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/*
* trans.c - Matrix transpose B = A^T
*
* Each transpose function must have a prototype of the form:
* void trans(int M, int N, int A[N][M], int B[M][N]);
*
* A transpose function is evaluated by counting the number of misses
* on a 1KB direct mapped cache with a block size of 32 bytes.
*/
#include <stdio.h>
#include "cachelab.h"
int is_transpose(int M, int N, int A[N][M], int B[M][N]);
void print_array(int M, int N, int A[N][M], int x, int y)
{
int i,j;
for (i = 0; i < N; i++) {
for (j = 0; j < M; j++) {
if(i == y && j == x)
printf("\x1B[31m # \x1B[0m");
else
printf(" # ");
}
printf("\n");
}
}
void diff_array(int M, int N, int A[N][M], int B[M][N], int x, int y)
{
printf("%d, %d\n", x, y);
print_array(M, N, A, x, y);
printf("\n");
print_array(N, M, B, y, x);
getchar();
}
/*
* transpose_submit - This is the solution transpose function that you
* will be graded on for Part B of the assignment. Do not change
* the description string "Transpose submission", as the driver
* searches for that string to identify the transpose function to
* be graded.
*/
char transpose_submit_desc[] = "Transpose submission";
void transpose_submit(int M, int N, int A[N][M], int B[M][N])
{
int blocksize = 8; //2^5/4
int i, j, k, diag, diagX, diagY;
for(i=0; i<M; i+=blocksize)
{
for(j=0; j<N; j++)
{
diag = 0;
for(k=0; k<blocksize && (i+k < M); k++)
{
if (j*N + i + k != (i + k)*M + j)
{
B[i+k][j] = A[j][i+k];
//diff_array(M, N, A, B, j, i+k);
}
else
{
diag = A[j][i+k];
diagX = i+k;
diagY = j;
}
}
if(diag)
B[diagX][diagY] = diag;
}
}
}
/*
* You can define additional transpose functions below. We've defined
* a simple one below to help you get started.
*/
/*
* trans - A simple baseline transpose function, not optimized for the cache.
*/
char trans_desc[] = "Simple row-wise scan transpose";
void trans(int M, int N, int A[N][M], int B[M][N])
{
int i, j, tmp;
for (i = 0; i < N; i++) {
for (j = 0; j < M; j++) {
tmp = A[i][j];
B[j][i] = tmp;
}
}
}
/*
* registerFunctions - This function registers your transpose
* functions with the driver. At runtime, the driver will
* evaluate each of the registered functions and summarize their
* performance. This is a handy way to experiment with different
* transpose strategies.
*/
void registerFunctions()
{
/* Register your solution function */
registerTransFunction(transpose_submit, transpose_submit_desc);
/* Register any additional transpose functions */
registerTransFunction(trans, trans_desc);
}
/*
* is_transpose - This helper function checks if B is the transpose of
* A. You can check the correctness of your transpose by calling
* it before returning from the transpose function.
*/
int is_transpose(int M, int N, int A[N][M], int B[M][N])
{
int i, j;
for (i = 0; i < N; i++) {
for (j = 0; j < M; ++j) {
if (A[i][j] != B[j][i]) {
return 0;
}
}
}
return 1;
}