assembly language programming 8086 examples

of the program Enter d ss:0 to display the data at the stack segment + offset 0 and back Enter G 001a so that the first loop is executed and then enter D ss:0 to display the data in the stack Looking at the first line, the data in the data segment has been pressed into the stack, then the result after the second loop has been executed. Enter G 0027, then D ds:0, then D ss:0 See, the data in the data segment has been reversed. Input T, single step to 21H interrupt Enter P, run interrupt, and

Common usage and examples of static keywords in C Language In embedded system development Programming Language Is C and assembly,C ++ already has a corresponding compiler, but it is still rarely used. In a slightly largerIn large-scale embedded software, such as operating sy

Examples of C and assembly languages or programming in ARM, and explanations of arm instances1. inline assembly Notes for Embedded Assembly in C language: You cannot assign values directly to the PC register. B or BL commands must

direct operation of the hardware, visual inspection is more, I estimate will be more boring, and " 80X86 assembly Language Programming Tutorial "from 8086, the latter 80386 will be related to the modern CPU working mode, this became the reason I chose it."Now let's talk about what you can learn from it:1) You can see

The main function of this program is to display the CPU vendor's vendor IDThe source code is as follows:. Section. Data#x是占位符Output:. ASCII"The processor Vendor ID is: ' xxxxxxxxxx ' \ n"#_start和output都是标签. section. Text#如果用gcc编译的话, _start will change to main. Global _START_START:MOVL $,%eaxCpuid#相当于把字符串output的地址传入到寄存器edi中.Movl$output,%edi#将调用cpuid命令的内容放入指定的内存Movl%EBX, -(%edi) MOVL%edx, +(%edi) MOVL%ecx, $(%edi)#4表示系统调用的值, 1 represents the file descriptor, and output is the input characterMovl$4

Interrupt Descriptor Descriptor Register instruction: SIDT DSTSimilar to "Sgdt DST".b) Storage machine status word directive: SMSW DSTDST can be a 16-bit universal register or storage unit. No effect on the flag bit.Note: In order to be compatible with the 80286 instruction set (386 CR0 low byte equals 286 machine state word), in 386, the storage machine word should use the stored CR0 Register instruction.2) Real mode and privilege level 0 executable commandsKey register settings directives, et

physical registers directlyD,generally do not specify physical registers directly, but let the compiler assigntags used in inline assembly are __asm or ASM keywordsSyntax format for inline assembly:__asm{instruction [; instruction]...[Instruction]}Program Examples:#include void my_strcpy (const char *SRC, char *dest){Char ch;__asm{LoopLDRB ch, [src], #1STRB ch,

gcc 3.5 GNU Debugger 3.5.1 download and install GDB 3.5.2 Use GDB 3.6 KDE Debugger 3.6.1 download and install kdbg 3.6.2 use kdbg 3.7 GNU objdump Program 3.7.1 use objdump 3.7.2 objdump example 3.8 GNU downloader program' 3.8.1 use the Configurator 3.8.2 example 3.9 complete compilation and Development System 3.9.1 Linux Basics 3.9.2 download and run mepis 3.9.3 New Development System Conclusion 3.10 Chapter 2 assembly

: address provided for Stack operations. 4. I/O control logic: function: input/output operations. 5. CPU program execution process:1. Run Command 1, Run Command 2, Run Command 2, and run command 3 .... Low CPU EfficiencyTo overcome these disadvantages:8086 of commands are executed in parallel. The command queue is used.--------- Biu Bus Interface Unit EU Execution UnitBiu is responsible for reading the command to the queue, and waits until it knows it

service program), one BIOS, one DOS, and the use of interrupt mode (INT) for execution. Higher DOS level, priority should be used. The second is the BIOS (write ROM, which is part of the hardware), and finally the peripheral calling program. 80x86 instruction system and addressing method One address (INC, DEC), two addresses (majority), three address directives (few, such as SHL). Operation code as long as the unique binary code can be determined, but in order to facilitate the

Assembly language programming, the 32-bit system should not have any problems, but under the 64-bit system, there will be some different places. Some program samples also compile errors or execute errors.The system used by the blogger is CentOS v6.4 x64. This article mainly solves the 32-bit assembler how to assemble and connect in 64-bit environment, but does no

compiling the source code, and then the machine code is run by the machine .Word has two bytes, one byte is 8 bit (bit)Complement, the original code is reversed by a bit and added 116 complement, the original number, the bitwise 15 minus the number in this position, and the last plus 1But the decimal number needs to be converted into binary and complementQuestion 7: If a Boolean function has 4 inputs, how many rows does its truth table need?Answer: There are title 5 and title 6 can be seen, 2 i

operate on the built-in TSC in the CPU. TSC is time Stamp Counter, a 64-bit timestamp counter provided for Pentium series CPUs, which is counted once per instruction cycle after the CPU is power up or reset, and Intel guarantees that the TSC overflow period is greater than 10. Like the 300MHz CPU we use, its TSC accuracy is about 33ns and the overflow period is about 19,303 years, which is perfectly in line with our project requirements.However, it is faint that the project cannot be compiled a

available (DWARF 2, stabs, or the native format ifNeither of those is supported), including GDB extensions if at all possible.Assembly Language Programming This book is the use of-gstabs this parameter. Why not use other parameters?Finally, the size of the test1 generated by test1.c with these four parameters is recorded as follows, and the resulting file size is not the same as the visible parameter. -G a

address, you can use R14 between calls for other purposes, and when the program returns, the R15 is the program counter PC. It cannot be used for any other purpose. Note: In the interrupt program, all registers must be protected and the compiler will automatically protect the R4~r11*/intC_add (intAintb) { returnA +b;}intC_sub (intAintb) { returnA-b;} unsignedintASM_C_CTL_CP15 (void) {unsignedinti =0; uart_printf ("asm_c_ctl_cp15\n"); __asm {MRC P15,0, R0,C1,C0,0MOV I,r0}returni;}