Typedef Questions and Answers

Q: In the following code what is 'P'? typedef char *charp; const charp P;

Correct answer: P is a constantSince charp is a single type meaning char *, the const qualifies that whole type, giving char * const P - a constant pointer to a modifiable character. It is not const char *, which is what textual substitution of the macro form would have produced. This is one of the clearest illustrations of why typedef and #define are not interchangeable.

Q: What will be the output of the following C code? #include <stdio.h> int main() { typedef union a { int i; char ch[2]; }hello; hello u; u.ch[0] = 3; u.ch[1] = 2; printf("%d, %d", u.ch[0], u.ch[1]); return 0; }

Correct answer: 3, 2The two writes go to ch[0] and ch[1], and the two reads come straight back from the same bytes, so the output is the values in the order they were assigned: 3 then 2. The union sharing storage with int i makes no difference here, because i is never read - had it been, the result would have depended on the machine’s byte order.

Q: What is the output of this C code? typedef struct p { int x, y; }k; int main() { struct p p = {1, 2}; k k1 = p; printf("%d\n", k1.x); }

Correct answer: 1Every element is used correctly. k is an alias for struct p, the variable named p shadows the tag p harmlessly because tags occupy a separate namespace, and k k1 = p; is an ordinary structure copy. So k1.x holds 1. Structure assignment copies every member, which is why no loop is needed.

Q: What will be the output of following program ? #include <stdio.h> int main() { typedef int AAA,BBB,CCC,DDD; AAA aaa=10; BBB bbb=20; CCC ccc=30; DDD ddd=40; printf("%d,%d,%d,%d",aaa,bbb,ccc,ddd); return 0; }

Correct answer: 10,20,30,40A single typedef can declare several names, so AAA, BBB, CCC and DDD all become aliases for int. They are four names for one type, not four distinct types, so the four variables are ordinary independent ints holding 10, 20, 30 and 40. Sharing an alias creates no relationship between the variables at all.

Q: In a C program, following variables are defined: float x = 2.17; double y = 2.17; long double z = 2.17; Which of the following is correct way for printing these variables via printf.

Correct answer: printf("%f %lf %Lf",x,y,z);A float passed to a variadic function is promoted to double, so %f serves for both float and double, and %lf is accepted as a synonym for the same. A long double is not promoted and needs its own specifier, %Lf with a capital L; using %f for it reads the wrong number of bytes.

Q: What will be the output of the program? #include <stdio.h> int main() { enum color{red, green, blue}; typedef enum color mycolor; mycolor m = red; printf("%d", m); return 0; }

Correct answer: 0Enumerators are numbered from zero unless given explicit values, so red is 0, green 1 and blue 2. An enumeration constant has integer type, so %d prints the numeric value rather than the name - C keeps no run-time record of the identifier. Printing red would require an array of strings indexed by the enumerator.

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: In the following code snippet can we declare a new typedef named ptr even though struct employee has not been completely declared while using typedef? typedef struct employee *ptr; struct employee { char name[20]; int age; ptr next; }

Correct answer: YesA pointer to an incomplete type is itself a complete type, because its size and representation do not depend on what it points at. So typedef struct employee *ptr; is legal before struct employee has been defined - the tag is merely declared by its mention. This is exactly the mechanism that lets a self-referential structure hold a ptr next member.

51.

In the following code what is 'P'?

typedef char *charp;
const charp P;

A.P is a constant
B.P is a character constant
C.P is character type
D.None of above

Answer: A

Correct answer: P is a constant

Since charp is a single type meaning char *, the const qualifies that whole type, giving char * const P - a constant pointer to a modifiable character. It is not const char *, which is what textual substitution of the macro form would have produced. This is one of the clearest illustrations of why typedef and #define are not interchangeable.

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52.

What will be the output of the following C code?

#include <stdio.h>
int main()
{
    typedef union a
    {
        int i;
        char ch[2];
    }hello;
    hello u;
    u.ch[0] = 3;
    u.ch[1] = 2;
    printf("%d, %d", u.ch[0], u.ch[1]);
    return 0;
}

A.2, 3
B.3, 2
C.32
D.error

Answer: B

Correct answer: 3, 2

The two writes go to ch[0] and ch[1], and the two reads come straight back from the same bytes, so the output is the values in the order they were assigned: 3 then 2. The union sharing storage with int i makes no difference here, because i is never read - had it been, the result would have depended on the machine’s byte order.

