zoukankan      html  css  js  c++  java
  • [转]关于alignment

    转自 https://stackoverflow.com/questions/41719845/memory-alignment-in-c-c

    #pragma pack(push, 1)
    struct ReallySlowStruct
    {
        char c : 6;
        __int64 d : 64;
        int b : 32;
        char a : 8;
    };
    
    struct SlowStruct
    {
        char c;
        __int64 d;
        int b;
        char a;
    };
    
    struct FastStruct
    {
       __int64 d;
       __int b;
       char a;
       char c;
       char unused[2];
    };
    #pragma pack(pop)

    The examples given in the book are highly dependent on the used compiler and computer architecture. If you test them in your own program you may get totally different results than the author. I will assume a 64-bit architecture, because the author does also, from what I've read in the description. Lets look at the examples one by one:

    ReallySlowStruct IF the used compiler supports non-byte aligned struct members, the start of "d" will be at the seventh bit of the first byte of the struct. Sounds very good for memory saving. The problem with this is, that C does not allow bit-adressing. So to save newValue to the "d" member, the compiler must do a whole lot of bit shifting operations: Save the first two bits of "newValue" in byte0, shifted 6 bits to the right. Then shift "newValue" two bits to the left and save it starting at byte 1. Byte 1 is a non-aligned memory location, that means the bulk memory transfer instructions won't work, the compiler must save every byte at a time.

    SlowStruct It gets better. The compiler can get rid of all the bit-fiddling. But writing "d" will still require writing every byte at a time, because it is not aligned to the native "int" size. The native size on a 64-bit system is 8. so every memory address not divisable by 8 can only be accessed one byte at a time. And worse, if I switch off packing, I will waste a lot of memory space: every member which is followed by an int will be padded with enough bytes to let the integer start at a memory location divisable by 8. In this case: char a and c will both take up 8 bytes.

    FastStruct this is aligned to the size of int on the target machine. "d" takes up 8 bytes as it should. Because the chars are all bundled at one place, the compiler does not pad them and does not waste space. chars are only 1 byte each, so we do not need to pad them. The complete structure adds up to an overall size of 16 bytes. Divisable by 8, so no padding needed.


    In most scenarios, you never have to be concerned with alignment because the default alignment is already optimal. In some cases however, you can achieve significant performance improvements, or memory savings, by specifying a custom alignment for your data stuctures.

    In terms of memory space, the compiler pads the structure in a way that naturally aligns each element of the structure.

    struct x_
    {
       char a;     // 1 byte
       int b;      // 4 bytes
       short c;    // 2 bytes
       char d;     // 1 byte
    } bar[3];
    

    struct x_ is padded by the compiler and thus becomes:

    // Shows the actual memory layout
    struct x_
    {
       char a;           // 1 byte
       char _pad0[3];    // padding to put 'b' on 4-byte boundary
       int b;            // 4 bytes
       short c;          // 2 bytes
       char d;           // 1 byte
       char _pad1[1];    // padding to make sizeof(x_) multiple of 4
    } bar[3];
    

    Source: https://docs.microsoft.com/en-us/cpp/cpp/alignment-cpp-declarations?view=vs-2019

  • 相关阅读:
    PL/SQL Developer连接Oracle
    Oracle 11g 监听命令
    Oracle 11g的登陆问题
    PL/SQL Developer 配置和使用
    KMP应用求两个字符串的最长公共子串
    msc pool概念
    nformix调优之执行计划取得
    lsof 与fuser
    informix onstat命令收集
    各类系统上查看占cpu最多的进程
  • 原文地址:https://www.cnblogs.com/yi-mu-xi/p/13851965.html
Copyright © 2011-2022 走看看