i'm trying to calculate the checksum of an array of ASCII hex values.
Say I have the following
char exArray[] = "3030422320303030373830434441453141542355";
It's 20 pairs of hex values that each represent an ASCII character (eg. 0x41 = A).
How can I split them up to calculate a checksum?
Alternatively, how can I merge two values in an array to be one value?
(eg. '4', '1' -> '41')
#pmg:
First step would be converting the string representation (in hex) to an integer.
For the 2nd part, try ('4' - '0') * 16 + ('1' - '0')
This ultimately did the trick, love how simple it is too,
My implementation now looks somewhat like this.
uint8_t t = 0, tem, tem2, sum;
uint32_t chksum = 0;
void checkSum(void)
{
while (t < 40)
{
asciiToDec(exArray[t]);
tem = global.DezAscii[0];
t++;
asciiToDec(exArray[t]);
tem2 = global.DezAscii[0];
t++;
sum = (tem) * 16 + (tem2);
chksum += sum;
}
}
void asciiToDec(uint8_t value)
{
if (value == 'A')
global.DezAscii[0] = 10;
else if (value == 'B')
global.DezAscii[0] = 11;
else if (value == 'C')
global.DezAscii[0] = 12;
else if (value == 'D')
global.DezAscii[0] = 13;
else if (value == 'E')
global.DezAscii[0] = 14;
else if (value == 'F')
global.DezAscii[0] = 15;
else
global.DezAscii[0] = value;
}
uint16_t exArray[] = "3030422320303030373830434441453141542355";
I don't think this does what you are trying to do. The string literal is treated as a pointer to a const char. It's not even compiling for me. What you want here is something like this:
const char * exArray = "3030422320303030373830434441453141542355";
It's 20 pairs of hex values that each represent an ASCII character
(eg. 0x41 = A). How can I split them up to calculate a checksum?
You could loop through the array, doing what you want to do with the two chars inside the loop:
for (int i = 0; exArray[i]; i+=2) {
printf("my chars are %c and %c\n", exArray[i], exArray[i+1]);
// do the calculations you need here using exArray[i] and exArray[i+1]
}
Alternatively, how can I merge two values in an array to be one value?
(eg. '4', '1' -> '41')
I'm not sure what you mean here. Do you mean "41", as in the string representing 41? To do that, allocate three chars, then copy over those two chars and a null terminator. Something like
char hexByte[3];
hexByte[2] = 0; // setting the null terminator
for (int i = 0; exArray[i]; i+=2) {
hexByte[0] = exArray[i];
hexByte[1] = exArray[i+1];
printf("the string \"hexByte\" is: %s\n", hexByte);
// do something with hexByte here
}
If you want to convert it to its integer representation, use strtol:
printf("int value: %ld\n", strtol(hexByte, 0, 16));
Related
I have an input data of 4 character string (alphanumeric) or 3 character string and I need to convert these ASCII character string to unique float in 2 digits each, separated by decimal.
Ex:
Input string = 5405, output data = 54.05
Input string = 53BC, output data = 53.199 ( B ascii value is ~ 0x42 in hex and C is 0x43 )
Issue is I am seeing the same output when input strings are 560B and 5618, as both results in same output as 56.18.
Is there a way to uniquely generate a float number in these cases?
Max value of float allowed is 99.999.
Simple math tells us that this is not possible. The number of unique alphanumeric strings of length 4 (case-insensitive) is 36^4 = 1,679,616 while the number of non-negative unique floating point numbers with at most 3 fractional digits and less than 100 is 10^5 = 100,000.
If the string were restricted to hexadecimal digits, there would only be 16^4 = 65,536 possibilities in which case a unique encoding would be possible.
Slightly off-topic: when a mapping is needed into a domain which is too small to accommodate the result of a unique mapping, a hash function is the "standard tool", but collisions must be handled.
Your encoding is somewhat confusing, but here is a simple solution:
use 2 digits for the integral part
use 2 digits for fractional parts 00 to 99
use a combination of 1 letter and 1 letter or digit for fractional parts 100 to 999. There are 26*36 = 936 such combinations, enough to cover the 900 possibilities.
all values from 00.00 to 99.999 can be encoded.
some 4 letter and digit combinations are not used.
the encoding is not unique. eg: 53A0 is 53.100, the same number as 53.10 encoded as 5310.
