Dword ptr assembly

Mark Anderson

May 11, 2025

10 min read

Assembly/Compiler Coding Rule 55.What is dword ptr mean? Basically, it means “the size of the target operand is 32 bits”, so this will bitwise-AND the 32-bit value at the address computed by taking the .XOR DWORD PTR DS:[ECX+EAX],1234567 This first add the values of ECX and EAX, then interprets the sum as a memory address and xors the value at that .

large has nothing to do with the width of the data (dword), but the address (segment+offset). (This address is usually assigned by the linker. Viewed 57k times. So this line copies the argument num to a local (offset -4 bytes from the frame pointer value stored in rbp). The memory operand form of the instruction allows .IDA Assembly Code:. If anyone could explain the r/m form and why imul is getting 2 I'd greatly appreciate it. In MOV WORD PTR [addr],AX the size of AX is known to be 16-bits to again WORD PTR [addr] is a verbose way of saying [addr] so is the same as . – Laurie Stearn.Balises :Assembly LanguageStack OverflowDword Ptr AssemblyI'm trying to understand the x64 assembly optimization that is done by the compiler. DWORD_PTR is also commonly used for general 32-bit parameters that have been extended to 64 bits in 64-bit Windows.The syntax is the following: = ptr .What does mov qword ptr ds: [rax+18], r8 mean? Ask Question. 在Assemble中通常会看到 WORD ptr 或者 DWORD ptr 或者 BYTE ptr 等关键字，这些关键字主要用来标识指令操作数的大小 (或者说长度) WORD ptr ：标识该操作数为1个字 DWORD ptr ：标识该操作数为2个字 BYTE ptr ：标识该操作数 .mov QWORD PTR [rbp-0x30],0x4020c5 means exactly move 0x4020c5 to a memory location rbp-0x30 and treat this number as qword (8 - byte number). The target register already defines the size.0x080483f3 : mov DWORD PTR [ebp-0x8],0x0 The 4 bytes at [ebp-0x8] are initialised to the value 0.

For ecx repetitions, stores the contents of eax into where edi points to, incrementing or decrementing edi (depending on the direction flag) by 4 bytes each time.However, I'm used to a different form of the assembly.[EBP + local_c] is an address, in an indirect addressing mode. For this assembly language (defined by the assembler not the target) to indicate what size this indirect address based instruction should operate on has dword ptr meaning the first operand is a 32 bit value at the address indicated.

PUSH DWORD PTR

The range is 0 through 4294967295 decimal. Ce résultat est sauvegardé dans la première case de la stackframe, puis est à nouveau assignée à EAX exactement comme la fin de la fonction add . 0x5555555545fe mov dword ptr [rbp - 4], edi 0x555555554601 mov qword ptr [rbp - 0x10], rsi 0x555555554605 mov eax, 0x2a 0x55555555460a pop rbp 0x55555555460b ret edi stores argc a 32 bit value - packed in a 64 bit . This is the opcode part of the instruction.As 0Ah is immediate value its size cannot be determined without using DWORD, WORD, BYTE or other similar qualifier. There are move op-codes that support 32bit operations but (most likely check .Balises :Dword Ptr in AssemblyMov Eax Dword Ptr Where is the structure mana derived from?Balises :Dword PointerDword Ptr Assembly In x86 parlance, a WORD is 16 bits, a DWORD 32 bits, and a QWORD 64 bits. The first operand on the mov instruction is a target . を意味する修飾子となります。. Yeah, the article in the answer's linked official list now has the ranges: Quote: DWORD: A 32-bit unsigned integer. For more information, see ULONG_PTR. and dword ptr [ebp-4], 0. Cái này dùng để chỉ ô nhớ kích cỡ 1, 2, 4, 8 hay 10 (TEN) byte; tương ứng với kí hiệu byte ptr, word ptr, dword ptr, qword ptr, tbyte ptr.Balises :Dword Ptr in AssemblyStack OverflowX86 AssemblyDword Pointer They're able to see the address the pointer is pointing to.If you declare a variable DWORD iNumAddr in a high-level language such as C++, then the symbol iNumAddr in assembly-language code takes the value of the address of the high-level language variable iNumAddr. Arithmetic opcodes can not operate with two memory operands. You need something like byte ptr when (for example) you move an immediate value to an indirect address: mov bx, some offset. Data needs to be moved to a register first before the . これは対象の値（やレジスタ）を何バイトとして扱うか、ということを修飾するもののようです。.

[MS-DTYP]: DWORD

What does arg

I know their usage -- size directives; but where had the term PTR . is the operand whose attribute is to be overridden. BYTEは1バイト.Alright so I have this line in my assembly.Balises :Assembly LanguageReverse Engineering Stack ExchangeSegment Registers So the answer is yes.

NOP — No Operation

mov DWORD PTR [rbp-4], edi ;not sure what happens here The x86_64 System V ABI passes function arguments via registers - the first integer argument is passed in the rdi/edi register. However, large doesn't really add any new information, that line simply means pop dword ptr .BYTE PTR, WORD PTR, DWORD PTR, and QWORD PTR are directives in x86 and x64 assembly that specify the referenced data is 8-bit, 16-bit, 32-bit, or 64-bit in .

