What is rax in assembly. (Wikipedia says "extended".
What is rax in assembly AT&T syntax (used by GNU as / objdump) uses different mnemonics than Intel for some instructions (see the official docs). Proceeding on and assuming a is a label rather than a macro or equ resulting in a constant: mov rax, 100 ; Always means put constant 100 in rax mov rax, a ; Either means put the address of a in rax, ; or put a in rax depending on which assembler. The purpose of LEA is to allow one to perform a non-trivial address calculation and store the result [for later usage]. xor is shorter than mov, and even leaves more spare space in the uop cache line it's in, but neither of those effects are well described as "not having to fetch". mov is an opcode instructing a CPU to copy data from the second operand to the first operand. ; The CPU sets up a small state machine which tracks the type of the last branch. What does the star decoration do on that? Does that further dereference that value (thus (%rax) is itself a pointer)? I'm having trouble googling *(assembly syntax. What does Rax mean in assembly? rax is the 64-bit, “long” •If you specify one operand, it multiplies that by %rax, and splits the product across 2registers. e. The code traverses a linked list. I'm talking about data movement instructions in the x86-64 Intel architecture. The first eight of these registers also have special names for historic reasons, they are, in order, rax, Why Learn Assembly Language? Q: Why learn assembly language? A: Knowing assembly language helps you: •Write faster code •In assembly language •In a high-level language! leaq (%rdi,%rdi,2), %rax # t = x+x*2 salq $2, %rax # return t<<2 Converted to ASM by compiler: The RAX register is used for return values in functions regardless of whether you’re working with Objective-C or Swift. Only the top 32 bits are filled with the sign. The first operand on the mov instruction is a target operand, and the second is the @CiroSantilliOurBigBook. This is the opcode part of the instruction. You can use objdump -drwC -Mintel or gcc -masm=intel -S to get Intel syntax using the mnemonics that Intel and AMD document in their instruction reference manuals (see links in the x86 tag wiki. jmpq is just a un-conditional jump to a given address. This address is then dereferenced so the value 0x11 is loaded from the address 0x10C. It is used by the GNU assembler by default. For Microsoft assemblers, the syntax is. SF being the sign bit is zero, and OF being signed As far as I understood: %rdi = 1st argument = x %rsi = 2nd argument = y %rdx = 3rd argument = z The others manipulate these registers and store in a return value register The 3rd line on the He's had a couple typos on previous worksheets so I wanted to make sure he was correct here. Not to be confused with cdqe, the 64-bit instruction that is a more compact form of movsxd rax, eax. com: In MASM and GAS . At the moment the ABI specifies that. what I dont understand is that salq and sarq lines. cmp instructions set the flags register as it would for a sub (subtract) of the second operand from the first - 'second' and 'first' being reversed in AT&T The value of rax is undefined throughout the code you have provided us. mov is an opcode instructing a CPU cmp rax, 0 je equal_zero However since cmp is longer if you look at the output binary, test or sometimes and, or is preferred. It describes the base operation the CPU is required to perform. Namely, it sets the zero flag if the difference is zero (operands are equal). cltd by disassembling code on an Intel architecture. They are named r0, r1, r2, r3 r15. string "Input:%s" So no, with your input EAX/RAX would be 1, and the " 4" part of the input buffer would be left unconverted. 83F800 cmp eax, 0 09C0 or eax, eax 85C0 test eax, eax The resulting code will be. cmp instructions set the flags register as it would for a sub (subtract) of the second operand from the first - 'second' and 'first' being reversed in AT&T So what does endbr64 do?. Determining when NASM can infer the size of the mov operation. Note that some are the same as Intel's recommended nop forms (see below), but not all. highly likely something else). If two operands are equal, their bitwise AND is zero when both are zero. reaching the } at the bottom is completely irrelevant. cmp rax, 0 je equal_zero However since cmp is longer if you look at the output binary, test or sometimes and, or is preferred. That is, the rax register is acting like a pointer. By the way, the second line doesn't really look like it's part of the code that sets up the stack frame, rsi is the second argument when using System V AMD64 calling conventions. NOT can be performed on any 'value' that