Typedef Questions and Answers

Answer: D

Correct answer: typedef int* intptr; intptr x,y,z;

The keyword comes first, followed by the existing type and then the new name: typedef int *intptr;. The alias is then used exactly like a built-in type, so intptr x, y, z; declares three pointers. Note that this is one place the alias behaves better than the raw type - int *x, y, z; would make only x a pointer.

Answer: C

Correct answer: typedef

typedef introduces a new name for an existing type; it creates no new type, only an alias, so the compiler treats the alias and the original as interchangeable. static and volatile are storage-class and type qualifiers, and default belongs to the switch statement.

Answer: B

Correct answer: False

The claim is false: aliasing a pointer type is one of the commonest uses of typedef. typedef int *intptr; makes intptr mean int *, and the same applies to pointers to functions and to structures. The one thing to watch is that intptr a, b; makes both pointers, which a plain int *a, b; would not.

Answer: B

Correct answer: It creates an alternative name for an existing type

typedef introduces a synonym; the underlying type is unchanged, so the compiler treats the alias and the original as the same type. It is used to shorten unwieldy declarations and to make the intent of a type clearer.

Answer: A

The syntax of the keyword typedef is: keyword ;
Hence if we want to create an alias name (ul) for an identifier of type unsigned long, the correct way to do this would be: typedef unsigned long ul;

Answer: B

Correct answer: typedef defined substitutes can be redefined again. (Eg: typedef char a; typedef int a;)

This is the false statement. Repeating a typedef name with a different type in the same scope is an error, because the compiler enforces the alias as a genuine type declaration. The other two statements are true: typedefs obey the usual scope rules, so one declared inside a function is local to it.

Answer: D

Correct answer: hey

Here the declaration uses the tag rather than the alias - struct student s; - and the tag student does exist, so this compiles. The member a is a char *, assigning a string literal to it is legal, and %s prints it: the output is hey. Compare the previous question, which used struct stu and failed for exactly this reason.

Answer: A

Correct answer: typedef char *STR;

A typedef is written exactly like a variable declaration with typedef placed in front, so the new name occupies the position the variable name would take. Reading char *STR as a declaration of STR makes the form easy to remember.

Answer: D

Correct answer: p is a pointer to an array of 5 integers

The parentheses are what matter. int (*p)[5] binds the asterisk to p first, so p is a pointer, and what it points to is an array of five int. Without them, int *p[5] would be an array of five pointers - the same tokens in a different grouping giving a completely different type.

Answer: C

Correct answer: compiler, pre-processor

#define is handled by the preprocessor, which performs blind textual substitution before compilation begins and understands nothing of types or scope. typedef is processed by the compiler itself, so it obeys scope rules and creates a genuine type alias. This is why #define intptr int * and typedef int *intptr; behave differently when several variables are declared on one line.