如何打印一个长整型，我在一个寄存器？ [英] How do I print a long integer that I have in a register?

问题描述

我已经定义了一个长整型如下：

  

    

memTotal：0。长期

  

然后，我加入到它的东西是这样的：

  

    

12 ADDL（％DI），ECX％

  

在哪里我再往前％ECX到memTotal。我的问题是，我将如何去在memTotal的MB计算的大小。我试图沿着线的东西：

  

    

SHRL $ 20％EAX

  

不过，我怎么会那么打印在INT为MB？

我在正确的轨道上吗？任何帮助是AP preciated。

解决方案

我假设你知道如何打印单个ASCII字符。所以，现在你需要一个算法来从一个整数提取数字 A - 我会提供一个，我觉得是很容易理解和容易扩展（这是不一定的最好的的）。

• 计算 B：=一％10 。 B 是你的号码的最后一位数字


• 设置 A：= A / 10 （整数除法）


• 从开始重复获得第二至最后一位数字，等停止时 A == 0 。

一旦你有一个数字的值，你可以添加一个固定常数来获得它的ASCII值，您可以使用打印相应的字符。

以上可以打印一个数字。不幸的是，在这一点上，你只能向后打印。为了解决这个问题，分配一个字符串，你可以把数字从右到左。通过分配一个字符串，将大到足以容纳开始的所有的长整型，然后拿出上面的算法，将跟踪的多少位有，并据此分配一个字符串的修改。

（是的， SHRL $ 20％EAX 听起来似乎在理车削字节到通常所说的MIB。约定有所不同，但我认为，通常一个是1 MB = 10 ^ 6个字节，1 MIB = 2 ^ 20字节）。

I have defined a long integer as follows:

memTotal: .long 0

Then, I am adding to it with something like this:

addl 12(%di), %ecx

Where I then move %ecx to memTotal. My question is, how would I go about calculating the size in MB of the memTotal. I tried something along the lines of:

shrl $20, %eax

But how would I then print that as in int for MB?

Am I on the right track? Any help is appreciated.

解决方案

I assume you know how to print a single ASCII character. So now you need an algorithm to extract the digits from an integer a - I will provide one that I think is easy to understand and easy to expand (it's not necessarily the best).

• calculate b := a%10. b is the last digit of your number
• set a := a/10 (integer division)
• repeat from beginning to get second-to-last digit, etc. Stop when a == 0.

Once you have the value of a digit, you can add a fixed constant to get its ASCII value, which you can use to print the corresponding character.

The above enables you to print a number. Unfortunately, at this point you can only print it backwards. In order to fix that, allocate a string in which you can put the digits from right to left. Start by allocating a string that will be large enough to hold all long ints, then come up with a modification of the above algorithm that will keep track of how many digits there are, and allocate a string accordingly.

(And yes, shrl $20, %eax sounds sensible for turning bytes into what is usually called MiBs. Conventions vary, but I think the usual one is 1 MB = 10^6 bytes, and 1 MiB = 2^20 bytes.)

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