1. The following functions are: ()
int fun(char *x)
{
char * y = x；
while(* y++){ }；
Returns y-x-1;
}
A. Find the length of the rope
B. Find the storage location of the string
C. compare the sizes of two strings.
D. concatenate the string x after the string y.
2. If there are the following descriptions and statements, and 0
int a[ ]={ 1，2，3，4，5，6，7，8，9，0}，*p，I；
p = a；
a.*(a+i)
B. One
c.p+i
d.*(& amp； a[i])
3. The output of the following program is: ()
# include & ltstdio.h & gt
void main()
{
int a[ 10]={ 1，2，3，4，5，6，7，8，9， 10}，* p = a；
printf(" % d " ,*(p+2))；
}
a.3
B.iv.
c. 1
d.2
4. If there are the following definitions and statements, and 0
int a[ ]={ 1，2，3，4，5，6，7，8，9，0}，*p，I；
p = a；
a.& amp(a+ 1)
b.a++
c.*p
d.& ampp[i]
5. It is known that: int a, x =1; The following correct assignment statement is: ()
a.a =(a[ 1]+a[2])/2；
b.a =(a[ 1]+a[2])/2；
c.a = x+2；
D.a= "good";
6. It is known that: int a[]={ 1, 2, 3, 4, 5, 6}, * p = a；; In the following expression, the value of 5 is: ()
a . p+= 3； *(p++)；
b . p+= 5； * p++；
c . p+= 4； * ++ p；
d . p+= 4； ++ * p；
7. If there are the following statements and 0
int x[]={ 1，9， 10，7，32，4}，*ptr=x，k = 1；
a.x++
B. & photoelectric tape reader
c.& ampptr[k]
d.& amp(x+ 1)
8. If: int i, j=7, * p； is specified; P = & Then the statement equivalent to i=j is ()
a . I = * p；
b.* p = * &j；
c.i = & ampj
d . I = * * p；
9. If p 1 and p2 are pointer variables pointing to the same one-dimensional array of type int, and k is an int variable, the statement that cannot be executed correctly is ().
a . k = * p 1+* p2；
b . p2 = k；
c . p 1 = p2；
d . k = * p 1 *(* p2)；
10. If the following function definition is set, the value returned by this function is ().
int * fun(int a)
{
int *t，n；
n = a； t = & ampn；
return t；
}
A. Unavailable storage unit address values
B. Available memory cell address values
The value expressed in c.n.
D. the value in formal parameter a
Two: fill in the blanks
1, give the output result of the following program.
# include & ltstdio.h & gt
void main()
{
int *v，b；
v = & ampb； b = 100； * v+ = b；
printf("%d\n "，b)；
}
2. Description statement "int(* p)()； ();" means
3. Known function statement: void * fun (); So what it means is
4, known: int I；; char * s = " a \ 045+045 \ ' b "； Execute the statement "fot(I = 0；; * s++； i++)； After that, the result of variable I is
5. The following process is to find the sum of two integers and return the result through formal parameters.
int add(intx，int y，z)
{ = x+y； }
6. The output of the following program is as follows
Invalid funds (int *x)
{ printf(" % d \ n " ,++ * x)'； }
Master ()
{
int a = 25
Fun (& AMPA);
}
7. Give the running results of the following programs.
# include & ltstdio.h & gt
void main()
{
int a[ ]={ 1，2，3，4，5 }；
int x，y，* p；
p = & ampa[0]；
x = *(p+2)；
y = *(p+4)；
printf("*p=%d，x=%d，y=%d\n "，*p，x，y)；
Return;
}
8. Give the running results of the following programs.
# include & ltiostream.h & gt
void callbyval(int a，int b，int c)
{
a = 3； b = 2； c = 1；
}
void callbypointer(int* a，int* b，int* c)
{
* a = 3； * b = 2； * c = 1；
}
void call by reference(int & amp； I int& international bank. c)
{
a = 1； b = 2； c = 3；
}
void main()
{
int a= 1，b=2，c = 3；
int & ampa 1 = a；
int & ampb 1 = a；
int & ampc 1 = a；
callbyval(a，b，c)；
cout & lt& lta & lt& ltb & lt& ltc & lt& ltendl
callby pointer(& amp； First, & ampb & amp；; c)；
cout & lt& lta & lt& ltb & lt& ltc & lt& ltendl
callbyreference(a 1，b 1，c 1)；
cout & lt& lta & lt& ltb & lt& ltc & lt& ltendl
}
9. Give the running results of the following programs.
# include & ltiostream.h & gt
void ast(int x，int y，int *cp，int *dp)
{
* CP = x * y；
* DP = x % y；
}
void main()
{
int a，b，c，d；
a = 2； b = 3；
Ast(a, B and ampc, & AMPD);
cout & lt& ltc & lt& ltd & lt& ltendl
}
10, giving the results of the following program.
