using printf to print out floating values

Question

#include
#include

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

  ptr_f = &f;
  ptr_x = &x;
  *ptr_x = 5;
  *ptr_f =         


        

Answer 1

printf() is not typesafe.

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

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

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.

Here's a picture of your arguments:

+-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------|

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

f = 3FF8000000000000

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  |

Note that 1073217536 is 0x3FF80000.