军事时间转常规时间：如何补全分钟位的前导零？

问题：实现C++程序将军事时间转为常规时间时，分钟为个位数无法显示前导零，例如输入0000得到12:0 AM，输入608得到6:8 AM，需要修复该问题。相关代码如下：

main.cc #

#include "time_converter.h"
#include <iostream>

int main() {
  int military_time;
  std::cout << "Please enter the time in military time: ";
  std::cin >> military_time;
  
  std::string regular_time;
  regular_time = MilitaryToRegularTime(military_time);

  std::cout << "The equivalent regular time is: " << regular_time << "\n";
  return 0;
}

time_converter.h #

#include <iostream>

// Converts the time in military format to regular format.
std::string MilitaryToRegularTime(int military_time);

time_converter.cc #

#include <iostream>

std::string amorpm;
std::string MilitaryToRegularTime(int military_time) {
  int regular_hr = military_time / 100;
  if (regular_hr >= 13){
     regular_hr = (military_time / 100) - 12;
  }
  
  if (regular_hr == 0){
    regular_hr = 12;
  }
 
  int regular_min = military_time % 100;
  
  if (military_time >= 1200 && military_time <= 2359){
    amorpm = " PM\n";
  }
  if (military_time >= 0000 && military_time <= 1159 ){
    amorpm = " AM\n";
  }

  std::string regular_hr_str = std::to_string(regular_hr);
  std::string regular_min_str = std::to_string(regular_min);

  return regular_hr_str + ":" + regular_min_str + amorpm;
}

修复方案 #

方法1：手动补全前导零

在生成分钟字符串时，判断其长度，若为1位则在前面添加'0'：

std::string regular_min_str = std::to_string(regular_min);
if (regular_min_str.length() == 1) {
    regular_min_str = "0" + regular_min_str;
}

替换原代码中对应的行即可，简单直接，适合快速修复。

方法2：使用字符串流格式化（更规范）

利用std::ostringstream结合格式化控制符输出固定长度的分钟，同时将全局变量amorpm改为局部变量，避免全局状态污染：

修改后的time_converter.cc：

#include <iostream>
#include <sstream>
#include <iomanip> // 用于setw和setfill

std::string MilitaryToRegularTime(int military_time) {
  int regular_hr = military_time / 100;
  if (regular_hr >= 13){
     regular_hr = regular_hr - 12;
  }
  
  if (regular_hr == 0){
    regular_hr = 12;
  }
 
  int regular_min = military_time % 100;
  
  std::string amorpm; // 改为局部变量
  if (military_time >= 1200 && military_time <= 2359){
    amorpm = " PM";
  } else { // 用else替代重复判断，提升效率
    amorpm = " AM";
  }

  std::ostringstream oss;
  oss << regular_hr << ":" << std::setw(2) << std::setfill('0') << regular_min << amorpm;
  
  return oss.str();
}

• std::setw(2)确保输出占2个字符宽度
• std::setfill('0')指定宽度不足时用'0'填充
• 将全局变量amorpm改为局部变量，避免多次调用函数时的状态异常

两种方法均可解决分钟位前导零缺失问题，方法2更符合C++格式化规范，同时优化了变量作用域。

内容的提问来源于stack exchange，提问作者Jose