How to fill a string with random (hex) characters? - c

Question

I have a string (unsigned char) and i want to fill it with only hex characters.
my code is
unsigned char str[STR_LEN] = {0};
for(i = 0;i<STR_LEN;i++) {
sprintf(str[i],"%x",rand()%16);
}
Of course, when running this I get segfaulted

Answer 1

string is an array of char-s not unsigned char-s
you are using str[i] (which is of type unsigned char) as a 1st argument to sprintf, but it requires type char * (pointer).
This should be a little better:
char str[STR_LEN + 1];
for(i = 0; i < STR_LEN; i++) {
sprintf(str + i, "%x", rand() % 16);
}

Answer 2

The first argument to sprintf() should be a char*, but str[i] is a char: this is the cause of the segmentation fault. The compiler should have emitted a warning about this. gcc main.c, without specifying a high warning level, emitted the following:
warning: passing argument 1 of sprintf makes pointer from integer without a cast
A hex representation of a character can be 1 or 2 characters (9 or AB for example). For formatting, set the precision to 2 and the fill character to 0. Also need to add one character for the terminating null to str and set the step of the for loop to 2 instead of 1 (to prevent overwriting previous value):
unsigned char str[STR_LEN + 1] = {0};
int i;
for (i = 0; i < STR_LEN; i += 2)
{
sprintf(&str[i], "%02X", rand() % 16);
}

Answer 3

You could try something like this:
#include <stdio.h>
#include <stdlib.h>
#define STR_LEN 20
int main(void)
{
unsigned char str[STR_LEN + 1] = {0};
const char *hex_digits = "0123456789ABCDEF";
int i;
for( i = 0 ; i < STR_LEN; i++ ) {
str[i] = hex_digits[ ( rand() % 16 ) ];
}
printf( "%s\n", str );
return 0;
}

Answer 4

There are several unclarities and problems in your code. I interpret "hex character" to mean "hex digit", i.e. a symbol from {0,1,2,3,4,5,6,7,8,9,a,b,c,d,e,f}, not "the hexadecimal value of an ascii character's code point". This might or might not be what you meant.
This should do it:
void hex_fill(char *buf, size_t max)
{
static const char hexdigit[16] = "0123456789abcdef";
if(max < 1)
return;
--max;
for(i = 0; i < max; ++i)
buf[i] = hexdigit[rand() % sizeof hexdigit];
buf[max] = '\0';
}
The above will always 0-terminate the string, so there's no requirement that you do so in advance. It will properly handle all buffer sizes.

Answer 5

My variation on some of answers below; note the time seeded rand function and instead of a char using a const size, I use a vector that is then converted to a string array.
Boost variate generator docs
std::string GetRandomHexString(unsigned int count)
{
std::vector<char> charVect = std::vector<char>(count);
//Rand generator
typedef boost::random::mt19937 RNGType;
RNGType rng(std::time(nullptr) + (unsigned int)clock());
//seeding rng
uniform_int<> range(0, 15); //Setting min max
boost::variate_generator<RNGType, boost::uniform_int<> >generate(rng, range); //Creating our generator
//Explicit chars to sample from
const char hexChars[16] = { '0','1','2','3','4','5','6','7','8','9','A','B','C','D','E','F' };
//
for (int i = 0; i < count; i++)
{
charVect[i] = hexChars[generate()];
}
//
return std::string(charVect.begin(), charVect.end());;
}
Examples (count = 32):
1B62C49C416A623398B89A55EBD3E9AC
26CFD2D1C14B9F475BF99E4D537E2283
B8709C1E87F673957927A7F752D0B82A
DFED20E9C957C4EEBF4661E7F7A58460
4F86A631AE5A05467BA416C4854609F8