This question already has answers here:
How to get fractions in an integer division?
(2 answers)
Closed 7 years ago.
I'm not getting the expected result when I do division. Can someone please tell me what I'm doing wrong?
int page_list_size = 20;
int page_fault_counter = 0;
double failure = 0.0;
double success = 0.0;
failure = page_list_size / page_fault_counter;
success = 1 - failure;
printf("failure Is %lf\n",failure);
printf("success Is %lf\n",success);
failure Is 1.000000
success Is 0.000000
Should be some decimal number between 0 and 1, and they should add up to 1.
You are not allowed to divide by 0. If the denominator is not 0, dividing ints results in an int, so you need to use floats or doubles to get a number between 0 and 1.
Related
This question already has answers here:
Is floating point math broken?
(31 answers)
Closed 2 years ago.
I have written this piece of code in my computer and the result is 7 instead of 8 (the correct result ... I think).
I don't know why... Can somebody help me?
#include <stdio.h>
int main() {
int num;
num = (68/10.0 - 68/10)*10;
printf("the result %d", num);
return 0;
}
double typically represents exactly about 264 different numbers. 68/10.0 is not one of them,
As a binary64, 68/10.0 is about
6.7999999999999998223643161..., the closest value to 6.8 that is a multiple of a dyadic rational. # AntonH
68/10 is an integer division with a quotient of 6.
(68/10.0 - 68/10)*10 is thus about 7.9999999999999982236431606...
Assigning that to an int is 7 not 8 as the fraction is discarded even though it is very close to 8.
When converting a floating point value consider round to the the closest, rather than truncating.
num = lround((68/10.0 - 68/10)*10);
This question already has answers here:
Why does dividing two int not yield the right value when assigned to double?
(10 answers)
Closed 6 years ago.
I have an array of double:
double theoretical_distribution[] = {1/21, 2/21, 3/21, 4/21, 5/21, 6/21};
And I am trying to computer it's entropy as:
double entropy = 0;
for (int i = 0; i < sizeof(theoretical_distribution)/sizeof(*theoretical_distribution); i++) {
entropy -= (theoretical_distribution[i] * (log10(theoretical_distribution[i])/log10(arity)));
}
However I am getting NaN, I have checked the part
(theoretical_distribution[i] * (log10(theoretical_distribution[i])/log10(arity)))
And found it to return NaN itself, so I assume it's the culprit, however all it's supposed to be is a simple base conversion of the log? Am I missing some detail about the maths of it?
Why is it evaluating to NaN.
You are passing 0 to the log10 function.
This is because your array theoretical_distribution is being populated with constant values that result from integer computations, all of which have a denominator larger than the numerator.
You probably intended floating computations, so make at least one of the numerator or denominator a floating constant.
This question already has answers here:
C program to convert Fahrenheit to Celsius always prints zero
(6 answers)
Closed 7 years ago.
I've written a code for the following program but the output seems to be wrong.
Question:
https://www.hackerrank.com/challenges/plus-minus
Code:
#include <stdio.h>
int main() {
int N,num,i,cp=0,cn=0,cz=0;
double fp,fn,fz;
scanf("%d",&N);
for(i=0;i<N;i++)
{
scanf("%d",&num);
if(num>0)
cp=cp+1;
else if(num==0)
cz=cz+1;
else
cn=cn+1;
}
fp=cp/N;
fn=cn/N;
fz=cz/N;
printf("%lf\n%lf\n%lf",fp,fn,fz);
return 0;
}
The Output comes as:
0.000000
0.000000
0.000000
Istructions:
fp=cp/N;
fn=cn/N;
fz=cz/N;
Are performed as integer division.
Change your code to:
fp=(double)(cp)/(double)(N);
fn=(double)(cn)/(double)(N);
fz=(double)(cz)/(double)(N);
you are doing an integer division which creates only integer results. If you want to calculate floating point results you need to perform floating point divisions.
int a = 1;
int b = 3;
int c = a / b;
// c is now 0 -> integer division
double i = 1.0;
double j = 3.0;
double k = i / j;
// k is now 0.3333333 -> floating point division
For correct result cast these expression's to double-
like this
fp=cp/N; // fp=(double)cp/N;
fn=cn/N; // fn=(double)cn/N;
fz=cz/N; // fz=(double)cz/N;
Working code
In previous case if cp(or cn or cz) is less than N then due to integer division you will get 0(fraction part will be discarded). Therefore m these casts .
Another method would be to use all variables as double .
This question already has answers here:
C program to convert Fahrenheit to Celsius always prints zero
(6 answers)
Closed 8 years ago.
I have this little program in C that calculates the square root x of a positive integer N using a recursive function (implemented using a while loop). If I calculate x using this:
x = (1/2)*(x + N/x) //x0 = 1.0
Then x keeps growing to inf and then nan. However if I use this:
x = (x + N/x)/2 //x0 = 1.0
It works fine, why? Thanks.
1/2 does integer division, its result is 0, change either or both operand to double, e.g:
1.0/2
This question already has answers here:
Modulo operation with negative numbers
(12 answers)
Problem using modulo with negative numbers in decryption program
(2 answers)
Closed 4 years ago.
I have couple line of C code testing the modulo operator as follows:
// line 1
printf("%d\n", 5 % (-3)); => output: 2
// line 2
printf("%d\n", -5 % 3); => output: -2
I know that the sign of the modulo depends on the sign of the numerator, but I am curious why not otherwise?
5/(-3) = -1;
(-5)/3 = -1;
If that is agreed then let's calculate the remainder
Remainder = Dividend - ( Divisor * Factor)
5 - (-3 * -1) = 2
-5 - (3 * -1) = -2