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4 bytes.) ow assume the same 1kB of memory but now, word size is 64 bits. The starting address of the first word is 0, the cond word starts at the third word starts at and the last word starts at

4 bytes.) ow assume the same 1kB of memory but now, word size is 64 bits. The starting address of the first word is 0, the cond word starts at the third word starts at and the last word starts at

C++ for Engineers and Scientists
C++ for Engineers and Scientists
4th Edition
ISBN:9781133187844
Author:Bronson, Gary J.
Publisher:Bronson, Gary J.
Chapter10: Pointers
Section10.1: Addresses And Pointers
Problem 2E
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A byte is always 8 bits; Each byte has its own address Addr Addr Contents (Bin) 000 1 0 0 0 1 10 1 001 0 000 o 0 00 010 1 1 1 0 1 0 10 011 0 1 1 1 1 1 1 1 100 1 0 0 00 100 101 1 0 10 10 1 1 1 Word size is 4 bytes Starting address 0 2 3 Word size is 4 bytes Starting address 4 6 110 0 0 1 1 0 0 10 111 0 1 1 0 o 0 1 1 Notice that each word starts with an address that is a multiple of 4. Some systems require that addresses be aligned to these boundaries, i.e. address provided must be a multiple of 4, while other systems allow unaligned memory accesses. It depends on the way the system is designed. Assume you now have 1kB of memory, i.e. the memory address space runs from 0 to 1023. The starting address of the first word is 0, the second word is 4, the third word is 8, and so on. The last word comprising 4 bytes resides in addresses 1020, 1021, 1022, 1023. Thus, the last word starts at 1020, which is a multiple of 4. (Recall that the word size is 4 bytes.) Now assume the same 1kB of memory but now, word size is 64 bits. The starting address of the first word is 0, the second word starts at the third word starts at and the last word starts at Check 4) 7.
Transcribed Image Text:A byte is always 8 bits; Each byte has its own address Addr Addr Contents (Bin) 000 1 0 0 0 1 10 1 001 0 000 o 0 00 010 1 1 1 0 1 0 10 011 0 1 1 1 1 1 1 1 100 1 0 0 00 100 101 1 0 10 10 1 1 1 Word size is 4 bytes Starting address 0 2 3 Word size is 4 bytes Starting address 4 6 110 0 0 1 1 0 0 10 111 0 1 1 0 o 0 1 1 Notice that each word starts with an address that is a multiple of 4. Some systems require that addresses be aligned to these boundaries, i.e. address provided must be a multiple of 4, while other systems allow unaligned memory accesses. It depends on the way the system is designed. Assume you now have 1kB of memory, i.e. the memory address space runs from 0 to 1023. The starting address of the first word is 0, the second word is 4, the third word is 8, and so on. The last word comprising 4 bytes resides in addresses 1020, 1021, 1022, 1023. Thus, the last word starts at 1020, which is a multiple of 4. (Recall that the word size is 4 bytes.) Now assume the same 1kB of memory but now, word size is 64 bits. The starting address of the first word is 0, the second word starts at the third word starts at and the last word starts at Check 4) 7.
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