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When the scope resolution operator is used before a variable name, the compiler is instructed to use a(n) ______ variable.

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User Matt Mc
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Answer:

HOPE THIS HELPS!!

Explanation:

When the scope resolution operator (::) is used before a variable name, the compiler is instructed to use a specific variable that is defined in a particular scope. This operator allows you to access variables that have the same name but are defined in different scopes.

For example, let's say we have a global variable called "x" and a local variable called "x" inside a function. If we want to access the global variable "x" instead of the local variable, we can use the scope resolution operator like this:

```cpp

int x = 10; // global variable

void myFunction() {

int x = 5; // local variable

cout << x << endl; // Output: 5

cout << ::x << endl; // Output: 10

}

```

In the above code, the first cout statement outputs the value of the local variable "x" which is 5. However, when we use the scope resolution operator "::x", we explicitly tell the compiler to use the global variable "x" instead of the local one, resulting in an output of 10.

So, when the scope resolution operator is used before a variable name, the compiler is instructed to use the variable defined in a specific scope, allowing us to disambiguate between variables with the same name in different scopes.

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User Barry Carlyon
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When the scope resolution operator (::) is used before a variable name, the compiler is instructed to use a global or namespace variable, depending on the context in which it is used.
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User David Harvey
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