The problem of using double type variables in if. . .

无风无尘

double ff = 1.0;
if (ff == 1) // precision
{
cout << "ok" << endl;
}
else
{
cout << "cancel" << endl;
}
Why is the output ok instead of cancel?
Thank you!

suizbs

Generally, it is compared with 0.9999999999 1.0000001 for size, no == comparison

orangedj

Example of comparison with zero:

const float EPSINON = 0.00001; // here is determined according to the accuracy
if ((x >=-EPSINON)&&(x <= EPSINON))

无风无尘

You misunderstood what I meant, I know that due to accuracy issues,
if ((x >=-EPSINON)&&(x <= EPSINON))

double ff=1.0;
if(ff==1)//Precision, but due to accuracy problems, this should not be true here! , But the procedure is considered to be established
{
cout<<"ok"<<endl;
}
else
{
cout<<" cancel"<<endl;
}
//Why is the output ok instead of cancel?
Thank you!

jjledu

It is normal to be established, and it is normal to not be established, so we should avoid that.

wanxiaohai

Double type data cannot be directly compared in this way. I also encountered this problem when writing programs.

food0012

You are like writing directly
if(0.1f == 0.1f)
This comparison can be true. The floating-point number is actually a binary data string, and of course it can be equal. It is said that its imprecision is reflected in
1/3 is such a numerical representation

twfx007

if(ff==1)
Comparison can only be performed between variables of the same type, so the compiler will first convert 1 to double, and then compare.

无风无尘

But I think if (ff==1) this sentence is definitely not true, because double can only hold 10 significant digits, how can all the digits after 10 be guaranteed to be 0?
However, the compiler thinks that if (ff==1) is valid after 100 cycles of comparison.

food0012

1.0f is expressed in memory as
00 00 80 3F Four bytes of data, of course, can have the same value, as long as the representation of another floating point number is also such four bytes.