Convert each character to an unsigned character (so the 0xc0 isn't negative), then convert it to an integer and output as a two digit hexadecimal value.

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


#define INT(x) ((int)(unsigned char)(x))

int main()
{
     char arr[4];
     arr[0] = 0x11;
     arr[1] = 0xc0;
     arr[2] = 0x0c;
     arr[3] = 0x00;

     char *str;

     str=malloc(32);
     sprintf(str, "%02x%02x%02x%02x",
         INT(arr[0]), INT(arr[1]), INT(arr[2]), INT(arr[3]));

     puts(str);
}

Output is:

11c00c00

Another simple way is to print "&#" + CharCode like this:

for(var i=9984; i<=10175; i++){
    document.write(i + "&nbsp;&nbsp;&nbsp;" + i.toString(16) + "&nbsp;&nbsp;&nbsp;&#" + i + "<br>");
}

OR

for(var i=0x2700; i<=0x27BF; i++){
    document.write(i + "&nbsp;&nbsp;&nbsp;" + i.toString(16) + "&nbsp;&nbsp;&nbsp;&#" + i + "<br>");
}

JSFIDDLE

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;
}

You do it this way:

var str = s.join();

Here is something:

char const hex_chars[16] = { '0', '1', '2', '3', '4', '5', '6', '7', '8', '9', 'A', 'B', 'C', 'D', 'E', 'F' };
string result = "";

for( int i = 0; i < data_length; ++i )
{
    char const byte = data[i];

    result += hex_chars[ ( byte & 0xF0 ) >> 4 ];
    result += hex_chars[ ( byte & 0x0F ) >> 0 ];
}

Use %x flag to print hexadecimal integer

example.c

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

int main(void) 
{
  char *string = "hello", *cursor;
  cursor = string;
  printf("string: %s\nhex: ", string);
  while(*cursor)
  {
    printf("%02x", *cursor);
    ++cursor;
  }
  printf("\n");
  return 0;
}

output

$ ./example 
string: hello
hex: 68656c6c6f

reference

• printf reference
• ascii table

Here is my take on it - the phex() function converts any data in memory into a newly allocated string containing the hex representation.

The main() function shows an example usage. The output is "31323334353637383930" for the example data.

#include <stdlib.h> /* malloc() */
#include <stdio.h>  /* sprintf() */
#include <string.h> /* strlen(), in the example main() */

/*
 * Return a hex string representing the data pointed to by `p`,
 * converting `n` bytes.
 *
 * The string should be deallocated using `free()` by the caller.
 */
char *phex(const void *p, size_t n)
{
    const unsigned char *cp = p;              /* Access as bytes. */
    char *s = malloc(2*n + 1);       /* 2*n hex digits, plus NUL. */
    size_t k;

    /*
     * Just in case - if allocation failed.
     */
    if (s == NULL)
        return s;

    for (k = 0; k < n; ++k) {
        /*
         * Convert one byte of data into two hex-digit characters.
         */
        sprintf(s + 2*k, "%02X", cp[k]);
    }

    /*
     * Terminate the string with a NUL character.
     */
    s[2*n] = '\0';

    return s;
}

/*
 * Sample use of `phex()`.
 */
int main(void)
{
    const char *data = "1234567890";               /* Sample data */
    char *h = phex(data, strlen(data));  /* Convert to hex string */

    if (h != NULL)
        puts(h);                                  /* Print result */

    free(h);                             /* Deallocate hex string */
    return 0;
}

For completude, you can also easily do it without calling any heavy library function (no snprintf, no strcat, not even memcpy). It can be useful, say if you are programming some microcontroller or OS kernel where libc is not available.

Nothing really fancy you can find similar code around if you google for it. Really it's not much more complicated than calling snprintf and much faster.

#include <stdio.h>

int main(){
    unsigned char buf[] = {0, 1, 10, 11};
    /* target buffer should be large enough */
    char str[12];

    unsigned char * pin = buf;
    const char * hex = "0123456789ABCDEF";
    char * pout = str;
    int i = 0;
    for(; i < sizeof(buf)-1; ++i){
        *pout++ = hex[(*pin>>4)&0xF];
        *pout++ = hex[(*pin++)&0xF];
        *pout++ = ':';
    }
    *pout++ = hex[(*pin>>4)&0xF];
    *pout++ = hex[(*pin)&0xF];
    *pout = 0;

    printf("%s\n", str);
}

Here is another slightly shorter version. It merely avoid intermediate index variable i and duplicating laste case code (but the terminating character is written two times).

#include <stdio.h>
int main(){
    unsigned char buf[] = {0, 1, 10, 11};
    /* target buffer should be large enough */
    char str[12];

    unsigned char * pin = buf;
    const char * hex = "0123456789ABCDEF";
    char * pout = str;
    for(; pin < buf+sizeof(buf); pout+=3, pin++){
        pout[0] = hex[(*pin>>4) & 0xF];
        pout[1] = hex[ *pin     & 0xF];
        pout[2] = ':';
    }
    pout[-1] = 0;

    printf("%s\n", str);
}

Below is yet another version to answer to a comment saying I used a "trick" to know the size of the input buffer. Actually it's not a trick but a necessary input knowledge (you need to know the size of the data that you are converting). I made this clearer by extracting the conversion code to a separate function. I also added boundary check code for target buffer, which is not really necessary if we know what we are doing.

#include <stdio.h>

void tohex(unsigned char * in, size_t insz, char * out, size_t outsz)
{
    unsigned char * pin = in;
    const char * hex = "0123456789ABCDEF";
    char * pout = out;
    for(; pin < in+insz; pout +=3, pin++){
        pout[0] = hex[(*pin>>4) & 0xF];
        pout[1] = hex[ *pin     & 0xF];
        pout[2] = ':';
        if (pout + 3 - out > outsz){
            /* Better to truncate output string than overflow buffer */
            /* it would be still better to either return a status */
            /* or ensure the target buffer is large enough and it never happen */
            break;
        }
    }
    pout[-1] = 0;
}

int main(){
    enum {insz = 4, outsz = 3*insz};
    unsigned char buf[] = {0, 1, 10, 11};
    char str[outsz];
    tohex(buf, insz, str, outsz);
    printf("%s\n", str);
}

Using map() won't work if the input is of a type like Uint8Array: the result of map() is also Uint8Array which can't hold the results of string conversion.

function toHexString(byteArray) {
  var s = '0x';
  byteArray.forEach(function(byte) {
    s += ('0' + (byte & 0xFF).toString(16)).slice(-2);
  });
  return s;
}

If you don't wanna use library functions, you'll have to build a simple lookup table yourself:

#include <stdio.h> // only for printing the result

const char table[] = "0123456789abcdef";

int main(void) {
    char src[5 + 1] = "hello";
    char dst[5 * 2 + 1];
    char *s, *d;

    for (s = src, d = dst; *s != '\0'; s++, d += 2) {
        const unsigned char lo = *s & 0xf;
        const unsigned char hi = *s >> 4;

        *d = table[hi];
        *(d + 1) = table[lo];
    }

    *d = '\0';

    puts(dst);
    return 0;
}