[英]C - Copy structures with dynamics arrays
拥有这种类型的结构,如何将一个结构复制到另一个结构中。 我Array first
声明了Array first
,然后声明了Array first
Array second
,然后我初始化了两者并在每个数据中放入了不同的数据。
然后将first
复制到second
我尝试了second = first
但它不起作用。
我该怎么做?
#include <stdio.h>
#include <stdlib.h>
#include <inttypes.h>
//////////////////////////////////////
typedef struct {
int group[8];
uint64_t points;
} BestGroup;
//////////////////////////////////////
typedef struct {
BestGroup *array;
size_t used;
size_t size;
} Array;
void initArray(Array *a, size_t initialSize) {
a->array = (BestGroup *)malloc(initialSize * sizeof(BestGroup));
a->used = 0;
a->size = initialSize;
}
void insertArray(Array *a, int *group_add, uint64_t points_add) {
// a->used is the number of used entries, because a->array[a->used++] updates a->used only *after* the array has been accessed.
// Therefore a->used can go up to a->size
if (a->used == a->size)
{
a->size *= 2;
a->array = (BestGroup *)realloc(a->array, a->size * sizeof(BestGroup));
}
int i;
for (i = 0; i < 8; i++)
{
a->array[a->used].group[i] = group_add[i];
}
a->array[a->used].points = points_add;
a->used++;
}
void freeArray(Array *a) {
free(a->array);
a->array = NULL;
a->used = a->size = 0;
}
void CopyArray(Array *a, Array *b)
{
b = a;
}
int main()
{
Array first;
Array second;
int first_data[8] = {0, 1, 2, 3, 4, 5, 6, 7};
int second_data[8] = {7, 6, 5, 4, 3, 2, 1, 0};
initArray(&first, 2);
initArray(&second, 2);
insertArray(&first, first_data, 5);
insertArray(&first, first_data, 5);
insertArray(&first, first_data, 5);
insertArray(&second, second_data, 2);
///////////////////////////////////////////
printf("Total Points: %" PRIu64 "\n", first.array->points);
printf("Number: %lu\n\n", first.used);
printf("\n");
int i;
int j;
for (i = 0; i < first.used; i++)
{
printf("[");
for (j = 0; j < 8; j++)
{
if (j) printf(", ");
printf("%d", first.array[i].group[j]);
}
printf("]\n");
}
////////////////////////////////////////////
printf("\n");
printf("Total Points: %" PRIu64 "\n", second.array->points);
printf("Number: %lu\n\n", second.used);
printf("\n");
for (i = 0; i < second.used; i++)
{
printf("[");
for (j = 0; j < 8; j++)
{
if (j) printf(", ");
printf("%d", second.array[i].group[j]);
}
printf("]\n");
}
/////////////////////////////////
CopyArray(&first, &second);
printf("\n");
printf("\n");
printf("\n");
printf("\n");
///////////////////////////////////////////
printf("Total Points: %" PRIu64 "\n", first.array->points);
printf("Number: %lu\n\n", first.used);
printf("\n");
for (i = 0; i < first.used; i++)
{
printf("[");
for (j = 0; j < 8; j++)
{
if (j) printf(", ");
printf("%d", first.array[i].group[j]);
}
printf("]\n");
}
////////////////////////////////////////////
printf("\n");
printf("Total Points: %" PRIu64 "\n", second.array->points);
printf("Number: %lu\n\n", second.used);
printf("\n");
for (i = 0; i < second.used; i++)
{
printf("[");
for (j = 0; j < 8; j++)
{
if (j) printf(", ");
printf("%d", second.array[i].group[j]);
}
printf("]\n");
}
return 0;
}
输出是:
Total Points: 5
Number: 3
[0, 1, 2, 3, 4, 5, 6, 7]
[0, 1, 2, 3, 4, 5, 6, 7]
[0, 1, 2, 3, 4, 5, 6, 7]
Total Points: 2
Number: 1
[7, 6, 5, 4, 3, 2, 1, 0]
