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Asked: 2026-05-14T19:15:24+00:00

Below is an example of a question given on my last test in a

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Below is an example of a question given on my last test in a Computer Engineering course. Anyone mind explaining to me how to get the start/end addresses of each? I have listed the correct answers at the bottom…

The MSP430F2410 device has an address space of 64 KB (the basic MSP430 architecture). Fill in the table below if we know the following. The first 16 bytes of the address space (starting at the address 0x0000) is reserved for special function registers (IE1, IE2, IFG1, IFG2, etc.), the next 240 bytes is reserved for 8-bit peripheral devices, and the next 256 bytes is reserved for 16-bit peripheral devices. The RAM memory capacity is 2 Kbytes and it starts at the address 0x1100. At the top of the address space is 56KB of flash memory reserved for code and interrupt vector table. 

What                            Start Address End Address
Special Function Registers (16 bytes) 0x0000 0x000F
8-bit peripheral devices (240 bytes)  0x0010 0x00FF
16-bit peripheral devices (256 bytes) 0x0100 0x01FF
RAM memory (2 Kbytes)                 0x1100 0x18FF
Flash Memory (56 Kbytes)              0x2000  0xFFFF
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  1. Editorial Team

    For starters, don’t get thrown off by what’s stored in each segment – that’s only going to confuse you. The problem is just asking you to figure out the hex numbering, and that’s not too difficult. Here are the requirements:

    • 64 KB total memory
    • The first 16 bytes of the address space (starting at the address 0x0000) is reserved for special function registers (IE1, IE2, IFG1, IFG2, etc.)
    • The next 240 bytes is reserved for 8-bit peripheral devices
    • The next 256 bytes is reserved for 16-bit peripheral devices
    • The RAM memory capacity is 2 Kbytes and it starts at the address 0x1100
    • At the top of the address space is 56KB of flash memory reserved for code and interrupt vector table.

    Since each hex digit in your memory address can handle 16 values (0-F), you’ll need 4 digits to display 64KB of memory (16 ^ 4 = 65536, or 64K).

    You start with 16 bytes, and that covers 0x0000 – 0x000F (one whole digit of your address). That means that the next segment, which starts immediately after it (8-bit devices), begins at 0x0010 (the next byte), and since it’s 240 bytes long, it ends at byte 255 (240 + 15), or 0x00FF.

    The next segment (16-bit devices) starts at the next byte, which is 0x0100, and is 256 bytes long – that puts the end at 0x01FF.

    Then comes 2KB (2048 bytes) of RAM, but it starts at 0x1100, as the description states, instead of immediately after the previous segment, so that’s your starting address. Add 2048 to that, and you get 0x18FF.

    The last segment covers the upper section of the memory, so you’ll have to work backward, You know it ends at 0xFFFF (the end of the available memory), and it’s 56KB long. If you convert the 56KB to hex, it’s 0xDFFF. If you imagine that this segment starts at 0, That leaves 2000 unused (0xE000-0xEFFF and 0xF000-0xFFFF), so you know that this segment has to start at 0x2000 to end at the upper end of the memory space.

    I hope that’s more clear, though when I read over it, I don’t know that it’s any help at all 🙁 Maybe that’s why I’ll leave teaching that concept to somebody more qualified…

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