将32位十六进制值拆分为4个字节[QB64]

问题描述：

我想问一下，如何将32位十六进制（例如：CEED6644）拆分为4个字节（var1 = CE，var2 = ED， var3 = 66，var4 = 44）。在QB64或QBasic中。我会用它来将几个数据字节存储到一个数组地址中。事情是这样的：将32位十六进制值拆分为4个字节[QB64]

DIM Array(&HFFFF&) AS _UNSIGNED LONG 
Array(&HAA00&) = &HCEED6644& 
addr = &HAA00& 
SUB PrintChar 
SHARED addr 

IF var1 = &HAA& THEN PRINT "A" 
IF var1 = &HBB& THEN PRINT "B" 
IF var1 = &HCC& THEN PRINT "C" 
IF var1 = &HDD& THEN PRINT "D" 
IF var1 = &HEE& THEN PRINT "E" 
IF var1 = &HFF& THEN PRINT "F" 
IF var1 = &H00& THEN PRINT "G" 
IF var1 = &H11& THEN PRINT "H"

等等......

答

你可以使用整数除法（\）和按位AND（AND）做到这一点。

DIM x(0 TO 3) AS _UNSIGNED _BYTE 
a& = &HCEED6644& 
x(0) = (a& AND &HFF000000&) \ 2^24 
x(1) = (a& AND &H00FF0000&) \ 2^16 
x(2) = (a& AND &H0000FF00&) \ 2^8 
x(3) = a& AND &HFF& 
PRINT HEX$(x(0)); HEX$(x(1)); HEX$(x(2)); HEX$(x(3))

需要注意的是，你可以选择使用一个通用RShift~&函数，而不是原始整数除法，因为你真的做什么换挡位：

x(0) = RShift~&(a& AND &HFF000000&, 18) 
... 

FUNCTION RShift~& (value AS _UNSIGNED LONG, shiftCount AS _UNSIGNED BYTE) 
    ' Raise illegal function call if the shift count is greater than the width of the type. 
    ' If shiftCount is not _UNSIGNED, then you must also check that it isn't less than 0. 
    IF shiftCount > 32 THEN ERROR 5 
    RShift~& = value/2^shiftCount 
END FUNCTION

在该大厦，您可以创建另一个功能：

FUNCTION ByteAt~%% (value AS _UNSIGNED LONG, position AS _UNSIGNED BYTE) 
    'position must be in the range [0, 3]. 
    IF (position AND 3) <> position THEN ERROR 5 
    ByteAt~%% = RShift~&(value AND LShift~&(&HFF&, 8*position), 8*position) 
END FUNCTION

注意的是使用一个LShift~&函数（由2的幂乘）转移位到左侧。一个潜在的更好的替代方法是首先执行右移，只是掩模的低8位，从而消除了LShift~&的需要：

FUNCTION ByteAt~%% (value AS _UNSIGNED LONG, position AS _UNSIGNED BYTE) 
    'position must be in the range [0, 3]. 
    IF (position AND 3) <> position THEN ERROR 5 
    ByteAt~%% = RShift~&(value, 8*position) AND 255 
END FUNCTION

顺便提及，另一种已知的作为FreeBASIC QB样实现具有实际SHR操作者，像MOD或AND一样使用，以直接执行移位操作而不是使用可能更快的分割。

你也可以使用QB64的DECLARE LIBRARY设施，以创建在C++中，将进行移位操作功能：

/* 
* Place in a separate "shift.h" file or something. 
*/ 
unsigned int LShift (unsigned int n, unsigned char count) 
{ 
    return n << count; 
} 

unsigned int RShift (unsigned int n, unsigned char count) 
{ 
    return n >> count; 
}

下面是完整的相应QB64代码：

DECLARE LIBRARY "shift" 
    FUNCTION LShift~& (value AS _UNSIGNED LONG, shiftCount AS _UNSIGNED _BYTE) 
    FUNCTION RShift~& (value AS _UNSIGNED LONG, shiftCount AS _UNSIGNED _BYTE) 
END DECLARE 

x(0) = ByteAt~%%(a&, 0) 
x(1) = ByteAt~%%(a&, 1) 
x(2) = ByteAt~%%(a&, 2) 
x(3) = ByteAt~%%(a&, 3) 
END 

FUNCTION ByteAt~%% (value AS _UNSIGNED LONG, position AS _UNSIGNED BYTE) 
    'position must be in the range [0, 3]. 
    IF (position AND 3) <> position THEN ERROR 5 
    ByteAt~%% = RShift~&(value, 8*position) AND 255 
END FUNCTION

如果QB64已文档化的API当移位计数过高时，可能会从C++代码中引发QB64错误，而不是依赖C++的行为来实质上忽略太高的移位计数。不幸的是，事实并非如此，它可能会导致更多的问题而不是价值。

哦，谢谢！这可能是我见过的最详细的答案！ – Aureal

答

这剪断得到一个十六进制值的字节对：

DIM Value AS _UNSIGNED LONG 
Value = &HCEED6644& 
S$ = RIGHT$("00000000" + HEX$(Value), 8) 
PRINT "Byte#1: "; MID$(S$, 1, 2) 
PRINT "Byte#2: "; MID$(S$, 3, 2) 
PRINT "Byte#3: "; MID$(S$, 5, 2) 
PRINT "Byte#4: "; MID$(S$, 7, 2)

将32位十六进制值拆分为4个字节[QB64]

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