assembly

问题 I am about to figure out how exactly a programm stack is set up. I have learned that calling the function with call pointer; Is effectively the same as: mov register, pc ;programcounter add register, 1 ; where 1 is one instruction not 1 byte ... push register jump pointer However, this would mean that when the Unix Kernel calls the main function that the stack base should point to reentry in the kernel function which calls main. Therefore jumping "*rbp-1" in the C - Code should reenter the

问题 I am about to figure out how exactly a programm stack is set up. I have learned that calling the function with call pointer; Is effectively the same as: mov register, pc ;programcounter add register, 1 ; where 1 is one instruction not 1 byte ... push register jump pointer However, this would mean that when the Unix Kernel calls the main function that the stack base should point to reentry in the kernel function which calls main. Therefore jumping "*rbp-1" in the C - Code should reenter the

问题 I am trying to assemble and link this tiny x86 assembly code in Linux (Ubuntu 18.04 LTS): ;hello.asm global _start extern scanf, printf, exit section .data read_name db '%255s', 0 msg db 'Hello, %s', 0 section .text _start: sub esp, 256 push esp push read_name call scanf add esp, 8 push esp push msg call printf add esp, 264 push dword 0 call exit I am using nasm to assemble and ld to link. As you can probably tell, the code uses C functions, so it has to be linked to glibc . Since my code is

问题 I am trying to assemble and link this tiny x86 assembly code in Linux (Ubuntu 18.04 LTS): ;hello.asm global _start extern scanf, printf, exit section .data read_name db '%255s', 0 msg db 'Hello, %s', 0 section .text _start: sub esp, 256 push esp push read_name call scanf add esp, 8 push esp push msg call printf add esp, 264 push dword 0 call exit I am using nasm to assemble and ld to link. As you can probably tell, the code uses C functions, so it has to be linked to glibc . Since my code is

问题 I am trying to assemble and link this tiny x86 assembly code in Linux (Ubuntu 18.04 LTS): ;hello.asm global _start extern scanf, printf, exit section .data read_name db '%255s', 0 msg db 'Hello, %s', 0 section .text _start: sub esp, 256 push esp push read_name call scanf add esp, 8 push esp push msg call printf add esp, 264 push dword 0 call exit I am using nasm to assemble and ld to link. As you can probably tell, the code uses C functions, so it has to be linked to glibc . Since my code is

问题 I am trying to assemble and link this tiny x86 assembly code in Linux (Ubuntu 18.04 LTS): ;hello.asm global _start extern scanf, printf, exit section .data read_name db '%255s', 0 msg db 'Hello, %s', 0 section .text _start: sub esp, 256 push esp push read_name call scanf add esp, 8 push esp push msg call printf add esp, 264 push dword 0 call exit I am using nasm to assemble and ld to link. As you can probably tell, the code uses C functions, so it has to be linked to glibc . Since my code is

问题 How do I round an integer to the nearest whole in assembly? The use of branching is not allowed here. For example, 195 * .85 = 165.75. Normally, I'd multiply 195 by a scale factor (100) and then multiply, then divide down the scale factor. This would give me 165. How do I get 166? I'm sorry if this is a terrible question - I'm new to assembly! Thank you. 回答1: For example, 195 * .85 = 165.75. Normally, I'd multiply 195 by a scale factor (100) and then multiply, then divide down the scale

问题 How do I round an integer to the nearest whole in assembly? The use of branching is not allowed here. For example, 195 * .85 = 165.75. Normally, I'd multiply 195 by a scale factor (100) and then multiply, then divide down the scale factor. This would give me 165. How do I get 166? I'm sorry if this is a terrible question - I'm new to assembly! Thank you. 回答1: For example, 195 * .85 = 165.75. Normally, I'd multiply 195 by a scale factor (100) and then multiply, then divide down the scale

问题 How do I round an integer to the nearest whole in assembly? The use of branching is not allowed here. For example, 195 * .85 = 165.75. Normally, I'd multiply 195 by a scale factor (100) and then multiply, then divide down the scale factor. This would give me 165. How do I get 166? I'm sorry if this is a terrible question - I'm new to assembly! Thank you. 回答1: For example, 195 * .85 = 165.75. Normally, I'd multiply 195 by a scale factor (100) and then multiply, then divide down the scale

问题 I am using NASM assembler on linux 64 bit. There is something with variables and registers I can't understand. I create a variable named "msg": msg db "hello, world" Now when I want to write to the stdout I move the msg to rsi register, however I don't understand the mov instruction bitwise ... the rsi register consists of 64 bit , while the msg variable has 12 symbols which is 8 bits each , which means the msg variable has a size of 12 * 8 bits , which is greater than 64 bits obviously. So