This question is quite clear. What are the differences between an int, an uint8_t, and an uint16_t. I know it has to do with bytes and memory but can someone clarify me a bit?

Things I want to know:

1- How much memory does each take.

2- When to use what.

3- In the end of the day, are they that different?

  • 1
    Beware that the compiler does not do arithmetic with anything smaller than int. E.g. (uint8_t) 200 + (uint8_t) 200 does not overflow: the terms are promoted to int before the addition and the result is (int) 400. Nov 1, 2016 at 21:55

1 Answer 1


You can decipher most of them yourself.

  • A u prefix means unsigned.
  • The number is the number of bits used. There's 8 bits to the byte.
  • The _t means it's a typedef.

So a uint8_t is an unsigned 8 bit value, so it takes 1 byte. A uint16_t is an unsigned 16 bit value, so it takes 2 bytes (16/8 = 2)

The only fuzzy one is int. That is "a signed integer value at the native size for the compiler". On an 8-bit system like the ATMega chips that is 16 bits, so 2 bytes. On 32-bit systems, like the ARM based Due, it's 32 bits, so 4 bytes. Of the three it is the only one that changes.

Personally I rarely use int and always use uint8_t etc., since the variable type is the same no matter what architecture you compile for. When you use int you can run into problems if you had a program that worked fine on a 32-bit ARM but then doesn't work right on an 8-bit ATMega, since the int can only store a fraction of the range of numbers on the 8-bit system compared to the 32-bit system.

  • Can I make an uint8_t array?
    – Dat Ha
    Nov 1, 2016 at 20:06
  • You can make an array of any data type.
    – Majenko
    Nov 1, 2016 at 20:07
  • Can I use it like this too: uint8_t function()
    – Dat Ha
    Nov 1, 2016 at 20:37
  • 2
    Yes. uint8_t is the same as "byte" or "unsigned char".
    – Majenko
    Nov 1, 2016 at 20:37
  • 2
    If you need negative values, you need to use signed integers. E.g. int16_t.
    – Gerben
    Nov 1, 2016 at 21:26

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