convert int to float to hex - c

Question

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
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;
}

Answer 1

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);

Answer 2

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

Answer 3

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.