You can't fit 5 bytes worth of data into a 4 byte array; that leads to buffer overflows.

If you have the hex digits in a string, you can use sscanf() and a loop:

#include <stdio.h>
#include <ctype.h>

int main()
{
    const char *src = "0011223344";
    char buffer[5];
    char *dst = buffer;
    char *end = buffer + sizeof(buffer);
    unsigned int u;

    while (dst < end && sscanf(src, "%2x", &u) == 1)
    {
        *dst++ = u;
        src += 2;
    }

    for (dst = buffer; dst < end; dst++)
        printf("%d: %c (%d, 0x%02x)\n", dst - buffer,
               (isprint(*dst) ? *dst : '.'), *dst, *dst);

    return(0);
}

Note that printing the string starting with a zero-byte requires care; most operations terminate on the first null byte. Note that this code did not null-terminate the buffer; it is not clear whether null-termination is desirable, and there isn't enough space in the buffer I declared to add a terminal null (but that is readily fixed). There's a decent chance that if the code was packaged as a subroutine, it would need to return the length of the converted string (though you could also argue it is the length of the source string divided by two).

Answer from Jonathan Leffler on Stack Overflow
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Ondrovo
bits.ondrovo.com › hexc.html
Hex to C array - Ondrovo.com
Paste hex string. Characters outside 0-9a-fA-F and anything after | is discarded (hexdump). Hex digit count must be even.
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Stack Overflow
stackoverflow.com › questions › 1557400 › hex-to-char-array-in-c
Hex to char array in C - Stack Overflow
Top answer
1 of 11
18

You can't fit 5 bytes worth of data into a 4 byte array; that leads to buffer overflows.

If you have the hex digits in a string, you can use sscanf() and a loop:

#include <stdio.h>
#include <ctype.h>

int main()
{
    const char *src = "0011223344";
    char buffer[5];
    char *dst = buffer;
    char *end = buffer + sizeof(buffer);
    unsigned int u;

    while (dst < end && sscanf(src, "%2x", &u) == 1)
    {
        *dst++ = u;
        src += 2;
    }

    for (dst = buffer; dst < end; dst++)
        printf("%d: %c (%d, 0x%02x)\n", dst - buffer,
               (isprint(*dst) ? *dst : '.'), *dst, *dst);

    return(0);
}

Note that printing the string starting with a zero-byte requires care; most operations terminate on the first null byte. Note that this code did not null-terminate the buffer; it is not clear whether null-termination is desirable, and there isn't enough space in the buffer I declared to add a terminal null (but that is readily fixed). There's a decent chance that if the code was packaged as a subroutine, it would need to return the length of the converted string (though you could also argue it is the length of the source string divided by two).

2 of 11
3

I would do something like this;

// Convert from ascii hex representation to binary
// Examples;
//   "00" -> 0
//   "2a" -> 42
//   "ff" -> 255
// Case insensitive, 2 characters of input required, no error checking
int hex2bin( const char *s )
{
    int ret=0;
    int i;
    for( i=0; i<2; i++ )
    {
        char c = *s++;
        int n=0;
        if( '0'<=c && c<='9' )
            n = c-'0';
        else if( 'a'<=c && c<='f' )
            n = 10 + c-'a';
        else if( 'A'<=c && c<='F' )
            n = 10 + c-'A';
        ret = n + ret*16;
    }
    return ret;
}

int main()
{
    const char *in = "0011223344";
    char out[5];
    int i;

