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C語言練習題：指標(C language exercise: Pointer)

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Pointer concepts:
2. Introduction to C Pointers

練習一：基本語法
設計一個C語言程式來呈現指標的語法，例如宣告、取址、取值等。

Exercise 1: Basic Syntax
Design a C program to demonstrate the basic syntax of pointer. Such as declaration, address and value. 

練習一參考解法：
Exercise 1 solution:
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/*
Pointer Basic Syntax
Author: Holan
*/

#include <stdio.h>
#include <stdlib.h>

int main()
{
    int n = 50;

    // declaration
    int *ip;

    // assignment
    ip = &n;

    printf("The value of &n:%X\n", &n);
    printf("The value of n:%i\n", n);

    printf("The value of &ip:%X\n", &ip);
    printf("The value of ip:%X\n", ip);
    printf("The value of *ip:%i\n", *ip);
    return 0;
}


練習二：動態記憶體配置
撰寫用 malloc() 與 free() 來配置記憶體的程式。

Exercise 2: Dynamic Memory Allocation
Write a C program to allocate memory.

練習二參考解法：
Exercise 2 solution:
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/*
Dynamic memory allocation
Author: Holan
*/

#include <stdio.h>
#include <stdlib.h>

int main()
{
    int n;
    int *pI;

    printf("How many integers? ");
    scanf("%d", &n);

    // using malloc to allocate memory
    pI = (int *) malloc(n*sizeof(int));

    if(pI == NULL) // if fail to allocate
    {
        printf("Couldn't allocate memory\n");
        return 0;   // exit
    }

    // Releasing the memory allocated by malloc
    free(pI);

    return 0;
}


練習三：兩數相乘
使用指標的方式，將兩個數字相乘。

Exercise 3: Multiplying two numbers
Multiplying two numbers with pointers.

練習三參考解法：
Exercise 3 solution:
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/*
Multiplying two numbers with pointers
Author: Holan
*/

#include <stdio.h>
#include <stdlib.h>

int main()
{
    int n1, n2;
    int *pn1, *pn2, product;

    printf("Enter num1:");
    scanf("%d", &n1);

    printf("Enter num2:");
    scanf("%d", &n2);

    pn1 = &n1;
    pn2 = &n2;

    product = *pn1 * *pn2;

    printf("The product of %d and %d is %d\n", *pn1, *pn2, product);
    return 0;
}

練習四：指標與陣列
使用指標的語法來取得整數陣列的元素。

Exercise 4: Pointer and array
Using a pointer to access the elements of an integer array.

練習四參考解法：
Exercise 4 solution:
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/*
Pointer and array
Author: Holan
*/

#include <stdio.h>
#include <stdlib.h>

#define N 5
int main()
{
    int a[N] = { 2, 3, 5, 7, 11};
    int *pI = a;

    for(int i = 0; i < N; i++)
        printf("a[%d] = %d\t", i, *(pI + i));
    return 0;
}

練習五：指標運算
在練習四時，使用到指標與整數相加的運算。而本題請使用遞增與遞減來達成與練習四一樣的功能。

Exercise 5: Pointer Arithmetic
In exercise 4, the operation: "a pointer plus a integer" is used. Could we use pointer increment and decrement to do exercise 4?

這篇連結文章可以幫助瞭解指標運算。
Here is a tutorial: "How pointer arithmetic works?"

練習五參考解法：
Exercise 5 solution:
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/*
Pointer Arithmetic: increment and decrement
Author: Holan
*/

#include <stdio.h>
#include <stdlib.h>

#define N 5
int main()
{
    int a[N] = { 2, 3, 5, 7, 11};
    int *pI = a;

    printf("Pointer increment:");
    // pointer increment
    for(int i = 0; i < N; i++)
        printf("a[%d] = %d\t", i, *(pI++));


    printf("\n\nPointer decrement:");
    // pointer decrement
    for(int i = N - 1; i >= 0; i--)
        printf("a[%d] = %d\t", i, *(--pI));

    return 0;
}

練習六：字串長度
使用指標語法來求出所輸入字串的長度。

Exercise 6: The length of a string
Using pointer to calculate the length of a user input string.

