使用 C 中的结构将数据存储到动态数组中

Question

我在将文件中的数据存储到我的动态数组时遇到问题。我知道我现在所拥有的是不正确的，但它只是暂时存在。我有一个文件，它的第一行基本上包含了数据行的数量。下面几行有两个并排的整数来表示一个有序对。我想将这两个整数存储到一个结构 point 中，它表示一个有序对。此外，还有一个数组，其结构位于另一个结构 list 内部，其中包含数组的大小，或当前存储在数组中的数据量以及容量，即总量数组中的 space 个。

我想将这两个整数存储到 int 类型的变量中，然后将它们存储到我的 list 结构中数组内的 point 中。我对拥有两个结构感到非常困惑，并且不确定这是否是正确的方法。欢迎任何反馈。

#include <stdio.h>
#include <math.h>
#include <stdlib.h>

typedef struct
{
    int x;
    int y;
} point;

typedef struct
{
    int size;
    int capacity;
    point *A;
} list;

// Compute the polar angle in radians formed
// by the line segment that runs from p0 to p
double polarAngle(point p, point p0)
{
    return atan2(p.y - p0.y, p.x - p0.x);
}

// Determine the turn direction around the corner
// formed by the points a, b, and c. Return a
// positive number for a left turn and negative
// for a right turn.
double direction(point a, point b, point c)
{
    return (b.x - a.x)*(c.y - a.y) - (c.x - a.x)*(b.y - a.y);
}

int whereSmallest(point A[], int begin, int end, point p0)
{
    point min = A[begin];
    int where = begin;
    int n;
    for (n = begin + 1; n < end; n++)
        if (polarAngle(A[n], p0) < polarAngle(min, p0))
        {
            min = A[n];
            where = n;
        }
    return where;
}
void selectionSort(point A[], int N, point p0)
{
    int n, s;
    point temp;
    for (n = 0; n < N; n++)
    {
        s = whereSmallest(A, n, N, p0);
        temp = A[n];
        A[n] = A[s];
        A[s] = temp;
    }
}

// Remove the last item from the list
void popBack(list *p)
{
    int x;
    x = p->size - 1;
    p->A[x] = p->A[x + 1];
}

// Return the last item from the list
point getLast(list *p)
{
    point value;
    value = p->A[p->size];
    return value;
}

// Return the next to the last item
point getNextToLast(list *p)
{
    point value;
    value = p->A[p->size - 1];
    return value;
}

int main(int argc, const char *argv[])
{
    point p0, P;
    FILE *input;
    list *p;
    int N, n, x, y;

    /*Assuming that the first piece of data in the array indicates the amount of numbers in the array then we record this number as a reference.*/
    N = 0;
    input = fopen("points.txt", "r");
    fscanf(input, "%d", &N);

    /*Now that we have an exact size requirement for our array we can use that information to create a dynamic array.*/
    p = (point*)malloc(N*sizeof(point));
    if (p == NULL)//As a safety precaution we want to terminate the program in case the dynamic array could not be successfully created.
        return -1;

    /*Now we want to collect all of the data from our file and store it in our array.*/
    for (n = 0; n < N; n++)
    {
        fscanf(input, "%d %d", &P.x, &P.y);
        p->A[n] = P.x;
        p->A[n] = P.y;
    }
    fclose(input);

    free(p);
    return 0;
}

Answer 1

首先，你的代码无法编译，因为这个

p->A[n] = P.x;
p->A[n] = P.y;

错了，应该是

p->A[n].x = P.x;
p->A[n].y = P.y;

因为 A 的类型为 point，您应该访问结构的成员以便为它们赋值。

但这只是问题的开始，您没有为 A 指针分配 space，所以这不会起作用。

您需要为类型list的实例分配space，这是这样完成的
```
p = malloc(sizeof(*p));
```
然后需要初始化p的成员，为此
```
p->values   = malloc(N * sizeof(point));
p->capacity = N;
p->size     = 0;
```
如您所见，space 已分配给 values 成员。
检查 fscanf() 以确保数据完整性并避免未定义的行为，如果 fscanf() 失败，您永远不会知道您的代码，并且您可能会访问导致未定义行为的未初始化变量.

