Question

Possible Duplicate:
Store an int in a char array?

i want to load 4 8-bit unsigned char to 32-bit integer. And store 32 bit integer to unsigned char pointer. How is this possible?Example usage below;

int 32bitint1= 0xff000000 | (uchar1<<16) | (uchar2<<8) | uchar3;
int 32bitint2= 0xff000000 | (uchar4<<16) | (uchar5<<8) | uchar6;
//then this 32-bit integer to uchar pointer;
ucharpointer[0] = 32bitint1;
ucharpointer[4] = 32bitint2;//is this possible?or how

Answer 1

Store: (store 4 chars into an unsigned int)

int store(uint32_t * reg, unsigned char c[4])
{
    *reg = 0;
        for(int i=0;i<4;i++)
        {
            *reg = (*reg<<8) | c[i];
        }
        return 0;
}

Load: (load 4 chars from an unsigned int)

int load(uint32_t * reg, unsigned char c[4])
{
        for(int i=0;i<4;i++)
        {
            c[i] = *reg;
            *reg = *reg>>8;
        }
        return 0;
}

Usage example:

int main ()
{
    unsigned char c[4] = {'a','b','c','d'};
    uint32_t reg;

    printf("%c",c[0]);  //it prints 'a'
    store(&reg,c);   

    c[0] = 'e';
    printf("%c",c[0]);  //it prints 'e'

    load(&reg,c);     //load
    printf("%c",c[0]);  //it prints 'a' again

    return 0;
}

If you don’t want to reload them into a char array, but to access them by a char pointer, then here’s an example:

int main (int argc, char const *argv[])
{
    unsigned char c[4] = {'a','b','c','d'};
    uint32_t reg;
    store(&reg,c);

    unsigned char *cpointer = (unsigned char *) ®

    for(int i=0;i<4;i++)
    {
        printf("%c",cpointer[i]);  //access the 4 chars by a char pointer
    }
    return 0;
}

Please note that the you will get an output ‘dcba’ in this way, as the memory address are made in the reverse order.