Here is an implementation:
#include <stdib.h>
double fdecode(const char *s) {
char a[7];
a[0] = s[0];
a[1] = s[1];
a[2] = '.';
if (s[2] >= '0' && s[2] <= '9') {
a[3] = s[3];
a[4] = s[4];
a[5] = '\0';
} else {
// assuming uppercase letters
int n = 100 + (s[3] - 'A') * 36;
if (s[4] >= '0' && s[4] <= '9') {
n += s[4] - '0';
} else {
n += 10 + (s[4] - 'A') % 26;
}
snprintf(&a[3], 4, "%d", n);
}
return strtod(a, NULL);
}
int fencode(char *s, double d) {
char a[7];
if (d >= 0 && snprintf(a, 7, "%06.3f", d) == 6) {
s[0] = a[0];
s[1] = a[1];
if (a[5] == '0') {
s[2] = a[3];
s[3] = a[4];
} else {
int n = atoi(a + 3);
s[2] = 'A' + (n / 36);
n %= 36;
if (n < 10) {
s[3] = '0' + n;
} else {
s[3] = 'A' + n - 10;
}
}
s[4] = '\0';
return 4;
} else {
s[0] = '\0';
return -1;
}
}
I can't figure out what's wrong with this code!
It returns 208 as a decimal
where it should be 0
typedef unsigned char uchar;
int CONVERTION_BinStrToDecimal(char* binstr) //transform a inary string to a decimal number
{
int cpts = 0;
unsigned char dec = 0;
uchar x = 0;
for (cpts = 0; cpts <= 7; cpts++) {
x = 7 - cpts;
dec += (binstr[cpts]*pow(2,x));
}
return dec;
}
int main()
{
uchar decimal = 0;
char bin[8] = "00000000"; //example
decimal = CONVERTION_BinStrToDecimal(bin);
printf("%d", decimal);
}
binstr[cpts] yields the ascii code of 0 or 1 (which is 0x30 or 0x31).
You need to use binstr[cpts] == '1' to convert a ascii '1' to the number 1 and everything else to 0 (assuming that no other characters may occur). Another option would be binstr[cpts] - '0'.
Btw, using the pow() function is disregarded for such cases, better substitute pow(2,x) by (1<<x).
for (cpts = 0; cpts <= 7; cpts++) {
x = 7 - cpts;
dec += ((binstr[cpts] == '1')*(1 << x));
}
There are many possibilities to make it look nicer, of course, the most obvious being (binstr[cpts] == '1') << x.
Furthermore, mind that your code expects exactly 8 binary digits to calculate the correct result.
Alternatively, if you zero-terminate your string you can use strtol function with base 2, e.g.:
char bin[9] = "00000000";
decimal = strtol(bin, NULL, 2);
I tried this to parse data given in a csv file into ID, AGE, and GPA fields in a "data" file, but I don't think I'm doing this right (when I tried printing the data, its printing weird numbers). What am I doing wrong?
char data[1000];
FILE *x = fopen("database.csv","rt");
char NAME[300];
int ID[300],AGE[300],GPA[300];
int i,j;
i = 0;
while(!feof(x)) {
fgets(data,999,x);
for (j = 0; j < 300 && data[i] != ','; j++, i++) {
ID[j] = data[i];
i++;
}
for (j = 0; j < 300 && data[i] != ','; j++, i++) {
NAME[j] = data[i];
i++;
}
for (j = 0; j < 300 && ( data[i] != '\0' || data[i] != '\r' || data[i] != data[i] != '\n'); j++, i++) {
GPA[j] = data[i];
}
}
First of all: for what you're doing, you probably want to look carefully at the function strtok and the atoi macro. But given the code you posted, that's perhaps still a bit too advanced, so I'm taking a longer way here.
Supposing that the line is something like
172,924,1182
then you need to parse those numbers. The number 172 is actually represented by two or four bytes in memory, in a very different format, and the byte "0" is nothing like the number 0. What you'll read is the ASCII code, which is 48 in decimal, or 0x30 in hex.
If you take the ASCII value of a single digit and subtract 48, you will get a number, because fortunately the numbers are stored in digit order, so "0" is 48, "1" is 49 and so on.
