How well do you know your TryParse()?

Ravi Bhavnani

What would you expect foo to be after this code fragment runs? Post your answer without first peeking at the docs! :-D

string badNumericString = "Bogus";
Decimal foo = Decimal.MinValue;
bool status = Decimal.TryParse (badNumericString, out foo); // status is false, as expected

Answer:

Decimal.MinValue (since the parse failed)
Something else (if so, what?)

/ravi

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Alan N

Well, ok I have peeked to see if the behaviour is the same as Int32.TryParse and it is. Importantly the out keyword requires that the variable passed as the argument is definitely assigned by the method and whether the method returns true or false has no bearing on that requirement. So my answer is I would expect foo to be assigned but whether or not that changes the value depends on whether the method assigns MinValue or something else. Alan.

Ravi Bhavnani

Alan N wrote:

the out keyword requires that the variable passed as the argument is definitely assigned by the method

Yes, absolutely. And I was hoping TryParse() would assign the incoming value of foo when the parse failed. When it failed to do so, I decided to write my own SafeTryParse() implementation that would do just that, only to be chastised by the compiler - an out parameter is always assumed to be uninitialized and therefore its value can't be used. :) /ravi

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Alan N

I did that test, only to get my wrist slapped too! Alan.

Ravi Bhavnani

Heh. So this is what I ended up creating:

public void SafeTryParse
(string stringValue,
Decimal? defaultValue,
out Decimal value)
{
if (!Decimal.TryParse (stringValue, out value) && defaultValue.HasValue) {
value = defaultValue.Value;
}
}

/ravi

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Alan N

Or change out to ref and do this:

private bool SafeTryParse(string s, ref decimal result) {
  decimal temp;
  bool ok = Decimal.TryParse(s, out temp);
  if (ok) result = temp;
  return ok;
}

Alan

Ravi Bhavnani

Great minds think alike. :) /ravi

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Luc Pattyn

Hmm. That isn't very safe, result could contain an unintended value:

decimal val;
// now first use of val, not meant to provide the default value
val=some_garbage;
someOperations(val);
// now second use, forgetting to set the actual default value
bool OK=SafeTryParse("foo", ref val);
// val still contains garbage

One shouldn't give two roles to a single variable, so I'd rather have two overloads both using the out keyword:

// result is zero when parse fails
private bool SafeTryParse(string s, out decimal result);

// result is dflt when parse fails
private bool SafeTryParse(string s, out decimal result, decimal dflt);

You can still get the "overwrite when parse succeeds" effect by passing val twice to the latter. :)

Luc Pattyn [My Articles] Nil Volentibus Arduum

Ravi Bhavnani

I would argue that forgetting to set a valid initial value is programmer error, however I now like the idea of an explicit default. That's what I initially implemented (link[^]), but like Alan, switched to the 2 parameter ref version. I'm going to switch back. Thanks, /ravi

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PIEBALDconsult

Well it's zero, right?

Lost User

Do we actually care, if the value isn't being used?

string badNumericString = "Bogus";
Decimal foo; // will be initialized to 0 by the runtime
if (!Decimal.TryParse (badNumericString, out foo)) foo = 1; // default value here

..and if the TryParse don't use the value, it'll probably initialize it with the same value it initializes an empty variable. Since the variable hasn't been set (according to application logic), we can't be sure about the value unless we explicitly set it. (Future versions of .NET might display other behavior)

Bastard Programmer from Hell :suss:

DaveyM69

Eddy Vluggen wrote:

it'll probably initialize it with the same value it initializes an empty variable

Double checked using Reflector...

public static bool TryParse(string s, out decimal result)
{
return Number.TryParseDecimal(s, NumberStyles.Number, NumberFormatInfo.CurrentInfo, out result);
}

internal static unsafe bool TryParseDecimal(string value, NumberStyles options, NumberFormatInfo numfmt, out decimal result)
{
byte* stackBuffer = stackalloc byte[0x72];
NumberBuffer number = new NumberBuffer(stackBuffer);
result = 0M; // result is set to zero here!
if (!TryStringToNumber(value, options, ref number, numfmt, true))
{
return false;
}
if (!NumberBufferToDecimal(number.PackForNative(), ref result))
{
return false;
}
return true;
}

... so it WILL be zero.

Dave
Binging is like googling, it just feels dirtier. Please take your VB.NET out of our nice case sensitive forum. Astonish us. Be exceptional. (Pete O'Hanlon)
BTW, in software, hope and pray is not a viable strategy. (Luc Pattyn)

Luc Pattyn

That is what the doc[^] says. This entire thread seams to suggest no-one reads or beliefs it. :)

Luc Pattyn [My Articles] Nil Volentibus Arduum

PIEBALDconsult

Luc Pattyn wrote:

no-one reads or beliefs it

Maybe I'm a one-percenter. :-D My understanding is that all TryParse methods are expected to set the value to zero/null/default(T) when they fail -- so that's what I do when I write a Tryxxx method.

    public bool
    TryParse
    (
        string Name
    ,
        out T  Value
    )
    {
        bool result = true ;

        Value = default(T) ;

public static bool
TryGetValue<T>
(
this object Source
,
out T Value
)
{
bool result = false ;

Value = default(T) ;

PIEBALDconsult

Then I prefer to provide a value; something like this:

    public virtual T
    ExecuteScalar<T>
    (
        T IfNull
    )
    {

you could do similar for TryParse.

PIEBALDconsult

Eddy Vluggen wrote:

Do we actually care

Not particularly, but we care whether or not it does what the documentation says.

Luc Pattyn

I agree. :)

Luc Pattyn [My Articles] Nil Volentibus Arduum

Ravi Bhavnani

Yessir, it is. /ravi

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Ravi Bhavnani

Eddy Vluggen wrote:

Do we actually care, if the value isn't being used?

It is being used. The example I gave was intentionally limited to focus on the question. In my app, foo (not its real name) is used after the TryParse() executes. Different things happen depending on whether foo is Decimal.MinValue. /ravi

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Ravi Bhavnani

You are right in assuming I didn't read the doc. :) /ravi

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