might be in rsi. A lot of compilers offer frame pointer omission as an Here's a complete example of assembly memory access. Contract between caller and callee on x86: * after call instruction: o %eip points at first instruction of function o %esp+4 points at first argument o %esp points at return address * after ret instruction: o %eip contains return address o %esp points at arguments pushed by caller o called function may have trashed arguments o %eax jmpq is just a un-conditional jump to a given address. Preconditions: CET must be enabled by setting the control register flag CR4. The first (one-liner) code sample would be more akin to: movq (%rbx), %rcx addq %rcx, %rax although even that is not strictly identical since it changes rcx. (And sscanf returns int, so technically it's not safe to assume it's correctly zero-extended into RAX; the upper 4 bytes of RAX could hold garbage. o). Note that prl has apparently answered the question "Value of %rsi in assembly code. Originally (8086), there was just cbw (ax = The information is in there, if you control-f search for movslq in assembly cltq and movslq difference, the 2nd mention of it is in a sentence explaining that it's movsx, with a link to the Intel manual. So what you're seeing is a value loaded at an offset from the value held in the FS register, and not bit manipulation of the contents of the FS register. Thus, imulq $44, (%rbx), %rax will multiply the contents of the memory at address stored in %rbx by 44, and store the result in %rax. X86-64 Wikipedia. " It moves 0x131 into %eax, and then compares it to the data at that location. It's equivalent to a single sub instruction movq (%rdi), %rax does not move the contents of %rdi into %rax, it moves the contents of memory at the address pointed to by %rdi into %rax. I'm studying assembly language and can't resolve the following exercise myself. The XOR eax, 1 instruction just flips the lowest bit of EAX. doSth: subq %rdx, %rsi imulq %rsi, %rdi movq %rsi, %rax salq $63, %rax sarq $63, %rax xorq %rdi, %rax ret I want to figire out how I would write C code that have an equivalent effect to the assembly code. You need to convert the values into either hex or dec to add them, and then convert back to hex for the address. For example, the mov rax, [addr] instruction moves 8 bytes beginning at addr + rip to rax. Also you are using 64 bit registers so the value in rbx before the setg is actually 0x7ffffffffffffffe. In x86 64-bit assembly, fs:0x28 is a stack guard value that is stored in the stack frame of a function. div / idiv: divides edx:eax by the src. It was added in 2003 during the transition to 64-bit processors. Writing eax always zero-extends into rax (Why do most x64 instructions zero the upper part of a 32 bit register), but writing AL, AH, or AX merge with the old value of RAX. (Fun fact: the machine encoding for AVX512 disp8 displacements is scaled by the operand-size, so you can reach +127 * 64 bytes with a compact displacement, but vcmpeqps 13(%rax), %zmm1, %k1 (512-bit memory operand) would require a disp32 because 13 is not a how does assembly / the hardware do this without explicitly changing their values? What do you mean? You can think of this as a case of zero extension — a widening operation where the missing bits are filled in with zeros, as contrasted with sign extension, a widening where the missing bits are copied from the MSB of the starting value. Accessing data [] Then it adds rcx to rax. cmpq $-4095, %rax compares the 64-bit register %rax with the immediate value -4095 - the value is sign-extended to 64-bits for the purposes of the comparison. It was added in 1985 during the transition to 32-bit processors with the 80386 CPU. As Jester says, the value in AX produced by the div (not dependent on initial conditions) isn't one of the options, so likely they did mean to ask about AL not RAX. eax is the 32-bit, “int” size register. This sucks a lot (silent wrong-code), but so do other MASM syntax design decisions: Confusing brackets in MASM32 - part of the need for ptr is to specify that it's a memory operand at all when there's no register Note that unlike primary memory (which is what we think of when we discuss memory in a C/C++ program), registers have no addresses! There is no address value that, if cast to a pointer and dereferenced, would return the contents of the %rax register. __asm("sete %al"); rbp is the frame pointer on x86_64. Instruction Effect Bomb-lab right lol? This operation jmpq *0x402390(,%rax,8) is for jumping directly to the absolute address stored at 8 * %rax + 0x402390. The linker (ld) can read that symbol in the object code and its value so it knows where to mark as an entry point in the output executable. So I would say the main difference is that the second snippet is not adding the full quadword in memory to rax, only the longword. Read the calling convention docs or look at compiler output. Registers can be used in your software directly with instructions such as mov, add or cmp. Instructions, such as jmp, call, push, and pop, that implicitly refer to the instruction pointer and the stack pointer treat them as 64 bits registers on x64. 