I'm reading data from a serial port from a hardware device which I need to increment by 1 and then send back out to the device however, I need to increment it as if it were a base10 number.
For example, if I read 0x09, I need to send back 0x10 rather than 0x0a. Or, if I receive 0x89, I should send back 0x90. If I receive 0x99, I send back 0x00 and carry the 1 up to the previous byte. It's actually a total of 5 bytes I have to run through.
I have this increment working in the following way. I'd like know know if there's a better way through some unique shifting and/or and/or'ing of bits.
Thank you for any pointers you can provide!
Stateful
#include <stdio.h>
#include <stdlib.h>
int main()
{
//start with 0x09 as byte
char input = 0x09;
printf("input is: 0x%02x\n", input);
//increment it by one
input++;
//turn it into a two char array as a base10 value, ignore overflow for now
char asString[3];
sprintf(asString, "%d", input);
//convert back to byte
unsigned char newI = ((asString[0]-0x30)*16)+((asString[1]-0x30));
printf ("newI is 0x%02x\n", newI);
return 0;
}
You compute the modulo 16 of the received number.
If it's 9, you add 7, else 1.
Convert the whole byte sequence from BCD to an integer type, add, then convert back. Something like these functions should work for the conversion, but be aware that you may need a longer type than unsigned if you need to support completely arbitrary 5-byte BCD sequences (but 5 bytes of BCD suspiciously coincides with the range of a 32-bit integer).
/*
* Decodes a BCD byte sequence to an unsigned integer. The bytes are assumed to be in
* order from most- to least-significant.
*/
unsigned bcd_to_int(unsigned char bytes[], int byte_count) {
unsigned result = 0;
int counter;
for (counter = 0; counter < byte_count; counter += 1) {
result = result * 100 + (bytes[counter] & 0xf0) * 10 + (bytes[counter] & 0x0f);
}
return result;
}
/*
* Encodes an unsigned integer into a BCD byte sequence. The bytes will be ordered
* from most- to least-significant.
*/
void int_to_bcd(unsigned char bytes[], int byte_count, unsigned value) {
int counter;
for (counter = byte_count; counter-- > 0; ) {
unsigned chunk = value % 100;
bytes[counter] = (chunk / 10) * 0x10 + (chunk % 10);
value /= 100;
}
}
You could also implement long-form addition directly on your byte sequence; that might perform as well or better, but if you want to perform more or different operations than a single add / increment then it will be to your advantage to use native arithmetic.
Related
i am again facing a formatting problem. I want to pass a port number (as integer) as parameter to a function (argv[]) and want to display it in a special format. In my actual case i want to display the port number 1234 in hexadecimal. I try it this way
int port = 1234;
char* _port = (char*)&port;
for (int i = 0; i < sizeof(port); i++) {
printf("\\x%02x", _port[i]);
}
but it shows
\xffffffd2\x04\x00\x00
But i want it in format with leading zeros and 2 digits like
\x04\xd2
can you help me please?
EDIT: I changed it to
sizeof(port)-2
and it shows only 2 digits but in the wrong endian :S
On most systems the size of int is four bytes, 32 bits. The hexadecimal representation of 1234 is 0x000004d2. On a little-endian system (like x86 and x86-64) it's stored in memory like the four bytes 0xd2, 0x04, 0x00 and 0x00 in that order.
If we look at it as an array of bytes, it looks like
+------+------+------+------+
| 0xd2 | 0x04 | 0x00 | 0x00 |
+------+------+------+------+
There are three problems you have:
You loop over all four bytes of the int, while you only want the significant bits
You don't consider the endianness
That char on your system is signed and when promoted to int it will be sign-extended (according to the two's complement rules)
To solve the first point you need to discard the "leading" zero bytes.
To solve the second point you need to loop from the end (but only on little-endian systems).
To solve the third point use a type which won't be sign-extended (i.e. uint8_t).
