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Hi

I have to work with float values but the micro i am using does not support floating point operations.

for exmple

x=y+z;

z=0.255

y=1;

then x should be 1.255 but i am getting 1 as the out put.

how to use the float values with out using the float data type in code.

I have to work with float values but the micro i am using does not support floating point operations.

for exmple

x=y+z;

z=0.255

y=1;

then x should be 1.255 but i am getting 1 as the out put.

how to use the float values with out using the float data type in code.

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I an expressuion a + b , b is coming as float value.

when i add a and b. the value of b is not retaied as float but taken as int .

I have to retain that value

Anyway, a typical way to do that used to be to scale the input values to allow the smallest value needed. Like @ozo suggested, scaling the values lets it work with integer arithmetic. You have to remember to display the results correctly on the output by putting a decimal where it needs to be.

Do you mean print? If so, what print operations does the micro you are using support?

What do you mean by "retain"?

What do you want to do with the value?

sum = y+x;

product = y*x/scale;

What do you want to do with the value?

sum = y+x;

product = y*x/scale;

x= (int)((a+b)*(value *(a-b)/100))

here a+b is coming as interger

value *(a-b)/100) is decimal

so the second part is discarded as i am trypecasting to int

but i dont want that value to be discarded.

It seems we have a bit of a language barrier. Maybe another voice will help?

x= (int)((a+b)*(value *(a-b)/100))

here a+b

value *(a-b)/100)

so

Because all of your variables are integers, all values, including the intermediate values will have an integer type. When you divide two values, the result is an integer. In your example above you divide by 100. If the divisor is between 1 and 99, you get the same result, 0. In effect, 1/100 = 99/100, for the purpose of this calculation as both result in integer 0.

To capture the fractional part of the calculation you need to determine how many decimal places you need. It's probably beyond the scope of your project to use true floating point (variable precision), so decide how many digits to the right of the decimal that you'll need. Are 2 enough? With your equation that becomes the simplest. Are 3 enough? That resolution is 1/1000.

x= (int) ((a+b)*(value *(a-b)/100))

becomes

2 decimal places -- x = ((a+b) * (value *(a-b)))

3 decimal places -- x = ((a+b) * 1000 * (value *(a-b)/100))

When you convert the value to a string, insert a decimal point 2 (or 3 places) from the right.

If you need the fractional part for future computations, just subtract it. But remember that you're using integer data types so the value is 1000 times the fraction's true value.

// Compute value * 1000

x = ((a+b) * 1000 * (value *(a-b)/100))

// Compute fractional part

frac = x % 1000;

// Restore x to the correct scale

x /= 1000;

Good Luck,

Kent

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x= (int)((a+b)*(value *(a-b)/100))

use

x=(((a+b)*value*(a-b)) + 50)/100

that will work as long as (a+b)*value*(a-b)+50 is not greater than maximum int. the +50 makes correct rounding, for example if the product of the 3 factors is 150 by adding 50 it would be 200 and divided by 100 is 2. so the decimal result of 1.5 would be correctly rounded to 2.

Sara

Using scaled arithmetic works fine for a small number of variables with limited range and precision. Otherwise it may be difficult to make workable.

C

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If it supports integer operations in the appropriate range, you might use

z=255

y=1000

x=1255