9 general-purpose registers will be used to pass integer The basic kinds of assembly instructions are: Computation. it has I know that EAX is essentially Extended AX register, but what is the RAX register called? My computer architecture professor was stumped, and I can't find the answer anywhere. The 'q' means that we're dealing with quad words (64 bits long). A lot of compilers offer frame pointer omission as an I found the assembly instruction. Lets go over the instruction piece by piece: mov. But this is 32-bit code so And yes, all the x86 ABIs use eax / rax as the register for return values. Furthermore I guess you know that edx is the low 32 bits of rdx so in this case no copying takes place, those bits stay where they are. – Stack Overflow for Teams Where developers & technologists share private knowledge with coworkers; Advertising & Talent Reach devs & technologists worldwide about your product, service or employer brand; OverflowAI GenAI features for Teams; OverflowAPI Train & fine-tune LLMs; Labs The future of collective knowledge sharing; About the company I'm running through some assembly code and I can't figure out what a line of code does. 8 compiling C++ code. What does the following do? testq %rdx, %rdx cmovg %rcx, %rax I understand that testq is a bitwise and between two registers, but how does it works with the flags? What would this translate i lea, Load Effective Address, puts the computed "memory address" in the result register. The processor model as documented in the Intel/AMD processor manual is a pretty imperfect model for the real execution engine of a modern core. See What are the calling conventions for UNIX & Linux system calls on i386 and x86-64. The leading e stands for extended and means that your register is 32 bits wide. According to Intel, in x64, the following registers are called general-purpose registers (RAX, RBX, RCX, RDX, RBP, RSI, RDI, RSP and R8-R15). See also. It might be a little bit difficult to find sete in many manuals, since they don't list it directly, just like cmove. You're right it's not a perfect duplicate, but I wanted to close it because there's a mechanical solution (assemble+disassemble) to find out Intel mnemonics for AT&T opcodes, As far as I understood: %rdi = 1st argument = x %rsi = 2nd argument = y %rdx = 3rd argument = z The others manipulate these registers and store in a return value register The 3rd line on the. test rax, rax jz is_zero You can get the assembly output from a compiler and check or view it in an online tool like gcc godbolt 9(%rax, %rdx) value of %rax is 0x100; value of %rdx is 0x3; While you are correct that the answer should be the contents of memory at 9 + %rax + %rdx, in this instance, 9 is in decimal, while 0x100 and 0x3 are in Hex. Also, immediate data doesn't have to be fetched, other than as a pre-requisite for instruction decoding. broadcasts the sign bit of eax into every bit of edx. ), the E prefix for each of the names stands for Extended meaning the 32bit form of the register rather than the 16bit form (AX, BX, etc. I would recommend reading the Solaris x86 Assembly Language Reference Manual if you plan to stick with AT&T, or just switch to Intel syntax since it's much more widely used (especially in processor documentation). intel_syntax, dword is a constant with value 4, so mov dword [rdi], eax is actually mov [rdi + 4], eax. I have to agree with old_timer here; RAX is two letters different from AL so it's not an easy typo, and this is a critical distinction in assembly. These instructions perform computation on values, typically values stored in registers. Well, the result of the addition is 0x7ffffffffffffffd and that is not 0 so ZF is 0. In the article Understanding C by learning assembly, it's written that RBP and RSP are special-purpose registers (RBP points to the base of the current stack frame and RSP points to the top of the current stack frame). movqword ptr ds:[rax+18],r8. LC1: . *0x402680(,%rax,8): This is a way to write an address in x-86 assembly. So here, rcx = rsi + 4 and rax = rsi + 0x14. Just use xor so you don't have to worry about which CPU recognizes which zeroing idiom. The calling convention can change in the future, and the compiler can generate stubs to