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NEW QUESTION: 1
Which of the following are common benefits of cloud computing? Select the correct answer.
A. Reduced Investment and Proportional Costs
B. Increased Scalability
C. All of the above.
D. Increased Availability and Reliability
Answer: C
NEW QUESTION: 2
Given the code fragment:
Path path1 = Paths.get("D:\\sales\\.\\quarterly\\..\\report");
path1 = path1.normalize();
Path path2 = path1.relativize(Paths.get("d:\\empdetails.dat"));
path2 = path2.resolve(path1);
System.out.println(path1);
System.out.println(path2);
}
What is the result?
A. \sales\report\empdetails.dat
B. D: \sales\report
C. \sales\report
D. \sales\report
E. \sales\report\empdetails.dat
F. D: \sales\quarterly\ . . . \report
G. D: \sales\report
H. D: \sales\quarterly\ . . .\report
Answer: G
Explanation:
Path1 is the normalized result of D:\\sales\\.\\quarterly\\..\\report
namely D: \sales\report.
The normalize method removes any redundant elements, which includes any "." or "directory/.."
occurrences.
Consider path2.
With the relativize line path2 is set to../../empdetails.dat
In this scenario the following applies to the resolve statement: Passing an absolute path to
the resolve method returns the passed-in path.So Path2 will be set to Path1 in the statementpath2
= path2.resolve(path1);
Note:
A common requirement when you are writing file I/O code is the capability to construct a path from
one location in the file system to another location. You can meet this using the relativizemethod.
This method constructs a path originating from the original path and ending at the location
specified by the passed-in path. The new path is relative to the original path.
You can combine paths by using the resolve method. You pass in a partial path , which is a path
that does not include a root element, and that partial path is appended to the original path.
Reference: The Java Tutorials,Path Operations
NEW QUESTION: 3
What happens when you attempt to compile and run the following code?
#include <iostream>
using namespace std;
int main()
{
int i = 0;
do {
i++;
if (i==3)
break;
cout<<i;
}
while(i < 5);
return 0;
}
A. No output
B. It prints: 12
C. It prints: 0
D. It prints: 1
Answer: B
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