Home
| | Sitemap
||Page number :6
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 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.
C Aptitude questions : 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18