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“C” Questions
1. Base class
has some virtual method and derived class has a method with the same name. If
we initialize the base class pointer with derived
object,. calling of that virtual method will result in which method being
called?
a. Base
method
b. Derived method..
Ans. b
2. For the
following C program
#define
AREA(x)(3.14*x*x)
main()
{float r1=6.25,r2=2.5,a;
a=AREA(r1);
printf("\n Area of the circle is %f", a);
a=AREA(r2);
printf("\n Area of the circle is %f", a);
}
What is the
output?
Ans. Area of
the circle is 122.656250
Area of the circle is
19.625000
3. What do
the following statements indicate. Explain.
·
int(*p)[10]
·
int*f()
·
int(*pf)()
·
int*p[10]
Refer to:
-- Kernighan & Ritchie page no. 122
-- Schaum series page no. 323
4.
void main()
{
int d=5;
printf("%f",d);
}
Ans:
Undefined
5.
void main()
{
int i;
for(i=1;i<4,i++)
switch(i)
case 1: printf("%d",i);break;
{
case 2:printf("%d",i);break;
case 3:printf("%d",i);break;
}
switch(i) case 4:printf("%d",i);
}
Ans: 1,2,3,4
6.
void main()
{
char *s="\12345s\n";
printf("%d",sizeof(s));
}
Ans: 6
7.
void main()
{
unsigned i=1; /* unsigned char k= -1 => k=255; */
signed j=-1; /* char k= -1 => k=65535 */
/* unsigned or signed int k= -1 =>k=65535 */
if(i
printf("less");
else
if(i>j)
printf("greater");
else
if(i==j)
printf("equal");
}
Ans: less
8.
void main()
{
float j;
j=1000*1000;
printf("%f",j);
}
1. 1000000
2. Overflow
3. Error
4. None
Ans: 4
9. How do you declare an array of N pointers to functions returning
pointers to functions returning pointers to
characters?
Ans: The first part of this question can be answered in at least
three ways:
1. char *(*(*a[N])())();
2. Build the declaration up incrementally, using typedefs:
typedef char *pc;
/* pointer to char */
typedef pc fpc();
/* function returning pointer to char */
typedef fpc *pfpc;
/* pointer to above */
typedef pfpc
fpfpc(); /* function returning... */
typedef fpfpc
*pfpfpc; /* pointer to... */
pfpfpc a[N];
/* array of... */
3. Use the cdecl program, which turns English into C and
vice
versa:
cdecl> declare a as array of
pointer to function returning
pointer to
function returning pointer to char
char *(*(*a[])())()
cdecl can also explain complicated declarations, help with
casts, and indicate which set of parentheses the arguments
go in (for complicated function definitions, like the one
above).
Any good book on C should explain how to read these
complicated
C declarations "inside out" to understand them
("declaration
mimics use").
The pointer-to-function declarations in the examples above
have
not included parameter type information. When the parameters
have complicated types, declarations can *really* get messy.
(Modern versions of cdecl can help here, too.)
10. A structure pointer is defined of the type time . With 3 fields min,sec
hours having pointers to intergers.
Write the way to initialize the 2nd element to 10.
11. In the above question an array of pointers is declared.
Write the statement to initialize the 3rd element of the 2
element to 10;
12.
int f()
void main()
{
f(1);
f(1,2);
f(1,2,3);
}
f(int i,int j,int k)
{
printf("%d %d %d",i,j,k);
}
What are the number of syntax errors in the above?
Ans: None.
13.
void main()
{
int i=7;
printf("%d",i++*i++);
}
Ans: 56
14.
#define one 0
#ifdef one
printf("one is defined ");
#ifndef one
printf("one is not defined ");
Ans: "one is defined"
15.
void main()
{
int count=10,*temp,sum=0;
temp=&count;
*temp=20;
temp=∑
*temp=count;
printf("%d %d %d ",count,*temp,sum);
}
Ans: 20 20 20
16. There was question in c working only on unix machine with pattern matching.
14. what is alloca()
Ans : It
allocates and frees memory after use/after getting out of scope
17.
main()
{
static i=3;
printf("%d",i--);
return i>0 ? main():0;
}
Ans: 321
18.
char *foo()
{
char result[100]);
strcpy(result,"anything is good");
return(result);
}
void main()
{
char *j;
j=foo()
printf("%s",j);
}
Ans: anything is good.
19.
void main()
{
char *s[]={
"dharma","hewlett-packard","siemens","ibm"};
char **p;
p=s;
printf("%s",++*p);
printf("%s",*p++);
printf("%s",++*p);
}
Ans:
"harma" (p->add(dharma) && (*p)->harma)
"harma" (after printing, p->add(hewlett-packard)
&&(*p)->harma)
"ewlett-packard"
20. Output of
the following program is
main()
{int i=0;
for(i=0;i<20;i++)
{switch(i)
case 0:i+=5;
case 1:i+=2;
case 5:i+=5;
default i+=4;
break;}
printf("%d,",i);
}
}
a)
0,5,9,13,17
b) 5,9,13,17
c) 12,17,22
d) 16,21
e) Syntax error
Ans. (d)
21. What is
the ouptut in the following program
main()
{char c=-64;
int i=-32
unsigned int u =-16;
if(c>i)
{printf("pass1,");
if(c
printf("pass2");
else
printf("Fail2");
}
else
printf("Fail1);
if(i
printf("pass2");
else
printf("Fail2")
}
a)
Pass1,Pass2
b) Pass1,Fail2
c) Fail1,Pass2
d) Fail1,Fail2
e) None of these
Ans. (c)
22. What will
the following program do?
