26. main(){
clrscr();
}
clrscr();
Answer:
No output/error
Explanation:
The first clrscr() occurs inside a function. So it becomes a
function call. In the second clrscr(); is a function declaration (because it is
not inside any function).
27. enum colors {BLACK,BLUE,GREEN}
main(){
printf("%d..%d..%d",BLACK,BLUE,GREEN);
return(1);
}
Answer:
0..1..2
Explanation:
enum assigns numbers starting from 0, if not
explicitly defined.
28. void main(){
char far *farther,*farthest;
printf("%d..%d",sizeof(farther),sizeof(farthest));
}
Answer:
4..2
Explanation:
The
second pointer is of char type and not a far pointer
29. main(){
int i=400,j=300;
printf("%d..%d");
}
Answer:
400..300
Explanation:
printf takes the values of the first two assignments
of the program. Any number of printf's may be given. All of them take
only the first two values. If more number of assignments given in the
program,then printf will take garbage values.
30. main(){
char *p;
p="Hello";
printf("%c\n",*&*p);
}
Answer:
H
Explanation:
* is a dereference operator & is a reference operator. They can be applied any number of times provided it is
meaningful. Here p points to the first character in the string
"Hello". *p dereferences it and so its value is H. Again & references it to an address and *
dereferences it to the value H.
31. main(){
int i=1;
while (i<=5){
printf("%d",i);
if (i>2)
goto here;
i++;
}
}
fun(){
here:
printf("PP");
}
Answer:
Compiler error: Undefined label 'here' in function main
Explanation:
Labels have functions scope, in other words The
scope of the labels is limited to functions . The label 'here' is
available in function fun() Hence it is not visible in function main.
32. main(){
static char names[5][20]={"pascal","ada","cobol","fortran","perl"};
int i;
char *t;
t=names[3];
names[3]=names[4];
names[4]=t;
for (i=0;i<=4;i++)
printf("%s",names[i]);
}
Answer:
Compiler error: Lvalue required in function main
Explanation:
Array names are pointer constants. So it cannot be
modified.
33. void main(){
int i=5;
printf("%d",i++ + ++i);
}
Answer:
Output Cannot be predicted
exactly.
Explanation:
Side effects are involved in the evaluation of i.
34. void main(){
int i=5;
printf("%d",i+++++i);
}
Answer:
Compiler Error
Explanation:
The expression i+++++i is parsed as i ++ ++ + i which is an
illegal combination of operators.
35. main(){
int i=1,j=2;
switch(i){
case 1:
printf("GOOD");
break;
case j:
printf("BAD");
break;
}
}
Answer:
Compiler Error: Constant expression required in function main.
Explanation:
The case statement can have only constant expressions (this
implies that we cannot use variable names directly so an error).
Note:
Enumerated types can be used in case statements.
36. main(){
int i;
printf("%d",scanf("%d",&i)); // value 10 is given as input here
}
Answer:
1
Explanation:
Scanf returns number of items successfully read.Here
10 is given as input which should have been scanned successfully. So number of
items read is 1.
37. #define f(g,g2) g##g2
main(){
int var12=100;
printf("%d",f(var,12));
}
Answer:
100
38. main(){
int i=0;
for(;i++;printf("%d",i)) ;
printf("%d",i);
}
Answer:
1
Explanation:
Before entering into the for loop the checking condition is
"evaluated". Here it evaluates to 0 (false) and comes out of the
loop, and i is incremented (note the semicolon after the for loop).
39. main(){
extern int i;
i=20;
printf("%d",sizeof(i));
}
Answer:
Linker error: undefined symbol '_i'.
Explanation:
extern declaration specifies that the variable i is
defined somewhere else. The compiler passes the external variable to be
resolved by the linker. So compiler doesn't find an error. During linking the
linker searches for the definition of i. Since it is not found the linker flags
an error.
40. main(){
printf("%d", out);
}
int
out=100;
Answer:
Compiler error: undefined symbol out in function main.
Explanation:
The rule is that a variable is available for use from the point of
declaration. Even though a is a global variable, it is not available for main.
Hence an error.
41. main(){
extern out;
printf("%d", out);
}
int
out=100;
Answer:
100
Explanation:
This is the correct way of writing the previous program.
42. 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().
43.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.
44. main( ){
int a[ ] = {10,20,30,40,50},j,*p;
for(j=0; j<5; j++){
printf(“%d” ,*;
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.
45. 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.
46. 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.
47. 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’.
48. 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’.
49.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.
50. 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.
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