Note : All the
programs are tested under Turbo C/C++ compilers.
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().
47)
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
Explanation:
The given array is a 3-D one. It can
also be viewed as a 1-D array.
2 4 7 8 3 4 2 2 2 3 3 4
100 102 104 106 108 110 112 114 116
118 120 122
thus, for the first printf statement
a, *a, **a give address of first
element . since the indirection ***a
gives the value. Hence, the
first line of the output.
for the second printf a+1 increases in
the third dimension thus
points to value at 114, *a+1
increments in second dimension thus
points to 104, **a +1 increments the
first dimension thus points to
102 and ***a+1 first gets the value at
first location and then
increments it by 1. Hence, the output.
48)
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 nonmodifiable
lvalue.**
49)
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
Explanation:
Let us consider the array and the two
pointers with some address
a
0 1 2 3 4
100 102 104 106 108
p
100 102 104 106 108
1000 1002 1004 1006 1008
ptr
1000
2000
After execution of the instruction
ptr++ value in ptr becomes 1002,
if scaling factor for integer is 2
bytes. Now ptr – p is value in ptr –
starting location of array p, (1002 –
1000) / (scaling factor) = 1,
*ptr – a = value at address pointed by
ptr – starting value of array
a, 1002 has a value 102 so the value
is (102 – 100)/(scaling
factor) = 1, **ptr is the value stored
in the location pointed by
the pointer of ptr = value pointed by
value pointed by 1002 =
value pointed by 102 = 1. Hence the
output of the firs printf is 1,
1, 1.
After execution of *ptr++ increments
value of the value in ptr by
scaling factor, so it becomes1004.
Hence, the outputs for the
second printf are ptr – p = 2, *ptr –
a = 2, **ptr = 2.
After execution of *++ptr increments
value of the value in ptr by
scaling factor, so it becomes1004.
Hence, the outputs for the third
printf are ptr – p = 3, *ptr – a = 3,
**ptr = 3.
After execution of ++*ptr value in ptr
remains the same, the value
pointed by the value is incremented by
the scaling factor. So the
value in array p at location 1006
changes from 106 10 108,.
Hence, the outputs for the fourth
printf are ptr – p = 1006 – 1000
= 3, *ptr – a = 108 – 100 = 4, **ptr =
4.
50)
main( )
{
char *q;
int j;
for (j=0; j<3; j++) scanf(“%s”
,(q+j));
for (j=0; j<3; j++) printf(“%c”
,*(q+j));
for (j=0; j<3; j++) printf(“%s”
,(q+j));
}
Explanation:
Here we have only one pointer to type
char and since we take input
in the same pointer thus we keep
writing over in the same location,
each time shifting the pointer value
by 1. Suppose the inputs are
MOUSE, TRACK and VIRTUAL. Then for the
first input suppose the
pointer starts at location 100 then
the input one is stored as
M O U S E \0
When the second input is given the
pointer is incremented as j
value becomes 1, so the input is
filled in memory starting from
101.
M T R A C K \0
The third input starts filling from
the location 102
M T V I R T U A L \0
This is the final value stored .
The first printf prints the values at
the position q, q+1 and q+2 =
M T V
The second printf prints three strings
starting from locations q,
q+1, q+2
i.e MTVIRTUAL, TVIRTUAL and VIRTUAL.
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 thefile.
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
Last Updated Date: March 23, 2012
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