虚拟地址 0x1FE0C0 是否位于“行”下方？还是线以上？
我正在研究 z/OS，有人问我虚拟地址 0x1FE0C0 是否是：
A. 线上方。
B. 柱线上方和线下方。
C. 在线上方和条线下方。
D. 栏下方。
我选择D作为我的答案，但我的老师说这是错误的。
我选择 D 的逻辑是，要以二进制表示 0x1FE0C0，您将需要 21 位。 该线标记了可以通过 24 位寻址的区域，因此我认为地址位于该线下方。由此我可以消除答案 A 和 C。 答案 B 被排除，因为条形图位于线条上方，因此如果地址位于条形图上方，它也将位于线条上方。所以我们留下答案 D，这对我来说似乎是正确的，因为如果地址低于该线，它也低于该条（31 位地址空间）。
那么正确答案是什么呢？
这让我发疯，我将感谢你的帮助。
I'm studying z/OS and I was asked if the virtual address 0x1FE0C0 is:
A. above the line.
B. above the bar and below the line.
C. above the line and below the bar.
D. below the bar.
I choose D as my answer, but my teacher wrote that it's wrong.
My logic for choosing D was that to represent 0x1FE0C0 in binary, you will need 21 bits.
The line marks the area that can be addressed by 24 bits, so that's why I believe the address is below the line. From that I could eliminate answer A and C.
Answer B was eliminated because the bar is above the line, so if the address is above the bar it will also be above the line. So we are left with answer D which seems correct to me, because if the address is below the line it is also below the bar (31 bit address space).
So what is the correct answer?
This is driving me crazy and I would appreciate your help.
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1FE0C0 是 6 个十六进制字符，因此它是一个 24 位地址，基于此 page 它位于线下方，也位于条线下方，因此 D 是正确答案。
1FE0C0 is 6 hex characters so it is a 24bit address, based on this page it is below the line which is also below the bar so D is the correct answer.
将 0x1FE0C0 转换为十进制 2089152，小于 2GB（2 147 483 648 字节），因此唯一可能的答案是 D。在条形下方。
B. 是不可能的，其他人在上面说，超过 2GB，但事实并非如此。
Converting 0x1FE0C0 to decimal, 2089152, is less than 2GB (2 147 483 648 bytes), so the only possible answer is D. Below the bar.
B. is impossible, and the others state above the line, more than 2GB which it is not.