What will be the output of the following C code?
#include <stdio.h>
enum sanfoundry
{
a=1,b,c,d,e
};
int main()
{
printf("%d",b*c+e-d);
}
Answer: B
Correct answer: 7
Values run on from the one given: a = 1, so b = 2, c = 3, d = 4, e = 5. Then
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What is the correct syntax to declare a function foo() which receives an array of structure in function?
Answer: A
Correct answer: void foo(struct *var);
An array argument decays to a pointer to its first element, so a function receiving an array of structures takes a pointer to structure. (In real code the tag must be named - void foo(struct tagname *var); - since struct alone is not a type.) Passing an array never copies it, so the length must be passed separately.
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What will be the output of the following C code?
main()
{
enum resut {pass, fail};
enum result s1,s2;
s1=pass;
s2=fail;
printf("%d",s1);
}
Answer: A
Correct answer: error
The enumeration is declared as enum resut - a typo - and the variables are then declared as enum result, a tag that does not exist. The compiler reports an incomplete type.
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Which operator connects the structure name to its member name?
Answer: C
Correct answer: .
The dot operator selects a member of a structure variable: s.member. The arrow -> does the same through a pointer, and is exactly equivalent to (*p).member.
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#include <stdio.h>
typedef enum{
Male, Female
}SEX;
int main()
{
SEX var = Male;
SEX var1 = Female;
printf("%s %s",var, var1);
return 0;
}
Answer: C
Correct answer: Runtime error (Segmentation fault)
%s tells printf to treat its argument as a pointer to a string, but the arguments are the integers 0 and 1. printf then dereferences address 0 and address 1, and the program is killed. Nothing in the program remembers the names Male and Female - printing them requires an array of strings indexed by the enumerator.
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Which operator connects the structure name to its member name?
Answer: C
Correct answer: .
The dot connects a structure variable to its member: s.member. Through a pointer the arrow is used instead, p->member, which is shorthand for (*p).member.
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What will be output for the following code?
Answer: A
The code shown above results in an error, stating : storage size of t1 and t2 unknown. There is an error in the declaration of these variables in the statement: enum test1 t1,t2;
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#include "stdio.h"
typedef enum{
Male=5, Female
}SEX;
int main()
{
SEX var = Male;
SEX var1 = Female;
printf("%d %d",var, var1);
return 0;
}
Answer: D
Correct answer: 5 6
Male is set to 5, and Female, having no value of its own, continues from it at 6. Enumerators always resume counting from the last value given.
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#include <stdio.h>
int main()
{
struct site
{
char name[] = "GeeksQuiz";
int no_of_pages = 200;
};
struct site *ptr;
printf("%d ", ptr->no_of_pages);
printf("%s", ptr->name);
getchar();
return 0;
}
Answer: D
Correct answer: Compiler Error
A structure declaration describes a type, not an object, so its members cannot carry initialisers - both char name[] = "GeeksQuiz"; and int no_of_pages = 200; are rejected. A member array also cannot have an unspecified size. The program never compiles, so the uninitialised ptr never becomes an issue.
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What will be the output of the following C code?
#include <stdio.h>
enum sanfoundry
{
a=1,b
};
enum sanfoundry1
{
c,d
};
int main()
{
enum sanfoundry1 s1=c;
enum sanfoundry1 s=a;
enum sanfoundry s2=d;
printf("%d",s);
printf("%d",s1);
printf("%d",s2);
}
Answer: D
Correct answer: 101
Take each in turn. s1 = c, and in sanfoundry1 c is 0. s = a, and a is 1. s2 = d, which is 1. The prints are in the order s, s1, s2, giving 101. Note that C does not object to assigning a value of one enumeration to a variable of another - they are all just ints.
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