C 语言指针学习：数组反转和遍历

反转一个数组

用指针法反转数组

int reverseArray() {
    int a[10] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10};
    int *p = a;
    int *q = a + 9;
    int temp;
    while (p < q) {
        temp = *p;
        *p = *q;
        *q = temp;
        p++;
        q--;
    }
    for (int i = 0; i < 10; i++) {
        std::cout << std::setw(3) << a[i];
    }
    std::cout << std::endl;
    return 0;
}

常规的反转数组

// Iterative C++ program to reverse an array
#include <bits/stdc++.h>
using namespace std;

/* Function to reverse arr[] from start to end*/
void rvereseArray(int arr[], int start, int end)
{
	while (start < end)
	{
		int temp = arr[start];
		arr[start] = arr[end];
		arr[end] = temp;
		start++;
		end--;
	}
}

/* Utility function to print an array */
void printArray(int arr[], int size)
{
for (int i = 0; i < size; i++)
cout << arr[i] << " ";

cout << endl;
}

/* Driver function to test above functions */
int main()
{
	int arr[] = {1, 2, 3, 4, 5, 6};

	int n = sizeof(arr) / sizeof(arr[0]);

	// To print original array
	printArray(arr, n);

	// Function calling
	rvereseArray(arr, 0, n-1);

	cout << "Reversed array is" << endl;

	// To print the Reversed array
	printArray(arr, n);

	return 0;
}

遍历数组

int loopArrayElement() {
    int a[3][4] = {{1, 2,  3,  4},
                   {5, 6,  7,  8},
                   {9, 10, 11, 12}};

    // 普通递归
    for (int i = 0; i < 3; i++) {
        for (int j = 0; j < 4; j++) {
            cout << a[i][j] << endl;
        }
    }

    // 指针递归
    for (int i = 0; i < 3; i++) {
        for (int j = 0; j < 4; j++) {
            cout << *(*(a + i) + j) << endl;
        }
    }

    // 更高级的指针递归
    int *p;
    for (p = a[0]; p < a[0] + 12; p++) {
        cout << *p << endl;
    }
}

了解指针

int TwoDimensionalArrayAddress() {
    int a[3][4] = {{1, 2,  3,  4},
                   {5, 6,  7,  8},
                   {9, 10, 11, 12}};
    cout << a << endl;
    cout << &a[0] << endl;

    cout << a + 1 << endl;
    cout << &a[0] + 1 << endl;

    cout << *a << endl;
    cout << a[0] << endl;
    cout << &a[0][0] << endl;

    cout << *a + 1 << endl;
    cout << a[0] + 1 << endl;
    cout << &a[0][0] + 1 << endl;
}