But you still have the problem of converting the three digits 1 7 2 into 172.
So once you have 'data':
(I have added commented code to deal with a unquoted, unescaped text field inside the CSV, since in your question you mention an AGE field, but then you seem to want to use a NAME field. The case when the text field is quoted or escaped is another can of worms entirely)
size_t i = 0;
int number = 0;
int c;
int field = 0; // Fields start at 0 (ID).
// size_t x = 0;
// A for loop that never ends until we issue a "break"
for(;;) {
c = data[i++];
// What character did we just read?
if ((',' == c) || (0x0c == c) || (0x0a == c) || (0x00 == c)) {
// We have completed read of a number field. Which field was it?
switch(field) {
case 0: ID[j] = number; break;
case 1: AGE[j] = number; break;
// case 1: NAME[j][x] = 0; break; // we have already read in NAME, but we need the ASCIIZ string terminator.
case 2: GPA[j] = number; break;
}
// Are we at the end of line?
if ((0x0a == c) || (0x0c == c)) {
// Yes, break the cycle and read the next line
break;
}
// Read the next field. Reinitialize number.
field++;
number = 0;
// x = 0; // if we had another text field
continue;
}
// Each time we get a digit, the old value of number is shifted one order of magnitude, and c gets added. This is called Horner's algorithm:
// Number Read You get
// 0 "1" 0*10+1 = 1
// 1 "7" 1*10+7 = 17
// 17 "2" 17*10+2 = 172
// 172 "," Finished. Store 172 in the appropriate place.
if (c >= '0' && c <= '9') {
number = number * 10 + (c - '0');
}
/*
switch (field) {
case 1:
NAME[j][x++] = c;
break;
}
*/
}
I have a char array say char value []={'0','2','0','c','0','3'};
I want to convert this into a byte array like unsigned char val[]={'02','0c','03'}
This is in an embedded application so i can't use string.h functions. How can i do this?
Sicne you talk about an embedded application I assume that you want to save the numbers as values and not as strings/characters. So if you just want to store your character data as numbers (for example in an integer), you can use sscanf.
This means you could do something like this:
char source_val[] = {'0','A','0','3','B','7'} // Represents the numbers 0x0A, 0x03 and 0xB7
uint8 dest_val[3]; // We want to save 3 numbers
for(int i = 0; i<3; i++)
{
sscanf(&source_val[i*2],"%x%x",&dest_val[i]); // Everytime we read two chars --> %x%x
}
// Now dest_val contains 0x0A, 0x03 and 0xB7
However if you want to store it as a string (like in your example), you can't use unsigned char
since this type is also just 8-Bit long, which means it can only store one character. Displaying 'B3' in a single (unsigned) char does not work.
edit: Ok according to comments, the goal is to save the passed data as a numerical value. Unfortunately the compiler from the opener does not support sscanf which would be the easiest way to do so. Anyhow, since this is (in my opinion) the simplest approach, I will leave this part of the answer at it is and try to add a more custom approach in this edit.
Regarding the data type, it actually doesn't matter if you have uint8. Even though I would advise to use some kind of integer data type, you can also store your data into an unsigned char. The problem here is, that the data you get passed, is a character/letter, that you want to interpret as a numerical value. However, the internal storage of your character differs. You can check the ASCII Table, where you can check the internal values for every character.
For example:
char letter = 'A'; // Internally 0x41
char number = 0x61; // Internally 0x64 - represents the letter 'a'
As you can see there is also a differnce between upper an lower case.
If you do something like this:
int myVal = letter; //
myVal won't represent the value 0xA (decimal 10), it will have the value 0x41.
The fact you can't use sscanf means you need a custom function. So first of all we need a way to conver one letter into an integer:
int charToInt(char letter)
{
int myNumerical;
// First we want to check if its 0-9, A-F, or a-f) --> See ASCII Table
if(letter > 47 && letter < 58)
{
// 0-9
myNumerical = letter-48;
// The Letter "0" is in the ASCII table at position 48 -> meaning if we subtract 48 we get 0 and so on...
}
else if(letter > 64 && letter < 71)
{
// A-F
myNumerical = letter-55
// The Letter "A" (dec 10) is at Pos 65 --> 65-55 = 10 and so on..
