我想通过 printf 在 C 中多次使用相同的字符串

Question

我正在做哈佛 CS50，在 C 的第二节课上，讲师说 %s 会自动连续打印每个额外的参数。

好的，这很酷。但是如果我想多次打印第一个字符串怎么办？

#include <stdio.h>

int main(void) {
   char firstname[5] = "Bruce";
   char lastname[5] = "Wayne";

   printf("Hi, %s!\nI am your virtual butler. I will call you Master %s from now on.\nWelcome to %s Manor!", firstname, lastname);
}

然而，它输出：

Hi, BruceWayne!
I am your virtual butler. I will call you Master Wayne from now on.
Welcome to  q��U Manor!

代替：

Hi, Bruce!
I am your virtual butler. I will call you Master Wayne from now on.
Welcome to Wayne Manor!

我尝试在谷歌上搜索，但没有找到相关信息来自行解决问题。

---

更新#1：

根据建议，我进行了更改。

#include <stdio.h>

int main(void) {
   char firstname[] = "Bruce";
   char lastname[] = "Wayne";

   printf("Hi, %s!\nI am your virtual butler. I will call you Master %s from now on.\nWelcome to %s Manor!", firstname, lastname);
}

它输出：

Hi, Bruce!
I am your virtual butler. I will call you Master Wayne from now on.
Welcome to  ���(V Manor!

韦恩庄园还是有问题

---

更新#2

printf("Hi, %s!\nI am your virtual butler. I will call you Master %s from now on.\nWelcome to %s Manor!", firstname, lastname, lastname);

按预期工作！

Answer 1

问题是这些数组

char firstname[5] = "Bruce";
char lastname[5] = "Wayne";

不包含字符串。您需要扩大数组以包含字符串文字的终止零字符 ' '，例如

char firstname[6] = "Bruce";
char lastname[6]  = "Wayne";

或者允许编译器自己计算数组的大小，例如

char firstname[] = "Bruce";
char lastname[]  = "Wayne";

另一个问题是格式字符串包含三个转换说明符

s

但您只提供了两个相应的参数

printf("Hi, %s\nI am your virtual butler. I will call you Master %s from now on.\nWelcome to %s Manor!", firstname, lastname);

你需要再写一个参数

printf("Hi, %s\nI am your virtual butler. I will call you Master "
       "%s from now on.\nWelcome to %s Manor!", 
       firstname, lastname, lastname);