C Aptitude Questions and Answers
Predict the output or error(s) for the following:
1. void main()
{
int const * p=5;
printf("%d",++(*p));
}
Answer:
Compiler error: Cannot modify a constant value.
Explanation:
p is a pointer to a "constant integer". But we tried to
change the value of the "constant integer".
2. main()
{
char s[ ]="man";
int i;
for(i=0;s[ i ];i++)
printf("\n%c%c%c%c",s[
i ],*(s+i),*(i+s),i[s]);
}
Answer:
mmmm
aaaa
nnnn
Explanation:
s[i], *(i+s), *(s+i), i[s] are all different ways of
expressing the same idea. Generally array name is the
base address for that array. Here s is the base address. i
is the index number/displacement from the base
address. So, indirecting it with * is same as s[i]. i[s] may
be surprising. But in the case of C it is same as s[i].
3. main()
{
float me = 1.1;
double you = 1.1;
if(me==you)
printf("I love U");
else
printf("I hate U");
}
Answer:
I hate U
Explanation:
For floating point numbers (float, double, long double)
the values cannot be predicted exactly. Depending on
the number of bytes, the precession with of the value
represented varies. Float takes 4 bytes and long double
takes 10 bytes. So float stores 0.9 with less precision
than long double.
Rule of Thumb:
Never compare or at-least be cautious when using
floating point numbers with relational operators (== , >,
<, <=, >=,!= ) .
4. main()
{
static int var = 5;
printf("%d ",var--);
if(var)
main();
}
Answer:
5 4 3 2 1
Explanation:
When static storage class is given, it is initialized once.
The change in the value of a static variable is retained
even between the function calls. Main is also treated like
any other ordinary function, which can be called
recursively.
5. main()
{
int c[ ]={2.8,3.4,4,6.7,5};
int j,*p=c,*q=c;
for(j=0;j<5;j++) {
printf(" %d ",*c);
++q; }
for(j=0;j<5;j++){
printf(" %d ",*p);
++p; }
}
Answer:
2 2 2 2 2 2 3 4 6 5
Explanation:
Initially pointer c is assigned to both p and q. In the first
loop, since only q is incremented and not c , the value 2
will be printed 5 times. In second loop p itself is
incremented. So the values 2 3 4 6 5 will be printed.
6. main()
{
extern int i;
i=20;
printf("%d",i);
}
Answer:
Linker Error : Undefined symbol '_i'
Explanation:
extern storage class in the following
declaration,
extern int i;
specifies to the compiler that the memory for i is
allocated in some other program and that address will
be given to the current program at the time of linking.
But linker finds that no other variable of name i is
available in any other program with memory space
allocated for it. Hence a linker error has occurred .
Predict the output or error(s) for the following:
7. main()
{
int i=-1,j=-1,k=0,l=2,m;
m=i++&&j++&&k++||l++;
printf("%d %d %d %d %d",i,j,k,l,m);
}
Answer:
0 0 1 3 1
Explanation :
Logical operations always give a result of 1 or 0 . And
also the logical AND (&&) operator has higher priority
over the logical OR (||) operator. So the expression ‘i++
&& j++ && k++’ is executed first. The result of this
expression is 0 (-1 && -1 && 0 = 0). Now the
expression is 0 || 2 which evaluates to 1 (because OR
operator always gives 1 except for ‘0 || 0’ combinationfor
which it gives 0). So the value of m is 1. The values
of other variables are also incremented by 1.
8. main()
{
char *p;
printf("%d %d ",sizeof(*p),sizeof(p));
}
Answer:
1 2
Explanation:
The sizeof() operator gives the number of bytes taken
by its operand. P is a character pointer, which needs one
byte for storing its value (a character). Hence sizeof(*p)
gives a value of 1. Since it needs two bytes to store the
address of the character pointer sizeof(p) gives 2.
9. main()
{
int i=3;
switch(i)
{
default:printf("zero");
case 1: printf("one");
break;
case 2:printf("two");
break;
case 3: printf("three");
break;
}
}
Answer :
three
Explanation :
The default case can be placed anywhere inside the
loop. It is executed only when all other cases doesn't
match.
10. main()
{
printf("%x",-1<<4);
}
Answer:
fff0
Explanation :
-1 is internally represented as all 1's. When left shifted
four times the least significant 4 bits are filled with
0's.The %x format specifier specifies that the integer
value be printed as a hexadecimal value.
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.
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.
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
14. 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).
Page Numbers : 1 2 3 4 5
6 7 8
15. #include
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).
16. #include
main()
{i
nt 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.
17. #include
main()
{s
truct 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
18. #include
main()
{s
truct xx
{i
nt 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.
