Question

我应该制作2个构建图形的代码。我已经有了1.它只是建立图片：

  double a = LabeledEdit1 -> Text.ToDouble();
  double S = LabeledEdit2 -> Text.ToDouble();
  Series1-> Clear();
  for( float i = -5; i <= 5; i+=S ) {
    float y = exp(i*log(a));
    Series1->AddXY( i , y ,"",clBlue);
  }

但是第二项任务对我来说要困难得多。我应该制作新结构

struct K {
  double x;
  double y;
  struct K *next;
};

然后制作链表。然后将点（x，y）放入StringGrid，然后构建grapfic。我制作了一些代码，但它无法正常工作。需要帮助。

  K* head = 0;
  K* curr = 0;
  K* vyv;
  double x = 1;
  double y;
  double a = LabeledEdit1 -> Text.ToDouble();
  double S = LabeledEdit2 -> Text.ToDouble();
  int i=0;
  for (i = 0; i < 500; ++i) {
    if (i==0) {
      y = exp(x*log(a));
      head = (K*) malloc(sizeof K);
      head->x = x;
      head->y = y;
      head->next = NULL;
    }
    else {
      y = exp(x*log(a));
      curr = (K*) malloc(sizeof K);
      curr->x = x;
      curr->y = y;
      curr->next = NULL;
    }
    x++;
  }
  vyv = head;
  i = 0;
  int l = 0, I = 0;
  while(vyv){
    x = vyv->x;
    StringGrid1->Cells[i][0] = x;
    y = vyv->y;
    StringGrid1->Cells[i][1] = y;
    Series1->AddXY( i , y ,"",clBlue);
    vyv = vyv->next;
    ++i;
  }

Answer 1

您的第一个循环没有将节点链接在一起，因此它们的next值都为NULL，因此您的第二个循环仅运行1次迭代（对于head节点）。

您还应该使用new代替malloc()。

请改为尝试：

K* head = 0;
K* curr = 0;
K* last = 0;
double x = 1;
double y;
double a = LabeledEdit1 -> Text.ToDouble();
double S = LabeledEdit2 -> Text.ToDouble();
int i;
for (i = 0; i < 500; ++i) {
    y = exp(x*log(a));
    curr = new K;
    curr->x = x;
    curr->y = y;
    curr->next = NULL;
    if (!head) head = curr;
    if (last) last->next = curr;
    last = curr;
    x++;
}
curr = head;
i = 0;
int l = 0, I = 0;
while(curr){
    x = curr->x;
    StringGrid1->Cells[i][0] = x;
    y = curr->y;
    StringGrid1->Cells[i][1] = y;
    Series1->AddXY( i , y ,"",clBlue);
    curr = curr->next;
    ++i;
}

不要忘记在完成使用后释放节点：

curr = head;
while(curr){
    K *next = curr->next;
    //...
    delete curr;
    curr = next;
}

话虽如此，您应该使用std::list类，让它为您处理内存管理：

#include <list>

struct K {
    double x;
    double y;
};

std::list<K> myList;

double x = 1;
double y;
double a = LabeledEdit1 -> Text.ToDouble();
double S = LabeledEdit2 -> Text.ToDouble();
int i;
for (i = 0; i < 500; ++i) {
    y = exp(x*log(a));
    K curr;
    curr.x = x;
    curr.y = y;
    myList.push_back(curr);
    x++;
}
i = 0;
int l = 0, I = 0;
for(std::list<K>::iterator iter = myList.begin(); iter != myList.end(); ++iter)
{
    x = iter->x;
    StringGrid1->Cells[i][0] = x;
    y = iter->y;
    StringGrid1->Cells[i][1] = y;
    Series1->AddXY( i , y ,"",clBlue);
    ++i;
}

构建图形并使用链表

1 个答案: