C 语言编程练习，实践，解决方案：指针
2017-01-03 09:54
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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：
#include <stdio.h>
void main(void)
{
int m=10,n,o;
int *z=&m ;
printf("\n\n Pointer : Show the basic declaration of pointer :\n");
printf("-------------------------------------------------------\n");
printf(" Here is m=10, n and o are two integer variable and *z is an integer");
printf("\n\n z stores the address of m = %p\n", z); // z is a pointer so %p would print the address
printf("\n *z stores the value of m = %i\n", *z);
printf("\n &m is the address of m = %p\n", &m); // &m gives the address of the integer variable m
// so %p is the specifier for that address
printf("\n &n stores the address of n = %p\n", &n);
printf("\n &o stores the address of o = %p\n", &o);
printf("\n &z stores the address of z = %p\n\n", &z); // &z gives the address, where the pointer z is
// stored -> still an address -> %p is the right
// specifier
}
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:
#include <stdio.h>
int main()
{
int* ab;
int m;
m=29;
printf("\n\n Pointer : How to handle the pointers in the program :\n");
printf("------------------------------------------------------------\n");
printf(" Here in the declaration ab = int pointer, int m= 29\n\n");
printf(" Address of m : %p\n",&m);
printf(" Value of m : %d\n\n",m);
ab=&m;
printf(" Now ab is assigned with the address of m.\n");
printf(" Address of pointer ab : %p\n",ab);
printf(" Content of pointer ab : %d\n\n",*ab);
m=34;
printf(" The value of m assigned to 34 now.\n");
printf(" Address of pointer ab : %p\n",ab);
printf(" Content of pointer ab : %d\n\n",*ab);
*ab=7;
printf(" The pointer variable ab is assigned the value 7 now.\n");
printf(" Address of m : %p\n",&m);//as ab contain the address of m
//so *ab changed the value of m and now m become 7
printf(" Value of m : %d\n\n",m);
return 0;
}
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:
#include <stdio.h>
void main()
{
int m=300;
float fx = 300.60;
char cht = 'z';
printf("\n\n Pointer : Demonstrate the use of & and * operator :\n");
printf("--------------------------------------------------------\n");
int *pt1;
float *pt2;
char *pt3;
pt1= &m;
pt2=&fx;
pt3=&cht;
printf ( " m = %d\n",m);
printf ( " fx = %f\n",fx);
printf ( " cht = %c\n",cht);
printf("\n Using & operator :\n");
printf("-----------------------\n");
printf ( " address of m = %p\n",&m);
printf ( " address of fx = %p\n",&fx);
printf ( " address of cht = %p\n",&cht);
printf("\n Using & and * operator :\n");
printf("-----------------------------\n");
printf ( " value at address of m = %d\n",*(&m));
printf ( " value at address of fx = %f\n",*(&fx));
printf ( " value at address of cht = %c\n",*(&cht));
printf("\n Using only pointer variable :\n");
printf("----------------------------------\n");
printf ( " address of m = %p\n",pt1);
printf ( " address of fx = %p\n",pt2);
printf ( " address of cht = %p\n",pt3);
printf("\n Using only pointer operator :\n");
printf("----------------------------------\n");
printf ( " value at address of m = %d\n",*pt1);
printf ( " value at address of fx= %f\n",*pt2);
printf ( " value at address of cht= %c\n\n",*pt3);
}
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:
#include <stdio.h>
int main()
{
int fno, sno, *ptr, *qtr, sum;
printf("\n\n Pointer : Add two numbers :\n");
printf("--------------------------------\n");
printf(" Input the first number : ");
scanf("%d", &fno);
printf(" Input the second number : ");
scanf("%d", &sno);
ptr = &fno;
qtr = &sno;
sum = *ptr + *qtr;
printf(" The sum of the entered numbers is : %d\n\n",sum);
return 0;
}
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:
#include <stdio.h>
long addTwoNumbers(long *, long *);
int main()
{
long fno, sno, *ptr, *qtr, sum;
printf("\n\n Pointer : Add two numbers using call by reference:\n");
printf("-------------------------------------------------------\n");
printf(" Input the first number : ");
scanf("%ld", &fno);
printf(" Input the second number : ");
scanf("%ld", &sno);
sum = addTwoNumbers(&fno, &sno);
printf(" The sum of %ld and %ld is %ld\n\n", fno, sno, sum);
return 0;
}
long addTwoNumbers(long *n1, long *n2)
{
long sum;
sum = *n1 + *n2;
return sum;
}
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:
#include <stdio.h>
int main()
{
int arr1[25], i,n;
printf("\n\n Pointer : Store and retrieve elements from an array :\n");
