以十六进制字节计算大小 [英] Calculate size in Hex Bytes

问题描述

计算代码段的十六进制字节大小的正确方法是什么？我得到：

pre code> IP = 0848 CS = 1488 DS = 1808 SS = 1C80 ES = 1F88

我正在研究的练习题要求代码段的大小（以十六进制字节为单位），并给出以下选择： p>

  A。 3800 B. 1488 C. 0830 D. 0380 E.以上都不是
  

正确答案是A 。3800，但我不知道该如何计算。

解决方案

如何计算长度： p>

• 注意CS。找到离它最近的段落寄存器，但更大。


• 取两者之间的差值，并乘以0x10（读取：0上的粘性）。
  

在您的示例中，DS最接近。 1808 - 1488 == 380.和380 x 10 = 3800。

顺便说一下，这只适用于8086和其他类似骨头的CPU，并且在x86上处于实模式。在x86上的保护模式下（也就是说，除非你正在编写引导扇区或简单的DOS程序），段寄存器的值与段的大小无关，因此上面的内容只是简单的不适用。

what is the proper way to calculate the size in hex bytes of a code segment. I am given:

IP = 0848    CS = 1488    DS = 1808   SS = 1C80    ES = 1F88

The practice exercise I am working on asks what is the size (in hex bytes) of the code segment and gives these choices:

A. 3800    B. 1488    C. 0830    D. 0380    E. none of the above

The correct answer is A. 3800, but I haven't a clue as to how to calculate this.

解决方案

How to calculate the length:

• Note CS. Find the segment register that's nearest to it, but greater.
• Take the difference between the two, and multiply by 0x10 (read: tack on a 0).

In your example, DS is closest. 1808 - 1488 == 380. And 380 x 10 = 3800.

BTW, this only works on the 8086 and other, similarly boneheaded CPUs, and in real mode on x86. In protected mode on x86 (which is to say, unless you're writing a boot sector or a simple DOS program), the value of the segment register has very little to do with the size of the segment, and thus the stuff above simply doesn't apply.

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