C Programming Exercises, Practice, Solution : Pointer
1.在C中编写一个程序以显示指针的基本声明。
期待输出：
z sotres the address of m = 0x7ffe97a39854
*z stores the value of m = 10
&m is the address of m = 0x7ffe97a39854
&n stores the address of n = 0x7ffe97a39858
&o stores the address of o = 0x7ffe97a3985c
&z stores the address of z = 0x7ffe97a39860
解决：
1. #include <stdio.h>
2. void main(void)
3. {
4. int m=10,n,o;
5. int *z=&m ;
6.
7. "\n\n Pointer : Show the basic declaration of pointer :\n");
8. "-------------------------------------------------------\n");
9. " Here is m=10, n and o are two integer variable and *z is an integer");
10. "\n\n z stores the address of m = %p\n", z); // z is a pointer so %p would print the address
11. "\n *z stores the value of m = %i\n", *z);
12. "\n &m is the address of m = %p\n", &m); // &m gives the address of the integer variable m
13. // so %p is the specifier for that address
14. "\n &n stores the address of n = %p\n", &n);
15. "\n &o stores the address of o = %p\n", &o);
16. "\n &z stores the address of z = %p\n\n", &z); // &z gives the address, where the pointer z is
17. // stored -> still an address -> %p is the right
18. // specifier
19. }
2.在C中编写一个程序，演示如何处理程序中的指针。
期待输出：
Address of m : 0x7ffcc3ad291c
Value of m : 29
Now ab is assigned with the address of m.
Address of pointer ab : 0x7ffcc3ad291c
Content of pointer ab : 29
The value of m assigned to 34 now.
Address of pointer ab : 0x7ffcc3ad291c
Content of pointer ab : 34
The pointer variable ab is assigned with the value 7 now.
Address of m : 0x7ffcc3ad291c
Value of m : 7
C Code:
1. #include <stdio.h>
2. int main()
3. {
4. int* ab;
5. int m;
6. m=29;
7. "\n\n Pointer : How to handle the pointers in the program :\n");
8. "------------------------------------------------------------\n");
9. " Here in the declaration ab = int pointer, int m= 29\n\n");
10.
11. " Address of m : %p\n",&m);
12. " Value of m : %d\n\n",m);
13. ab=&m;
14. " Now ab is assigned with the address of m.\n");
15. " Address of pointer ab : %p\n",ab);
16. " Content of pointer ab : %d\n\n",*ab);
17. m=34;
18. " The value of m assigned to 34 now.\n");
19. " Address of pointer ab : %p\n",ab);
20. " Content of pointer ab : %d\n\n",*ab);
21. *ab=7;
22. " The pointer variable ab is assigned the value 7 now.\n");
23. " Address of m : %p\n",&m);//as ab contain the address of m
24. //so *ab changed the value of m and now m become 7
25. " Value of m : %d\n\n",m);
26. return 0;
27. }
3.在C中编写程序以演示使用＆（地址of）和*（地址处的值）运算符。
期待输出：
Using & operator :
-----------------------
address of m = 0x7ffea3610bb8
address of fx = 0x7ffea3610bbc
address of cht = 0x7ffea3610bb7
Using &
and * operator :
-----------------------------
value at address of m = 300
value at address of fx = 300.600006
value at address of cht = z
Using only pointer variable :
----------------------------------
address of m = 0x7ffea3610bb8
address of fx = 0x7ffea3610bbc
address of cht = 0x7ffea3610bb7
Using only pointer operator :
----------------------------------
value at address of m = 300
value at address of fx= 300.600006
value at address of cht= z
C Code:
1. #include <stdio.h>
2. void main()
3. {
4. int m=300;
5. float fx = 300.60;
6. char cht = 'z';
7.
8. "\n\n Pointer : Demonstrate the use of & and * operator :\n");
9. "--------------------------------------------------------\n");
10.
11.
