Consider the following segment of a C++ code for...

Consider the following segment of a C++ code for multiplication of two matrices A and B of size (NxN) each: for (int i=0;i<N;i++) { for (int j=0;j<N;j++) { temp = 0; for (int k=0;k<N; k++) { temp = temp + A[i][k]*B[k][j]; C[i][j] = temp; } } } Thus, matrix C is the product of matrix A and matrix B. For example, for N=3: $\begin{bmatrix} 1 & 3 & 2 \\ 1 & 3 & 2 \\ 1 & 3 & 2 \end{bmatrix}$, $B = \begin{bmatrix} 0 & 1 & 2 \\ 0 & 1 & 2 \\ 0 & 1 & 2 \end{bmatrix}$ The result C=AxB is given by: $\begin{bmatrix} 0 & 6 & 12 \\ 0 & 6 & 12 \\ 0 & 6 & 12 \end{bmatrix}$ Write an assembly program for matrix multiplication assuming all elements are integers and the matrices are stored in main memory. Initially, you may hard code the input matrices A and B of size 3x3 each in your program. You can choose either storing them in row-major or column-major order. Observe the cache performance (e.g., hit- rate, miss- rate etc.) for your code.

Transcript

00:01 Hello student, this program uses 26 real mips instructions that is less than maximum of 28 instructions.

00:07 So, it works by first initializing the variable a, b and c, then it calculates x, y and store them to the memory location 0x20 and 0x24 respectively.

00:19 Next it shifts y right by the 16 bit.

00:23 The while loop begins by checking if the c is greater than or equal to 1665 if it is the loop ends.

00:29 Otherwise, the loop calculates y divided by c divided by 2 and add it to the c.

00:34 Then it shifts c left by the 8 bits and store it to the memory location 0x30 and the loop then jumps back to the beginning of itself.

00:43 The program ends by jumping to the end label.

00:46 So here c+ + instructions pseudocodes are given.

00:50 So i have to write their corresponding assembly codes mips assembly code.

00:56 So first here variable initialization.

01:00 So here or immediate with this mnemonics i have to store in a note i have to assign this value 0x8000 and in with the help of load immediate instruction i have to assign the a1 means b value to this one 0x00 a9 and after this in s note i have to save the c value which will be 1974.

01:31 So these are the register i have taken $a0 $a1 $s0.

01:37 So arithmetic computations now i have to do the multiplication.

01:41 So here i what i am doing i am multiplying a into a.

01:45 So my a is in a note.

01:47 So a note with a note will be multiplied with this and the value will be saved to the $s1.

01:56 So $s1 means x.

01:59 So here it will a note into a note.

02:05 So this will be the calculation.

02:07 After this sw store a word now store x to the memory location at the address of 0x20.

02:14 So my now this s1 will be saved at the memory location 0x20 and after this i have to do this calculation y equal to x into b.

02:25 This is my a this is b and this is c.

02:29 So here for x into b this is x.

02:32 So x into b x value is in s1 and b value is in $a1.

02:39 So both of these will be multiplied and the value y $s2 will carry the value of y and y is equal to x into x into b.

02:51 So $s2 will carry the value of y and y calculation is x into b according to our program it will be $s1 into $a1.

03:04 So $a1 or $s1.

03:08 Now i have to save this into the memory location 0x24...

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