printf("------------------------------------------------------------\n");
printf(" Input the number of elements to store in the array :");
scanf("%d",&n);
printf(" Input %d number of elements in the array :\n",n);
for(i=0;i<n;i++)
{
printf(" element - %d : ",i);
scanf("%d",arr1+i);
}
printf(" The elements you entered are : \n");
for(i=0;i<n;i++)
{
printf(" element - %d : %d \n",i,*(arr1+i));
}
return 0;
}
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:
#include <stdio.h>
#include <stdlib.h>
int main()
{
int i,n;
float *element;
printf("\n\n Pointer : Find the largest element using Dynamic Memory Allocation :\n");
printf("-------------------------------------------------------------------------\n");
printf(" Input total number of elements(1 to 100): ");
scanf("%d",&n);
element=(float*)calloc(n,sizeof(float)); // Memory is allocated for 'n' elements
if(element==NULL)
{
printf(" No memory is allocated.");
exit(0);
}
printf("\n");
for(i=0;i<n;++i)
{
printf(" Number %d: ",i+1);
scanf("%f",element+i);
}
for(i=1;i<n;++i)
{
if(*element<*(element+i))
*element=*(element+i);
}
printf(" The Largest element is : %.2f \n\n",*element);
return 0;
}
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:
#include <stdio.h>
void swapNumbers(int *x,int *y,int *z);
int main()
{
int e1,e2,e3;
printf("\n\n Pointer : Swap elements using call by reference :\n");
printf("------------------------------------------------------\n");
printf(" Input the value of 1st element : ");
scanf("%d",&e1);
printf(" Input the value of 2nd element : ");
scanf("%d",&e2);
printf(" Input the value of 3rd element : ");
scanf("%d",&e3);
printf("\n The value before swapping are :\n");
printf(" element 1 = %d\n element 2 = %d\n element 3 = %d\n",e1,e2,e3);
swapNumbers(&e1,&e2,&e3);
printf("\n The value after swapping are :\n");
printf(" element 1 = %d\n element 2 = %d\n element 3 = %d\n\n",e1,e2,e3);
return 0;
}
void swapNumbers(int *x,int *y,int *z)
{
int tmp;
tmp=*y;
*y=*x;
*x=*z;
*z=tmp;
}
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:
#include <stdio.h>
void findFact(int,int*);
int main()
{
int fact;
int num1;
printf("\n\n Pointer : Find the factorial of a given number :\n");
printf("------------------------------------------------------\n");
printf(" Input a number : ");
scanf("%d",&num1);
findFact(num1,&fact);
printf(" The Factorial of %d is : %d \n\n",num1,fact);
return 0;
}
void findFact(int n,int *f)
{
int i;
*f =1;
for(i=1;i<=n;i++)
*f=*f*i;
}
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:
#include <stdio.h>
int main()
{
char str1[50];
char *pt;
int ctrV,ctrC;
printf("\n\n Pointer : Count the number of vowels and consonants :\n");
printf("----------------------------------------------------------\n");
printf(" Input a string: ");
fgets(str1, sizeof str1, stdin);
//assign address of str1 to pt
pt=str1;
ctrV=ctrC=0;
while(*pt!='\0')
{
if(*pt=='A' ||*pt=='E' ||*pt=='I' ||*pt=='O' ||*pt=='U' ||*pt=='a' ||*pt=='e' ||*pt=='i' ||*pt=='o' ||*pt=='u')
ctrV++;
else
ctrC++;
pt++; //pointer is increasing for searching the next character
}
printf(" Number of vowels : %d\n Number of constant : %d\n",ctrV,ctrC);
return 0;
}
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:
#include <stdio.h>
void main()
{
int *a,i,j,tmp,n;
printf("\n\n Pointer : Sort an array using pointer :\n");
printf("--------------------------------------------\n");
printf(" Input the number of elements to store in the array : ");
scanf("%d",&n);
printf(" Input %d number of elements in the array : \n",n);
for(i=0;i<n;i++)
{
printf(" element - %d : ",i+1);
scanf("%d",a+i);
}
for(i=0;i<n;i++)
{
for(j=i+1;j<n;j++)
{
if( *(a+i) > *(a+j))
{
tmp = *(a+i);
*(a+i) = *(a+j);
*(a+j) = tmp;
}
}
}
printf("\n The elements in the array after sorting : \n");
for(i=0;i<n;i++)
{
printf(" element - %d : %d \n",i+1,*(a+i));
}
printf("\n");
}
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:
#include <stdio.h>
int* findLarger(int*, int*);
void main()
{
int numa=0;
int numb=0;
int *result;
printf("\n\n Pointer : Show a function returning pointer :\n");
printf("--------------------------------------------------\n");
printf(" Input the first number : ");
scanf("%d", &numa);
printf(" Input the second number : ");
scanf("%d", &numb);
result=findLarger(&numa, &numb);
printf(" The number %d is larger. \n\n",*result);
}
int* findLarger(int *n1, int *n2)
{
if(*n1 > *n2)
return n1;
