如何获得唯一的方法标识符? [英] How to get a unique method identifier?

问题描述

我需要获取一个唯一的方法标识符以用作HashMap上的键.

I'm needing to get a unique method identifier to use as a key on a HashMap.

我正在尝试使用堆栈跟踪和反射来做一些事情，并使用方法签名.但是问题是我没有找到一种方法来检索完整的方法签名(以避免方法重载).

I'm trying to do something using stacktrace and reflection and user the method signature. But the problem is I didn´t find a way to retrive the complete method signature (to avoid methods overload).

已编辑

我希望这种想法可行:

public class Class1 {

    HashMap<String, Object> hm;

    public Class1() {
        hm = new HashMap<String, Object>();
    }

    public Object method() {
        if (!containsKey()) {
            Object value;
            ...
            put(value);
        }

        return get();
    }

    public Object method(String arg1) {
        if (!containsKey()) {
            Object value;
            ...
            put(value);
        }

        return get();
    }

    public Boolean containsKey() {
        if (hm.containsKey(Util.getUniqueID(2)) {
            return true;
        } else {
            return false;
        }
    }

    public void put(Object value) {
        hm.put(Util.getUniqueID(2), value);
    }

    public Object get() {
        String key = Util.getUniqueID(2);
        if (hm.containsKey(key) {
            return hm.get(key);
        } else {
            return null;
        }
    }
}

class Util {
    public static String getUniqueID(Integer depth) {
        StackTraceElement element = Thread.currentThread().getStackTrace()[depth];
        return element.getClassName() + ":" + element.getMethodName();
    }
}

但是问题在于，采用这种策略的两种方法将具有相同的ID.

But the problem is the two methods, with this strategy, will have the same ID.

我该如何解决?

推荐答案

您可以附加+ ":" + element.getLineNumber()，但是您仍然需要担心将两个重载方法放在一行上的情况.

You can append + ":" + element.getLineNumber() but you'd still have to worry about the case where two overloaded methods are put on one long line.

查看StackTraceElement方法，似乎不可能以这种方式获得唯一的方法标识符.此外，我认为该代码不太可读.

Looking at the StackTraceElement methods, it doesn't seem possible to get a unique method identifier this way. Besides, the code is not very readable in my opinion.

我建议您尝试更明确些，然后做

I'd suggest you try to be more explicit and do

if (hm.containsKey("getValue(int)") {
    ...
}

或类似的东西.

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