Can a reference to an array be written without a typedef?

[John Nagle]

This is valid C++:

float s0[3];

typedef float array3[3];
array3& s1 = s0;

This creates a reference to an array.
Can the two statements be written without a typedef? Thanks.

John Nagle

[kwikius]

"John Nagle" <na***@animats.comwrote in message
news:48***********************@news.sonic.net...

This is valid C++:

float s0[3];

typedef float array3[3];
array3& s1 = s0;

This creates a reference to an array.
Can the two statements be written without a typedef? Thanks.

float(& s1)[3] = s0;

regards
Andy Little

[Chris M. Thomasson]

"John Nagle" <na***@animats.comwrote in message
news:48***********************@news.sonic.net...

This is valid C++:

float s0[3];

typedef float array3[3];
array3& s1 = s0;

This creates a reference to an array.
Can the two statements be written without a typedef? Thanks.

Try:
float (&s1)[3] = s0;

[Francis Litterio]

John Nagle wrote:

This is valid C++:

float s0[3];

typedef float array3[3];
array3& s1 = s0;

This creates a reference to an array.
Can the two statements be written without a typedef? Thanks.

In addition to responses else-thread, I think that in C++0x you will be
able to do this:

float s0[3];

decltype(s0) & s1 = s0;
--
Fran

[Greg Comeau]

In article <u3***********@not.available>,
Francis Litterio <em***@not.availablewrote:

>John Nagle wrote:

>This is valid C++:

float s0[3];

typedef float array3[3];
array3& s1 = s0;

This creates a reference to an array.
Can the two statements be written without a typedef? Thanks.

In addition to responses else-thread, I think that in C++0x you will be
able to do this:

float s0[3];

decltype(s0) & s1 = s0;

Yes, that's fine too.
--
Greg Comeau / 4.3.10.1 with C++0xisms now in beta!
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