4. Consider the memory contents as below. The lw instruction (below) loads the word at 0x0060500C...
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4. Consider the memory contents as below. The lw instruction (below) loads the word at 0x0060500C into $12. Memory Addresses 00000005 00605014 00000004 00605010 00000003 0060500C 00000002 00605008 00000001 00605004 00605000 00000000 Complete the instruction sequence so that the base register (register $13) contains the address 0x00605000, which will be used with a displacement of Ox0C. lui $13, Ox ori $13, $13, 0xl lw $12, 0xC($13) 4. Consider the memory contents as below. The lw instruction (below) loads the word at 0x0060500C into $12. Memory Addresses 00000005 00605014 00000004 00605010 00000003 0060500C 00000002 00605008 00000001 00605004 00605000 00000000 Complete the instruction sequence so that the base register (register $13) contains the address 0x00605000, which will be used with a displacement of Ox0C. lui $13, Ox ori $13, $13, 0xl lw $12, 0xC($13)
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Computer Organization and Design The Hardware Software Interface
ISBN: 978-0124077263
5th edition
Authors: David A. Patterson, John L. Hennessy
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