英文:
Why does switch to long mode cause VGA text output to behave strangely?
问题
I'm developing a small demo that boots a x86_64 machine. During early init (real mode), I set videomode 3 via int 10h. I then write to memory-mapped text at 0xb8000. My second stage already is high-level C code. This worked perfectly in protected mode, 32 bit, with paging.
I changed the bootloader to also enable PAE and then set LME, then jump to the second stage (which then has been compiled as x86_64 already). This is where my display fell apart and I have no idea what is going on. I've been debugging small samples and have something that works reliably even in 64 bit mode:
for (uint32_t i = 0xb8000; i < 0xb8000 + (25 * 80 * 2); i += 2) {
*((volatile uint16_t*)i) = 0x0741;
}
As expected, this fills the screen with all "A"s. Here's the generated assembly:
000000000000843f <main>:
843f: f3 0f 1e fa endbr64
8443: 55 push %rbp
8444: 48 89 e5 mov %rsp,%rbp
8447: c7 45 fc 00 80 0b 00 movl $0xb8000,-0x4(%rbp)
844e: eb 0c jmp 845c <main+0x1d>
8450: 8b 45 fc mov -0x4(%rbp),%eax
8453: 66 c7 00 41 07 movw $0x741,(%rax)
8458: 83 45 fc 02 addl $0x2,-0x4(%rbp)
845c: 81 7d fc 9f 8f 0b 00 cmpl $0xb8f9f,-0x4(%rbp)
8463: 76 eb jbe 8450 <main+0x11>
8465: 90 nop
8466: eb fd jmp 8465 <main+0x26>
However, when I change my code to this:
volatile uint16_t *screen_base = (volatile uint16_t*)0xb8000;
for (uint32_t i = 0; i < 25 * 80; i++) {
screen_base[i] = 0x0741;
}
It stops working; it outputs pink control characters (indicating the "0x07" is the character and the "0x41" is the color code), but does not even fill the whole screen (last two characters at the lower right end are not filled). Here's the generated assembly:
000000000000843f <main>:
843f: f3 0f 1e fa endbr64
8443: 55 push %rbp
8444: 48 89 e5 mov %rsp,%rbp
8447: 48 c7 45 f0 00 80 0b movq $0xb8000,-0x10(%rbp)
844e: 00
844f: c7 45 fc 00 00 00 00 movl $0x0,-0x4(%rbp)
8456: eb 17 jmp 846f <main+0x30>
8458: 8b 45 fc mov -0x4(%rbp),%eax
845b: 48 8d 14 00 lea (%rax,%rax,1),%rdx
845f: 48 8b 45 f0 mov -0x10(%rbp),%rax
8463: 48 01 d0 add %rdx,%rax
8466: 66 c7 00 41 07 movw $0x741,(%rax)
846b: 83 45 fc 01 addl $0x1,-0x4(%rbp)
846f: 81 7d fc cf 07 00 00 cmpl $0x7cf,-0x4(%rbp)
8476: 76 e0 jbe 8458 <main+0x19>
8478: 90 nop
8479: eb fd jmp 8478 <main+0x39>
Weirdly enough I can mask the issue by just botching the pointer to 0xb8003, but this is obviously incorrect. I cannot figure out what is going on here, does anyone have an idea what could be happening?
英文:
I'm developing a small demo that boots a x86_64 machine. During early init (real mode), I set videomode 3 via int 10h. I then write to memory-mapped text at 0xb8000. My second stage already is high-level C code. This worked perfectly in protected mode, 32 bit, with paging.
