what is the meaning of this error in C#.net?

The Math.Floor method has two overrides - one that takes a double argument and one that takes a decimal.

I would imagine that the variable nDegree is some kind of integer and therefore so too is nDegree/2. Consequently, the compiler doesn't know which override to use.

However, unless nDegree can be negative, you don't actually need to use Math.Floor here.If nDegree is even, then nDegree/2 will be exactly one half of it and, if nDegree is odd (45 say), then nDegree/2 will be automatically rounded down to the lower integer (i.e. 22).

So you could just use:

short nHalf = (short)(nDegree/2);

However, if nDegree could be negative, you could use:

short nHalf = (short)Math.Floor((double)nDegree/2.0);

The cast and use of a double literal (i.e. 2.0) forces the compiler to use floating point arithmetic instead of integer arithmetic.

1 Reply

Know the answer? Post it — somebody with the same question will find it here.

Alan19y ago
The Math.Floor method has two overrides - one that takes a double argument and one that takes a decimal.

I would imagine that the variable nDegree is some kind of integer and therefore so too is nDegree/2. Consequently, the compiler doesn't know which override to use.

However, unless nDegree can be negative, you don't actually need to use Math.Floor here.If nDegree is even, then nDegree/2 will be exactly one half of it and, if nDegree is odd (45 say), then nDegree/2 will be automatically rounded down to the lower integer (i.e. 22).

So you could just use:

short nHalf = (short)(nDegree/2);

However, if nDegree could be negative, you could use:

short nHalf = (short)Math.Floor((double)nDegree/2.0);

The cast and use of a double literal (i.e. 2.0) forces the compiler to use floating point arithmetic instead of integer arithmetic.
0