C结构与初始化字符数组 [英] C struct initialization with char array

问题描述

我有一个C结构定义如下：

I have a C struct defined as follows:

struct Guest {
   int age;
   char name[20];
};

当我创建了一个访客变量并使用以下初始化它：

When I created a Guest variable and initialized it using the following:

int guest_age = 30;
char guest_name[20] = "Mike";
struct Guest mike = {guest_age, guest_name};

我对第二个参数初始化它告诉我， guest_name 不能用于初始化成员变量的误差字符名称[20] 。

I got the error about the second parameter initialization which tells me that guest_name cannot be used to initialize member variable char name[20].

我能做到这一点，则所有：

I could do this to initialize all:

struct Guest mike = {guest_age, "Mike"};

但是，这不是我想要的。我想通过变量初始化所有领域。如何做到这在C？

But this is not I want. I want to initialize all fields by variables. How to do this in C?

推荐答案

mike.name 是20个字节的结构内的保留内存。 guest_name 是一个指针到另一个存储位置。通过试图分配 guest_name 来结构体的成员，你尝试一些不可能的。

mike.name is 20 bytes of reserved memory inside the struct. guest_name is a pointer to another memory location. By trying to assign guest_name to the struct's member you try something impossible.

如果你将数据复制到你必须使用的memcpy 和朋友的结构。在这种情况下，你需要处理 \\ 0 终止。

If you have to copy data into the struct you have to use memcpy and friends. In this case you need to handle the \0 terminator.

memcpy(mike.name, guest_name, 20);
mike.name[19] = 0; // ensure termination

如果您有 \\ 0 结尾的字符串，你也可以使用的strcpy ，但因为名称的大小为20，我建议函数strncpy 。

If you have \0 terminated strings you can also use strcpy, but since the name's size is 20, I'd suggest strncpy.

strncpy(mike.name, guest_name, 19);
mike.name[19] = 0; // ensure termination

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