[英]Reading then adding large number of integers from binary file fast in C/C++
我正在編寫代碼,用於在32位Linux操作系統上使用C / C ++從二進制文件中讀取無符號整數,該操作系統旨在在8核x86系統上運行。 應用程序接受一個輸入文件,該文件包含一個接一個的小端格式的無符號整數。 因此輸入文件大小(以字節為單位)是4的倍數。文件中可能有十億個整數。 讀取和添加所有整數並以64位精度返回總和的最快方法是什么?
以下是我的實施。 檢查損壞數據的錯誤不是這里的主要問題,在這種情況下輸入文件被認為沒有任何問題。
#include <iostream>
#include <fstream>
#include <pthread.h>
#include <string>
#include <string.h>
using namespace std;
string filepath;
unsigned int READBLOCKSIZE = 1024*1024;
unsigned long long nFileLength = 0;
unsigned long long accumulator = 0; // assuming 32 bit OS running on X86-64
unsigned int seekIndex[8] = {};
unsigned int threadBlockSize = 0;
unsigned long long acc[8] = {};
pthread_t thread[8];
void* threadFunc(void* pThreadNum);
//time_t seconds1;
//time_t seconds2;
int main(int argc, char *argv[])
{
if (argc < 2)
{
cout << "Please enter a file path\n";
return -1;
}
//seconds1 = time (NULL);
//cout << "Start Time in seconds since January 1, 1970 -> " << seconds1 << "\n";
string path(argv[1]);
filepath = path;
ifstream ifsReadFile(filepath.c_str(), ifstream::binary); // Create FileStream for the file to be read
if(0 == ifsReadFile.is_open())
{
cout << "Could not find/open input file\n";
return -1;
}
ifsReadFile.seekg (0, ios::end);
nFileLength = ifsReadFile.tellg(); // get file size
ifsReadFile.seekg (0, ios::beg);
if(nFileLength < 16*READBLOCKSIZE)
{
//cout << "Using One Thread\n"; //**
char* readBuf = new char[READBLOCKSIZE];
if(0 == readBuf) return -1;
unsigned int startOffset = 0;
if(nFileLength > READBLOCKSIZE)
{
while(startOffset + READBLOCKSIZE < nFileLength)
{
//ifsReadFile.flush();
ifsReadFile.read(readBuf, READBLOCKSIZE); // At this point ifsReadFile is open
int* num = reinterpret_cast<int*>(readBuf);
for(unsigned int i = 0 ; i < (READBLOCKSIZE/4) ; i++)
{
accumulator += *(num + i);
}
startOffset += READBLOCKSIZE;
}
}
if(nFileLength - (startOffset) > 0)
{
ifsReadFile.read(readBuf, nFileLength - (startOffset));
int* num = reinterpret_cast<int*>(readBuf);
for(unsigned int i = 0 ; i < ((nFileLength - startOffset)/4) ; ++i)
{
accumulator += *(num + i);
}
}
delete[] readBuf; readBuf = 0;
}
else
{
//cout << "Using 8 Threads\n"; //**
unsigned int currthreadnum[8] = {0,1,2,3,4,5,6,7};
if(nFileLength > 200000000) READBLOCKSIZE *= 16; // read larger blocks
//cout << "Read Block Size -> " << READBLOCKSIZE << "\n";
if(nFileLength % 28)
{
threadBlockSize = (nFileLength / 28);
threadBlockSize *= 4;
}
else
{
threadBlockSize = (nFileLength / 7);
}
for(int i = 0; i < 8 ; ++i)
{
seekIndex[i] = i*threadBlockSize;
//cout << seekIndex[i] << "\n";
}
pthread_create(&thread[0], NULL, threadFunc, (void*)(currthreadnum + 0));
pthread_create(&thread[1], NULL, threadFunc, (void*)(currthreadnum + 1));
pthread_create(&thread[2], NULL, threadFunc, (void*)(currthreadnum + 2));
pthread_create(&thread[3], NULL, threadFunc, (void*)(currthreadnum + 3));
pthread_create(&thread[4], NULL, threadFunc, (void*)(currthreadnum + 4));
pthread_create(&thread[5], NULL, threadFunc, (void*)(currthreadnum + 5));
pthread_create(&thread[6], NULL, threadFunc, (void*)(currthreadnum + 6));
pthread_create(&thread[7], NULL, threadFunc, (void*)(currthreadnum + 7));
pthread_join(thread[0], NULL);
pthread_join(thread[1], NULL);
pthread_join(thread[2], NULL);
