Hi I have a base 10 number for example 3198, and the hex representation is 0x0C7E
How do I convert that number to hex and put that hex value in a byte array in the format of [00][0C][7E], assuming the biggest hex value i can have is 0xffffff.
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What C data type do you mean by "base 10 number"? C has integer types, and they don't differentiate between 3198 and 0x0C7E.Klas Lindbäck– Klas Lindbäck2011-12-02 08:26:40 +00:00Commented Dec 2, 2011 at 8:26
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3 Answers
Maybe this will work ?
uint32_t x = 0x0C7E;
uint8_t bytes[3];
bytes[0] = (x >> 0) & 0xFF;
bytes[1] = (x >> 8) & 0xFF;
bytes[2] = (x >> 16) & 0xFF;
/* Go back. */
x = (bytes[2] << 16) | (bytes[1] << 8) | (bytes[0] << 0);
3 Comments
kkh
ah nice! thanks sorry i am very noob at C. But how do I convert 3198 to hex?
kkh
oh another question, if i want o revert the byte array back to the hex format how do i do it?
Number is already a continuous memory block - no need to convert it to yet ANOTHER array ! Just fetch separate bytes by using pointer arithmetic:
EDIT: Edited to be endianness-independent
#define FAST_ABS(x) ((x ^ (x>>31)) - (x>>31))
int is_big_endian(void)
{
union {
uint32_t i;
char c[4];
} bint = {0x01020304};
return bint.c[0] == 1;
}
uint32_t num = 0xAABBCCDD;
uint32_t N = is_big_endian() * 3;
printf("first byte 0x%02X\n"
"second byte 0x%02X\n"
"third byte 0x%02X\n"
"fourth byte 0x%02X\n",
((unsigned char *) &num)[FAST_ABS(3 - N)],
((unsigned char *) &num)[FAST_ABS(2 - N)],
((unsigned char *) &num)[FAST_ABS(1 - N)],
((unsigned char *) &num)[FAST_ABS(0 - N)]
);
4 Comments
FrankH.
Endianness-dependent. If you run this on a SPARC or PPC system (big-endian), you're getting the reverse.
Agnius Vasiliauskas
See edit. I'v changed it to be endianness-invariant. Endianess check idea is taken from here
FrankH.
A complex solution to a problem that doesn't exist if you program in a portable way to start with, re - cnicutar's answer. I give you a +1 for shrewed persistence ;-)
Agnius Vasiliauskas
I totally agree :-) I just wanted to show that in C - everything is bits and bytes - only interpretation of them is what really matters for compilers. That's the main power of C which i love most {and while others hate it and call it "non-existance of type system" in C :-) }
#include <stdio.h>
union uint32_value {
unsigned int value;
struct little_endian {
unsigned char fou;
unsigned char thi;
unsigned char sec;
unsigned char fir;
} le;
struct big_endian {
unsigned char fir;
unsigned char sec;
unsigned char thi;
unsigned char fou;
} be;
};
int main(void)
{
union uint32_value foo;
foo.value = 3198;
printf("%02x %02x %02x %02x\n", foo.le.fir, foo.le.sec, foo.le.thi, foo.le.fou);
return 0;
}