# include & ltstdio.h & gt
void main()
{
int a= 10，b=0，*pa，* pb
pa = & ampa； pb = & ampb；
printf("%d，%d\n "，a，b)；
printf("%d，%d\n "，*pa，* Pb)；
a = 20b = 30
* pa = a++； * Pb = b++；
printf("%d，%d\n "，a，b)；
printf("%d，%d\n "，*pa，* Pb)；
(* pa)++；
(* Pb)++；
printf("%d，%d\n "，a，b)；
printf("%d，%d\n "，*pa，* Pb)；
}
Three: Short answer questions and programming questions
1. After executing the following statement, determine the value of ab.
int *var，ab；
ab = 100；
var = & ampab；
ab = * var+ 10；
2. What is the value of the expression *-p when the pointer p points as shown in the figure below? (Here, one [0], one [1], one [2] ... are only used as code names of continuous storage units, with low-order address [0] and high-order address [1]).
3. there is a definition: double var, so,
A) What is the definition statement that makes the pointer p point to a variable of type double?
B) What is the assignment statement that makes the pointer p point to the variable var?
C) What is the scanf function call statement that reads data into the variable var through the pointer p?
4. Write a function, find the length of a string, input the string in the main function, and output its length.
5. Programming: Input any character string from the keyboard, and then input the character to be searched. If it exists, it returns the position where it first appears in the string; Otherwise, the output is "not found in the string!" . To realize the same character string, you can input the characters you want to find continuously.
6. Programming: Delete substring from string. Enter a string from the keyboard, and then enter the substring to delete. Finally, the new string after substring is deleted is output.
One: multiple choice questions:
1.A
2.C
3.A
4.D
5.C
6.A
7.C
8.B
9.B
10.A
Two: fill in the blanks
1.200
2.p is a pointer to a function whose return value is an integer.
3. The return value of the 3.fun function is a universal pointer.
4.8
Because the for statement is to find the length of the string, we can know that the result of I is 8.
5.( 1)int * (2)*z
6.26
7.*p= 1 x=3 y=5
8. 123
32 1
32 1
9.6 2
10. 10,0
10,0
2 1,3 1
2 1,3 1
22,32
22,32
Third, Q&A and programming problems.
1. This example defines two variables, the variable ab of type int and the pointer variable var point to type int; Statement var = &；; ab； The address of ab is assigned to var. In the next sentence, *var indicates the content of the unit referred to by var, that is, 100. Then, *var+ 10 is assigned to the variable ab; Then, the value of ab should be 1 10.
2. According to the priority of the operator, the expression *-p = * (-p). Therefore, firstly, the -P operation is performed, so that the pointer P moves to the low address direction by one memory cell, pointing to a[ 1]. Then use the operator "*" to get the data stored at this address, which is 20.
3.( 1) requires that p be defined as a pointer variable pointing to double type, so the definition statement is: double * P；;
(2) To make P point to the variable var, assign the address of var to the pointer P, and the assignment statement is: P = &；; var
(3) The statement of assigning the read data to the variable var with the scanf function is: scanf ("%lf ",&; var)； Since the address of var has been assigned to p in (2), this statement is equivalent to scanf("%lf ",p);
4. The procedure is:
# include & ltstdio.h & gt
int length(char * a)；
void main()
{
int len
char string[20]；
Printf ("Please enter a string:");
scanf("%s "，str)；
Len = length (str);
Printf ("string length is %d", len);
Return;
}
Int length(char* a)/* Find string length function */
{
char * p；
p = a；
int n；
n = 0；
while(*p！ ='\0')
{
n++；
p++；
}
Return (n);
}
5. The procedure is:
# include & ltiostream.h & gt
Char *strchr( char *string, int letter)
{
While((* string! = letters)&&(* string))
string++；
Returns a string;
}
void main()
{
int TT = 1；
while(tt){
char title[64]；
Cout & lt& lt Please enter a string: ";
CIN & gt； & gt title;
char * p；
Cout & lt& lt Please enter the character you are looking for:;
char ch
CIN & gt； & gtch；
p=strchr(title，ch)；
If (*p)
The first occurrence of the search character in the cout & lt& lt "string is:"
other
Cout & lt& lt "There are no searched characters in the string!" & lt& ltendl
Cout & lt& lt "Be your coward! 1-count, 0-exit "< < endl
CIN & gt； & gtTT；
}
}
6. The procedure is:
# include & ltstring.h & gt
# include & ltiostream.h & gt
char* strstr_rep(char *source，char *old)
{
int i，j，k，loc =- 1；
for(I = 0； Source [I]&; & amp(loc = =- 1)； i++)
for(j=i，k = 0； source[j]= = old[k]； j++，k++)
If (! Old [k+ 1])
loc = I；
If (loc! =- 1)
{
for(k = 0； Old [k]; k++)；
for(j=loc，I = loc+k； Source [I]; j++，i++)
source[j]= source[I]；
source[I]= NULL；
}
Return (source);
}
void main()
{
char a[50]；
Cout & lt& lt Please enter a string: ";
CIN & gt； & gta；
char b[30]；
Cout & lt& lt Please enter the substring of the above string: ";
CIN & gt； & gtb；
cout & lt& ltstrstr_rep(a，b)& lt； & ltendl
}