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pozzari
Joined: 30 Oct 2010
Posts: 3
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| float 1 div by 2 == 0 ?? |
 Posted: Mon May 28, 2012 4:09 pm |
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Very single program running on pic18f452
This displays 0.00 on my debug... why??
| Code: |
void main()
{
float div_result;
int part1;
int part2;
part1 = 1;
part2 = 2;
div_result = part1 / part2;
fprintf(usart_debug,"Ini=%f\r\n",div_result);
while(1){//loop forever
}
}
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dyeatman
Joined: 06 Sep 2003
Posts: 1934
Location: Norman, OK
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| Posted: Mon May 28, 2012 4:19 pm |
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Try this...
div_result = (float) part1 / part2;
_________________
Google and Forum Search are some of your best tools!!!! |
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pozzari
Joined: 30 Oct 2010
Posts: 3
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| Posted: Mon May 28, 2012 4:59 pm |
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| div_result = (float)part1 / (float)part2 worked fine... thanks a lot |
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RF_Developer
Joined: 07 Feb 2011
Posts: 839
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| Posted: Tue May 29, 2012 1:19 am |
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Pozzari, do you understand why that worked?
RF Developer |
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Ttelmah
Joined: 11 Mar 2010
Posts: 19537
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| Posted: Tue May 29, 2012 1:32 am |
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Worth also explaining 'why'.
In C, the maths operations, when compiled, look at the 'types' of the value/values _involved in each part of the calculation_ (not the result), and select the maths 'type', required by the 'highest type' of these. Types are ordered:
| Code: |
Float, signed int32, int32, signed int16, int16, signed int8, int8.
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So in the original sum, the compiler says "int8/int8, use int8 arithmetic".
Only one of the values (either), actually needs to be 'cast' to the higher type, since the highest type involved is selected. So:
| Code: |
div_result = (float)part1 / part2;
div_result = part1 / (float)part2;
div_result = (float)part1 / (float)part2;
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will all give the same result. Once one value in the sum is raised to the higher type, the others are automatically raised as needed.
Now, this is very useful, since it means you can elect to use 'lower' maths types (which are much faster/smaller), until a higher type is needed.
So (for instance),
| Code: |
result = (int16)part1*part2;
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Uses int16 arithmetic to multiply the two int8 values - takes about 17 instructions, versus 180 for the 'float' version.
Then:
| Code: |
result=((int16)part1*part2)/(float)part3;
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Uses int16 for the multiplication, and then switches to float for the division. Still saving perhaps 150 instructions.....
Historically, C was written to do this, since it allowed efficient use of the chip's resources. However on things like the PC, with the addition of a hardware maths co-processor, this has become slightly 'bypassed', with all the sums using this hardware feature, and therefore not worrying about the extra performance involved....
Best Wishes |
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