HI-TECH Software Forums: type mis-interpretation??

mov a, 0
cpl a
mov [(??_BootLoader^0+2)], a
mov a, [(_Checksum+1)^0]
cpl a
mov [(??_BootLoader^0+3)], a
mov a, reg[46]
cmp a, [(??_BootLoader^0+3)]
jnz l34
mov a, [(??_BootLoader^0+2)]
jnz l34

Rolf,

The code generated is correct. Even though the value is a byte, the C standard specifies that it is converted to an integer before applying the complement operator. So the value of ~x, where x is an unsigned byte, is always in the range 0xFF00-0xFFFF.

Applying the typecast to the result of the ~ operator as you have done is the correct way to deal with this to achieve the result you wanted.

Incidentally, this treatment of data (promoting to integer before performing arithmetic) applies to all the standard operators. The compiler optimizes the resultant code where possible to achieve the correct result. The C standard allows arithmetic to be done in any way as long as the result is the same as if it were done strictly according to the standard (this is known as the "as-if" rule).

In this case if the complement and subsequent comparison were done in 8 bits, the result would not be the same as if they were done in the correct 16 bit length. Adding the typecast to the result of the complement ensures that the top 8 bits are always zero, so the comparison can then be done in 8 bits, which in turn means the complement only needs to be done on the 8 bits of the variable.

An alternative way of achieving what you wanted would be:
Code:


	if (RX_Buffer == (Checksum.Byte.Low ^ 0xFF)) RXstate++;

Clyde



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