C++复制文件的最可靠方法

好的，所以我知道磁盘写入错误非常罕见，所以请回顾一下，因为我正在使用的数据非常重要（如SSID有点重要）。因此，我想使用绝对最少的内存以绝对最强大的方式复制文件。到目前为止，这是我所得到的。它消耗了很多内存，但我找不到来源。它的工作方式是重新检查大量次，直到得到确认的结果（它可能会使错误的误报数量增加很多，但它可能会大大减少实际错误的机会）。此外，底部的睡眠是这样你有时间使用 Windows 任务管理器分析程序的整体性能。

#include <cstdio>   // fopen, fclose, fread, fwrite, BUFSIZ
#include <cstdlib>
#include <unistd.h>
#include <iostream>
using namespace std;
__inline__ bool copy_file(const char* From, const char* To)
{
    FILE infile = (*fopen(From, "rb"));
    FILE outfile = (*fopen(To, "rwb+"));
    setvbuf( &infile, nullptr, _IONBF, 0);
    setvbuf( &outfile, nullptr, _IONBF, 0);
    fseek(&infile,0,SEEK_END);
    long int size = ftell(&infile);
    fseek(&infile,0,SEEK_SET);
    unsigned short error_amount;
    bool success;
    char c;
    char w;
    char l;
    for ( fpos_t i=0; (i != size); ++i ) {
        error_amount=0;
        fsetpos( &infile, &i );
        c = fgetc(&infile);
        fsetpos( &infile, &i );
        success=true;
        for ( l=0; (l != 126); ++l ) {
            fsetpos( &infile, &i );
            success = ( success == ( fgetc(&infile)==c ) );
        }
        while (success==false) {
            fsetpos( &infile, &i );
            if (error_amount==32767) {
                cerr << "There were 32768 failed attemps at accessing a part of the file! exiting the program...";
                return false;
            }
            ++error_amount;
            //cout << "an error has occured at position ";
            //printf("%d in the file.n", (int)i);
            c = fgetc(&infile);
            fsetpos( &infile, &i );
            success=true;
            for ( l=0; (l != 126); ++l ) {
                fsetpos( &infile, &i );
                success = ( success == ( fgetc(&infile)==c ) );
            }
        }

        fsetpos( &infile, &i );
        fputc( c, &outfile);
        fsetpos( &outfile, &i );

        error_amount=0;
        w = fgetc(&infile);
        fsetpos( &outfile, &i );
        success=true;
        for ( l=0; (l != 126); ++l ) {
            fsetpos( &outfile, &i );
            success = ( success == ( fgetc(&outfile)==w ) );
        }
        while (success==false) {
            fsetpos( &outfile, &i );
            fputc( c, &outfile);
            if (error_amount==32767) {
                cerr << "There were 32768 failed attemps at writing to a part of the file! exiting the program...";
                return false;
            }
            ++error_amount;
            w = fgetc(&infile);
            fsetpos( &infile, &i );
            success=true;
            for ( l=0; (l != 126); ++l ) {
                fsetpos( &outfile, &i );
                success = ( success == ( fgetc(&outfile)==w ) );
            }
        }
        fsetpos( &infile, &i );
    }
    fclose(&infile);
    fclose(&outfile);
    return true;
}
int main( void )
{
    int CopyResult = copy_file("C:\Users\Admin\Desktop\example file.txt","C:\Users\Admin\Desktop\example copy.txt");
    std::cout << "Could it copy the file? " << CopyResult << 'n';
    sleep(65535);
    return 1;
}

那么，如果我的代码以最好的方式走在正确的轨道上，那么我的代码可以做些什么来改进它呢？但是，如果我的代码完全不符合最佳解决方案，那么最佳解决方案是什么？请注意，这个问题本质上是关于检测罕见的磁盘写入错误，用于复制非常非常非常（等）重要数据的应用。