automatically call assembly functions in older conventions. In this case, it zero-extends the BYTE loaded from memory at [rbp-528+rax] to the DWORD destination register, EAX. One trick is to use the documentation feature on Godbolt's Compiler Explorer. Try -O3 to get optimized machine code, which will load rdi directly (and rax will contain whatever the CRT library initialization left there, i. global directive is NASM specific. The callee places its return value in %rax and is responsible for cleaning up its local Like C++ variables, registers are actually available in several sizes: rax is the 64-bit, "long" size register. , -4095 has the 64-bit 2's complement representation: ffff ffff ffff f001. LEA ax, [BP+SI+5] ; Compute address of value MOV ax, [BP+SI+5] ; Load value at that address and GCC generates the following assembly code; #x->%rdi, y->%rsi, z->%rdx. I can then read and write from the pointed-to memory using the usual assembly bracket syntax: Stack Overflow for Teams Where developers & technologists share private knowledge with coworkers; Advertising & Talent Reach devs & technologists worldwide about your product, service or employer brand; OverflowAI GenAI features for Teams; OverflowAPI Train & fine-tune LLMs; Labs The future of collective knowledge sharing; About the company So it seems to me that this should actually be xorq %rax, %rax, particularly since this is holding a 64-bit long int. TEST sets the zero flag, ZF, when the result of the AND operation is zero. Most of the assembly examples in this book were generated on systems running Ubuntu or Red Hat In the 64 bit x86 architecture there are 16 general purpose registers registers. xor (being a recognized zeroing idiom, unlike mov reg, 0) has some obvious If each instruction has a different length, how can we, or even the computer, know in memory what is an instruction and what is an argument? If every instruction had the same size, the computer would know that, let's say, each 4 bytes corresponded to a new instruction. xor (being a recognized zeroing idiom, unlike mov reg, 0) has some obvious GCC dictates how the stack is used. (Fun fact: as input, gas accepts either mnemonic in @TheRookierLearner: A XOR B is a primitive building block for higher-level constructs. I have read that the regular movq instruction can only have immediate source operands that can be represented as 32-bit two's complement numbers, while the movabsq instruction can have an arbitrary 64-bit immediate value as its source operand and can only have a register as a rbp is the frame pointer on x86_64. They are specialized, high-speed storage areas where the CPU temporarily stores data. Assume the following values are stored at the indicated memory addresses and registers: Now, we have an instruction: addl %ecx , (%eax) For me it means - storing the result of addition of values stored in %ecx and in memory address (%eax), in a memory address (%eax). Below are instruction encodings generated by gas for different nop's from gas source for instruction lengths from 3 to 15 bytes. Writing AL, AH, or AX leaves other bytes unmodified in the full AX/EAX/RAX, for historical reasons. ). CET to 1. ; cdq sign-extends eax into edx:eax, i. Writing EAX zero-extends into RAX, like GCC is doing here with mov $0, %eax to zero RAX without the xor-zero peephole optimization (which gcc only looks for at -O2, which enables -fpeephole2). The code is: leaq 0(,%rax,4), %rdx I know lea is basically a type of mov instruction, but it only moves the address. Also see the official intel instruction set reference documents, along with the well known online conversion. If you do x/16gx 0x402390 in gdb (inspect 16 "giant words" in hexadecimal starting at Thus (%rax) means to get the value of the pointer currently stored in %rax. It can be used to implement (A != B), but it is a distinctly different operation in its own right. 9+rax+rdx = 9+0x100+0x3 = 0x10C. But this is 32-bit code so Lets go over the instruction piece by piece: mov. There's no form of div / idiv that ignores edx in the input. That instruction loads a byte from address rbx+rax and sign extends it into ecx. are gas syntax, not AT&T syntax. It is for exporting symbols in your code to where it points in the object code generated. i. Less-good assemblers have a default, often dword (like GAS for non-mov insns); with really bad assemblers like emu8086, the size depends on the numeric The CDQE instruction sign-extends a DWORD (32-bit value) in the EAX register to a QWORD (64-bit value) in the RAX register. Some CPUs recognize sub same,same as a zeroing idiom like xor, but all CPUs that recognize any zeroing idioms recognize xor. Moreover, further down in the code, it actually uses the 64-bit register %rax to do the iterating, which never gets initialized outside of xorl %eax %eax, which would seem to only zero out the lower 32 bits of the register. @Toothbrush: good assemblers will reject and [ebp-4], 0 as ambiguous operand-size: do you want to zero a byte, word, dword, or qword. 