C语言 “错误：分配给具有数组类型错误的表达式”当我分配结构字段 (C)
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"error: assignment to expression with array type error" when I assign a struct field (C)
提问by Chief096
I'm a beginner C programmer, yesterday I learned the use of C structs and the possible application of these ones about the resolution of specific problems. However when I was experimenting with my C IDE (Codeblocks 16.01) in order to learn this aspect of C programming, I've encountered a strange issue. The code is the following:
我是一名初学 C 程序员，昨天我学习了 C 结构体的使用以及这些结构体在解决特定问题方面的可能应用。然而，当我尝试使用我的 C IDE（Codeblocks 16.01）来学习 C 编程的这方面时，我遇到了一个奇怪的问题。代码如下：
#include <stdio.h>
#define N 30
typedef struct{
char name[N];
char surname[N];
int age;
} data;
int main() {
data s1;
s1.name="Paolo";
s1.surname = "Rossi";
s1.age = 19;
getchar();
return 0;
}
During the compilation, the compiler (GCC 4.9.3-1 under Windows) reported me an error that says
在编译过程中，编译器（Windows下的GCC 4.9.3-1）向我报告了一个错误，说
"error: assignment to expression with array type error"
“错误：赋值给具有数组类型错误的表达式”
on instruction
根据指示
s1.name="Paolo"
s1.surname="Rossi"
while if I do
如果我这样做
data s1 = {"Paolo", "Rossi", 19};
it works. What am I doing wrong?
有用。我究竟做错了什么？
回答by Sourav Ghosh
You are facing issue in
您正面临以下问题
s1.name="Paolo";
because, in the LHS, you're using an arraytype, which is not assignable.
因为，在 LHS 中，您使用的是不可分配的数组类型。
To elaborate, from C11, chapter §6.5.16
详细说明，从C11第 6.5.16 章开始
assignment operator shall have a modifiable lvalue as its left operand.
赋值运算符应有一个可修改的左值作为其左操作数。
and, regarding the modifiable lvalue, from chapter §6.3.2.1
并且，关于可修改的 lvalue，来自第 §6.3.2.1 章
A modifiable lvalue is an lvalue that does not have array type, [...]
可修改的左值是没有数组类型的左值，[...]
You need to use strcpy()to copy intothe array.
您需要使用strcpy()到复制到阵列中。
That said, data s1 = {"Paolo", "Rossi", 19};works fine, because this is not a direct assignmentinvolving assignment operator. There we're using a brace-enclosed initializer listto provide the initial values of the object. That follows the law of initialization, as mentioned in chapter §6.7.9
也就是说，data s1 = {"Paolo", "Rossi", 19};工作正常，因为这不是涉及赋值运算符的直接赋值。在那里，我们使用大括号括起来的初始化列表来提供object的初始值。这遵循初始化定律，如第 §6.7.9 章所述
Each brace-enclosed initializer list has an associated current object. When no designations are present, subobjects of the current object are initialized in order according to the type of the current object: array elements in increasing subscript order, structure members in declaration order, and the first named member of a union.[....]
每个花括号括起来的初始化列表都有一个关联的当前对象。当不存在指定时，当前对象的子对象根据当前对象的类型按顺序初始化：数组元素按递增下标顺序，结构成员按声明顺序，以及联合的第一个命名成员。[... .]
回答by teknopaul
typedef struct{
char name[30];
char surname[30];
int age;
} data;
defines that datashould be a block of memory that fits 60 chars plus 4 for the int (see note)
定义data应该是一个适合 60 个字符加 4 个用于 int 的内存块（见注释）
[----------------------------,------------------------------,----]
^ this is name ^ this is surname ^ this is age
This allocates the memory on the stack.
这会在堆栈上分配内存。
data s1;
Assignments just copies numbers, sometimes pointers.
赋值只是复制数字，有时是指针。
This fails
这失败了
s1.name = "Paulo";
because the compiler knows that s1.nameis the start of a struct 64 bytes long, and "Paulo"is a char[] 6 bytes long (6 because of the trailing \0 in C strings)
Thus, trying to assign a pointer to a string into a string.
因为编译器知道这s1.name是一个 64 字节长的 struct 的开始，并且"Paulo"是一个 char[] 6 字节长（6 因为在 C 字符串中尾随 \0）
因此，尝试将指向字符串的指针分配到字符串中。
To copy "Paulo" intothe struct at the point nameand "Rossi" intothe struct at point surname.
以“保罗”复制到该结构在点name和“罗西”入点的结构surname。
memcpy(s1.name, "Paulo", 6);
memcpy(s1.surname, "Rossi", 6);
s1.age = 1;
You end up with
你最终得到
[Paulo0----------------------,Rossi0-------------------------,0001]
strcpydoes the same thing but it knows about \0termination so does not need the length hardcoded.
strcpy做同样的事情，但它知道\0终止，所以不需要硬编码的长度。
Alternatively you can define a struct which points tochar arrays of any length.
或者，您可以定义一个指向任何长度的 char 数组的结构。
typedef struct {
char *name;
char *surname;
int age;
} data;
This will create
这将创建
[----,----,----]
This will now work because you are filling the struct with pointers.
现在这将起作用，因为您正在用指针填充结构。
s1.name = "Paulo";
s1.surname = "Rossi";
s1.age = 1;
Something like this
像这样的东西
[---4,--10,---1]
Where 4 and 10 are pointers.
其中 4 和 10 是指针。
Note: the ints and pointers can be different sizes, the sizes 4 above are 32bit as an example.
注意：int 和指针可以有不同的大小，上面的大小 4 以 32bit 为例。
回答by teknopaul
Please check this example here: Accessing Structure Members
请在此处查看此示例：访问结构成员
There is explained that the right way to do it is like this:
有解释说正确的做法是这样的：
strcpy(s1.name , "Egzona");
printf( "Name : %s\n", s1.name);
