尝试将维度为 1000*1000 的两个矩阵相乘。但是，尝试这样做会导致 Segmentation fault。可能是什么原因造成的，如何解决？
#include <stdio.h>
#include<stdlib.h>
#include<time.h>
int main()
{
clock_t t;
t = clock();
long int a[1000][1000], b[1000][1000], result[1000][1000], r1=1000, c1=1000, r2=1000, c2=1000, i, j, k;
// Column of first matrix should be equal to column of second matrix and
/* while (c1 != r2)
{
printf("Error! column of first matrix not equal to row of second.\n\n");
printf("Enter rows and column for first matrix: ");
scanf("%d %d", &r1, &c1);
printf("Enter rows and column for second matrix: ");
scanf("%d %d",&r2, &c2);
}
*/
// Storing elements of first matrix.
printf("\nEnter elements of matrix 1:\n");
for(i=0; i<r1; ++i)
for(j=0; j<c1; ++j)
{
a[i][j]=rand()%20;
}
// Storing elements of second matrix.
printf("\nEnter elements of matrix 2:\n");
for(i=0; i<r2; ++i)
for(j=0; j<c2; ++j)
{
b[i][j]=rand()%20;
}
// Initializing all elements of result matrix to 0
for(i=0; i<r1; ++i)
for(j=0; j<c2; ++j)
{
result[i][j] = 0;
}
// Multiplying matrices a and b and
// storing result in result matrix
for(i=0; i<r1; ++i)
for(j=0; j<c2; ++j)
for(k=0; k<c1; ++k)
{
result[i][j]+=a[i][k]*b[k][j];
}
// Displaying the result
printf("\nOutput Matrix:\n");
for(i=0; i<r1; ++i)
for(j=0; j<c2; ++j)
{
printf("%ld ", result[i][j]);
if(j == c2-1)
printf("\n\n");
}
t = clock() - t;
double time_taken = ((double)t)/CLOCKS_PER_SEC; // in seconds
printf("\n \nfunction took %f seconds to execute \n", time_taken);
return 0;
}
在大内存分配的情况下，您需要使用动态内存分配。堆栈内存无法处理这些大内存需求。
您可以使用动态内存分配来解决这个问题。尝试:-
int (*a)[r1][c1] = malloc(sizeof *a);
int (*b)[r2][c2] = malloc(sizeof *b);
int (*result)[r1][c2] = malloc(sizeof *result);
并使用以下方式访问元素:-
(*result)[i][j] ;
(*a)[i][k] ;
(*b)[k][j] ;
试试这段代码:-
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
int main()
{
clock_t t;
t = clock();
int r1 = 1000, c1 = 1000, r2 = 1000, c2 = 1000, i, j, k;
// Dynamic allocation.
int(*a)[r1][c1] = malloc(sizeof *a);
int(*b)[r2][c2] = malloc(sizeof *b);
int(*result)[r1][c2] = malloc(sizeof *result);
// Storing elements of first matrix.
printf("\nEnter elements of matrix 1:\n");
for (i = 0; i < r1; ++i)
{
for (j = 0; j < c1; ++j)
{
(*a)[i][j] = rand() % 20;
}
}
// Storing elements of second matrix.
printf("\nEnter elements of matrix 2:\n");
for (i = 0; i < r2; ++i)
{
for (j = 0; j < c2; ++j)
{
(*b)[i][j] = rand() % 20;
}
}
// Initializing all elements of result matrix to 0
for (i = 0; i < r1; ++i)
{
for (j = 0; j < c2; ++j)
{
(*result)[i][j] = 0;
}
}
// Multiplying matrices a and b and
// storing result in result matrix
for (i = 0; i < r1; ++i)
for (j = 0; j < c2; ++j)
for (k = 0; k < c1; ++k)
{
(*result)[i][j] += (*a)[i][k] * (*b)[k][j];
}
// Displaying the result
printf("\nOutput Matrix:\n");
for (i = 0; i < r1; ++i)
for (j = 0; j < c2; ++j)
{
printf("%d ", (*result)[i][j]);
if (j == c2 - 1)
printf("\n\n");
}
t = clock() - t;
double time_taken = ((double)t) / CLOCKS_PER_SEC; // in seconds
printf("\n \nfunction took %f seconds to execute \n", time_taken);
free(a);
free(b);
free(result);
return 0;
}
输出:-
.......................................................................
91717 92211 96529 90328 89167 88774 90433 88320 93834 89054 92225 92226 89919 88005 90772 90436 89091 92446 88477 94143 95777 88805 88487 89082 92528 88899 93436 90423 88637 90254 91569 87516 89079 91309 93554 86422 90069 91096 86981 95437 92805 88638 89828 88568 89607 88025 91700 88144 90401 91147 88284 92998 90959 85520 92640 92247 95616 90006 87248 89726 91751 90077 90543 91489 92399 90828 89026 92866 91548 87131 88450 93247 87748 90734 90228 91972 93300 92444 91592 85842 91167 89554 91144 90536 91256 89646 92815 91476 91863 94836 95462 87122 91735 96059 91312 90480 93306
function took 6.060788 seconds to execute
注意:由于太大，无法包含完整输出。
不要忘记释放分配的内存。