Is allocating a dynamic array without specifying size well formed code?

Question

The following simple program snippet gives compilation errors with gcc-4.3.4.

Program:

int main() 
{   
    char *ptr = new char[10];     
    char *ptr1 = new char[];      
    return 0; 
}

Compilation errors:

prog.cpp: In function ‘int main()’:
prog.cpp:4: error: expected primary-expression before ‘]’ token
prog.cpp:3: warning: unused variable ‘ptr’
prog.cpp:4: warning: unused variable ‘ptr1’

But the same compiles cleanly with MSVC without any diagnostic message.

So my question is:
Does the Standard allow an new [] to be called without specifying the size? Or this a bug in MSVC?
Can someone provide a reference from the standard which will conclusively say that the above code example is ill-formed or well-formed?

I have had a look at:

5.3.4 New [expr.new] &
18.4.1.2 Array forms [lib.new.delete.array]

but couldnt find any conclusive evidence about the behavior.

EDIT:
Adding the Language Lawyer tag.
I am expecting the answer for an observed behavior regardless of whether it is useful or not, I am fully aware it is not useful nor recommended.

I don't see how it could be well formed code? What is the compiler going to do, guess the size? That seems rather difficult (if not impossible). — Corbin
– Corbin, Commented Mar 20, 2012 at 5:27
does new char[] allocate memory? i think it creates only a pointer — Rohit Vipin Mathews
– Rohit Vipin Mathews, Commented Mar 20, 2012 at 5:31

Mankarse · Accepted Answer · 2012-03-20 05:38:01Z

4

This is not syntactically correct.

Have a look at the syntax for a new-expression.

A noptr-new-declarator must contain an expression between the square brackets, and an expression must have a token in it.

answered Mar 20, 2012 at 5:38

Mankarse

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BЈовић · Accepted Answer · 2012-03-21 13:08:41Z

4

That is not legal c++.

5.3.4 New [expr.new] shows what are legal ways to call new in a big list, which contains this line :

noptr-new-declarator:
        [ expression ] attribute-speciﬁer-seqopt
        noptr-new-declarator [ constant-expression ] attribute-speciﬁer-seqopt

and later it explains what the constant-expression can be (in 5.4.3/6 and 5.4.3/7) :

Every constant-expression in a noptr-new-declarator shall be an integral constant expression (5.19) and evaluate to a strictly positive value.

After some thoughts, next items should be relavant :

8.3.4/1 [dcl.array], these parts :

In a declaration T D where D has the form

    D1 [ constant-expressionopt ] attribute-speciﬁer-seqopt

and the type of the identiﬁer in the declaration T D1 is “derived-declarator-type-list T”, then the type of the identiﬁer of D is an array type;

and

if the constant expression is omitted, the type of the identiﬁer of D is “derived-declarator-type- list array of unknown bound of T”, an incomplete object type.

5.3.4/1 tells :

This type shall be a complete object type, but not an abstract class type or array thereof

Since you omitted the array size, the type is not complete, and your program is not valid c++.

edited Mar 21, 2012 at 13:08

answered Mar 20, 2012 at 5:44

BЈовић

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11 Comments

Loki Astari Over a year ago

What about <expression> in the first line. The <constant-expression> is for multi-dimensional arrays. eg x = new char[plop*5 /*expression */][10 /*const-expression*/][100 /* const-expression */]

Loki Astari Over a year ago

With a quick glance I can not see <expression> being empty but ?

BЈовић Over a year ago

@LokiAstari That is why I added what the constant-expression can be in edit : Every constant-expression in a noptr-new-declarator shall be an integral constant expression. Since it is omitted in the question, then it is not a standard c++, but a compiler extension. It's not an UB, since it compiles, but it shouldn't.

Keith Thompson Over a year ago

@VJovic: Since the program violates a syntax rule, a diagnostic is required; a compiler that doesn't issue a diagnostic is non-conforming. But a conforming compiler could still accept it after issuing a diagnostic (a warning is sufficient). Once it's accepted it, the program's behavior is undefined (because nothing in the Standard defines its behavior).

Keith Thompson Over a year ago

@LokiAstari: Only if it's a documented extension. By issuing the diagnostic, the compiler has fulfilled its commitment to the standard. ("Implementation defined" means that the implementation must document the behavior.)

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