将结构的成员声明为uint32_t时的额外字节 [英] Extra bytes when declaring a member of a struct as uint32_t
问题描述
使用stdint.h库中的uint32_t类型时遇到问题.如果我运行以下代码(在Ubuntu linux 11.10 x86_64，g ++版本4.6.1上):
I have a problem when using the uint32_t type from the stdint.h library. If I run the following code (on Ubuntu linux 11.10 x86_64, g++ version 4.6.1):
#include "stdint.h"
#include <iostream>
using std::cout;
typedef struct{
// api identifier
uint8_t api_id;
uint8_t frame_id;
uint32_t dest_addr_64_h;
uint32_t dest_addr_64_l;
uint16_t dest_addr_16;
uint8_t broadcast_radius;
uint8_t options;
// packet fragmentation
uint16_t order_index;
uint16_t total_packets;
uint8_t rf_data[];
} xbee_tx_a;
typedef struct{
// api identifier
uint8_t api_id;
uint8_t frame_id;
uint16_t dest_addr_64_h;
uint16_t dest_addr_64_l;
uint16_t dest_addr_16;
uint8_t broadcast_radius;
uint8_t options;
// packet fragmentation
uint16_t order_index;
uint16_t total_packets;
uint8_t rf_data[];
} xbee_tx_b;
int main(int argc, char**argv){
xbee_tx_a a;
cout<<"size of xbee_tx_a "<<sizeof (xbee_tx_a)<<std::endl;
cout<<"size of xbee_tx_a.api_id "<<sizeof (a.api_id)<<std::endl;
cout<<"size of xbee_tx_a.frame_id "<<sizeof (a.frame_id)<<std::endl;
cout<<"size of xbee_tx_a.dest_addr_64_h "<<sizeof (a.dest_addr_64_h)<<std::endl;
cout<<"size of xbee_tx_a.dest_addr_64_l "<<sizeof (a.dest_addr_64_l)<<std::endl;
cout<<"size of xbee_tx_a.dest_addr_16 "<<sizeof (a.dest_addr_16)<<std::endl;
cout<<"size of xbee_tx_a.broadcast_radius "<<sizeof (a.broadcast_radius)<<std::endl;
cout<<"size of xbee_tx_a.options "<<sizeof (a.options)<<std::endl;
cout<<"size of xbee_tx_a.order_index "<<sizeof (a.order_index)<<std::endl;
cout<<"size of xbee_tx_a.total_packets "<<sizeof (a.total_packets)<<std::endl;
cout<<"size of xbee_tx_a.rf_data "<<sizeof (a.rf_data)<<std::endl;
cout<<"----------------------------------------------------------\n";
xbee_tx_b b;
cout<<"size of xbee_tx_b "<<sizeof (xbee_tx_b)<<std::endl;
cout<<"size of xbee_tx_b.api_id "<<sizeof (b.api_id)<<std::endl;
cout<<"size of xbee_tx_b.frame_id "<<sizeof (b.frame_id)<<std::endl;
cout<<"size of xbee_tx_b.dest_addr_64_h "<<sizeof (b.dest_addr_64_h)<<std::endl;
cout<<"size of xbee_tx_b.dest_addr_64_l "<<sizeof (b.dest_addr_64_l)<<std::endl;
cout<<"size of xbee_tx_b.dest_addr_16 "<<sizeof (b.dest_addr_16)<<std::endl;
cout<<"size of xbee_tx_b.broadcast_radius "<<sizeof (b.broadcast_radius)<<std::endl;
cout<<"size of xbee_tx_b.options "<<sizeof (b.options)<<std::endl;
cout<<"size of xbee_tx_b.order_index "<<sizeof (b.order_index)<<std::endl;
cout<<"size of xbee_tx_b.total_packets "<<sizeof (b.total_packets)<<std::endl;
cout<<"size of xbee_tx_b.rf_data "<<sizeof (b.rf_data)<<std::endl;
}
然后我得到以下输出:
size of xbee_tx_a 20
size of xbee_tx_a.api_id 1
size of xbee_tx_a.frame_id 1
size of xbee_tx_a.dest_addr_64_h 4
size of xbee_tx_a.dest_addr_64_l 4
size of xbee_tx_a.dest_addr_16 2
size of xbee_tx_a.broadcast_radius 1
size of xbee_tx_a.options 1
size of xbee_tx_a.order_index 2
size of xbee_tx_a.total_packets 2
size of xbee_tx_a.rf_data 0
----------------------------------------------------------
size of xbee_tx_b 14
size of xbee_tx_b.api_id 1
size of xbee_tx_b.frame_id 1
size of xbee_tx_b.dest_addr_64_h 2
size of xbee_tx_b.dest_addr_64_l 2
size of xbee_tx_b.dest_addr_16 2
size of xbee_tx_b.broadcast_radius 1
size of xbee_tx_b.options 1
size of xbee_tx_b.order_index 2
size of xbee_tx_b.total_packets 2
size of xbee_tx_b.rf_data 0
我正在做的是打印出一个结构的总大小和该结构的每个成员的大小.
What I'm doing is printing out the total size of a struct and the size of each member of the struct.
对于xbee_tx_b，成员的大小之和等于结构(14)的大小
In the case of xbee_tx_b the sizes of the members add up to the size of the struct (14)
对于xbee_tx_a，成员的大小总计为18个字节...但是结构的大小为20个字节！
In the case of xbee_tx_a the sizes of the members add up to 18 bytes... but the size of the struct is 20 bytes!
xbee_tx_a和xbee_tx_b之间的唯一区别在于dest_addr_64_X成员的类型.它们是xbee_tx_a中的uint32_t和xbee_tx_b中的uint16_t.使用uint32_t时，为什么结构的大小大于其成员的大小之和?这两个额外的字节从哪里来?
The only difference between xbee_tx_a and xbee_tx_b is in the type of the dest_addr_64_X members. They are uint32_t in xbee_tx_a and uint16_t in xbee_tx_b. Why is the size of the structure bigger than the sum of the sizes of its members when I use uint32_t? Where do those 2 extra bytes come from?
谢谢！
推荐答案
将结构填充为4字节的整数倍 1 ，以便它们按字对齐. http://en.wikipedia.org/wiki/Data_structure_alignment#Data_structure_padding
Structs are padded to be an integer multiple of 4 bytes1 so that they are word-aligned. http://en.wikipedia.org/wiki/Data_structure_alignment#Data_structure_padding
另请参阅:
1 正如 @Mooing Duck 所说，这不是总是 true:
1 As @Mooing Duck commented, this isn't always true:
它并不总是4字节的倍数，它根据成员的不同(略有不同).另一方面，99％的时间是4字节的倍数.
It's not always a multiple of 4 bytes, it varies (slightly) depending on the members. On the other hand, 99% of the time it's a multiple of 4 bytes.
这篇关于将结构的成员声明为uint32_t时的额外字节的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
