[英]typedef required in struct declaration
I'm trying to create an array of struct elements, as shown below: 我正在尝试创建一个struct元素数组,如下所示:
#include <stdio.h>
#include <stdlib.h>
struct termstr{
double coeff;
double exp;
};
int main(){
termstr* lptr = malloc(sizeof(termstr)*5);
return 0;
}
When i compile this, i get errors as follows: 当我编译它时,出现如下错误:
term.c: In function ‘main’:
term.c:11:1: error: unknown type name ‘termstr’
term.c:11:31: error: ‘termstr’ undeclared (first use in this function)
However, when i change my code to the following, it compiles as usual: 但是,当我将代码更改为以下代码时,它会照常编译:
#include <stdio.h>
#include <stdlib.h>
typedef struct termstr{
double coeff;
double exp;
}term;
int main(){
term* lptr = malloc(sizeof(term)*5);
return 0;
}
I've added typedef (with type name as term), changed the name of struct to termstr and am allocating memory with term* as the type of pointer. 我添加了typedef(类型名称为term),将struct的名称更改为termstr,并以term *作为指针的类型分配内存。
Is typedef always required for such a situation ie for creating arrays of structs? 在这种情况下(即创建结构数组)是否总是需要typedef? If not, why was the first code giving errors?
如果不是,为什么第一个代码给出错误? Is typedef also required to create and use a single instance of a struct?
是否还需要typedef创建和使用结构的单个实例?
First type is not working because you have forgot struct
keyword before termstr
. 第一种类型不起作用,因为您在
termstr
之前忘记了struct
关键字。 Your data type is struct termstr
but not just termstr
. 您的数据类型是
struct termstr
,而不仅仅是termstr
。 When you typedef
, the resulting name is used as an alias for struct termstr
. 当您输入
typedef
,结果名称将用作struct termstr
的别名。
Even you don't need to do that. 甚至您也不需要这样做。 Using
typedef
is better: 使用
typedef
更好:
By the way don't forget to free the memory: 顺便说一句,不要忘记释放内存:
read why to use typedef? 阅读为什么要使用typedef?
Your working code should be: 您的工作代码应为:
#include <stdio.h>
#include <stdlib.h>
struct termstr{
double coeff;
double exp;
};
int main(){
struct termstr* lptr = malloc(sizeof(struct termstr)*5);
free(lptr);
return 0;
}
It should be: 它应该是:
struct termstr * lptr = malloc(sizeof(struct termstr)*5);
or even better: 甚至更好:
struct termstr * lptr = malloc(sizeof(*lptr)*5);
在C语言中,数据类型的名称是“ struct termstr”,而不仅仅是“ termstr”。
You can do something like this: 您可以执行以下操作:
typedef struct termstr{
double coeff;
double exp;
} termstrStruct;
And then you can use only termstrStruct
as the struct's name: 然后,您只能使用
termstrStruct
作为结构的名称:
termstrStruct* lptr = malloc(sizeof(termstrStruct)*5);
It is not always required, you can simply write struct termstr
. 它并不总是必需的,您只需编写
struct termstr
。
Don't forget to free
the allocated memory! 不要忘记
free
分配的内存!
Typedef is a convenient way of shortening this: Typedef是简化此操作的便捷方法:
struct termstr* lptr = (struct termstr*)malloc(sizeof(struct termstr)*5);
to this: 对此:
typedef struct termstr* term;
term* lptr = (term*)malloc(sizeof(term)*5);
Casting the malloc is also a good idea! 强制转换malloc也是一个好主意!
If you want to use the typename termstr on it's own you can use typedef: typedef struct { double a; 如果您想单独使用typename termstr,则可以使用typedef:typedef struct {double a; double b;
双b; } termstr;
} termstr;
In C you need to add the struct keyword as well, so either you use a typedef to link an alias with 'struct termstr' or you need to write something like 在C语言中,您还需要添加struct关键字,因此您可以使用typedef将别名与“ struct termstr”链接起来,或者需要编写类似
struct termstr* lptr = malloc(sizeof(struct termstr)*5);
In C++ however you can reference it as 'termstr' directly (read: the struct keyword isn't required there anymore). 但是,在C ++中,您可以直接将其引用为'termstr'(请阅读:在那里不再需要struct关键字)。
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.