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I underdstand that this is extremely stupid quiestion, but I can't figure an answer for some time How do I correctly declare and define "variables" in GAS AT&T assembly language? For example, I want buffer for 5 bytes, two 1-byte variables (initially with 0 value), 2-byte variable with 0 and 2-byte variable with 10. This code doesn't work correctly, at least debugger says (on the first line of the program, after these declarations, just nop instruction) that b and c are big numbers instead of zeros. .bss .lcomm r, 5 .data a: .byte 0 b: .byte 0 c: .word 0 d: .word 10 Here's what you see in your

If one of my command lines says: jmp *0x804a180(,%eax,4) what does that mean? I ask specifically because there is no value before the first comma and I'm not sure exactly what the * before the address means. This instruction jumps to the location whose value is located at the address calculated as %eax * 4 + 0x804a180 . The * is used in AT&T syntax to indicate indirect jumps and calls. It basically means "jump to the location pointed to by this, instead of the value of this". It is useful to differentiate the following instructions: jmp myAddress # Jumps to the location myAddress jmp

I am trying to load the address of 'main' into a register (R10) in the GNU Assembler. I am unable to. Here I what I have and the error message I receive. main: lea main, %r10 I also tried the following syntax (this time using mov) main: movq $main, %r10 With both of the above I get the following error: /usr/bin/ld: /tmp/ccxZ8pWr.o: relocation R_X86_64_32S against symbol `main' can not be used when making a shared object; recompile with -fPIC /usr/bin/ld: final link failed: Nonrepresentable section on output collect2: error: ld returned 1 exit status Compiling with -fPIC does not resolve the

I have the following x86 assembly code: movl 8(%ebp), %edx //get an argument from the caller movl $0, %eax testl %edx, %edx je .L1 .L2: // what's the purpose of this loop body? xorl %edx, %eax shrl $1, %edx jne .L2 .L1: andl $1, %eax The corresponding C code that the textbook gives as follows int f1(unsigned x) { int y = 0; while(x != 0) { __________; } return __________; } The book asks readers to fill the blank and answer the question of "What does it do?" I can't combine the loop body in one C expression. I can tell what the loop body does, but I have no idea about its purpose. The textbook

Chapter 3 of Computer Systems A Programmer's Perspective (2nd Edition) mentions that cltq is equivalent to movslq %eax, %rax . Why did they create a new instruction ( cltq ) instead of just using movslq %eax,%rax ? Isn't that redundant? TL;DR : use cltq when possible, because it's one byte shorter than the exactly-equivalent movslq %eax, %rax . That's a very minor advantage (so don't sacrifice anything else to make this happen) but choose eax if you're going to want to sign-extend it a lot. This is mostly relevant for compiler-writers (compiling signed-integer loop counters indexing arrays);

接前面笔记: 关系抽取总结 中最后说的，在关系抽取基于Distant Supervision的NYT+Freebase的数据集有两个版本，目前大部分文章都是在这两份数据集上做的。通过纵向实验发现，同一个模型在不同版本数据集上的表现有不少差异，这篇笔记是基于自己使用Pytorch复现的PCNN(Zeng 2015)与PCNN+ATT(Lin 2016)的实验结果来简单对比。 代码地址: pytorch-relation-extraction ，关于实现的细节，调参，踩过的一些坑等，会记录在readme中，这里不再赘述。 个人属于入门级别, 复现代码可能有问题, 结果供参考. 数据集 首先描述一下这两份数据集： 27类关系，Zeng2015 发布的数据集，做了一些过滤，如删除了两个实体之间距离大于40个词的句子，以及去掉了实例少的关系，相对较小，以SMALL表示：详细数据如下所示: 训练数据集: 实体对: 65726; 句子数: 112941; 关系数: 27 测试数据集: 实体对: 93574; 句子数: 152416 53类关系，Lin2016 发布的数据集，没有做过滤，相对较大，训练数据大概是小数据的4倍，以LARGE表示，详细如下: 训练数据集: 实体对: 281270; 句子数: 522611; 关系数: 53 测试数据集: 实体对: 96678; 句子数: 172448;

After reading this stack overflow answer , and this document , I still don't understand the difference between movq and movabsq . My current understanding is that in movabsq , the first operand is a 64-bit immediate operand whereas movq sign-extends a 32-bit immediate operand. From the 2nd document referenced above: Moving immediate data to a 64-bit register can be done either with the movq instruction, which will sign extend a 32-bit immediate value, or with the movabsq instruction, when a full 64-bit immediate is required. In the first reference, Peter states: Interesting experiment: movq