12. int *pt1;
13. float *pt2;
14. char *pt3;
15. pt1= &m;
16. pt2=&fx;
17. pt3=&cht;
18. " m = %d\n",m);
19. " fx = %f\n",fx);
20. " cht = %c\n",cht);
21. "\n Using & operator :\n");
22. "-----------------------\n");
23. " address of m = %p\n",&m);
24. " address of fx = %p\n",&fx);
25. " address of cht = %p\n",&cht);
26. "\n Using & and * operator :\n");
27. "-----------------------------\n");
28. " value at address of m = %d\n",*(&m));
29. " value at address of fx = %f\n",*(&fx));
30. " value at address of cht = %c\n",*(&cht));
31. "\n Using only pointer variable :\n");
32. "----------------------------------\n");
33. " address of m = %p\n",pt1);
34. " address of fx = %p\n",pt2);
35. " address of cht = %p\n",pt3);
36. "\n Using only pointer operator :\n");
37. "----------------------------------\n");
38. " value at address of m = %d\n",*pt1);
39. " value at address of fx= %f\n",*pt2);
40. " value at address of cht= %c\n\n",*pt3);
41. }
4.在C中编写一个程序，使用指针添加两个数字。
测试数据：
输入第一个数字：5
输入第二个数字：6
期待输出：
The sum of the entered numbers is : 11
C Code:
1. #include <stdio.h>
2. int main()
3. {
4. int fno, sno, *ptr, *qtr, sum;
5.
6. "\n\n Pointer : Add two numbers :\n");
7. "--------------------------------\n");
8.
9. " Input the first number : ");
10. "%d", &fno);
11. " Input the second number : ");
12. "%d", &sno);
13.
14. ptr = &fno;
15. qtr = &sno;
16.
17. sum = *ptr + *qtr;
18.
19. " The sum of the entered numbers is : %d\n\n",sum);
20.
21. return 0;
22. }
5.在C中编写程序，使用调用引用添加数字。
测试数据：
输入第一个数字：5
输入第二个数字：6
期待输出：
The sum of 5 and 6 is 11
C Code:
1. #include <stdio.h>
2. long addTwoNumbers(long *, long *);
3.
4. int main()
5. {
6. long fno, sno, *ptr, *qtr, sum;
7.
8. "\n\n Pointer : Add two numbers using call by reference:\n");
9. "-------------------------------------------------------\n");
10.
11. " Input the first number : ");
12. "%ld", &fno);
13. " Input the second number : ");
14. "%ld", &sno);
15. sum = addTwoNumbers(&fno, &sno);
16. " The sum of %ld and %ld is %ld\n\n", fno, sno, sum);
17. return 0;
18. }
19. long addTwoNumbers(long *n1, long *n2)
20. {
21. long sum;
22. sum = *n1 + *n2;
23. return sum;
24. }
6.在C中编写一个程序，使用指针找到两个数字之间的最大数目。 转到编辑器
测试数据：
输入第一个数字：5
输入第二个数字：6
期待输出：
6 is the maximum number.
C Code:
#include <stdio.h>
#include <stdlib.h>
void main()
{
int fno,sno,*ptr1=&fno,*ptr2=&sno;
printf("\n\n Pointer : Find the maximum number between two numbers :\n");
printf("------------------------------------------------------------\n");
printf(" Input the first number : ");
scanf("%d", ptr1);
printf(" Input the second number : ");
scanf("%d", ptr2);
if(*ptr1>*ptr2)
{
printf("\n\n %d is the maximum number.\n\n",*ptr1);
}
else
{
printf("\n\n %d is the maximum number.\n\n",*ptr2);
}
}
7.在C中编写一个程序，将n个元素存储在数组中，并使用指针打印元素。
测试数据：
输入要存储在数组中的元素数量：5
输入5数组中的元素数量：
元素 - 0：5
元素-1：7
元素-2：2
元素-3：9
元素-4：8
期待输出：
The elements you entered are :
element - 0 : 5
element - 1 : 7
element - 2 : 2
element - 3 : 9
element - 4 : 8
C Code:
1. #include <stdio.h>
2. int main()
3. {
4. int arr1[25], i,n;
5. "\n\n Pointer : Store and retrieve elements from an array :\n");
6. "------------------------------------------------------------\n");
7. " Input the number of elements to store in the array :");
8. "%d",&n);
9.
