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  • salq and sarq in x86-64 Assembly Language - Stack Overflow
    It certainly has an effect because it fills the register with the least significant bit The salq moves the LSB into the sign bit, and sarq fills with the sign bit
  • Shift (sal, shl, sar, shr) (IA-32 Assembly Language Reference . . . - Oracle
    shr right shifts (unsigned divides) a byte, word, or long value for a count specified by an immediate value and stores the quotient in that byte, word, or long respectively The second variation divides by a count value specified in the CL register shr sets the high-order bit to 0
  • Machine Level – Assembly (x86-64) basics - New York University
    Sets the lower byte of dest based on combinations of condition codes and does not alter remaining 7 bytes Destination can also be memory location These instructions are usually used after a comparison cmpq %rsi, %rdi setg %al movzbq %al, %rax # Compare x:y # Set when > # Zero rest of %rax ret
  • Assembly x86_64 question! [salq sarq] : r learnprogramming - Reddit
    salq is a shift-left operation Basically it's taking a binary number (y) and shifting all it's bits to the left, thereby making it's least significant bit into it's most significant bit
  • CS356: A short guide to x86-64 assembly - HackMD
    I'll try to improve this guide over time The goal is to collect the essential knowledge about x86-64 assembly and examples of very common patterns (if, while, for, procedure calls) in BombLab We have plenty of registers inside the CPU; they can store 8 bytes each and they are called:
  • SAL SAR SHL SHR — Shift - felixcloutier. com
    Multiply r m8 by 2, CL times Multiply r m8 by 2, imm8 times Multiply r m16 by 2, once Multiply r m16 by 2, CL times Multiply r m16 by 2, imm8 times Multiply r m32 by 2, once Multiply r m64 by 2, once Multiply r m32 by 2, CL times Multiply r m64 by 2, CL times Multiply r m32 by 2, imm8 times Multiply r m64 by 2, imm8 times
  • CS 107 Lecture 11: Assembly Part II - Stanford University
    It has the form of an instruction that reads from memory to a register, but it does not reference memory at all It's first operand appears to be a memory reference, but instead of reading from the designated location, the instruction copies the effective address to the destination
  • assembly - Why is %cl the only register the sal-operation will accept . . .
    Today while working on an x86 Assembly-project I tried left-shifting the value of the %rax register by the value of %rbx using the salq -operation, but couldn't get it to work Shifting with an immediate value worked fine, of course, but why I couldn't use the register I had no idea
  • Soc2040 SP Week6 Lecture1 Slides On Machine Level Programming Part2 . . .
    The document discusses x86-64 assembly language instructions including address computation, arithmetic operations, and an example of translating an arithmetic expression to assembly code Key instructions covered are leaq for address computation, imulq for multiplication, and salq for shifting
  • SAL|SAR|SHL|SHR - x86-64 Simplified - Treeniks
    SAL and SHL are synonymous Shifts the bits in the first operand (destination operand) to the left or right by the number of bits specified in the second operand (count operand) Bits shifted beyond the destination operand boundary are first shifted into the CF flag, then discarded





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