我正在研究“Sams教你自己C编程”第66页上这个非常简单和基本的脚本。
基本上，脚本会要求你几秒钟 输入一个数字，然后计算它们构成的小时数和秒数。
然后我稍微修改了一下脚本来探索代码并使它变得有点不那么沉闷 修改包括扩大可接受的输入和输出范围。
在下面提供的代码中，我使用了注释来指出原始代码，修改和添加。
最后，在最后，我添加了评论，以说明几个非常古怪的输出。
****表示对计数的增量检查
**** ****表示输入中的输入发生了巨大变化（即1位数）。
当我随机输入85000000时，我首先注意到了这个问题。
我已尽一切努力简化代码，以便更容易专注于数学 我不确定这是数学问题还是错字。
无论如何，这是代码：
/* Illustrates the modulus operator. */
/* inputs a number of seconds, and converts to hours, minutes, and seconds. */
// minute = 60 seconds : 60
// hour = 60 * 60 seconds : 3600
// day = 24 * 60 * 60 seconds : 86400
// year = 365 * 24 * 60 * 60 seconds : 31536000
#include <stdio.h> // from original script
/* Define constants */
#define SECS_PER_MIN 60 // from original script
#define MIN_PER_HOUR 60
#define HOURS_PER_DAY 24
#define DAYS_PER_YEAR 365
#define SECS_PER_YEAR 31536000
#define SECS_PER_HOUR 3600 // from original script
#define SECS_PER_DAY 86400
#define LIMIT 2147438647
unsigned int seconds, minutes, hours, days, years, secs_remain, mins_remain, hours_remain, days_remain; // modified from original script
int main(void) // from original script
{
seconds = minutes = hours = days = years = secs_remain = mins_remain = hours_remain = days_remain = 0;
/* Input the number of seconds. */
printf( "Enter the number of seconds ( > 0, < %u ): \n", LIMIT ); // modified from original script
scanf( "%d", &seconds ); // from original script
years = seconds /SECS_PER_YEAR;
days = seconds / SECS_PER_DAY;
hours = seconds / SECS_PER_HOUR; // from original script
minutes = seconds / SECS_PER_MIN; // from original script
days_remain = days / DAYS_PER_YEAR; // ***missing 22643200 seconds, given input: 85000000***
hours_remain = hours % HOURS_PER_DAY;
mins_remain = minutes % MIN_PER_HOUR; // modified from original script
secs_remain = seconds % SECS_PER_MIN; // from original script
printf( "%u seconds is equal to ", seconds ); // from original script
if ( seconds < SECS_PER_HOUR )
{
printf( "%u m, and %u s\n", minutes, secs_remain );
return 0;
}
else if((seconds >= SECS_PER_HOUR ) && (seconds < SECS_PER_DAY ))
{
printf( "%u h, %u m, and %u s\n", hours, mins_remain, secs_remain ); // from original script
return 0; // from original script
}
else if((seconds >= SECS_PER_DAY ) && (seconds < SECS_PER_YEAR ))
{
printf( "%u d, %u h, %u m, and %u s\n", days, hours_remain, mins_remain, secs_remain );
return 0;
}
else if((seconds >= SECS_PER_YEAR ) && (seconds < LIMIT ))
{
printf( "%u y %u d, %u h, %u m, and %u s\n", years, days_remain, hours_remain, mins_remain, secs_remain ); // calculation day's inaccurate
return 0;
}
else if(seconds > LIMIT )
{
printf("error: excessive amount of seconds.\n");
printf("range for seconds must be between 0 and %u!\n", LIMIT );
printf("If number of seconds exceeds %u, then it is beyond range for type 'int'.\n", LIMIT );
printf("EXITING seconds program. \n");
