#include <stdio.h>
typedef enum{
Male, Female=0
}SEX;
int main()
{
SEX var = Male;
SEX var1 = Female;
printf("%d %d",var, var1);
return 0;
}
Answer: A
Correct answer: 0 0
Male takes the default first value, 0, and Female is explicitly set to 0 as well. Duplicate values are legal - the restriction is on duplicate names, not on duplicate values - so both print 0.
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What is the output of this C code?
#include <stdio.h>
struct student
{
int no;
char name[20];
}
void main()
{
struct student s;
s.no = 8;
printf("hello");
}
Answer: A
Correct answer: Compile time error
The structure definition is missing its terminating semicolon after the closing brace. Without it the compiler reads on and tries to make sense of void main() as part of the declaration, and the program fails to compile.
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Pick the incorrect statement with respect to enums.
Answer: A
Correct answer: Two enum symbols cannot have the same value
The false statement is that two enumerators cannot share a value: they can. enum {a = 1, b = 1}; is perfectly legal, and it is sometimes useful for giving one constant two names. The other three are true: enumerator values must be integer constant expressions, they cannot be assigned to at run time, and any you leave unspecified are filled in automatically, continuing from the last value given.
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Arithmetic operations such as addition, subtraction, multiplication and division are allowed on enumerated constants.
Answer: A
Correct answer: True
An enumerator is an integer constant and an enum variable an integer, so all the arithmetic operators apply to them. That is a strength and a weakness: convenient, but it means the compiler will not stop you adding a colour to a weekday.
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What will be the output of the following C code?
#include <stdio.h>
enum colour
{
blue, red, yellow
};
main()
{
enum colour c;
c=yellow;
printf("%d",c);
}
Answer: B
Correct answer: 2
Enumerators are numbered from 0 in order unless given explicit values, so blue is 0, red 1 and yellow 2. An enum variable is an integer, so %d prints the number, not the name.
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Assume that this code is running on 64 bit system
#include <stdio.h>
enum Xenum{
C, CPP, Java
};
int main()
{
enum Xenum var;
printf("%d",sizeof(var));
return 0;
}
Answer: C
Correct answer: 4
An enumeration variable has the size of an int - 4 bytes on the usual platforms - because the compiler stores the constant itself, not a list of them. How many enumerators the type has makes no difference. So on this system it prints 4 - and it would still be 4 if the enumeration had thirty members.
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Which of the following are themselves a collection of different data types?
Answer: B
Correct answer: Structures
A structure groups members of different types under one name - an int, a float and a character array together. An array requires every element to be the same type, and a string is an array of characters.
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Arithmetic operations are not allowed on enumerated constants.
Answer: B
Correct answer: FALSE
The statement is false. Enumerated constants are integers, so arithmetic on them is perfectly legal - b * c + e - d compiles and computes. C provides no type safety here at all.
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What is the output of this program?
#include <stdio.h>
struct student
{
char *c;
};
void main()
{
struct student s[2];
printf("%d", sizeof(s));
}
Answer: D
Correct answer: 8
s is an array of two structures, each containing a single pointer. On a machine with 4-byte pointers that is \(2 \times 4 = 8\) bytes. Note that sizeof reports the size of the array itself, not the length of anything the pointers reach.
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#include <stdio.h>
struct st
{
int x;
struct st next;
};
int main()
{
struct st temp;
temp.x = 10;
temp.next = temp;
printf("%d", temp.next.x);
return 0;
}
Answer: A
Correct answer: Compiler Error
A structure cannot contain an instance of itself - the type would be infinitely large, since each copy would contain another. The compiler rejects struct st next;. What is allowed is a pointer to the same type, struct st *next;, since a pointer has a fixed size; that is how linked lists are built.
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