Question:Refer to the following code snippet:
...
1. XmlSerializer ser = new XmlSerializer( typeof(ServiceDescription) );
2. FileStream objFile = new FileStream( txtWSDL.Text, FileMode.Open );
3. ServiceDescription objSD = ser.Deserialize( file );
4. file.Close();
...

The code given above opens a WSDL file in the fileStream object 'objFile' from the location given in txtWSDL.Text. But on opening WSDL in the ServiceDescription object 'objSD', the code generates an error. Identify the line number that contains the error. 

A 1 
B 2 
C 3 
D None of these 

+ Answer
C 

+ Explanation
Answer not sure

+ Report

Mohammad Towhidul Islam

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