将字符串2012-12-05字符串的年月日存在int变量中
1 string str(2012-12-05) 2 3 int year, month, day; 4 5 //暴力赋值 6 7 year = (str[0] - '0') * 1000 + (str[1] - '0') * 100; 8 year += (str[2] - '0') * 10 + (str[3] - '0'); 9 month = (str[5] - '0') * 10 + (str[6] - '0'); 10 day = (str[8] - '0') * 10 + (str[9] - '0');

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