C: Converting unsigned char array to signed int (vice versa)

Doe Joe Published at Dev

Doe Joe

I'm trying to convert an unsigned char array buffer into a signed int (vice versa).

Below is a demo code:

int main(int argv, char* argc[])
{
    int original = 1054;
    unsigned int i = 1054;
    unsigned char c[4];
    int num;

    memcpy(c, (char*)&i, sizeof(int));

    //num  = *(int*) c;                          // method 1 get
    memcpy((char *)&num, c, sizeof(int));        // method 2 get
    printf("%d\n", num);

    return 0;
}

1) Which method should I use to get from unsigned char[] to int?

method 1 get or method 2 get? (or any suggestion)

2) How do I convert the int original into an unsigned char[]?

I need to send this integer via a buffer that only accepts unsigned char[]

Currently what i'm doing is converting the int to unsigned int then to char[], example :

int g = 1054;
unsigned char buf[4];
unsigned int n;
n = g;
memcpy(buf, (char*)&n, sizeof(int));

Although it works fine but i'm not sure if its the correct way or is it safe?

PS. I'm trying to send data between 2 devices via USB serial communication (between Raspberry Pi & Arduino)

Grzegorz Szpetkowski

You could store both int (or unsigned int) and unsigned char array as union. This method is called type punning and it is fully sanitized by standard since C99 (it was common practice earlier, though). Assuming that sizeof(int) == 4:

#include <stdio.h>

union device_buffer {
    int i;
    unsigned char c[4];
};

int main(int argv, char* argc[])
{
    int original = 1054;

    union device_buffer db;
    db.i = original;

    for (int i = 0; i < 4; i++) {
        printf("c[i] = 0x%x\n", db.c[i]);
    }
}

Note that values in array are stored due to byte order, i.e. endianess.

Collected from the Internet

Please contact [email protected] to delete if infringement.