assembly

Guide to x86 Assembly

Would be the same as saying LEA AX, [d] which moves the address of d to AX. The multi-byte NOP instruction performs no operation on supported processors and generates undefined opcode exception on processors that do not support the multi-byte NOP instruction. It is used when casting a pointer to an unsigned long type to perform pointer arithmetic. The most important and frequent use for PTR is to assure that MACRO-86 understands what . The value that is provided is a 32bit value (8 hex-digits => 8 times a hex-digit/letter => 8 * 4bits => 32bit). So in your example, IDA identified the first argument to the function . A DWORD_PTR is an unsigned long type used for pointer precision. Although I'm aware of what dword ptr means, I do not get the last part that is added, :es . DWORD is on the windows libraries equal to an unsigned long.Balises :Assembly LanguageDword Ptr in AssemblyDword Ptr 4+2Dword Ptr EaxQword Ptr vs Dword Ptr MOV EAX, DWORD PTR DS:[ESI] where ESI is 00402050 (ascii, 123456789012) After this instruction: EAX = 34333231. movqword ptr ds:[rax+18],r8.pop large dword ptr fs:0 is IDA's way of bringing to your attention that fs:0 is a far pointer: a regular offset ( 0) but with a segment selector ( fs ). 由于只是指出了目的操作数的内存基址，但是没有指出寄存器到底要操作一个BYTE，还是一个WORD，还是DWORD，所以需要加上 dword ptr 来表示进行与操作的时候，读取从ss . This is your x local variable. PBYTE typedef PTR BYTE. the instruction number (called opcode),; what its operands are (if .What does 'dword ptr [this]' mean in VS disassembly of C++ code? Asked 2 years, 4 months ago. is the new attribute; the new type or new distance. WORD PTR [d] is verbose way of saying [d] since the size of the data doesn't matter in getting its address. Also, please note that the EAX register is always 4-byte sized.0x0804840c : mov DWORD PTR [ebp-0x4],eax 0x0804840f : mov eax,DWORD PTR [ebp-0x4] Nous avons vu précédemment que le résultat de add était retourné dans EAX .Balises :Dword Ptr in AssemblyX86 AssemblyAssembly Qword Ptr64-bitBalises :Dword Ptr in AssemblyMov Eax Dword PtrX86 Assembly qword ptr is a hint for the assembler to create a move op-code (machine code) using a 64bit constant address.In fact, in assembly a common bug is accidentally changing a pointer instead of the value, or messing up your array indexing so you wind up pointing halfway inside an array . The one-byte NOP instruction is an alias mnemonic for the XCHG (E)AX, (E)AX instruction.I do not get why they use dword ptr es:[edi] and word ptr ds:[0B65740h]. vì 1 này là 1 gì, 01h hay 0001h. 0x080483f3 : mov ebp,esp. Asked 8 years, 7 months ago. Checking the value, they get the address. It describes the base operation the CPU is required to perform. EBP plus some offset. While harmless, the assembler already knows that al and dl are byte-sized, and that bx is word-sized.

Why would we need two .) So the instruction add dword ptr [iNumAddr], 50 will increment the variable .Declaring Static Data Regions

What does 'dword ptr[this]' mean in VS disassembly of C++ code?

text:0807EC41 call dword ptr [mana+0Ch] I cannot figure to what address is the call going to, and I cannot seem to jump to that either, why is this happening? how can I solve it? I'm very newbie with IDA but I've been managing lately, and I just encountered this one and I'm clueless on what to do.Assemble指令中标识大小. All binary numbers which not have set the 13th bit on memory address [eax], or all numbers present like b'xxxxxxxx xxxxxxxx xx0xxxxx xxxxxxxx', where x is 0 or 1, there is exactly 2^31 numbers.3e 8b 06 mov eax,DWORD PTR ds:[esi] 8b 06 mov eax,DWORD PTR [esi] In my opinion it does the same, I also tried this program: int main() { __asm { mov esi, ebx mov eax,DWORD PTR [esi] mov ebx,DWORD PTR ds:[esi] } return 0; } And it confirmed my guess, so asking for you guys, if there is any difference. 0x080483f5 : sub esp,0x18.CPUs are controlled with machine language, which is just a stream of binary-encoded instructions that specify. The instruction mov dword [rbp-4H], edi is 4 bytes less . レジスタは何バイト . MOV EAX, DWORD PTR I.

Modified 4 years, 4 months ago. The biggest thing I don't understand is all this var_28 = dword ptr -28h stuff at the top. And one of the lines contained the following assembly code: B8 31 00 00 00 mov eax,31h 0F 1F 44 00 00 nop dword ptr [rax+rax] And here's a screenshot:Lets go over the instruction piece by piece: mov. Read about segment registers here: https://en.Balises :Dword Ptr 4Reverse Engineering Stack Exchange For less bytes, you may use ax, al or ah and for higher (8-byte) values you may use RAX (on 64 bit .The PTR operator overrides the type (BYTE, WORD, DWORD) or the distance (NEAR, FAR) of an operand. So, a QWORD consists of 2 DWORDS, one of them we will refer to as the LOW DWORD (this one goes to a lower memory address, since the x86 is little endian), and the other we will refer as the HIGH DWORD. Then running x/s on that address returns the string value, which in this case should be the location of the program '/root/Desktop/00byte'.Balises :Assembly LanguageCS 301Ptr Rsp DWORDは4バイト. Cảm ơn trước.Regarder la vidéo4:04I cover the meaning of PUSH DWORD PTR in x86 assembly.アセンブラのコードを見ていると見かけるこの DWORD PTR 。.Is this correct? I don't see any instruction to move the stack-pointer up the stack, so I am assuming this is done implicitly.

The second operand is an . WORDは2バイト.XOR DWORD PTR DS:[ECX+EAX],1234567 This first add the values of ECX and EAX, then interprets the sum as a memory address and xors the value at that location with 1234567. Dec 27, 2017 at 3:02. In the cases you're looking at, the byte ptr and word ptr don't accomplish much. Not familiar with that assembler, but dword ptr in your example is redundant. Experienced assembly coders know all the details mentioned above just by . Usually, that instruction is simply written rep stosd. In MASM the size directive is .