Mips Data layout calculation

I am self studying computer architect开发者_高级运维ure offer at Michigan university. I do not understand why the memory layout for d starts at 312 to 319 instead of 308 http://www.flickr.com/photos/45412920@N03/4442695706/. ( http://www.flickr.com/photos/45412920@N03/4442695706/ ) Maybe I did not understanding the Golden rule specified http://www.flickr.com/photos/45412920@N03/4441916461/sizes/l/ here( http://www.flickr.com/photos/45412920@N03/4441916461/sizes/l/ ) well.

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The second link shows that MIPS cannot pack variables, therefore the addresses they take up must fall on word boundaries.

if short is half word aligned, it takes up two bytes, int, is word aligned so it takes up 4 bytes, double must be doubleword aligned so it takes up 8 bytes.

In order to align at these places..

A zero in the Least Significant bit (LSB) would indicate every other or every 2 bytes (half word aligned), 2 Zeros indicates every 4th byte, and 3 zeros every 8th byte.

Address (4 LSBs)
    XXX0 - half word aligned (2 bytes)
    XX00 - Word aligned (4 bytes)
    X000 - Double word aligned (8 bytes)

The double must be double word aligned so it can not start at 308 (100110100) because it is only word aligned (2 LSBs = 0) it must start at the next double word alignment 312 (100111000)

[Addr]   [Binary]    [Alignment]  
300      100101100   Word, Half-Word
301      100101101
302      100101110   Half-Word
303      100101111   
304      100110000   Double-Word, Word, Half-Word
305      100110001
306      100110010   Half-Word
307      100110011 
308      100110100   Word, Half-Word
309      100110101   
310      100110110   Half-Word
311      100110111  
312      100111000   Double-Word, Word, Half-Word

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Memory access on the MIPS is word aligned which means that is reads the memory 32 bits/4 bytes at a time. Since variable "b" is a single byte, it actually reads addresses 300-303. If variable "c" were to start at 301, the processor would have to know that "b" is only a byte and zero out the other bytes and possibly shift it to the LSB position (or the compiler would have to do it). Either way, it's more efficient to just alight all memory access on 4 byte boundaries (multiple of 4).

See Data Structure Alignment for more info.