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.
2 Answers
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.
2 Comments
Bartek Banachewicz
How is
expression defined?
druckermanly
@BartekBanachewicz It's anything that's not a
statement. As that isn't very helpful: stackoverflow.com/questions/3846727/…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';
1 Comment
didierc
Although this feature indeed exists in some implementations, this is not a dynamic allocation of an array using
new.
std::vector