在x86汇编中声明数组 [英] Declaring Arrays In x86 Assembly

问题描述

我正在学习Assembly，我需要做很多事情。我已经看过如何声明数组，并且遇到了这个问题。

I am learning Assembly and I need to make a large array. I have looked around at how to declare arrays and I have come across this.

array db 10 dup(?)

其中声明了10个未初始化字节的数组。我尝试过并尝试对其进行汇编，并得到错误：操作数1之后应加逗号。我意识到x86不支持？，因此我将其设置为常量并得到了相同的错误。我最后做了这个。

Where an array of 10 uninitialized bytes is declared. I tried this and tried to assemble it and get "error: comma expected after operand 1". I realized that the '?' is not supported in x86 so I made it a constant and got the same error. I ended up doing this.

array db 1, 2, 3, 4, 5, 6, 7, 8, 9, 10

这有效！但是问题是，我需要大数组（〜100-400个整数），并且它们的值不会总是已知的。我可以写出400 0，但我认为必须有一种更简单的方法。那么，有没有更好的方法来声明大数组呢？

This worked! But the problem is, I need large arrays (~100-400 integers) and their values will not always be known. I could write out 400 0's but I figured there must be an easier way. So is there a better way to declare large arrays?

我正在基于AT& T语法的基于Intel的Macbook Pro上使用x86_64 Assembly。

I am using x86_64 Assembly on an Intel-Based Macbook Pro with AT&T syntax.

推荐答案

您尝试使用 TIMES指令。使用此代码声明一个

Did you try TIMES directive.Use this code for declaring an array of a given size.

array TIMES 8 DB 0

这将创建大小为8的数组

This will create an array of size 8

请参阅此链接以获取更多信息。

Refer to this link for more.

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