[英]Converting struct to byte and back to struct
我目前正在使用Arduino Uno ,9DOF和XBee ,我正在嘗試創建一個結構,可以通過串行,逐字節發送,然后重新構造成結構。
到目前為止,我有以下代碼:
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.
}
資料來源: http : //codepad.org/xMgxGY9Q
這段代碼似乎不起作用,我不確定我做錯了什么。
我該如何解決這個問題?
看來我用以下代碼解決了我的問題。
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
}
警告:此代碼僅在發送和接收使用相同的endian體系結構時才有效。
你以錯誤的順序做事,表達式
&struct_data+i
獲取struct_data
的地址並將其增加i
倍於結構的大小 。
試試這個:
*((char *) &struct_data + i)
這會將struct_data
的地址轉換為char *
, 然后添加索引,然后使用解除引用運算符(一元*
)來獲取該地址的“char”。
始終充分利用數據結構..
union AMG_ANGLES {
struct {
float yaw;
float pitch;
float roll;
}data;
char size8[3*8];
int size32[3*4];
float size64[3*1];
};
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;
}
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