[C++] Converting float to int

[sixb0nes]

Hello. I've come across a rather interesting problem involving float to int conversions. If I have a float, say, 98.85, and I assign an int to this float times 100...

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float myfloat = 98.85;
int myint = myfloat * 100;

myint becomes 9884, which is not what I want. Is there anyway to do this float->int conversion without losing precision and without making myint a float? I've figured a rather hack-like workaround by simply adding one to myint, but I would much rather a solution which did not involve this step.
Thanks.

[rabbit]

Floats are weird in C++.  Your '98.85' is actually '98.849999...'  Adding 0.01 and then multiplying should work.

[Maddox]

Hello. I've come across a rather interesting problem involving float to int conversions. If I have a float, say, 98.85, and I assign an int to this float times 100...

Code Select Expand


float myfloat = 98.85;
int myint = myfloat * 100;

myint becomes 9884, which is not what I want. Is there anyway to do this float->int conversion without losing precision and without making myint a float? I've figured a rather hack-like workaround by simply adding one to myint, but I would much rather a solution which did not involve this step.
Thanks.

Try this...

Code Select Expand


float myfloat = 98.85f;
int myint = (int)myfloat * 100;


[MyndFyre]

Hello. I've come across a rather interesting problem involving float to int conversions. If I have a float, say, 98.85, and I assign an int to this float times 100...

Code Select Expand


float myfloat = 98.85;
int myint = myfloat * 100;

myint becomes 9884, which is not what I want. Is there anyway to do this float->int conversion without losing precision and without making myint a float? I've figured a rather hack-like workaround by simply adding one to myint, but I would much rather a solution which did not involve this step.
Thanks.

Try this...

Code Select Expand


float myfloat = 98.85f;
int myint = (int)myfloat * 100;


That would result in myint == 9800.

[Joe[x86]]

I think he may have meant int myint = (int)(myfloat * 100); so that the multiplication would process before the casting.

[rabbit]

It would automatically be cast to int because he is multiplying by an int and storing as an int.  A typecast is not needed.