转换十六进制浮点常数 [英] Convert Hexadecimal Floating-Point Constant

问题描述

我有一些要转换的常量，它们的定义如下:

I have some constants to convert, they are defined as follow :

0x1.0p-126f

0x1.0p-126f

0x1.fffffep127f

0x1.fffffep127f

0x1.0p-23f

0x1.0p-23f

如何准确转换?

顺便说一句，这是因为Visual Studio不支持该表示法！！

BTW, it is because Visual Studio does not support this notation !!

推荐答案

在不符合C99的编译器中，您可以编写:

In a non-C99 compliant compiler, you can write:

0x1.0p-126f -> ldexp (1.0, -126) 

0x1.fffffep127f -> ldexp(0x1fffffe,  127 - 4 * 6)

0x1.0p-23f -> ldexp(1.0, -23)

在第二个示例中， 6 是点后的十六进制数字的数量.原始文件具有 0x1.fffffe ，因此当我写整数 0x1fffffe 时，我写出的数字对于从后移出的每个数字要大2 ⁴ 倍指向该点之前的点，应使用第二个参数进行补偿.

In the second example, the 6 is the number of hexadecimal digits that were after the dot. The original had 0x1.fffffe so when I wrote the integer 0x1fffffe, I wrote a quantity 2⁴ times larger for each digit I shifted from after the point to before the point, which should be compensated for with the second argument.

函数 ldexp()是标准的:

简介

 #include <math.h>

 double
 ldexp(double x, int n);

说明

 The ldexp() functions multiply x by 2 to the power n.

如果这些表达式出现在需要常量表达式的地方，则函数调用不是可接受的替代方法.在这种情况下，只需使用C程序以％.16e 格式打印 ldexp()调用的结果，然后使用打印的值代替原始值.

If these expressions appear in a place where constant expressions are expected, then function calls are not an acceptable substitute. In this case, simply print the result of the ldexp() calls with a C program and the format %.16e, and use the printed value in substitution for the original.

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