Question: For lab 3 part 2, you will be given four different word-length numbers stored sequentially in a DATA area of memory. This data area of

 For lab 3 part 2, you will be given four different

For lab 3 part 2, you will be given four different word-length numbers stored sequentially in a DATA area of memory. This data area of memory will start at the address labeled "TEST_DATA". Your program will be responsible for concatenating the first (lowest significant digits) byte of the first word with the second lowest byte from the second word with the third lowest byte from 3181: Microcontrollers and Embedded Systems Lab ng 2020 Professor Joe Camp the third word with the highest byte from the fourth word. Your program will then store the newly created word in the following area of memory labeled "NEW". For example, if you are given the following strings: OxFDFFEF1A, OxDC184F28, 0x897FDAB1, 0xFFFF8473 The answer stored in the next word of memory should be: 0xFF7F4F1A. Tip: remember, every address corresponds to a byte in memory. Also, be careful about the endian-ness of your program. Given: AREA asm_data, DATA, READONLY TEST_DATA DCD OxEFD76541 ;test data 0 - 0x000 offset DCD 0x876FDCBB test data 1 - 0x004 offset DCD OXA9875432 ;test data 2 - 0x008 offset DCD OxBBBOOBBB ;test data 3 - 0x00c offset NEW AREA storage, DATA, READWRITE SPACE 20 END For lab 3 part 2, you will be given four different word-length numbers stored sequentially in a DATA area of memory. This data area of memory will start at the address labeled "TEST_DATA". Your program will be responsible for concatenating the first (lowest significant digits) byte of the first word with the second lowest byte from the second word with the third lowest byte from 3181: Microcontrollers and Embedded Systems Lab ng 2020 Professor Joe Camp the third word with the highest byte from the fourth word. Your program will then store the newly created word in the following area of memory labeled "NEW". For example, if you are given the following strings: OxFDFFEF1A, OxDC184F28, 0x897FDAB1, 0xFFFF8473 The answer stored in the next word of memory should be: 0xFF7F4F1A. Tip: remember, every address corresponds to a byte in memory. Also, be careful about the endian-ness of your program. Given: AREA asm_data, DATA, READONLY TEST_DATA DCD OxEFD76541 ;test data 0 - 0x000 offset DCD 0x876FDCBB test data 1 - 0x004 offset DCD OXA9875432 ;test data 2 - 0x008 offset DCD OxBBBOOBBB ;test data 3 - 0x00c offset NEW AREA storage, DATA, READWRITE SPACE 20 END

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