C-Aptitude Questions 2
Continued..
Continued..
51) 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’.
52) 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’.
53) main()
{
int i,
n;
char *x = “girl”;
n = strlen(x);
*x = x[n];
for(i=0; i<n; ++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.
54) int i,j;
for(i=0;i<=10;i++)
{
j+=5;
assert(i<5);
}
Answer:
Runtime
error: Abnormal program termination.
assert
failed (i<5), <file name>,<line number>
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.
55) 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).
56) What are the files which are
automatically opened when a C file is executed?
Answer:
stdin,
stdout, stderr (standard input,standard output,standard error).
57) 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.
58) 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.
59) 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.
60) 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.
61) 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.
62) 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.
63) 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.
64) #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.
65) 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".
66) 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.
67) #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.
68) 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.
69) 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.
70) 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.
71) #include<stdio.h>
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
72) #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:
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.
73) #include<stdio.h>
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.
74) main()
{
int i=5,j=6,z;
printf("%d",i+++j);
}
Answer:
11
Explanation:
the
expression i+++j is treated as (i++ + j)
76) struct aaa{
struct
aaa *prev;
int
i;
struct
aaa *next;
};
main()
{
struct aaa abc,def,ghi,jkl;
int x=100;
abc.i=0;abc.prev=&jkl;
abc.next=&def;
def.i=1;def.prev=&abc;def.next=&ghi;
ghi.i=2;ghi.prev=&def;
ghi.next=&jkl;
jkl.i=3;jkl.prev=&ghi;jkl.next=&abc;
x=abc.next->next->prev->next->i;
printf("%d",x);
}
Answer:
2
Explanation:
above all statements
form a double circular linked list;
abc.next->next->prev->next->i
this
one points to "ghi" node the value of at particular node is 2.
77) struct point
{
int x;
int y;
};
struct
point origin,*pp;
main()
{
pp=&origin;
printf("origin
is(%d%d)\n",(*pp).x,(*pp).y);
printf("origin
is (%d%d)\n",pp->x,pp->y);
}
Answer:
origin
is(0,0)
origin
is(0,0)
Explanation:
pp
is a pointer to structure. we can access the elements of the structure either
with arrow mark or with indirection operator.
Note:
Since
structure point is globally declared x
& y are initialized as zeroes
78) main()
{
int i=_l_abc(10);
printf("%d\n",--i);
}
int
_l_abc(int i)
{
return(i++);
}
Answer:
9
Explanation:
return(i++)
it will first return i and then increments. i.e. 10 will be returned.
79) main()
{
char *p;
int *q;
long *r;
p=q=r=0;
p++;
q++;
r++;
printf("%p...%p...%p",p,q,r);
}
Answer:
0001...0002...0004
Explanation:
++
operator when applied to pointers
increments address according to their corresponding data-types.
80) main()
{
char c=' ',x,convert(z);
getc(c);
if((c>='a') && (c<='z'))
x=convert(c);
printf("%c",x);
}
convert(z)
{
return z-32;
}
Answer:
Compiler
error
Explanation:
declaration
of convert and format of getc() are wrong.
81) main(int argc, char **argv)
{
printf("enter the character");
getchar();
sum(argv[1],argv[2]);
}
sum(num1,num2)
int
num1,num2;
{
return num1+num2;
}
Answer:
Compiler
error.
Explanation:
argv[1]
& argv[2] are strings. They are passed to the function sum without
converting it to integer values.
82) # include <stdio.h>
int
one_d[]={1,2,3};
main()
{
int *ptr;
ptr=one_d;
ptr+=3;
printf("%d",*ptr);
}
Answer:
garbage
value
Explanation:
ptr
pointer is pointing to out of the array range of one_d.
83) # include<stdio.h>
aaa()
{
printf("hi");
}
bbb(){
printf("hello");
}
ccc(){
printf("bye");
}
main()
{
int (*ptr[3])();
ptr[0]=aaa;
ptr[1]=bbb;
ptr[2]=ccc;
ptr[2]();
}
Answer:
bye
Explanation:
ptr
is array of pointers to functions of return type int.ptr[0] is assigned to
address of the function aaa. Similarly ptr[1] and ptr[2] for bbb and ccc
respectively. ptr[2]() is in effect of writing ccc(), since ptr[2] points to
ccc.
85) #include<stdio.h>
main()
{
FILE
*ptr;
char
i;
ptr=fopen("zzz.c","r");
while((i=fgetch(ptr))!=EOF)
printf("%c",i);
}
Answer:
contents
of zzz.c followed by an infinite loop
Explanation:
The
condition is checked against EOF, it should be checked against NULL.
86) main()
{
int i =0;j=0;
if(i && j++)
printf("%d..%d",i++,j);
printf("%d..%d,i,j);
}
Answer:
0..0
Explanation:
The
value of i is 0. Since this information is enough to determine the truth value
of the boolean expression. So the statement following the if statement is not
executed. The values of i and j remain
unchanged and get printed.
87) main()
{
int i;
i = abc();
printf("%d",i);
}
abc()
{
_AX = 1000;
}
Answer:
1000
Explanation:
Normally
the return value from the function is through the information from the
accumulator. Here _AH is the pseudo global variable denoting the accumulator.
