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How to shift a binary number by filling MSB with 1?


How to shift a binary number by filling MSB with 1?

By : Raoulavelo
Date : November 22 2020, 07:01 PM
should help you out If the sign bit is set, the Arithmetic Shift Right will shift in 1's on the left (for sign-extension presumably)
sra Shift right arithmetic by a constant number of bits
code :


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count the the number of 1s in its binary representation -- shift right don't work?

count the the number of 1s in its binary representation -- shift right don't work?


By : Rohan Saraf
Date : March 29 2020, 07:55 AM
I wish this help you If we're talking about python's unlimited-precision integers, then any negative number has an infinite number of 1's! So regardless of sign-filling (which you'll also get in C), counting the bits in a negative number is non-sensical except for a fixed bit length.
For a 32-bit or 64-bit int, just shift this many times and stop.
code :
>>> n = -4
>>> for bit in reversed([ (n>>shift)&1 for shift in range(32) ]):
...    print bit,
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0
sum( (n>>shift)&1 for shift in range(32) )
How to shift binary number to the left

How to shift binary number to the left


By : Eve Yac
Date : March 29 2020, 07:55 AM
With these it helps Just tell the formatter that you need to print exactly N (11) digits in any case, and that 0 should be used as a padding symbol:
code :
my $num = 0b00000010001;
for (0..6) {
    printf("%011b\n", $num);
    $num <<= 1;
}
Shift value to some point in binary number

Shift value to some point in binary number


By : gvanderclay
Date : March 29 2020, 07:55 AM
wish help you to fix your issue I wonder if it is possible to do the following.
code :
Instruction  Carry  Reg value
-----------  -----  ---------
PUSH           x    01110000
ROL            0    1110000x
ROL            1    110000x0    b6 now in carry
POP            1    01110000
ROL            0    11100001    b6 now to b0
AND $7Fh       x    01100001
MOV     r1,r0       ;copy r0
ROL     r0
ROL     r0          ;b6 in Carry
MOV     r0,r1       ;restore r0
ROL     r0          ;shift in original b6 to b0
ROL     r0          ;replace b7 in r0 with b7 in r1
ROL     r1
ROR     r0
How to interpret hex number byte array from left shift binary sum?

How to interpret hex number byte array from left shift binary sum?


By : William Culp
Date : March 29 2020, 07:55 AM
around this issue To check if a bit is set, you can do a bitwise AND with that bit. Then check if the result is equal to 0. If it isn't, the bit was set.
e.g.
code :
00100110
00000010     // checks the second bit
-------- &
00000010     // result != 0, so the bit was set
byte[] flag = new byte[4];      
flag[0] = 0x30;
flag[1] = 0x30;
flag[2] = 0x32;
flag[3] = 0x30;

// Bytes to char, using the 'oversized' short so the numbers won't be out of range
short b1 = Short.parseShort(new String(new byte[]{flag[0], flag[1]}), 16);
short b2 = Short.parseShort(new String(new byte[]{flag[2], flag[3]}), 16);
char i = (char) (b1 | (b2 << 8));

// Print contents as binary string
System.out.println(String.format("%16s", Integer.toBinaryString(i)).replace(' ', '0'));

// Output: 0010000000000000

// Check if 14'th bit is set (at index 13)
boolean isSet = ((i & (1 << 13)) != 0);

System.out.println(isSet); // true
What is the limit I can shift a binary number?

What is the limit I can shift a binary number?


By : Sunil Kumar Pasumart
Date : March 29 2020, 07:55 AM
around this issue I think your assignment has a typo. The MIPS sll instruction only supports 5 bits worth of shifting. That is, the field in the instruction encoding is only 5 bits long, so only shift values in [0,31] are legal. In fact, if I try to assemble this simple program:
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