練習六參考解法：
Exercise 6 solution:
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/*
Length of a string
Author: Holan
*/

#include <stdio.h>
#include <stdlib.h>

int main()
{
    char str[100];
    int len = 0;

    printf("Enter a string:");

    // Input a string with whitespaces
    scanf("%[^\n]s", str);

    char *pC = str;

    while(*pC != '\0')
    {
        len++;
        pC++;
    }

    printf("The length of the given string[%s] is:%d", str, len);

    return 0;
}


練習七：交換兩數
使用 call by reference 的方式，設計一個可以交換兩數的函式。

Exercise 7: Swap two numbers
Using call by reference to design a function that swaps two numbers.

練習七參考解法：
Exercise 7 solution:
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/*
Swap two numbers
Author: Holan
*/

#include <stdio.h>
#include <stdlib.h>

void swapV(double *x, double *y);

int main()
{
    double n1, n2;

    printf("Enter number 1:");
    scanf("%lf", &n1);

    printf("Enter number 2:");
    scanf("%lf", &n2);

    printf("Before swap n1:%lf, n2:%lf", n1, n2);
    swapV(&n1, &n2);
    printf("\nAfter swap n1:%f, n2:%f", n1, n2);

    return 0;
}

void swapV(double *x, double *y)
{
    double t = *x;
    *x = *y;
    *y = t;
}

練習八：字串合併
使用指標語法來合併兩字串。

Exercise 8: String concatenation
Using pointer to concatenate two strings.

練習八參考解法：
Exercise 8 solution:
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/*
String concatenation
Author: Holan
*/

#include <stdio.h>
#include <stdlib.h>

#define LEN 256
int main()
{
    char str1[LEN];
    char str2[LEN];
    char str3[LEN*2];

    printf("Enter str1:");
    scanf("%[^\n]%*c", str1);

    printf("Enter str2:");
    scanf("%[^\n]%*c", str2);

    char *pC = str1;
    char *pStr = str3;

    while(*pC)
    {
        *pStr = *pC;
        pC++;
        pStr++;
    }

    pC = str2;
    while(*pC)
    {
        *pStr = *pC;
        pC++;
        pStr++;
    }

    // end with '\0'
    *pStr = '\0';
    printf("Str1:%s, Str2:%s, Str3:%s", str1, str2, str3);

    return 0;
}

練習九：空指標
常見的觀念可參考「空指標與NULL」。
請撰寫一程式來練習空指標的語法。

Exercise 9: null pointer 
Please refer to this tutorial: NULL pointer in C.
Design a C program to practice the null pointer syntax.

練習九參考解法：
Exercise 9 solution:
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/*
NULL pointer
Author: Holan
*/

#include <stdio.h>
#include <stdlib.h>

int main()
{
    int *iP = NULL;

    if(iP == NULL)
    {
        printf("The value of iP is %u", iP);

        // Couldn't access to *iP
        // The following code will crush
        printf("The value of *iP is %d", *iP);
    }
    else
    {
        printf("The value of iP is %u", iP);
        printf("The value of *iP is %d", *iP);
    }

    return 0;
}

練習十：函式指標
可參考此篇文章：[C語言] function pointer介紹來了解 function pointer。
請設計一程式來練習 function pointer，例如兩數字的加、減、乘、除等運算。

Exercise 10: Function Pointer
Please refer to this tutorial: Function pointers in C.
Design a C program to practice function pointer. For example, addition, subtraction, multiplication and division of two numbers. 

練習十參考解法：
Exercise 10 solution:
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/*
Function pointers
Author: Holan
*/

#include <stdio.h>
#include <stdlib.h>

double add(double n1, double n2);
double sub(double n1, double n2);
double mul(double n1, double n2);
double divi(double n1, double n2);

int main()
{
    double a = 3.44, b = 9.999;

    double (*op)(double, double) = add;

    printf("The addition of %f and %f is %f\n", a, b, op(a, b));

    op = sub;
    printf("The subtraction of %f and %f is %f\n", a, b, op(a, b));

    op = mul;
    printf("The multiplication of %f and %f is %f\n", a, b, op(a, b));

    op = divi;
    printf("The division of %f and %f is %f\n", a, b, op(a, b));

    return 0;
}

double add(double a, double b) { return a + b;}
double sub(double a, double b) { return a - b;}
double mul(double a, double b) { return a * b;}
double divi(double a, double b) { return a / b;}

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