quick guide

1.What would be the output of the following program?
main()
{
printf(5+"fascimile");
}
A. error B. fascimile C. mile D. none of the above

Ans: C

2.What would be the output of the following program?
main()
{
char str1[] = "Hello";
char str2[] = "Hello";
if(str1 == str2)
printf("\n Equal");
else
printf("\nUnequal");
}

A. Equal B. Unequal C. Error D. None of the above

Ans: B

3.What would be the output of the following program?

main()
{
printf("%c","abcdefgh"[4]);
}

A. Error B. d C. e D. abcdefgh

Ans: C

4.How would you output \n on the following screen?

Ans: printf("\\n");

5.What would be the output of the following program?

main()
{
char str[7] = "strings";
printf("%s", str);
}
A. error B. strings C. cannot predict D. none of the above

Ans: C
Here str[] has been declared as a 7 character array and
into it a 8 character string has been stored . This would
result into overwriting of the byte beyond the seventh
byte reserved for the array with a '\0'.There is always a
possibility that something important gets overwritten
which would be unsafe .

6.What would be the output of the following program?

main()
{
char ch = 'A';
printf("%d%d", sizeof(ch), sizeof('A'));
}

A. 1 1 B. 1 2 C. 2 2 D. 2 1

Ans: B

7. What would be the output of the following program?

main()
{
printf("\n%d%d%d", sizeof('3'), sizeof("3"), sizeof(3));
}
A 1 1 1 B. 2 2 2 C. 1 2 2 D. 1 1 1

Ans:B

8. What would be the output of the following program?

main()
{
char *str[] = {"Frogs", "Do", "Not", "Die",
"They", "Croak"};
printf("%d %d", sizeof(str), sizeof(str[0]));
}

Ans: 12 2

9. What would be the output of the following program?

main()
{
static char s[] = "Rendezvous";
printf("%d", *(s+strlen(s)));
}

Ans: 0

Explanation: No Rendezvous, but a zero is printed out.
Mentioning the name of the string gives the base address
of the string. The function strlen(s) returns the length
of the string s[], which in this case is 12. In the
printf(), using the 'contents of' operator , we are
trying to print out the contents of 12th address from the
base address of the string. At this address there is a
'\0' ,which is automatically stored to mark the end of the
string. The ASCII value of '\0' is 0, which is what being
printed by the printf().

10. What would be the output of the following program?
main()
{
char ch[20];
int i;
for(i=0; i<=19; i++)
*(ch+i) = 67;
*(ch+i) = '\0';
printf("%s" , ch);
}

Ans: CCCCCCCCCCCCCCCCCCC

Explanation:Mentioning the name of the array always gives
its base address . Therefore (ch+i) would give the address
of the ith element from the base address , and *(ch+i)
would give the value at the address , i.e the value of the
ith element . Through the for loop we store 67, which is the
ASCII value of upper case 'C' , in all locations of the
string.Once the control reaches outside the for loop the
value of i would be 19 , and in the 19th location from the
base address we store a '\0' to mark the end of the string.
This is essential , as the compiler has no other way of
knowing where the string is terminated .In the printf() that
follows , %s is the format specification for printing a string,
and ch gives the base address of the string .Hence starting
from the first element ,the complete string is printed out.

11. What would be the output of the following program?
main()
{
char str[20];
int i;
for(1=0;1<=18; i++)
i[str] = 'C';
i[str] = '\0';
printf("%s",str);
}

Ans:CCCCCCCCCCCCCCCCCCC

Explanation: If your concept of arrays is fool-proof , you
should find the above o/p only natural. If not, here's your
chance to make it so. C makes no secret of the fact that it
uses pointers internally to access array elements .With the
knowledge of how array elements are accessed using pointers,
we can think of str[i] as *(str+i). Basic maths tells us
that *(str+i) would be same as *(i+str). And if str[i] is
same as *(str+i), then naturally *(i+str) would be same as
i[str].Thus, we can conclude that all the following
expressions are different ways of referring the ith element
of the string:
str[i]
*(str+i)
*(i+str)
i[str]
Hence, through the for loop upper case C is stored in all
the elements of the string. A '\0' is stored to mark the end
of the string, and then the string is printed out using
printf().

