我在将数据从文件存储到动态数组时遇到问题。我知道我现在所拥有的是不正确的，但它现在就在那里。我有一个文件，在第一行基本上包含数据行。以下行并排有两个整数来表示有序对。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，还有一个带有这样结构的数组，该结构位于另一个结构list内，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空间。
我想将两个整数存储到int类型的变量中，然后将它们存储到我的point结构中的数组内list。
我对两个结构感到非常困惑，我不确定这是否是正确的方法。任何反馈都会受到欢迎。
#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;
}
答案 0 :(得分：2)
首先，您的代码无法编译，因为
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，你没有为A指针分配空间，所以这不起作用。
您需要为类型为list的实例分配空间，这样做
p = malloc(sizeof(*p));
然后你需要初始化p的成员
p->values = malloc(N * sizeof(point));
p->capacity = N;
p->size = 0;
如您所见，已为values成员分配了空间。
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，您将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。
从两个int变量中捕获从文件中扫描的值，只有在读取的地方才能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;
}
检查文件是否已打开。
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;
}
正如您所看到的，您可以对代码进行另一项改进，但它并不太重要，但它会避免使用不必要的N和n变量。
注意：在使用某个功能之前，请尝试仔细阅读它的文档，以防止出现各种意外结果，例如fscanf()，这将有助于您更多地了解我的修复程序。< / p>
答案 1 :(得分：1)
变量p应为list p。
点数组分配为p.A = (point*)malloc(N*sizeof(point));
在填充循环中，由于A [n]是一个点，你不能将它分配给int P.x或P.y.您可以直接将值放入A [n]点，如下所示：
for (n = 0; n < N; n++)
{
fscanf(input, "%d %d", &(p.A[n].x), &(p.A[N].y));
}
列表的大小和容量应该在成功的内存分配后立即初始化：p.capacity = N;和填充数组后的p.capacity = n;
最后，您应该拨打free(p.A)而不是free(p)。