10. " Input %d number of elements in the array :\n",n);
11. for(i=0;i<n;i++)
12. {
13. " element - %d : ",i);
14. "%d",arr1+i);
15. }
16. " The elements you entered are : \n");
17. for(i=0;i<n;i++)
18. {
19. " element - %d : %d \n",i,*(arr1+i));
20. }
21. return 0;
22. }
8.在C中编写一个程序，使用指针打印给定字符串的所有排列。
期待输出：
The permutations of the string are :
abcd abdc acbd acdb adcb adbc bacd badc bcad bcda bdca bdac cbad cbda cabd cadb cdab cdba db
ca dbac dcba dcab dacb dabc
C Code:
#include <stdio.h>
#include <string.h>
void changePosition(char *ch1, char *ch2)
{
char tmp;
tmp = *ch1;
*ch1 = *ch2;
*ch2 = tmp;
}
void charPermu(char *cht, int stno, int endno)
{
int i;
if (stno == endno)
printf("%s ", cht);
else
{
for (i = stno; i <= endno; i++)
{
changePosition((cht+stno), (cht+i));
charPermu(cht, stno+1, endno);
changePosition((cht+stno), (cht+i));
}
}
}
int main()
{
char str[] = "abcd";
printf("\n\n Pointer : Generate permutations of a given string :\n");
printf("--------------------------------------------------------\n");
int n = strlen(str);
printf(" The permutations of the string are : \n");
charPermu(str, 0, n-1);
printf("\n\n");
return 0;
}
9.在C中编写一个程序，使用动态内存分配找到最大的元素。
测试数据：
Input total number of elements(1 to 100): 5
Number 1: 5
Number 2: 7
Number 3: 2
Number 4: 9
Number 5: 8
期待输出：
The Largest element is : 9.00
C Code:
1. #include <stdio.h>
2. #include <stdlib.h>
3. int main()
4. {
5. int i,n;
6. float *element;
7. "\n\n Pointer : Find the largest element using Dynamic Memory Allocation :\n");
8. "-------------------------------------------------------------------------\n");
9. " Input total number of elements(1 to 100): ");
10. "%d",&n);
11. float*)calloc(n,sizeof(float)); // Memory is allocated for 'n' elements
12. if(element==NULL)
13. {
14. " No memory is allocated.");
15. exit(0);
16. }
17. "\n");
18. for(i=0;i<n;++i)
19. {
20. " Number %d: ",i+1);
21. "%f",element+i);
22. }
23. for(i=1;i<n;++i)
24. {
25. if(*element<*(element+i))
26. *element=*(element+i);
27. }
28. " The Largest element is : %.2f \n\n",*element);
29. return 0;
30. }
10.在C中编写一个程序，使用指针计算字符串的长度。
测试数据：
输入字符串：w3resource
期待输出：
The length of the given string w3resource
is : 10
C Code:
#include <stdio.h>
int calculateLength(char*);
void main()
{
char str1[25];
int l;
printf("\n\n Pointer : Calculate the length of the string :\n");
printf("---------------------------------------------------\n");
printf(" Input a string : ");
fgets(str1, sizeof str1, stdin);
l = calculateLength(str1);
printf(" The length of the given string %s is : %d ", str1, l-1);
printf("\n\n");
}
int calculateLength(char* ch) // ch = base address of array str1 ( &str1[0] )
{
int ctr = 0;
while (*ch != '\0')
{
ctr++;
ch++;
}
return ctr;
}
11. Write a program in C to swap elements using call by reference. Go to the editor Test Data :
Input the value of 1st element : 5
Input the value of 2nd element : 6
Input the value of 3rd element : 7
期待输出：
The value before swapping are :
element 1 = 5
element 2 = 6
element 3 = 7
The value after swapping are :
element 1 = 7
element 2 = 5
element 3 = 6
C Code:
1. #include <stdio.h>
2. void swapNumbers(int *x,int *y,int *z);
3. int main()
4. {
5. int e1,e2,e3;
6. "\n\n Pointer : Swap elements using call by reference :\n");
7. "------------------------------------------------------\n");
8. " Input the value of 1st element : ");
9. "%d",&e1);
10. " Input the value of 2nd element : ");
11. "%d",&e2);
12. " Input the value of 3rd element : ");
13. "%d",&e3);
14.