Hence, the value of the accumulator is set 1000 so the function returns value
1000.
88) int i;
main(){
int
t;
for
( t=4;scanf("%d",&i)-t;printf("%d\n",i))
printf("%d--",t--);
}
// If the inputs are 0,1,2,3 find
the o/p
Answer:
4--0
3--1
2--2
Explanation:
Let
us assume some x= scanf("%d",&i)-t the values during execution
will
be,
t i
x
4 0
-4
3 1
-2
2 2
0
89) main(){
int a= 0;int b = 20;char x =1;char y =10;
if(a,b,x,y)
printf("hello");
}
Answer:
hello
Explanation:
The
comma operator has associativity from left to right. Only the rightmost value
is returned and the other values are evaluated and ignored. Thus the value of
last variable y is returned to check in if. Since it is a non zero value if
becomes true so, "hello" will be printed.
90) main(){
unsigned int i;
for(i=1;i>-2;i--)
printf("c
aptitude");
}
Explanation:
i
is an unsigned integer. It is compared with a signed value. Since the both
types doesn't match, signed is promoted to unsigned value. The unsigned
equivalent of -2 is a huge value so condition becomes false and control comes
out of the loop.
91) In the following pgm add a stmt in the function fun such that the address of
'a'
gets stored in 'j'.
main(){
int *
j;
void fun(int **);
fun(&j);
}
void fun(int **k) {
int a =0;
/* add a stmt here*/
}
Answer:
*k = &a
Explanation:
The
argument of the function is a pointer to a pointer.
92) What are the following notations of
defining functions known as?
i. int abc(int a,float b)
{
/* some code */
}
ii. int abc(a,b)
int a; float b;
{
/*
some code*/
}
Answer:
i. ANSI C notation
ii.
Kernighan & Ritche notation
93) main()
{
char
*p;
p="%d\n";
p++;
p++;
printf(p-2,300);
}
Answer:
300
Explanation:
The
pointer points to % since it is incremented twice and again decremented by 2,
it points to '%d\n' and 300 is printed.
94) main(){
char a[100];
a[0]='a';a[1]]='b';a[2]='c';a[4]='d';
abc(a);
}
abc(char
a[]){
a++;
printf("%c",*a);
a++;
printf("%c",*a);
}
Explanation:
The
base address is modified only in function and as a result a points to 'b' then
after incrementing to 'c' so bc will be printed.
95) func(a,b)
int
a,b;
{
return( a= (a==b) );
}
main()
{
int
process(),func();
printf("The
value of process is %d !\n ",process(func,3,6));
}
process(pf,val1,val2)
int
(*pf) ();
int
val1,val2;
{
return((*pf)
(val1,val2));
}
Answer:
The
value if process is 0 !
Explanation:
The
function 'process' has 3 parameters - 1, a pointer to another function 2 and 3, integers. When this function is
invoked from main, the following substitutions for formal parameters take place:
func for pf, 3 for val1 and 6 for val2. This function returns the result of the
operation performed by the function 'func'. The function func has two integer
parameters. The formal parameters are substituted as 3 for a and 6 for b. since
3 is not equal to 6, a==b returns 0. therefore the function returns 0 which in
turn is returned by the function 'process'.
96) void main()
{
static int i=5;
if(--i){
main();
printf("%d
",i);
}
}
Answer:
0 0 0 0
Explanation:
The variable "I" is
declared as static, hence memory for I will be allocated for only once, as it
encounters the statement. The function main() will be called recursively unless
I becomes equal to 0, and since main() is recursively called, so the value of
static I ie., 0 will be printed every time the control is returned.
97) void main()
{
int k=ret(sizeof(float));
printf("\n here value is
%d",++k);
}
int ret(int ret)
{
ret += 2.5;
return(ret);
}
Answer:
Here value is 7
Explanation:
The
int ret(int ret), ie., the function name and the argument name can be the same.
Firstly, the function ret() is
called in which the sizeof(float) ie., 4 is passed, after the first expression the value in ret
will be 6, as ret is integer hence the value stored in ret will have implicit
type conversion from float to int. The ret is returned in main() it is printed
after and preincrement.
98) void main()
{
char a[]="12345\0";
int i=strlen(a);
printf("here in 3
%d\n",++i);
}
Answer:
here
in 3 6
Explanation:
The char array 'a' will hold the
initialized string, whose length will be counted from 0 till the null
character. Hence the 'I' will hold the value equal to 5, after the
pre-increment in the printf statement, the 6 will be printed.
99) void main()
{
unsigned giveit=-1;
int gotit;
printf("%u ",++giveit);
printf("%u
\n",gotit=--giveit);
}
Answer:
0 65535
Explanation:
100) void main()
{
int i;
char a[]="\0";
if(printf("%s\n",a))
printf("Ok here
\n");
else
printf("Forget
it\n");
}
Answer:
Ok here
Explanation:
Printf will
return how many characters does it print. Hence printing a null character
returns 1 which makes the if statement true, thus "Ok here" is
printed.
Al D Best..!!
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