In a MIPS32 system suppose a 4byte signed integer variable n

In a MIPS32 system, suppose a 4-byte signed integer variable named x is allocated in memory at 0x1001_4822? Would there be any issues with this scenario? In a MIPS32 system, suppose a 4-byte signed integer variable named z with value 0xAABBCCDD is allocated in memory at 0x1001_0000? The system is big-endian. Draw a memory diagram showing the four bytes of x stored in memory and clearly label the addresses of each byte of x. A third degree polynomial equation (a cubic equation) is of the form p(x)=c_3x^3+ c_2x^2+c_1x+ c_0, where x and the four coefficients are integers for this exercise. Suppose the values of the coefficients c_o, c_1 c_2, and c_3 have been loaded into registers $t0, $t1, St2, and St3, respectively. Suppose the value of x is in St7. Write the MIPS32 instructions that would evaluate this polynomial, placing the result in $t9.

Solution

13)The purpose of memory is to store groups of bits, and deliver them to the processor for loading into registers upon demand. Most present-day computers store information in multiples of 8 bits, called a byte.Most also assign a numeric address to each byte. This is convenient because characters can be stored in bytes.
Memory addresses are 32-bit numbers, ranging from 0x00000000 to 0xFFFFFFFF. This is a large amount of memory, most computers do not have actual memory for all of this address space.
Memory can hold both program instructions and data. One function of the operating system is to assign blocks of
memory for the instructions and data of each process .

MIPS is little-endian. One result of this is that character data appear to be stored backwards within words.

In the case of MIPS, a word is 32 bits, that is, 4 bytes.Words are always stored in consecutive bytes, starting with an address that is divisible by 4.In the MIPS, all registers hold 32 bits.