Put together you could do something like this:
// The number we want to print
int port = 1234;
// Copy the raw binary data to a buffer
// This buffer is to not break strict aliasing
uint8_t _port[sizeof port];
memcpy(_port, &port, sizeof port);
// Skip leading zeroes in the buffer
// This is done by looping from the end of the buffer to the beginning,
// and loop as long as the current byte is zero
uint8_t *current;
for (current = _port + sizeof _port - 1; current > _port && *current == 0; --current)
{
// Empty
}
// Print the remaining bytes
for (; current >= _port; --current)
{
printf("\\x%02x", *current); // Print with zero-padding, so e.g. \x4 becomes \x04
}
Proof of concept
Get rid of signess and modify the formats.
void foo(int port, int endianess)
{
unsigned char * _port = (unsigned char*)&port;
if(endianess)
{
for (size_t i = 0; i < 2; i++)
{
printf("\\x%02hhx", _port[i]);
}
}
else
{
for (size_t i = sizeof(port) - 1; i >= sizeof(port) - 2; i--)
{
printf("\\x%02hhx", _port[i]);
}
}
}
I'm doing an assignment for school to swap the bytes in an unsigned long, and return the swapped unsigned long. ex. 0x12345678 -> 0x34127856.
I figured I'll make a char array, use sprintf to insert the long into a char array, and then do the swapping, stepping through the array. I'm pretty familiar with c++, but C seems a little more low level. I researched a few topics on sprintf, and I tried to make an array, but I'm not sure why it's not working.
unsigned long swap_bytes(unsigned long n) {
char new[64];
sprintf(new, "%l", n);
printf("Char array is now: %s\n", new);
}
TLDR; The correct approach is at the bottom
Preamble
Issues with what you're doing
First off using sprintf for byte swapping is the wrong approach because
it is a MUCH MUCH slower process than using the mathematical properties of bit operations to perform the byte swapping.
A byte is not a digit in a number. (a wrong assumption that you've made in your approach)
It's even more painful when you don't know the size of your integer (is it 32-bits, 64 bits or what)
The correct approach
Use bit manipulation to swap the bytes (see way way below)
The absolutely incorrect implementation with wrong output (because we're ignoring issue #2 above)
There are many technical reasons why sprintf is much slower but suffice it to say that it's so because moving contents of memory around is a slow operation, and of course more data you're moving around the slower it gets:
In your case, by changing a number (which sits in one manipulatable 'word' (think of it as a cell)) into its human readable string-equivalence you are doing two things:
You are converting (let's assume a 64-bit CPU) a single number represented by 8 bytes in a single CPU cell (officially a register) into a human equivalence string and putting it in RAM (memory). Now, each character in the string now takes up at least a byte: So a 16 digit number takes up 16 bytes (rather than 8)
You are then moving these characters around using memory operations (which are slow compared do doing something directly on CPU, by factor of a 1000)
Then you're converting the characters back to integers, which is a long and tedious operation
However, since that's the solution that you came up with let's first look at it.
The really wrong code with a really wrong answer
Starting (somewhat) with your code:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
unsigned long swap_bytes(unsigned long n) {
int i, l;
char new[64]; /* the fact that you used 64 here told me that you made assumption 2 */
sprintf(new, "%lu", n); /* you forgot the `u` here */
printf("The number is: %s\n", new); /* well it shows up :) */
l = strlen(new);
for(i = 0; i < l; i+=4) {
char tmp[2];
tmp[0] = new[i+2]; /* get next two characters */
tmp[1] = new[i+3];
new[i+2] = new[i];
new[i+3] = new[i+1];
new[i] = tmp[0];
new[i+1] = tmp[1];
}
return strtoul(new, NULL, 10); /* convert new back */
}
/* testing swap byte */
int main() {
/* seems to work: */
printf("Swapping 12345678: %lu\n", swap_bytes(12345678));
/* how about 432? (err not) */
printf("Swapping 432: %lu\n", swap_bytes(432));
}
As you can see the above is not really byte swapping but character swapping. And any attempt to try and "fix" the above code is nonsensical. For example,how do we deal with odd number of digits?