15.
16. "\n The value before swapping are :\n");
17. " element 1 = %d\n element 2 = %d\n element 3 = %d\n",e1,e2,e3);
18. swapNumbers(&e1,&e2,&e3);
19. "\n The value after swapping are :\n");
20. " element 1 = %d\n element 2 = %d\n element 3 = %d\n\n",e1,e2,e3);
21. return 0;
22. }
23. void swapNumbers(int *x,int *y,int *z)
24. {
25. int tmp;
26. tmp=*y;
27. *y=*x;
28. *x=*z;
29. *z=tmp;
30. }
12.在C中编写一个程序，使用指针查找给定数字的阶乘。
测试数据：
输入数字：5
期待输出：
The Factorial of 5 is : 120
C Code:
1. #include <stdio.h>
2. void findFact(int,int*);
3. int main()
4. {
5. int fact;
6. int num1;
7. "\n\n Pointer : Find the factorial of a given number :\n");
8. "------------------------------------------------------\n");
9. " Input a number : ");
10. "%d",&num1);
11.
12. findFact(num1,&fact);
13. " The Factorial of %d is : %d \n\n",num1,fact);
14. return 0;
15. }
16.
17. void findFact(int n,int *f)
18. {
19. int i;
20.
21. *f =1;
22. for(i=1;i<=n;i++)
23. *f=*f*i;
24. }
13.在C中编写一个程序，使用指针来计算字符串中元音和辅音的数量。
测试数据：
输入字符串：string
期待输出：
Number of vowels : 1
Number of constant : 5
C Code:
1. #include <stdio.h>
2. int main()
3. {
4. char str1[50];
5. char *pt;
6. int ctrV,ctrC;
7. "\n\n Pointer : Count the number of vowels and consonants :\n");
8. "----------------------------------------------------------\n");
9. " Input a string: ");
10. sizeof str1, stdin);
11.
12. //assign address of str1 to pt
13. pt=str1;
14.
15. ctrV=ctrC=0;
16. while(*pt!='\0')
17. {
18. if(*pt=='A' ||*pt=='E' ||*pt=='I' ||*pt=='O' ||*pt=='U' ||*pt=='a' ||*pt=='e' ||*pt=='i' ||*pt=='o' ||*pt=='u')
19. ctrV++;
20. else
21. ctrC++;
22. //pointer is increasing for searching the next character
23. }
24.
25. " Number of vowels : %d\n Number of constant : %d\n",ctrV,ctrC);
26. return 0;
27. }
14.在C中编写一个程序，使用指针对数组进行排序。
测试数据：
testdata
期待输出：
Test Data :
Input the number of elements to store in the array : 5
Input 5 number of elements in the array :
element - 1 : 25
element - 2 : 45
element - 3 : 89
element - 4 : 15
element - 5 : 82
期待输出：
The elements in the array after sorting :
element - 1 : 15
element - 2 : 25
element - 3 : 45
element - 4 : 82
element - 5 : 89
C Code:
1. #include <stdio.h>
2. void main()
3. {
4. int *a,i,j,tmp,n;
5. "\n\n Pointer : Sort an array using pointer :\n");
6. "--------------------------------------------\n");
7.
8. " Input the number of elements to store in the array : ");
9. "%d",&n);
10.
