What will be the size of the following structure?
struct demo{
int a;
char b;
float c;
}
Answer: A
Correct answer: 12
Members are laid out with padding so that each is properly aligned:
| int a | 4 bytes |
|---|---|
| char b | 1 byte + 3 padding |
| float c | 4 bytes |
| total | 12 bytes |
The three padding bytes exist so that the float begins on a 4-byte boundary. Without alignment the total would be 9.
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Which of the following operators can be applied on structure variables?
Answer: B
Correct answer: Assignment ( = )
A structure may be assigned as a whole, which copies it member by member. It may not be compared with ==: the compiler could not simply compare the bytes, because padding between members holds unspecified values, and comparing member by member is left to the programmer.
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How many bytes does the structure below occupy?
#include <stdio.h>
struct test
{
int k;
char c;
};
Answer: A
Correct answer: Multiple of integer size
The members are an int and a char, but the compiler adds padding so that the whole structure aligns to its strictest member. The size therefore comes out a multiple of the int size - 8 bytes on a typical machine, not the 5 the members occupy. Padding is what makes sizeof a structure unpredictable from its members alone.
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What will be the output of the following C code?
#include <stdio.h>
#define MAX 4
enum sanfoundry
{
a,b=3,c
};
main()
{
if(MAX!=c)
printf("hello");
else
printf("welcome");
}
Answer: C
Correct answer: welcome
The enumerators are a = 0, b = 3 and therefore c = 4 - values continue from the last one given. MAX is also 4, so the condition MAX != c is false and the else branch prints welcome.
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What will be the output of the following C code if input given is 2?
#include <stdio.h>
enum day
{
a,b,c=5,d,e
};
main()
{
printf("Enter the value for a");
scanf("%d",a);
printf("%d",a);
}
Answer: C
The code shown above results in an error. This is because memory is not allocated for constants in enum. Thereore ampersand(&) cannot be used with them. Also, a value has already been assigned to ‘a’, which cannot be changed.
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How do you free allocated memory?
Answer: B
Correct answer: free(var-name);
free() returns memory obtained from malloc, calloc or realloc to the heap. delete is the C++ operator, and the other two names do not exist. Freeing the same block twice, or using a pointer after freeing it, is undefined behaviour.
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The above C declaration define 's' to be
struct node
{
int i;
float j;
};
struct node *s[10];
Answer: A
Correct answer: An array, each element of which is a pointer to a structure of type node
Subscript binds tighter than the dereference, so struct node *s[10] groups as struct node *(s[10]) - an array of 10 pointers, each to a struct node. A pointer to an array would need parentheses in the other place: struct node (*s)[10].
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What is the output of this program?
#include <stdio.h>
void main()
{
struct demo{
char * a;
int n;
};
struct demo p = {"hello", 2015};
struct demo q = p;
printf("%d", printf("%s",q.a));
}
Answer: C
Correct answer: hello5
The inner printf runs first, writing hello and returning 5 - the number of characters it wrote. The outer printf then prints that return value. So the output is hello5, in that order. (The assignment q = p copies the structure, including the pointer, so both point at the same string.)
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What will be the output of the program if it is executed like below?
cmd> sample
/* sample.c */
#include <stdio.h>
int main(int argc, char **argv)
{
printf("%s\n", argv[argc-1]);
return 0;
}
Answer: A
Correct answer: sample
argv[0] is the program name as invoked, and with no further arguments argc is 1 - so argv[argc-1] is argv[0], which prints sample. The array always has argc entries plus a null pointer at argv[argc].
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Members of a union are accessed as ________________
Answer: C
Correct answer: Both a & b
A union member is reached exactly as a structure member is: with the dot through a variable (u.member) and with the arrow through a pointer (p->member). The syntax is identical; only the storage layout differs.
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