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unsigned char Mid;
if( (data[2]) == 0x9A){
Mid = data[5];
if( (Mid == 1) || (Mid == 2) || (Mid == 3) )
return(Mid);
The code above gives:
warning: comparison is always false due to limited range of data type
error as I expected.
It’s ok if I type
if( (unsigned char)data[2] == 0x9A){
or
if( (data[2]&0xFF) == 0x9A){
I understand the first one, but what happens in the second one?
What is the effect of masking with 0xFF?
When casting the value to unsigned char, you are doing an explicit type conversion.
When you mask the value with
0xFF, an implicit type conversion takes place.You have the signed char
data[2]and0xFFas operands to the bitwise AND operator. Because one of the operands (0xFF) cannot fit inside a signed char, both operands are implicitly converted into an unsigned type.