addressing-mode

问题 I got this code: BITS 16 data: bytemap: db 0x0, 0x1, 0x4; pixel_x: db 2; to return the 0x4 value main: ; code... mov al, [bytemap+[pixel_x]]; i need that byte in al register ; more code... jmp main; but nasm returns "expression syntax error", i tryed using mov bl, [pixel_x]; mov al, [bytemap+bl] , but don't work, how the right way to do it? ( if it exists )... 回答1: You need to use pointer-width registers in addressing modes . x86 doesn't have memory-indirect addressing modes, only register

问题 This question already has answers here : Can't subtract registers in an addressing mode? (3 answers) Subtracting registers with an LEA instruction? (1 answer) Referencing the contents of a memory location. (x86 addressing modes) (2 answers) Printing an integer as a string with AT&T syntax, with Linux system calls instead of printf (2 answers) Closed 4 months ago . I am trying to do a negative offset for memory addressing. This is to convert a number to a string and make sure that it doesn't

问题 Quick question. This code will not compile: mov eax, dword [rbx+rsi*5] I don't expect it to, with the explaination that mov and multiplication are two different CPU operations. The only reason it can be achieved is through bit-shifting. However, this does compile: mov eax, dword [lst+rsi*5] With "lst" being a variable array. It also produces output when used in context (so the code compiles AND runs). What's the explanation for why this works? yasm -Worphan-labels -g dwarf2 -f elf64 NAME.asm

问题 In assmebly the square brackets seem to have the same meaning as * in C. They are used to dereference a pointer. Dereferencing a pointer means going to refer to a specific memory location to read or write it. So it is quite logical to use square brackets in the case of a MOV. But what is the logical reason why they also use it for LEA. LEA EAX, [EBP -4], looks like dereferencing a pointer, ebp - 4, to refer to the pointed memory location but it will not read the value contained in the

问题 This question already has answers here : What does the bracket in `movl (%eax), %eax` mean? (3 answers) What does a hexadecimal number, with a register in parenthesis mean in Assembly? (1 answer) Closed 22 days ago . int x=1; int y=2; int z=3; turns into movl $1, -4(%rbp) movl $2, -8(%rbp) movl $3, -12(%rbp) What is the -4,-8,-12 for ? Why is it going by 4's? 4 bytes = 32 bits? 回答1: -4 / -8 / -12 bytes relative to the address held in rbp , which is the pointer to the top of the stack (which

问题 I am solving a fundamental question of Assembly Language Programming to add BCD numbers and two ASCII numbers, for that I got that I have to use DAA and AAA instructions respectively, Now I am trying to store the result stored in AX register into my desirable memory location, but not getting that why the following code is giving me error Immediate mode Illegal Below is the code that I have coded till now, please help me that how to eradicate this error PS: I want to move my result into my

问题 Does the LEA instruction support negative displacement? mov rax, 1 lea rsi, [rsp - rax] When I use the above code in my asm file I got the error: $ nasm -f macho64 test.asm $ error: invalid effective address I Know that we can do pointer arithmetic like this in C: void foo(char *a, size_t b) { *(a - b) = 1; } then I assume that: lea rsi, [rsp - rax] will work. And I also try to see what the GCC compiler do by using: $ gcc -S foo.c // foo.c has the function foo(above) in it but my asm

问题 Does the LEA instruction support negative displacement? mov rax, 1 lea rsi, [rsp - rax] When I use the above code in my asm file I got the error: $ nasm -f macho64 test.asm $ error: invalid effective address I Know that we can do pointer arithmetic like this in C: void foo(char *a, size_t b) { *(a - b) = 1; } then I assume that: lea rsi, [rsp - rax] will work. And I also try to see what the GCC compiler do by using: $ gcc -S foo.c // foo.c has the function foo(above) in it but my asm

问题 Does the LEA instruction support negative displacement? mov rax, 1 lea rsi, [rsp - rax] When I use the above code in my asm file I got the error: $ nasm -f macho64 test.asm $ error: invalid effective address I Know that we can do pointer arithmetic like this in C: void foo(char *a, size_t b) { *(a - b) = 1; } then I assume that: lea rsi, [rsp - rax] will work. And I also try to see what the GCC compiler do by using: $ gcc -S foo.c // foo.c has the function foo(above) in it but my asm

问题 Does the LEA instruction support negative displacement? mov rax, 1 lea rsi, [rsp - rax] When I use the above code in my asm file I got the error: $ nasm -f macho64 test.asm $ error: invalid effective address I Know that we can do pointer arithmetic like this in C: void foo(char *a, size_t b) { *(a - b) = 1; } then I assume that: lea rsi, [rsp - rax] will work. And I also try to see what the GCC compiler do by using: $ gcc -S foo.c // foo.c has the function foo(above) in it but my asm