C Aptitude Part II

Note : All the programs are tested under Turbo C/C++ compilers.
It is assumed that,
Programs run under DOS environment,
The underlying machine is an x86 system,
Program is compiled using Turbo C/C++ compiler.
The program output may depend on the information based on this assumptions
(for example sizeof(int) == 2 may be assumed).
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Predict the output or error(s) for the following:
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11. 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.
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12. 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. ” }
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13. #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
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14. main()
{
int i=10;
i=!i>14; // Here Not of I is compared with 14,that’s how result is executed
Printf ("i=%d",i);
}
Answer:
14i=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).
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15. #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).
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16. #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.
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17. #include<stdio.h>
main()
{
struct xx
{
int x=3; // canot initialized variable at deecleration
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
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18. #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.
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19. main()
{
printf("\nab");
printf("\bsi");
printf("\rha");
}
Answer:
16hai
Explanation:
\n - newline
\b - backspace
\r - linefeed
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20. 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.
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21. #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
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22. 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
*p that is value at the location currently pointed by p will be taken
++*p the retrieved value will be incremented
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.
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23. #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.
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24. #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
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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.