convert int to float to hex

Using scanf, each number typed in, i would like my program to print out two lines: for example

byte order: little-endian

> 2
     2 0x00000002
  2.00 0x40000000

> -2
    -2 0xFFFFFFFE
 -2.00 0xC0000000
开发者_Python百科

I can get it to print out the 2 in hex but i also need a float and of course i cant scanf as one when i need to also scan as an int

If i cast as a float when i try to printf i get a zero. If i scan in as a float i get the correct output. I have tried to convert the int to a float but it still comes out as zero.

here is my output so far

Int - float - hex

byte order: little-endian

>2

         2  0x000002
      2.00  00000000

it looks like i am converting to a float fine why wont it print as a hex? if i scan in as a float i get the correct hex representation like the first example. this should be something simple. i do need to scan in as a decimal keep in mind i am running this in cygwin

here is what i have so far..

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

int main(int argc, char *argv[])
{


int HexNumber;
    float convert;
printf("Int - float - hex\n");



int a = 0x12345678;
unsigned char *c = (unsigned char*)(&a);
if (*c == 0x78)
{
    printf("\nbyte order: little-endian\n");
}
else
{
    printf("\nbyte order: big-endian\n");
}

printf("\n>");
scanf("%d", &HexNumber);
printf("\n%10d  ",HexNumber);
printf("%#08x",HexNumber);





convert =  (float)HexNumber; // converts but prints a zero

printf("\n%10.2f  ", convert); 
printf("%#08x", convert); // prints zeros


return 0;
}

try this:

int i = 2;
float f = (float)i;
printf("%#08X", *( (int*) &f ));

[EDIT]

@Corey:

let's parse it from inside out:

&  f = address of f = say address 0x5ca1ab1e
(int*)  &f = interpret the address 0x5ca1ab1e as integer pointer
*  ((int*)&f) = get the integer at address 0x5ca1ab1e

the following is more concise, but it's hard to remember the C language's operator associativity and operator precedence(i prefer the extra clarity of some added parenthesis and whitespace provides):

printf("%#08X", *(int*)&f);

printf("%#08x", convert); // prints zeros

This line is not going to work because you are telling printf that you are passing in an int (by using the %x) but infact you are passing it in a float.

What is your intention with this line? To show the binary representation of the floating point number in hex? If so, you may want to try something like this:

printf("%lx\n", *(unsigned long *)(&convert));

What this line is doing is taking the address of convert (&convert) which is a pointer to a float and casting it into a pointer to an unsigned long (note: that the type you cast into here may be different depending on the size of float and long on your system). The last * is dereferencing the pointer to an unsigned long into an unsigned long which is passed to printf

Given an int x, converting to float, then printing out the bytes of that float in hex could be done something like this:

show_as_float(int x) { 
   float xx = x;

   //Edit: note that this really prints the value as a double.
   printf("%f\t", xx);

   unsigned char *ptr = (unsigned char *)&xx;

   for (i=0; i<sizeof(float); i++)
       printf("%2.2x", ptr[i]);
}

The standards (C++ and C99) give "special dispensation" for unsigned char, so it's safe to use them to view the bytes of any object. C89/90 didn't guarantee that, but it was reasonably portable nonetheless.