uint8_t not printing the right value when casted to unsigned int

Question

I defined an uint8_t variable x and gave it the value 11110000 , but when I print it (casting it to an unsigned integer so it's not invisible), it prints out 112 when it should be 240 . What did I do wrong ?

Also, I only need 4 bits, but I don't know any types which could provides me only 4 bits, which is why I use an uint8_t .

Brief code :

std::uint8_t walls;
walls=11110000;
cout << (unsigned int) walls <<endl;

Answer 1

11110000 is being interpreted as a decimal number.

Which its binary representation is:

1010 1001 1000 0110 0111 0000

Since your variable is unsigned and overflow causes wrap around, the lower 8 bit are being assigned

0111 0000

Which is 112.

To use binary constants , simply add 0b as suggested in comments.

walls=0b11110000

And apparently this is part of C++14. Thanks to @Daniel Kamil Kozar.

Answer 2

What did I do wrong ?

You didn't assign a base 2 value but a base 10 as you thought here:

walls=11110000;

To assign base 2 values use (at least with c++14 standard)

walls=0b11110000;

For older standards you can use

walls=0xF0;

Also, I only need 4 bits, but I don't know any types which could provides me only 4 bits, which is why I use an uint8_t.

In fact uint8_t is the smallest number of bits you can have.

But you probably want to mask and only use the lower nibble:

walls=0b000011111;

or again for pre c++14:

walls=0x0F;