Predict the output or error(s)
for the following:
1. main()
{
char
string[]="Hello World";
display(string);
}
void display(char *string)
{
printf("%s",string);
}
Answer:
Compiler Error
: Type mismatch in redeclaration of function display
Explanation
:
In
third line, when the function display
is encountered, the compiler doesn't know anything about the function display.
It assumes the arguments and return types to be integers, (which is the default
type). When it sees the actual function display,
the arguments and type contradicts with what it has assumed previously. Hence a
compile time error occurs.
2. main()
{
int
c=- -2;
printf("c=%d",c);
}
Answer:
c=2;
Explanation:
Here
unary minus (or negation) operator is used twice. Same maths rules applies, ie. minus * minus= plus.
Note:
However
you cannot give like --2. Because -- operator can only be applied to variables as a decrement operator (eg., i--). 2 is a
constant and not a variable.
3. #define int char
main()
{
int
i=65;
printf("sizeof(i)=%d",sizeof(i));
}
Answer:
sizeof(i)=1
Explanation:
Since
the #define replaces the string int by the macro char
4. main()
{
int i=10;
i=!i>14;
printf("i=%d",i);
}
Answer:
i=0
Explanation:
In
the expression !i>14 , NOT (!)
operator has more precedence than ‘ >’ symbol. !
is a unary logical operator. !i (!10) is 0 (not of true is false). 0>14 is false (zero).
5. #include<stdio.h>
main()
{
char
s[]={'a','b','c','\n','c','\0'};
char *p,*str,*str1;
p=&s[3];
str=p;
str1=s;
printf("%d",++*p +
++*str1-32);
}
Answer:
77
Explanation:
p
is pointing to character '\n'. str1 is pointing to character 'a' ++*p. "p
is pointing to '\n' and that is incremented by one." the ASCII value of
'\n' is 10, which is then incremented to 11. The value of ++*p is 11. ++*str1,
str1 is pointing to 'a' that is incremented by 1 and it becomes 'b'. ASCII
value of 'b' is 98.
Now performing (11 + 98 – 32), we get
77("M");
So we get the output 77 :: "M"
(Ascii is 77).
6. #include<stdio.h>
main()
{
int a[2][2][2] = { {10,2,3,4},
{5,6,7,8} };
int *p,*q;
p=&a[2][2][2];
*q=***a;
printf("%d----%d",*p,*q);
}
Answer:
SomeGarbageValue---1
Explanation:
p=&a[2][2][2] you declare only two 2D arrays, but you are
trying to access the third 2D(which you are not declared) it will print garbage
values. *q=***a starting address of a is assigned integer pointer. Now q is
pointing to starting address of a. If you print *q, it will print first element
of 3D array.
7. #include<stdio.h>
main()
{
struct xx
{
int x=3;
char name[]="hello";
};
struct xx *s;
printf("%d",s->x);
printf("%s",s->name);
}
Answer:
Compiler
Error
Explanation:
You
should not initialize variables in declaration
8. #include<stdio.h>
main()
{
struct xx
{
int x;
struct yy
{
char s;
struct
xx *p;
};
struct yy *q;
};
}
Answer:
Compiler
Error
Explanation:
The
structure yy is nested within structure xx. Hence, the elements are of yy are
to be accessed through the instance of structure xx, which needs an instance of
yy to be known. If the instance is created after defining the structure the
compiler will not know about the instance relative to xx. Hence for nested
structure yy you have to declare member.
9. main()
{
printf("\nab");
printf("\bsi");
printf("\rha");
}
Answer:
hai
Explanation:
\n - newline
\b - backspace
\r - linefeed
10. main()
{
int i=5;
printf("%d%d%d%d%d%d",i++,i--,++i,--i,i);
}
Answer:
45545
Explanation:
The
arguments in a function call are pushed into the stack from left to right. The
evaluation is by popping out from the stack. and the evaluation is from right to left, hence the
result.
11. #define square(x) x*x
main()
{
int i;
i = 64/square(4);
printf("%d",i);
}
Answer:
64
Explanation:
the
macro call square(4) will substituted by 4*4 so the expression becomes i =
64/4*4 . Since / and * has equal priority the expression will be evaluated as
(64/4)*4 i.e. 16*4 = 64
12. main()
{
char *p="hai friends",*p1;
p1=p;
while(*p!='\0') ++*p++;
printf("%s %s",p,p1);
}
Answer:
ibj!gsjfoet
Explanation:
++*p++
will be parse in the given order
Ø
*p that is value at the location currently
pointed by p will be taken
Ø
++*p the retrieved value will be incremented
Ø
when ; is encountered the location will be
incremented that is p++ will be executed
Hence, in the while loop initial
value pointed by p is ‘h’, which is changed to ‘i’ by executing ++*p and
pointer moves to point, ‘a’ which is similarly changed to ‘b’ and so on.
Similarly blank space is converted to ‘!’. Thus, we obtain value in p becomes
“ibj!gsjfoet” and since p reaches ‘\0’ and p1 points to p thus p1doesnot print
anything.
13. #include <stdio.h>
#define a 10
main()
{
#define a 50
printf("%d",a);
}
Answer:
50
Explanation:
The
preprocessor directives can be redefined anywhere in the program. So the most
recently assigned value will be taken.
14. #define clrscr() 100
main()
{
clrscr();
printf("%d\n",clrscr());
}
Answer:
100
Explanation:
Preprocessor
executes as a seperate pass before the execution of the compiler. So textual
replacement of clrscr() to 100 occurs.The input
program to compiler looks like this :
main()
{
100;
printf("%d\n",100);
}
Note:
100;
is an executable statement but with no action. So it doesn't give any problem
15. main()
{
printf("%p",main);
}
Answer:
Some address will be
printed.
Explanation:
Function
names are just addresses (just like array names are addresses).
main() is also
a function. So the address of function main will be printed. %p in printf
specifies that the argument is an address. They are printed as hexadecimal
numbers.
27) main()
{
clrscr();
}
clrscr();
Answer:
No
output/error
Explanation:
The
first clrscr() occurs inside a function. So it becomes a function call. In the
second clrscr(); is a function declaration (because it is not inside any
function).
28) enum colors {BLACK,BLUE,GREEN}
main()
{
printf("%d..%d..%d",BLACK,BLUE,GREEN);
return(1);
}
Answer:
0..1..2
Explanation:
enum
assigns numbers starting from 0, if not explicitly defined.
29) void main()
{
char far *farther,*farthest;
printf("%d..%d",sizeof(farther),sizeof(farthest));
}
Answer:
4..2
Explanation:
the
second pointer is of char type and not a far pointer
30) main()
{
int i=400,j=300;
printf("%d..%d");
}
Answer:
400..300
Explanation:
printf
takes the values of the first two assignments of the program. Any number of
printf's may be given. All of them take only the first two values. If more
number of assignments given in the program,then printf will take garbage
values.
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