Why does my C# decimal value get rounded to an integer
Date : March 29 2020, 07:55 AM
To fix this issue I have a class that includes a Decimal public property (generated by Entity Framework, code shown lower down), and am having problems with it being rounded to an integer. , The database has the field defined as decimal(18,0)
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Why integer result is being rounded - no decimal? Xcode Swift
Date : March 29 2020, 07:55 AM
it fixes the issue Integer is always a round number without decimals. If you need a number with decimals, you need something like a Double. This extension will help you to get a double value from your String: extension String{
func toDouble() ->Double{
return (self as NSString).doubleValue
}
}
labelPassedData.text.toDouble() //Your text as Double
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Correct number `x` divided into integer `y` parts and rounded to `2` decimal places so the sum is `x` again
Date : March 29 2020, 07:55 AM
will help you I want to show the values in the excel sheet such that , your question is pure math problem that can be rewritten as this: x/y = z = (floor(100*x/y)/100)+d
d=(x/y)-(floor(100*x/y)/100)
x/y = z = (floor(100*x/y)/100)+(x/y)-(floor(100*x/y)/100)
D=d*y=d(0)+d(1)+d(2)+...+d(y-1)
D=x-((floor(100*x/y)/100)*y)
d(i)= floor(100*D/y)/100 i = { 1,2,3,...,y-1 }
d(0)=D-(y-1)*(floor(100*D/y)/100)
{ 0.01, 0.02, 0.03, ... , 0.11 }
D=x-((floor(100*x/y)/100)*y)
d(i)= floor(100*D/y)/100+0.01 i = <0 , y-1> AND i < 100*(D%y)
d(i)= floor(100*D/y)/100 i = <0 , y-1> AND i >= 100*(D%y)
int i,n,y=12; double x=10000.00,fy=y,z,D,d,s;
z=floor((100.0*x)/fy)*0.01; // cell value rounded
D=x-(z*fy); // global correction
n=floor((100.0*D)+0.5); // iregular remainder
for (s=0.0,i=0;i<y;i++)
{
d=0.01*floor(100.0*D/y); // regular local correction
if (i<n) d+=0.01; // iregular local correction
// z+d is the new cell value and s is the sum for check ...
s+=z+d;
mm_log->Lines->Add(AnsiString().sprintf("%3i: %5.2f",i,z+d));
}
mm_log->Lines->Add("--------------");
mm_log->Lines->Add(AnsiString().sprintf("sum: %5.2f",s));
0: 833.34
1: 833.34
2: 833.34
3: 833.34
4: 833.33
5: 833.33
6: 833.33
7: 833.33
8: 833.33
9: 833.33
10: 833.33
11: 833.33
--------------
sum: 10000.00
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In bash how do I divide two variables and output the answer rounded upto 5 decimal digits?
Tag : bash , By : wiznick
Date : March 29 2020, 07:55 AM
wish of those help The problem here is that you missed the echo (or printf or any other thing) to provide the data to bc: $ echo "scale=5; 12/7" | bc
1.71428
sum=12
n=7
output=$(echo "scale=5;$sum/$n" | bc)
echo "$output"
$ "scale=5;sum/n"|bc
bash: scale=5;sum/n: No such file or directory
$ sum=3345699
$ n=1000000
$ echo "scale=5;($sum/$n)" | bc
3.34569
$ printf "%.5f" "$(echo "scale=10;$sum/$n" | bc)"
3.34570
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How do I modify a function to return true if the given decimal is even when it is rounded to an integer and false otherw
Date : March 29 2020, 07:55 AM
it fixes the issue I'm working on a Javascript exercise. I am trying to modify a function to return true when the given decimal is rounded to an even number and false when it is not. , You have described two steps.
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