I have a uint8_t array which I want to be able to convert to a char array.
i.e. uint_8 myInput [4] = {0, 233, 3, 1};
Which in bits it corresponds to: 00000000 00000011 11101001 00000001
And I would like to have an array of characters: char myChar [10] ={2,5,6,2,5,7}; which is the equivalent number.
I was able to create a solution: MySolution:
#include <stdio.h>
#include <math.h>
#include <stdlib.h> /* strtol */
#include <stdint.h>
#include <string.h>
int my_uint8_t_to_char (uint8_t givenData[]);
int main ()
{
uint8_t testArray[4];
uint32_t monitor_counter;
char sendBuffer[10];
monitor_counter = 256257;
// Here we create our array of uint8_t
memcpy (testArray, &monitor_counter, sizeof (monitor_counter));
// We flip the order of the bits
for (int i = 0; i < 4 / 2; i++)
{
int temp = testArray[i];
testArray[i] = testArray[4 - 1 - i];
testArray[4 - 1 - i] = temp;
}
//Call the function that converts the uint8_t to an integer value
int beforeSend = my_uint8_t_to_char(testArray);
printf ("\nHere it is: %d",beforeSend);
sprintf(sendBuffer, "%d", beforeSend);
// for ( int i=0; i< 10;i++)
printf("\nConverted character: %s",sendBuffer);
return 0;
}
int my_uint8_t_to_char (uint8_t givenData[])
{
char biggerArrayToSaveOctades[32];
int n, c, k, i, w;
unsigned int arr1[8];
int arrTemp[8];
long long bigBinaryString;
int j;
int index;
const int length_A = 8;
char astr[length_A*10];
char astr2[length_A*10];
char astr3[length_A*10];
char astr4[length_A*10];
printf("\nSIZE: %d",sizeof(astr4));
// In every loop we get one of the 4 integers from the uint8_t array and convert it to binary character
for (i = 0; i < 4; i++)
{
n = givenData[i];
for (c = 7; c >= 0; c--)
{
k = n >> c;
if (k & 1)
{
arr1[c] = 1;
}
else
{
arr1[c] = 0;
}
}
/*Flip the bits order */
for (int w = 0; w < 8 / 2; w++)
{
int temp = arr1[w];
arr1[w] = arr1[8 - 1 - w];
arr1[8 - 1 - w] = temp;
}
// in each case we append into the biggerArrayToSaveOctades the bitstrings
switch (i)
{
case 0:
j = 0;
index = 0;
for (j = 0; j < length_A; j++)
{
index += sprintf (&astr[index], "%d", arr1[j]);
}
for (w = 0; w < 8; w++)
biggerArrayToSaveOctades[w] = astr[w];
break;
case 1:
j = 0;
index = 0;
for (j = 0; j < length_A; j++)
{
index += sprintf (&astr2[index], "%d", arr1[j]);
}
j = 0;
for (w = 8; w < 16; w++)
{
biggerArrayToSaveOctades[w] = astr2[j];
j++;
}
break;
case 2:
j = 0;
index = 0;
for (j = 0; j < length_A; j++)
{
index += sprintf (&astr3[index], "%d", arr1[j]);
}
j = 0;
for (w = 16; w < 24; w++)
{
biggerArrayToSaveOctades[w] = astr3[j];
j++;
}
break;
case 3:
j = 0;
index = 0;
for (j = 0; j < length_A; j++)
{
index += sprintf (&astr4[index], "%d", arr1[j]);
}
j = 0;
for (w = 24; w < 32; w++)
{
biggerArrayToSaveOctades[w] = astr4[j];
j++;
}
break;
}
}
//Convert the bitstring to integer
int result = (int) strtol (biggerArrayToSaveOctades, NULL, 2);
return result;
}
but unfortunately I have some issues with the creation of the char astr[length_A*10]; with the ISO since I am using an IoT device and also I believe that my solution it's pretty complicated and not efficient.
Is there any way of doing that ?
The bytes represent one of the following:
A very odd byte order is being used:
This is neither big-endian nor little-endian byte order.
First convert the bytes into the appropriate number, then convert the resulting number into its decimal text representation (e.g. using sprintf/snprintf).
32-bit unsigned integer
Convert to number:
#include <stdint.h>
uint32_t num =
( (uint32_t)(myInput[0]) << 24 ) |
( (uint32_t)(myInput[2]) << 16 ) |
( (uint32_t)(myInput[1]) << 8 ) |
( (uint32_t)(myInput[3]) << 0 );
For your input, this would be equivalent to
uint32_t num = 256257;
Convert to its decimal representation:
#include <inttypes.h>
#include <stdio.h>
char decimal[11]; // Up to 10 digits plus final NUL.
sprintf(decimal, "%" PRIu32, num);
For your input, this would be equivalent to
char decimal[11] = { '2', '5', '6', '2', '5', '7', 0 };
Test:
#include <inttypes.h>
#include <stdio.h>
#include <stdint.h>
typedef uint8_t uint_8; // Why not just use uint8_t??
int main(void) {
uint_8 myInput[4] = {0, 233, 3, 1};
uint32_t num =
( (uint32_t)(myInput[0]) << 24 ) |
( (uint32_t)(myInput[2]) << 16 ) |
( (uint32_t)(myInput[1]) << 8 ) |
( (uint32_t)(myInput[3]) << 0 );
char decimal[11]; // Up to 10 digits plus final NUL.
sprintf(decimal, "%" PRIu32, num);
// If all you want to do is print the number,
// `printf("%" PRIu32, num);` would do the trick.
printf("%s\n", decimal); // 256257
}