System.Convert.ToInt vs (int)

Question

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.

Answer 1

Casting will throw an exception if it fails or the input is null. The Convert class will return 0 if the input is null. This has been covered numerous times (search for cast).

So when deciding which one to use, you need to weigh the risk of an exception and the readability of what you're trying to accomplish.

Answer 2

You can use Convert when you have a string that you want to convert to an int

int i = Convert.ToInt32("1234");

Convert and casting will both throw an exception if they fail.

i.e, this will still throw an exception, it will not return 0:

Convert.ToInt32("1234NonNumber");

In many cases Convert and casting will have the same result, but a cast is often times easier to read.

Answer 3

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.

Answer 4

Convert.ToInt32() is used on strings (http://msdn.microsoft.com/en-us/library/sf1aw27b.aspx) while casting can only be used on types that have internal converters (numeric types). The real trick comes in deciding between Int32.Parse and Convert.ToInt32(). Convert.ToInt32() is tolerant of a null parameter and returns 0 while Int32.Parse() will throw an ArgumentNullException.

Answer 5

Rounding is also handled differently:

x=-2.5 (int)x=-2 Convert.ToInt32(x)=-2
x=-1.5 (int)x=-1 Convert.ToInt32(x)=-2
x=-0.5 (int)x= 0 Convert.ToInt32(x)= 0
x= 0.5 (int)x= 0 Convert.ToInt32(x)= 0
x= 1.5 (int)x= 1 Convert.ToInt32(x)= 2
x= 2.5 (int)x= 2 Convert.ToInt32(x)= 2

Notice the x=-1.5 and x=1.5 cases.
In some algorithms, the rounding method used is critical to getting the right answer.