整个测试遵循以下的约定：
u
假定在所有的程序中必须的头文件都已经被正确包含。
考虑如下的数据类型：
u
char
为
1
个字节
u
int
为
4
个字节
u
long int
为
4
个字节
u
float
为
4
个字节
u
double
为个
8
字节
u
long double
为
8
个字节
u
指针为
4
个字节
1. Consider the following program:
1
#include
<
setjmp.h
>
2
static
jmp_buf buf;
3
main()
4
{
5
volatile
int
b;
6
b
=
3
;
7
if
(setjmp(buf)
!=
0
)
8
{
9
printf(
"
%d
"
, b);
10
exit(
0
);
11
}
12
b
=
5
;
13
longjmp(buf ,
1
);
14
}
The output for this program is:
(a) 3
(b) 5
(c) 0
(d) None of the above
2. Consider the following program:
1main()
2{
3 struct node
4 {
5 int a;
6 int b;
7 int c;
8 };
9 struct node s= { 3, 5,6 };
10 struct node *pt = &s;
11 printf("%d" , *(int*)pt);
12}
The output for this program is:
(a) 3
(b) 5
(c) 6
(d) 7
3. Consider the following code segment:
1int foo ( int x , int n)
2{
3 int val;
4 val =1;
5 if (n>0)
6 {
7 if (n%2 == 1) val = val *x;
8 val = val * foo(x*x , n/2);
9 }
10 return val;
11}
What function of x and n is compute by this code segment?
(a) x^n
(b) x*n
(c) n^x
(d) None of the above
4. Consider the following program:
1
main()
2
{
3
int
a[
5
]
=
{
1
,
2
,
3
,
4
,
5
}
;
4
int
*
ptr
=
(
int
*
)(
&
a
+
1
);
5
printf(
"
%d %d
"
,
*
(a
+
1
),
*
(ptr
-
1
) );
6
}
The output for this program is:
(a) 2 2
(b) 2 1
(c) 2 5
(d) None of the above
5. Consider the following program:
1
void
foo(
int
[][
3
] );
2
main()
3
{
4
int
a [
3
][
3
]
=
{
{
1
,
2
,
3
}
,
{
4
,
5
,
6
}
,
{
7
,
8
,
9
}
}
;
5
foo(a);
6
printf(
"
%d
"
, a[
2
][
1
]);
7
}
8
9
void
foo(
int
b[][
3
])
10
{
11
++
b;
12
b[
1
][
1
]
=
9
;
13
}
The output for this program is:
(a) 8
(b) 9
(c) 7
(d) None of the above
6. Consider the following program:
1
main()
2
{
3
int
a, b,c, d;
4
a
=
3
;
5
b
=
5
;
6
c
=
a,b;
7
d
=
(a,b);
8
printf(
"
c=%d
"
,c);
9
printf(
"
d=%d
"
,d);
10
}
11
The output for this program is:
(a) c=3 d=3
(b) c=5 d=3
(c) c=3 d=5
(d) c=5 d=5
7. Consider the following program:
1
main()
2
{
3
int
a[][
3
]
=
{
1
,
2
,
3
,
4
,
5
,
6
}
;
4
int
(
*
ptr)[
3
]
=
a;
5
printf(
"
%d %d
"
,(
*
ptr)[
1
], (
*
ptr)[
2
] );
6
++
ptr;
7
printf(
"
%d %d
"
,(
*
ptr)[
1
], (
*
ptr)[
2
] );
8
}
9
The output for this program is:
(a) 2 3 5 6
(b) 2 3 4 5
(c) 4 5 0 0
(d) None of the above
8. Consider following function
1int *f1(void)
2{
3 int x =10;
4 return(&x);
5}
6
7int *f2(void)
8{
9 int*ptr;
10 *ptr =10;
11 return ptr;
12}
13
14int *f3(void)
15{
16 int *ptr;
17 ptr=(int*) malloc(sizeof(int));
18 return ptr;
19}
20
Which of the above three functions are likely to cause problem with pointers
(a) Only f3
(b) Only f1 and f3
(c) Only f1 and f2
(d) f1 , f2 ,f3
9. Consider the following program:
1main()
2{
3 int i=3;
4 int j;
5 j = sizeof(++i+ ++i);
6 printf("i=%d j=%d", i ,j);
7}
8
The output for this program is:
(a) i=4 j=2
(b) i=3 j=2
(c) i=3 j=4
(d) i=3 j=6
10. Consider the following program:
1void f1(int *, int);
2void f2(int *, int);
3void(*p[2]) ( int *, int);
4main()
5{
6 int a;
7 int b;
8 p[0] = f1;
9 p[1] = f2;
10 a=3;
11 b=5;
12 p[0](&a , b);
13 printf("%d\t %d\t" , a ,b);
14 p[1](&a , b);
15 printf("%d\t %d\t" , a ,b);
16}
17
18void f1( int* p , int q)
19{
20 int tmp;
21 tmp =*p;
22 *p = q;
23 q= tmp;
24}
25
26void f2( int* p , int q)
27{
28 int tmp;
