CODEWITHSUNDEEP
memoryassemblyintegermipsendianness
Johnny Rocket
Johnny Rocket

Reputation: 1434

Understanding the memory layout of a 4-byte integer

I am working with the MIPS architecture (not sure if this is relevant since we are dealing with memory).

I am told that A 32-bit integer is in memory at physical address 0x00A0CE48.
I assume that number is 00000000111111110000000011111111.

The system is byte-addressable, what value would be at memory address 0x00A0?

I wasn't sure if the first 8 bits were at address 0x00, next 8 bits at 0x00A0, next 8 bits at 0x00A0CE, and the last 8 bits at 0x00A0CE48? I'm asking because I have to manipulate a value in 0x00A0, but im not sure what's there.

Part of the problem is to 1st assume big endian is used, then little endian.

A 32-bit integer resides in memory at physical address 0x00A0CE48. The bits within the 32-bit word are numbered 0 to 31 from least significant bit to most significant bit. The code below extracts a single bit from this 32 bit pattern and places the bit into $t4.

lui    $t0,0x00A0
ori    $t0,$t0,0xCE48
lbu    $t4,2($t0)
srl    $t4,$t4,5
andi   $t4,$t4,1

The next question in my assignment is to indicate the number of the bit (0 through 31) within the 32-bit word that is left in $t4 if the memory order used is little-endian or big-endian.

Upvotes: 0

Views: 2924

Answers (1)

Ray
Ray

Reputation: 1869

On a big endian system, the most significant byte is stored first. So, assuming the value is 0x12345678, 0x12 will be stored at address 0x00A0CE48, 0x34 will be stored at address 0x00A0CE49, 0x56 will be stored at address 0x00A0CE4A, and 0x78 will be stored at address 0x00A0CE4B.

On the other hand, on a little endian system, the least significant byte will be stored first. So, 0x78 would be stored at 0x00A0CE48, and so on.

Note that if a 32-bit word is stored at address 0x00A0CE48, the next word will be four bytes later, at address 0x00A0CE4C. The arithmetic should be performed on the address as a whole. You cannot consider the bytes making up the address separately when reading from memory.

In the assembly you've posted, lui (which stands for "load upper immediate") will shift the immediate value 16 bits to the left and store it in $t0. After that instruction, the value in $t0 will be 0x00A00000. The next instruction will OR the contents of $t0 with 0xCE48 and store the results in $t0. After that, $t0 will contain your full address, 0x00A0CE48.

Upvotes: 5

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