GCBASIC documentation
Variables Operations

Explaination

Using and accessing bytes within word and long numbers etc may be required when you are creating your solution. This can be done with some ease.

Example 1:

You can access the bytes within word and longs variables using the following as a guide using the Suffixes _H, _U and _E

    Dim workvariable as word
    workvariable = 21845
    Dim lowb as byte
    Dim highb as byte
    Dim upperb as byte
    Dim lastb as byte

    lowb = workvariable
    highb = workvariable_H
    upperb = workvariable_U
    lastb = workvariable_E

To further explain, where

    Dim rB as Byte
    Dim sW as Word

To extract the bytes from a WORD of 16 bits use the Suffix _H

    'To use the bits 7-0 [lower byte] in the Word variable sW
    rB = sW

    'For bits 15-8 [upper byte] in the Word variable sW, use sw_H
    rB = sW_H

To extract the bytes from a LONG of 32 bits use the Suffixes _H, _U and _E, where

    Dim rB as Byte
    Dim tL as Long

    ‘ For bits 7-0 [lowest byte #0] in Long variable tL
    rB = tL

    ‘ For bits 15-8 [lower middle byte #1] in Long variable tL
    rB = tL_H

    ‘ For bits 23-16 [upper middle byte #2] in Long variable tL
    rB = tL_U

    ‘ For bits 31-24 [highest byte #3] in Long variable tL
    rB = tL_E

To extract nibbles from the variable rB

    lower_nibble = rB & 0x0F
    upper_nibble = (rB & 0xF0) / 16

Example 2:

Assigning values to Word and Long variables via the the Byte variable (the Least Significant Byte [.lsb]) of the same Word and Long variable.

Because a Long (or Word) variable and the Least Significant Byte, of the variable, have the same variable assignments to specific byte elements (_e, _u and _h) assignment must be appropriate to the element.

The code below uses a Long variable but the same principle is used for a Word.

Assigning two values, a byte and a word constant value, to the variable tL to compare resulting impact on Long variable.

    Dim tL as Long

    tL = 255  'All bits of the value 255 will reside in the lowest byte of the Long variable tL
    tL = 286   'This assignment will flow into tL_H where tL_H =1 and tl=30.

Assigning values to specific elements of a Long variable.

    'Assign value to specific elements
    tL_E = 0xF7
    tL_U = 0xC5
    tL_H = 0xE3

    'is same as the following assignment, we show the use of casting for clarification only.
    [Long] tL = 0xF7C5E300   The lower byte (tL) is empty (zero).

    'or, treat the Long as a byte and assign a byte.
    [byte]tL = [byte]0xA4

Assigning values to the byte element of a long variable.

    'This will assign the lowest byte with 0xA4 but this assignment will also clear the upper 3 byte elements of the long variable.
    tL = 0xA4

    'To assign the lowest byte
    tL = ( tL and 0xffffff00 ) + 0xA4  'Wwill preserve the upper bytes and ensure the lowest byte is assigned correctly.

A method to check a variable is assigned as expected is to use HserPrint and HserPrint hex(), as follows:

    ' HserPrint hex() only prints one byte so we need to handle the four elements
    HserPrint " Print tL _E, tL_U, tL_H & tL as hex"
    HserPrint hex (tL_E)
    HserPrint hex (tL_U)
    HserPrint hex (tL_H)
    HserPrint hex (tL)
    HserPrintCRLF
    HserPrint "Variable tL = "
    HserPrint tL

The user code above will result in an output as follows:

    Print tL _E, tL_U, tL_H & tL as hexF7C5E3A4
    Variable tL = 4156941220