首页 \ 问答 \ 从@encoded类型字符串中解码类(Decode Class from @encoded type string)

从@encoded类型字符串中解码类(Decode Class from @encoded type string)

Objective-C的@encode生成C字符串以表示任何类型,包括基本类和类,如下所示:

NSLog(@"%s", @encode(int));       // i
NSLog(@"%s", @encode(float));     // f
NSLog(@"%s", @encode(CGRect));    // {CGRect={CGPoint=ff}{CGSize=ff}}
NSLog(@"%s", @encode(NSString));  // {NSString=#}
NSLog(@"%s", @encode(UIView));    // {UIView=#@@@@fi@@I{?=b1b1b1b1b1b1b1b1b1b1b1b1b1b1b1b6b1b1b1b1b1b1b1b1b1b1b1b1b1b1b1b1b1b1b1b1b1b1b1b1b1b1b1b1b1b1b1b1b1b1b1b1b1b1b1b1b1b1b3b1b1b1b2b2b1}}

所以,我可以使用@encode(ClassName)得到一个有意义的类的编码(一个包含类名),但它的格式与一般struct的编码相同(如上例)。

现在,我的问题是,鉴于任何 (当然有效)类型编码,是否有可能找出编码是否为Objective-C类,如果是,获取与该编码相对应的Class对象?

当然,我可能只是尝试从类型编码中解析类名,并使用NSClassFromString从类中获取类,但这听起来不像是正确的方式,或者特别是性能高效。 这是否是实现这一目标的最佳方式?


Objective-C's @encode produces C strings to represent any type, including primitives, and classes, like so:

NSLog(@"%s", @encode(int));       // i
NSLog(@"%s", @encode(float));     // f
NSLog(@"%s", @encode(CGRect));    // {CGRect={CGPoint=ff}{CGSize=ff}}
NSLog(@"%s", @encode(NSString));  // {NSString=#}
NSLog(@"%s", @encode(UIView));    // {UIView=#@@@@fi@@I{?=b1b1b1b1b1b1b1b1b1b1b1b1b1b1b1b6b1b1b1b1b1b1b1b1b1b1b1b1b1b1b1b1b1b1b1b1b1b1b1b1b1b1b1b1b1b1b1b1b1b1b1b1b1b1b1b1b1b1b3b1b1b1b2b2b1}}

So, I can get a meaningful encoding of a class (one that contains the class name) using @encode(ClassName), but it's also in the same format as the encoding of a generic struct (as in the above example).

Now, my question is, given any (valid of course) type encoding, is it possible to find out whether the encoding is of an Objective-C class, and if so, to get the Class object that corresponds to that encoding?

Of course, I probably could just try to parse the class name out of the type encoding, and get the class from that using NSClassFromString, but that just doesn't sound like the right way to do it, or particularly performance-efficient. Is this really the best way to achieve this?


原文:https://stackoverflow.com/questions/18161519
更新时间:2023-06-01 16:06

最满意答案

不, is一个直接的指针比较,并且id只是将对象的地址转换为long

来自ceval.c

case PyCmp_IS:
    res = (v == w);
    break;
case PyCmp_IS_NOT:
    res = (v != w);
    break;

vw在这里只是PyObject *

来自bltinmodule.c

static PyObject *
builtin_id(PyObject *self, PyObject *v)
{
    return PyLong_FromVoidPtr(v);
}

PyDoc_STRVAR(id_doc,
"id(object) -> integer\n\
\n\
Return the identity of an object. This is guaranteed to be unique among\n\
simultaneously existing objects. (Hint: it's the object's memory address.)");

No, is is a straight pointer comparison, and id just returns the address of the object cast to a long.

From ceval.c:

case PyCmp_IS:
    res = (v == w);
    break;
case PyCmp_IS_NOT:
    res = (v != w);
    break;

v and w here are simply PyObject *.

From bltinmodule.c:

static PyObject *
builtin_id(PyObject *self, PyObject *v)
{
    return PyLong_FromVoidPtr(v);
}

PyDoc_STRVAR(id_doc,
"id(object) -> integer\n\
\n\
Return the identity of an object. This is guaranteed to be unique among\n\
simultaneously existing objects. (Hint: it's the object's memory address.)");

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