c++ - 指针，将 char ** 转换为 char *

Question

我正在研究指针，但我被下面的示例程序难住了。它应该是将 char** 转换为 char*，但我不明白程序背后的逻辑。程序在做什么？

#include <iostream>
using namespace std;

int main() {

    char *notes[] = {"cpp","python","java","mariadb"};
    void * base = notes; // notes and base, holds the address of note's first element
    void * elemAddr = (char*) base + 3* sizeof(char *); // i didn't understand this line???
    cout << *(char **)elemAddr;   // and this line

    return 0;
}

Answer 1

这些行:

char *notes[] = {"cpp","python","java","mariadb"};
void * base = notes; 
void * elemAddr = (char*) base + 3* sizeof(char *);
cout << *(char **)elemAddr;

是混淆后的等价物:

char *notes[] = {"cpp","python","java","mariadb"};
cout << notes[3];

解释:

void * base = notes; 
void * elemAddr = (char*) base + 3* sizeof(char *);

等同于:

char * base = (char*)notes; 
char * elemAddr = base + 3 * sizeof(char *);

由于指针通常大小相同，因此这些行类似于:

char ** base = notes; 
char ** elemAddr = base + 3;

这使得 elemAddr == &notes[3]。这导致线

cout << *(char **)elemAddr;

与

相同

cout << notes[3];