  C RUBY-ON-RAILS MYSQL ASP.NET DEVELOPMENT RUBY .NET LINUX SQL-SERVER REGEX WINDOWS ALGORITHM ECLIPSE VISUAL-STUDIO STRING SVN PERFORMANCE APACHE-FLEX UNIT-TESTING SECURITY LINQ UNIX MATH EMAIL OOP LANGUAGE-AGNOSTIC VB6 MSBUILD # How to shift a binary number by filling MSB with 1?  » assembly » 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 : ## 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

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

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?

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;
flag = 0x30;
flag = 0x30;
flag = 0x32;
flag = 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, flag}), 16);
short b2 = Short.parseShort(new String(new byte[]{flag, flag}), 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?

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: 