C++有一串01字符序列,如何把它们转化为对应的二进制位的形式写入文件
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参考技术A #include<fstream.h>char c[3] = '0','1','1';
string s("1010");
fstream f;
f.open("out.txt", ios::out | ios::trunc);
for(int i=0;i<3;++i)
out<<c[i];
for(i=0;i<s.size();++i)
out<<s[i];
f.close();本回答被提问者采纳
有一串json字符串，我需要将其中的时间格式从yyyy-M-d转换为yyyy/M/d，用c#的正则表达式替换的方法怎么写
如下2015-9-24转换为2015/9/24
["Data":null,"Time":"2015-9-24 0:00:00","Data":null,"Time":"2015-9-24 1:00:00","Data":null,"Time":"2015-9-24 2:00:00","Data":null,"Time":"2015-9-24 3:00:00","Data":null,"Time":"2015-9-24 4:00:00","Data":null,"Time":"2015-9-24 5:00:00","Data":null,"Time":"2015-9-24 6:00:00","Data":null,"Time":"2015-9-24 7:00:00","Data":null,"Time":"2015-9-24 8:00:00","Data":null,"Time":"2015-9-24 9:00:00","Data":null,"Time":"2015-9-24 10:00:00","Data":null,"Time":"2015-9-24 11:00:00","Data":null,"Time":"2015-9-24 12:00:00","Data":null,"Time":"2015-9-24 13:00:00","Data":null,"Time":"2015-9-24 14:00:00","Data":null,"Time":"2015-9-24 15:00:00","Data":null,"Time":"2015-9-24 16:00:00","Data":null,"Time":"2015-9-24 17:00:00","Data":null,"Time":"2015-9-24 18:00:00","Data":null,"Time":"2015-9-24 19:00:00","Data":null,"Time":"2015-9-24 20:00:00","Data":null,"Time":"2015-9-24 21:00:00","Data":null,"Time":"2015-9-24 22:00:00","Data":null,"Time":"2015-9-24 23:00:00","Data":null,"Time":"2015-9-25 0:00:00"]
Console.WriteLine(s);
String s0 = Regex.Replace(s, "([0-9]4)-([0-9]1,2)-([0-9]1,2)", "$1/$2/$3");
Console.WriteLine(s0); 参考技术A
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201\\d-\\d-[123]\\d [\\d:]7以上是关于C++有一串01字符序列,如何把它们转化为对应的二进制位的形式写入文件的主要内容，如果未能解决你的问题，请参考以下文章
有一串json字符串，我需要将其中的时间格式从yyyy-M-d转换为yyyy/M/d，用c#的正则表达式替换的方法怎么写