Let v be an vector; for instance:
std::vector<doublev;
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
3  1292 
Alex Vinokur wrote:
Let v be an vector; for instance:
std::vector<doublev;
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
Alex Vinokur wrote:
Let v be an vector; for instance:
std::vector<doublev;
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
Alex Vinokur wrote:
Let v be an vector; for instance:
std::vector<doublev;
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. This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics
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