- 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().
- 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,***;
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.
- 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 non-modifiable lvalue.
- 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.
- 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.
- 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’.
- 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’.
- 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.
- 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.
- 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).
- What are the files which are automatically opened when a C file is executed?
Answer:
stdin, stdout, stderr
(standard input,standard output,standard error).
- What will be the position of the file marker?
fseek(ptr,0,SEEK_SET);
fseek(ptr,0,SEEK_CUR);
Answer :
The SEEK_SET sets the file
position marker to the starting of the file.
The SEEK_CUR sets the file
position marker to the current position of the file.
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