Well, I suppose we can pad odd digit counts with a zero:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
unsigned long swap_bytes(unsigned long n) {
int i, l;
char new[64]; /* the fact that you used 64 here told me that you made assumption 2 */
sprintf(new, "%lu", n); /* you forgot the `u` here */
printf("The number is: %s\n", new); /* well it shows up :) */
l = strlen(new);
if(l % 2 == 1) { /* check if l is odd */
printf("adding a pad to make n even digit count");
sprintf(new, "0%lu", n);
l++; /* length has increased */
}
for(i = 0; i < l; i+=4) {
char tmp[2];
tmp[0] = new[i+2]; /* get next two characters */
tmp[1] = new[i+3];
new[i+2] = new[i];
new[i+3] = new[i+1];
new[i] = tmp[0];
new[i+1] = tmp[1];
}
return strtoul(new, NULL, 10); /* convert new back */
}
/* testing swap byte */
int main() {
/* seems to work: */
printf("Swapping 12345678: %lu\n", swap_bytes(12345678));
printf("Swapping 432: %lu\n", swap_bytes(432));
/* how about 432516? (err not) */
printf("Swapping 432: %lu\n", swap_bytes(432));
}
Now we run into an issue with numbers which are not divisible by 4... Do we pad them with zeros on the right or the left or the middle? err NOT REALLY.
In any event this entire approach is wrong because we're not swapping bytes anyhow, we're swapping characters.
Now what?
So you may be asking
what the heck is my assignment talking about?
Well numbers are represented as bytes in memory, and what the assignment is asking for is for you to get that representation and swap it.
So for example, if we took a number like 12345678 it's actually stored as some sequence of bytes (1 byte == 8 bits). So let's look at the normal math way of representing 12345678 (base 10) in bits (base 2) and bytes (base 8):
(12345678)10 = (101111000110000101001110)2
Splitting the binary bits into groups of 4 for visual ease gives:
(12345678)10 = (1011 1100 0110 0001 0100 1110)2
But 4 bits are equal to 1 hex number (0, 1, 2, 3... 9, A, B...F), so we can convert the bits into nibbles (4-bit hex numbers) easily:
(12345678)10 = 1011 | 1100 | 0110 | 0001 | 0100 | 1110
(12345678)10 = B | C | 6 | 1 | 4 | E
But each byte (8-bits) is two nibbles (4-bits) so if we squish this a bit:
(12345678)10 = (BC 61 4E)16
So 12345678 is actually representable in 3 bytes;
However CPUs have specific sizes for integers, usually these are multiples of 2 and divisible by 4. This is so because of a variety of reasons that are beyond the scope of this discussion, suffice it to say that you will get things like 16-bit, 32-bit, 64-bit, 128-bit etc... And most often the CPU of a particular bit-size (say a 64bit CPU) will be able to manipulate unsigned integers representable in that bit-size directly without having to store parts of the number in RAM.
Slight Digression
So let's say we have a 32-bit CPU, and somewhere at byte number α in RAM. The CPU could store the number 12345678 as:
> 00 BC 61 4E
> ↑ α ↑ α+1 ↑ α+2 ↑ α+3
(Figure 1)
Here the most significant part of the number, is sitting at the lowest memory address index α
Or the CPU could store it differently, where the least significant part of the number is sitting at the lowest memory.
> 4E 61 BC 00
> ↑ α ↑ α+1 ↑ α+2 ↑ α+3
(Figure 2)
The way a CPU stores a number is called Endianness (of the CPU). Where, if the most significant part is on the left then it's called Big-Endian CPU (Figure 1), or Little-Endian if it stores it as in (Figure 2)
Getting the correct answer (the wrong way)
Now that we have an idea of how things may be stored, let's try and pull this out still using sprintf.
We're going to use a couple of tricks here:
we'll convert the numbers to hexadecimal and then pad the number to 8 bytes
we'll use printf's (therefore sprintf) format string capability that if we want to use a variable to specify the width of an argument then we can use a * after the % sign like so:
printf("%*d", width, num);
If we set our format string to %0*x we get a hex number that's zero padded in output automatically, so:
sprintf(new, "%0*llx", sizeof(n), n);
Our program then becomes:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
unsigned long swap_bytes(unsigned long n) {
int i, l;
char new[64] = "";
sprintf(new, "%0*llx", sizeof(n), n);
printf("The number is: %s\n", new);
l = strlen(new);
for(i = 0; i < l; i+=4) {
char tmp[2];
tmp[0] = new[i+2]; /* get next two characters */
tmp[1] = new[i+3];
new[i+2] = new[i];
new[i+3] = new[i+1];
new[i] = tmp[0];
new[i+1] = tmp[1];
}
return strtoul(new, NULL, 16); /* convert new back */
}
/* testing swap byte */
int main() {
printf("size of unsigned long is %ld\n", sizeof(unsigned long));
printf("Swapping 12345678: %llx\n", swap_bytes(12345678));
/* how about 123456? */
printf("Swapping 123456: %llx\n", swap_bytes(123456));
printf("Swapping 123456: %llx\n", swap_bytes(98899));
}
The output would look something like:
size of unsigned long is 8
The number is: 00bc614e
Swapping 12345678: bc004e61
The number is: 0001e240
Swapping 123456: 10040e2
The number is: 00018253
Swapping 123456: 1005382
Obviously we can change our outputs by using %ld and print the base 10 versions of the numbers, rather than base 16 as is happening above. I'll leave that to you.
Now let's do it the right way
This is however rather terrible, since byte swapping can be done much faster without ever doing the integer to string and string to integer conversion.
Let's see how that's done:
The rather explicit way
Before we go on, just a bit on bit shifting in C:
If I have a number, say 6 (=1102) and I shift all the bits to the left by 1 I would get 12 (11002) (we simply shifted everything to the left adding zeros on the right as needed)
This is written in C as 6 << 1.
A right shift is similar and can be expressed in C with >> so if I have a number say 240 = (11110000)2 and I right-shift it 4 times I would get 15 = (1111)2 this is expressed as 240 >> 3
Now we have unsigned long integers which are (in my case at least) 64 bits long, or 8 bytes long.
Let's say my number is 12345678 which is (00 00 00 00 00 bc 61 4e)16 in hex at 8 bytes long. If I want to get the value of byte number 3 I can extract it by taking the number 0xFF (1111 1111) all bits of a byte set to 1 and left shifting it until i get to the byte 3 (so left shift 3*8 = 24 times) performing a bitwise and with the number and then right shifting the results to get rid of the zeros. This is what it looks like:
0xFF << (3 * 8) = 0xFF0000 & 0000 0000 00bc 614e = 0000 0000 00bc 0000
Now right shift:
0xFF0000 & 0000 0000 00bc 0000 >> (3 * 8) = bc
Another (better) way to do it would be to right shift first and then perform bitwise and with 0xFF to drop all higher bits:
0000 0000 00bc 614e >> 24 = 0000 0000 0000 00bc & 0xFF = bc
We will use the second way, and make a macro using #define now we can add the bytes back at the right location by right shifting each kth byte k+1 times and each k+1st byte k times.
Here is a sample implementation of this:
#define GET_BYTE(N, B) ((N >> (8 * (B))) & 0xFFUL)
unsigned long swap_bytes(unsigned long n)
{
unsigned long long rv = 0ULL;
int k;
printf("number is %016llx\n", n);
for(k =0 ; k < sizeof(n); k+=2) {
printf("swapping bytes %d[%016lx] and %d[%016lx]\n", k, GET_BYTE(n, k),
k+1, GET_BYTE(n, k+1));
rv += GET_BYTE(n, k) << 8*(k+1);
rv += GET_BYTE(n, k+1) << 8*k;
}
return rv;
}
/* testing swap byte */
int main() {
printf("size of unsigned long is: %ld\n", sizeof(unsigned long));
printf("Swapping 12345678: %llx\n", swap_bytes(12345678));
/* how about 123456? */
printf("Swapping 123456: %llx\n", swap_bytes(123456));
printf("Swapping 123456: %llx\n", swap_bytes(98899));
}
But this can be done so much more efficiently. I leave it here for now. We'll come back to using bit blitting and xor swapping later.
Update with GET_BYTE as a function instead of a macro:
#define GET_BYTE(N, B) ((N >> (8 * (B))) & 0xFFUL)
Just for fun we also use a shift operator for multiplying by 8. You can note that left shifting a number by 1 is like multiplying it by 2 (makes sense since in binary 2 is 10 and multiplying by 10 adds a zero to the end and therefore is the same as shifting something left by one space) So multiplying by 8 (1000)2 is like shifting something three spaces over or basically tacking on 3 zeros (overflows notwithstanding):
unsigned long __inline__ get_byte(const unsigned long n, const unsigned char idx) {
return ((n >> (idx << 3)) & 0xFFUL);
}
Now the really really fun and correct way to do this
Okay so a fast way to swap integers around is to realize that if we have two integers x, and y we can use properties of xor function to swap their values. The basic algorithm is this:
X := X XOR Y
Y := Y XOR X
X := X XOR Y
Now we know that a char is one byte in C. So we can force the compiler to treat the 8 byte integer as a sequence of 1-byte chars (hehe it's a bit of a mind bender considering everything I said about not doing it in sprintf) but this is different. You have to just think about it a bit.
We'll take the memory address of our integer, cast it to a char pointer (char *) and treat the result as an array of chars. Then we'll use the xor function property above to swap the two consecutive array values.
To do this I am going to use a macro (although we could use a function) but using a function will make the code uglier.
One thing you'll note is that there is the use of ?: in XORSWAP below. That's like an if-then-else in C but with expressions rather than statements, so basically (conditional_expression) ? (value_if_true) : (value_if_false) means if conditional_expression is non-zero the result will be value_if_true, otherwise it will be value_if_false. AND it's important not to xor a value with itself because you will always get 0 as a result and clobber the content. So we use the conditional to check if the addresses of the values we are changing are DIFFERENT from each other. If the addresses are the same (&a == &b) we simply return the value at the address (&a == &b) ? a : (otherwise_do_xor)
So let's do it:
#include <stdio.h>
/* this macro swaps any two non floating C values that are at
* DIFFERENT memory addresses. That's the entire &a == &b ? a : ... business
*/
#define XORSWAP(a, b) ((&(a) == &(b)) ? (a) : ((a)^=(b),(b)^=(a),(a)^=(b)))
unsigned long swap_bytes(const unsigned long n) {
unsigned long rv = n; /* we are not messing with original value */
int k;
for(k = 0; k < sizeof(rv); k+=2) {
/* swap k'th byte with k+1st byte */
XORSWAP(((char *)&rv)[k], ((char *)&rv)[k+1]);
}
return rv;
}
int main()
{
printf("swapped: %lx", swap_bytes(12345678));
return 0;
}
Here endeth the lesson. I hope that you will go through all the examples. If you have any more questions just ask in comments and I'll try to elaborate.
unsigned long swap_bytes(unsigned long n) {
char new[64];
sprintf(new, "%lu", n);
printf("Char array is now: %s\n", new);
}
You need to use %lu - long unsigned, for format in sprintf(), the compiler should also given you conversion lacks type warning because of this.
To get it to print you need to use %lu (for unsigned)
It doesn't seem like you attempted the swap, could I see your try?
I'm writing some code for an old 8-bit microprocessor (the Hu6280 - a WDC 65C02 derivative in the old NEC PC-Engine console) with 32kb of ram and up to 2.5mbytes of data/code rom. The language is a variant of Small-C but is limited to just the two following basic types:
char (1 byte)
int (2 byte)
It has no struct support and no long int support.
I'm writing a FAT filesystem library to interface with a SD card reader that was primarily developed for loading game ROM images, however an enterprising hacker has written some assembly to allow raw sector reading from the console side. He achieves this by stuffing the 4 8bit values of a 32bit sector address into 4 consecutive memory addresses (char address[4];).
My C code leverages his work to read (for the moment) the dos MBR boot sector and partition type information off the SD card. I've got MBR checksum verifcation and FAT partition detection working.
However, as I need to support FAT32 (which is what the FPGA on the SD card device supports), most of the sector and cluster arithmetic to look up directory entries and files will be based on 32bit LBA sector values.
What easy mechanisms do I have to do add/subtract/multiply 8/16/32bit integers, based on the above limitations? Does anyone have any ready made C routines to handle this? Maybe something along the lines of:
char int1[4], int2[4], int3[4];
int1[0] = 1;
int1[1] = 2;
int1[2] = 3;
int1[3] = 4;
int2[0] = 4;
int2[1] = 3;
int2[2] = 2;
int2[3] = 1;
int3 = mul_32(int1, int2);
int3 = add_32(int1, int2);
int3 = sub_32(int1, int2);`
EDIT: Based on the above replies, this is what I've come up with so far - this is untested as yet and I'll need to do similar for multiplication and subtraction:
char_to_int32(int32_result, int8)
char* int32_result;
char int8;
{
/*
Takes an unsigned 8bit number
and converts to a packed 4 byte array
*/
int32_result[0] = 0x00;
int32_result[1] = 0x00;
int32_result[2] = 0x00;
int32_result[3] = int8;
return 0;
}
int_to_int32(int32_result, int16)
char* int32_result;
int int16;
{
/*
Takes an unsigned 16bit number
and converts to a packed 4 byte array
*/
int32_result[0] = 0x00;
int32_result[1] = 0x00;
int32_result[2] = (int16 >> 8);
int32_result[3] = (int16 & 0xff);
return 0;
}
int32_is_zero(int32)
char* int32;
{
/*
Is a packed 4 byte array == 0
returns 1 if true, otherwise 0
*/
if ((int32[0] == 0) & (int32[1] == 0) & (int32[2] == 0) & (int32[3] == 0)) {
return 1;
} else {
return 0;
}
}
add_32(int32_result, int32_a, int32_b)
char* int32_result;
char* int32_a;
char* int32_b;
{
/*
Takes two 32bit values, stored as 4 bytes each -
adds and stores the result.
