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C Question Paper-7
Posted Date: 05 Jun 2008 Resource Type: Articles/Knowledge Sharing Category: Placement Papers
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Posted By: ADITYA Member Level: Gold
Rating:  Points: 1
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121) void main()
{
if(~0 == (unsigned int)-1)
printf(“You can answer this if you know how values are represented in
memory”);
}
Answer
You can answer this if you know how values are represented in
memory
Explanation
~ (tilde operator or bit-wise negation operator) operates on 0 to
produce all ones to fill the space for an integer. –1 is represented in
unsigned value as all 1’s and so both are equal.
122) int swap(int *a,int *b)
{
*a=*a+*b;*b=*a-*b;*a=*a-*b;
}
main()
{
int x=10,y=20;
swap(&x,&y);
printf("x= %d y = %d\n",x,y);
}
Answer
x = 20 y = 10
Explanation
This is one way of swapping two values. Simple checking will help
understand this.
123) main()
{
char *p = “ayqm”;
printf(“%c”,++*(p++));
}
Answer:
b
124) main()
{
int i=5;
printf("%d",++i++);
}
Answer:
Compiler error: Lvalue required in function main
Explanation:
++i yields an rvalue. For postfix ++ to operate an lvalue is required.
125) main()
{
char *p = “ayqm”;
char c;
c = ++*p++;
printf(“%c”,c);
}
Answer:b
Explanation:
There is no difference between the expression ++*(p++) and ++*p++.
Parenthesis just works as a visual clue for the reader to see which
expression is first evaluated.
126)
int aaa() {printf(“Hi”);}
int bbb(){printf(“hello”);}
iny ccc(){printf(“bye”);}
main()
{
int ( * ptr[3]) ();
ptr[0] = aaa;
ptr[1] = bbb;
ptr[2] =ccc;
ptr[2]();
}
Answer:
bye
Explanation:
int (* ptr[3])() says that ptr is an array of pointers to functions that takes no
arguments and returns the type int. By the assignment ptr[0] = aaa; it means
that the first function pointer in the array is initialized with the address of the
function aaa. Similarly, the other two array elements also get initialized with
the addresses of the functions bbb and ccc. Since ptr[2] contains the address
of the function ccc, the call to the function ptr[2]() is same as calling ccc(). So
it results in printing "bye".
127)
main()
{
int i=5;
printf(“%d”,i=++i ==6);
}
Answer:
1
Explanation:
The expression can be treated as i = (++i==6), because == is of higher
precedence than = operator. In the inner expression, ++i is equal to 6
yielding true(1). Hence the result.
128) main()
{
char p[ ]="%d\n";
p[1] = 'c';
printf(p,65);
}
Answer:
A
Explanation:
Due to the assignment p[1] = ‘c’ the string becomes, “%c\n”. Since this string
becomes the f ormat string for printf and ASCII value of 65 is ‘A’, the same
gets printed.
129) void ( * abc( int, void ( *def) () ) ) ();
Answer::
abc is a ptr to a function which takes 2 parameters .(a). an integer
variable.(b). a ptrto a funtion which returns void. the return type of the
function is void.
Explanation:
Apply the clock-wise rule to find the result.
130) main()
{
while (strcmp(“some”,”some\0”))
printf(“Strings are not equal\n”);
}
Answer:
No output
Explanation:
Ending the string constant with \0 explicitly makes no difference. So “some”
and “some\0” are equivalent. So, strcmp returns 0 (false) hence breaking out
of the while loop.
131) main()
{
char str1[] = {‘s’,’o’,’m’,’e’};
char str2[] = {‘s’,’o’,’m’,’e’,’\0’};
while (strcmp(str1,str2))
printf(“Strings are not equal\n”);
}
Answer:
“Strings are not equal”
“Strings are not equal”
….
Explanation:
If a string constant is initialized explicitly with characters, ‘\0’ is not appended
automatically to the string. Since str1 doesn’t have null termination, it treats
whatever the values that are in the following positions as part of the string
until it randomly reaches a ‘\0’. So str1 and str2 are not the same, hence the
result.
132) main()
{
int i = 3;
for (;i++=0;) printf(“%d”,i);
}
Answer:
Compiler Error: Lvalue required.
Explanation:
As we know that increment operators return rvalues and hence it
cannot appear on the left hand side of an assignment operation.
133) void main()
{
int *mptr, *cptr;
mptr = (int*)malloc(sizeof(int));
printf(“%d”,*mptr);
int *cptr = (int*)calloc(sizeof(int),1);
printf(“%d”,*cptr);
}
Answer:
garbage-value 0
Explanation:
The memory space allocated by malloc is uninitialized, whereas calloc
returns the allocated memory space initialized to zeros.
134) void main()
{
static int i;
while(i<=10)
(i>2)?i++:i--;
printf(“%d”, i);
}
Answer:
32767
Explanation:
Since i is static it is initialized to 0. Inside the while loop the conditional
operator evaluates to false, executing i--. This continues till the integer value
rotates to positive value (32767). The while condition becomes false and
hence, comes out of the while loop, printing the i value.
135) main()
{
int i=10,j=20;
j = i, j?(i,j)?i:j:j;
printf("%d %d",i,j);
}
Answer:
10 10
Explanation:The Ternary operator ( ? : ) is equivalent f or if-then-else statement. So the
question can be written as:
if(i,j)
{
if(i,j)
j = i;
else
j = j;
}
else
j = j;
136) 1. const char *a;
2. char* const a;
3. char const *a;
-Differentiate the above declarations.
Answer:
1. 'const' applies to char * rather than 'a' ( pointer to a constant char )
*a='F' : illegal
a="Hi" : legal
2. 'const' applies to 'a' rather than to the value of a (constant pointer to char )
*a='F' : legal
a="Hi" : illegal
3. Same as 1.
137) main()
{
int i=5,j=10;
i=i&=j&&10;
printf("%d %d",i,j);
}
Answer:1 10
Explanation:
The expression can be written as i=(i&=(j&&10)); The inner expression
(j&&10) evaluates to 1 because j==10. i is 5. i = 5&1 is 1. Hence the result.
138) main()
{
int i=4,j=7;
j = j || i++ && printf("YOU CAN");
printf("%d %d", i, j);
}
Answer:
4 1
Explanation:
The boolean expression needs to be evaluated only till the truth value of the
expression is not known. j is not equal to zero itself means that the
expression’s truth value is 1. Because it is followed by || and true || (anything)
=> true where (anything) will not be evaluated. So the remaining expression
is not evaluated and so the value of i remains the same.
Similarly when && operator is involved in an expression, when any of the
operands become false, the whole expression’s truth value becomes false
and hence the remaining expression will not be evaluated.
false && (anything) => false where (anything) will not be evaluated.
139) main()
{
register int a=2;
printf("Address of a = %d",&a);
printf("Value of a = %d",a);
}
Answer:
Compier Error: '&' on register variable
Rule to Remember:
& (address of ) operator cannot be applied on register variables.
140) main()
{
float i=1.5;
switch(i)
{
case 1: printf("1");
case 2: printf("2");
default : printf("0");
}
}
Answer:
Compiler Error: switch expression not integral
Explanation:
Switch statements can be applied only to integral types.
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Responses
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| Author: Lenin 05 Jun 2008 | Member Level: Diamond Points : 2 | Thanks for sharing these type of information. Surely the students of computer studies will use this.
The beauty is you have given the answer , also have explained the answers i mean with solutions, it is delightful
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