Hi I need help understanding some code. The code is used to convert a byte to a hexadecimal.
When converting from a byte to an int why do you have to do a logical AND with 0xFF.
For example: int i = 'A' & 0xff;
Also what will the following operation do. I know it does a right shift of 4 bits but what is the resulting. The code converts the character to hex but I am unclear as to how it does it.
(i >> 4); (i & 0xf);
examples using bit patterns would be helpful tanx.
When converting from a byte to an int why do you have to do a logical AND with 0xFF.
For example: int i = 'A' & 0xff;
First of all, in that line of code there is no conversion from a byte to an int. The 'A' is a character literal (its type is char, not byte). The 0xff is an integer literal. Also, the AND operator & is not a logical AND, it is a bitwise AND. The logical AND operator is &&.
Doing a bitwise AND with 0xff means keeping the lower 8 bits and clearing all higher bits.
About the second question: You first shift i four bits to the right and with the & operation you keep the lower 4 bits and mask the rest off.
In hexadecimal, each digit represents 4 bits in the value. Take for example the value 156 (decimal). In binary, this is: 10011100. It is easy to convert this to hexadecimal: split the binary number into groups of 4 bits: 1001 1100. Each group of 4 bits corresponds to 1 hexadecimal digit. 1001 = 9 and 1100 = C, so 156 = 10011100 = 0x9C. See this: Hexadecimal (Wikipedia)
Notice the following: The result is 1 when both bits are 1, otherwise the result is 0.
Suppose you have a value, and you want to keep the lower 8 bits of the value, and set all the other bits of the value to 0. You can do this by doing a bitwise AND operation, where you AND the value with a mask in which you set the bits you want to keep to 1, and the bits you want to clear to 0.
For example:
Note that bits 8-31 are 0 in the result, and bits 0-7 contain the lower 8 bits of the original value.
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