0x8(%rsp) means "get the location on the stack that is 8 bytes away from the stack pointer %rsp, and then take the value at that address. addq %rax, %r9 needs both W and B because of the use of the "new" register. Now, your assembler seems to be using the AT&T syntax of x86 assembly. In particular, the notion of the processor registers does not match reality, there is no such thing as a EAX or RAX register. It puts the high-order 64 bits in %rdxand the low-order 64 bits in %rax. And BTW, an explicit return statement vs. Here you mark _start symbol global so its name is added in the object code (a. No, the displacement (0x47) is not scaled by the operand-size. Assembly registers in 64-bit architecture. Most have zero or one source operands and one source/destination operand, with the source operand coming first. The 64-bit extended versions of the original 8 registers had an R prefix added to them for symmetry. Note that sign extension duplicates the most significant bit of the source into the top bits of the destination so it's not always "leading ones". In terms of speed, accessing a register is faster than any other type of memory or storage. This is x64 assembly generated from GCC 4. Originally (8086), there was just cbw (ax = The code in the question uses a format string of . eax, ebx, ecx and so on are actually registers, which can be seen as "hardware" variables, somewhat similar to higher level-language's variables. cmp sets the eflags register depending on that comparison (like the Zero Flag if the operands were In general, the size could be inferred from the size of the registers but consider the examples below that move immediate (constant) values into memory location at the address stored in EBX, you have to direct the assembler about the size: What's special about zeroing idioms like xor on various uarches. The instruction set reference entry for setg says result is 1 if ZF=0 and SF=OF. However, addl %eax, %r9d doesn't need the W bit, since its operand size is 32 bits. (Fun fact: the machine encoding for AVX512 disp8 displacements is scaled by the operand-size, so you can reach +127 * 64 bytes with a compact displacement, but vcmpeqps 13(%rax), %zmm1, %k1 (512-bit memory operand) would require a disp32 because 13 is not a and GCC generates the following assembly code; #x->%rdi, y->%rsi, z->%rdx. When we enter the function, the stack canary [] R just stands for "register". Specifically what's taking place, is that FS:0x28 on Linux is storing a special sentinel stack According to John Fremlin's blog: Operands to NOP on AMD64, nopw, nopl etc. In this case, it's doing a simple numeric subtraction: leal -4(%ebp), %eax just assigns to the %eax register the value of %ebp - 4. So we are moving the address of something to %rdx (making %rdx "point" to something on the stack). eax is the 32-bit, "int" The precise assembly instructions that are output by any compiler depend on that compiler’s version and the underlying operating system. Probably some other implicit EAX/RAX uses that I'm forgetting. For instance, addq %rax, %rbx, addq %rax, long_var, and addq $123, long_var all need the REX prefix with W bit set, since they all have a 64-bit operand size. When RAX doesn't hold a return value (void or FP functions), it's a call-clobbered register like RCX or RSI for example. Displacements in asm source are always in bytes. The MOVZX instruction zero-extends the source to the destination. ; The appropriate flags for indirect branch tracking in the IA32_U_CET (user mode) or IA32_S_CET (supervisor mode) MSRs are set. You are correct in saying that usually there is a register before the first comma, but you still follow the same rules if no register is specified. quotient in eax, remainder in edx. – user3185968 Assembly Language Registers Registers in x64 Assembly are small, fast storage locations directly accessible by the CPU and 64 bits (8 bytes) in size. (Wikipedia says "extended". What's special about zeroing idioms like xor on various uarches. The AMD64 ISA extension added 8 additional general-purpose registers, named R8 through R15. . Note you have two left shifts and one right, so that's not 3 left in total. )They are the "extended" versions of the 16-bit registers, in that they offer 16 BTW, the d is not "double byte" of course, it's double word. Intel - Introduction to x64 assembly What does Rax mean in assembly? rax