}
else if(letter > 96 && letter < 103)
{
// a-f
myNumerical = letter-87
// The Letter "a" (dec 10) is at Pos 97--> 97-87 = 10 and so on...
}
else
{
// Not supported letter...
myNumerical = -1;
}
return myNumerical;
}
Now we have a way to convert every single character into a number. The other problem, is to always append two characters together, but this is rather easy:
int appendNumbers(int higherNibble, int lowerNibble)
{
int myNumber = higherNibble << 4;
myNumber |= lowerNibbler;
return myNumber;
// Example: higherNibble = 0x0A, lowerNibble = 0x03; -> myNumber 0 0xA3
// Of course you have to ensure that the parameters are not bigger than 0x0F
}
Now everything together would be something like this:
char source_val[] = {'0','A','0','3','B','7'} // Represents the numbers 0x0A, 0x03 and 0xB7
int dest_val[3]; // We want to save 3 numbers
int temp_low, temp_high;
for(int i = 0; i<3; i++)
{
temp_high = charToInt(source_val[i*2]);
temp_low = charToInt(source_val[i*2+1]);
dest_val[i] = appendNumbers(temp_high , temp_low);
}
I hope that I understood your problem right, and this helps..
If you have a "proper" array, like value as declared in the question, then you loop over the size of it to get each character. If you're on a system which uses the ASCII alphabet (which is most likely) then you can convert a hexadecimal digit in character form to a decimal value by subtracting '0' for digits (see the linked ASCII table to understand why), and subtracting 'A' or 'a' for letters (make sure no letters are higher than 'F' of course) and add ten.
When you have the value from the first hexadeximal digit, then convert the second hexadecimal digit the same way. Multiply the first value by 16 and add the second value. You now have single byte value corresponding to two hexadecimal digits in character form.
Time for some code examples:
/* Function which converts a hexadecimal digit character to its integer value */
int hex_to_val(const char ch)
{
if (ch >= '0' && ch <= '9')
return ch - '0'; /* Simple ASCII arithmetic */
else if (ch >= 'a' && ch <= 'f')
return 10 + ch - 'a'; /* Because hex-digit a is ten */
else if (ch >= 'A' && ch <= 'F')
return 10 + ch - 'A'; /* Because hex-digit A is ten */
else
return -1; /* Not a valid hexadecimal digit */
}
...
/* Source character array */
char value []={'0','2','0','c','0','3'};
/* Destination "byte" array */
char val[3];
/* `i < sizeof(value)` works because `sizeof(char)` is always 1 */
/* `i += 2` because there is two digits per value */
/* NOTE: This loop can only handle an array of even number of entries */
for (size_t i = 0, j = 0; i < sizeof(value); i += 2, ++j)
{
int digit1 = hex_to_val(value[i]); /* Get value of first digit */
int digit2 = hex_to_val(value[i + 1]); /* Get value of second digit */
if (digit1 == -1 || digit2 == -1)
continue; /* Not a valid hexadecimal digit */
/* The first digit is multiplied with the base */
/* Cast to the destination type */
val[j] = (char) (digit1 * 16 + digit2);
}
for (size_t i = 0; i < 3; ++i)
printf("Hex value %lu = %02x\n", i + 1, val[i]);
The output from the code above is
Hex value 1 = 02
Hex value 2 = 0c
Hex value 3 = 03
A note about the ASCII arithmetic: The ASCII value for the character '0' is 48, and the ASCII value for the character '1' is 49. Therefore '1' - '0' will result in 1.
