Let us now analyse the sample assembly language program, which is to be found in the main.c file. It is specified one byte at a time, as follows:
0: LDI 1: 128 2: 12 3: XCHR 4: 0 5: LDI 6: 0 7: 0xA0 8: LDX 9: XCHR 10: 1 11: LDI 12: 65 13: 131 14: STX 15: 0 16: 0 17: XCHR 18: 0 19: SBI 20: 1 21: 0 22: CMPI 23: 0 24: 0 25: JNA 26: 38 27: 0 28: XCHR 29: 0 30: XCHR 31: 1 32: ADI 33: 2 34: 0 35: JMP 36: 8 37: 0 38: HLT
There are a total of 39 bytes of code, taking up memory locations 0 through 38 in our machine.
Each instruction takes up one byte of memory, and each operand is 8 bits or 16 bits, requiring either one or two bytes of memory.
For each 16 bit operand, the low 8 bits are specified first and the high 8 bits are specified second. This is standard for machines which are known as little endian. Thus, to store the number 15 as a 16 bit number, we first note that the binary for 15 is 0000 0000 0000 1111 (i.e. 8 + 4 + 2 + 1 = 15). Now the lower 8 bits are stored first, followed by the high order 8 bits. So the number is stored as two bytes, 15 then 0. To work out what the value of any 2 byte combination is, multiply the second byte by 256 and add it to the first byte. Thus in our example 0*256 + 15 = 15.
