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세 개의 작은 숫자를 하나의 두 배에 저장하는 방법은 무엇입니까? (How to store three small numbers into one double?)
I have int A, B, C. And A is in range 0‑9999, B is 0‑99, C is 0‑99.
Because the function must return only one double, I think of putting them all into one number. Otherwise I need to call function three times. But I cannot write an efficient code to do this. This will be called millions times, so it should be quite effective, but no ASM.
I need a function double pack3int_to_double(int A, int B, int C) {}
참조 솔루션
방법 1:
Couldn't you just store A + 1000B + 100000C?
For example, if you wanted to store A = 1234, B = 6, and C = 89, you'd just store
89061234
CCBAAAA
You can then extract the numbers by casting the double to an int and using standard integer division and modulus tricks to recover the individual values.
Hope this helps!
방법 2:
If A<10,000 and B & C <100, A can be expressed with 14 bits, and B & C with 8 bits. Thus you need 30 bits in total. You could therefore pack/unpack the integers by shifting it to the right place:
int packed = A + B<<14 + C<<22;
A = packed & 0x3FFF; B = (packed >> 14) & 0xFF; C = (packed >> 22) & 0xFF;
Bit shifting is of course MUCH faster than multiply/divide, and you can cast the int to a double and vice versa.
방법 3:
This is technically not legal C code, so you would use this at your own risk:
typedef union {
double x;
struct {
unsigned a : 14;
unsigned b : 7;
unsigned c : 7;
} y;
} result_t;
The C standard doesn't allow using a union member to write a value and a different one to read it out, but I am not aware of a compiler that does the static analysis to diagnose such a problem (it doesn't mean one won't do so in the future). Also, using certain int values may result in a trap representation for a double. But, if you know your system will not generate any trap representations, you can consider using this.
double pack3int_to_double(int A, int B, int C) {
result_t r;
r.y.a = A;
r.y.b = B;
r.y.c = C;
return r.x;
}
void unpack3int_from_double (double X, int *A, int *B, int *C) {
result_t r = { X };
*A = r.y.a;
*B = r.y.b;
*C = r.y.c;
}
방법 4:
You can use out parameters in function call and retrieve all 3 int variables.
방법 5:
You could return a NaN double with the data stored in the mantissa. That gives you 53 bits to utilize. Should be plenty.
http://en.m.wikipedia.org/wiki/NaN
(by exebook、templatetypedef、Reinhard Männer、jxh、Pritesh Tayade、StilesCrisis)