Total Points: 5
Number: 3
[0, 1, 2, 3, 4, 5, 6, 7]
[0, 1, 2, 3, 4, 5, 6, 7]
[0, 1, 2, 3, 4, 5, 6, 7]
Total Points: 2
Number: 1
[7, 6, 5, 4, 3, 2, 1, 0]
当它应该以最后一个列表结束时:
Total Points: 5
Number: 3
[0, 1, 2, 3, 4, 5, 6, 7]
[0, 1, 2, 3, 4, 5, 6, 7]
[0, 1, 2, 3, 4, 5, 6, 7]
Total Points: 5
Number: 3
[0, 1, 2, 3, 4, 5, 6, 7]
[0, 1, 2, 3, 4, 5, 6, 7]
[0, 1, 2, 3, 4, 5, 6, 7]
编辑
按照建议我使用了 memcpy(),所以我将CopyArray()
函数更改为:
void CopyArray(Array *a, Array *b)
{
// b = a;
memcpy(b, a, a->size * sizeof(BestGroup));
}
看起来结果会很好,直到它到达程序结束并输出
*** stack smashing detected ***
此代码生成您想要的答案。 CopyArray()
函数现在释放b
已有的数据,然后初始化b
,最后将数据从a
复制到b
。 有些地方代码使用了8
但可能应该使用不同的大小变量——我在这段代码中对它们进行了注释。 dump_array()
函数是我在调试复杂结构时经常编写和使用的函数示例。 它需要一个标记字符串(它允许您识别正在打印的调用)加上要打印的对象。 我也经常使用文件流参数,如有疑问,为输出流添加一个fflush()
。
/* SO 4639-4467 */
#include <inttypes.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
typedef struct
{
int group[8];
uint64_t points;
} BestGroup;
typedef struct
{
BestGroup *array;
size_t used;
size_t size;
} Array;
static void initArray(Array *a, size_t initialSize)
{
a->array = (BestGroup *)malloc(initialSize * sizeof(BestGroup));
a->used = 0;
a->size = initialSize;
}
static void insertArray(Array *a, const int *group_add, uint64_t points_add)
{
if (a->used == a->size)
{
a->size *= 2;
a->array = (BestGroup *)realloc(a->array, a->size * sizeof(BestGroup));
}
for (int i = 0; i < 8; i++) // Why 8 and not points_add?
{
a->array[a->used].group[i] = group_add[i];
}
a->array[a->used].points = points_add;
a->used++;
}
static void freeArray(Array *a)
{
free(a->array);
a->array = NULL;
a->used = a->size = 0;
}
static void CopyArray(const Array *a, Array *b)
{
freeArray(b);
initArray(b, a->used);
memmove(b->array, a->array, a->used * sizeof(a->array[0]));
b->used = a->used;
}
static void dump_array(const char *tag, const Array *arr)
{
printf("Array: %s\n", tag);
printf("Total Points: %" PRIu64 "\n", arr->array->points);
printf("Number: %lu\n", arr->used);
for (size_t i = 0; i < arr->used; i++)
{
printf("[");
for (size_t j = 0; j < 8; j++) // Why 8 and not arr->array[i].points?
{
if (j)
printf(", ");
printf("%d", arr->array[i].group[j]);
}
printf("]\n");
}
putchar('\n');
}
int main(void)
{
Array first;
Array second;
int first_data[8] = {0, 1, 2, 3, 4, 5, 6, 7};
int second_data[8] = {7, 6, 5, 4, 3, 2, 1, 0};
initArray(&first, 2);
initArray(&second, 2);
insertArray(&first, first_data, 5);
insertArray(&first, first_data, 5);
insertArray(&first, first_data, 5);
insertArray(&second, second_data, 2);
dump_array("first", &first);
dump_array("second", &second);
CopyArray(&first, &second);
printf("\n");
dump_array("first", &first);
dump_array("second", &second);
return 0;
}
运行它的输出是:
Array: first
Total Points: 5
Number: 3
[0, 1, 2, 3, 4, 5, 6, 7]
[0, 1, 2, 3, 4, 5, 6, 7]
[0, 1, 2, 3, 4, 5, 6, 7]
Array: second
Total Points: 2
Number: 1
[7, 6, 5, 4, 3, 2, 1, 0]
Array: first
Total Points: 5
Number: 3
[0, 1, 2, 3, 4, 5, 6, 7]
[0, 1, 2, 3, 4, 5, 6, 7]
[0, 1, 2, 3, 4, 5, 6, 7]
Array: second
Total Points: 5
Number: 3
[0, 1, 2, 3, 4, 5, 6, 7]
[0, 1, 2, 3, 4, 5, 6, 7]
[0, 1, 2, 3, 4, 5, 6, 7]
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