assembly

问题 Alright, Im going about this, with what is probably a really complicated solution, but its the first thing that popped into my head. I need to write an assembly language program that reverses a "source" string, without using a "target" string ( a temp variable ) .. this is my attempt at it. INCLUDE Irvine32.inc .data source BYTE "This is the source string", 0 count DWORD ? .code main PROC mov ecx, LENGTHOF source L1: mov count, ecx ; save our outer loop count mov al,[source+0] ; get the first

问题 Alright, Im going about this, with what is probably a really complicated solution, but its the first thing that popped into my head. I need to write an assembly language program that reverses a "source" string, without using a "target" string ( a temp variable ) .. this is my attempt at it. INCLUDE Irvine32.inc .data source BYTE "This is the source string", 0 count DWORD ? .code main PROC mov ecx, LENGTHOF source L1: mov count, ecx ; save our outer loop count mov al,[source+0] ; get the first

问题 Alright, Im going about this, with what is probably a really complicated solution, but its the first thing that popped into my head. I need to write an assembly language program that reverses a "source" string, without using a "target" string ( a temp variable ) .. this is my attempt at it. INCLUDE Irvine32.inc .data source BYTE "This is the source string", 0 count DWORD ? .code main PROC mov ecx, LENGTHOF source L1: mov count, ecx ; save our outer loop count mov al,[source+0] ; get the first

问题 I have two UInt64 (i.e. 64-bit quadword) integers. they are aligned to an 8-byte ( sizeof(UInt64) ) boundary (i could also align them to 16-byte if that's useful for anything) they are packed together so they are side-by-side in memory How do i load them into an xmm register, e.g. xmm0 : I've found: movq xmm0, v[0] but that only moves v[0] , and sets the upper 64-bits in xmm0 to zeros: xmm0 0000000000000000 24FC18D93B2C9D8F Bonus Questions How do i get them back out? What if they're not side

问题 Duntemann says 8086 could address 16 times as much memory as 8080 and has gone on to elaborate upon this as 16 * 64K = 1MB . So I am assuming 16 bits at a time (16-bit processor) * 64K because 16 address lines (2^16). But, by that logic for calculation shouldn't 8080 (8-bit processor) also having 16 address lines have been able to address 8 * 64K = 0.5MB? shouldn't the 8086 then be able to address twice as much as 8080? 回答1: 16x is the difference between 2^20 and 2^16 . The 8080 has 16 bits

问题 First I am a little bit confused with the differences between movq and movabsq , my text book says: The regular movq instruction can only have immediate source operands that can be represented as 32-bit two’s-complement numbers. This value is then sign extended to produce the 64-bit value for the destination. The movabsq instruction can have an arbitrary 64-bit immediate value as its source operand and can only have a register as a destination. I have two questions to this. Question 1 The

问题 First I am a little bit confused with the differences between movq and movabsq , my text book says: The regular movq instruction can only have immediate source operands that can be represented as 32-bit two’s-complement numbers. This value is then sign extended to produce the 64-bit value for the destination. The movabsq instruction can have an arbitrary 64-bit immediate value as its source operand and can only have a register as a destination. I have two questions to this. Question 1 The

问题 When attempting to build Jonesforth (32-bit GNU Assembler program) on Ubuntu 16.04.4 64-bit (Xenial Xerus), I'm seeing the following: ~/src/jonesforth $ make gcc -m32 -nostdlib -static -o jonesforth jonesforth.S jonesforth.S:1154:24: fatal error: asm/unistd.h: No such file or directory compilation terminated. Makefile:11: recipe for target 'jonesforth' failed Looking in the file jonesforth.S , I noticed the following lines: //#include <asm-i386/unistd.h> // You might need this instead

问题 I was reading a chapter related to I/O using the MC68HC11 ; this book showed a suggested exercise (not really a hard one) but i was not able to solve it by using assembly: I've been thinking and i can do it by using some basic logic (i do program in C and C++ ) but i got stuck while trying to do it in assembly. Logic goes like this: Loop: value = ReadSensorValue() if (value condition 1) // do action else if (value condition 2) // do something end if go to Loop Can help me to solve it but

问题 I was reading a chapter related to I/O using the MC68HC11 ; this book showed a suggested exercise (not really a hard one) but i was not able to solve it by using assembly: I've been thinking and i can do it by using some basic logic (i do program in C and C++ ) but i got stuck while trying to do it in assembly. Logic goes like this: Loop: value = ReadSensorValue() if (value condition 1) // do action else if (value condition 2) // do something end if go to Loop Can help me to solve it but