Reputation: 13
I have a function that returns the bits of a short (inspired from Converting integer to a bit representation):
bool* bitsToShort(short value) {
bool* bits = new bool[15];
int count = 0;
while(value) {
if (value&1)
bits[count] = 1;
else
bits[count] = 0;
value>>=1;
count++;
}
return bits;
}
How can I do the reverse? Converte the array of bits in the short?
Upvotes: 0
Views: 2401
Reputation: 726639
Like this:
bool* bits = ... // some bits here
short res = 0;
for (int i = 14 ; i >= 0 ; i--) {
res <<= 1; // Shift left unconditionally
if (bits[i]) res |= 1; // OR in a 1 into LSB when bits[i] is set
}
return res;
Upvotes: 1
Reputation: 76360
Essentially:
unsigned short value = 0;
for (int i = sizeof(unsigned short) * CHAR_BIT - 1; 0 <= i; --i) {
value *= 2;
if (bits[i)
++value;
}
This assumes that bits
points to an array of bool
with at least sizeof(unsigned short)
elements. I have not tested it. There could be an off-by-one error somewhere. But maybe not.
Upvotes: -1
Reputation: 4435
short shortFromBits(bool* bits) {
short res = 0;
for (int i = 0; i < 15; ++i) {
if (bits[i]) {
res |= 1 << i;
}
}
return res;
}
res |= (1<<i)
sets the i-th bit in res
to one.
Upvotes: 2