copying 2 8 bit values into 1 16 bit value

[califguy]

Hey guys,
here's what I am looking to do.
Have an array thats 8 bit.
char [32].

I have a qty, say temp which is 16 bit.

I need qty to hold char [0] in first 8 bits and char[1] in next 8 bits.
Memset is givin me weird errors, so was hoping someone here could show me howto approach this or give me sample memset code.
Thanks for the help.

[jsmith]

So why not

// assume
uint16_t qty;
char array[ /*some number >= 2 */ ];
qty = *static_cast<uint16_t*>( array );

[jmc]

Depending on your matter you can just define the address of your 16bit variable as the start of your 8 bit array and you will be able to change it by using both variables without to copy the hole thing.

[Duthomhas]

Those solutions will only work on little-endian machines.

When playing with bits you should generally just give-in and do some bit-shifting.

char eights[ 2 ] = { 0xA2, 0x4F };  /* 0x4FA2 */

short temp = eights[ 0 ] +(eights[ 1 ] << 8);

printf( "%X\n", temp );  /* 4FA2 */

Then when your hardware changes you'll never have to wonder where your code broke.

Hope this helps.

[califguy]

@jsmith
Didnt know about the static_cast operator. Did'nt want to use typecasting as it simply converts the whole var to the typecasted qty.
What i did was cast a struct, typecasted the array to this struc and that way I got my 16 bit structure.
Thanks @jsmith,Duoas and JMC too.