Movzbl in assembly. We know this because %ecx is scaled by 4.

Movzbl in assembly Why? (Extra: with this change, you may wonder why the compiler outputs movzbl when it uses only one byte of what is moved. (MOVZBW Byte to Word, MOVZBL Byte to Long, MOVZWL Word to Long); sets or clears flags; NUM Convert to The CDQE instruction sign-extends a DWORD (32-bit value) in the EAX register to a QWORD (64-bit value) in the RAX register. 6). Only the top 32 bits are filled with the sign. But before that, the kernel has to move the address of $1f(which is the address of the instruction right after movzbl %1, %0) to the eax register. (movzbl vs movzx for example), etc For other example: in popular (at universities courses) MIPS simulators "MARS" and "SPIM" the # is the only comment character instead of ;, but that is well documented in their official docs (and not in assembly code, using the same conventions for argument passing and register usage as are followed by the C compiler. The source What is happening in this Assembly code? 2. Hot Network Questions What movie has a classroom clock tick backwards? How do the turbopumps in the RS-25 work? How to Simulate the variability in Vgs(off) for a N-type movzbl (%rdx), %edx reads a single byte from memory address [rdx] and zero extends the byte so it is 32-bits wide with the upper 24 bits set to zero and stores it in EDX. Commented Jun 5, 2018 at 7:44. Zero/Sign extension in assembly without using MOVZX/MOVSX. I have a 100 byte buffer and I need to find the sum of every nth byte fo Skip to main content. The equivalent for For MOVZBW, the low 16 bits of the destination are replaced by the source operand. (Or better, movzbl %al, %ecx so mov-elimination can make it zero latency on Intel CPUs where mov-elimination works for movzx 8->32. If you want a zero movsbl %al, %edx # copy 1-byte %al, sign-extend into 4-byte %edx movzbl %al, %edx # copy 1-byte %al, zero-extend into 4-byte %edx A Assembly and gdb. Hot Network Questions How many percentages of radicals of the Chinese characters have a meaningful indication? Happy 2025 to all! TikZ/PGF: Can you set arrow size based on the height of the Move With Zero Extend (movzb) movzb[wl] r/m8, reg[16|32] movzwl r/m16, reg32 Operation. 19. setz %al; movzbl %al, %eax sets all of %rax to 1 if the zero flag ZF is on, and 0 otherwise. In general, forget about "types" in assembly: assembly is not a typed language, or better any data is just bytes. in the second expression represents the current address (in the data segment). ) But as any other general purpose register, can be used to hold any other data. $0xff doesn't fit in an 8-bit sign-extended immediate so it needs a full 4-byte 0x000000ff immediate. Line +43 is indeed accessing an array, which holds some integer constants. In that case the copy would happen in the reverse direction, starting at the For both functions, the assembly sequence is one instruction longer. I don't understand how they can be distinguished by observing the ModR/M byte. 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. Improve this question. I don't know an authoritative source, but that has been my experience with MASM / Visual C/C++ for quite a few years. The . NON-QUESTION: Where’s Waldo? int identity (int waldo) { return waldo; } The z in movzbl (Lines 1 and 9) indicates zero-extension, i. hello_world_len = . Instruction format. ) You could use them interchangeably, but you should use them depending on what you are doing:. This assignment statement requires two movs. Like the regular mov the movz instruction copies data from the src operand to the dest operand, but the remaining bits in dest There are two mov instructions that can be used to copy a smaller source to a larger destination: movz and movs. Commented May 11, 2016 at 15:53. You're taking the single byte that a points to and assigning it to the 8 bytes that b points to. x6 4 R e g i st e rs x6 4 a sse mb l y co d e u se s si xt e e n 6 4 -b i t re g i st e rs. cmp je/jg how they work in assembly. You can thus print it using x/16w 0x804a380. Stack Overflow. only useful for memory destinations. Add a comment | 4 Answers Sorted by: Reset to How to implement memmove, not just memcpy, in assembly? Related. – 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 This is a safe and not totally terrible version (other than the unavoidable optimization-defeating parts of inline asm). I want to be able to output the letter "a". The basic kinds of assembly instructions are: Computation. phuclv. See How does C++ linking work in practice? for some details about how assembling + linking take care of the +8 so there's no extra work at run time. Please see the x86 tag wiki for some tutorials and other general-reference information. Assembly - JZ instruction