51.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).
52. What are the files which are automatically
opened when a C file is executed?
Answer:
stdin, stdout, stderr (standard input,standard output,standard
error).
53. 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.
54. 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.
55. 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.
56. 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.
57. 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.
58. 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.
59. 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.
60. #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.
61. 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".
62. 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.
63. #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.
64. 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.
65. 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.
66. 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.
67. main() {
const int i=4;
float j;
j = ++i;
printf("%d %f",
i,++j);
}
Answer:
Compiler error
Explanation:
i is a constant. you cannot change the value of constant
68. main(){
int a[2][2][2] = { {10,2,3,4}, {5,6,7,8} };
int *p,*q;
p=&a[2][2][2];
*q=***a;
printf("%d..%d",*p,*q);
}
Answer:
garbagevalue..1
Explanation:
p=&a[2][2][2] you
declare only two 2D arrays. but you are trying to access the third 2D(which you
are not declared) it will print garbage values. *q=***a starting address of a
is assigned integer pointer. now q is pointing to starting address of a.if you
print *q meAnswer:it will print first element of 3D array.
69. main() {
register i=5;
char j[]= "hello";
printf("%s
%d",j,i);
}
Answer:
hello 5
Explanation:
if you declare i as register
compiler will treat it as ordinary integer and it will take integer
value. i value may be stored either in register or in memory.
70. main(){
int i=5,j=6,z;
printf("%d",i+++j);
}
Answer:
11
Explanation:
The expression i+++j is treated as (i++ + j).
71. struct aaa{
struct aaa *prev;
int i;
struct aaa *next;
};
main(){
struct aaa abc,def,ghi,jkl;
int x=100;
abc.i=0;abc.prev=&jkl;
abc.next=&def;
def.i=1;def.prev=&abc;def.next=&ghi;
ghi.i=2;ghi.prev=&def;
ghi.next=&jkl;
jkl.i=3;jkl.prev=&ghi;jkl.next=&abc;
x=abc.next->next->prev->next->i;
printf("%d",x);
}
Answer:
2
Explanation:
above all statements form a double circular linked
list;
abc.next->next->prev->next->i
this one points to "ghi" node the value of at particular
node is 2.
72. struct point{
int x;
int y;
};
struct
point origin,*pp;
main(){
pp=&origin;
printf("origin is(%d%d)\n",(*pp).x,(*pp).y);
printf("origin is (%d%d)\n",pp->x,pp->y);
}
Answer:
origin is(0,0)
origin is(0,0)
Explanation:
pp is a pointer to structure. we can access the elements of the
structure either with arrow mark or with indirection operator.
Note:
Since structure point is
globally declared x & y are initialized as zeroes
73. main(){
int i=_l_abc(10);
printf("%d\n",--i);
}
int
_l_abc(int i){
return(i++);
}
Answer:
9
Explanation:
return(i++) it will first return i and then increments. i.e. 10
will be returned.
74. main(){
char *p;
int *q;
long *r;
p=q=r=0;
p++;
q++;
r++;
printf("%p...%p...%p",p,q,r);
}
Answer:
0001...0002...0004
Explanation:
++ operator when applied to
pointers increments address according to their corresponding data-types.
75. main(){
char c=' ',x,convert(z);
getc(c);
if((c>='a') && (c<='z'))
x=convert(c);
printf("%c",x);
}
convert(z){
return z-32;
}
Answer:
Compiler error
Explanation:
Declaration of convert and format of getc() are wrong.
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