Hello
I am trying to convert an array of 8 uint8_t to 1 uint64_t.
But when I try to do the following it fails. Because I try to shift with 32.
But uint64_t has 64 bits and uint8_t has only 8 so this is possible?
So wy can't I do this?
Thank youCode:int main(void){ uint8_t bla[8]; uint64_t temp; temp = ((bla[3] << 32) | (bla[4] << 24) | (bla[5] << 16) | (bla[6] << 8) | (bla[7])); printf("0x%x\n", temp); }
I would do that with a union...
Since the variables in a union overlay one another you can load your convert.charvar array with the values and read it back by accessing convert.bigvar.Code:union t_convert { uint64_t bigvar; uint8_t charvar[8]; } convert;
Last edited by CommonTater; 10-05-2011 at 02:49 PM.
You would need to cast your bytes to uint64_t before trying to shift more than 32 bits.
Eg
(uint64_t)bla[3] << 32
Lookup "endian" to find out why the union cast would not work on all machines.
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That's easily compensated for by the order of the values placed in the array... it's even easy to test for at run time by seeding known values and seeing what comes back... But you are correct, without logic to compensate it does become platform dependent.Originally Posted by Salem
Here's the easy way:Code:uint8_t bla[8] = {0x01, 0x02, 0x03, 0x04, 0x05, 0x06, 0x07, 0x08 }; uint64_t temp; for (int i=0; i<8; ++i) temp = (temp << 8) | bla[i]; printf("0x%x\n", temp);
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