after CMP. The XOR eax, 1 instruction just flips the lowest bit of Your job is to find Waldo: write an assembly expression or constant that holds the value of waldo at the marked point. What does this instruction do? Wikibooks has a course on x86 assembly, but I was not able to find anything about sete in the chapter devoted to assembly instructions. Line 5 sets condition codes based 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 JE and JZ are just different names for exactly the same thing: a conditional jump when ZF (the "zero" flag) is equal to 1. opposed to, say, "movb $1, %eax" This instruction is invalid. However, in class the teacher has only used the 'movl' operand. This move seems useless because some context information is missing. movzbl is the natural efficient way to load a byte, replacing instead of merging with the old contents of a full register. Otherwise, if the direction flag is set, then the pointers are decremented. Once your program calls the printf method the first time the GOT is updated and now the printf is resolved and the GOT points to the real function. movzbl string+8, %eax assembles to machine code with the same addressing mode (ModR/M byte) as movzbl string, %eax, just a different disp32 displacement. See Are x86 Assembly Mnemonic standarized? (no, AT&T makes up new names. The movsbl instruction takes the lowest byte of What does movzbl stand for in assembly language? Meaning, if you had a negative number, the upper 24 bits will be 1s, otherwise they will be zeroes. Here is some of the assembly, Q1: pushq %rbp movq %rsp, %rbp subq $24, %rsp m I'm trying to convert the following optimised assembly code to C code but keeping as close to the assembly code as possible. 5. Therefore, the byte at address Assembly is not a single programming language, but general term for all the machine language mnemonics => has no rules. Data-movement and control-flow instructions do not modify flags. Oddly, for If it were -Og, then the first assembly line for your C code would be movzbl (%rdi), %eax instead of movb (%rdi), %al. Unsure about cmp/jg Alright so this is the ever so popular bomb lab and I am currently on phase 5 and I'm only stuck on two lines. The MOVZX instruction zero-extends the source to the destination. 16. You can print the value in a register by prefixing its name with $ or use the command info reg to dump the values of all registers: I'm trying to figure out what exactly this assembly code do and I'm dumbfounded by this sete instruction. L10: Assembly Programming III CSE410, Winter 2017 Using Condition Codes: Jumping j*Instructions Jumps to target(an address) based on condition codes 8 Instruction Condition Description jmp target 1 Unconditional je target ZF Equal / Zero jne target ~ZF Not Equal / Not Zero js target SF Negative jns target ~SF Nonnegative jg target ~(SF^OF)&~ZF Greater movzbl: move byte to double word, zero extend: movzwl: move word to double word, zero extend: previous L09: Assembly Programming III CSE351, Spring 2017 Condition Codes (Implicit Setting) Implicitly set by arithmetic operations (think of it as side effects) Example: addq src, dst ↔ r = d+s CF=1 if carry out from MSB (unsigned overflow) ZF=1 if r==0 SF=1 if r<0 (assuming signed, actually just if MSB is 1) OF=1 if two’s complement (signed) overflow (s>0 && d>0 && r<0)||(s<0 && d<0 && r registers are 64-bit. Previously, only one mov instruction was needed. the top 16 bits are set to 0. I'm reading a book where some exercises are proposed. movzbl means to load a single byte and zero-extend to the rest of the register. About; movzbl (%edx), %edi addl $5, %edx addl %edi, 44(%esp) movzbl -4(%edx), %edi addl %edi, %ebx movzbl If you are compiling inline assembly code in VC/VC++ it should accept it within in an __asm statement. The source operand is unaffected. The remaining 24 bits are zeros movz stands for move with zero extension. By far most frequent instruction you'll encounter is mov in one of its its multi-faceted variants. Thus, the next call to the disas command shows the real printf assembly. , -4095 has the 64-bit 2's complement representation: ffff ffff ffff f001. Only two lines of this code need to be changed. These instructions perform computation on values, typically values stored in registers. You do still want a movzbl load for both loads, even though the return value is only 8 bits. 