System.Convert.ToInt vs (int) in C# – Key Differences

c++types

I noticed in another post, someone had done something like:

double d = 3.1415;
int i = Convert.ToInt32(Math.Floor(d));

Why did they use the convert function, rather than:

double d = 3.1415;
int i = (int)d;

which has an implicit floor and convert.

Also, more concerning, I noticed in some production code I was reading:

double d = 3.1415;
float f = Convert.ToSingle(d);

Is that the same as:

float f = (float)d;

Are all those otherwise implicit conversions just in the Convert class for completeness, or do they serve a purpose? I can understand a need for .ToString(), but not the rest.

Best Answer

Casting to int is implicit truncation, not implicit flooring:

double d = -3.14;
int i = (int)d;
// i == -3

I choose Math.Floor or Math.Round to make my intentions more explicit.

Related Solutions

C# Type Conversion – Difference Between Convert.ToInt32 and (int)

(int)foo is simply a cast to the Int32 (int in C#) type. This is built into the CLR and requires that foo be a numeric variable (e.g. float, long, etc.) In this sense, it is very similar to a cast in C.

Convert.ToInt32 is designed to be a general conversion function. It does a good deal more than casting; namely, it can convert from any primitive type to a int (most notably, parsing a string). You can see the full list of overloads for this method here on MSDN.

And as Stefan Steiger mentions in a comment:

Also, note that on a numerical level, (int) foo truncates foo (ifoo = Math.Floor(foo)), while Convert.ToInt32(foo) uses half to even rounding (rounds x.5 to the nearest EVEN integer, meaning ifoo = Math.Round(foo)). The result is thus not just implementation-wise, but also numerically not the same.

C# – Difference Between Convert.ToInt16 and (Int16)

The numeric literal 10 is treated as an integer, and more specifically an Int32. Though you typed your variable as object, under the covers it is still the integer. You can only unbox a value type to its same type, or to a nullable version of that type, directly.

For example, this code:

int i = 10;
object o = i;
short j = (short)o;

Will not execute, because the original value of i is not a short, it is an integer. You have to first unbox to integer, then you can cast to short.

short j = (short)(int)o;

Convert.ToInt16 sidesteps that issue, and the way it does it is an implementation detail. However, that method has multiple overloads that accepts multiple types, including strings, so it is not the equivalent of code using a direct cast.

Edit: I noticed I'm mixing terms here, so just so it's clear for a novice C# reader, the names short and Int16 are interchangeable for a 16 bit integer, as are the names int and Int32 for 32 bit integers. In C#, short and int are aliases for the .NET types Int16 and Int32, respectively.

Best Answer

Related Solutions

C# Type Conversion – Difference Between Convert.ToInt32 and (int)

C# – Difference Between Convert.ToInt16 and (Int16)

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