CODEWITHSUNDEEP
c++arrayslanguage-lawyernew-operator
FrozenHeart
FrozenHeart

Reputation: 20746

Allocate array without specifying size

Does the following construction valid according to the C++ standards and what can I do with the arr after this statement?

char* arr = new char[];

Thanks in advance.

Upvotes: 2

Views: 219

Answers (2)

Columbo
Columbo

Reputation: 60979

No, this is not allowed. The grammar for new-expressions in [expr.new]/1 specifies

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

Clearly there is no expression between the brackets in your code. And it woudn't make sense either: How could you allocate an array of unknown length?

If you need an array whose size is to be changed dynamically, use std::vector.

Upvotes: 5

Anonymous
Anonymous

Reputation: 2172

C++ compiler offen defines extension allowing to declare array of size = 0. Usualy it can be useful for declaring last field in a structure so you can choose length of such array during structure allocation.

struct A
{
    float something;
    char arr[];
};

So if you like to allocate such A with let say arr to have 7 elements, you do:

A* p = (A*)malloc( sizeof(A) + sizeof(char)*7) ;

You should note that sizeof(A) is equal to sizeof(float), so for compiler your arr field is 0 sized.

Now you can use your arr field up to 7 indexes:

p->arr[3]='d';

Upvotes: 1

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