arrays - 修复char格式错误——Linux系统C程序

Question

我试图摆脱我的程序中的错误。我不确定数组有什么问题。
这是示例代码:

typedef struct listNoNodes
{
 char FirstName[50];
 char LastName[60];
 int ID;
 int age;
 struct listNoNodes* next;
} listNoNodes;


listNoNodes* listNodes(char FirstName[], char LastName[], int ID, int age,
listNoNodes* next)
{
 listNoNodes* tmp = (listNoNodes*)malloc(sizeof(listNoNodes));
 tmp->FirstName[50] = FirstName;
 tmp->LastName[60] = LastName;
 tmp->ID = ID;
 tmp->age = age;
 tmp->next = next;
 return tmp;
};

这是错误

lab3.c: In function ‘listNodes’:
lab3.c:21:21: warning: assignment to ‘char’ from ‘char *’ makes integer from pointer without a cast [-Wint-conversion]
   21 |  tmp->FirstName[50] = FirstName;
      |                     ^
lab3.c:22:20: warning: assignment to ‘char’ from ‘char *’ makes integer from pointer without a cast [-Wint-conversion]
   22 |  tmp->LastName[60] = LastName;

Answer 1

这两行是错误的:

tmp->FirstName[50] = FirstName;
tmp->LastName[60] = LastName;

您正在尝试分配 char*指向单个 char 的指针元素。
您需要复制数组的内容，例如:

strncpy(tmp->FirstName, FirstName, 50);
strncpy(tmp->LastName, LastName, 60);