C Programming Interview Questions with answers, interview questions and answers, c programming interview question part six.learn c programming online, c programs, online c questions
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 datatypes.
80)
main()
{
char c=' ',x,convert(z);
getc(c);
if((c>='a')
&& (c<='z'))
x=convert(c);
printf("%c",x);
}
convert(z)
{
return z32;
}
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:
40 31 22
Explanation:
Let us assume some
x= scanf("%d",&i)t the values during
execution
will be, t ix 4 04 3 12 220
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(p2,300);
}
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:
0000
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 preincrement 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
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.
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