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;
}
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);