C# 对增量求和并返回下一个数字
是否有方法计算该值的时间增量并返回下一个值C# 对增量求和并返回下一个数字,c#,C#,是否有方法计算该值的时间增量并返回下一个值 int num = 77; int sum = 0; for (int n = num; n > 0; sum += n % 10, n /= 10) ; 例如0001+0002+0003 int num = 77; int sum = 0; for (int n = num; n > 0; sum += n % 10, n /= 10) ; 当它返回时,结果为0004 int num = 77; int sum = 0; for (i
int num = 77;
int sum = 0;
for (int n = num; n > 0; sum += n % 10, n /= 10) ;
例如0001+0002+0003
int num = 77;
int sum = 0;
for (int n = num; n > 0; sum += n % 10, n /= 10) ;
当它返回时,结果为0004
int num = 77;
int sum = 0;
for (int n = num; n > 0; sum += n % 10, n /= 10) ;
是我干的
int num = 77;
int sum = 0;
for (int n = num; n > 0; sum += n % 10, n /= 10) ;
但此方法将int num相加为'14'
int num = 77;
int sum = 0;
for (int n = num; n > 0; sum += n % 10, n /= 10) ;
我想要的是int num'0001',然后继续增加到'0002','0003'不是直接增加,但您可以使用属性换行:
int num = 77;
int sum = 0;
for (int n = num; n > 0; sum += n % 10, n /= 10) ;
public int Counter { get; private set; }
private int _number;
public int Number
{
get { return _number; }
set
{
_number = value;
Counter++;
}
}
让该方法给出下一个结果,定义如下:
int num = 77;
int sum = 0;
for (int n = num; n > 0; sum += n % 10, n /= 10) ;
public static string GetTheNextResult(ref int number)
{
return (++number).ToString("0000");
}
让numberInput
为数字,然后您可以使用以下调用获得下一个结果:
int num = 77;
int sum = 0;
for (int n = num; n > 0; sum += n % 10, n /= 10) ;
int numberInput = 1;
var result1 = GetTheNextResult(ref numberInput); // Gives you 0002
var result2 = GetTheNextResult(ref numberInput); // Gives you 0003
var result3 = GetTheNextResult(ref numberInput); // Gives you 0004
你能为你的问题发布一些代码吗?很大程度上取决于你的问题不清楚,添加一些代码或适当的例子。我已经编辑了这个问题。它是ok@David-如果任何答案解决了您的问题,请将问题标记为已解决:)