How to convert a struct varible to uint8_t array in C?

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jack1998

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How to convert msg to uint8_t array msg_arr[8]?
Code: 
typedef struct message_t
{
    uint16_t time;
    uint16_t lat; 
    uint8_t ns;
    uint16_t lon;
    uint8_t ew;
} message;

message msg;
msg.time = 0x1234;
msg.lat = 0x2122;
msg.ns = 'n';
msg.lon = 0x1834;
msg.ew = 'e';

uint8_t msg_arr[8];
 
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Hi,

general recommendation is to start with the biggest variables.
--> first all 32 bit variables then all 16 bit variables, then all 8 bit variables. This is to avoid "align" bytes.

Now it depends
* if you want to copy the struct into an 8 bit array
* or you just want to access the struct data

...( I´m no C specialist, thus - in doubt - rely on the answers of others)
COPY:
needs to be done byte by byte like:
Code: 
byte0 = msg.time >>8;
byte1 = msg.time & 0xFF;
....

ACCESS:
could be done with pointers.
I guess something like this:

Code: 
uint8_t * databyte;
uint8_t i;
databyte = (uint8_t *)&msg;

for (i = 0; i < sizeof(msg)8 ; i++)
{
   dosomethingwith databyte[i];
}

Klaus
 
Last edited:
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You need to consider that depending on compler settings and processor hardware requirements the structure message_t isn't necessarily respectively even can't be packed. Thus type conversions of the discussed kind aren't safe and not necessarily portable.
 
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Another option may be to use a union that contains both the struct and the array.
All of the usual caveats apply regarding compiler options, alignments, struct padding etc..
Susan
 
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