else
return n2;
}
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:
#include <stdio.h>
void main()
{
int arr1[10];
int i,n, sum = 0;
int *pt;
printf("\n\n Pointer : Sum of all elements in an array :\n");
printf("------------------------------------------------\n");
printf(" Input the number of elements to store in the array (max 10) : ");
scanf("%d",&n);
printf(" Input %d number of elements in the array : \n",n);
for(i=0;i<n;i++)
{
printf(" element - %d : ",i+1);
scanf("%d",&arr1[i]);
}
pt = arr1; // pt store the base address of array arr1
for (i = 0; i < n; i++) {
sum = sum + *pt;
pt++;
}
printf(" The sum of array is : %d\n\n", sum);
}
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:
#include <stdio.h>
void main()
{
int n, i, arr1[15];
int *pt;
printf("\n\n Pointer : Print the elements of an array in reverse order :\n");
printf("----------------------------------------------------------------\n");
printf(" Input the number of elements to store in the array (max 15) : ");
scanf("%d",&n);
pt = &arr1[0]; // pt stores the address of base array arr1
printf(" Input %d number of elements in the array : \n",n);
for(i=0;i<n;i++)
{
printf(" element - %d : ",i+1);
scanf("%d",pt);//accept the address of the value
pt++;
}
pt = &arr1[n - 1];
printf("\n The elements of array in reverse order are :");
for (i = n; i > 0; i--)
{
printf("\n element - %d : %d ", i, *pt);
pt--;
}
printf("\n\n");
}
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:
#include <stdio.h>
struct EmpAddress
{
char *ename;
char stname[20];
int pincode;
}
employee={"John Alter","Court Street \n",654134},*pt=&employee;
int main()
{
printf("\n\n Pointer : Show the usage of pointer to structure :\n");
printf("--------------------------------------------------------\n");
printf(" %s from %s \n\n",pt->ename,(*pt).stname);
return 0;
}
19.在C中编写一个程序以显示指向union的指针。
期待输出：
Jhon Mc Jhon Mc
C Code:
#include <stdio.h>
union empAdd
{
char *ename;
char stname[20];
int pincode;
};
int main()
{
printf("\n\n Pointer : Show a pointer to union :\n");
printf("----------------------------------------\n");
union empAdd employee,*pt;
employee.ename="Jhon Mc\0Donald";//assign the string up to null character i.e. '\0'
pt=&employee;
printf(" %s %s\n\n",pt->ename,(*pt).ename);
return 0;
}
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:
#include <stdio.h>
struct employee
{
char *empname;
int empid;
};
int main()
{
printf("\n\n Pointer : Show a pointer to an array which contents are pointer to structure :\n");
printf("-----------------------------------------------------------------------------------\n");
static struct employee emp1={"Jhon",1001},emp2={"Alex",1002},emp3={"Taylor",1003};
struct employee(*arr[])={&emp1,&emp2,&emp3};
struct employee(*(*pt)[3])=&arr;
printf(" Exmployee Name : %s \n",(**(*pt+1)).empname);
printf("---------------- Explanation --------------------\n");
printf("(**(*pt+1)).empname\n");
printf("= (**(*&arr+1)).empname as pt=&arr\n");
printf("= (**(arr+1)).empname from rule *&pt = pt\n");
printf("= (*arr[1]).empname from rule *(pt+i) = pt[i]\n");
printf("= (*&emp2).empname as arr[1] = &emp2\n");
printf("= e2.empname = Alex from rule *&pt = pt\n\n");
printf(" Employee ID : %d\n",(*(*pt+1))->empid);
printf("---------------- Explanation --------------------\n");
printf("(*(*pt+1))-> empid\n");
printf("= (**(*pt+1)).empid from rule -> = (*).\n");
printf("= emp2.empid = 1002\n");
printf("\n\n");
return 0;
}
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:
#include <stdio.h>
int main()
{
char alph[27];
int x;
char *ptr;
printf("\n\n Pointer : Print all the alphabets:\n");
printf("----------------------------------------\n");
ptr = alph;
for(x=0;x<26;x++)
{
*ptr=x+'A';
ptr++;
}
ptr = alph;
printf(" The Alphabets are : \n");
for(x=0;x<26;x++)
{
printf(" %c ", *ptr);
ptr++;
}
printf("\n\n");
return(0);
}
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:
#include <stdio.h>
int main()
{
char str1[50];
char revstr[50];
char *stptr = str1;
char *rvptr = revstr;
int i=-1;
printf("\n\n Pointer : Print a string in reverse order :\n");
printf("------------------------------------------------\n");
printf(" Input a string : ");
scanf("%s",str1);
while(*stptr)
{
stptr++;
i++;
}
while(i>=0)
{
stptr--;
*rvptr = *stptr;
rvptr++;
--i;
}
*rvptr='\0';
printf(" Reverse of the string is : %s\n\n",revstr);
return 0;
}
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