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Error converting double to int

Hi everyone !
I am facing a strange issue converting a double to int.
Here is what I do :qDebug() << 0.04*double(9)/0.0016 << int(0.04*double(9)/0.0016);
And the result I have is :
225 224
It gives me the same result (224) with :
qDebug() << int(0.04*9/0.0016);
I first thought that the double value of 225 was an approximation of the real value that may be closer to 224 but when I print it, why would it print exactly 225 and not 224,3257 for example ?
Moreover, if the debugger chose to round the double value to 225, why is it rounded to 224 when converting to int then ?
When a test this computation on my calculator it gives me exactly 225...
Thanks for your help !

@HB76
Alldouble
s (orfloat
s) in C/C++ are only approximate values. You cannot guarantee that any number, whole/integer or with fraction/decimal places has any prefect representation, other than the number0
.int(doubleValue)
always rounds down. Use another method for control over rounding, or lazilyint(doubleValue + 0.5)
. Don't rely on the output ofqDebug()
to outputdouble
s to any particular decimal places, use something likeQString::number(double n, char format = 'g', int precision = 6)
to explicitly pick your output format.Moreover, if the debugger chose to round the double value to 225
qDebug()
is not "the debugger" :) Examine the value inside the actual debugger for an accurate value.

You are simply dealing with rounding issues on the output.
qDebug() is an output channel which is intended for debugging purposes, but it is not the debugger.
@HB76 said in Error converting double to int:
why would it print exactly 225 and not 224,3257 for example ?
How do come with this example? It has nothing to do with the actual numbers you are using.
Try
#include <iostream> ... qDebug() << 0.04*double(9)/0.0016 << int(0.04*double(9)/0.0016); cout << 0.04*double(9)/0.0016 << int(0.04*double(9)/0.0016); cerr << 0.04*double(9)/0.0016 << int(0.04*double(9)/0.0016);
It should show the same as qDebug() output

@JonB I know that double values are approximated but normaly all 32 bits integers values should be perfectly represented by 64 bits double, and when I try :
double a = 0.04*double(9)/0.0016; qDebug() << QString::number(a,'g',6);
it returns me "225"...
And the most amazing fact is when I try :
qDebug() << QString::number(a,'g',6) << QString::number(a,'g',6).toInt();
It returns "225" 225

@koahnig Yes I just invented these number to illustrate my point

@HB76 said in Error converting double to int:
double a = 0.04*double(9)/0.0016;
I told you to try this in the debugger if you really want to know what the
double
value is. If you had bothered to try, you would have found224.99999999999997
. And what isint(224.99999999999997)
, or to 6 decimal places? TryqDebug() << QString::number(a,'g',20)
.normaly all 32 bits integers values should be perfectly represented by 64 bits double
How does that have any relevance to the code you show? It doesn't.

Ok I tried to run this computation on python and it gaves me :
a = 0.04 * 9 / 0.0016 a
224.99999999999997
int (a)
224
round(a)
225
It was just that Qt didn't showed the real value of a even with this output format
QString::number(a,'g',16)
Thanks for your help !

@HB76 said in Error converting double to int:
It was just that Qt didn't showed the real value of a even with this output format
QString::number(a,'g',16)
FGS. Count the number of digits. Yes, that is all digits, not just the ones after the decimal place, as per the documentation:
For the 'g' and 'G' formats, the precision represents the maximum number of significant digits (trailing zeroes are omitted).
Is 16 enough?