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[email protected]
Joined: 28 Jul 2014
Posts: 38
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| 10 bits conversion |
 Posted: Wed Jan 06, 2021 3:44 am |
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i have a 10 bits in an array
bits[]={1,1,1,1,1,1,1,1,1,1};
the value of the above is = 1023
now how can i convert the bits in the array and store the result in int16 bit variable. |
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Ttelmah
Joined: 11 Mar 2010
Posts: 19499
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| Posted: Wed Jan 06, 2021 6:55 am |
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There are lots of different ways. However by far the most efficient
(depending on how the 'bits' are stored), is to use a union or a #locate.
Now I say 'depending on how the bits are stored'. This approach only works
if they are stored as int1 values.
| Code: |
union {
unsigned int16 word;
int1 bits[10];
} combiner;
//Then if combiner.bits is used as the array, combiner.word contains
//the 16bit value.
//You can also do the same effective thing using #locate.
int1 bits[] = {1,1,1,1,1,1,1,1,1,1};
unsigned int16 word;
#locate word=bits;
//Then word is the int equivalent of the int1 array.
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If the values are not int1, then you have to build the result manually.
Unfortunately, this then wastes a lot of processor time. The most efficient
way is:
| Code: |
unsigned int16 from_bits(void)
{
unsigned int16 mask=1, result=0;
unsigned int8 ctr;
for (ctr=0;ctr<10;ctr++)
{
if (bits[ctr])
result+=mask;
mask<<=1;
}
return result;
}
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[email protected]
Joined: 28 Jul 2014
Posts: 38
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| Posted: Wed Jan 06, 2021 7:33 am |
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| Ttelmah wrote: | There are lots of different ways. However by far the most efficient
(depending on how the 'bits' are stored), is to use a union or a #locate.
Now I say 'depending on how the bits are stored'. This approach only works
if they are stored as int1 values.
| Code: |
union {
unsigned int16 word;
int1 bits[10];
} combiner;
//Then if combiner.bits is used as the array, combiner.word contains
//the 16bit value.
thank you for your reply your code is working.
this is my code this is also working but yours is simple!
while(true)
{
unsigned int8 dd[]={0,0,0,0,0,0,0,0,0,1};
unsigned int16 val=0;
unsigned int16 mul=0;
for(int8 i=0;i<10;i++)
{
mul=mul*2;
if(mul==0)
{
mul=1;
}
if(dd[i]==1)
{
val=val+mul;
}
}
lcd_gotoxy(1, 1);
printf(lcd_putc1, "%04lu",val);
}
//You can also do the same effective thing using #locate.
int1 bits[] = {1,1,1,1,1,1,1,1,1,1};
unsigned int16 word;
#locate word=bits;
//Then word is the int equivalent of the int1 array.
|
If the values are not int1, then you have to build the result manually.
Unfortunately, this then wastes a lot of processor time. The most efficient
way is:
| Code: |
unsigned int16 from_bits(void)
{
unsigned int16 mask=1, result=0;
unsigned int8 ctr;
for (ctr=0;ctr<10;ctr++)
{
if (bits[ctr])
result+=mask;
mask<<=1;
}
return result;
}
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temtronic
Joined: 01 Jul 2010
Posts: 9221
Location: Greensville,Ontario
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| Posted: Wed Jan 06, 2021 7:58 am |
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yeesh, I was looking and thinking about this while coffee slowly dripped,dripped, drip...
Mr. T's union is perfect !!
I was looking at a loop with bit_test to make the sum then sorry, got distracted with fresh coffee...
Providing the 'order' of the bits is correct, the union is the fastest,easiest way to do it.
Jay |
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