如何使用两个 return 不同类型的函数作为 C 中的函数指针？

Question

我有一个创建函数，我想传递一个函数指针，但是我希望函数指针能够调用下面的任何一个函数，我遇到的问题是每个 return一种不同的类型。但是当我传递函数指针时，我必须指定函数指针的 return 类型。这个对吗？我以为在里面用一个void指针就可以解决return类型的问题，但是现在我不能调用函数指针了。为了澄清，我想将 populateMonsters 或 populateClassesTaken 传递给 create 并能够在 create 中调用它们。

typedef struct node{
   void * data;
   struct node* next;
} node;

typedef struct MonsterAttacks{
    unsigned int attackID;
    char monsterName[41];
    char attackLocation[41];
    unsigned int numOfVictims;
}MonsterAttacks;

typedef struct ClassesTaken{
    unsigned int classID;
    char semester[41];
    unsigned int year;
    unsigned int numberOfUnits;
    char grade[10];
}ClassesTaken;


  unsigned int idM = 1;

  MonsterAttacks * populateMonsters(){

    MonsterAttacks *m = NULL;

    m->attackID = idM;
    printf("Enter the name for the Monster \n");
    scanf("%40s",m->monsterName);
    puts("What is his/her attack location?");
    scanf("%40s",m->attackLocation);
    puts("What are the number of victims this monster has demolished?");
    scanf("%ud", &m->numOfVictims);      
    //attackID is assigned and id when node is created
    idM++;
    return m;
}

  unsigned int idC = 1;

  ClassesTaken* populateClassesTaken(){

    ClassesTaken *c = NULL;
    c->classID = idC;
    scanf("What semester was the class taken in? %40s",c->semester);
    scanf("In what year? %ud", &c->year);  
    scanf("How many units was the class? %ud", &c->numberOfUnits);
    scanf("What was your grade in the class? %9s", c->grade);     


    idC++;
    return c;
}

node* create(node* next, void *fp)
{
   node* new_node = (node*)malloc(sizeof(node));
   if(new_node == NULL)
   {
       printf("Error creating a new node.\n");
       exit(0);
   }
    void *data = (*fp)();
}



int main(void)
{

   node* tmp = NULL;
   MonsterAttacks* (*fp)() = &populateMonsters;

   create(tmp, *fp);
}

Answer 1

替换：

node* create(node* next, void *fp)
{

与：

node* create(node* next, void* (*fp)())
{

和

create(tmp, *fp);

和

create(tmp, (void*)fp); //[updated]

您的 populateMonsters() 导致分段错误。我用 fputs("In populateMonsters\n", stderr);

替换了它的代码

它似乎起作用了：

$ ./a.out

In populateMonsters

Answer 2

populateMonsters() 和 populateClassesTaken() 中仍有问题。你想用用户输入填充结构，但你从不分配结构（既不在堆上也不在堆栈上）。因此，您会遇到分段错误。这是我没有 typedef 的代码，包括，...

typedef void*(*myFunctionInterface)(void);

MonsterAttacks * populateMonsters (){

  MonsterAttacks *m = (MonsterAttacks*)malloc(sizeof(MonsterAttacks));
  if (m == NULL){
      printf("Error creating MonsterAttacks");
      exit (-1);
  }

  m->attackID = idM;
  printf ("Enter the name for the Monster \n");
  scanf ("%40s", m->monsterName);
  puts ("What is his/her attack location?");
  scanf ("%40s", m->attackLocation);
  puts ("What are the number of victims this monster has demolished?");
  scanf ("%ud", &m->numOfVictims);
  //attackID is assigned and id when node is created
  idM++;
  return m;
}

unsigned int idC = 1;

ClassesTaken * populateClassesTaken (){

  ClassesTaken *c = (ClassesTaken*)malloc(sizeof(ClassesTaken));
  if (c == NULL){
      printf("Error creating ClassesTaken");
      exit(-2);
  }
  c->classID = idC;
  scanf ("What semester was the class taken in? %40s", c->semester);
  scanf ("In what year? %ud", &c->year);
  scanf ("How many units was the class? %ud", &c->numberOfUnits);
  scanf ("What was your grade in the class? %9s", c->grade);


  idC++;
  return c;
}

node * create (node * next, void *(*fp) ()){
  node *new_node = (node *) malloc (sizeof (node));
  if (new_node == NULL)
    {
      printf ("Error creating a new node.\n");
      exit (-3);
    }
  void *data = fp();
  new_node->data = data;
  return new_node;
}



int main (void) {    
  node *tmp = NULL;
  MonsterAttacks *(*fp_M) () = &populateMonsters;
  ClassesTaken *(*fp_C) () = &populateClassesTaken;
  create (tmp, (myFunctionInterface)fp_M);
}

如果出现错误，return0 也不是一个好的样式。 0 通常表示一切正常。因此，我将退出值更改为一些不同的负数以区分错误

Answer 3

首先，不言而喻，你不能写出ClassesTaken *c = NULL; c->classID = idC;这样的代码。如果你不明白为什么你不能，你需要回去学习指针的基础知识，如下所述：

至于不同的功能 - 它们是不同的功能类型这一事实清楚地表明这种设计没有意义。如果你希望使用通用函数指针，你需要定义一个统一的函数格式，例如void* func (void).

您不能使用void*作为通用函数指针。它仅定义为与指向对象的指针一起使用，而不是与函数一起使用。诸如 create(tmp, *fp); 之类的代码，其中 fp 是一个函数指针并且该函数期望 void* 是无效的标准 C，不应编译，除非您使用的是非标准编译器扩展。

值得注意的是，您永远不应该在 C 中用空括号 () 声明函数。这是一种过时的语言功能。（与 C++ 不同，在 C++ 中这是很好的并且受到鼓励。）

如何使用两个 return 不同类型的函数作为 C 中的函数指针？

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