Returns 0 on success, 1 on error or overflow.
*/
int sum;
char i;
char carry;
carry = 0x00;
/* loop over each byte of the 4byte array */
for (i = 4; i != 0; i--) {
/* sum the two 1 byte numbers as a 2 byte int */
sum = int32_a[i-1] + int32_b[i-1] + carry;
/* would integer overflow occur with this sum? */
if (sum > 0x00ff) {
/* store the most significant byte for next loop */
carry = (sum >> 8);
} else {
/* no carry needed */
carry = 0x00
}
/* store the least significant byte */
int32_result[i+1] = (sum & 0xff);
}
/* Has overflow occured (ie number > 32bit) */
if (carry != 0) {
return 1;
} else {
return 0;
}
}
EDIT 2: Here's an updated and tested version of the emulated 32bit + 32bit integer add code. It works with all values I've tried so far. Overflow for values bigger than a 32bit unsigned integer is not handled (will not be required for my purposes):
add_int32(int32_result, int32_a, int32_b)
char* int32_result;
char* int32_a;
char* int32_b;
{
/*
Takes two 32bit values, stored as 4 bytes each -
adds and stores the result.
Returns 0 on success, 1 on error or overflow.
*/
int sum;
char i, pos;
char carry;
zero_int32(int32_result);
carry = 0x00;
/* loop over each byte of the 4byte array from lsb to msb */
for (i = 1; i < 5; i++) {
pos = 4 - i;
/* sum the two 1 byte numbers as a 2 byte int */
sum = int32_a[pos] + int32_b[pos] + carry;
/* would integer overflow occur with this sum? */
if (sum > 0x00ff) {
/* store the most significant byte for next loop */
carry = (sum >> 8);
} else {
/* no carry needed */
carry = 0x00;
}
/* store the least significant byte */
int32_result[pos] = (sum & 0x00ff);
}
/* Has overflow occured (ie number > 32bit) */
if (carry != 0) {
return 1;
} else {
return 0;
}
}
I also found some references to 32bit arithmetic on some PIC controllers after searching SO a bit more:
http://web.media.mit.edu/~stefanm/yano/picc_Math32.html
Although there is some PIC assembly inline in their add/subtract code, there are some useful platform agnostic char-based C functions there that have already implemented shifts, comparisons, increment/decrement etc, which will be very useful. I will look into subtract and multiply next - thanks for the info; I guess I was looking at things and thinking they were much harder than they needed to be.
I know you know how to do this. go back to your grade school math...
When you multiply to numbers, base 10
12
x34
====
You do four multiplications right and then add four numbers together right?
4x2 = 8
4x1 = 4
3x2 = 6
3x1 = 3
then
12
x34
====
0008
0040
0060
+0300
======
Now what about addition
12
+34
===
We learned to break that down into two additions
2+4 = 6 carry a zero
1+3+carryin of 0 = 4
With that knowledge that you already have from childhood, you then simply apply that. remember that basic math works whether we have 2 digits operated on 2 digits or 2 million digits operated on 2 million digits.
The above uses single decimal numbers, but the math works if it were single base 16 numbers or single bits or octal or bytes, etc.
Your C compiler should already be handling these things for you but if you need to synthesize them you can, multiplication at the easiest form for digital is to use bits.
addition is easier with bytes using assembly because the carry out is right there, C does not have a carry out so you have to do the exercise of figuring out the carry out using 8 bit math (it can be determined) without needing a 9th bit. or you can just do something less than 8 bit math, 7 or 4 or whatever.
