如何将包含时间的字符串变量转换为 C++ 中的 time_t 类型?
我有一个包含 hh:mm:ss 格式时间的字符串变量.如何将其转换为 time_t 类型?例如:string time_details = "16:35:12"
I have a string variable containing time in hh:mm:ss format. How to convert it into time_t type? eg: string time_details = "16:35:12"
另外，如何比较两个包含时间的变量来决定哪个是最早的?例如:string curr_time = "18:35:21"string user_time = "22:45:31"
Also, how to compare two variables containing time so as to decide which is the earliest? eg : string curr_time = "18:35:21" string user_time = "22:45:31"
推荐答案
您可以使用 strptime(3) 解析时间，然后mktime(3) 将其转换为time_t:
const char *time_details = "16:35:12";
struct tm tm;
strptime(time_details, "%H:%M:%S", &tm);
time_t t = mktime(&tm); // t is now your desired time_t
相关文章