我看不出什么功能错在什么被执行条款，但我只是浪费了一个小时，证明我的code和编译器正在做什么，他们应该。

I can't see anything functionally wrong in terms of what gets executed, but I've just wasted an hour proving my code and the compiler are doing what they're supposed to.

考虑这个code：

public int getAnswer(int a) {

    int retval = 18;

    int x = 25;
    int y = 50;

    while (a-- > 0) {
        if (a == 3) {
            retval = a;
            return retval;
        }
    }

    x = 10;
    y = 20;

    return 0;
}

调用方式

int theAnswer = getAnswer(6);

断点既包含返回和运行code之前，尝试和predict的结果，所以这行会被打线。

Breakpoint both of the lines containing return and before you run the code, try and predict the result, and therefore which line will be hit.

正如你可能想象，从我的问题的基调，错线被突出显示，但正确的返回结果。如果通过code步骤，它周围的舞蹈颇有几分。

As you might imagine from the tone of my question, the wrong line is highlighted, but the right result is returned. If you step through the code, it dances around quite a bit.

有没有什么可以配置不同使得正确的行显示？ （我使用的是最新的Andr​​oid工作室，我AP preciate可能是不稳定的，但我敢肯定我见过类似的行为在Eclipse上一段时间回来，但我从来没有花的时间，然后跟踪下来，也许这是一个java的东西）。

Is there anything that can be configured differently such that the correct lines are displayed? (I'm using the latest Android Studio, which I appreciate is potentially unstable, but I'm sure I've seen similar behaviour on Eclipse a while back but I never spent the time then tracking it down; perhaps it's a java thing).

推荐答案

这是DX，这是一个将你的Java的.class文件转化为包装成安卓.dx文件生成的部分问题。根据这样的：

This is a problem with dx, which is the part of the build that turns your Java .class files into .dx files for packaging into Android. According to this:

https://$c$c.google.com/p/android/issues/detail?id=34193

如果一个函数有多个返回路径，DX合并返回指令到一个单一的返回指令，所以在调试过程中，调试器无法分辨出哪行回归属于事物四处跳动。通过它的若（a == 3）检查，跳转到返回0 结尾，然后再跳回循环。你看到的最后一个返回0 获得与合并了返回RETVAL 在循环的中间。

if a function has multiple return paths, dx merges the return instruction into a single return instruction, so during debugging, the debugger can't tell which line a return belongs to and things jump around. This corresponds to what I see when I try to reproduce your problem: each time through the loop it does the if (a == 3) check, jumps to the return 0 at the end, and then jumps back into the loop. You're seeing that last return 0 get merged with the return retval in the middle of the loop.

我怀疑这会尽快解决的任何时间，所以你可能只需要学会适应它。对不起，我知道这有点疯狂。

I doubt this will be fixed any time soon, so you may just have to learn to live with it. Sorry, I know it's kinda crazy.