Assume the contents of memory are shown below. All values are in hexadecimal. Th

Question

Assume the contents of memory are shown below. All values are in hexadecimal. The table shows four bytes per line; the given address is the starting address of each line. Each block in the table contains a single byte, with the low and high bytes per line indicated as shown. Each byte has its own address, so the byte at address 0x81114 is 0x69, address 0x81 115 is 0xC8, address 0x81116 is 0xA l, and address 0x81117 is 0x54. You should assume all multi-byte values are stored in little-endian format A word is 16 bits (2 bytes). For each address and amount of data listed, answer the following: What data are stored at that address? Would an access to the given amount of data at that address be aligned? If the data represents a signed integer, what is the sign of that value? For example, given "Address: 0x81114, Data size: word, " your response would be that the word at 0x81114 is 0xC869, the access is aligned, and the data represents a negative integer. Address: 0x8111C, Data size: word Address: 0x81125, Data size: byte Address: 0x8111B, Data size: double word Address: 0x81 126, Data size: word Address: 0x81 117, Data size: double word

Explanation / Answer

if the word size is 16 bit

0x81114             0xA154

0x81116              0x69C8

and so on

0x8112C 4536

0x8112E     0xBBB5

The little-endian system has the property that the same value can be read from memory at different lengths without using different addresses (even when alignment restrictions are imposed). For example, a 32-bit memory location with content 4A 00 00 00 can be read at the same address as either 8-bit (value = 4A), 16-bit (004A), 24-bit (00004A), or 32-bit (0000004A).

if the sign of data is negative then msb must be one

F58A binary conversion is 1111 0101 1000 1010 number is negative

69C8 0110 1001 1100 1000 number is positive

Address : 0x8111C i assume word(16 bit) little indian FEA3 negative

0x81125 byte(8 bit) A4 negative

0x8111B double word(32 bit) EA positive because of some other leading data

0x81126 45 negative

0x81117 54

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