The C function prototype

double mySqrt(int x);

Question

The C function prototype

double mySqrt(int x);

BlackTom AI · Accepted Answer

We begin by identifying what the given snippet represents in C: double mySqrt(int x); This is a function prototype that declares a function named mySqrt. The return type specified is double, and the function takes a single parameter of type int.

Option analysis:
- Provided option: "declares a function called mySqrt which takes an integer as an argument and returns a double". This accurately describes the meaning of the prototype: the function's name is mySqrt, the parameter type is int (the parameter can be named x, as shown, but the name is optional in a prototype), and the return type is double. The wording captures the essential elements: function name, parameter type, and return type. There is no mention of any additional qualifiers or defaults beyond what the prototype declares, so this option aligns with standard C syntax for a function prototype.

Notes on potential misconceptions:
- Some students might think a prototype must include a parameter name; in C, parameter names in the prototype are optional. Here, the name x is present, but even without it, the meaning would remain the same.
- A common confusion is about whether the prototype defines the function body; it does not. It merely declares the function signature so calls can be type-checked before the function is defined.
- The distinction between a declaration (prototype) and a definition is important: a definition would also provide the function body, whereas the prototype alone does not.

In summary, the option correctly identifies the function’s name, parameter type, and return type as expressed by the prototype.

25_SP_FT_CPS_444_01 CPS 444 Final Exam

The C function prototype double mySqrt(int x);

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