I'm unsure if this is possible due to structure padding and alignment but, assuming you take care of that by aligning your structures to 4/8 bytes, is it possible to bit shift on a structure as if it was a single variable?
What I'd like to do is take a string (max 8 bytes) and shift it into the high order bits of a 64-bit variable.
Like if I do this:
#include <stdint.h>
#include <string.h>
void shiftstr(uint64_t* t,char* c,size_t len){
memcpy(t, c, len);
//now *t==0x000000617369616b
*t<<=(sizeof(uint64_t)-len)*8;
//now *t==0x617369616b000000
}
int main(){
uint64_t k = 0;
char n[] = "kaisa";
shiftstr(&k, n,strlen(n));
return 0;
}
This works just fine, but what if I had, instead of a uint64_t
, two uint32_t
, either as individual variables or a structure.
#include <stdint.h>
#include <string.h>
struct U64{
uint32_t x;
uint32_t y;
};
void shiftstrstruct(struct U64* t, char* c, size_t len){
memcpy(t, c, len);
/*
At this point I think
x == 0x7369616b
y == 0x00000061
But I could be wrong
*/
//but how can I perform the bit shift?
//Where
//x==0x0000006b
//y==0x61697361
}
int main(){
char n[] = "kaisa";
struct U64 m = {0};
shiftstrstruct(&m, n, strlen(n));
return 0;
}
Up to the memcpy
part, it should be the same as if I were performing it on a single variable. I believe the values of x
and y
are correct in such situations. But, if that's the case that means the values need to be shifted away from x
towards y
.
I know I can cast but what if I wanted to deal with a 16 byte string that needed to be shifted into two 64 bit variables, or even larger?
Is shifting structures like this possible? Is there a better alternative?