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.
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