我有一个设计用于malloc数组的函数,然后用文件中的值填充它(n维坐标,虽然现在在2d工作)。
#include <stdio.h>
#include <stdlib.h>
#define dim 2
typedef struct {
double **array; /*store the coordinates*/
int num; /* store the number of coordinates*/
/* store some other things too */
} foo;
void read_particles(foo *bar);
int main(void)
{
foo bar;
read_particles(&bar);
printf("\n");
for(int i = 0; i < bar.num; i++)
printf("%f %f\n", bar.array[0][i], bar.array[1][i]);
/* printf here does not output the array properly.
* Some values are correct, some are not.
* Specifically, the first column bar.array[0][i] is correct,
* the second column bar.array[1][i] is not, some values from the
* first column are appearing in the second.
*/
return 0;
}
void read_particles(foo *bar)
{
FILE *f = fopen("xy.dat", "r");
/* read number of coordinates from file first*/
fscanf(f, "%d", &bar->num);
bar->array = (double**) malloc(bar->num * sizeof(double*));
for(int i = 0; i < bar->num; i++)
bar->array[i] = (double*) malloc(dim * sizeof(double));
for(int i = 0; i < bar->num; i++)
{
for(int j = 0; j < dim; j++)
fscanf(f, "%lf", &(bar->array[j][i]));
/* For now, coordinates are just 2d, print them out
* The values are displayed correctly when printing here*/
printf("%f %f\n", bar->array[0][i], bar->array[1][i]);
}
fclose(f);
}
Some sample data is available here.
当从函数内部打印值时,它们很好,当在函数外部打印时,它们不是。所以我一定不能正确处理指针。它可能(或可能不)值得注意,我最初没有使用结构并将函数定义为double **read_and_malloc(num)
,返回指向数组的指针,并且生成的输出是相同的。
那是怎么回事?
如果需要,我可以包含一些示例数据或任何其他信息。
答案 0 :(得分:3)
你的第二个循环不正确:
for(int i = 0; i < dim; i++)
bar->array[i] = (double*) malloc(dim * sizeof(double));
您创建了bar->num
个元素,但是您在dim
元素上进行了迭代:
bar->array = (double**) malloc(bar->num * sizeof(double*))
循环应迭代第一维中的元素数量:bar->num
答案 1 :(得分:2)
在更新的代码中,您要分配bar->num
行和2
列。但是,您的fscanf
和printf
代码会尝试处理包含2
行和bar->num
列的数组。
为了保持读/写代码的完整性,分配代码为:
bar->array = malloc(dim * sizeof *bar->array);
for (int i = 0; i < dim; ++i)
bar->array[j] = malloc(bar->num * sizeof *bar->array[j]);
NB。如果你不熟悉这个malloc习语,see here