Simple Character Interpretation In C

Question

Here is my code

 #include

 void main()
 {
     char ch = 129;
     printf(\"%d\", ch);
 }

I get the output as -127. What d

Answer 1

The type char can be either signed or unsigned, it's up to the compiler. Most compilers have it as `signed.

In your case, the compiler silently converts the integer 129 to its signed variant, and puts it in an 8-bit field, which yields -127.

Answer 2

The char type is a 8-bit signed integer. If you interpret the representation of unsigned byte 129 in the two's complement signed representation, you get -127.

Answer 3

This comes from the fact that a char is coded on one byte, so 8 bits of data.

In fact char has a value coded on 7 bits and have one bit for the sign, unsigned char have 8 bits of data for its value.

This means:

Taking abcdefgh as 8 bits respectively (a being the leftmost bit, and h the rightmost), the value is encoded with a for the sign and bcdefgh in binary format for the real value:

42(decimal) = 101010(binary) stored as : abcdefgh 00101010

When using this value from the memory : a is 0 : the number is positive, bcdefgh = 0101010 : the value is 42

What happens when you put 129 :

129(decimal) = 10000001(binary) stored as : abcdefgh 10000001

When using this value from the memory : a is 0 : the number is negative, we should substract one and invert all bits in the value, so (bcdefgh - 1) inverted = 1111111 : the value is 127 The number is -127