在两个int变量中捕获从文件中扫描的值，并且仅当where成功读取时才将它们复制到数组中

for (n = 0 ; ((n < N) && (fscanf(input, "%d%d", &x, &y) == 2)) ; n++)
/* check that the values were read from the file _______^ */
{
    /* store them in the array */
    p->values[n].x = x;
    p->values[n].y = y;
    p->size       += 1;
}

检查文件是否打开。

我建议使用以下代码

#include <stdio.h>
#include <math.h>
#include <stdlib.h>

typedef struct
{
    int x;
    int y;
} point;

typedef struct
{
    int size;
    int capacity;
    point *values;
} list;

// Compute the polar angle in radians formed
// by the line segment that runs from p0 to p
double polarAngle(point p, point p0)
{
    return atan2(p.y - p0.y, p.x - p0.x);
}

// Determine the turn direction around the corner
// formed by the points a, b, and c. Return a
// positive number for a left turn and negative
// for a right turn.
double direction(point a, point b, point c)
{
    return (b.x - a.x)*(c.y - a.y) - (c.x - a.x)*(b.y - a.y);
}

int whereSmallest(point values[], int begin, int end, point p0)
{
    point min = values[begin];
    int where = begin;
    int n;
    for (n = begin + 1; n < end; n++)
        if (polarAngle(values[n], p0) < polarAngle(min, p0))
        {
            min = values[n];
            where = n;
        }
    return where;
}
void selectionSort(point values[], int N, point p0)
{
    int n, s;
    point temp;
    for (n = 0; n < N; n++)
    {
        s         = whereSmallest(values, n, N, p0);
        temp      = values[n];
        values[n] = values[s];
        values[s] = temp;
    }
}

// Remove the last item from the list
void popBack(list *p)
{
    int x;
    x = p->size - 1;
    p->values[x] = p->values[x + 1];
}

// Return the last item from the list
point getLast(list *p)
{
    point value;
    value = p->values[p->size];
    return value;
}

// Return the next to the last item
point getNextToLast(list *p)
{
    point value;
    value = p->values[p->size - 1];
    return value;
}

int main(int argc, const char *argv[])
{
    FILE *input;
    list *p;
    int   N, n, x, y;

    /*Assuming that the first piece of data in the array indicates the amount of numbers in the array then we record this number as a reference.*/
    N     = 0;
    input = fopen("points.txt", "r");
    if (input == NULL)
        return -1;
    if (fscanf(input, "%d", &N) != 1)
    {
        fclose(input);
        return -1;
    }

    p = malloc(sizeof(*p));
    if (p == NULL)
        return -1;

    /*Now that we have an exact size requirement for our array we can use that information to create a dynamic array.*/
    p->values   = malloc(N * sizeof(point));
    p->capacity = N;
    p->size     = 0;
    if (p->values == NULL)//As a safety precaution we want to terminate the program in case the dynamic array could not be successfully created.
    {
        free(p);
        fclose(input);

        return -1;
    }

    /*Now we want to collect all of the data from our file and store it in our array.*/
    for (n = 0 ; ((n < N) && (fscanf(input, "%d%d", &x, &y) == 2)) ; n++)
    {
        p->values[n].x = x;
        p->values[n].y = y;
        p->size       += 1;
    }
    fclose(input);

    free(p->values);
    free(p);
    return 0;
}

如您所见，您可以对代码进行另一项改进，它不是太重要，但可以避免使用不必要的 N 和 n 变量。

注意：在使用一个函数之前，尽量通读它的文档，这将防止各种意想不到的结果，例如fscanf()，将帮助您理解我的修复更多。

Answer 2

变量p应该是list p。

积分数组分配为p.A = (point*)malloc(N*sizeof(point));

在填充循环中，由于 A[n] 是一个点，因此您不能将其分配给 int P.x 或 P.y。您可以直接将值放入 A[n] 点：

for (n = 0; n < N; n++)
{
    fscanf(input, "%d %d", &(p.A[n].x), &(p.A[N].y));
}

应该初始化列表的大小和容量：p.capacity = N; 在成功分配内存之后 p.capacity = n; 在填充数组之后

最后你应该调用 free(p.A) 而不是 free(p)。

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