It's easy with strtol():
#include <stdlib.h>
#include <assert.h>
void parse_bytes(unsigned char *dest, const char *src, size_t n)
{
/** size 3 is important to make sure tmp is \0-terminated and
the initialization guarantees that the array is filled with zeros */
char tmp[3] = "";
while (n--) {
tmp[0] = *src++;
tmp[1] = *src++;
*dest++ = strtol(tmp, NULL, 16);
}
}
int main(void)
{
unsigned char d[3];
parse_bytes(d, "0a1bca", 3);
assert(d[0] == 0x0a);
assert(d[1] == 0x1b);
assert(d[2] == 0xca);
return EXIT_SUCCESS;
}
If that is not available (even though it is NOT from string.h), you could do something like:
int ctohex(char c)
{
if (c >= '0' && c <= '9') {
return c - '0';
}
switch (c) {
case 'a':
case 'A':
return 0xa;
case 'b':
case 'B':
return 0xb;
/**
* and so on
*/
}
return -1;
}
void parse_bytes(unsigned char *dest, const char *src, size_t n)
{
while (n--) {
*dest = ctohex(*src++) * 16;
*dest++ += ctohex(*src++);
}
}
Assuming 8-bit bytes (not actually guaranteed by the C standard, but ubiquitous), the range of `unsigned char` is 0..255, and the range of `signed char` is -128..127. ASCII was developed as a 7-bit code using values in the range 0-127, so the same value can be represented by both `char` types.
For the now discovered task of converting a counted hex-string from ascii to unsigned bytes, here's my take:
unsigned int atob(char a){
register int b;
b = a - '0'; // subtract '0' so '0' goes to 0 .. '9' goes to 9
if (b > 9) b = b - ('A' - '0') + 10; // too high! try 'A'..'F'
if (b > 15) b = b - ('a' - 'A); // too high! try 'a'..'f'
return b;
}
void myfunc(const char *in, int n){
int i;
unsigned char *ba;
ba=malloc(n/2);
for (i=0; i < n; i+=2){
ba[i/2] = (atob(in[i]) << 4) | atob(in[i+1]);
}
// ... do something with ba
}
Can anyone tell me what is wrong with the following code?
__inline__
char* ut_byte_to_long (ulint nb) {
char* a = malloc(sizeof(nb));
int i = 0;
for (i=0;i<sizeof(nb);i++) {
a[i] = (nb>>(i*8)) & 0xFF;
}
return a;
}
This string is then concatenated as part of a larger one using strcat. The string prints fine but for the integers which are represented as character symbols. I'm using %s and fprintf to check the result.
Thanks a lot.
EDIT
I took one of the comments below (I was adding the terminating \0 separately, before calling fprintf, but after strcat. Modifying my initial function...
__inline__
char* ut_byte_to_long (ulint nb) {
char* a = malloc(sizeof(nb) + 1);
int i = 0;
for (i=0;i<sizeof(nb);i++) {
a[i] = (nb>>(i*8)) & 0xFF;
}
a[nb] = '\0' ;
return a;
}
This sample code still isn't printing out a number...
char* tmp;
tmp = ut_byte_to_long(start->id);
fprintf(stderr, "Value of node is %s \n ", tmp);
strcat is expecting a null byte terminating the string.
Change your malloc size to sizeof(nb) + 1 and append '\0' to the end.
You have two problems.
The first is that the character array a contains numbers, such as 2, instead of ASCII codes representing those numbers, such as '2' (=50 on ASCII, might be different in other systems). Try modifying your code to
a[i] = (nb>>(i*8)) & 0xFF + '0';
The second problem is that the result of the above computation can be anything between 0 and 255, or in other words, a number which requires more than one digit to print.
If you want to print hexadecimal numbers (0-9, A-F), two digits per such computation will be enough, and you can write something like
a[2*i + 0] = int2hex( (nb>>(i*8)) & 0x0F ); //right hexa digit
a[2*i + 1] = int2hex( (nb>>(i*8+4)) & 0x0F ); //left hexa digit
where
char int2hex(int n) {
if (n <= 9 && n >= 0)
return n + '0';
else
return (n-10) + 'A';
}
if you dont want to use sprintf(target_string,"%lu",source_int) or the non standard itoa(), here is a version of the function that transform a long to a string :
__inline__
char* ut_byte_to_long (ulint nb) {
char* a = (char*) malloc(22*sizeof(char));
int i=21;
int j;
do
{
i--;
a[i] = nb % 10 + '0';
nb = nb/10;
}while (nb > 0);
// the number is stored from a[i] to a[21]
//shifting the string to a[0] : a[21-i]
for(j = 0 ; j < 21 && i < 21 ; j++ , i++)
{
a[j] = a[i];
}
a[j] = '\0';
return a;
}
I assumed that an unsigned long contain less than 21 digits. (biggest number is 18,446,744,073,709,551,615 which equals 2^64 − 1 : 20 digits)