29 tmp =*p;
30 *p = q;
31 q= tmp;
32}
33
The output for this program is:
(a) 5 5 5 5
(b) 3 5 3 5
(c) 5 3 5 3
(d) 3 3 3 3
11. Consider the following program:
1void e(int );
2main()
3{
4 int a;
5 a=3;
6 e(a);
7}
8
9void e(int n)
10{
11 if(n>0)
12 {
13 e(--n);
14 printf("%d" , n);
15 e(--n);
16 }
17}
18
19
The output for this program is:
(a) 0 1 2 0
(b) 0 1 2 1
(c) 1 2 0 1
(d) 0 2 1 1
12. Consider following declaration
1
typedef
int
(
*
test) (
float
*
,
float
*
)
2
test tmp;
3
type of tmp is
(a) Pointer to function of having two arguments that is pointer to float
(b) int
(c) Pointer to function having two argument that is pointer to float and return int
(d) None of the above
13. Consider the following program:
1
main()
2
{
3
char
*
p;
4
char
buf[
10
]
=
{
1
,
2
,
3
,
4
,
5
,
6
,
9
,
8
}
;
5
p
=
(buf
+
1
)[
5
];
6
printf(
"
%d
"
, p);
7
}
8
The output for this program is:
(a) 5
(b) 6
(c) 9
(d) None of the above
14. Consider the following program:
1Void f(char**);
2main()
3{
4 char * argv[] ={ "ab" ,"cd" , "ef" ,"gh", "ij" ,"kl" };
5 f( argv );
6}
7
8void f( char **p )
9{
10 char* t;
11 t= (p+= sizeof(int))[-1];
12 printf( "%s" , t);
13}
14
The output for this program is:
(a) ab
(b) cd
(c) ef
(d) gh
15. Consider the following program:
1#include<stdarg.h>
2int ripple ( int , );
3main()
4{
5 int num;
6 num = ripple ( 3, 5,7);
7 printf( " %d" , num);
8}
9
10int ripple (int n, )
11{
12 int i , j;
13 int k;
14 va_list p;
15 k= 0;
16 j = 1;
17 va_start( p , n);
18 for (; j<n; ++j)
19 {
20 i = va_arg( p , int);
21 for (; i; i &=i-1 )
22 ++k;
23 }
24 return k;
25}
26
27
The output for this program is:
(a) 7
(b) 6
(c) 5
(d) 3
16. Consider the following program:
1int counter (int i)
2{
3 static int count =0;
4 count = count +i;
5 return (count );
6}
7
8main()
9{
10 int i , j;
11 for (i=0; i <=5; i++)
12 j = counter(i);
13}
14
The value of j at the end of the execution of the this program is:
(a) 10
(b) 15
(c) 6
(d) 7
Answer With Detailed Explanation
_____________________________________________________________
Answer 1.
The answer is (b)
volatile variable isn't affected by the optimization. Its value after the longjump is the last value variable assumed.
b last value is 5 hence 5 is printed.
setjmp : Sets up for nonlocal goto /* setjmp.h*/
Stores context information such as register values so that the lomgjmp function can return control to the statement following the one calling setjmp.Returns 0 when it is initially called.
Lonjjmp: longjmp Performs nonlocal goto /* setjmp.h*/
Transfers control to the statement where the call to setjmp (which initialized buf) was made. Execution continues at this point as if longjmp cannot return the value 0.A nonvolatile automatic variable might be changed by a call to longjmp.When you use setjmp and longjmp, the only automatic variables guaranteed to remain valid are those declared volatile.
Note: Test program without volatile qualifier (result may very)
Answer 2.
The answer is (a)
The members of structures have address in increasing order of their declaration. If a pointer to a structure is cast to the type of a pointer to its first member, the result refers to the first member.