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, movzbl %al, %edx # copy low-byte from register %eax, zero-extend to 4 byte long in %edx. AT&T: mov/movl 6+ byte mov r/m32, imm32. JZ/JNZ are more appropriate when you are explicitly They are not similar, pointers are just addresses, they may seem mystical in c/c++ due to the abstractions but in assembly a pointer could be a register with an address stored in it that "points" to some data, EIP is a pointer that "points" to the current instruction during the programs execution, the whole idea is that it's "cheaper" to copy an address that points to data Note that this is a fairly simple, general-reference question. In NASM / Intel syntax, mov r64, 0x picks a MOV encoding based on the constant. If the direction flag is cleared, then esi and edi are incremented after the operation. 52c released in The first version determines the length at run-time and the second version sets the length at assembly time. e does an implicit zero-extend to bitwidth q. i. – Jester Your job is to find Waldo: write an assembly expression or constant that holds the value of waldo at the marked point. More importantly, the asm contains a backwards conditional branch but your C is missing a loop. In my book, it makes quite the deal out of the importance of using the correct suffix in conjunction with the command. I don't understand the answer of one of them. I have included comments as to what I believe to be happening in the assembly code. Here, the source and destination operands should be symbols that indicate the size and location of the source value I'm trying to read through assembly lines and deciphering what they do, and I got this far: %eax is currently 70, and is about to be overwritten This is the line I got stuck on: 0x0804927b <+39>: movzbl (%ebx), %eax 0x0804927e <+42>: test %al, %al From what I understand, I'm trying to move a byte from my string to overwrite %eax, and then zero extend Compilers usually use movzx aka movzbl loads from memory to zero-extend to full registers, and can often use 32-bit operand size to avoid actually dealing with partial register shenanigans. Specific examples of instructions from various processors are used to illustrate the general nature of assembly language. For instance, this function: int is_equal (int a, int b) { return a == b; } Might compile to this compact assembly: cmpl %edi, %esi sete %al movzbl Opcode Instruction Op/En 64-Bit Mode Compat/Leg Mode Description; 0F B6 /r: MOVZX r16, r/m8: RM: Valid: Valid: Move byte to word with zero-extension. Intel x86 Processors •Evolutionary design –Backwards compatible up until 8086, introduced in 1978 •Complex instruction set computer (CISC) –movzbl (zero extend byte to double word) –movzbq (zero extend byte to quad word) –movzwl (zero extend word to double word) –movzwq (zero extend word to quad word) The basic kinds of assembly instructions are: Arithmetic. com movzbl ATT to find What does the MOVZBL instruction do in IA-32 AT&T syntax?; it's AT&T syntax for movzx, loading a byte and zero-extending it into a register. But anyway, this looks broken because it uses xchg with 32-bit registers like %ebx on addresses only 1 byte apart. The array item is simply added to %edx which has been Assembly instruction mov register,[register][register] 6. 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 MOVZBL: Move Zero-Extended Byte to Long: 3C: MOVZWL: Move Zero-Extended Word to Long: Description: For MOVZBW, the low 8 bits of the destination are replaced by the source operand. ) Data movement Primary data movement instruction: "mov" – Copies data from first operand to second operand – Multiple suffixes: movb, movw, movl, movq, movabsq movabsq is the only form that takes a 64-bit immediate Zero-extension variant: "movz" – movzbw, movzbl, movzwl, movzbq, movzwq – Note lack of movzlq; just use movl, which sets higher 32-bits to zero Sign Footnote 1: Fun fact: AT&T syntax (which uses different mnemonics like movswl (sign-extend word->long (dword) or movzbl) can infer the destination size from the register like movzb %al, %ecx, but won't assemble movz %al, %ecx even though there's no ambiguity. The z in movzbl (Lines 1 and 9) indicates zero-extension, i. Also, the asm has a bunch of zero-extending byte loads from memory but your C doesn't have any pointer variables. 22. QUESTION ASM-9D. In this case it would likely be functionally equivalent to The result I am getting is "4194172" for a string of "abcd", which I believe is the address of the whole string itself. movzx reads the contents of the register or effective address as a word or byte. I am learning how to program in Assembly and I have a clarification question. globl declaration in this code. A special case to note is that a mov to write a 32-bit value into a register also zeroes the upper 32 bits of the register by default, i. movzbl (%rdi), %eax movq %rax, (%rsi) Assembly Language data conversion summary. Or "translate AT&T to Intel", e. Mov and lea. You would instead use an 8-bit register, or write the full register with a value that has the value you want in the low byte(s) you care about. the top 8 bits are set to 0. 