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53.

What is the output of this C code?

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

Every element is used correctly. k is an alias for struct p, the variable named p shadows the tag p harmlessly because tags occupy a separate namespace, and k k1 = p; is an ordinary structure copy. So k1.x holds 1. Structure assignment copies every member, which is why no loop is needed.

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54.

What will be the output of following program ?

#include <stdio.h>
int main()
{
    typedef int AAA,BBB,CCC,DDD;
    AAA aaa=10;
    BBB bbb=20;
    CCC ccc=30;
    DDD ddd=40;
    printf("%d,%d,%d,%d",aaa,bbb,ccc,ddd);
    return 0;
}

A.Error
B.10,10,10,10
C.10,20,30,40
D.AAA,BBB,CCC,DDD

Answer: C

Correct answer: 10,20,30,40

A single typedef can declare several names, so AAA, BBB, CCC and DDD all become aliases for int. They are four names for one type, not four distinct types, so the four variables are ordinary independent ints holding 10, 20, 30 and 40. Sharing an alias creates no relationship between the variables at all.

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55.

In a C program, following variables are defined:

float x = 2.17;

double y = 2.17;

long double z = 2.17;

Which of the following is correct way for printing these variables via printf.

A.printf("%f %lf %Lf",x,y,z);
B.printf(“%f %f %f”,x,y,z);
C.printf("%f ? ?f",x,y,z);
D.printf("%f %lf %llf",x,y,z);

Answer: A

Correct answer: printf("%f %lf %Lf",x,y,z);

A float passed to a variadic function is promoted to double, so %f serves for both float and double, and %lf is accepted as a synonym for the same. A long double is not promoted and needs its own specifier, %Lf with a capital L; using %f for it reads the wrong number of bytes.

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56.

What will be the output of the program?

#include <stdio.h>

int main()
{
    enum color{red, green, blue};
    typedef enum color mycolor;
    mycolor m = red;
    printf("%d", m);
    return 0;
}

A.1
B.2
C.red
D.0

Answer: D

Correct answer: 0

Enumerators are numbered from zero unless given explicit values, so red is 0, green 1 and blue 2. An enumeration constant has integer type, so %d prints the numeric value rather than the name - C keeps no run-time record of the identifier. Printing red would require an array of strings indexed by the enumerator.

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57.

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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58.

Assuming int size is 4 bytes, what is going to happen when we compile and run the following program?

#include "stdio.h"
int main()
{
  printf("GeeksQuiz\n");
  main();
  return 0;
}

A.We can’t use main() inside main() and compiler will catch it by showing compiler error.
B.GeeksQuiz would be printed in 2147483647 times i.e. (2 to the power 31) - 1.
C.It’ll print GeeksQuiz infinite times i.e. the program will continue to run forever until it’s terminated by other means such as CTRL+C or CTRL+Z etc.
D.GeeksQuiz would be printed until stack overflow happens for this program.

Answer: D

Correct answer: GeeksQuiz would be printed until stack overflow happens for this program.

C permits main to call itself, and each call adds a frame to the stack that is never popped, because the recursive call precedes the return. The program therefore prints until the stack is exhausted and it crashes - typically after a few hundred thousand iterations. It is not infinite, since the stack is finite, and no compiler error arises.

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59.

In the following code snippet can we declare a new typedef named ptr even though struct employee has not been completely declared while using typedef?

typedef struct employee *ptr;
struct employee
{
    char name[20];
    int age;
    ptr next;
}

A.Yes
B.No

Answer: A

Correct answer: Yes

A pointer to an incomplete type is itself a complete type, because its size and representation do not depend on what it points at. So typedef struct employee *ptr; is legal before struct employee has been defined - the tag is merely declared by its mention. This is exactly the mechanism that lets a self-referential structure hold a ptr next member.

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60.

What is typedef declaration?

A.Does not create a new type
B.It merely adds a new name for some existing type
C.Does not create a new type, It merely adds a new name for some existing type
D.None of the mentioned

Answer: C

Correct answer: Does not create a new type, It merely adds a new name for some existing type

Both halves are right, which is why the combined option is correct. A typedef name is a synonym, fully interchangeable with the original: no new type is created, no type checking distinguishes them, and passing one where the other is expected raises no complaint. This is the essential difference between typedef and a genuine new type such as a distinct struct.

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