Thread: 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:

MapIn/64

is a char, then perhaps you actually want:

Code:

MapIn/64.0

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

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Mats

A cast from float to char is something like this:

Code:

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

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

--
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

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

Originally Posted by username101

hello Todd Burch,

what do you meen by ???

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

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Mats

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?

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

Code:

char c = (char)3.9f;

will result in 3 in c.

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Mats

Originally Posted by username101

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

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

Originally Posted by matsp

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

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Mats

Actually, it is

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

Todd