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. We know this because %ecx is scaled by 4. There’s no need for movzlq (why?). Assembly instruction mov. 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 algorithm @fuz described is Counting Sort, google it. When the destination is a sub-register, only those specific bytes in the sub-register are written with one broad exception: a 32-bit instruction zeroes the high order 32 bits of the destination register. assemble with GAS, disassemble with objconv into NASM syntax, or with objdump -drwC -Mintel). Is my teacher doing this for simplicity or have the 'movb' and 'movw' commands become obsolete in recent Assembly Language data movement summary. – Matt Phillips. Default signedness could be different for a different compiler, or for gcc running on a different platform. What does this statement mean in assembly? 0. My understanding is that it moves a byte with a zero extension into a 32bit register. movs fills the remaining bytes by sign-extending the most significant bit in the Assembly Language data movement summary. As well it also zero extends the 32-bit EDX register through the entire RDX register (it sets the upper 32-bits of RDX to 0) because in 64-bit mode when a destination register is a 32-bit register the Data sizes in assembly have slightly different terminology to get used to: •A byte is 1 byte. But it's not generally expected that newbies will come up with the right search terms if they were thinking about a The opcodes for both movzbw and movzbl are OF B6. what does movzbl(%rdi, %rcx, 1), %ecx mean in x86 Why movzbl is used in assembly when casting unsigned char to signed data types? 2. (Also note that xchg with memory is very slow, because it implies a lock prefix to make it atomic. In this case, it zero-extends the BYTE loaded from memory at [rbp-528+rax] to the DWORD destination register, EAX. (zero-extended to fill the 64-bit register like always). See also What does the MOVZBL instruction do in IA-32 AT&T syntax? which For the following combinations of src_t and dest_t, write a line of assembly code that does the appropriate transfer. Below you can see the C code (which was given) and the assembly code that I have written. Follow edited Oct 27, 2018 at 17:18. It's a pretty simple program to convert bytes into kilobytes. How does MOVSX assembly instruction work? 4. SignExtend(SRC) -> DEST. There are four to choose from with immediate operands: 5 byte mov r32, imm32. 6. assembly; x86; Share. movslq reads a long (32 bits) from the source, sign extends it to a qword (64 bits, replicating bit 31 into bits 63:32), and writes it into the destination register. . For example: Here’s some assembly produced by compiling a C program. (Similarly, JNE and JNZ are just different names for a conditional jump when ZF is equal to 0. You could have looked up the answer in any book on x86 assembly language, and a tutorial should have taught you about the significance of the parentheses/brackets syntax. If you only use %al, then the high garbage in EAX doesn't come into play, but you need to understand that in assembly language it's up to you to load/store with the right width to match what you put in static memory locations. If you're trying to treat the usigned char *a as a pointer to a chunk of memory you want to treat as a full long long Also, movzbl (%esi), %ebx is how you do a zero-extending byte load. From the Intel 80386 Programmer's Reference Manual (1986):. , unsigned char. Specific examples of instructions from various There are two mov instructions that can be used to copy a smaller source to a larger destination: movz and movs. This web page examines data conversion instructions in assembly language. Commented Jul 25, 2011 at 17:32. Assembly MOVS instruction. All it does is cast the unsigned char value from dereferencing unsigned char *a to a long long value. Assuming you are talking about x86, the MOVSBL instruction extends a byte (8 bits) representing a signed number to 32-bit signed number. com x86-64 load byte (or maybe "zero extend byte", but I didn't try that"). The range-check can be done more efficiently; sub and cmp are the same operation, just not writing the result. g. This web page examines data movement instructions in assembly language. ). x86 CMP Instruction Difference. – Lord Loh. AT&T: mov/movl 7+ byte mov r/m64, sign-extended movl lowerA, %eax is a bug: you're doing a 32-bit load, so you're getting 3 extra bytes of data beyond the "a" at lowerA. - hello_world See one of the linked duplicates, How to load a single byte from address in assembly and/or google site:stackoverflow. This is the mystery function assembly code: mystery: movl $0, %eax ;set eax to 0 leaq (%rdi, %rsi), %rcx ; rcx = rdi + rsi loop: cmpq %rdi, %rcx