As Joachim pointed out, this topic hsa been beat to death decades/centuries ago. At the same time it is so simple that it often doesnt warrent a lot of discussion. StackOverflow certainly has this topic covered several times over.
I have some code below that is supposed to be converting a C (Arduino) 8-bit byte array to a 16-bit int array, but it only seems to partially work. I'm not sure what I'm doing wrong.
The byte array is in little endian byte order. How do I convert it to an int (two bytes per enty) array?
In layman's terms, I want to merge every two bytes.
Currently it is outputting for an input BYTE ARRAY of: {0x10, 0x00, 0x00, 0x00, 0x30, 0x00}. The output INT ARRAY is: {1,0,0}. The output should be an INT ARRAY is: {1,0,3}.
The code below is what I currently have:
I wrote this function based on a solution in Stack Overflow question Convert bytes in a C array as longs.
I also have this solution based off the same code which works fine for byte array to long (32-bits) array http://pastebin.com/TQzyTU2j.
/**
* Convert the retrieved bytes into a set of 16 bit ints
**/
int * byteA2IntA(byte * byte_slice, int sizeOfB, int * ret_array){
//Variable that stores the addressed int to be stored in SRAM
int currentInt;
int sizeOfI = sizeOfB / 2;
if(sizeOfB % 2 != 0) ++sizeOfI;
for(int i = 0; i < sizeOfB; i+=2){
currentInt = 0;
if(byte_slice[i]=='\0') {
break;
}
if(i + 1 < sizeOfB)
currentInt = (currentInt << 8) + byte_slice[i+1];
currentInt = (currentInt << 8) + byte_slice[i+0];
*ret_array = currentInt;
ret_array++;
}
//Pointer to the return array in the parent scope.
return ret_array;
}
What is the meaning of this line of code?
if(i + 1 < sizeOfB) currentInt = (currentInt << 8) + byte_slice[i+1];
Here currentInt is always 0 and 0 << 8 = 0.
Also what you do is, for each couple of bytes (let me call them uint8_t from now on), you pack an int (let me call it uint16_t from now on) by doing the following:
You take the rightmost uint8_t
You shift it 8 positions to the left
You add the leftmost uint8_t
Is this really what you want?
Supposing you have byte_slice[] = {1, 2}, you pack a 16 bit integer with the value 513 (2<<8 + 1)!
Also, you don't need to return the pointer to the array of uint16_t as the caller has already provided it to the function.
If you use the return of your function, as Joachim said, you get a pointer starting from a position of the uint16_t array which is not position [0].
Vincenzo has a point (or two), you need to be clear what you're trying to do;
Combine two bytes to one 16-bit int, one byte being the MSB and one byte being the LSB
int16 result = (byteMSB << 8) | byteLSB;
Convert an array of bytes into 16-bit
for(i = 0; i < num_of_bytes; i++)
{
myint16array[i] = mybytearray[i];
}
Copy an array of data into another one
memcpy(dest, src, num_bytes);
That will (probably, platform/compiler dependent) have the same effect as my 1st example.
Also, beware of using ints as that suggests signed values, use uints, safer and probably faster.
The problem is most likely that you increase ret_array and then return it. When you return it, it will point to one place beyond the destination array.
Save the pointer at the start of the function, and use that pointer instead.
Consider using a struct. This is kind of a hack, though.
Off the top of my head it would look like this.
struct customINT16 {
byte ByteHigh;
byte ByteLow;
}
So in your case you would write:
struct customINT16 myINT16;
myINT16.ByteHigh = BYTEARRAY[0];
myINT16.ByteLow = BYTEARRAY[1];
You'll have to go through a pointer to cast it, though:
intpointer = (int*)(&myINT16);
INTARRAY[0] = *intpointer;
As part of my CS course I've been given some functions to use. One of these functions takes a pointer to unsigned chars to write some data to a file (I have to use this function, so I can't just make my own purpose built function that works differently BTW). I need to write an array of integers whose values can be up to 4095 using this function (that only takes unsigned chars).
However am I right in thinking that an unsigned char can only have a max value of 256 because it is 1 byte long? I therefore need to use 4 unsigned chars for every integer? But casting doesn't seem to work with larger values for the integer. Does anyone have any idea how best to convert an array of integers to unsigned chars?
