quick question,
I want to cast a 3-bytes signed two's complement array into a little-endian int32
I tried setting the size of the memcpy to 3 and +1 to the destination address (without forgetting to zero initialize my int32)
Here is my code:
int main(void)
{
// little endian 3 bytes
__uint8_t n1[]={0,0xFF,0xDE,0x81};
__uint8_t n2[]={0xFF,0xDE,0x81};
// temp var
__int32_t tmp1=0;
__int32_t tmp2=0;
// cast array to int
memcpy(&tmp1,n1,4);
memcpy((&tmp2)+1,n2,3);
// printf
printf("n1 : %d\n",tmp1);
printf("n2 : %d\n",tmp2);
}
The output I get:
n1 : -2122195201
n2 : 0
The output I want:
n1 : -2122195201
n2 : -2122195201
How do I fix this? Perhaps it is better to use union?
By doing (&tmp2)+1
you take the adress of tmp2
, which is a pointer to int32_t
, and you increment this pointer. So you add 4 to its adress. This means that the destination adress of your memcpy is the adress of tmp2
+ 4. You actually didn't modify tmp2
but instead overlap some other variable or memory.
It's not a clean way to convert or cast data, but if you really want to use memcpy do something like (not tested) memcpy((void *)(((char *)&tmp2)+1),n2,3);
I'd avoid casting the value, and make the conversion explicit (this will probably be inlined)
I am assuming that your character-array is little-endian, too. (otherwise swap the indexes)
static int three2four(__uint8_t three[3] )
{
int val;
val = ((unsigned)three[2] << 16) | ((unsigned)three[1] << 8) | (unsigned)three[0] ;
if (three[2] & 0x80) val |= (0xffu << 24);
return val;
}
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