整测试遵循约定：
u 假定程序中必须头文件已正确包含
考虑数类型：
u char 1字节
u int 4字节
u long int 4字节
u float 4字节
u double 8字节
u long double 8字节
u 指针4字节
1 Consider the following program
#include
static jmp_buf buf
main()
{
volatile int b
b 3
if(setjmp(buf)0)
{
printf(d b)
exit(0)
}
b5
longjmp(buf 1)
}
The output for this program is
(a) 3
(b) 5
(c) 0
(d) None of the above
2 Consider the following program
main()
{
struct node
{
int a
int b
int c
}
struct node s { 3 56 }
struct node *pt &s
printf(d *(int*)pt)
}
The output for this program is
(a) 3
(b) 5
(c) 6
(d) 7
3 Consider the following code segment
int foo ( int x int n)
{
int val
val 1
if (n>0)
{
if (n2 1) val val *x
val val * foo(x*x n2)
}
return val
}
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
main()
{
int a[5]
int *ptr (int*)(&a+1)
printf(d d *(a+1) *(ptr1) )
}
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
void foo(int [][3] )
main()
{
int a [3][3] { { 123} { 456}}
foo(a)
printf(d a[2][1])
}
void foo( int b[][3])
{
++ b
b[1][1] 9
}
The output for this program is
(a) 8
(b) 9
(c) 7
(d) None of the above
6 Consider the following program
main()
{
int a bc d
a3
b5
cab
d(ab)
printf(cd c)
printf(dd d)
}
The output for this program is
(a) c3 d3
(b) c5 d3
(c) c3 d5
(d) c5 d5
7 Consider the following program
main()
{
int a[][3] { 123 456}
int (*ptr)[3] a
printf(d d (*ptr)[1] (*ptr)[2] )
++ptr
printf(d d (*ptr)[1] (*ptr)[2] )
}
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
int *f1(void)
{
int x 10
return(&x)
}
int *f2(void)
{
int*ptr
*ptr 10
return ptr
}
int *f3(void)
{
int *ptr
ptr(int*) malloc(sizeof(int))
return ptr
}
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
main()
{
int i3
int j
j sizeof(++i+ ++i)
printf(id jd i j)
}
The output for this program is
(a) i4 j2
(b) i3 j2
(c) i3 j4
(d) i3 j6
10 Consider the following program
void f1(int * int)
void f2(int * int)
void(*p[2]) ( int * int)
main()
{
int a
int b
p[0] f1
p[1] f2
a3
b5
p[0](&a b)
printf(dt dt a b)
p[1](&a b)
printf(dt dt a b)
}
void f1( int* p int q)
{
int tmp
tmp *p
*p q
q tmp
}
void f2( int* p int q)
{
int tmp
tmp *p
*p q
q tmp
}
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
void e(int )
main()
{
int a
a3
e(a)
}
void e(int n)
{
if(n>0)
{
e(n)
printf(d n)
e(n)
}
}
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
typedef int (*test) ( float * float*)
test tmp
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
main()
{
char *p
char buf[10] { 12345698}
p (buf+1)[5]
printf(d p)
}
The output for this program is
(a) 5
(b) 6
(c) 9
(d) None of the above
14 Consider the following program
Void f(char**)
main()
{
char * argv[] { ab cd ef gh ij kl }
f( argv )
}
void f( char **p )
{
char* t
t (p+ sizeof(int))[1]
printf( s t)
}
The output for this program is
(a) ab
(b) cd
(c) ef
(d) gh
15 Consider the following program
#include
int ripple ( int )
main()
{
int num
num ripple ( 3 57)
printf( d num)
}
int ripple (int n )
{
int i j
int k
va_list p
k 0
j 1
va_start( p n)
for ( j {
i va_arg( p int)
for ( i i &i1 )
++k
}
return k
}
The output for this program is
(a) 7
(b) 6
(c) 5
(d) 3
16 Consider the following program
int counter (int i)
{
static int count 0
count count +i
return (count )
}
main()
{
int i j
for (i0 i <5 i++)
j counter(i)
}
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 * setjmph*
Stores context information such as register values so that the lomgjmp function can return control to the statement following the one calling setjmpReturns 0 when it is initially called
Lonjjmp longjmp Performs nonlocal goto * setjmph*
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 0A nonvolatile automatic variable might be changed by a call to longjmpWhen 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
int what ( int x int n)
{
int val
int product
product 1
val x
while(n>0)
{
if (n2 1)
product product*val
n n2
val val* val
}
}
* Code raise a number (x) to a large power (n) using binary doubling strategy *
Algorithm description
(while n>0)
{
if next most significant binary digit of n( power) is one
then multiply accumulated product by current val
reduce n(power) sequence by a factor of two using integer division
get next val by multiply current value of itself
}
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 lefttoright evaluation of each Ei with the value and type of En giving the result of the whole expression
cab *yields ca*
d(ab) * d b *
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)
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
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
( i i&i1) 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
let i 9 1001
i1 1000
(i1) +1 i
1000
+1
1 001
The right most 1 bit of i has corresponding 0 bit in i1 this way i & i1 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
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u 假定程序中必须头文件已正确包含
考虑数类型：