19. main()
{
printf("\nab");
printf("\bsi");
printf("\rha");
}
Answer:
hai
Explanation:
\n - newline
\b - backspace
\r - linefeed
20. main()
{i
nt 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.
21. #define square(x) x*x
main()
{i
nt 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
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
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.
23. #include
#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.
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
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++;
}
}f
un()
{
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()
{i
nt 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.
Predict the output or error(s) for the following:
36. main()
{i
nt i;
printf("%d",scanf("%d",&i)); // value 10 is given as input here
}
Answer:
1
Explanation:
Scanf returns number of items successfully read and not 1/0. Here 10 is given as input
which should have been scanned successfully. So number of items read is 1.
37. #define f(g,g2) g##g2
main()
{i
nt var12=100;
printf("%d",f(var,12));
}
Answer:
100
38. main()
{i
nt i=0;
for(;i++;printf("%d",i)) ;
printf("%d",i);
}
Answer:
1
Explanation:
before entering into the for loop the checking condition is "evaluated". Here it evaluates
to 0 (false) and comes out of the loop, and i is incremented (note the semicolon after
the for loop).
39. #include
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:
M
Explanation:
p is pointing to character '\n'.str1 is pointing to character 'a' ++*p meAnswer:"p is
pointing to '\n' and that is incremented by one." the ASCII value of '\n' is 10. then it is
incremented to 11. the value of ++*p is 11. ++*str1 meAnswer:"str1 is pointing to 'a'
that is incremented by 1 and it becomes 'b'. ASCII value of 'b' is 98. both 11 and 98 is
added and result is subtracted from 32.
i.e. (11+98-32)=77("M");
40. #include
main()
{
struct xx
{
int x=3;
char name[]="hello";
};
struct xx *s=malloc(sizeof(struct xx));
printf("%d",s->x);
printf("%s",s->name);
}
Answer:
Compiler Error
Explanation:
Initialization should not be done for structure members inside the structure declaration
41. #include
main()
{s
truct xx
{
int x;
struct yy
{
char s;
struct xx *p;
};
struct yy *q;
};
}
Answer:
Compiler Error
Explanation:
in the end of nested structure yy a member have to be declared
42. main()
{
extern int i;
i=20;
printf("%d",sizeof(i));
}
Answer:
Linker error: undefined symbol '_i'.
Explanation:
extern declaration specifies that the variable i is defined somewhere else. The compiler
passes the external variable to be resolved by the linker. So compiler doesn't find an
error. During linking the linker searches for the definition of i. Since it is not found the
linker flags an error.
43. main()
{
printf("%d", out);
}
int out=100;
Answer:
Compiler error: undefined symbol out in function main.
Explanation:
The rule is that a variable is available for use from the point of declaration. Even though
a is a global variable, it is not available for main. Hence an error
Predict the output or error(s) for the following:
44. main()
{
extern out;
printf("%d", out);
}
int out=100;
Answer:
100
Explanation:
This is the correct way of writing the previous program.
45. main()
{
show();
}
void show()
{
printf("I'm the greatest");
}
Answer:
Compier error: Type mismatch in redeclaration of show.
Explanation:
When the compiler sees the function show it doesn't know anything about it. So the
default return type (ie, int) is assumed. But when compiler sees the actual definition of
show mismatch occurs since it is declared as void. Hence the error.
The solutions are as follows:
1. declare void show() in main() .
2. define show() before main().
3. declare extern void show() before the use of show().
46. main( )
{
int a[2][3][2] = {{{2,4},{7,8},{3,4}},{{2,2},{2,3},{3,4}}};
printf(“%u %u %u %d \n”,a,*a,**a,***a);
printf(“%u %u %u %d \n”,a+1,*a+1,**a+1,***a+1);
}
Answer:
100, 100, 100, 2
114, 104, 102, 3
47. main( )
{
int a[ ] = {10,20,30,40,50},j,*p;
for(j=0; j<5; j++)
{
printf(“%d” ,*a);
a++;
}
p = a;
for(j=0; j<5; j++)
{
printf(“%d ” ,*p);
p++;
}
}
Answer:
Compiler error: lvalue required.
Explanation:
Error is in line with statement a++. The operand must be an lvalue and may be of any
of scalar type for the any operator, array name only when subscripted is an lvalue.
Simply array name is a non-modifiable lvalue.
