Map integers, shorts and char to bitpattern

Question

hi

Imagine that I have the following variables:

unsigned long a = 1;    //32-bit value
unsigned short b = 1;   //16-bit value
unsigned char c ='\x01' //8-bit value

unsigned char buffer[7];

Now I would like to map or combine those variables above in the buffer as follows: first four bytes should be occupied by value of int a, next 2 bytes by value b and the last byte by c.

==> buffer = 0x 00000001 0001 01

Is here anyone aware of an easy way to do that in C++? At the moment I am having help function that return the byte pattern for shorts and integers but I guess in C++ there must be a better way to do that?

Thanks

Answer 1

*(unsigned long*)(buffer+0) = a;
*(unsigned short*)(buffer+4) = b;
*(unsigned char*)(buffer+6) = c;

Note however, that on some CPU architectures (though not x86), this will fail due to alignment issues. On x86 it will just cause some inefficiency in memory transactions.

Answer 2

It's messy, but:

Going into buffer

buffer[ 0 ] = ( unsigned char ) a;
buffer[ 1 ] = ( unsigned char )( a >> 8 );
buffer[ 2 ] = ( unsigned char )( a >> 16 );
buffer[ 3 ] = ( unsigned char )( a >> 24 );
buffer[ 4 ] = ( unsigned char ) b;
buffer[ 5 ] = ( unsigned char )( b >> 8 );
buffer[ 6 ] = c;

Coming out of buffer

a = buffer[ 0 ] | ( buffer[ 1 ] << 8 ) | ( buffer[ 2 ] << 16 ) | ( buffer[ 3 ] << 24 );
b = buffer[ 4 ] | ( buffer[ 5 ] << 8 );
c = buffer[ 6 ];

Answer 3

Use a union

#include <iostream>
struct MyStructData   
{
    unsigned long  a;
    unsigned short b;
    unsigned char  c; 
};
union Swap
{
     MyStructData   data;
     char           buffer[7];// Can use [sizeof(MyStructData)]
};

int main()
{
    Swap  data;
    data.data.a = 1;
    data.data.b = 1;
    data.data.c = 1;

    for(int loop=0;loop < 7;++loop)
    {
        std::cout << "buffer(" << loop <<")  = (" << (int)data.buffer[loop] << ")\n";
    }
}

Answer 4

in addition to the other answers (which look fine), you can do something like this (note depending on arch, you may need to reorder the contents of the struct. you may also need to enable packing to ensure no padding is inserted between members):

union {
    struct {
        int   int_value;
        short short_value;
        char  char_value;
    } values;
    char buffer[sizeof(values)];
} x;

x.values.int_value   = a;
x.values.short_value = b;
x.values.char_value  = c;