Typedef Questions and Answers

Q: What will be the output of the following C code? #include <stdio.h> typedef struct student { char *a; }stu; void main() { struct student s; s.a = "hey"; printf("%s", s.a); }

Correct answer: heyHere 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.

Q: What will be the output of the following C code? #include <stdio.h> typedef int integer; int main() { int i = 10, *ptr; float f = 20; integer j = i; ptr = &j; printf("%d\n", *ptr); return 0; }

Correct answer: 10typedef int integer; makes integer an alias for int - the same type, not a distinct one. So integer j = i; copies 10 into an ordinary int, and ptr = &j; is a perfectly ordinary int * assignment with no conversion involved. Dereferencing it prints 10. The unused float f is there only as a distraction.

Q: What will be the output of the following C code? #include <stdio.h> int (*(x()))[2]; typedef int (*(*ptr)())[2] ptrfoo; int main() { ptrfoo ptr1; ptr1 = x; ptr1(); return 0; } int (*(x()))[2] { int (*ary)[2] = malloc(sizeof*ary); return &ary; }

Correct answer: Compile time errorThe line typedef int (*(*ptr)())[2] ptrfoo; is malformed. A typedef declares the new name inside the declarator, in the position a variable name would occupy - here that position is already taken by ptr, so the trailing ptrfoo is a second declarator with nothing joining it. The compiler rejects the declaration before any of the rest matters.

Q: What will be the output of the following C code? #include <stdio.h> int *(*(x()))[2]; typedef int **(*ptrfoo)())[2]; int main() { ptrfoo ptr1; ptr1 = x; ptr1(); return 0; } int *(*(x()))[2] { int (*ary)[2] = malloc(sizeof * ary); return &ary; }

Correct answer: Compile time errorThe typedef line is syntactically invalid: typedef int **(*ptrfoo)())[2]; has unbalanced parentheses, the () and the [2] both trying to attach after the declarator has already been closed. Compiling it produces an error at that line before execution is ever reached, and a second error follows on the incompatible assignment ptr1 = x;.

Q: What will be the output of the following C code? #include <stdio.h> typedef struct p { int x, y; }; int main() { p k1 = {1, 2}; printf("%d\n", k1.x); }

Correct answer: Compile time errorTwo things go wrong. The declaration typedef struct p { int x, y; }; supplies no new name at all, so the typedef declares nothing; and p k1 then uses the struct tag as though it were a type name, which C does not allow. Writing struct p k1, or giving the typedef a name, would fix it.

Q: What will be the output of the following C code? #include <stdio.h> typedef struct p { int x, y; }k = {1, 2}; int main() { p k1 = k; printf("%d\n", k1.x); }

Correct answer: Compile time errorBecause the declaration begins with typedef, the k after the closing brace is a type name, and a type name cannot be given an initialiser - = {1, 2} is meaningless there. The later p k1 = k; compounds the error by using the tag p as a type. Remove the typedef and k becomes an initialised variable.

Q: What will be the output of the following C code? #include <stdio.h> typedef struct p { int x, y; }k; int main() { struct p p = {1, 2}; k k1 = p; printf("%d\n", k1.x); }

Correct answer: 1Here everything is used correctly. k is an alias for struct p; struct p p = {1, 2}; declares and initialises a structure, the variable name p shadowing the tag harmlessly since tags live in their own namespace; and k k1 = p; copies it, structure assignment being a member-by-member copy. So k1.x is 1.

Q: Which option should be selected to work the following C expression? string p = "HELLO";

Correct answer: typedef char *string;A string in C is a pointer to its first character, so the alias must be a pointer type: typedef char *string; lets string p = "HELLO"; assign the literal’s address to p. The array form is not merely wrong syntax - even typedef char string[]; would give an incomplete type that cannot be initialised this way.

Q: Which of the given option is the correct method for initialization? typedef char *string;

Correct answer: string p = “Hello”;Once string means char *, a variable of that type is declared exactly as any other: string p = "Hello";, with no extra asterisk. Writing *string p would apply the dereference operator to a type name, which is not valid syntax, and assigning a single character constant would be a type mismatch as well.

11.

What will be the output of the following C code?

#include <stdio.h>
    typedef struct student
    {
        char *a;
    }stu;
    void main()
    {
        struct student s;
        s.a = "hey";
        printf("%s", s.a);
    }

A.Compile time error
B.Varies
C.he
D.hey

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.

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12.

What will be the output of the following C code?

#include <stdio.h>
    typedef int integer;
    int main()
    {
        int i = 10, *ptr;
        float f = 20;
        integer j = i;
        ptr = &j;
        printf("%d\n", *ptr);
        return 0;
    }

A.Compile time error
B.Undefined behaviour
C.Depends on the standard
D.10

Answer: D

Correct answer: 10

typedef int integer; makes integer an alias for int - the same type, not a distinct one. So integer j = i; copies 10 into an ordinary int, and ptr = &j; is a perfectly ordinary int * assignment with no conversion involved. Dereferencing it prints 10. The unused float f is there only as a distraction.

