C++ specify array indexes in initializer, like C

Question

I used C before (embedded stuff), and I can initialize my arrays like that:

int widths[] = { [0 ... 9] = 1, [10 ... 99] = 2, [100] = 3 };

i.e. I can specify indexes inside initializer. Currently I'm learning Qt/C++, and I can't believe this isn't supported in C++.

I have this option: -std=gnu++0x, but anyway it isn't supported. (I don't know if it is supported in C++11, because Qt works buggy with gcc 4.7.x)

So, is it really not supported in C++? Or maybe there's a way to enable it?

UPD: currently I want to initialize const array, so std::fill won't work.

"Or maybe there's a way to enable it?" - use C? (:D) Good question, +1. — , Nov 26 '12 at 10:30
Your code is neither C (including C11). C11 does not support initializing ranges this way (`[a...b]=1`), only single elements (`[a]=1`). — Yakov Galka, Nov 26 '12 at 10:33
hmm... can you use `extern "C" { ................ }` around declaration? But I believe this syntax is your embedded compiler's addition. — anishsane, Nov 26 '12 at 10:33
@ybungalobill, not C11, but C99 (if i haven't mixed up anything) — Dmitry Frank, Nov 26 '12 at 10:38
@anishsane "I believe this syntax is your embedded compiler's addition" - no, it's a GNU extension to C99. — , Nov 26 '12 at 10:38
@H2CO3: I know, but the OP calls its "C" and then wanders why it is not in C++...\ — Yakov Galka, Nov 26 '12 at 10:40
[The same thing was asked just the other day](http://stackoverflow.com/questions/13477281/initializing-an-array-of-ints). — Lundin, Nov 26 '12 at 10:40
@ybungalobill No relation there. C is **not** a strict subset of C++. You better learn this rule well. — , Nov 26 '12 at 10:42
@ybungalobill yes you did. "OP calls its "C" and then wanders why it is not in C++" means you implied that everything that is in C is also present in C++. — , Nov 26 '12 at 10:45
@H2CO3: "it's a GNU extension to C99" - Oh, interesting... :-) — anishsane, Nov 26 '12 at 11:48

score 7 · Answer 1 · edited May 23 '17 at 12:19

7

hm, you should use std::fill_n() for that task...

as stated here http://gcc.gnu.org/onlinedocs/gcc/Designated-Inits.html the designated inits (extention) are not implemented in GNU C++

as a comment said, you can use std:vector to get the result desired. You could still enforce the const another way around and use fill_n.

int* a = new int[N];
// fill a

class C {
  const std::vector<int> v;
public:
  C():v(a, a+N) {}
};

edited May 23 '17 at 12:19

Community

answered Nov 26 '12 at 10:32

Najzero

Thanks for the answer. Actually, I need to initialize const array, so, `std::fill_n()` won't work for me. – Dmitry Frank Nov 26 '12 at 10:35
1

@DmitryFrank: Use `std::vector`, write a function that initializes the vector using `fill_n` and returns the result. Use this function to initialize a `const vector`. – Yakov Galka Nov 26 '12 at 10:37
1

You need `delete a;` also. Two dynamic allocation for filling array, isn't it an overkill? – iammilind Nov 26 '12 at 10:50
1

After years, it works in g++ 4.8.2. :) when compiling with `-std=c++11` – Dmitry Frank Aug 23 '15 at 14:35

score 3 · Accepted Answer · answered Aug 23 '15 at 14:34

3

After years, I tested it just by chance, and I can confirm that it works in g++ -std=c++11, g++ version is 4.8.2.

answered Aug 23 '15 at 14:34

Dmitry Frank

score 2 · Answer 3 · answered Nov 26 '12 at 10:33

No, it is not possible to do it in C++ like that. But you can use std::fill algorithm to assign the values:

int widths[101];
fill(widths, widths+10, 1);
fill(widths+10, widths+100, 2);
fill(widths+100, widths+101, 3);

It is not as elegant, but it works.

3 Answers3