Usually an unsigned char holds 8 bits, with a max value of 255. If you want to know this for your particular compiler, print out CHAR_BIT and UCHAR_MAX from <limits.h> You could extract the individual bytes of a 32 bit int,
#include <stdint.h>
void
pack32(uint32_t val,uint8_t *dest)
{
dest[0] = (val & 0xff000000) >> 24;
dest[1] = (val & 0x00ff0000) >> 16;
dest[2] = (val & 0x0000ff00) >> 8;
dest[3] = (val & 0x000000ff) ;
}
uint32_t
unpack32(uint8_t *src)
{
uint32_t val;
val = src[0] << 24;
val |= src[1] << 16;
val |= src[2] << 8;
val |= src[3] ;
return val;
}
Unsigned char generally has a value of 1 byte, therefore you can decompose any other type to an array of unsigned chars (eg. for a 4 byte int you can use an array of 4 unsigned chars). Your exercise is probably about generics. You should write the file as a binary file using the fwrite() function, and just write byte after byte in the file.
The following example should write a number (of any data type) to the file. I am not sure if it works since you are forcing the cast to unsigned char * instead of void *.
int homework(unsigned char *foo, size_t size)
{
int i;
// open file for binary writing
FILE *f = fopen("work.txt", "wb");
if(f == NULL)
return 1;
// should write byte by byte the data to the file
fwrite(foo+i, sizeof(char), size, f);
fclose(f);
return 0;
}
I hope the given example at least gives you a starting point.
Yes, you're right; a char/byte only allows up to 8 distinct bits, so that is 2^8 distinct numbers, which is zero to 2^8 - 1, or zero to 255. Do something like this to get the bytes:
int x = 0;
char* p = (char*)&x;
for (int i = 0; i < sizeof(x); i++)
{
//Do something with p[i]
}
(This isn't officially C because of the order of declaration but whatever... it's more readable. :) )
Do note that this code may not be portable, since it depends on the processor's internal storage of an int.
If you have to write an array of integers then just convert the array into a pointer to char then run through the array.
int main()
{
int data[] = { 1, 2, 3, 4 ,5 };
size_t size = sizeof(data)/sizeof(data[0]); // Number of integers.
unsigned char* out = (unsigned char*)data;
for(size_t loop =0; loop < (size * sizeof(int)); ++loop)
{
MyProfSuperWrite(out + loop); // Write 1 unsigned char
}
}
Now people have mentioned that 4096 will fit in less bits than a normal integer. Probably true. Thus you can save space and not write out the top bits of each integer. Personally I think this is not worth the effort. The extra code to write the value and processes the incoming data is not worth the savings you would get (Maybe if the data was the size of the library of congress). Rule one do as little work as possible (its easier to maintain). Rule two optimize if asked (but ask why first). You may save space but it will cost in processing time and maintenance costs.
The part of the assignment of: integers whose values can be up to 4095 using this function (that only takes unsigned chars should be giving you a huge hint. 4095 unsigned is 12 bits.
You can store the 12 bits in a 16 bit short, but that is somewhat wasteful of space -- you are only using 12 of 16 bits of the short. Since you are dealing with more than 1 byte in the conversion of characters, you may need to deal with endianess of the result. Easiest.
You could also do a bit field or some packed binary structure if you are concerned about space. More work.
It sounds like what you really want to do is call sprintf to get a string representation of your integers. This is a standard way to convert from a numeric type to its string representation. Something like the following might get you started:
char num[5]; // Room for 4095
// Array is the array of integers, and arrayLen is its length
for (i = 0; i < arrayLen; i++)
{
sprintf (num, "%d", array[i]);
// Call your function that expects a pointer to chars
printfunc (num);
}
Without information on the function you are directed to use regarding its arguments, return value and semantics (i.e. the definition of its behaviour) it is hard to answer. One possibility is:
Given:
void theFunction(unsigned char* data, int size);
then
int array[SIZE_OF_ARRAY];
theFunction((insigned char*)array, sizeof(array));
or
theFunction((insigned char*)array, SIZE_OF_ARRAY * sizeof(*array));
or
theFunction((insigned char*)array, SIZE_OF_ARRAY * sizeof(int));
All of which will pass all of the data to theFunction(), but whether than makes any sense will depend on what theFunction() does.