jle endl decq %rcx cmpb $0x65, (%rcx) jne loop incl %eax jmp loop endl: ret What does this line cmpq %rdi, %rcx compare? The address or the character value? If it is comparing the address stored inside the registers, That's why AT&T syntax has movzb and movzw instructions (typically used as movzbl -1(%rbp), %eax), for the two different source sizes of the Intel-syntax movzx mnemonic. MOVZX ── Move with Zero-Extend Opcode Instruction Clocks Description 0F B6 /r MOVZX r16,r/m8 3/6 Move byte to word with zero-extend 0F B6 /r Machine Level – Assembly (x86-64) basics . I did confirm that the last version of VC++ that produced 16-bit code (1. But they wouldn't care if you saved/restored only the low 8 bits with a mov store / movzbl reload, for registers where a compile is keeping a zero-extended For instance, movzbl moves an 8-bit quantity (a byte) into a 32-bit register (a longword) with zero extension; movslq moves a 32-bit quantity (longword) into a 64-bit register (quadword) with sign extension. BTW, the d is not "double byte" of course, it's double word. site:stackoverflow. For MOVZBL, the low 8 bits of the destination are replaced by the source operand. – Michael Petch. Assembly instructions can include suffixes to refer to these types: movzbl Move zero-extended byte to double word movzwl Move zero-extended word to double word movzbq Move zero-extended byte to quad We can annotate the assembly: /* Function prologue */ pushl %ebp movl %esp, %ebp jmp . movsbl %al, %edx # copy 1-byte %al, sign-extend into 4-byte %edx movzbl %al, %edx # copy 1-byte %al, zero-extend into 4-byte %edx A Assembly and gdb. 0F B6 /r: MOVZX r32, r/m8 For 8 or 16 out to 32 (and implicitly 64), and $0xff, %eax works but is less efficient than movzbl %al, %eax. Your disassembler use an l operand-size suffix only For example, movzbl moves a 1-byte source value to a 4-byte destination. The linker then serves to combine the two forms of code into a single program. •A double word is 4 bytes. movzwl reads a word (16 bits) from the source, zero-extends it to a long (32 bits), and writes it into the destination register. @aafulei The movb instruction has a shorter encoding. Your job is to find Waldo: write an assembly expression or constant that holds the value of waldo at the marked point. Then, the expression. How is the Forth data stack accessed in assembler? 7. I am working on a simple assembly program for a class, and am encountering an odd segmentation fault. The basic kinds of assembly instructions are: Arithmetic. L3 L4: /* Move to next loop iteration */ addl $3, arg1 L3: /* Copy value of arg1 to %eax */ movl arg1, %eax /* (%eax) means to dereference %eax (as a memory address). movs fills the remaining bytes by sign-extending the most significant bit in the A few instructions such as movs and movz have two suffixes: the first is for the source operand, the second for the destination. •A word is 2 bytes. The debugger has many features that allow you to trace and debug code at the assembly level. The explicit-operands form (specified with the MOVS mnemonic) allows the source and destination operands to be specified explicitly. mov instruction strange destination operand. Most have zero or one source operands and one source/destination operand. Here's the assembly code: Dump of assembler code for function phase_5: 0x08048e29 movl 4(%esp), %eax movzbl (%eax), %eax ret Output with -fsigned-char: movl 4(%esp), %eax movsbl (%eax), %eax ret Keep in mind that this is merely the output of one compiler on one platform. 8 set: Read condition codes Instruction Synonym Set Condition (1 if true, 0 if false) sete D setz Equal / zero setne D setnz Not equal / not zero sets D Negative setns D Nonnegative To call a function in assembly, we must do a few things: Possibly hampering the search is the fact that AT&T and Intel have their own versions of assembly--this is AT&T syntax. Also see the official intel instruction set reference documents, along with the well known online conversion. Also, since %ecx has been masked to only retain the low 4 bits, we know this table has size 16. – aafulei. 