无法连续两次从变量输出协处理器浮点数 [英] Can't output coprocessor float from variable two times in a row

问题描述

下午好！在此示例中，我仅用逗号添加两个数字，将变量保存在 tbyte 中，并在屏幕上连续两次显示相同的变量，但这是我第一次获得 11.1 ，并且第二次 4.667261E-062 .为什么会这样?

Good afternoon! In this example, I simply add two numbers with a comma, save the variable in tbyte and display the same variable two times in a row on the screen, but the first time I get 11.1, as it should be, and the second time 4.667261E-062. Why is this happening?

还有一个问题，是否可以在 tbyte 中以某种方式按数组类型保存和访问数字?例如，将数字存储在 dd 中，我只能以4的增量进行保存和读取，例如，结果[0] ，结果[4] 等.是否可以将其与 tbyte 一起使用，如何使用?如果我理解正确-应该是 10 的一个步骤.

And one more question, is it possible in tbyte to somehow save and access numbers by array type? for example, storing numbers in dd, I just could save and read them in increments of 4, for example, result [0], result [4], etc. Is it possible to use the same with tbyte and how? If I understand right - it should be a step of 10.

.386
.model flat,stdcall
option casemap:none

include \masm32\include\masm32rt.inc

.data
titletext db  'Title',0
frmt db 'Result1 = %.7G',10
     db 'Result2 = %.7G',0
buff db 1024 dup (?)
result tbyte ?
num1 qword 5.5
num2 qword 5.6

.code
start:
    finit
    fld qword ptr [num1]
    fld qword ptr [num2]
    fadd
    fstp qword ptr [result]

    invoke crt_sprintf,addr buff,addr frmt, result, result
    invoke MessageBox,0,addr buff,addr titletext,MB_OK
    invoke ExitProcess,0
end start

推荐答案

为什么要对一个tbyte( long double)?

Why are you doing a fstp qword (double) into a tbyte (long double)?

哦，这可能是您的错误.大概 invoke 宏为每个 result 宏args压入12个字节.(由于将10字节的 tbyte 填充到4字节的堆栈插槽的倍数中，因此为12).

Oh, that's probably your bug. Presumably the invoke macro pushes 12 bytes for each of the result macro args. (Because a 10-byte tbyte padded to a multiple of 4-byte stack slots is 12).

但是您的格式字符串仅告诉sprintf查找8字节宽的 double args.由于仅将一个qword double 存储到 result 的低8个字节中，因此 crt_sprintf 可以正确读取第一个可变参数arg作为 double.(x86是little-endian，因此低8个字节位于sprintf正在查看的堆栈地址处.)

But your format string only tells sprintf to look for double args, which are 8 bytes wide. Since you only stored a qword double to the low 8 bytes of result, crt_sprintf can correctly read the first variadic arg as a double. (x86 is little-endian so the low 8 bytes are at the stack address sprintf is looking at.)

但是第二个％G 转换将在上一个arg结束后立即寻找另一个 double .根据格式字符串，哪个应该晚8个字节.但是您的 invoke 实际推送的内容与此不匹配.因此，第二个％G 读取8个字节，与两个12字节的请求重叠.

But the 2nd %G conversion will be looking for another double right after the end of the previous arg. Which according to the format string should be 8 bytes later. But what your invoke actually pushed didn't match that. So the 2nd %G reads 8 bytes that overlap the two 12-byte pushes.

它的高4个字节(包括指数和符号位)可能为 0 ，仅在尾数的低31位中为非零，这给您一个非常小的非正常数.您可以使用调试器将内存作为 double 进行检查，并查看其是否代表sprintf read的值.

It's probably 0 in the upper 4 bytes (including exponent and sign bit), and non-zero only in the low 31 bits of the mantissa, giving you a very small subnormal number. You can use a debugger to examine memory as a double and see that it represents the value sprintf read.

如果该库中的 long double 是10字节x87类型，请使用％LG 并使用 fstptbyte .

If long double in that C library is the 10 byte x87 type, use %LG and use fstp tbyte.

如果 sizeof(long double)只有8，则它与 double 相同，并且您无法使用该C库打印x87个tbyte值.(除非它具有一些非标准扩展名.)在这种情况下，您只需将 result 更改为 qword ，以匹配您正在做的商店

If sizeof(long double) is only 8, then it's the same as double and you can't printf x87 tbyte values with that C library. (Unless it has some non-standard extension for it.) In that case You just change result to also be a qword, matching the store you're doing.

此外，您没有平衡x87堆栈；使用 faddp ，因此在 fstp .(如果汇编器需要操作数，请使用 faddp st(1)或 st1 ，但是它喜欢拼写x87寄存器名称.)

Also, you didn't balance the x87 stack; use faddp so it's empty after the fstp. (If your assembler requires an operand, use faddp st(1) or st1, however it likes to spell x87 register names.)

从技术上讲，您通过对x87堆栈进行非空函数调用而违反了调用约定，但显然crt_sprintf并未使用全部8个 st0..7 ，因此它不会通过溢出x87堆栈获得NaN.

You're technically violating the calling convention by making a function call with the x87 stack non-empty, but apparently crt_sprintf doesn't use all 8 of st0..7 so it doesn't get a NaN from overflowing the x87 stack.

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