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多次元配列 in C
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#include <stdio.h> | |
#include <stdlib.h> | |
#include <string.h> | |
#include <assert.h> | |
#include <sys/time.h> | |
double getETime() { | |
struct timeval tv; | |
gettimeofday(&tv, NULL); | |
return tv.tv_sec + (double) tv.tv_usec * 1e-6; | |
} | |
int main( int argv, char **argc ) { | |
int NT = 10000; | |
int NX = 1024; | |
int NY = 1024; | |
int x, y, t; | |
double ts, te; | |
// version1 | |
ts = getETime(); | |
double **u1, **p1; | |
u1 = (double**)malloc(sizeof(double*)*(NY+2)); | |
p1 = (double**)malloc(sizeof(double*)*(NY+2)); | |
for (y = 0 ; y < NY+2; y++) { | |
u1[y] = (double*)malloc(sizeof(double)*(NX+2)); | |
p1[y] = (double*)malloc(sizeof(double)*(NX+2)); | |
} | |
for (y = 0; y < NY+2; y++) { | |
for (x = 0; x < NX+2; x++) { | |
u1[y][x] = x + y; | |
} | |
} | |
for (t = 0; t < NT; t++) { | |
for (y = 1; y < NY+1; y++) { | |
for (x = 1; x < NX+1; x++) { | |
p1[y][x] = u1[y][x] + u1[y+1][x] + u1[y-1][x] + u1[y][x+1] + u1[y][x-1]; | |
} | |
} | |
} | |
te = getETime(); | |
double time1 = te - ts; | |
printf("1: elapsed time %g\n", time1); | |
// version2 | |
ts = getETime(); | |
double (*u2)[NX+2]; | |
double (*p2)[NX+2]; | |
u2 = (double(*)[NX+2])malloc(sizeof(double)*(NX+2)*(NY+2)); | |
p2 = (double(*)[NX+2])malloc(sizeof(double)*(NX+2)*(NY+2)); | |
for (y = 0; y < NY+2; y++) { | |
for (x = 0; x < NX+2; x++) { | |
u2[y][x] = x + y; | |
} | |
} | |
for (t = 0; t < NT; t++) { | |
for (y = 1; y < NY+1; y++) { | |
for (x = 1; x < NX+1; x++) { | |
p2[y][x] = u2[y][x] + u2[y+1][x] + u2[y-1][x] + u2[y][x+1] + u2[y][x-1]; | |
} | |
} | |
} | |
te = getETime(); | |
double time2 = te - ts; | |
printf("2: elapsed time %g\n", time2); | |
// version3 | |
ts = getETime(); | |
double **u3, **p3; | |
u3 = (double**)malloc(sizeof(double*)*(NY+2)); | |
p3 = (double**)malloc(sizeof(double*)*(NY+2)); | |
for (y = 0 ; y < NY+2; y++) { | |
u3[y] = (double*)malloc(sizeof(double)*(NX+2)); | |
p3[y] = (double*)malloc(sizeof(double)*(NX+2)); | |
} | |
for (y = 0; y < NY+2; y++) { | |
for (x = 0; x < NX+2; x++) { | |
u3[y][x] = x + y; | |
} | |
} | |
for (t = 0; t < NT; t++) { | |
for (y = 1; y < NY+1; y++) { | |
for (x = 1; x < NX+1; x++) { | |
p3[y][x] = u3[y][x] + u3[y+1][x] + u3[y-1][x] + u3[y][x+1] + u3[y][x-1]; | |
} | |
} | |
} | |
te = getETime(); | |
double time3 = te - ts; | |
printf("3: elapsed time %g\n", time3); | |
// version4 | |
ts = getETime(); | |
int ind, ny; | |
double *u4, *p4; | |
u4 = (double*)malloc(sizeof(double)*(NY+2)*(NX+2)); | |
p4 = (double*)malloc(sizeof(double)*(NY+2)*(NX+2)); | |
for (y = 0; y < NY+2; y++) { | |
ny = y * (NX+2); | |
for (x = 0 ; x < NX+2; x++) { | |
ind = ny + x; | |
u4[ind] = x + y; | |
} | |
} | |
int nyu; | |
for (t = 0; t < NT; t++) { | |
nyu = NX+2; | |
for (y = 1; y < NY; y++) { | |
for (x = 1; x < NX; x++) { | |
ind = nyu + x; | |
p4[ind] = u4[ind] + u4[ind+NX+2] + u4[ind-NX-2] + u4[ind+1] + u4[ind-1]; | |
} | |
nyu += NX+2; | |
} | |
} | |
te = getETime(); | |
double time4 = te - ts; | |
printf("4: elapsed time %g\n", time4); | |
// check | |
nyu = NX+2; | |
for (y = 1; y < NY; y++) { | |
for (x = 1; x < NX; x++) { | |
assert(p1[y][x] == p2[y][x]); | |
assert(p1[y][x] == p3[y][x]); | |
assert(p1[y][x] == p4[nyu+x]); | |
} | |
nyu += NX+2; | |
} | |
return 0; | |
} |
1: elapsed time 21.5483
2: elapsed time 20.7368
3: elapsed time 21.0106
4: elapsed time 20.8277
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version1: ポインタの配列へのポインタ(メモリはバラバラに確保)
version2: 配列へのポインタ
version3: ポインタの配列へのポインタ(メモリは一括確保)
version4: 一次元配列