Make up for the next step of hook learning: how to extract one of the 32 bits

The integer type is 32-bit and has 4 bytes. Now we need to be able to extract one of its 32-bit values.
But the smallest integer type of Delphi is also a byte (8 bits): byte (unsigned), struct int (Signed ).
Otherwise, start with extracting a byte:

VaR I: integer; B: byte; begin I: = maxint; {maximum integer} showmessage (inttostr (I); {2147483647} {the binary representation of I is: 01111111 11111111 11111111 11111111} {the highest bit of interger 0 indicates that this is a positive number (1 indicates a negative number)} {suppose:} I: = 2146439167; {now the binary representation of I is: 01111111 11110000 00001111 11111111} {now its hexadecimal representation is: $7 F 0 0 f} {implement it, from right to left, the four bytes are: $ ff, $ 0f, $ F0, $ 7f} {if you need to extract one of the four bytes separately, we provide two functions for the Delphi bit:} B: = Lo (I ); {extract low byte} showmessage (Format ('%. 2x ', [B]); {ff; this is the first byte from the right} B: = Hi (I); {extract high byte} showmessage (format (' %. 2x ', [B]); {0f; this is the second byte from the right} {so how do we extract the third and fourth byte? Method 1:} {shifted right 16 bits, and then extracted with Lo and HI} {01111111 11110000 00001111 11111111 shifted right 16 bits, it will become:} {01111111 11110000; try:} B: = Lo (I SHR 16); showmessage (format ('%. 2x ', [B]); {F0; this is the third byte from the right} B: = Hi (I SHR 16); showmessage (format (' %. 2x ', [B]); {7f; this is the fourth byte from the right} {of course, the fourth byte of I can also be extracted as follows:} B: = Lo (I SHR 24); showmessage (format ('%. 2x ', [B]); {7f; this is the fourth byte from the right} {now let's change our mind. If there are no lo or hi functions, can we do it? Certainly yes:} B: = (I and $ ff); showmessage (format ('%. 2x ', [B]); {ff; this is the first byte from the right} B: = (I SHR 8 and $ ff); showmessage (format (' %. 2x ', [B]); {0f; this is the second byte from the right} B: = (I SHR 16 and $ ff); showmessage (format (' %. 2x ', [B]); {F0; this is the third byte from the right} B: = (I SHR 24 and $ ff); showmessage (format (' %. 2x ', [B]); {7f; this is the fourth byte from the right} {why? Change the statement block for careful analysis} end;
 
   
 

 
// Supplement the above example: var B: byte; begin {we know that the maximum value of byte is 255, that is, the hexadecimal $ ff, binary 11111111} {This $ FF has such a special purpose:} B: = 0 and $ ff; showmessage (inttostr (B); {0} B: = 1 and $ ff; showmessage (inttostr (B); {1} B: = 100 and $ ff; showmessage (inttostr (B); {100} B: = 255 and $ ff; showmessage (inttostr (B); {255} {0 .. 255 After performing the and operation on any number and $ FF directly, the value remains unchanged (this is not difficult to understand from the binary perspective)} {In addition, byte is a byte. When you give it more, it only needs one byte (low byte), such as:} B: = byte (maxint); showmessage (inttostr (B )); {255} {the above example should be understandable} end;

  
 

Back to topic:

// In this example, we call the percentile bit 0th bits and the maximum integer bits 31st bits. vaR I: integer; B: byte; begin {use the value in the first example:} I: = 2146439167; {the binary representation of I is: 01111111 11110000 00001111 11111111} {the maximum value of a single byte is $ ff; the maximum value of a binary bit is of course 1, written as hexadecimal or $1} {extract 0th bits:} B: = I and 1; showmessage (inttostr (B); {1} {extract 0th digits can also be written (0 digits to the right is not moved):} B: = I shr 0 and 1; showmessage (inttostr (B); {1} {extract 1st bits:} B: = I SHR 1 and 1; showmessage (inttostr (B); {1} {extract 13th bits:} B: = I shr 13 and 1; showmessage (inttostr (B )); {0} {extract the highest bit (31st bits):} B: = I shr 31 and 1; showmessage (inttostr (B); {0} end;

  
 

 
// If you only judge the highest bit, whether it is 0 or 1 (that is, whether it is positive or negative), you can also: var I: integer; begin I: = maxint; {this must be a positive number} If I shr 31 = 0 then showmessage ('position'); {positive} I: =-1; if I SHR 31 = 1 then showmessage ('negative '); {negative} end;