C语言 C 结构体初始化与字符数组
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C struct initialization with char array
提问by tonga
I have a C struct defined as follows:
我有一个 C 结构定义如下：
struct Guest {
int age;
char name[20];
};
When I created a Guestvariable and initialized it using the following:
当我创建一个Guest变量并使用以下内容对其进行初始化时：
int guest_age = 30;
char guest_name[20] = "Mike";
struct Guest mike = {guest_age, guest_name};
I got the error about the second parameter initialization which tells me that guest_namecannot be used to initialize member variable char name[20].
我收到关于第二个参数初始化的错误，它告诉我guest_name不能用于初始化成员变量char name[20]。
I could do this to initialize all:
我可以这样做来初始化所有：
struct Guest mike = {guest_age, "Mike"};
But this is not I want. I want to initialize all fields by variables. How to do this in C?
但这不是我想要的。我想通过变量初始化所有字段。如何在 C 中做到这一点？
回答by Scolytus
mike.nameis 20 bytes of reserved memory inside the struct. guest_nameis a pointer to another memory location. By trying to assign guest_nameto the struct's member you try something impossible.
mike.name是结构内 20 字节的保留内存。guest_name是指向另一个内存位置的指针。通过尝试分配guest_name给结构的成员，您尝试了一些不可能的事情。
If you have to copy data into the struct you have to use memcpyand friends. In this case you need to handle the \0terminator.
如果您必须将数据复制到结构中，您必须使用memcpy和朋友。在这种情况下，您需要处理\0终止符。
memcpy(mike.name, guest_name, 20);
mike.name[19] = 0; // ensure termination
If you have \0terminated strings you can also use strcpy, but since the name's size is 20, I'd suggest strncpy.
如果您有\0终止的字符串，您也可以使用strcpy，但由于name的大小为 20，我建议strncpy.
strncpy(mike.name, guest_name, 19);
mike.name[19] = 0; // ensure termination
回答by Steve Howard
mike.name is a character array. You can't copy arrays by just using the = operator.
Mike.name 是一个字符数组。您不能仅使用 = 运算符来复制数组。
Instead, you'll need to use strncpyor something similar to copy the data.
相反，您需要使用strncpy或类似的东西来复制数据。
int guest_age = 30;
char guest_name[20] = "Mike";
struct Guest mike = { guest_age };
strncpy(mike.name, guest_name, sizeof(mike.name) - 1);
You've tagged this question as C++, so I'd like to point out that in that case you should almost always use std::stringin preference to char[].
您已将此问题标记为 C++，因此我想指出，在这种情况下，您几乎应该始终std::string优先使用char[].
回答by Daniel
You can statically allocate a struct with a fixed char[] array in C. For example, gcc allows the following:
您可以在 C 中静态分配具有固定 char[] 数组的结构体。例如，gcc 允许以下操作：
#include <stdio.h>
typedef struct {
int num;
char str[];
} test;
int main(void) {
static test x={.num=sizeof("hello"),.str="hello"};
printf("sizeof=%zu num=%d str=%s\n",sizeof(x),x.num,x.str);
return 0;
}
And it does the right thing (though beware of the sizeof(x): it returns 4 on my machine; not the length of the total statically allocated memory).
它做了正确的事情（尽管要注意 sizeof(x)：它在我的机器上返回 4；而不是静态分配的总内存的长度）。
This does not work for structs allocated from the stack, as you might suspect.
正如您可能怀疑的那样，这不适用于从堆栈分配的结构。
