将 unsigned char * 转换为 char * 字符串 C

Question

I have code to convert an int into a unsigned char * as such:

int i = //a number <= 15;
unsigned char * byte = (unsigned char *) &i;

I do this because I need the byte value of of that int. Later I want to attach this byte value to a char * , so I need to convert it to a string.

Ex. if: int i = 15;
then: unsigned char * byte = F;

However I need to convert this to get:

char * string = "F";

I have tried:

char * string = (char *) &byte;
//and
char * string = (char *) byte;
//or just straight up treating it like a char * by going:
printf("%s", byte);

and several other combinations, but they've all resulted in seg faults. Does anyone know how I can do this? Or even if there's an easier way of converting an int to the hexidecimal representation in a char?

I'm very new to C and appreciate any answers thank you.

Answer 1

I actually want to append the converted char to an existing char *

1. Insure the destination is big enough.

    char existing[100]; // existing is populated somehow, now to add the `unsigned char` size_t len = strlen(existing); // 2 is the max size of a 8-bit value printed in hex. Adjust as needed. if (len + 2 >= sizeof existing) Error_BufferTooSmall();

2. Write to the end of the existing char * .

    // Only use the least significant `char` portion with `hh` sprintf(existing + len , "%hhX", (unsigned) i);

hh Specifies that a following d , i , o , u , x , or X conversion specifier applies to a signed char or unsigned char argument (the argument will have been promoted according to the integer promotions, but its value shall be converted to signed char or unsigned char before printing); ... C11 §7.21.6.1 7

Answer 2

int i = //a number <= 15;
unsigned char * byte = (unsigned char *) &i;

Is plain wrong.

What you are doing is taking the address of the integer variable i and putting into a pointer to an unsigned char .

From what I read I think you want to convert an integer to a character (or string).

You can do this with itoa or even easier with sprintf .

For example:

#include <stdio.h> // printf, snprintf

int main(void) {
    int i = 255;
    char theIntegerAsString[5] = {0};

    if (snprintf(theIntegerAsString, sizeof(theIntegerAsString), "%X", i) > 0) {
        printf("The number %d in Hexadecimal is: %s\n", i, theIntegerAsString);
    }
}

You might wonder why I used snprintf instead of sprintf .

This is because sprintf does not always check the buffer size, whereas snprintf always does. See buffer overflow .

Please note that %X is specific to the type unsigned int for larger or smaller types you need another specifier. I strongly recommend using <stdint.h> and <inttypes.h> like this:

#include <stdio.h> // printf, snprintf
#include <stdint.h> // intX_t
#include <inttypes.h> // PRIx..

int main(void) {
    int32_t i = 255; // signed 32 bit integer
    char theIntegerAsString[5] = {0};

    if (snprintf(theIntegerAsString, sizeof(theIntegerAsString), "%" PRIX32, i) > 0) {
        printf("The number %" PRId32 " in Hexadecimal is: %s\n", i, theIntegerAsString);
    }
}