Maths in C

I am trying to perform 2 math equations in C and send the result as a char

x = 64*(pow(2,(conversionH/256))-1); // this is 64*(2^(conversionh/256)-1)

ConversionH is a char

the next one i need to use log base 2

y = 256*(log10(1+(MapIn/64))*3.321928); // so idealy this needs to be: 256*log2(1+(mapin/64)

MapIn is a char

so my inputs and outputs are char, but this dose not seen to work. Please can you help me.
thank you.

if MapIn in

Code:

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MapIn/64

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is a char, then perhaps you actually want:

Code:

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MapIn/64.0

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to force the compiler to make the calculation using floating point, rather than an integer calculation.

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Mats

I shal try that now. thanks

also how would i go about converting the float (result) into a char, unless the is another way to do log base 2 in C?

A cast from float to char is something like this:

Code:

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char c = (char) (4.2f / 1.4f);

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Which should give you a value of 3 in the char.

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Mats

All log operations in C must be in base 10 ( log10() ) or log e ( log() ).

Just use the change of base logic for your log base 2 to get into log base 10.

Todd

this is what i've got, an ADC that dose this

Code:

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conversionH = ADC0L;          // read ADC0H data
               
            x = (char) (64*(pow(2,(conversionH/256))-1));
           
            MapOut = x;
            SBUF0 = MapOut;          // send data H to UART

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then a DAC that dose this

Code:

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        MapIn = SBUF0;              // read data H from UART
        RI0 = 0;                    // clear Receive Interrupt Flag

        y =  (char) (256*(log10(1+(MapIn/64.0))*3.321928));
        dacH = y;

        DAC0H = dacH;              // convert digital to analogue

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it still doesn't seem to work?

Cast all your numeric constants, like matsp said, to float, ie, 256.0, not 256.

hello Todd Burch,

what do you meen by

Quote:

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Just use the change of base logic for your log base 2 to get into log base 10

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???

Quote:

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Originally Posted by username101

hello Todd Burch,

what do you meen by ???

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Like you did, log2(x) == log10(x)*log10(2)

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Mats

that seems to of worked. thank you.

Just one more thing. how dose the float round to the char? ie up or down or what?

Quote:

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Originally Posted by username101

that seems to of worked. thank you.

Just one more thing. how dose the float round to the char? ie up or down or what?

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If you just cast a float to char, then it just chops of the decimals, so for example

Code:

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char c = (char)3.9f;

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will result in 3 in c.

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Mats

so how could i round it, so that 3.5 goes to 4 and 3.49 goes to 3?

Quote:

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Originally Posted by username101

so how could i round it, so that 3.5 goes to 4 and 3.49 goes to 3?

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You would add 0.5 to the result, then 3.5 will become 4.0 which becomes 4. , and 3.49 becomes 3.99 - which rounds down to 3.

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Mats

thank you very much for all of your help

Quote:

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Originally Posted by matsp

Like you did, log2(x) == log10(x)*log10(2)

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Mats

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Actually, it is

log2(x) == log10(x) / log10(2)

Todd

Quote:

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Originally Posted by Todd Burch

Actually, it is

log2(x) == log10(x) / log10(2)

Todd

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Doh! Thanks for fixing.


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Mats

np - just took my Algebra final last night. PASSED!