return 1;
}
}
// ERROR: 31536000 seconds is equal to 1 y 1 d, 0 h, 0 m, and 0 s
// ERROR: 63072000 seconds is equal to 2 y 2 d, 0 h, 0 m, and 0 s
// ERROR: 70000000 seconds is equal to 2 y 2 d, 4 h, 26 m, and 40 s ****
// ERROR: 78840000 seconds is equal to 2 y 2 d, 12 h, 0 m, and 0 s ****
// ERROR: 90000000 seconds is equal to 2 y 2 d, 16 h, 0 m, and 0 s ****
// ERROR: 92500000 seconds is equal to 2 y 2 d, 14 h, 26 m, and 40 s ****
// ERROR: 93750000 seconds is equal to 2 y 2 d, 1 h, 40 m, and 0 s ****
// ERROR: 94380000 seconds is equal to 2 y 2 d, 8 h, 40 m, and 0 s ****
// ERROR: 94537500 seconds is equal to 2 y 2 d, 4 h, 25 m, and 0 s ****
// ERROR: 94576875 seconds is equal to 2 y 2 d, 15 h, 21 m, and 15 s ****
// ERROR: 94596562 seconds is equal to 2 y 2 d, 20 h, 49 m, and 22 s ****
// ERROR: 94606406 seconds is equal to 2 y 2 d, 23 h, 33 m, and 26 s ****
// ERROR: 94607636 seconds is equal to 2 y 2 d, 23 h, 53 m, and 56 s ****
// ERROR: 94607943 seconds is equal to 2 y 2 d, 23 h, 59 m, and 3 s ****
// ERROR: 94607981 seconds is equal to 2 y 2 d, 23 h, 59 m, and 41 s ****
// ERROR: 94607990 seconds is equal to 2 y 2 d, 23 h, 59 m, and 50 s ****
// ERROR: 94607995 seconds is equal to 2 y 2 d, 23 h, 59 m, and 55 s ****
// ERROR: 94607998 seconds is equal to 2 y 2 d, 23 h, 59 m, and 58 s ****
// ERROR: 94607999 seconds is equal to 2 y 2 d, 23 h, 59 m, and 59 s **** ****
// ERROR: 94608000 seconds is equal to 3 y 3 d, 0 h, 0 m, and 0 s **** ****
// ERROR: 94608174 seconds is equal to 3 y 3 d, 0 h, 2 m, and 54 s ****
// ERROR: 94608097 seconds is equal to 3 y 3 d, 0 h, 1 m, and 37 s ****
// ERROR: 94608251 seconds is equal to 3 y 3 d, 0 h, 4 m, and 11 s ****
// ERROR: 94608867 seconds is equal to 3 y 3 d, 0 h, 14 m, and 27 s ****
// ERROR: 94611328 seconds is equal to 3 y 3 d, 0 h, 55 m, and 28 s ****
// ERROR: 94616250 seconds is equal to 3 y 3 d, 2 h, 17 m, and 30 s ****
// ERROR: 94695000 seconds is equal to 3 y 3 d, 0 h, 10 m, and 0 s ****
// ERROR: 95000000 seconds is equal to 3 y 3 d, 12 h, 53 m, and 20 s ****
// ERROR: 100000000 seconds is equal to 3 y 3 d, 9 h, 46 m, and 40 s ****
// ERROR: 122990400 seconds is equal to 3 y 3 d, 12 h, 0 m, and 0 s
// ERROR: 126144000 seconds is equal to 4 y 4 d, 0 h, 0 m, and 0 s
答案 0 :(得分：2)
在您的代码行中，
days_remain = days / DAYS_PER_YEAR; // ***missing 22643200 seconds, given input: 85000000***
这里应该是%而不是/，你错误地写错了。
当我看到你的代码的这一部分时，我怀疑也许这不会起作用
else if(seconds > LIMIT )
{
printf("error: excessive amount of seconds.\n");
printf("range for seconds must be between 0 and %u!\n", LIMIT );
printf("If number of seconds exceeds %u, then it is beyond range for type 'int'.\n", LIMIT );
printf("EXITING seconds program. \n");
return 1;
}
请参阅此Validate max integer in scanf了解如何验证用户提供的输入。
答案 1 :(得分：1)
像这样编辑你的代码： days_remain = days％DAYS_PER_YEAR;