Question

我有这个代码，但是我不确定如何在传递之后访问struct指针通过引用结构化为一个函数，程序在这一行崩溃，访问指针不起作用。

    scanf("(%lf,%lf)",polygon->xvals[i],polygon->yvals[i]);

固定代码，感谢所有回答

的人

    struct Polygon{
    double *xvals, *yvals;
    int numverts;
    };
    typedef struct Polygon pol;
    pol getpoly(pol *polygon);


    int main(){
     pol polygon;
     getpoly(&polygon);  
            }


    pol getpoly(pol *polygon){
      polygon->xvals = (double * )malloc(sizeof(double)*polygon->numverts);
      polygon->yvals = (double * )malloc(sizeof(double)*polygon->numverts); 

      check=0;
      int i;

      for(i=0;i<10;i++){

       while(check !=2 ){
        cout<<"enter vertices "<<i<<" (x,y)\n";
        check = scanf("(%lf,%lf)",&polygon->xvals[i],&polygon->yvals[i]);
        _flushall();
       }
      check=0; 
      }
      polygon->xvals[polygon->numverts-1] = polygon->xvals[0];
      polygon->yvals[polygon->numverts-1] = polygon->yvals[0]; 

    return *polygon;    
    }

Answer 1

没有为xvals和yvals分配内存，它们是未初始化的指针。 numverts也是未初始化的。 malloc()和xvals需要yvals空格并初始化numverts：

polygon->numverts = 10;
polygon->xvals = malloc(polygon->numverts * sizeof(double));
polygon->yvals = malloc(polygon->numverts * sizeof(double));

并防止超出这些数组的末尾，正如此代码所做的那样：

polygon->xvals[polygon->numverts] = polygon->xvals[0];
polygon->yvals[polygon->numverts] = polygon->yvals[0];

应该是：

polygon->xvals[polygon->numverts - 1] = polygon->xvals[0];
polygon->yvals[polygon->numverts - 1] = polygon->yvals[0];

当不再需要时，请记得free() xvals和yvals。

通过引用函数传递struct后访问struct指针

1 个答案: