关于 ARM 上的 Objective C 调用约定和参数传递的问题 [英] Question about Objective C calling convention and argument passing on ARM

问题描述

当我调用像这样的目标 C 方法时，我想知道目标 C 运行时如何处理参数

I want to know how objective C runtime handle arguments when I call a objective C method like

[NSString stringWithFomat:@"%@, %@", @"Hello", @"World"]

这个目标 C 调用有三个参数，与 ARM 系统上的典型方式相比，它是如何工作的.我知道寄存器 r0, r1, r2, r3 将保存前 4 个参数，还有其他参数如何?它如何将它们放在堆栈上并稍后弹出它们?

There are three arguments for this objective C call, how does it work compared to typical way on a ARM system. I have known register r0, r1, r2, r3 will hold first 4 arguments, how about there are additional arguments? How does it put them on a stack and pop them later?

推荐答案

对于返回简单类型的函数:

For functions that returns a simple type:

r0 = self (NSString)
r1 = _cmd (@selector(stringWithFormat:))
r2 = 1st argument (@"%@, %@")
r3 = 2nd argument (@"Hello")

然后剩下的放在栈上:

[sp,#0] = 3rd argument (@"World")
[sp,#4] = 4th argument (does not exist in your example)
...

当然，这里的参数"是指一个 4 字节的对象.如果参数有 >4 个字节，那么它将被拆分，例如

Of course, "argument" here means a 4-byte object. If the argument has >4 bytes then it will be split out, e.g.

-[UIView initWithFrame:rect];

r0 = self
r1 = _cmd
r2 = rect.origin.x
r3 = rect.origin.y
[sp,#0] = rect.size.width
[sp,#4] = rect.size.height

返回的值(最多 16 个字节)会放在 r0、r1、r2、r3 中.

The returned value (up to 16 bytes) will be placed in r0, r1, r2, r3.

对于返回结构体的函数:r0用于存储返回值的指针.

For functions that returns a struct: r0 is used to store the pointer of the return value.

NSRange retval = [self rangeOfString:string options:options range:range]

r0 = &retval (of type NSRange*)
r1 = self
r2 = _cmd (@selector(rangeOfString:options:range:))
r3 = string
[sp,#0] = options
[sp,#4] = range.location
[sp,#8] = range.length

这篇关于关于 ARM 上的 Objective C 调用约定和参数传递的问题的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！