70) 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.
71) #include<stdio.h>
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
72) #include<stdio.h>
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.
73) #include<stdio.h>
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.
74) main()
{
int i=5,j=6,z;
printf("%d",i+++j);
}
Answer:
11
Explanation:
the expression i+++j is treated as (i++
+ j)
76) 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.
77) 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
78) 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.
79) 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.
80) 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.
81) main(int argc, char **argv)
{
printf("enter the character");
getchar();
sum(argv[1],argv[2]);
}
sum(num1,num2)
int
num1,num2;
{
return num1+num2;
}
Answer:
Compiler error.
Explanation:
argv[1]
& argv[2] are strings. They are passed to the function sum without
converting it to integer values.
82) # include <stdio.h>
int
one_d[]={1,2,3};
main()
{
int *ptr;
ptr=one_d;
ptr+=3;
printf("%d",*ptr);
}
Answer:
garbage value
Explanation:
ptr pointer is pointing to out of the
array range of one_d.
83) # include<stdio.h>
aaa()
{
printf("hi");
}
bbb(){
printf("hello");
}
ccc(){
printf("bye");
}
main()
{
int (*ptr[3])();
ptr[0]=aaa;
ptr[1]=bbb;
ptr[2]=ccc;
ptr[2]();
}
Answer:
bye
Explanation:
ptr
is array of pointers to functions of return type int.ptr[0] is assigned to
address of the function aaa. Similarly ptr[1] and ptr[2] for bbb and ccc
respectively. ptr[2]() is in effect of writing ccc(), since ptr[2] points to
ccc.
85) #include<stdio.h>
main()
{
FILE
*ptr;
char
i;
ptr=fopen("zzz.c","r");
while((i=fgetch(ptr))!=EOF)
printf("%c",i);
}
Answer:
contents of zzz.c followed by an
infinite loop
Explanation:
The
condition is checked against EOF, it should be checked against NULL.
86) main()
{
int i =0;j=0;
if(i && j++)
printf("%d..%d",i++,j);
printf("%d..%d,i,j);
}
Answer:
0..0
Explanation:
The
value of i is 0. Since this information is enough to determine the truth value
of the boolean expression. So the statement following the if statement is not
executed. The values of i and j remain
unchanged and get printed.
87) main()
{
int i;
i = abc();
printf("%d",i);
}
abc()
{
_AX = 1000;
}
Answer:
1000
Explanation:
Normally
the return value from the function is through the information from the
accumulator. Here _AH is the pseudo global variable denoting the accumulator.
Hence, the value of the accumulator is set 1000 so the function returns value
1000.
88) int i;
main(){
int
t;
for
( t=4;scanf("%d",&i)-t;printf("%d\n",i))
printf("%d--",t--);
}
// If the inputs are 0,1,2,3 find
the o/p
Answer:
4--0
3--1
2--2
Explanation:
Let us assume some x=
scanf("%d",&i)-t the values during execution
will
be,
t i
x
4 0
-4
3 1
-2
2 2
0
89) main(){
int a= 0;int b = 20;char x =1;char y =10;
if(a,b,x,y)
printf("hello");
}
Answer:
hello
Explanation:
The
comma operator has associativity from left to right. Only the rightmost value
is returned and the other values are evaluated and ignored. Thus the value of
last variable y is returned to check in if. Since it is a non zero value if
becomes true so, "hello" will be printed.
90) main(){
unsigned int i;
for(i=1;i>-2;i--)
printf("c
aptitude");
}
Explanation:
i
is an unsigned integer. It is compared with a signed value. Since the both
types doesn't match, signed is promoted to unsigned value. The unsigned
equivalent of -2 is a huge value so condition becomes false and control comes
out of the loop.
91) In the following pgm add a stmt in the function fun such that the address of
'a'
gets stored in 'j'.
main(){
int *
j;
void fun(int **);
fun(&j);
}
void fun(int **k) {
int a =0;
/* add a stmt here*/
}
Answer:
*k = &a
Explanation:
The
argument of the function is a pointer to a pointer.
92) What are the following notations of
defining functions known as?
i. int abc(int a,float b)
{
/* some code */
}
ii. int abc(a,b)
int a; float b;
{
/*
some code*/
}
Answer:
i.
ANSI C notation
ii. Kernighan & Ritche notation
93) main()
{
char
*p;
p="%d\n";
p++;
p++;
printf(p-2,300);
}
Answer:
300
Explanation:
The
pointer points to % since it is incremented twice and again decremented by 2,
it points to '%d\n' and 300 is printed.
94) main(){
char a[100];
a[0]='a';a[1]]='b';a[2]='c';a[4]='d';
abc(a);
}
abc(char
a[]){
a++;
printf("%c",*a);
a++;
printf("%c",*a);
}
Explanation:
The
base address is modified only in function and as a result a points to 'b' then
after incrementing to 'c' so bc will be printed.
95) func(a,b)
int
a,b;
{
return( a= (a==b) );
}
main()
{
int
process(),func();
printf("The
value of process is %d !\n ",process(func,3,6));
}
process(pf,val1,val2)
int
(*pf) ();
int
val1,val2;
{
return((*pf)
(val1,val2));
}
Answer:
The value if process is 0 !
Explanation:
The
function 'process' has 3 parameters - 1, a pointer to another function 2 and 3, integers. When this function is
invoked from main, the following substitutions for formal parameters take place:
func for pf, 3 for val1 and 6 for val2. This function returns the result of the
operation performed by the function 'func'. The function func has two integer
parameters. The formal parameters are substituted as 3 for a and 6 for b. since
3 is not equal to 6, a==b returns 0. therefore the function returns 0 which in
turn is returned by the function 'process'.
96) void main()
{
static int i=5;
if(--i){
main();
printf("%d
",i);
}
}
Answer:
0 0 0 0
Explanation:
The variable
"I" is declared as static, hence memory for I will be allocated for
only once, as it encounters the statement. The function main() will be called
recursively unless I becomes equal to 0, and since main() is recursively
called, so the value of static I ie., 0 will be printed every time the control
is returned.
97) void main()
{
int
k=ret(sizeof(float));
printf("\n
here value is %d",++k);
}
int ret(int ret)
{
ret += 2.5;
return(ret);
}
Answer:
Here value is 7
Explanation:
The
int ret(int ret), ie., the function name and the argument name can be the same.
Firstly, the
function ret() is called in which the sizeof(float) ie., 4 is passed, after the first expression the value in ret
will be 6, as ret is integer hence the value stored in ret will have implicit
type conversion from float to int. The ret is returned in main() it is printed
after and preincrement.
98) void main()
{
char
a[]="12345\0";
int i=strlen(a);
printf("here
in 3 %d\n",++i);
}
Answer:
here in 3 6
Explanation:
The char array 'a'
will hold the initialized string, whose length will be counted from 0 till the
null character. Hence the 'I' will hold the value equal to 5, after the
pre-increment in the printf statement, the 6 will be printed.
99) void main()
{
unsigned giveit=-1;
int gotit;
printf("%u
",++giveit);
printf("%u
\n",gotit=--giveit);
}
Answer:
0 65535
Explanation:
100) void main()
{
int i;
char
a[]="\0";
if(printf("%s\n",a))
printf("Ok
here \n");
else
printf("Forget
it\n");
}
Answer:
Ok here
Explanation:
Printf will return how many characters does it
print. Hence printing a null character returns 1 which makes the if statement
true, thus "Ok here" is printed
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