使用 sprintf 将字符串居中 [英] Center a string using sprintf

问题描述

我有一个代码想要在条形之间的中心显示消息.我查看了 C 函数，但没有发现任何允许我这样做的内容.

I have a code that would like to display the message in the center between the bars. I looked at the C functions and found nothing that would allow me this.

sprintf(message,"============================================================");
send_message(RED, message);
sprintf(message, "[ Welcome %s ]", p->client_name);
send_message(RED, message);
sprintf(message,"============================================================");
send_message(RED, message);

我正在寻找一种通过计算用户名的大小来显示欢迎消息的方法，始终集中显示.示例:

I am looking for a way to show the Welcome message by counting the size of the user name always show centralized. Example:

示例 1:

=============================================
                Welcome Carol                
=============================================

示例 2:

=============================================
               Welcome Giovanna               
=============================================

推荐答案

它没有特殊功能，所以你应该自己算一下.

There is no special function for it, so you should do the math.

• 计算条的数量和消息的长度.
• 将它们相减并除以 2.
• 如果消息的长度是偶数，则将商加 1.
• 将消息的长度与商相加.

示例代码:

#include <stdio.h>
#include <string.h>

int main(void) {
    char* message = "Welcome Giovanna";
    int len = (int)strlen(message);
    printf("===============================================\n"); // 45 chars
    printf("%*s\n", (45-len)/2 + ((len % 2 == 0) ? 1 : 0) + len, message);
    printf("===============================================\n");

    return 0;
}

输出:

=============================================
               Welcome Giovanna               
=============================================

<小时>

PS:您可以将 (45-len)/2 + ((len % 2 == 0) ? 1 : 0) 替换为 (46-len)/2code>，为了得到同样的结果，因为后者更短.

---

PS: You could replace (45-len)/2 + ((len % 2 == 0) ? 1 : 0) with (46-len)/2, in order to get the same result, since the latter is shorter.

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