• Read4096


    Given API:
    int Read4096(char* buf);
    It reads data from a file and records the position so that the next time when it is called it read the next 4k chars (or the rest of the file, whichever is smaller) from the file.
    The return is the number of chars read.

    Todo: Use above API to Implement API
    "int Read(char* buf, int n)" which reads any number of chars from the file.

    我大概会这么写。

    第二种写法是注意到尽量避免内存拷贝。

     1 #include <iostream>
     2 #include <cstdio>
     3 #include <cstring>
     4 using namespace std;
     5 
     6 int read4096(char* buf) {
     7     static int count = 9999;
     8     if (count >= 4096) {
     9         count -= 4096;
    10         return 4096;
    11     } else {
    12         int tmp = count;
    13         count = 0;
    14         return tmp;
    15     }    
    16 }
    17 
    18 class Reader {
    19     public: 
    20         Reader():remain(0), size(0) {}
    21         int readN(char* buf, int n) {
    22             int readCount = 0;
    23             while (readCount < n) {
    24                 if (remain >= n - readCount) {
    25                     memcpy(buf + readCount, tmp + size - remain, n - readCount);
    26                     remain -= (n - readCount);
    27                     readCount = n;
    28                 } else {
    29                     if (remain > 0) memcpy(buf, tmp + size - remain, remain);
    30                     readCount += remain;
    31                     remain = size = read4096(tmp);
    32                     if (size == 0) break;
    33                 }
    34             }
    35             return readCount;
    36         }
    37 
    38         // should be more efficient, avoid some memory copy
    39         int readN2(char* buf, int n) {
    40             if (remain >= n) {
    41                 memcpy(buf, tmp + size - remain, n);
    42                 remain -= n;
    43                 return n;
    44             }
    45             int readCount = 0;
    46             if (remain > 0) {
    47                 memcpy(buf, tmp + size - remain, remain);
    48                 readCount += remain;
    49                 remain = 0;
    50             }
    51             while (readCount + 4096 <= n) {
    52                 int count = read4096(buf + readCount);
    53                 readCount += count;
    54                 if (count < 4096) {
    55                     return readCount;
    56                 }
    57             }
    58             if (readCount < n) {
    59                 size = read4096(tmp);
    60                 if (size >= n - readCount) {
    61                     memcpy(buf + readCount, tmp, n - readCount);
    62                     remain = size - n + readCount;
    63                     readCount = n;
    64                 } else {
    65                     memcpy(buf + readCount, tmp, size);
    66                     readCount += size;
    67                 }
    68             }
    69             return readCount;
    70         }
    71     private:
    72         int remain;
    73         int size;
    74         char tmp[4096];
    75 };
    76 int main(int argc, char** argv) {
    77     freopen("input.txt", "r", stdin);
    78     Reader reader;
    79 
    80     char buff[4096];
    81     int n = 1024;
    82     while (true) {
    83         int count = reader.readN2(buff, n);
    84         cout << n << " " << count << endl;
    85         if (count == 0) break;
    86     }
    87 
    88     return 0;
    89 }
  • 相关阅读:
    给定中序和后序遍历,求前序序列(C++递归方式实现)
    myeclipse2014删除antlr-2.7.2.jar--解决struts和hibernate包冲突
    hadoop1.2.1配置与运行子串统计程序
    任务计划crontab
    建NTP
    vnc下运行runInstall报java错误
    rpm软件安装
    redis
    rpm包和deb包转换
    新老版本centos下载
  • 原文地址:https://www.cnblogs.com/linyx/p/4069386.html
Copyright © 2020-2023  润新知