Next lesson for hooks: How to extract one of the 32 digits

The Integer type is 32-bit and has 4 bytes, and now we need to be able to extract one of its 32-bit digits.

But the smallest integer type of Delphi is also a byte (8 bits): Byte (unsigned), Shortint (signed).

Start by extracting a byte first:

var


I:integer;


B:byte;


begin


I: = Maxint; {Integer maximum}


ShowMessage (IntToStr (i)); {2147483647}


{Now I binary representation is: 01111111 11111111 11111111 11111111}


{Interger's highest bit of 0 means that this is a positive number (1 is a negative number)}





{if:}


I: = 2146439167;


{Now I binary representation is: 01111111 11110000 00001111 11111111}


{Now its hexadecimal representation is: $ f F 0 0 F f F}


{implement, four bytes from right to left are: $FF, $0f, $F 0, $7f}





{If you need to extract one byte of four bytes separately, Delphi bit we provide two functions:}





B: = Lo (i); {Extract Low Byte}


showmessage (Format ('%.2x ', [b])); {FF; This is from the right of the first byte}





B: = Hi (i); {Extract High Byte}


ShowMessage (Format ('%.2x ', [b])); {0F; This is from the right number two bytes}





{So how do we extract the third and fourth bytes? Method one:}


{Move 16 digits to the right and then use Lo and Hi to extract}


{01111111 11110000 00001111 11111111 Right-shift 16-bit will become:}


{01111111 11110000; try it:}





B: = Lo (i shr 16);


showmessage (Format ('%.2x ', [b])); {F0; This is from the right number three bytes}


B: = Hi (i shr 16);


showmessage (Format ('%.2x ', [b])); {7F; This is from the right number four bytes}





{Of course i's fourth byte can also be extracted like this:}


B: = Lo (i shr 24);


showmessage (Format ('%.2x ', [b])); {7F; This is from the right number four bytes}








{Now, if there is no Lo and Hi function, can we do it? Of course:}


B: = (I and $FF);


showmessage (Format ('%.2x ', [b])); {FF; This is from the right of the first byte}


B: = (i shr 8 and $FF);


showmessage (Format ('%.2x ', [b])); {0F; This is from the right number two bytes}


B: = (i shr and $FF);


ShowMessage (Format ('%.2x ', [b])); {F0; This is from the right number three bytes}


B: = (i shr and $FF);


showmessage (Format ('%.2x ', [b])); {7F; This is from the right number four bytes}





{This is why?) to change the block of statements to carefully analyze}


end;





//Additions to the above examples:


var


B:byte;


begin


{We know that the maximum Byte value is 255, which is hexadecimal $FF, binary 11111111}


{This $FF has a special purpose:}


B: = 0 and $FF;


ShowMessage (IntToStr (b)); {0}


B: = 1 and $FF;


ShowMessage (IntToStr (b)); {1}


B: = $FF;


ShowMessage (IntToStr (b)); {100}


B: = 255 and $FF;


ShowMessage (IntToStr (b)); {255}


{0..255 any number immediately after and with the $FF, the value is unchanged (this is not difficult to understand from a binary perspective)}





{Additionally, byte is a byte, and when you give it more, it also takes a byte (low byte), for example:}


B: = Byte (maxint);


ShowMessage (IntToStr (b)); {255}





{The above example should be understood now}


end;

Back to topic:

//本例中我们把最低位叫第　0　位;　把　Integer　的最高位叫第　31　位.
var
i:　Integer;
b:　Byte;
begin
{还是用第一个例子中的值吧:}
i　:=　2146439167;
{现在　i　的二进制表示是:　01111111　11110000　00001111　11111111}

{一个字节的最大值是　$FF;　一个二进制位的最大值当然是　1,　写成十六进制还是　$1　}
{提取第　0　位:}
b　:=　i　and　1;　ShowMessage(IntToStr(b));　　　　{1}
{提取第　0　位也可以写作(右移0位就是没动):}
b　:=　i　shr　0　and　1;　ShowMessage(IntToStr(b));　{1}

{提取第　1　位:}
b　:=　i　shr　1　and　1;　ShowMessage(IntToStr(b));　{1}

{提取第　13　位:}
b　:=　i　shr　13　and　1;　ShowMessage(IntToStr(b));　{0}

{提取最高位(第　31　位):}
b　:=　i　shr　31　and　1;　ShowMessage(IntToStr(b));　{0}
end;

//假如只判断最高位,　是　0　还是　1(也就是判断正负),　还可以这样:
var
i:　Integer;
begin
i　:=　MaxInt;　{这肯定是个正数}
if　i　shr　31　=　0　then　ShowMessage('正');　{正}

i　:=　-1;
if　i　shr　31　=　1　then　ShowMessage('负');　{负}
end;