This question already has answers here:
Why does floating-point arithmetic not give exact results when adding decimal fractions?
(31 answers)
Closed 3 years ago.
I am trying to store coordinate points in a double variable. It is very simple. I have x and y coordinates. My double variable stores them like that x.y . When I tried to convert this value into separate coordinates, I had some troubles.
I have tried these codes but still get same error.
//First try
double temp=memory.pop();
int x=(int)temp;
int y=(int)((temp-(int)temp)*100);
//Second try
double temp=memory.pop();
int x=(int)temp;
int y=100.0f*temp-(((int)temp)*100.0f);
In temp variable, I have 5.14 double number. After calculations, x should be 5 and y should be 14. However, x become 5, y become 13.
The issue is related to integer casting. Specfically (int)temp will floor to the first smaller integer. For insance, the value 5,14 can be (and probably is) not exaclty that but a bit smaller/larger for a matter of precision (since, as said in comments, the floating point numbers don't have infinite precision). So without loss of generality let's say that is 5,13999999999999, you can see that when you perform your operation you will obtain 13 instead of 14 for the second part of the coordinate.
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This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
C programming division
probably my question is very simple and stupid.
I would like to store the value of a division, in particular 1 / x where x is a
integer value.
int x = 17;
double result = 1/x;
I try to do it but I always get 0.000000 ... I try to enter a value fixed in x, for example 1/17 but always get the same value .. what's Wrong?
You are doing integer division.
Try the following and it will work as expected:
int x = 17;
double result = 1.0 / x;
The type of the 1 in the expression you have above is int, and the type of x is int. When you do int / int, you get an int back. You need at least one of the types involved to be floating point (float or double) in order for floating point division to occur.
Unlike in Mathematics, division in C++ can either refer to truncated integer division (what you did) or floating point division (what I did in my example). Be careful of this!
In my example, explicitly what we have is double / int -> double.
This question already has answers here:
Is floating point math broken?
(31 answers)
Closed 4 years ago.
Below you can find a basic C++ exercise with a for loop, the question is: what is the value of i after the loop runs. (Apparently, the value is 5.)
My question, on the 5th run (i=5), the value of a is 100, shouldn't the loop stop since the condition is a<100? Isn't the condition checked after a is multiplied with 10? (so after a is already 100)
If I check the value of a after the loop exit, it will be 1000. I am confused. Why does the loop not stop at 100?
#include <iostream>
using namespace std;
int main(void) {
float a;
int i = 0;
for(a = 0.01; a < 1e2; a *= 1e1)
++i;
return 0;
}
The exercise is designed to show that float (and double etc., floating point numbers in general) do not have the exact decimal value as what you see in the program text.
0.01 (decimal) cannot be exactly represented in the binary floating point format used, therefore multiplying 10 five times will not yield exactly 100.0.
See also: Floating point inaccuracy examples
You can't tell exactly what will happen with this code without knowing the details of the floating point implementation on your platform; the C++ standard leaves this intentionally flexible.
In essence 0.01 could be an approximation for the closest float to that double literal. So the stopping conditional a < 1e2 could be met prematurely or otherwise.
Note that along with 0.01, 1e1 is a literal of type double.
So the conversions to float are complicating things further. Then there is floating point strictness to consider.
Cut a long story short: don't use a floating point as the "index" in a loop.
This question already has answers here:
Division not outputting correct answer c++
(2 answers)
Closed 7 years ago.
I'm currently working on a C++ program where I have a bankAccount class and I need to calculate interest. Problem is, my function is rounding off my numbers to the whole number, even though I'm using a float as my variable type. So if I had code like this:
float BankAccount::CalculateInterest(int Time)
{
float thing;
thing = (967 / 365);
return thing;
}
thing ends up equaling 2.00000000, when it should equal 2.649315068.
Any ideas? I had my interest equation in there before, and the result was always 1.00000000 no matter what, so I put this in as a test and I saw that it's rounding to the "floor".
Thanks for your help!
It's not rounding.
You are dividing two integers and the result will be an integer.
You can remedy this by simply adding ".0" to both numbers, or explicitly defining them as floats.
967 and 365 are integers, so integer division is used and the remainder discarded. Make one a floating point number and floating point division will be done: 967 / 365.0.
Alternatively:
float thing = 967; // integer converted to float during assignment
thing /= 365; // division of float and converted integer
This question already has answers here:
Closed 11 years ago.
Possible Duplicate:
C programming division
probably my question is very simple and stupid.
I would like to store the value of a division, in particular 1 / x where x is a
integer value.
int x = 17;
double result = 1/x;
I try to do it but I always get 0.000000 ... I try to enter a value fixed in x, for example 1/17 but always get the same value .. what's Wrong?
You are doing integer division.
Try the following and it will work as expected:
int x = 17;
double result = 1.0 / x;
The type of the 1 in the expression you have above is int, and the type of x is int. When you do int / int, you get an int back. You need at least one of the types involved to be floating point (float or double) in order for floating point division to occur.
Unlike in Mathematics, division in C++ can either refer to truncated integer division (what you did) or floating point division (what I did in my example). Be careful of this!
In my example, explicitly what we have is double / int -> double.
This question already has answers here:
Why are floating point numbers inaccurate?
(5 answers)
Closed 8 years ago.
Now I understand floats are less accurate than double, but does this explain when I have the std::string:
"7.6317"
and I do:
float x = atof(myString.c_str());
getting 7.63170004 is expected? Is there any way I can tell the assignment of x to only read the first 4 decimal places? Or is this because of the way the float representation stores the number 7.6317?
Yes. It is expected. It is so-called floating point error.
Some floating point literals do not have an accurate representation in the computer, even if -- in decimal notation -- the number seems harmless. This is because the computer uses 2 as a base. So even if a number might have a finite representation in base 10, it might not have on in base 2.
you can do it like:
float x = floorf(val * 10000) / 10000;
i think it should work! see See