[Solved] read file string and store in uint8_t array in c [closed]

by


Is that what you wanted? (to test give binary 01 combination as an first argument)

#include <stdio.h>
#include <stdint.h>

uint8_t charToBin(char c)
{
    switch(c)
    {
        case '0': return 0;
        case '1': return 1;
    }
    return 0;
}

uint8_t CstringToU8(const char * ptr)
{
    uint8_t value = 0;
    for(int i = 0; (ptr[i] != '\0') && (i<8); ++i)
    {
        value = (value<<1) | charToBin(ptr[i]);
    }
    return value;
}


int main(int argc,const char *argv[])
{
    printf("%d\n",CstringToU8(argv[1]));
    return 0;
}

You can use CstringToU8() to convert 8 characters to one u8 number. Read whole data to char array (e.g. char * text) and then convert 8 characters to u8 number, store it and move you pointer 8 bytes further until array won’t end.

Because after edit question was changed so here is my new solution. Code reading hex numbers from file and storing it into array.

#include <stdio.h>
#include <stdint.h>
#include <stdlib.h>

int main()
{
     FILE * fp;
    uint8_t number = 0;
    size_t fileSize = 0;
    uint8_t * array = NULL;
    size_t readedBytes = 0;
    size_t iterator = 0;

    fp = fopen ("hexNumbers.txt", "r");

    // check file size
    fseek(fp, 0L, SEEK_END);
    fileSize = ftell(fp);
    fseek(fp, 0L, SEEK_SET);

    /// allocate max possible array size = fileSize/2
    array = malloc(fileSize/2 * sizeof(uint8_t));

    /// read data into array
    while(!feof(fp))
    {
        if (fscanf(fp,"%2hhx",&number) == 1)
        {
            array[readedBytes++] = number;
        }
    }
    fclose(fp);


    /// print array output
    for (iterator=0; iterator<readedBytes; ++iterator)
    {
        printf("%02x ", array[iterator]);
    }

    free(array);
    return 0;
}

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solved read file string and store in uint8_t array in c [closed]