Vector as array

[Alex Vinokur]

Let v be an vector; for instance:
std::vector<dou blev;

Is sequence &v[0], &v[1], ..., &v[v.size() - 1] a C/C++-array?

In other words,
Let foo() have the following prototype:
void (double* p, std::size_t arrarySize);

Is this calling is correct:
foo (&v[0], v.size()) ?

Alex Vinokur

[Kai-Uwe Bux]

Re: Vector as array

Alex Vinokur wrote:

Let v be an vector; for instance:
std::vector<dou blev;
>
Is sequence &v[0], &v[1], ..., &v[v.size() - 1] a C/C++-array?
>
In other words,
Let foo() have the following prototype:
void (double* p, std::size_t arrarySize);
>
Is this calling is correct:
foo (&v[0], v.size()) ?

I think that is what the standard means by calling the memory of a
vector "contiguous ". The precise guarantee is for types other that bool
[23.2.4/1]:

... The elements of a vector are stored contiguously, meaning that if v is
a vector<T, Allocatorwhere T is some type other than bool, then it obeys
the identity &v[n] == &v[0] + n for all 0 <= n < v.size().


Best

Kai-Uwe Bux

[Hendrik Schober]

Re: Vector as array

Alex Vinokur wrote:

Let v be an vector; for instance:
std::vector<dou blev;
>
Is sequence &v[0], &v[1], ..., &v[v.size() - 1] a C/C++-array?
>
In other words,
Let foo() have the following prototype:
void (double* p, std::size_t arrarySize);
>
Is this calling is correct:
foo (&v[0], v.size()) ?

It is, except when 'v.size() == 0', in which case 'v[0]' is
undefined (and produces runt-time errors on some popular
platforms).

Alex Vinokur

Schobi

[Rolf Magnus]

Re: Vector as array

Alex Vinokur wrote:

Let v be an vector; for instance:
std::vector<dou blev;
>
Is sequence &v[0], &v[1], ..., &v[v.size() - 1] a C/C++-array?

Yes.

In other words,
Let foo() have the following prototype:
void (double* p, std::size_t arrarySize);
>
Is this calling is correct:
foo (&v[0], v.size()) ?

Yes.