Convert char * to unsigned char array in C

Question

I would like to optimize my code and avoid errors, I have this function that does the "work" but I think I can improve and avoid memory problems.

void function(char* message)
{
    char * pointer;
    unsigned char buffer[2048] = {0};
    int buffer_len = 0;

    memset(buffer, 0, sizeof(buffer));
    strcpy(buffer, message);

    buffer_len = strlen(buffer);
    memset(buffer, 0, sizeof(&buffer));

    for(int i = 0, pointer = message; i < (buffer_len / 2); i++, pointer += 2)
    {
        sscanf(pointer, "%02hhX", &buffer[i]);
    }
}

The idea of the function is to receive a string of this style "0123456789" and pass it to 0x01, 0x23, 0x45, ... in an unsigned char array. Any tip, good practice or improvement would be very useful.

The ussage is something like this:

function("0123456789");

In the function buffer ends like:

buffer[0] = 0x01
buffer[1] = 0x23
...

Might be good to show the target function too. Right now, there are quite a few potential issues with just the code you've shown (strcpy is not safe, multiple memset to same data, probably not needing to memset anyway. — Michael Dorgan
– Michael Dorgan, Commented Dec 15, 2017 at 18:44
If the string is gigantic, and copying it is a problem, consider using a real compression method. Something like a Huffman encoder can do better than 4-bits per character as it will take into account the usage frequency of each value, — Michael Dorgan
– Michael Dorgan, Commented Dec 15, 2017 at 18:48
I'm voting to close this question as off-topic because the code is working. Optimizing/reviewing questions can be asked on Code Review. — Jongware
– Jongware, Commented Dec 16, 2017 at 1:49

sg7 · Accepted Answer · 2017-12-16 00:52:34Z

3

There are a few optimizations possible. The biggest optimization comes from avoiding doing 2x memset and strcpy,

No need to:

// memset(buffer, 0, sizeof(buffer)); 
// strcpy(buffer, message);
// memset(buffer, 0, sizeof(&buffer));

which drastically simplifies the code:

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

void function(char* message)
{
    unsigned char * pointer;
    unsigned char buffer[2048]; // not needed = {0}; 
    int buffer_half_len; // not needed = 0;

    buffer_half_len = strlen(message)/2;  // avoiding division in the for loop;
    pointer = message;

    for(int i = 0;  i < buffer_half_len; i++, pointer += 2)
    {
        sscanf(pointer, "%02hhX", &buffer[i]);
        printf("%02hhX\n", buffer[i] );
    }
}

OUTPUT:

edited Dec 16, 2017 at 0:52

answered Dec 15, 2017 at 22:38

sg7

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1 Comment

J.Doe Over a year ago

buffer store the A B C D E F,how to change to store lower case?

amr.s Gr · Accepted Answer · 2018-09-12 07:28:42Z

0

char * a= -80; // let supposed you get returned value( i represented in int) from whatever source.
unsigned char b = * a; // now b equal the complemntry of -80 which will be 176
std::cout << b ;

answered Sep 12, 2018 at 7:28

amr.s Gr

1

1 Comment

hellow Over a year ago

While this might answer the authors question, it lacks some explaining words and links to documentation. Raw code snippets are not very helpful without some phrases around it. You may also find how to write a good answer very helpful. Please edit your answer.

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