void main()
{
int i;
char a[]="String";
char *p="New Sring";
char *Temp;
Temp=a;
a=malloc(strlen(p) + 1);
strcpy(a,p); //Line number:9//
p = malloc(strlen(Temp) + 1);
strcpy(p,Temp);
printf("(%s, %s)",a,p);
free(p);
free(a);
} //Line number 15//
a) Swap
contents of p & a and print:(New string, string)
b) Generate compilation error in line number 8
c) Generate compilation error in line number 5
d) Generate compilation error in line number 7
e) Generate compilation error in line number 1
Ans. (b)
23. In the
following code segment what will be the result of the function,
value of x ,
value of y
{unsigned int x=-1;
int y;
y = ~0;
if(x == y)
printf("same");
else
printf("not same");
}
a) same,
MAXINT, -1
b) not same, MAXINT, -MAXINT
c) same , MAXUNIT, -1
d) same, MAXUNIT, MAXUNIT
e) not same, MAXINT, MAXUNIT
Ans. (a)
24. What will
be the result of the following program ?
char *gxxx()
{static char xxx[1024];
return xxx;
}
main()
{char *g="string";
strcpy(gxxx(),g);
g = gxxx();
strcpy(g,"oldstring");
printf("The string is : %s",gxxx());
}
a) The string
is : string
b) The string is :Oldstring
c) Run time error/Core dump
d) Syntax error during compilation
e) None of these
Ans. (b)
25. Find the output for the following C program
main()
{
char *p1="Name";
char *p2;
p2=(char *)malloc(20);
while(*p2++=*p1++);
printf("%s\n",p2);
}
Ans. An
empty string
26. Find the output for the following C program
main()
{
int x=20,y=35;
x = y++ + x++;
y = ++y + ++x;
printf("%d %d\n",x,y);
}
Ans. 57 94
27. Find the output for the following C program
main()
{
int x=5;
printf("%d %d %d\n",x,x<<2,x>>2);
}
Ans. 5 20 1
28 Find the
output for the following C program
#define
swap1(a,b) a=a+b;b=a-b;a=a-b;
main()
{
int x=5,y=10;
swap1(x,y);
printf("%d %d\n",x,y);
swap2(x,y);
printf("%d %d\n",x,y);
}
int swap2(int a,int b)
{
int temp;
temp=a;
b=a;
a=temp;
return;
}
Ans. 10 5
29 Find the output for the following C program
main()
{
char *ptr = "Ramco Systems";
(*ptr)++;
printf("%s\n",ptr);
ptr++;
printf("%s\n",ptr);
}
Ans. Samco
Systems
30 Find the output for the following C program
#include
main()
{
char s1[]="Ramco";
char s2[]="Systems";
s1=s2;
printf("%s",s1);
}
Ans.
Compilation error giving it cannot be an modifiable 'lvalue'
31 Find the output for the following C program
#include
main()
{
char *p1;
char *p2;
p1=(char *) malloc(25);
p2=(char *) malloc(25);
strcpy(p1,"Ramco");
strcpy(p2,"Systems");
strcat(p1,p2);
printf("%s",p1);
}
Ans.
RamcoSystems
32. Find the output for the following C
program given that
[1]. The following variable is available in file1.c
static int average_float;
Ans. All the
functions in the file1.c can access the variable
33. Find the output for the following C
program
# define TRUE
0
some code
while(TRUE)
{
some code
}
Ans. This
won't go into the loop as TRUE is defined as 0
34.
struct list{
int x;
struct list *next;
}*head;
the struct head.x =100
Is the above assignment to pointer is
correct or wrong ?
Ans. Wrong
35.What is
the output of the following ?
int i;
i=1;
i=i+2*i++;
printf(%d,i);
Ans. 4
36. FILE
*fp1,*fp2;
fp1=fopen("one","w")
fp2=fopen("one","w")
fputc('A',fp1)
fputc('B',fp2)
fclose(fp1)
fclose(fp2)
}
Find the Error, If Any?
Ans. no
error. But It will over writes on same file.
37. What are
the output(s) for the following ?
38.
#include
char *f()
{char *s=malloc(8);
strcpy(s,"goodbye");
}
main()
{
char *f();
printf("%c",*f()='A'); }
39. #define
MAN(x,y) (x)>(y)?(x):(y)
{int i=10;
j=5;
k=0;
k=MAX(i++,++j);
printf(%d %d %d %d,i,j,k);
}
Ans. 10 5 0
40.
void main()
{
int i=7;
printf("%d",i++*i++);
}
Ans: 56
Top Of Page
C
Questions
Note : All the
programs are tested under Turbo C/C++ compilers.
It is assumed that,
Ř
Programs run under DOS
environment,
Ř
The underlying machine
is an x86 system,
Ř
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).
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 .
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’ combination- for 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";
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