48. main( )
{
static int a[ ] = {0,1,2,3,4};
int *p[ ] = {a,a+1,a+2,a+3,a+4};
int **ptr = p;
ptr++;
printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr);
*ptr++;
printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr);
*++ptr;
printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr);
++*ptr;
printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr);
}
Answer:
111
222
333
344
49. main( )
{
void *vp;
char ch = ‘g’, *cp = “goofy”;
int j = 20;
vp = &ch;
printf(“%c”, *(char *)vp);
vp = &j;
printf(“%d”,*(int *)vp);
vp = cp;
printf(“%s”,(char *)vp + 3);
}
Answer:
g20fy
Explanation:
Since a void pointer is used it can be type casted to any other type pointer. vp = &ch
stores address of char ch and the next statement prints the value stored in vp after type
casting it to the proper data type pointer. the output is ‘g’. Similarly the output from
second printf is ‘20’. The third printf statement type casts it to print the string from the
4th value hence the output is ‘fy’.
50. main ( )
{
static char *s[ ] = {“black”, “white”, “yellow”, “violet”};
char **ptr[ ] = {s+3, s+2, s+1, s}, ***p;
p = ptr;
**++p;
printf(“%s”,*--*++p + 3);
}
Answer:
ck
Explanation:
In this problem we have an array of char pointers pointing to start of 4 strings. Then we
have ptr which is a pointer to a pointer of type char and a variable p which is a pointer
to a pointer to a pointer of type char. p hold the initial value of ptr, i.e. p = s+3. The
next statement increment value in p by 1 , thus now value of p = s+2. In the printf
statement the expression is evaluated *++p causes gets value s+1 then the pre
decrement is executed and we get s+1 – 1 = s . the indirection operator now gets the
value from the array of s and adds 3 to the starting address. The string is printed
starting from this position. Thus, the output is ‘ck’.
51. main()
{
int i, n;
char *x = “girl”;
n = strlen(x);
*x = x[n];
for(i=0; i
{
printf(“%s\n”,x);
x++;
}
}
Answer:
(blank space)
irl
rl
l
Explanation:
Here a string (a pointer to char) is initialized with a value “girl”. The strlen function
returns the length of the string, thus n has a value 4. The next statement assigns value
at the nth location (‘\0’) to the first location. Now the string becomes “\0irl” . Now the
printf statement prints the string after each iteration it increments it starting position.
Loop starts from 0 to 4. The first time x[0] = ‘\0’ hence it prints nothing and pointer
value is incremented. The second time it prints from x[1] i.e “irl” and the third time it
prints “rl” and the last time it prints “l” and the loop terminates.
Predict the output or error(s) for the following:
52. int i,j;
for(i=0;i<=10;i++)
{
j+=5;
assert(i<5);
}
Answer:
Runtime error: Abnormal program termination.
assert failed (i<5), ,
Explanation:
asserts are used during debugging to make sure that certain conditions are satisfied. If
assertion fails, the program will terminate reporting the same. After debugging use,
#undef NDEBUG
and this will disable all the assertions from the source code. Assertion
is a good debugging tool to make use of.
53. main()
{
int i=-1;
+i;
printf("i = %d, +i = %d \n",i,+i);
}
Answer:
i = -1, +i = -1
Explanation:
Unary + is the only dummy operator in C. Where-ever it comes you can just ignore it
just because it has no effect in the expressions (hence the name dummy operator).
54. What are the files which are automatically opened when a C file is
executed?
Answer:
stdin, stdout, stderr (standard input,standard output,standard error).
55. what will be the position of the file marker?
a: fseek(ptr,0,SEEK_SET);
b: fseek(ptr,0,SEEK_CUR);
Answer :
a: The SEEK_SET sets the file position marker to the starting of the file.
b: The SEEK_CUR sets the file position marker to the current position
of the file.
56. main()
{
char name[10],s[12];
scanf(" \"%[^\"]\"",s);
}
How scanf will execute?
Answer:
First it checks for the leading white space and discards it.Then it matches with a
quotation mark and then it reads all character upto another quotation mark.
57. What is the problem with the following code segment?
while ((fgets(receiving array,50,file_ptr)) != EOF)
;
Answer & Explanation:
fgets returns a pointer. So the correct end of file check is checking for != NULL.
58. main()
{
main();
}
Answer:
Runtime error : Stack overflow.
Explanation:
main function calls itself again and again. Each time the function is called its return
address is stored in the call stack. Since there is no condition to terminate the function
call, the call stack overflows at runtime. So it terminates the program and results in an
error.
59. main()
{
char *cptr,c;
void *vptr,v;
c=10; v=0;
cptr=&c; vptr=&v;
printf("%c%v",c,v);
}
Answer:
Compiler error (at line number 4): size of v is Unknown.
Explanation:
You can create a variable of type void * but not of type void, since void is an empty
type. In the second line you are creating variable vptr of type void * and v of type void
hence an error.