pthread_join(thread[3], NULL);
pthread_join(thread[4], NULL);
pthread_join(thread[5], NULL);
pthread_join(thread[6], NULL);
pthread_join(thread[7], NULL);
for(int i = 0; i < 8; ++i)
{
accumulator += acc[i];
}
}
//seconds2 = time (NULL);
//cout << "End Time in seconds since January 1, 1970 -> " << seconds2 << "\n";
//cout << "Total time to add " << nFileLength/4 << " integers -> " << seconds2 - seconds1 << " seconds\n";
cout << accumulator << "\n";
return 0;
}
void* threadFunc(void* pThreadNum)
{
unsigned int threadNum = *reinterpret_cast<int*>(pThreadNum);
char* localReadBuf = new char[READBLOCKSIZE];
unsigned int startOffset = seekIndex[threadNum];
ifstream ifs(filepath.c_str(), ifstream::binary); // Create FileStream for the file to be read
if(0 == ifs.is_open())
{
cout << "Could not find/open input file\n";
return 0;
}
ifs.seekg (startOffset, ios::beg); // Seek to the correct offset for this thread
acc[threadNum] = 0;
unsigned int endOffset = startOffset + threadBlockSize;
if(endOffset > nFileLength) endOffset = nFileLength; // for last thread
//cout << threadNum << "-" << startOffset << "-" << endOffset << "\n";
if((endOffset - startOffset) > READBLOCKSIZE)
{
while(startOffset + READBLOCKSIZE < endOffset)
{
ifs.read(localReadBuf, READBLOCKSIZE); // At this point ifs is open
int* num = reinterpret_cast<int*>(localReadBuf);
for(unsigned int i = 0 ; i < (READBLOCKSIZE/4) ; i++)
{
acc[threadNum] += *(num + i);
}
startOffset += READBLOCKSIZE;
}
}
if(endOffset - startOffset > 0)
{
ifs.read(localReadBuf, endOffset - startOffset);
int* num = reinterpret_cast<int*>(localReadBuf);
for(unsigned int i = 0 ; i < ((endOffset - startOffset)/4) ; ++i)
{
acc[threadNum] += *(num + i);
}
}
//cout << "Thread " << threadNum + 1 << " subsum = " << acc[threadNum] << "\n"; //**
delete[] localReadBuf; localReadBuf = 0;
return 0;
}
我寫了一個小的C#程序來生成用於測試的輸入二進制文件。
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.IO;
namespace BinaryNumWriter
{
class Program
{
static UInt64 total = 0;
static void Main(string[] args)
{
BinaryWriter bw = new BinaryWriter(File.Open("test.txt", FileMode.Create));
Random rn = new Random();
for (UInt32 i = 1; i <= 500000000; ++i)
{
UInt32 num = (UInt32)rn.Next(0, 0xffff);
bw.Write(num);
total += num;
}
bw.Flush();
bw.Close();
}
}
}
在Core i5機器上運行程序@ 3.33 Ghz(它的四核,但我現在得到的)2 GB RAM和Ubuntu 9.10 32位具有以下性能數字
100個整數~0秒(否則我真的要吮吸)100000整數<0秒100000000整數~7秒500000000整數~29秒(1.86 GB輸入文件)
我不確定硬盤是5400RPM還是7200RPM。 我嘗試了不同的緩沖區大小進行讀取,一次讀取16 MB的大輸入文件就是最佳點。
有沒有更好的方法從文件中更快地讀取以提高整體性能? 是否有更智能的方法可以更快地添加大型整數數組並重復折疊? 我編寫代碼的方式是否有任何重大的障礙/我做了一些明顯錯誤的事情,這花費了很多時間?
我該怎么做才能更快地讀取和添加數據?
謝謝。
Chinmay
以您的方式從多個線程訪問機械硬盤將采取一些頭部移動(讀取速度慢)。 你幾乎可以肯定IO綁定(1.86GB文件為65MBps)。 嘗試通過以下方式更改策略:
你需要相當多的同步才能讓它完美運行,我認為通過執行順序文件訪問可以最大限度地提高你的硬盤/文件系統IO功能。 YMMV上的小文件,可以以閃電般的速度從緩存中緩存和提供。
你可以嘗試的另一件事是只啟動7個線程,為主線程和系統的其余部分留一個空閑CPU。
..或獲得SSD :)
編輯:
為簡單起見,請參閱沒有處理的單線程讀取文件(丟棄緩沖區)的速度有多快。 加上epsilon是你完成這項工作的速度的理論極限。
如果要快速讀取(或寫入)大量數據,並且不希望對該數據進行大量處理,則需要避免在緩沖區之間使用額外的數據副本。 這意味着你想要避免fstream或FILE抽象(因為它們引入了需要復制的額外緩沖區),並避免在內核和用戶緩沖區之間復制內容的讀/寫類型調用。
相反,在linux上,你想使用mmap(2)。 在64位操作系統上,只需將整個文件madvise(MADV_SEQUENTIAL)
到內存中,使用madvise(MADV_SEQUENTIAL)
告訴內核你將主要按順序訪問它,並擁有它。 對於32位操作系統,您需要以塊的形式進行mmap,每次都取消映射前一個塊。 與您當前的結構非常相似,每個線程一次只能映射一個固定大小的塊,應該可以正常工作。
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