Question
1.
#include
2.
using namespace std;
3.
namespace extra
4.
{
5.
int i;
6.
}
7.
void i()
8.
{
9.
using namespace extra;
10.
int i;
11.
i = 9;
12.
cout
What is the output of this program?
1.
#include
2.
using namespace std;
3.
namespace extra
4.
{
5.
int i;
6.
}
7.
void i()
8.
{
9.
using namespace extra;
10.
int i;
11.
i = 9;
12.
cout
Answer: Option A
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A scope resolution operator without a scope qualifier refers to the global namespace.
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