Please provide me the logic to print the result using PRINTF [closed] - c++

Question

Closed. This question does not meet Stack Overflow guidelines. It is not currently accepting answers.
We don’t allow questions seeking recommendations for books, tools, software libraries, and more. You can edit the question so it can be answered with facts and citations.
Closed 5 years ago.
Improve this question
need an programming logic to print an 4 decimal points
EX: scalar should be 0 to -5
value = 10006 , scalar = -3 then print result = 10.0060 (4 decimals)
value = 123 ,scalar = -5 then print result = 0.0012 (4 decimals)**
required value/divisor = 10 , value%divisor = 0060 (required logic after decimals )
I tried like this:
divisor = std::pow(10,std::abs(scalar));
**Result = snprintf(X,Y,"%d.%0*d",value/scalar,4,value%scalar);**
I'm not allowed to use float , setprecision() .
It does not necessarily represent the actual value , but we can format that value to print with logic like the original one (by using the logic , add ...subtract...pow etc)
std::int32_t divisor = static_cast(std::pow( 10.0F, std::abs( Scalar)) );
but int the above result modulus scalar value with 0 are not considering.
**Please provide me the logic to print the above result with scalar condition

Answer 1

In order to print decimals (easily), you need floating point:
printf("%10.6f", static_cast<double>(1) / 3);
If one of the arguments to division is floating point, the compiler will promote the expression to floating point.
Integral or scalar division will lose decimals.
You are always welcome to write your own division function.
Edit 1: Shifting
You don't need to use the pow function (especially since it's floating point).
Use a loop, in the loop multiply your divisor by 10 each time.
double result = (double) scalar / value;
int divisor = 10;
int i;
for (i = 0; i < NUMBER_OF_DECIMALS; ++)
{
// Isolate a digit using math.
// print the digit
divisor += 10;
}
The math part is left as an exercise for the OP.

Answer 2

In this homework exercise you are expected to perform your own number formatting, rather than using that of a library.
For instance, to format and output 100 as "100"
int accum = 100;
int position = 0;
while (accum > 0)
{
printf("%d", accum % 10);
accum /= 10;
position += 1;
}
For your homework assignment, you need to modify the above loop so that it puts a printf(".") in the correct place in the output number. Your answer is likely to involve multiplying accum before the loop and testing position relative to scalar