Why does Int32.TryParse reset the out parameter when not able to convert?

Question

if I run this C# code

int realInt = 3;  
string foo = "bar";  
Int32.TryParse(foo, out realInt); 

Console.WriteLine(realInt);  
Console.Read();

I get 0. And I would like to know why! Cause I cannot find any reason why it would. This forces me to make temp variables for every parsing. So please! Great coders of the universe, enlighten me!

Answer 1

It is "out", not "ref". Inside the method it has to assign it (without reading it first) to satisfy the meaning of "out".

Actually, "out" is a language concern (not a framework one) - so a managed C++ implementation could probably ignore this... but it is more consistent to follow it.

In reality; if the method returns false you simply shouldn't look at the value; treat it as garbage until it is next assigned. It is stated to return 0, but that is rarely useful.

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Also - if it didn't do this (i.e. if it preserved the value); what would this print:

int i;
int.TryParse("gibber", out i);
Console.WriteLine(i);

That is perfectly valid C#... so what does it print?

Answer 2

The Int32.TryParse Method (String, Int32) doc says:

Converts the string representation of a number to its 32-bit signed integer equivalent. A return value indicates whether the conversion succeeded.

result

Type: System.Int32

When this method returns, contains the 32-bit signed integer value equivalent to the number contained in s, if the conversion succeeded, or zero if the conversion failed. The conversion fails if the s parameter is null reference (Nothing in Visual Basic), is not of the correct format, or represents a number less than MinValue or greater than MaxValue. This parameter is passed uninitialized.