11. " Input %d number of elements in the array : \n",n);
12. for(i=0;i<n;i++)
13. {
14. " element - %d : ",i+1);
15. "%d",a+i);
16. }
17. for(i=0;i<n;i++)
18. {
19. for(j=i+1;j<n;j++)
20. {
21. if( *(a+i) > *(a+j))
22. {
23. tmp = *(a+i);
24. *(a+i) = *(a+j);
25. *(a+j) = tmp;
26. }
27. }
28. }
29. "\n The elements in the array after sorting : \n");
30. for(i=0;i<n;i++)
31. {
32. " element - %d : %d \n",i+1,*(a+i));
33. }
34. printf("\n");
35. }
15.在C中编写一个程序，显示一个函数返回指针。
测试数据：
Input the first number : 5
Input the second number : 6
期待输出：
The number 6 is larger.
C Code:
1. #include <stdio.h>
2. int* findLarger(int*, int*);
3. void main()
4. {
5. int numa=0;
6. int numb=0;
7. int *result;
8. "\n\n Pointer : Show a function returning pointer :\n");
9. "--------------------------------------------------\n");
10. " Input the first number : ");
11. "%d", &numa);
12. " Input the second number : ");
13. "%d", &numb);
14.
15. result=findLarger(&numa, &numb);
16. " The number %d is larger. \n\n",*result);
17. }
18.
19. int* findLarger(int *n1, int *n2)
20. {
21. if(*n1 > *n2)
22. return n1;
23. else
24. return n2;
25. }
16.在C中编写一个程序，使用指针计算数组中所有元素的总和。
测试数据：
Input the number of elements to store in the array (max 10) : 5
Input 5 number of elements in the array :
element - 1 : 2
element - 2 : 3
element - 3 : 4
element - 4 : 5
element - 5 : 6
期待输出：
The sum of array is : 20
C Code:
1. #include <stdio.h>
2. void main()
3. {
4. int arr1[10];
5. int i,n, sum = 0;
6. int *pt;
7.
8. "\n\n Pointer : Sum of all elements in an array :\n");
9. "------------------------------------------------\n");
10.
11. " Input the number of elements to store in the array (max 10) : ");
12. "%d",&n);
13.
14. " Input %d number of elements in the array : \n",n);
15. for(i=0;i<n;i++)
16. {
17. " element - %d : ",i+1);
18. "%d",&arr1[i]);
19. }
20.
21. // pt store the base address of array arr1
22.
23. for (i = 0; i < n; i++) {
24. sum = sum + *pt;
25. pt++;
26. }
27.
28. " The sum of array is : %d\n\n", sum);
29. }
17.在C中编写一个程序，以相反的顺序打印数组的元素。
测试数据：
Input the number of elements to store in the array (max 15) : 5
Input 5 number of elements in the array :
element - 1 : 2
element - 2 : 3
element - 3 : 4
element - 4 : 5
element - 5 : 6
期待输出：
The elements of array in reverse order are :
element - 5 : 6
element - 4 : 5
element - 3 : 4
element - 2 : 3
element - 1 : 2
C Code:
1. #include <stdio.h>
2. void main()
3. {
4. int n, i, arr1[15];
5. int *pt;
6. "\n\n Pointer : Print the elements of an array in reverse order :\n");
7. "----------------------------------------------------------------\n");
8.
9. " Input the number of elements to store in the array (max 15) : ");
10. "%d",&n);
11. // pt stores the address of base array arr1
12. " Input %d number of elements in the array : \n",n);
13. for(i=0;i<n;i++)
14. {
15. " element - %d : ",i+1);
16. "%d",pt);//accept the address of the value
17. pt++;
18. }
19.
20. pt = &arr1[n - 1];
21.
22. "\n The elements of array in reverse order are :");
23.
24. for (i = n; i > 0; i--)
25. {
26. "\n element - %d : %d ", i, *pt);
27. pt--;
28. }
29. printf("\n\n");
30. }
18.在C中编写一个程序以显示指针对结构的使用。
期待输出：
John Alter from Court Street
C Code:
19.在C中编写一个程序以显示指向union的指针。
期待输出：
1. #include <stdio.h>
2. struct EmpAddress
3. {
4. char *ename;
5. char stname[20];
6. int pincode;
7. }
8. employee={"John Alter","Court Street \n",654134},*pt=&employee;
9.