12.What would be the output of the following program?

main()
{
char str[20];
static int i;
for(; ;)
{
i++[str] = 'A' + 2;
if(i == 19)
break;
}
i[str] = '\0';
printf("%s",str);
}

Ans:CCCCCCCCCCCCCCCCCCC

13. What would be the output of the following program?

main()
{
static char str[]={48,48,48,48,48,48,48,48,48,48};
char *s;
int i;
s = str;
for( i = 0; i<=9; i++)
{
if(*s)
printf("%c", *s);
s++;
}
}

Ans: 0000000000

Explanation: In all 10 elements of str[], an integer , 48
is stored. Wondering whether a char string can hold ints?
The answer is yes, as 48 does not get stored literally in
the elements. 48 is interpreted as the ASCII value of the
character to be stored in the string. the character corr-
esponding to ASCII value 48 happens to be 0, which assig-
ned to all the locations of the string.'s', a character
pointer, is assigned the base address of the string str[].
Next , in the if condition,the value at address contained
in s is checked for truth/falsity.As 0 represents ASCII 48,
the condition evaluates to true every time, until the end
of the string is reached. At the end of the string a '\0',
i.e ASCII 0 is encountered , and the if condition fails.
Irrespective of whether the condition is satisfied or not,
's' is incremented so that each time it points to the
subsequent array element .This entire logic is repeated
in the for loop , printing out 10 zeros in the process.

14. What would be the output of the following program?

main()
{
static char str[]={0,0,0,0,0,0,0,0,0,0};
char *s;
int i;
s = str;
for( i = 0; i<=9; i++)
{
if(*s)
printf("%c", *s);
s++;
}
}

Ans: No output

Explanation: Though you may not have expected zeros to be
outputted this time,you surely did expect some output! We
stored the character corresponding to ASCII 0 in all 10
elements of the string. Next, we assigns s, a char pointer,
the base address of the string . Though the for loop , we are
attempting to print out all elements one by one , but not
before imposing the if condition.The if is made to test the
value at address contained in 's' before the execution of the
printf( ) .The first time , *s yields ASCII 0. Therefore the
if condition reduces to if(0) , and as 0 stands for falsity,
the condition fails. Hence ,'s'is incremented and control
loops back to for without executing the printf(). The same
thing happens the next time around , and the next, and so on,
till the for loop ends, resulting in no output at all.

15. What would be the output of the following program?

main()
{
static char s[]="C Smart!!";
int i;
for(i =0; s[i]; i++)
printf("%c%c%c%c\n",s[i],*(s+i),i[s],*(i+s));
}

Ans: C C C C
S S S S
m m m m
a a a a
r r r r
t t t t

! ! ! !
Explanation:The above program rubs in the point that s[i],
i[s],*(s+i) and *(i+s) are various ways of referring to the
same element, that is the ith element of the string s. Each
element of the string is printed out for four times,till the
end of the string is encountered. Note that in the for loop
there is an expression s[i] in the condition part. This means
the loop would continue to get executed till s[i] is not equal
to zero. We can afford to say this because a string always ends
with a '\0', whose ASCII value is 0. Thus the for loop will be
terminated when the expression s[i] yields a '\0'.

16. What would be the output of the following program?
main()
{
static char s[] = "Oinks Grunts and Guffaws";
printf("%c\n", *(&s[2]));
printf("%s\n", s+5);
printf("%s\n", s);
printf("%c\n", *(s+2));
printf(%s\n", s);
}

Ans: n
Grunts and Guffaws
Oinks Grunts and Guffaws
n
404
Explanation: In the first printf() the address of operator, &
gives the address of the second element of the string . Value
at this address is 'n' , which is printed out by the printf()
using %c. since s gives the base address of the array , (s+5)
would give the address of the fifth element from the base
address . This address is passed to the second printf() .
Using the format specification %s , the contents of the string
are printed out the 5th element onwards.The third printf()
prints the entire string , as the base address of the string
is being passed to it. The fourth printf() is made to print
the second character of the string , as *(s+2) is nothing but
s[2]. Thus 'n'gets printed.Does the o/p of the final printf()
surprise you by printing out a number , 404?

Note that the format specification %d is used with s, which
gives the base address of the string. It happened to be 404
when we executed the program, which got printed out. On
executing the same yourself, you may get any other address,
depending on what address is allotted to the string by the
compiler.

17.What would be the output of the following program?
main()
{
static char s[25] = "The cocaine man";
int i =0;
char ch;
ch = s[++i];
printf("%c%d\n",ch,i);
ch = s[i++];
printf("%c%d\n",ch,i);
ch = i++[s];
printf("%c%d\n",ch,i);
ch = ++i[s];
printf("%c%d\n",ch,i);
}
Ans: h 1
h 2
e 3
! 3

18.What would be the output of the following program?

main()
{
static char arr[] = "pickpocketing my piece of mind";
int i:
printf("%c\n", *arr);
arr++;
printf("%c\n", *arr);
}

Ans: Lvalue required in function main