I changed the bootloader to also enable PAE and then set LME, then jump to the second stage (which then has been compiled as x86_64 already). This is where my display fell apart and I have no idea what is going on. I've been debugging small samples and have something that works reliably even in 64 bit mode:
for (uint32_t i = 0xb8000; i < 0xb8000 + (25 * 80 * 2); i += 2) {
*((volatile uint16_t*)i) = 0x0741;
}
As expected, this fills the screen with all "A"s. Here's the generated assembly:
000000000000843f <main>:
843f: f3 0f 1e fa endbr64
8443: 55 push %rbp
8444: 48 89 e5 mov %rsp,%rbp
8447: c7 45 fc 00 80 0b 00 movl $0xb8000,-0x4(%rbp)
844e: eb 0c jmp 845c <main+0x1d>
8450: 8b 45 fc mov -0x4(%rbp),%eax
8453: 66 c7 00 41 07 movw $0x741,(%rax)
8458: 83 45 fc 02 addl $0x2,-0x4(%rbp)
845c: 81 7d fc 9f 8f 0b 00 cmpl $0xb8f9f,-0x4(%rbp)
8463: 76 eb jbe 8450 <main+0x11>
8465: 90 nop
8466: eb fd jmp 8465 <main+0x26>
However, when I change my code to this:
volatile uint16_t *screen_base = (volatile uint16_t*)0xb8000;
for (uint32_t i = 0; i < 25 * 80; i++) {
screen_base[i] = 0x0741;
}
It stops working; it outputs pink control characters (indicating the "0x07" is the character and the "0x41" is the color code), but does not even fill the whole screen (last two characters at the lower right end are not filled). Here's the generated assembly:
000000000000843f <main>:
843f: f3 0f 1e fa endbr64
8443: 55 push %rbp
8444: 48 89 e5 mov %rsp,%rbp
8447: 48 c7 45 f0 00 80 0b movq $0xb8000,-0x10(%rbp)
844e: 00
844f: c7 45 fc 00 00 00 00 movl $0x0,-0x4(%rbp)
8456: eb 17 jmp 846f <main+0x30>
8458: 8b 45 fc mov -0x4(%rbp),%eax
845b: 48 8d 14 00 lea (%rax,%rax,1),%rdx
845f: 48 8b 45 f0 mov -0x10(%rbp),%rax
8463: 48 01 d0 add %rdx,%rax
8466: 66 c7 00 41 07 movw $0x741,(%rax)
846b: 83 45 fc 01 addl $0x1,-0x4(%rbp)
846f: 81 7d fc cf 07 00 00 cmpl $0x7cf,-0x4(%rbp)
8476: 76 e0 jbe 8458 <main+0x19>
8478: 90 nop
8479: eb fd jmp 8478 <main+0x39>
Weirdly enough I can mask the issue by just botching the pointer to 0xb8003, but this is obviously incorrect. I cannot figure out what is going on here, does anyone have an idea what could be happening?
答案1
得分: 3
汇编代码在调试构建中有点难以理解,但看起来正常。您是否100%确定在执行jmp far
或其他导致CS描述符的L位被设置后,处于64位模式?因为有强烈的证据表明您可能没有。
在32位模式下,0x48
是dec eax
的操作码(而不是REX.W前缀),这似乎可以解释偏移量为-3字节的情况。add %rdx, %rax
变成了dec %eax
;add %edx, %eax
。而在LEA之前有一个dec %eax
,然后是将其加倍的操作,以及在mov -0x10(%ebp), %eax
存储之前。
您的可行版本通过将uint32_t
强制转换为指针来避免了任何REX前缀。请注意,没有指令使用64位操作大小、R8-R15或BPL-DIL,因此它们的机器码中没有以40
到4F
字节开头的部分。(除了最初的mov %rsp, %rbp
之外,但在那一点上EAX并没有存活;对EAX的下一次访问是只写的。)
因此,这相当强烈地表明CPU并没有处于完全的64位模式。使用Bochs单步执行您的切换到长模式的操作,并检查您实际处于的模式。然后单步执行不工作的代码中的指令;您将看到48
字节被解码为单独的指令。(您也可以在QEMU + GDB中执行此操作;GDB可能不确定CPU处于哪种模式,但通过陷阱标志TF进行单步执行将反映CPU实际执行的操作。)
顺便说一下,GCC调试构建更喜欢首先使用EAX/RAX来评估任何表达式,可能是因为它是返回值寄存器。如果GCC恰好选择了不同的寄存器,递减EAX就不会有影响。但在某些情况下,您肯定会遇到问题,例如当GCC使用RAX因为它是返回值寄存器,或者当它尝试使用DIL字节寄存器时,带有0x40
REX前缀(在32位模式下为inc eax
)时。
英文:
The asm is fairly painful to follow in a debug build but it looks normal. Are you 100% sure you are in 64-bit mode after a jmp far
or whatever to a CS descriptor with the L bit set? Because there's strong evidence you aren't.
0x48
in 32-bit mode is the opcode for dec eax
(instead of a REX.W prefix), which looks like it might explain an offset of -3 bytes. add %rdx, %rax
becomes dec %eax
; add %edx, %eax
. And earlier there's a dec %eax
before the LEA that doubles it, and before the mov -0x10(%ebp),%eax
store.
Your version that works avoids any REX prefixes by casting uint32_t
to a pointer. Note that none of the instructions use 64-bit operand-size, R8-R15, or BPL-DIL, so none of them start with a 40
to 4F
byte in machine code. (Except the initial mov %rsp, %rbp
, but EAX isn't live at that point; the next access to EAX is write-only.)
So that's pretty strong evidence the CPU's not in full 64-bit mode. Use Bochs to single-step your switch to long mode and check what mode you're actually in. And single-step by instructions in the not-working code; you'll see the 48
bytes decode as separate instructions. (You can do that in QEMU + GDB as well; GDB might not be sure what mode the CPU is in, but single-stepping via the trap flag TF will reflect what the CPU is actually doing.)
BTW, GCC debug builds prefer using EAX/RAX first for evaluating any expression, perhaps because it's the return value register. If GCC had happened to pick different registers, decrementing EAX wouldn't have mattered. But you'd definitely run into problems at some point, e.g. when GCC used RAX because it's the return value register, or when it tried to use the DIL byte register with a 0x40
REX prefix (inc eax
in 32-bit mode.)
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