Nasm 打印到下一行 [英] Nasm print to next line

问题描述

我有以下用 nasm Assembly 编写的程序:

section .text全局_start:_开始:;输入变量mov edx, inLenmov ecx, inMsgmov ebx, 1移动轴，4整数 0x80mov edx, 2mov ecx, num1移动 ebx, 0移动轴，3整数 0x80mov edx, inLenmov ecx, inMsgmov ebx, 1移动轴，4整数 0x80mov edx, 2mov ecx, num2移动 ebx, 0移动轴，3整数 0x80;将输入值放入正确的寄存器中mov eax, [num1]子 eax, '0' ;将字符转换为数字mov ebx, [num2]子 ebx, '0' ;将字符转换为数字;执行加法添加 eax, ebx添加 eax, '0' ;将数字转换为字符;在 res 中设置总和mov [res], eax;输出结果mov edx, resLenmov ecx, resMsgmov ebx, 1移动轴，4整数 0x80mov edx, 1mov ecx, resmov ebx, 1移动轴，4整数 0x80;退出程序移动轴，1整数 0x80.bss 节num1 resb 2num2 resb 2resb 2.data 节inMsg db "输入数字:", 0xA, 0xDinLen 等于 $-inMsgresMsg db "结果:", 0xA, 0xDresLen 等于 $-resMsg

当我运行它时，控制台看起来像这样:

tyler@ubuntu:~/ASM/Addition$ ./Add输入编号:3输入编号:2结果:5tyler@ubuntu:~/ASM/Addition$

我怎样才能得到它，这样 5 将在它自己的行上打印，而不是在它之后直接打印 cmd?IE.它应该是这样的:

tyler@ubuntu:~/ASM/Addition$ ./Add输入编号:3输入编号:2结果:5tyler@ubuntu:~/ASM/Addition$

解决方案

您已经掌握了所有信息，只是还没有看到:

resMsg db "结果:", 0xA, 0xD

你知道这到底是什么意思吗?它定义了一个由以下字符组成的字符串:

结果:XY

...其中 X 和 Y 实际上是不可见字符(数值为 0xA=10 和 0xD=13，也称为换行 (LF) 和回车return (CR)) 导致输出换行.由于它们的不可见性，它们被指定在双引号之外 - 您不能仅仅将它们包含在双引号中，因此您必须写下它们的数值.

当然你也可以单独使用它们:

newLineMsg db 0xA, 0xDnewLineLen equ $-newLineMsg

(newLineLen 当然会是 2，但我把它留在这里是为了让系统和你目前使用的一样，以便于理解.)

所以只输出一个没有任何其他文本的换行符(你想在 5 之后做什么)，你可以使用:

mov edx, newLineLenmov ecx, newLineMsgmov ebx, 1移动轴，4整数 0x80

...就像 resMsg/resLen 一样.

<小时>

但是，正如 Jester 所指出的，在 Linux 上，您还应该能够仅输出一个换行符 (0xA)(并且还可以从代码中删除现有的 0xD).

I have the following program written in nasm Assembly:

section .text
    global _start:

_start:
    ; Input variables
    mov edx, inLen
    mov ecx, inMsg
    mov ebx, 1
    mov eax, 4
    int 0x80

    mov edx, 2
    mov ecx, num1
    mov ebx, 0
    mov eax, 3
    int 0x80

    mov edx, inLen
    mov ecx, inMsg
    mov ebx, 1
    mov eax, 4
    int 0x80

    mov edx, 2
    mov ecx, num2
    mov ebx, 0
    mov eax, 3
    int 0x80

    ; Put input values in correct registers
    mov eax, [num1]
    sub eax, '0'    ; convert char to num
    mov ebx, [num2]
    sub ebx, '0'    ; convert char to num

    ; Perform addition
    add eax, ebx
    add eax, '0'    ; convert num to char

    ; Set sum in res
    mov [res], eax

    ; Output result
    mov edx, resLen
    mov ecx, resMsg
    mov ebx, 1
    mov eax, 4
    int 0x80

    mov edx, 1
    mov ecx, res
    mov ebx, 1
    mov eax, 4
    int 0x80

    ; Exit program
    mov eax, 1
    int 0x80

    section .bss
    num1    resb 2
    num2    resb 2
    res resb 2

section .data
    inMsg db "Input number: ", 0xA, 0xD
    inLen equ $-inMsg
    resMsg db "Result: ", 0xA, 0xD
    resLen equ $-resMsg

When I run it the console looks like this:

tyler@ubuntu:~/ASM/Addition$ ./Add 
Input number: 
3
Input number: 
2
Result: 
5tyler@ubuntu:~/ASM/Addition$ 

How can I get it so the 5 will print on its own line and not have the cmd print directly after it? I.E. it should look like this:

tyler@ubuntu:~/ASM/Addition$ ./Add 
Input number: 
3
Input number: 
2
Result: 
5
tyler@ubuntu:~/ASM/Addition$ 

解决方案

You have all the info already, you just don't see it yet:

resMsg db "Result: ", 0xA, 0xD

Do you know what this means exactly? It defines a string made of the following characters:

Result: XY

...where X and Y are actually invisible characters (with numerical values 0xA=10 and 0xD=13, also known as line feed (LF) and carriage return (CR)) which cause the output to wrap to a new line. They are specified outside of the doublequotes because of their invisible nature - you can't just include them there so you have to write their numerical values instead.

But of course you can use them alone as well:

newLineMsg db 0xA, 0xD
newLineLen equ $-newLineMsg

(newLineLen will be 2 of course but I left it here to keep the system the same as you currently use, for easier understanding.)

So to just output a line break without any other text (what you want to do after the 5), you can then use:

mov edx, newLineLen
mov ecx, newLineMsg
mov ebx, 1
mov eax, 4
int 0x80

...just like with resMsg/resLen.

---

However, as Jester pointed out, on Linux you should also be able to output just a single line feed (0xA) (and also remove the existing 0xD's from your code).

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