60. main()
{
char *str1="abcd";
char str2[]="abcd";
printf("%d %d %d",sizeof(str1),sizeof(str2),sizeof("abcd"));
}
Answer:
2 5 5
Explanation:
In first sizeof, str1 is a character pointer so it gives you the size of the pointer variable.
In second sizeof the name str2 indicates the name of the array whose size is 5
(including the '\0' termination character). The third sizeof is similar to the second one.
61. main()
{
char not;
not=!2;
printf("%d",not);
}
Answer:
0
Explanation:
! is a logical operator. In C the value 0 is considered to be the boolean value FALSE, and
any non-zero value is considered to be the boolean value TRUE. Here 2 is a non-zero
value so TRUE. !TRUE is FALSE (0) so it prints 0.
62. #define FALSE -1
#define TRUE 1
#define NULL 0
main() {
if(NULL)
puts("NULL");
else if(FALSE)
puts("TRUE");
else
puts("FALSE");
}
Answer:
TRUE
Explanation:
The input program to the compiler after processing by the preprocessor is,
main(){
if(0)
puts("NULL");
else if(-1)
puts("TRUE");
else
puts("FALSE");
}
Preprocessor doesn't replace the values given inside the double quotes. The check by if
condition is boolean value false so it goes to else. In second if -1 is boolean value true
hence "TRUE" is printed.
Predict the output or error(s) for the following:
63. main()
{
int k=1;
printf("%d==1 is ""%s",k,k==1?"TRUE":"FALSE");
}
Answer:
1==1 is TRUE
Explanation:
When two strings are placed together (or separated by white-space) they are
concatenated (this is called as "stringization" operation). So the string is as if it is given
as "%d==1 is %s". The conditional operator( ?: ) evaluates to "TRUE".
64. main()
{
int y;
scanf("%d",&y); // input given is 2000
if( (y%4==0 && y%100 != 0) || y%100 == 0 )
printf("%d is a leap year");
else
printf("%d is not a leap year");
}
Answer:
2000 is a leap year
Explanation:
An ordinary program to check if leap year or not.
65. #define max 5
#define int arr1[max]
main()
{
typedef char arr2[max];
arr1 list={0,1,2,3,4};
arr2 name="name";
printf("%d %s",list[0],name);
}
Answer:
Compiler error (in the line arr1 list = {0,1,2,3,4})
Explanation:
arr2 is declared of type array of size 5 of characters. So it can be used to declare the
variable name of the type arr2. But it is not the case of arr1. Hence an error.
Rule of Thumb:
#defines are used for textual replacement whereas typedefs are used for declaring new
types.
66. int i=10;
main()
{
extern int i;
{
int i=20;
{
const volatile unsigned i=30;
printf("%d",i);
}
printf("%d",i);
}
printf("%d",i);
}
Answer:
30,20,10
Explanation:
'{' introduces new block and thus new scope. In the innermost block i is declared as,
const volatile unsigned
which is a valid declaration. i is assumed of type int. So printf prints 30. In the next
block, i has value 20 and so printf prints 20. In the outermost block, i is declared as
extern, so no storage space is allocated for it. After compilation is over the linker
resolves it to global variable i (since it is the only variable visible there). So it prints i's
value as 10.
67. main()
{
int *j;
{
int i=10;
j=&i;
}
printf("%d",*j);
}
Answer:
10
Explanation:
The variable i is a block level variable and the visibility is inside that block only. But the
lifetime of i is lifetime of the function so it lives upto the exit of main function. Since the
i is still allocated space, *j prints the value stored in i since j points i.
68. main()
{
int i=-1;
-i;
printf("i = %d, -i = %d \n",i,-i);
}
Answer:
i = -1, -i = 1
Explanation:
-i is executed and this execution doesn't affect the value of i. In printf first you just print
the value of i. After that the value of the expression -i = -(-1) is printed.
69. #include
main()
{
const int i=4;
float j;
j = ++i;
printf("%d %f", i,++j);
}
Answer:
Compiler error
Explanation:
i is a constant. you cannot change the value of constant
70. #include
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:
garbagevalue..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
meAnswer:it will print first element of 3D array.
71. #include
main()
{
register i=5;
char j[]= "hello";
printf("%s %d",j,i);
}
Answer:
hello 5
Explanation:
if you declare i as register compiler will treat it as ordinary integer and it will take
integer value. i value may be stored either in register or in memory.
72. main()
{
int i=5,j=6,z;
printf("%d",i+++j);
}
Answer:
11
Explanation:
the expression i+++j is treated as (i++ + j)
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