What does &~0x0F mean?
Author
I was reading the newest sweet snippet "SSE2 for Dummies (who know C/C++)" at http://gamedev.net/reference/articles/article1987.asp, and I need a little help understanding the following function please:
inline void *AllignData(void *data)
{
return (void *)(((int)data + 15) &~ 0x0F);
}
Is &~0x0F the expression for the start of a memory block??? I am assuming from the article that this function adds 15 bytes to the address of data from the start of the memory block. Is this correct? Thanks.
data is the pointer you want to align
doing "return (void *)((int)(data) &~0x0f))" will align the pointer
to the nearest place where it divides by 16.
0x0f = 15 , or in binary 00000000000000000000000000001111 (hope its 32 bit),now ~ox0f is 11111111111111111111111111110000
doing a (void *)((int)(data) & 0~0xf)) that is logical and , will just clear the 4 lowest bits and and that pointer is aligned.
one problems is that you doing want the pointer to be aligned backwards , because that will give you a pointer to a place you did not malloc, you want it to go only forward , that is align to the nearest 16 only in the forward direction , that why you need to add the +15 , this solved this little problem
doing "return (void *)((int)(data) &~0x0f))" will align the pointer
to the nearest place where it divides by 16.
0x0f = 15 , or in binary 00000000000000000000000000001111 (hope its 32 bit),now ~ox0f is 11111111111111111111111111110000
doing a (void *)((int)(data) & 0~0xf)) that is logical and , will just clear the 4 lowest bits and and that pointer is aligned.
one problems is that you doing want the pointer to be aligned backwards , because that will give you a pointer to a place you did not malloc, you want it to go only forward , that is align to the nearest 16 only in the forward direction , that why you need to add the +15 , this solved this little problem
"&" is the binary AND operator. In the expression c = a & b, each bit in c is set according the corresponding bits in a and b using this table:
~0x0F == 0xFFFFFFF0 for 32-bit values.
The function takes the value of the pointer data and returns the smallest address greater than or equal to data that is a multiple of 16. To get an idea of how it works, take a look at this table:
[edited by - JohnBolton on September 2, 2003 7:15:08 PM]
a b c- - -0 0 00 1 01 0 01 1 1 a c- -0 11 0 ~0x0F == 0xFFFFFFF0 for 32-bit values.
The function takes the value of the pointer data and returns the smallest address greater than or equal to data that is a multiple of 16. To get an idea of how it works, take a look at this table:
data data+15 (data + 15) & ~0x0F---- ------- -------------------0 15 001 16 162 17 163 18 164 19 165 20 166 21 167 22 168 23 169 24 1610 25 1611 26 1612 27 1613 28 1614 29 1615 30 1616 31 1617 32 3218 33 32... ... ...31 46 3232 47 3233 48 4834 49 48... ... ...47 62 4848 63 4849 64 6450 65 64... ... ...63 78 6464 79 6465 80 8066 81 80... ... ...[edited by - JohnBolton on September 2, 2003 7:15:08 PM]
quote:Original post by gonen
one problems is that you doing want the pointer to be aligned backwards , because that will give you a pointer to a place you did not malloc, you want it to go only forward , that is align to the nearest 16 only in the forward direction , that why you need to add the +15 , this solved this little problem
I suspose I should post this in the article discussion thread, but since you brought it up
: You also have to add 15 to the size of the memory allocated, in the article he adds sixteen, by-way-of two additional doubles. [edited by - Magmai Kai Holmlor on September 2, 2003 7:42:40 PM]
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