10. int main()
11. {
12. "\n\n Pointer : Show the usage of pointer to structure :\n");
13. "--------------------------------------------------------\n");
14. " %s from %s \n\n",pt->ename,(*pt).stname);
15. return 0;
16. }
Jhon Mc Jhon Mc
C Code:
1. #include <stdio.h>
2. union empAdd
3. {
4. char *ename;
5. char stname[20];
6. int pincode;
7. };
8.
9. int main()
10. {
11. "\n\n Pointer : Show a pointer to union :\n");
12. "----------------------------------------\n");
13. union empAdd employee,*pt;
14. "Jhon Mc\0Donald";//assign the string up to null character i.e. '\0'
15.
16. pt=&employee;
17.
18. " %s %s\n\n",pt->ename,(*pt).ename);
19.
20. return 0;
21. }
20.在C中编写一个程序，以显示一个指向数组的指针，内容是指向结构的指针。
期待输出：
Exmployee Name : Alex
Employee ID : 1002
C Code:
1. #include <stdio.h>
2. struct employee
3. {
4. char *empname;
5. int empid;
6. };
7.
8. int main()
9. {
10. "\n\n Pointer : Show a pointer to an array which contents are pointer to structure :\n");
11. "-----------------------------------------------------------------------------------\n");
12.
13. static struct employee emp1={"Jhon",1001},emp2={"Alex",1002},emp3={"Taylor",1003};
14. struct employee(*arr[])={&emp1,&emp2,&emp3};
15. struct employee(*(*pt)[3])=&arr;
16.
17. " Exmployee Name : %s \n",(**(*pt+1)).empname);
18. "---------------- Explanation --------------------\n");
19. "(**(*pt+1)).empname\n");
20. "= (**(*&arr+1)).empname as pt=&arr\n");
21. "= (**(arr+1)).empname from rule *&pt = pt\n");
22. "= (*arr[1]).empname from rule *(pt+i) = pt[i]\n");
23. "= (*&emp2).empname as arr[1] = &emp2\n");
24. "= e2.empname = Alex from rule *&pt = pt\n\n");
25. " Employee ID : %d\n",(*(*pt+1))->empid);
26. "---------------- Explanation --------------------\n");
27. "(*(*pt+1))-> empid\n");
28. "= (**(*pt+1)).empid from rule -> = (*).\n");
29. "= emp2.empid = 1002\n");
30. "\n\n");
31. return 0;
32. }
21.在C中编写一个程序，使用指针打印所有的字母表。
期待输出：
The Alphabets are :
A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
C Code:
1. #include <stdio.h>
2.
3. int main()
4. {
5. char alph[27];
6. int x;
7. char *ptr;
8. "\n\n Pointer : Print all the alphabets:\n");
9. "----------------------------------------\n");
10. ptr = alph;
11.
12. for(x=0;x<26;x++)
13. {
14. 'A';
15. ptr++;
16. }
17. ptr = alph;
18.
19. printf(" The Alphabets are : \n");
20. for(x=0;x<26;x++)
21. {
22. " %c ", *ptr);
23. ptr++;
24. }
25. "\n\n");
26. return(0);
27. }
22.在C中编写一个程序，使用指针反向打印一个字符串。
测试数据：
Input a string : w3resource
期待输出：
Reverse of the string is : ecruoser3w
C Code:
1. #include <stdio.h>
2. int main()
3. {
4. char str1[50];
5. char revstr[50];
6. char *stptr = str1;
7. char *rvptr = revstr;
8. int i=-1;
9. "\n\n Pointer : Print a string in reverse order :\n");
10. "------------------------------------------------\n");
11. " Input a string : ");
12. "%s",str1);
13. while(*stptr)
14. {
15. stptr++;
16. i++;
17. }
18. while(i>=0)
19. {
20. stptr--;
21. *rvptr = *stptr;
22. rvptr++;
23. --i;
24. }
25. '\0';
26. " Reverse of the string is : %s\n\n",revstr);
27. return 0;
28. }
