从“无效*”为指针，以非'void“转换需要显式转换（第17行） [英] Conversion from 'void*' to pointer to non-'void' requires an explicit cast (Line 17)

问题描述

我按照书学习C坚硬方式，当我尝试运行这个程序，我收到此错误信息：

  

这是'无效*'的指针的非'void转换需要显式的转换。

我不知道如何解决这个问题，我必须改变返回变量的结构？

看看反正这里的code：（编译基于Visual C ++ 2010，没有尝试过GCC还）

  //学习C的在Hardway  ＃包括LT＆;＆ASSERT.H GT;
  ＃包括LT＆;＆stdlib.h中GT;
  ＃包括LT＆;＆string.h中GT;
  ＃包括LT＆;＆stdio.h中GT;  结构{人
      字符*名称;
      INT年龄;
      INT高度;
      诠释权重;
  };  结构人* Person_create（字符*名称，INT年龄，身高INT，INT重量）
  {
      结构人*谁=的malloc（sizeof的（结构人））;
      断言（谁！= NULL）;      如─＆GT;名称=的strdup（名）;
      如─＆GT;年龄=年龄;
      如─＆GT;高度=高度;
      如─＆GT;体重=体重;      返回谁;
  }  无效Person_destroy（结构人*谁）
  {
      断言（谁！= NULL）;      免费（如 - ＆GT;名）;
      免费的（谁）;
  }  无效Person_print（结构人*谁）
  {
      的printf（名称：％s \\ n，如─＆GT;名）;
      的printf（\\踏歌数：％d \\ n，如─＆GT;年龄）;
      的printf（\\ tHeight数：％d \\ n，如─＆GT;高度）;
      的printf（\\ tWeight数：％d \\ n，如─＆GT;重）;
  }  INT主（INT ARGC，CHAR *的argv []）
  {
      //让两个人结构
      结构人*乔= Person_create（
              乔亚历克斯，32，64，140）;      结构人*坦诚= Person_create（
              弗兰克空白，20，72，180）;      //打印出来，他们是在内存中
      的printf（乔是在内存位置％P：\\ n，JOE）;
      Person_print（JOE）;      的printf（弗兰克在内存位置％P：\\ n，坦率的）;
      Person_print（坦诚）;      //让每个人都20岁，再打印出来
      joe-＆GT;年龄+ = 20;
      joe-＆GT;高度 -  = 2;
      joe-＆GT;重+ = 40;
      Person_print（JOE）;      frank-＆GT;年龄+ = 20;
      frank-＆GT;重+ = 20;
      Person_print（坦诚）;      //消灭他们两个，所以我们清理
      Person_destroy（JOE）;
      Person_destroy（坦诚）;      返回0;
  }
 

解决方案

这行需要转换：

 结构人*谁=的malloc（sizeof的（结构人））;
 

应该是：

 结构人*谁=（结构人*）malloc的（的sizeof（结构人））;
 

这是唯一的，因为你正在编译此code和C ++，而不是C.在C，中投是不必要的，并为你做的隐含

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

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

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

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;
  }

解决方案

This line requires a cast:

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

Should be:

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

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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