C++ 编译器可以重新排序结构中的元素吗

Can a C++ compiler re-order elements in a struct(C++ 编译器可以重新排序结构中的元素吗)

问题描述

C++ 编译器(特别是 g++)能否对结构的内部元素重新排序?

Can a C++ compiler (specifically g++) re-order the internal elements of a struct?

我看到一些奇怪的行为，其中我的结构包含如下内容:

I'm seeing some strange behaviour where I have a structure that contains something like the following:

Struct SomeStruct{
   ...
   ...
   long someLong;
   long someLongArray[25];
   unsigned long someUnsignedLong;
   unsigned long someUnsignedLongArray[8];
   unsigned long int someUnsignedLongInt;
   ...
   ...
};

当我将输出写入文件时，someUnsignedLongArray 和 someLongArray 的顺序似乎颠倒了(即 someLongArray[] 中的元素出现在 someUnsignedLong 之后，someUnsignedLongArray[] 的元素出现在 someLong 之后).这可能吗??

When I write output this to file, the order of someUnsignedLongArray and someLongArray seem to be reversed (i.e. the elements in someLongArray[] appear after someUnsignedLong and the elements of someUnsignedLongArray[] appear after someLong). Is this possible??

谢谢

更新:根据要求，我正在使用以下内容写出结构:

Update: As requested, I am writing out the structure using the following:

int fd = open(fspec,O_RDWR|O_CREAT|O_TRUNC,0666);
int writeRes =  write(fd,(char *)&someStruct,sizeof(SomeStruct));

为了完整性，这里是完整的结构:

For completeness, here is the full struct:

struct SomeStruct{
byte someByte;
byte someByteArray[6];
char someChar;
char someCharArray[5];
char someCharArrayArray[3][5];
short someShort;
signed short someShortArray[2];
unsigned short someUnsignedShort;
unsigned short someUnsignedShortArray[8];
int someInt;
int someIntArray[3];
int someIntArrayArrayArrayArray[4][3][2][6];
int *pSomeInt;
unsigned int someUnsignedInt;
unsigned int someUnsignedIntArray[9];
long someLong;
long someLongArray[25];
unsigned long someUnsignedLong;
unsigned long someUnsignedLongArray[8];
unsigned long int someUnsignedLongInt;
long long someLongLong;
long long someLongLongArray[5];
bool someBool;
bool someBoolArray[3];
unsigned long long someUnsignedLongLong;
unsigned long long someUnsignedLongLongArray[5];
unsigned long long someUnsignedLongLongArrayArray[5][2];
unsigned long long int *pSomeUnsignedLongLongInt;
};

推荐答案

它通常不能重新排序元素，不行.

It normally can't reorder elements, no.

一个例外是如果有访问说明符将它们分开:

An exception is if there's an access specifier separating them:

struct Foo {    
  A a;
  B b;
  C c;
private:
  D d;
  E e;
  F f;
};

a、b 和 c 保证按此顺序存储，而d、e 和 f 保证按顺序存储.但是不能保证 a、b 和 c 相对于 d、e 的存储位置 和 f.

a, b and c are guaranteed to be stored in this order, and d, e and f are guaranteed to be stored in order. But there is no guarantees about where a, b and c are stored relative to d, e and f.

要记住的另一件事是编译器可以插入任意多的填充，即使它不重新排序任何内容.

Another thing to keep in mind is that the compiler can insert as much padding as it likes, even if it doesn't reorder anything.

这是标准的相关部分:

第 9.2.12 节:

Section 9.2.12:

a 的非静态数据成员(非联合)类声明没有干预访问说明符是分配，以便后来的成员有一个类中的更高地址目的.分配顺序非静态数据成员由一个分隔未指定访问说明符(11.1)"

Nonstatic data members of a (non-union) class declared without an intervening access-specifier are allocated so that later members have higher addresses within a class object. The order of allocation of nonstatic data members separated by an access-specifier is unspecified (11.1)"

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