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基本上，我正在尝试创建自己的工具来快速格式化 ext4 分区。

据我了解，其背后的想法是用空ext2_inode结构查找和覆盖现有的 inode。我认为我可以通过简单地计算和之间的差异来做到这一点，sb.s_inodes_count所以sb.s_free_inodes_count我会知道我必须覆盖多少个 inode。

但实际上它不起作用（毕竟只是无法安装），我很好奇为什么。如果是这样，那么我如何找到正确的 inode 位置（如果我发现它不正确）以及如何正确编写空结构？

#include <iostream>
#include <ext2fs/ext2_fs.h>
#include <unistd.h>
#include <fcntl.h>
#include <stdlib.h>
#include "Disk.h"

using namespace std;

Disk::Disk(string path){
    try{
        this->fd = open(path.c_str(), O_RDWR);
        if(this->fd == -1) throw new CantOpenDevice;
    } catch (CantOpenDevice& openError){
        logger->error(openError.what());
    }
}

ext2_super_block Disk::readSuperBlock(){
    lseek(this->fd, 1024, SEEK_SET); // 1024 - Group 0 Padding 
    read(this->fd, &this->sb, sizeof(ext2_super_block));
    this->logger->debug("Succesfully read superblock.");
    return this->sb;
}

ext4_group_desc Disk::readFirstBGD(){
    int block_size_offset = this->getBlockSizeOffset();
    lseek(this->fd, block_size_offset, SEEK_SET);
    read(this->fd, &this->bg, sizeof(ext4_group_desc));
    this->logger->debug("Succesfully read a group descriptor.");
    return this->bg;
}

void Disk::writeEmptyInodes(){
    ext2_inode inode = {0};
    int block_size_offset = this->getBlockSizeOffset();
    int inodes_to_clean = this->sb.s_inodes_count - this->sb.s_free_inodes_count;
    for(int i=1; i < inodes_to_clean; i++){
        write(this->fd, &inode, sizeof(inode));
    }
}

int Disk::getBlockSizeOffset(){
    return 1024 << this->sb.s_log_block_size;
}

Disk::~Disk(){
    close(this->fd);
    delete logger;
}
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