Here is the example,
Thanks.Code:string ex = @"This has a carriage return\r\n"
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Here is the example,
Thanks.Code:string ex = @"This has a carriage return\r\n"
It declares the string to be literal. It will therefore NOT interpret \r as a carriage return escape sequence and \n as a newline escape sequence. Read more at http://msdn.microsoft.com/en-us/libr...(v=vs.71).aspx