u char 1字节
u int 4字节
u long int 4字节
u float 4字节
u double 8字节
u long double 8字节
u 指针4字节
1 Consider the following program
#include
static jmp_buf buf
main()
{
volatile int b
b 3
if(setjmp(buf)0)
{
printf(d b)
exit(0)
}
b5
longjmp(buf 1)
}
The output for this program is
(a) 3
(b) 5
(c) 0
(d) None of the above
2 Consider the following program
main()
{
struct node
{
int a
int b
int c
}
struct node s { 3 56 }
struct node *pt &s
printf(d *(int*)pt)
}
The output for this program is
(a) 3
(b) 5
(c) 6
(d) 7
3 Consider the following code segment
int foo ( int x int n)
{
int val
val 1
if (n>0)
{
if (n2 1) val val *x
val val * foo(x*x n2)
}
return val
}
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
main()
{
int a[5]
int *ptr (int*)(&a+1)
printf(d d *(a+1) *(ptr1) )
}
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
void foo(int [][3] )
main()
{
int a [3][3] { { 123} { 456}}
foo(a)
printf(d a[2][1])
}
void foo( int b[][3])
{
++ b
b[1][1] 9
}
The output for this program is
(a) 8
(b) 9
(c) 7
(d) None of the above
6 Consider the following program
main()
{
int a bc d
a3
b5
cab
d(ab)
printf(cd c)
printf(dd d)
}
The output for this program is
(a) c3 d3
(b) c5 d3
(c) c3 d5
(d) c5 d5
7 Consider the following program
main()
{
int a[][3] { 123 456}
int (*ptr)[3] a
printf(d d (*ptr)[1] (*ptr)[2] )
++ptr
printf(d d (*ptr)[1] (*ptr)[2] )
}
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
int *f1(void)
{
int x 10
return(&x)
}
int *f2(void)
{
int*ptr
*ptr 10
return ptr
}
int *f3(void)
{
int *ptr
ptr(int*) malloc(sizeof(int))
return ptr
}
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
main()
{
int i3
int j
j sizeof(++i+ ++i)
printf(id jd i j)
}
The output for this program is
(a) i4 j2
(b) i3 j2
(c) i3 j4
(d) i3 j6
10 Consider the following program
void f1(int * int)
void f2(int * int)
void(*p[2]) ( int * int)
main()
{
int a
int b
p[0] f1
p[1] f2
a3
b5
p[0](&a b)
printf(dt dt a b)
p[1](&a b)
printf(dt dt a b)
}
void f1( int* p int q)
{
int tmp
tmp *p
*p q
q tmp
}
void f2( int* p int q)
{
int tmp
tmp *p
*p q
q tmp
}
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
void e(int )
main()
{
int a
a3
e(a)
}
void e(int n)
{
if(n>0)
{
e(n)
printf(d n)
e(n)
}
}
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
typedef int (*test) ( float * float*)
test tmp
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
main()
{
char *p
char buf[10] { 12345698}
p (buf+1)[5]
printf(d p)
}
The output for this program is
(a) 5
(b) 6
(c) 9
(d) None of the above
14 Consider the following program
Void f(char**)
main()
{
char * argv[] { ab cd ef gh ij kl }
f( argv )
}
void f( char **p )
{
char* t
t (p+ sizeof(int))[1]
printf( s t)
}
The output for this program is
(a) ab
(b) cd
(c) ef
(d) gh
15 Consider the following program
#include
int ripple ( int )
main()
{
int num
num ripple ( 3 57)
printf( d num)
}
int ripple (int n )
{
int i j
int k
va_list p
k 0
j 1
va_start( p n)
for ( j
i va_arg( p int)
for ( i i &i1 )
++k
}
return k
}
The output for this program is
(a) 7
(b) 6
(c) 5
(d) 3
16 Consider the following program
int counter (int i)
{
static int count 0
count count +i
return (count )
}
main()
{
int i j
for (i0 i <5 i++)
j counter(i)
}
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 * setjmph*
Stores context information such as register values so that the lomgjmp function can return control to the statement following the one calling setjmpReturns 0 when it is initially called
Lonjjmp longjmp Performs nonlocal goto * setjmph*
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 0A nonvolatile automatic variable might be changed by a call to longjmpWhen 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
int what ( int x int n)
{
int val
int product
product 1
val x
while(n>0)
{
if (n2 1)
product product*val
n n2
val val* val
}
}
* Code raise a number (x) to a large power (n) using binary doubling strategy *
Algorithm description
(while n>0)
{
if next most significant binary digit of n( power) is one
then multiply accumulated product by current val
reduce n(power) sequence by a factor of two using integer division
get next val by multiply current value of itself
}
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 lefttoright evaluation of each Ei with the value and type of En giving the result of the whole expression
cab *yields ca*
d(ab) * d b *
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)
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
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
( i i&i1) 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
let i 9 1001
i1 1000
(i1) +1 i
1000
+1
1 001
The right most 1 bit of i has corresponding 0 bit in i1 this way i & i1 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
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《香当网》用户分享的内容，不代表《香当网》观点或立场，请自行判断内容的真实性和可靠性！
该内容是文档的文本内容，更好的格式请下载文档