从'void *'到指向非'void'的转换需要一个显式转换（第17行） [英] Conversion from 'void*' to pointer to non-'void' requires an explicit cast (Line 17)

问题描述

我跟着这本书学习困难的方式，当我尝试运行这个程序，我得到这个错误消息：

I'm following the book Learn C the Hard Way and when I try to run this program I get this error message:

从void *转换为非void的指针需要显式转换。

Conversion from 'void*' to pointer to non-'void' requires an explicit cast.

我不知道如何解决这个，我必须改变结构中的返回变量？

I'm not sure how to resolve this, do I have to change the return variable in the struct?

看看反正，这里的代码：（在Visual C ++ 2010上编译，没有尝试过GCC

Take a look anyway, here the code: (Compiling on Visual C++ 2010, haven't tried GCC yet).

   //learn c the hardway

  #include <assert.h>
  #include <stdlib.h>
  #include <string.h> 
  #include <stdio.h>

  struct Person {
      char *name;
      int age;
      int height;
      int weight; 
  };

  struct Person *Person_create(char *name, int age, int height, int weight) 
  {
      struct Person *who = malloc(sizeof(struct Person)); 
      assert(who != NULL);

      who->name = strdup(name);
      who->age = age; 
      who->height = height;
      who->weight = weight;

      return who;
  } 

  void Person_destroy(struct Person *who)
  {
      assert(who != NULL);

      free(who->name);
      free(who);
  }

  void Person_print(struct Person *who) 
  {
      printf("Name: %s\n", who->name);
      printf("\tAge: %d\n", who->age); 
      printf("\tHeight: %d\n", who->height);
      printf("\tWeight: %d\n", who->weight); 
  }

  int main(int argc, char *argv[])
  {
      // make two people structures 
      struct Person *joe = Person_create(
              "Joe Alex", 32, 64, 140);

      struct Person *frank = Person_create(
              "Frank Blank", 20, 72, 180); 

      // print them out and where they are in memory 
      printf("Joe is at memory location %p:\n", joe);
      Person_print(joe);

      printf("Frank is at memory location %p:\n", frank);
      Person_print(frank); 

      // make everyone age 20 years and print them again 
      joe->age += 20;
      joe->height -= 2;
      joe->weight += 40; 
      Person_print(joe);

      frank->age += 20;
      frank->weight += 20; 
      Person_print(frank);

      // destroy them both so we clean up 
      Person_destroy(joe);
      Person_destroy(frank);

      return 0;
  }

推荐答案

  struct Person *who = malloc(sizeof(struct Person)); 

应为：

  struct Person *who = (struct Person *)malloc(sizeof(struct Person)); 

这是因为你正在编译这段代码为C ++，而不是C。强制转换是不需要的，并为您隐性完成。

This is only because you're compiling this code as C++, and not C. In C, the cast is unneeded, and implicitly done for you.

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