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时间:2022-11-07 17:03:09浏览次数:39  
标签:function code int ll transpose && include

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//Zhang Kaitai 2100012922
#include <stdio.h>
#include "cachelab.h"
#include <stdlib.h>
#include <unistd.h>
#include <getopt.h>
#include <string.h>
#include <limits.h>
#define ll long
char s[3],s2[100],s3,c;
ll f[555];
ll ls,S,E,lb,B,C,t;
char pth[100];
int hit_count, miss_count, eviction_count;
struct cache{
	ll t,st;
}**cash;
ll altime;
//cache cash[32][5][32];
void init_cash(){
	cash=(struct cache**) malloc(sizeof(struct cache*)*S);
	for(ll i=0;i<S;++i)
		cash[i]=(struct cache*)malloc(sizeof(struct cache)*E);
}
void end_cash(){
	for(int i=1;i<=S;++i)free(cash[i]);
	free(cash);
}
ll getv(char *s){
	ll len=strlen(s),a=0;
	for(ll i=0;i<len;++i)
		a=a*10+s[i]-'0';
	return a;
}
void getval2(char *a,char *s){
	int len=strlen(s);
	for(int i=0;i<len;++i)a[i]=s[i];
	return;
}

void init(){
	for(int i='0';i<='9';++i)f[i]=i-'0';
	for(int i='a';i<='f';++i)f[i]=i-'a'+10;
}

ll pos;int sz;
void check(ll pos){
	ll k=((1ll<<ls)-1);k=k&(pos>>lb);
	ll tag=((1ll<<t)-1);tag=tag&(pos>>(ls+lb));
	ll bs=(1ll<<lb)-1;bs=bs&pos;
	int fr=0;
	for(int i=0;i<E;++i){
		if(cash[k][i].t&&cash[k][i].st==tag){
			++hit_count;cash[k][i].t=++altime;
			return;
		}
		if(cash[k][i].t<cash[k][fr].t)fr=i;
	}
	++miss_count;
	if(cash[k][fr].t)++eviction_count;
	cash[k][fr].t=++altime;
	cash[k][fr].st=tag;
}
void work(){
	if(s[0]=='I')return;
	check(pos);
	if(s[0]=='M')check(pos);
}
int main(int argc,char *argv[]){
	init();
	char ch;
	while((ch=getopt(argc,argv,"s:E:b:t:"))!=-1){
		switch(ch){
			case 's':{

						 ls=getv(optarg);
						 break;
					 }
			case 'E':{
						 E=getv(optarg);
						 break;
					 }
			case 'b':{
						 lb=getv(optarg);
						 break;
					 }
			case 't':{
						 getval2(pth,optarg);
					 }
			default: break;
		}
	}
	t=64-ls-lb;
	S=1<<ls;B=1<<lb;C=S*E*B;
	init_cash();
	freopen(pth,"r",stdin);
	while(scanf("%s",s)!=EOF){
		pos=sz=0;
		c=getchar();
		while((c<'a'||c>'f')&&(c>'9'||c<'0'))c=getchar();
		while((c>='a'&&c<='f')||(c>='0'&&c<='9')){
			pos=(pos<<4)+f[(int)c];
			c=getchar();
		}
		while((c<'a'||c>'f')&&(c>'9'||c<'0'))c=getchar();
		while((c>='a'&&c<='f')||(c>='0'&&c<='9')){
			sz=sz*10+c-'0';
			c=getchar();
		}
		work();
	}
	end_cash();
	printSummary(hit_count, miss_count, eviction_count);
	return 0;
}
点击查看代码
/*
 * Zhang Kaitai
 * 2100012922
 * 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"
#include "contracts.h"

int is_transpose(int M, int N, int A[N][M], int B[M][N]);

/*
 * 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. The REQUIRES and ENSURES from 15-122 are included
 *     for your convenience. They can be removed if you like.
 */
char transpose_submit_desc[] = "Transpose submission";
void transpose_submit(int M, int N, int A[N][M], int B[M][N])
{
    REQUIRES(M > 0);
    REQUIRES(N > 0);
    int i,j,k,o,tmp;//t1,t2;
#define pad 2
    if(N==M){
//		for(i=0;i<N;++i)for(j=0;j<M;++j)B[i][j]=A[i][j];
		for(o=0;o<N;o+=8){
			for(k=0;k<N;k+=pad){
				for(i=o;i<o+8;++i){
					for(j=k;j<k+pad;++j){
						B[j][i]=A[i][j];
//						B[j+1][i]=A[i][j+1];
//						B[j+2][i]=A[i][j+2];
//						B[j+3][i]=A[i][j+3];
					}
				}
			}
		}
	}
	else{
		for (i = 0; i < N; i++) {
		    for (j = 0; j < M; j++) {
	        	tmp = A[i][j];
	    	    B[j][i] = tmp;
	 	   }
		}
	}
    ENSURES(is_transpose(M, N, A, B));
}

/*
 * 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;

    REQUIRES(M > 0);
    REQUIRES(N > 0);
   	for (i = 0; i < N; i++) {
	    for (j = 0; j < M; j++) {
	        tmp = A[i][j];
	        B[j][i] = tmp;
	    }
	}
    ENSURES(is_transpose(M, N, A, B));
}

/*
 * 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;
}

标签:function,code,int,ll,transpose,&&,include
From: https://www.cnblogs.com/loadingkkk/p/16866542.html

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