mistaken, this should result in an infinite loop. — Oliver Charlesworth
– Oliver Charlesworth, Commented Jan 20, 2018 at 18:27
But note that "what makes this a good interview question?" is off-topic here. — Oliver Charlesworth
– Oliver Charlesworth, Commented Jan 20, 2018 at 18:30
"if (i <= 255)" is obviously always true since i is unsigned char. — Jesper Juhl
– Jesper Juhl, Commented Jan 20, 2018 at 18:42

taskinoor · Accepted Answer · 2018-01-20 18:39:07Z

8

This is an infinite loop. Let's take a step-by-step look:

2 * half_limit

half_limit is unsigned char but 2 is int. So half_limit is promoted to int and the result is integer 300.

i < 2 * half_limit

Since right hand size is int, i is also promoted to int. There is no problem here.

But in ++i no promotion is involved and i is an unsigned char as declared. So after reaching 255 it goes back to 0 and the condition i < 2 * half_limit is always true as i will never reach to 300.

The situation is same for your second condition if (i <= 255) as i goes back to 0 before this check is made.

edited Jan 20, 2018 at 18:39

answered Jan 20, 2018 at 18:33

taskinoor

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2 Comments

Fureeish Over a year ago

"and i is an unsigned int as declared" it's actually unsigned char. Small typo ;>

paulsm4 Over a year ago

Good answer - and thank you for correcting the typo :)

Minor Threat · Accepted Answer · 2018-01-20 18:51:41Z

2

C/C++ extends expression's bit capacity if necessary. It's called Integer type promotion.

According to these rules, half_limit * 2 is getting promoted to integer value of 300. Assuming CHAR_BIT == 8, i is a single octet, and therefore it will never reach value of 300, so the cycle will continue forever. Also, an unsigned octet is always less or equal than 255, so it's your if expression is redundant.

edited Jan 20, 2018 at 18:51

answered Jan 20, 2018 at 18:27

Minor Threat

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1 Comment

Oliver Charlesworth Over a year ago

How does this address the question?

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