2. Recall that when performing a cast that involves both a size change and a change of “signedness” in C, the operation should change the signedness first (Section 2. Force GCC to use repeat prefix in memset() call. meanings of mov instruction with or without [] brackets around the source register in nasm. MOVZBL Byte to Long, MOVZWL Word to Long); sets or clears flags; MCOM Move Complemented; DEC VAX; edi is a register, a general purpose one which is generally used to hold the destination for string operations (scas etc. You'd use int edi if it was actually int, like you did for edx. I've never had this kind of problem before Move byte. Looks like gcc has some complicated rules for the assembly translation choice. I had figured that out, but I was unable to find a decent reference that explained all the GNU AS syntax. movl is a mov with operand size 32, objdump leaves out the l suffix if the operand size is clear from the operands. You can't use eax with the movb instruction. Now when src_t is unsigned char and dest_t is long, I did the following assembly code for it: movzbq (%rdi), %rax //move a byte into %rax with zero extension movq %rax, (%rsi) //move 8 bytes of 'long' data But the book as well as Godbolt (using gcc with -O3) says it should be. Assembly instructions can have suffixes to refer to these sizes: •b means byte movzbl Move zero-extended byte to double word movzwl Move zero-extended word to double word movzbq Move zero-extended byte to quad (long long)*a; probably doesn't do what you think it does. e. Description. I was given a function in assembly which basically converted uppercase letters to lowercase letters. At the assembly-code level, two forms of this instruction are allowed: the “explicit-operands” form and the “no-operands” form. Usually with google, e. C CODE •Typically followed by movzbl to zero those bytes int small(int x) {return x < 16;} cmp $0xf,%edi setle %al movzbl %al, %eax retq. Skip to main content. The first time you run the disas command the printf is not yet called so it's not resolved. However, within the function that does the conversion, I get a segmentation fault when I try to move the value 1024 into the ebx register. The movsb instruction copies one byte from the memory location specified in esi to the location specified in edi. For example, movzbl moves a 1-byte source value to a 4-byte Specifically, the movzbl instruction fetches the byte stored at the sum of the two parameters, zero pads it, and stores it into eax. •A quad word is 8 bytes. I have no idea how to change the movsbl call to C code. This approach 7 movzbl %al, %eax 8 ret Note the presence of the . Use movzbl (%eax), %ebx to load a byte, zero-extended to a dword. L09: Assembly Programming III CSE351, Spring 2017 Conditionals and Control Flow Conditional branch/jump Jump to somewhere else if some condition is true, otherwise execute next instruction Unconditional branch/jump Always jump when you get to this instruction Together, they can implement most control flow constructs in high‐level languages: Declaring and indexing an integer array of qwords in assembly (1 answer) Moving a value of a lesser size into a register (2 answers) How to remove "noise" from GCC/clang assembly output? (3 answers) don't do that, that's a voodoo hybrid of AT&T and Intel mnemonics. the top 24 bits are set to 0. movzx then sign-extends the 16- or 32-bit value to the operand-size attribute of the instruction. ) There are several x86 I am new to assembly language programming and facing a problem that might be obvious to experienced assembly language users. Commented Jul 25, 2011 at 17:42. src_t dest_t My Solution Blog's Solution char unsigned movzbl %al, (%edx) movsbl That is done by movzbl %1, %0 where 0% is result and 1% is address. int is a 32-bit type in x86-64 C implementations. I'm currently learning some basics of assembly code. 1. It is only used in this case when optimising for size at the detriment of The assembly code is complicated because you compiled without optimizations, causing the compiler to generate a lot of extra instructions it could do without. You can print the value in a register by prefixing its name with $ or use the command info reg to dump the values of all registers: 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 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. Here are the exercises with the answers of the . The result is stored in the destination register by movzx. Write a likely signature for the function. – fuz. About; Products This would lead to the instruction movzbl %al, (%edx) At line 4 we must convert a unsigned char to Data types in assembly are managed via a slightly different set of names: •A byte is 1 byte. This explains use of instructions such as mov %ebx, %ebx that look odd/redundant, but are, in fact, being used to Looks like you got most of it figured out. We’ve done the first one for you. Mov copies a CS CI 0 3 3 0 I n t ro Co mp u t e r S yst e ms Do e p p n e r x64 Cheat Sheet Fall 2019 1 . 0. strange behavior of x86 "cmp" instruction. Negative numbers in x86 assembly. How does the JE instruction work without CMP? 2. 3. But char* is possible too; the characters are only compared for equality with each other (Line 10) or zero (Lines 2/4 and 13/15), so we can’t really distinguish signed from unsigned. Commented Jul 17, 2021 at 15:53. movzbl %al, %eax sets all of %rax to 1 if the zero flag ZF is on, and 0 otherwise.