I've run the following code s开发者_如何转开发nippet on the MIPS MARS simulator. That simulator is little endian. So the results are as follows:
lui $t0,0x1DE # $t0 = 0x01DE0000
ori $t0,$t0,0xCADE # $t0 = 0x01DECADE
lui $t1,0x1001 # $t1 = 0x10010000
sw $t0,200($t1) # $t1 + 200 bytes = 0x01DECADE
lw $t2,200($t1) # $t2 = 0x01DECADE
So on a little endian MIPS simulator, the value of $t2 at the end of the program is 0x01DECADE. If this simulator was big endian, what would the value be? Would it be 0xDECADE01 or would it still be 0x01DECADE?
It would be the same -- the order of the bytes in memory would be different, but you would only see that if you loaded single bytes from 200($t1), 201($t1), 202($t1) and 203($t1)
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