Seeking clarification on apparent contradictions regarding weakly typed languages

Question

I think I understand strong typing, but every time I look for examples for what is weak typing I end up finding examples of programming languages that simply coerce/convert types automatically.

For instance, in this article named Typing: Strong vs. Weak, Static vs. Dynamic says that Python is strongly typed because you get an exception if you try to:

Python

1 + "1" Traceback (most recent call last): File "", line 1, in ?  TypeError: unsupported operand type(s) for +: 'int' and 'str' 

However, such thing is possible in Java and in C#, and we do not consider them weakly typed just for that.

Java

  int a = 10;   String b = "b";   String result = a + b;   System.out.println(result); 

C#

int a = 10; string b = "b"; string c = a + b; Console.WriteLine(c); 

In this another article named Weakly Type Languages the author says that Perl is weakly typed simply because I can concatenate a string to a number and viceversa without any explicit conversion.

Perl

$a=10; $b="a"; $c=$a.$b; print $c; #10a 

So the same example makes Perl weakly typed, but not Java and C#?.

Gee, this is confusing

The authors seem to imply that a language that prevents the application of certain operations on values of different types is strongly typed and the contrary means weakly typed.

Therefore, at some point I have felt prompted to believe that if a language provides a lot of automatic conversions or coercion between types (as perl) may end up being considered weakly typed, whereas other languages that provide only a few conversions may end up being considered strongly typed.

I am inclined to believe, though, that I must be wrong in this interepretation, I just do not know why or how to explain it.

So, my questions are:

• What does it really mean for a language to be truly weakly typed?
• Could you mention any good examples of weakly typing that are not related to automatic conversion/automatic coercion done by the language?
• Can a language be weakly typed and strongly typed at the same time?

Answer 1

UPDATE: This question was the subject of my blog on the 15th of October, 2012. Thanks for the great question!

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What does it really mean for a language to be "weakly typed"?

It means "this language uses a type system that I find distasteful". A "strongly typed" language by contrast is a language with a type system that I find pleasant.

The terms are essentially meaningless and you should avoid them. Wikipedia lists eleven different meanings for "strongly typed", several of which are contradictory. This indicates that the odds of confusion being created are high in any conversation involving the term "strongly typed" or "weakly typed".

All that you can really say with any certainty is that a "strongly typed" language under discussion has some additional restriction in the type system, either at runtime or compile time, that a "weakly typed" language under discussion lacks. What that restriction might be cannot be determined without further context.

Instead of using "strongly typed" and "weakly typed", you should describe in detail what kind of type safety you mean. For example, C# is a statically typed language and a type safe language and a memory safe language, for the most part. C# allows all three of those forms of "strong" typing to be violated. The cast operator violates static typing; it says to the compiler "I know more about the runtime type of this expression than you do". If the developer is wrong, then the runtime will throw an exception in order to protect type safety. If the developer wishes to break type safety or memory safety, they can do so by turning off the type safety system by making an "unsafe" block. In an unsafe block you can use pointer magic to treat an int as a float (violating type safety) or to write to memory you do not own. (Violating memory safety.)

C# imposes type restrictions that are checked at both compile-time and at runtime, thereby making it a "strongly typed" language compared to languages that do less compile-time checking or less runtime checking. C# also allows you to in special circumstances do an end-run around those restrictions, making it a "weakly typed" language compared with languages which do not allow you to do such an end-run.

Which is it really? It is impossible to say; it depends on the point of view of the speaker and their attitude towards the various language features.

Answer 2

As others have noted, the terms "strongly typed" and "weakly typed" have so many different meanings that there's no single answer to your question. However, since you specifically mentioned Perl in your question, let me try to explain in what sense Perl is weakly typed.

The point is that, in Perl, there is no such thing as an "integer variable", a "float variable", a "string variable" or a "boolean variable". In fact, as far as the user can (usually) tell, there aren't even integer, float, string or boolean values: all you have are "scalars", which are all of these things at the same time. So you can, for example, write:

$foo = "123" + "456";           # $foo = 579 $bar = substr($foo, 2, 1);      # $bar = 9 $bar .= " lives";               # $bar = "9 lives" $foo -= $bar;                   # $foo = 579 - 9 = 570 

Of course, as you correctly note, all of this can be seen as just type coercion. But the point is that, in Perl, types are always coerced. In fact, it's quite hard for a user to tell what the internal "type" of a variable might be: at line 2 in my example above, asking whether the value of $bar is the string "9" or the number 9 is pretty much meaningless, since, as far as Perl is concerned, those are the same thing. Indeed, it's even possible for a Perl scalar to internally have both a string and a numeric value at the same time, as is e.g. the case for $foo after line 2 above.

The flip side of all this is that, since Perl variables are untyped (or, rather, don't expose their internal type to the user), operators cannot be overloaded to do different things for different types of arguments; you can't just say "this operator will do X for numbers and Y for strings", because the operator can't (won't) tell which kind of values its arguments are.

Thus, for example, Perl has and needs both a numeric addition operator (+) and a string concatenation operator (.): as you saw above, it's perfectly fine to add strings ("1" + "2" == "3") or to concatenate numbers (1 . 2 == 12). Similarly, the numeric comparison operators ==, !=, <, >, <=, >= and <=> compare the numeric values of their arguments, while the string comparison operators eq, ne, lt, gt, le, ge and cmp compare them lexicographically as strings. So 2 < 10, but 2 gt 10 (but "02" lt 10, while "02" == 2). (Mind you, certain other languages, like JavaScript, try to accommodate Perl-like weak typing while also doing operator overloading. This often leads to ugliness, like the loss of associativity for +.)

(The fly in the ointment here is that, for historical reasons, Perl 5 does have a few corner cases, like the bitwise logical operators, whose behavior depends on the internal representation of their arguments. Those are generally considered an annoying design flaw, since the internal representation can change for surprising reasons, and so predicting just what those operators do in a given situation can be tricky.)

All that said, one could argue that Perl does have strong types; they're just not the kind of types you might expect. Specifically, in addition to the "scalar" type discussed above, Perl also has two structured types: "array" and "hash". Those are very distinct from scalars, to the point where Perl variables have different sigils indicating their type ($ for scalars, @ for arrays, % for hashes)¹. There are coercion rules between these types, so you can write e.g. %foo = @bar, but many of them are quite lossy: for example, $foo = @bar assigns the length of the array @bar to $foo, not its contents. (Also, there are a few other strange types, like typeglobs and I/O handles, that you don't often see exposed.)

Also, a slight chink in this nice design is the existence of reference types, which are a special kind of scalars (and which can be distinguished from normal scalars, using the ref operator). It's possible to use references as normal scalars, but their string/numeric values are not particularly useful, and they tend to lose their special reference-ness if you modify them using normal scalar operations. Also, any Perl variable² can be blessed to a class, turning it into an object of that class; the OO class system in Perl is somewhat orthogonal to the primitive type (or typelessness) system described above, although it's also "weak" in the sense of following the duck typing paradigm. The general opinion is that, if you find yourself checking the class of an object in Perl, you're doing something wrong.

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¹ Actually, the sigil denotes the type of the value being accessed, so that e.g. the first scalar in the array @foo is denoted $foo[0]. See perlfaq4 for more details.

² Objects in Perl are (normally) accessed through references to them, but what actually gets blessed is the (possibly anonymous) variable the reference points to. However, the blessing is indeed a property of the variable, not of its value, so e.g. that assigning the actual blessed variable to another one just gives you a shallow, unblessed copy of it. See perlobj for more details.