Your program wishes to open a file named C:\java\myProg\input.txt on a Windows system.Which of the following is the correct code to do this?
A)inputFile = new File("c:\java\myProg\input.txt");
B)inputFile = new File.open("c:\java\myProg\input.txt");
C)inputFile = new File("c:\java\myProg\input.txt");
D)inputFile = new File.open("c:\java\myProg\input.txt");

Question

Quizplus · Accepted Answer

Option C is the correct code to open a file named input.txt located in the directory C:\java\myProg\ on a Windows system. In Java, to represent a backslash character in a path, we need to use two backslashes (\) instead of a single one (\). Hence, option C uses the correct syntax to represent the path C:\java\myProg\input.txt. Option A is incorrect because it uses a single backslash instead of two backslashes to represent the path. Option B is incorrect because there is no method called "open()" in the File class. Option D is incorrect because it uses a single backslash instead of two backslashes and it also uses the non-existent "open()" method.

Your Program Wishes to Open a File Named C:\java\myProg\input