    // Hex to binary conversion loop. For example;
    // If in="0011223344" set out[] to {0x00,0x11,0x22,0x33,0x44}
    for( i=0; i<5; i++ )
    {
        out[i] = hex2bin( in );
        in += 2;
    }
    return 0;
}
Hex to char array in C
hash - Convert hex values to char array in C
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Hexadecimal string to byte array in C
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Hexadecimal string to byte array in C - Stack Overflow
Sign up to request clarification or add additional context in comments. ... This is a good approach, but be aware that it will give an incorrect result if there is an odd number of digits in the hex string (the implicit zero will prefix the last digit, not the first, so "5ab5c" will be printed as 0x5ab50c rather than 0x05ab5c). 2010-08-05T07:02:58.773Z+00:00 ... @Geremia nope. Endianness only comes into play when you're reinterpreting byte arrays ... More on stackoverflow.com
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ASCII to C-like array converter
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stackoverflow.com › questions › 3408706 › hexadecimal-string-to-byte-array-in-c
Hexadecimal string to byte array in C - Stack Overflow
Top answer
1 of 16
83

As far as I know, there's no standard function to do so, but it's simple to achieve in the following manner:

#include <stdio.h>

int main(int argc, char **argv) {
    const char hexstring[] = "DEadbeef10203040b00b1e50", *pos = hexstring;
    unsigned char val[12];

     /* WARNING: no sanitization or error-checking whatsoever */
    for (size_t count = 0; count < sizeof val/sizeof *val; count++) {
        sscanf(pos, "%2hhx", &val[count]);
        pos += 2;
    }

    printf("0x");
    for(size_t count = 0; count < sizeof val/sizeof *val; count++)
        printf("%02x", val[count]);
    printf("\n");

    return 0;
}

Edit

As Al pointed out, in case of an odd number of hex digits in the string, you have to make sure you prefix it with a starting 0. For example, the string "f00f5" will be evaluated as {0xf0, 0x0f, 0x05} erroneously by the above example, instead of the proper {0x0f, 0x00, 0xf5}.

2 of 16
25

I found this question by Googling for the same thing. I don't like the idea of calling sscanf() or strtol() since it feels like overkill. I wrote a quick function which does not validate that the text is indeed the hexadecimal presentation of a byte stream, but will handle odd number of hex digits:

uint8_t tallymarker_hextobin(const char * str, uint8_t * bytes, size_t blen)
{
   uint8_t  pos;
   uint8_t  idx0;
   uint8_t  idx1;

   // mapping of ASCII characters to hex values
   const uint8_t hashmap[] =
   {
     0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // ........
     0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // ........
     0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // ........
     0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // ........
     0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, //  !"#$%&'
     0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // ()*+,-./
     0x00, 0x01, 0x02, 0x03, 0x04, 0x05, 0x06, 0x07, // 01234567
     0x08, 0x09, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // 89:;<=>?
     0x00, 0x0a, 0x0b, 0x0c, 0x0d, 0x0e, 0x0f, 0x00, // @ABCDEFG
     0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // HIJKLMNO
     0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // PQRSTUVW
     0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // XYZ[\]^_
     0x00, 0x0a, 0x0b, 0x0c, 0x0d, 0x0e, 0x0f, 0x00, // `abcdefg
     0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // hijklmno
     0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // pqrstuvw
     0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // xyz{|}~.
     0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // ........
     0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // ........
     0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // ........
     0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // ........
     0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // ........
     0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // ........
     0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // ........
     0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // ........
     0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // ........
     0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // ........
     0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // ........
     0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // ........
     0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // ........
     0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // ........
     0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // ........
     0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00  // ........
   };

   bzero(bytes, blen);
   for (pos = 0; ((pos < (blen*2)) && (pos < strlen(str))); pos += 2)
   {
      idx0 = (uint8_t)str[pos+0];
      idx1 = (uint8_t)str[pos+1];
      bytes[pos/2] = (uint8_t)(hashmap[idx0] << 4) | hashmap[idx1];
   };

