how 256 stored in char variable and unsigned char

Question

Up to 255, I can understand how the integers are stored in char and unsigned char ;

#include<stdio.h>
int main()
{
        unsigned char a = 256;
        printf("%d\n",a);
        return(0);
}

In the code above I have an output of 0 for unsigned char as well as char.

For 256 I think this is the way the integer stored in the code (this is just a guess):

First 256 converted to binary representation which is 100000000 (totally 9 bits).

Then they remove the remove the leftmost bit (the bit which is set) because the char datatype only have 8 bits of memory.

So its storing in the memory as 00000000 , that's why its printing 0 as output.

Is the guess correct or any other explanation is there?

Answer 1

Your guess is correct. Conversion to an unsigned type uses modular arithmetic: if the value is out of range (either too large, or negative) then it is reduced modulo 2^N, where N is the number of bits in the target type. So, if (as is often the case) char has 8 bits, the value is reduced modulo 256, so that 256 becomes zero.

Note that there is no such rule for conversion to a signed type - out-of-range values give implementation-defined results. Also note that char is not specified to have exactly 8 bits, and can be larger on less mainstream platforms.