Answer 3.
The answer is (a)
Non recursive version of the program
1
int
what (
int
x ,
int
n)
2
{
3
int
val;
4
int
product;
5
product
=
1
;
6
val
=
x;
7
while
(n
>
0
)
8
{
9
if
(n
%
2
==
1
)
10
product
=
product
*
val;
11
n
=
n
/
2
;
12
val
=
val
*
val;
13
}
14
}
15
/* Code raise a number (x) to a large power (n) using binary doubling strategy */
Algorithm description
1
(
while
n
>
0
)
2
{
3
if
next most significant binary digit of n( power)
is
one
4
then multiply accumulated product by current val ,
5
reduce n(power) sequence by a factor of two
using
integer division .
6
get
next val by multiply current value of itself
7
}
8
Answer 4.
The answer is (c)
type of a is array of int
type of &a is pointer to array of int
Taking a pointer to the element one beyond the end of an array is sure to work.
Answer 5.
The answer is (b)
Answer 6.
The answer is (c)
The comma separates the elements of a function argument list. The comma is also used as an operator in comma expressions. Mixing the two uses of comma is legal, but you must use parentheses to distinguish them. the left operand E1 is evaluated as a void expression, then E2 is evaluated to give the result and type of the comma expression. By recursion, the expression
E1, E2, ..., En
results in the left-to-right evaluation of each Ei, with the value and type of En giving the result of the whole expression.
1
c
=
a,b;
/
*
yields c
=
a
*
/
2
d
=
(a,b);
/**/
/*
d =b
*/
3
Answer 7.
The answer is (a)
/* ptr is pointer to array of 3 int */
Answer 8.
The answer is (c)
f1 and f2 return address of local variable ,when function exit local variable disappeared
Answer 9.
The answer is (c)
sizeof operator gives the number of bytes required to store an object of the type of its operand . The operands is either an expression, which is not evaluated ( (++i + ++ i ) is not evaluated so i remain 3 and j is sizeof int that is 2) or a parenthesized type name.
Answer 10.
The answer is (a)
1
void
(
*
p[
2
]) (
int
*
,
int
);
define array of pointer to function accept two argument that is pointer to int and return int. p[0] = f1; p[1] = f2 contain address of function .function name without parenthesis represent address of function Value and address of variable is passed to function only argument that is effected is a (address is passed). Because of call by value f1, f2 can not effect b
Answer 11.
The answer is (a)
Answer 12.
The answer is (c)
C provide a facility called typedef for creating new data type names, for example declaration
1
typedef
char
string
Makes the name string a synonym for int .The type string can be used in declaration, cast, etc, exactly the same way that the type int can be. Notice that the type being declared in a typedef appears in the position of a variable name not after the word typedef.
Answer 13.
The answer is (c)
If the type of an expression is "array of T" for some type T, then the value of the expression is a pointer to the first object in the array, and the type of the expression is altered to "pointer to T"
So (buf+1)[5] is equvalent to *(buf +6) or buf[6]
Answer 14.
The answer is (d)
p+=sizeof(int) point to argv[2]
(p+=sizeof(int))[-1] points to argv[1]
Answer 15.
The answer is (c)
When we call ripple value of the first argument passed to ripple is collected in the n that is 3. va_start initialize p to point to first unnamed argument that is 5 (first argument).Each call of va_arg return an argument and step p to the next argument. va_arg uses a type name to determine what type to return and how big a step to take Consider inner loop
1
(; i; i
&=
i
-
1
) k
++
/**/
/*
count number of 1 bit in i *
in five number of 1 bits is (101) 2
in seven number of 1 bits is (111) 3
hence k return 5
example
1
let i
=
9
=
1001
2
i
-
1
=
1000
3
(i
-
1
)
+
1
=
i
4
1000
5
+
1
6
1
001
The right most 1 bit of i has corresponding 0 bit in i-1 this way i & i-1, in a two complement number system will delete the right most 1 bit I(repeat until I become 0 gives number of 1 bits)
Answer 16.
The answer is (b)
Static variable count remain in existence rather than coming and going each time function is called
so first call counter(0) count =0
second call counter(1) count = 0+1;
third call counter(2) count = 1+2; /* count = count +i */
fourth call counter(3) count = 3+3;
fifth call counter(4) count = 6+4;
sixth call counter(5) count = 10+5;