   return(0);
}
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github.com › AhmedYousriSobhi › Hex-To-array-txt
GitHub - AhmedYousriSobhi/Hex-To-array-txt: C code to convert Hex file to array Txt file
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dhinakg.github.io › hex.html
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reddit.com › r/c_programming › reading hex values from external file into an array
Reading hex values from external file into an array : r/C_Programming
February 15, 2024 - if (fscanf(file, "%x%c", &array[i], &dummy_char) == 2 && *dummy_char == ',') should do roughly what you want. If you don't want to use that, you could also use strtoul().
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reddit.com › r/askprogramming › storing hex values in an int array c
Storing Hex values in an int array C : r/AskProgramming
March 23, 2018 - A single hex digit ("0" through "F") is exactly four binary bits. If you know your "array" won't contain more than 8 hex digits, just keep them in a 32-bit integer (uint32_t) and use shifting and bitwise masking to access them: elem = (ar >> (index * 4)) && 0xF;
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GitHub
gist.github.com › xsleonard › 7341172
hex string to byte array, C · GitHub
I added a small check to fix that. int HexStringToBytes(const char *hexStr, unsigned char *output, unsigned int *outputLen) { size_t len = strlen(hexStr); if (len % 2 != 0) { return -1; } size_t finalLen = len / 2; *outputLen = finalLen; for (size_t inIdx = 0, outIdx = 0; outIdx < finalLen; inIdx += 2, outIdx++) { if ((hexStr[inIdx] - 48) <= 9 && (hexStr[inIdx + 1] - 48) <= 9) { goto convert; } else { if (((hexStr[inIdx] - 65) <= 5 && (hexStr[inIdx + 1] - 65) <= 5) || ((hexStr[inIdx] - 97) <= 5 && (hexStr[inIdx + 1] - 97) <= 5)) { goto convert; } else { *outputLen = 0; return -1; } } convert: output[outIdx] = (hexStr[inIdx] % 32 + 9) % 25 * 16 + (hexStr[inIdx + 1] % 32 + 9) % 25; } output[finalLen] = '\0'; return 0; }
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Yes, that's understandable. If you want to use hexadecimal literals in your code, they must have an 0x prefix. That's how the compiler can tell the difference and set the actual value accordingly. The equivalent decimal array would be:
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Stack Overflow
stackoverflow.com › questions › 16464278 › convert-array-of-numbers-into-single-hex-value-in-c
Convert array of numbers into single Hex value in c - Stack Overflow
Below is the simple logic i implemented.... void DecToHex(unsigned char* dst, unsigned char* src); unsigned char digits[12] = {0x01,0x02,0x03,0x04,0x05,0x06,0x07,0x08,0x09,0x01,0x02,0x03}; unsigned char result[10]= {0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00}; void main(void) { /* digits[12] = {0x01,0x02,0x03,0x04,0x05,0x06,0x07,0x08,0x09,0x01,0x02,0x03}; */ DecToHex(result, digits); /* result[10] : 0x1C,0xBE,0x99,0x1A,0x83,......... */ while(1); } void DecToHex(unsigned char* dst, unsigned char* src) { unsigned char tmp; unsigned char i, j=10; while(j!=0) { j -= 1; /* reinitialize for
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Stack Overflow
stackoverflow.com › questions › 33454988 › c-initialize-array-in-hexadecimal-values
C initialize array in hexadecimal values - Stack Overflow
Top answer
1 of 6
8

Defining this, for example

unsigned char a[16] = {0x20, 0x41, 0x42, };

will initialise the first three elements as shown, and the remaining elements to 0.

Your second way

unsigned char a[16] = {"0x20"};

won't do what you want: it just defines a nul-terminated string with the four characters 0x20, the compiler won't treat it as a hexadecimal value.

2 of 6
7

There is a GNU extension called designated initializers. This is enabled by default with gcc

With this you can initialize your array in the form 

unsigned char a[16] = {[0 ... 15] = 0x20};
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LinuxQuestions.org
linuxquestions.org › questions › programming-9 › how-to-define-hex-character-array-in-c-668213
How to define hex character array in C?
September 7, 2008 - Hi group, I have a simple question. I want to define a hex character array in C, but I don't want to have to define the elements one at a time.
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