C语言 在 C 中为字符数组（字符串）赋值
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Assigning a value to a char array (String) in C
提问by Ayibogan
I tried to assign a String a value in C, but for me it doesn't work... This is, what I tried to do:
我试图在 C 中为 String 分配一个值，但对我来说它不起作用......这就是我试图做的：
#include <stdio.h>
#include <string.h>
int main()
{
char k[25];
k == "Dennis"
printf("My Name is %s", k);
}
Sample Output would be: My Name is Dennis
示例输出将是：我的名字是丹尼斯
However, I receive the a warning: warning: comparison of distinct pointer types lacks a cast k == "Dennis";
但是，我收到警告：警告：不同指针类型的比较缺少强制转换 k == "Dennis";
I tried to find a solution on this website, but could not find one, where it was the same error with assigning a value to a char array (a string) in C
我试图在这个网站上找到一个解决方案，但找不到一个，在 C 中为字符数组（字符串）赋值时出现同样的错误
Also tried initializing my char as
还尝试将我的字符初始化为
char *k[25];
still didn't work...
还是没用...
回答by usr
You are indeed doing a comparison here:
k == "Dennis". So the compiler correctly warns you.You probably meant
k = "Dennis";(fixed the missing semi-colon there). But that won't work either. Because in C arrays are not modifiable lvalues.
您确实在这里进行了比较：
k == "Dennis"。所以编译器会正确警告你。您可能的意思是
k = "Dennis";（在那里修复了丢失的分号）。但这也行不通。因为在 C 数组中是不可修改的左值。
So you can either initializethe array:
所以你可以初始化数组：
char k[25] = "Dennis";
or, use strcpyto copy:
或者，用于strcpy复制：
strcpy(k, "Dennis");
If you actually have no need for an array, you can simply use a pointer that points to the string literal. The following is valid:
如果您实际上不需要array，则可以简单地使用指向字符串文字的指针。以下内容有效：
char *k;
k = "Dennis";
printf("My Name is %s", k);
