什么float值使得sprintf_s（）产生＆QUOT; 1＃QO＆QUOT;？ [英] What float value makes sprintf_s() produce "1.#QO"?

问题描述

我有一些（旧的嵌入式C）code这一些的sprintf 通话装置产生.csv文件。偶尔我看到 1＃QO 的值。我已经试过这应该给负无穷，正无穷大和NaN，但他们都不似乎给我的神奇 1＃QO 结果重现条件的值。那么，什么是它产生的价值呢？

I have some (legacy embedded c) code which produces a .csv file by means of some sprintf calls. Occasionally I see values of 1.#QO. I've tried reproducing those values with conditions which should give negative infinity, positive infinity and NaN but none of them appear to give me the magical 1.#QO result. So what is it that produces that value?

...是的，我知道有一些很明显在其中产生价值数学会错，但了解这意味着什么将有助于调试工作。

...and yes, I know there's obviously something going wrong in the maths which produce that value, but understanding what it means would assist in the debugging effort.

由它来完成转换的实际行是：

The actual line which does the conversion is:

sprintf_s(txt, CSV_HEADER_SIZE, "%.3f", value);

其中：

#define CSV_HEADER_SIZE (100)
char txt[CSV_HEADER_SIZE];

我与微软的Visual Studio 2008的编译。

I'm compiling with MS Visual Studio 2008.

多一点挖节目 0xFFFFFFFF的给 -1＃QO ：

unsigned int i = 0xFFFFFFFF;
float* f = (float*)&i;
printf("%.3f", *f); // gives -1.#QO

..看着，在Visual Studio调试器它扩展为 -1＃QNAN00 所以它看起来像这可能是微软特有的再$ P $的psentation NaN的？

..and looking at that in the Visual Studio debugger expands it to -1.#QNAN00 so it looks like this is probably a Microsoft-specific representation of NaN?

推荐答案

- 1＃QO是-1＃QNAN。四舍五入为小数点后3位之后。 N个回合到O为'A'> ='5'和'N'+ 1 =='O'。

"-1.#QO" is "-1.#QNAN" after "rounding" for 3 places after the decimal. The N rounds to an O as 'A' >= '5' and 'N' + 1 == 'O'.

这同样是为什么你的调试器显示-1。＃QNAN00，它打印与7位和添加填充零到最后。

This is similarly why your debugger shows "-1.#QNAN00", as it prints with 7 places and adds padding zeros to the end.

QNAN是静态NaN 。

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