用printf打印出的浮动值 [英] using printf to print out floating values

问题描述

#include<stdio.h>
#include<stdlib.h>

int main(void)
{
  int x, *ptr_x;
  float f , *ptr_f;

  ptr_f = &f;
  ptr_x = &x;
  *ptr_x = 5;
  *ptr_f = 1.5; //printf("%d %f\n", f,x);

  printf ("\n\nxd = %d \t xf = %f \n ff = %f \t fd = %d", x,x,f,f);
  return 0;
}

预计不会对FF =％f显示输出。

The output for ff = %f is not expected.

XD = 5 XF = 0.000000
结果 FF = 0.000000 的FD = 1073217536

xd = 5 xf = 0.000000
ff = 0.000000 fd = 1073217536

这在code的一点​​是要显示，如果一个浮动值印有％D和如果一个int值印制％F会发生什么。

The point of the this code is to show what would happen if a floating value is printed with %d and if a int value is printed %f.

为什么不被打印的浮点值正常，即使我用％F？

Why is the float value not being printed properly even if i use %f ?

推荐答案

的printf（）不是类型安全的。

传递给的printf（）的参数根据是什么您的承诺编译器进行处理。

The arguments that you pass to printf() are treated according to what you promise the compiler.

此外，浮动，则晋升为双击当通过可变参数去世了。

Also, floats are promoted to doubles when passed through variadic arguments.

所以，当你答应编译器％F 第一时间（ XF ），编译器消耗了一个整个双击（通常为8个字节）从参数，吞咽你漂浮在过程中。然后第二个％F 削减对入零尾数的第二个双。

So when you promise the compiler %f the first time (for xf), the compiler gobbles up an entire double (usually 8 byte) from the arguments, swallowing your float in the process. Then the second %f cuts right into the zero mantissa of the second double.

下面是你的论点图片：

+-0-1-2-3-+-0-1-2-3-+-0-1-2-3-4-5-6-7-+-0-1-2-3-4-5-6-7-+
|    x    |    x    |        f        |        f        |
+---------+---------+-----------------+-----------------+

%d--------|%f----------------|%f---------------|%d------|

但˚F看起来像这样（已晋升为双击）

But f looks like this (having been promoted to double):

f = 3FF8000000000000

让我们从价值观再次战平它，猜测你的机器字节顺序：

Let's draw it again with values, and speculating about your machine endianness:

| 05000000 | 05000000 | 00000000 0000F83F | 00000000 0000F83F |
| %d, OK   | %f, denormal...    | %f, denormal...   | %d, OK  |

注意，1073217536是0x3FF80000。

Note that 1073217536 is 0x3FF80000.

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