[edit]
From OP in comment to one of the solutions (explains more thoroughly):
"thanx for replying but i want to separate both bytes of uint16 and then the total 10 bytes which , i will get should be in array of uint8"
[/edit] uint8 arr2[10] = { 0 }; uint8* ptr = (uint8 *) &arr1; // cast the 16bit pointer to an 8bit pointer for ( int i= 0 ; i<10; i++) arr2[i] = *ptr; // pass data to other array ptr++; // move your pointer MB_DATA data = { 2 , 3 , 20 , 90 , 3857 }; // this will already contain both 16 and 8 bit representations
Should be noted that with this solution, the two methods of accessing the data are dependent, meaning you change something in one and it changes on both.... unions don't allocate memory independently for members.

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