Predict the output or
error(s) for the following:
25. 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.
26. 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).
27. 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.
28. 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
29. 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.
30. main()
{
char *p;
p="Hello";
printf("%c\n",*&*p);
}
Answer:
H
Explanation:
* is a dereference operator & is a reference operator. They can be
applied any number of times provided it is meaningful. Here p points to
the first character in the string "Hello". *p dereferences it and so its value
is H. Again & references it to an address and * dereferences it to the
value H.
31. main()
{
int i=1;
while (i<=5)
{
printf("%d",i);
if (i>2)
goto here;
i++;
}
}
fun()
{
here:
printf("PP");
}
Answer:
Compiler error: Undefined label 'here' in function main
Explanation:
Labels have functions scope, in other words The scope of the labels is limited
to functions . The label 'here' is available in function fun() Hence it is not
visible in function main.
32. main()
{
static char names[5][20]={"pascal","ada","cobol","fortran","perl"};
int i;
char *t;
t=names[3];
names[3]=names[4];
names[4]=t;
for (i=0;i<=4;i++)
printf("%s",names[i]);
}
Answer:
Compiler error: Lvalue required in function main
Explanation:
Array names are pointer constants. So it cannot be modified.
33. void main()
{
int i=5;
printf("%d",i++
+ ++i);
}
Answer:
Output Cannot be predicted exactly.
Explanation:
Side effects are involved in the evaluation of i
34. void main()
{
int i=5;
printf("%d",i+++++i);
}
Answer:
Compiler Error
Explanation:
The expression i+++++i is parsed as i ++ ++ + i which is an illegal combination
of operators.
35. #include
main()
{
int i=1,j=2;
switch(i)
{
case 1: printf("GOOD");
break;
case j: printf("BAD");
break;
}
}
Answer:
Compiler Error: Constant expression required in function main.
Explanation:
The case statement can have only constant expressions (this implies that we
cannot use variable names directly so an error).
Note:
Enumerated types can be used in case statements.
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