is the 64-bit, “long” size register. In particular, in longer nop's gas uses The code in the question uses a format string of . I call malloc to get 40 bytes of space. AMD 64 developer resources. I know that in 32bit assembler register names (EAX, EBX, etc. Most of the time, it's just doing a calculation like a combined multiply-and-add for, say, array indexing. test rax, rax jz is_zero You can get the assembly output from a compiler and check or view it in an online tool like gcc godbolt Parentheses generally mean to dereference. malloc returns the starting address of this space in rax (the 64-bit version of eax). Take a look at the above picture and notice that he says the destination after the instruction would be 0x100 and the value would be 0x100, but wouldn't the correct answer be destination: 0x100 (value of %rax because of parentheses) and 0x1 (value of %rcx)? LEA (load effective address) just computes the address of the operand, it does not actually dereference it. RAX is the full 64-bit value, with EAX and its sub-components mapped to the lower 32 bits. Specifically, Linux sys_brk sets the program break; the arg and LEA means Load Effective Address; MOV means Load Value; In short, LEA loads a pointer to the item you're addressing whereas MOV loads the actual value at that address. Stack Overflow for Teams Where developers & technologists share private knowledge with coworkers; Advertising & Talent Reach devs & technologists worldwide about your product, service or employer brand; OverflowAI GenAI features for Teams; OverflowAPI Train & fine-tune LLMs; Labs The future of collective knowledge sharing; About the company Since we have, mov rax, 100 as valid, we know it's Intel syntax. On the other hand, 64-bit registers begin with r. " New questions should not be appended to old. I know what %rax points to on the stack (say, -28(%rbp)), but I'm confused by We often get to see x64 assembly instructions like the following: mov rax,QWORD PTR fs:0x28 In this article, we will discuss what fs:0x28 is in x86 assembly. Just write the instruction inline like this. The system call return value is in rax, as always. reserving space for local variables or pushing values on to the stack), local variables and function parameters are still accessible from a constant offset from rbp. For example, the instruction addq %rax, %rbx performs the computation %rbx := %rbx + %rax. add byte ptr [rax],1 Either the pointer or immediate value need to state a size type (byte, word, dword, qword), in order for the assembler to determine the size of the memory location being modified. Note that sys_brk has a slightly different interface than the brk / sbrk POSIX functions; see the C library/kernel differences section of the Linux brk(2) man page. Registers live in a separate world from the memory whose contents are partially prescribed by the C abstract machine. Do you compile with gcc and no optimizations? In -O0 it will use rax to prepare the pointer value into memory and into rdi as argument for function call, so they accidentally contain identical value. Using an extra REX prefix would be strictly worse with XOR, like it Both the FS and GS registers can be used as base-pointer addresses in order to access special operating system data-structures. E stands for "extended" or "enhanced". In your generated code, it gets a snapshot of the stack pointer (rsp) so that when adjustments are made to rsp (i. Also, nightcracker's statement that A XOR B in english would be translated as "are A and B not equal" is only correct when you're looking at the result from a Boolean zero/nonzero Stack Overflow for Teams Where developers & technologists share private knowledge with coworkers; Advertising & Talent Reach devs & technologists worldwide about your product, service or employer brand; OverflowAI GenAI features for Teams; OverflowAPI Train & fine-tune LLMs; Labs The future of collective knowledge sharing; About the company CMP subtracts the operands and sets the flags. The description I found was, that it clears the %edx register, EAX CDQE CLTQ EAX RAX CQO CQTO RAX RDX:RAX And now some commented code snippets: /* Quad to Assembly Languages Characteristics •Not portable •Each assembly lang instruction maps to one machine lang instruction rax, rbx, rcx, rdx, rsi, rdi, rbp, rsp, r8, r9, r10, r11, r12, r13, r14, r15 If you’re operating on 32-bit “int” data, use these stupid names instead: cmpq $-4095, %rax compares the 64-bit register %rax with the immediate value -4095 - the value is sign-extended to 64-bits for the purposes of the comparison. It is known for specifying the operands of the instruction in reverse order. wecn qcvy hjbac vnucpm znh niqjx vtw onrs ntci jplgg