C++ 将结构体转换为字节并返回结构体
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Converting struct to byte and back to struct
提问by Steven10172
I'm currently working with Arduino Unos, 9DOFs, and XBees, and I was trying to create a struct that could be sent over serial, byte by byte, and then re-constructed into a struct.
我目前正在使用Arduino Unos、9DOFs 和XBees，我试图创建一个可以通过串行、逐字节发送的结构，然后重新构造成一个结构。
So far I have the following code:
到目前为止，我有以下代码：
struct AMG_ANGLES {
float yaw;
float pitch;
float roll;
};
int main() {
AMG_ANGLES struct_data;
struct_data.yaw = 87.96;
struct_data.pitch = -114.58;
struct_data.roll = 100.50;
char* data = new char[sizeof(struct_data)];
for(unsigned int i = 0; i<sizeof(struct_data); i++){
// cout << (char*)(&struct_data+i) << endl;
data[i] = (char*)(&struct_data+i); //Store the bytes of the struct to an array.
}
AMG_ANGLES* tmp = (AMG_ANGLES*)data; //Re-make the struct
cout << tmp.yaw; //Display the yaw to see if it's correct.
}
Source: http://codepad.org/xMgxGY9Q
来源：http: //codepad.org/xMgxGY9Q
This code doesn't seem to work, and I'm not sure what I'm doing wrong.
这段代码似乎不起作用，我不确定我做错了什么。
How do I solve this?
我该如何解决这个问题？
回答by Steven10172
It seems I've solved my issue with the following code.
看来我已经用下面的代码解决了我的问题。
struct AMG_ANGLES {
float yaw;
float pitch;
float roll;
};
int main() {
AMG_ANGLES struct_data;
struct_data.yaw = 87.96;
struct_data.pitch = -114.58;
struct_data.roll = 100.50;
//Sending Side
char b[sizeof(struct_data)];
memcpy(b, &struct_data, sizeof(struct_data));
//Receiving Side
AMG_ANGLES tmp; //Re-make the struct
memcpy(&tmp, b, sizeof(tmp));
cout << tmp.yaw; //Display the yaw to see if it's correct
}
WARNING: This code will only work if sending and receiving are using the same endianarchitecture.
警告：此代码仅在发送和接收使用相同的字节序架构时才有效。
回答by Some programmer dude
You do things in the wrong order, the expression
你以错误的顺序做事，表达
&struct_data+i
takes the address of struct_dataand increases it by itimes the size of the structure.
获取 的地址struct_data并将其增加i为结构大小的倍数。
Try this instead:
试试这个：
*((char *) &struct_data + i)
This converts the address of struct_datato a char *and thenadds the index, and then uses the dereference operator (unary *) to get the "char" at that address.
这会将 的地址转换struct_data为 a char *，然后添加索引，然后使用解引用运算符 (unary *) 获取该地址处的“char”。
回答by CharLess
Always utilize data structures to its fullest..
始终充分利用数据结构。
union AMG_ANGLES {
struct {
float yaw;
float pitch;
float roll;
}data;
char size8[3*8];
int size32[3*4];
float size64[3*1];
};
回答by jtepe
for(unsigned int i = 0; i<sizeof(struct_data); i++){
// +i has to be outside of the parentheses in order to increment the address
// by the size of a char. Otherwise you would increment by the size of
// struct_data. You also have to dereference the whole thing, or you will
// assign an address to data[i]
data[i] = *((char*)(&struct_data) + i);
}
AMG_ANGLES* tmp = (AMG_ANGLES*)data; //Re-Make the struct
//tmp is a pointer so you have to use -> which is shorthand for (*tmp).yaw
cout << tmp->yaw;
}