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13.

What will be the output of the following C code?

#include <stdio.h>
    int (*(x()))[2];
    typedef int (*(*ptr)())[2] ptrfoo;
    int main()
    {
        ptrfoo ptr1;
        ptr1 = x;
        ptr1();
        return 0;
    }
    int (*(x()))[2]
    {
        int (*ary)[2] = malloc(sizeof*ary);
        return &ary;
    }

A.Compile time error
B.Nothing
C.Undefined behaviour
D.Depends on the standard

Answer: A

Correct answer: Compile time error

The line typedef int (*(*ptr)())[2] ptrfoo; is malformed. A typedef declares the new name inside the declarator, in the position a variable name would occupy - here that position is already taken by ptr, so the trailing ptrfoo is a second declarator with nothing joining it. The compiler rejects the declaration before any of the rest matters.

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14.

What will be the output of the following C code?

#include <stdio.h>
    int *(*(x()))[2];
    typedef int **(*ptrfoo)())[2];
    int main()
    {
        ptrfoo ptr1;
        ptr1 = x;
        ptr1();
        return 0;
    }
    int *(*(x()))[2]
    {
        int (*ary)[2] = malloc(sizeof * ary);
        return &ary;
    }

A.Compile time error
B.Nothing
C.Undefined behaviour
D.Depends on the standard

Answer: B

Correct answer: Compile time error

The typedef line is syntactically invalid: typedef int **(*ptrfoo)())[2]; has unbalanced parentheses, the () and the [2] both trying to attach after the declarator has already been closed. Compiling it produces an error at that line before execution is ever reached, and a second error follows on the incompatible assignment ptr1 = x;.

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15.

What will be the output of the following C code?

#include <stdio.h>
    typedef struct p
    {
        int x, y;
    };
    int main()
    {
        p k1 = {1, 2};
        printf("%d\n", k1.x);
    }

A.Compile time error
B.1
C.Depends on the standard

Answer: A

Correct answer: Compile time error

Two things go wrong. The declaration typedef struct p { int x, y; }; supplies no new name at all, so the typedef declares nothing; and p k1 then uses the struct tag as though it were a type name, which C does not allow. Writing struct p k1, or giving the typedef a name, would fix it.

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16.

What will be the output of the following C code?

#include <stdio.h>
    typedef struct p
    {
        int x, y;
    }k = {1, 2};
    int main()
    {
        p k1 = k;
        printf("%d\n", k1.x);
    }

A.Compile time error
B.1
C.Depends on the standard

Answer: A

Correct answer: Compile time error

Because the declaration begins with typedef, the k after the closing brace is a type name, and a type name cannot be given an initialiser - = {1, 2} is meaningless there. The later p k1 = k; compounds the error by using the tag p as a type. Remove the typedef and k becomes an initialised variable.

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17.

What will be the output of the following C code?

#include <stdio.h>
    typedef struct p
    {
        int x, y;
    }k;
    int main()
    {
        struct p p = {1, 2};
        k k1 = p;
        printf("%d\n", k1.x);
    }

A.Compile time error
B.1
C.Depends on the standard

Answer: B

Correct answer: 1

Here everything is used correctly. k is an alias for struct p; struct p p = {1, 2}; declares and initialises a structure, the variable name p shadowing the tag harmlessly since tags live in their own namespace; and k k1 = p; copies it, structure assignment being a member-by-member copy. So k1.x is 1.

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18.

Which option should be selected to work the following C expression?

string p = "HELLO";

A.typedef char [] string;
B.typedef char *string;
C.typedef char [] string; and typedef char *string;
D.Such expression cannot be generated in C

Answer: B

Correct answer: typedef char *string;

A string in C is a pointer to its first character, so the alias must be a pointer type: typedef char *string; lets string p = "HELLO"; assign the literal’s address to p. The array form is not merely wrong syntax - even typedef char string[]; would give an incomplete type that cannot be initialised this way.

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19.

Which of the given option is the correct method for initialization?

typedef char *string;

A.*string *p = “Hello”;
B.string p = “Hello”;
C.*string p = ‘A’;
D.Not more than one space should be given when using typedef

Answer: B

Correct answer: string p = “Hello”;

Once string means char *, a variable of that type is declared exactly as any other: string p = "Hello";, with no extra asterisk. Writing *string p would apply the dereference operator to a type name, which is not valid syntax, and assigning a single character constant would be a type mismatch as well.

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20.

Which of the following is false about typedef?

A.typedef follow scope rules
B.typedef defined substitutes can be redefined again. (Eg: typedef char a; typedef int a;)
C.You cannot typedef a typedef with other term
D.All of the mentioned

Answer: B

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

This is the false statement, and hence the answer. A typedef name may be repeated only if every declaration gives it the same type; redefining it as a different type in the same scope is an error, because typedef is handled by the compiler and subject to its rules. A #define macro, being pure text substitution, can be redefined freely.

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