如何打印在大会8086多少？ [英] How to print a number in Assembly 8086?

问题描述

我想写一个接收数​​字（我提前了），并打印的功能。我该怎么办呢？

我到目前为止有：

 组织100H推10
调用print_numprint_num：推基点
MOV BP，SP
MOV AX，[BP + 2 * 2]
MOV BX，CS
MOV ES，BX
MOV DX，串
MOV DI，DX
STOSW
MOV啊，09H
INT 21H
流行基点
RET串：
 

解决方案

你在字符串的地址将是什么数值，​​而不是字符串重新$该值的对$ psentation。

的值12和字符串12是两个不同的事情。看作是一个16位的十六进制值，12是0x000C而12是0x3231（0x32 =='2'，0X31 =='1'）。

您需要的数值转换成它的字符串重新presentation，然后打印生成的字符串。结果而不是仅仅粘贴完成的解决方案，我将展示如何这可以在C做了一个简单的方法，这应该是够你在基地8086实现：

 字符字符串[8]，* stringptr;
短NUM = 123;字符串[7] =$; // DOS字符串结束//字符串会倒着填满
stringptr =字符串+ 6;而（stringptr＆GT; =字符串）{
    * stringptr ='0'+（NUM％10）;在第一次循环//'3'，'2'在第二，等等
    NUM / = 10; // 123 =＆GT; 12 =＆GT; 1 =＆GT; 0
    如果（NUM == 0）打破;
    stringptr--;
}
 

I'm trying to write a function that receives a number (which I pushed earlier), and prints it. How can I do it?

What I have so far:

org 100h

push 10
call print_num

print_num:

push bp
mov bp, sp
mov ax, [bp+2*2]
mov bx, cs
mov es, bx
mov dx, string
mov di, dx
stosw
mov ah, 09h
int 21h
pop bp
ret

string:

解决方案

What you're placing at the address of string is a numerical value, not the string representation of that value.

The value 12 and the string "12" are two separate things. Seen as a 16-bit hexadecimal value, 12 would be 0x000C while "12" would be 0x3231 (0x32 == '2', 0x31 == '1').

You need to convert the numerical value into its string representation and then print the resulting string.
Rather than just pasting a finished solution I'll show a simple way of how this could be done in C, which should be enough for you to base an 8086 implementation on:

char string[8], *stringptr;
short num = 123;

string[7] = '$';  // DOS string terminator    

// The string will be filled up backwards
stringptr = string + 6;

while (stringptr >= string) {
    *stringptr = '0' + (num % 10);  // '3' on the first iteration, '2' on the second, etc
    num /= 10;  // 123 => 12 => 1 => 0
    if (num == 0) break;
    stringptr--;
}        

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