In C++ (reference declaration syntax), spot the error and determine the outcome. What happens when compiling the following code?
#include
int main()
{
int x = 80;
int y& = x; // attempt to declare a reference
x++;
cout << x << " " << --y;
return 0;
}
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AThe program will print the output 80 80.
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BThe program will print the output 81 80.
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CThe program will print the output 81 81.
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DIt will result in a compile time error.
Answer
Correct Answer: It will result in a compile time error.
Explanation
Introduction / Context: C++ reference declarations must place the ampersand on the declarator, not after the identifier name. The correct form is int &y = x; not int y& = x;.
Given Data / Assumptions:
- The line int y& = x; is ill-formed.
Concept / Approach: The compiler parses int y& as an attempt to suffix a symbol with an ampersand, which is not valid C++ syntax. References are part of the type specifier and must precede the variable name (as in pointers).
Step-by-Step Solution:
Compilation fails with a syntax error at '&'.Verification / Alternative check: Fixing to int &y = x; then executing would yield "81 81" because y aliases x (after x++ both are 81, and --y prints 80 then decrements, but note the exact sequence). However, as given, it does not compile.
Why Other Options Are Wrong: They assume successful compilation and execution.
Common Pitfalls: Misplacing & or * when declaring references or pointers.
Final Answer: It will result in a compile time error.