在C ++中将静态数组声明为指针

Question

Here are a few lines of C++ code I recently saw.

double a[p][p];
for(i=0;i<p;i++)
{
    for(j=0;j<p;j++)
    {
        a[i][j] = m[i][j]; //m is sth we are given.
    }
}
dgeev_(&jobvl,&jobvr,&p,(double*)a,&lda,(double*)wr,(double*)wi,(double*)vl,
      &ldvl,(double*)vr,&ldvr,(double*)work,&lwork,&info);
// sth else..

dgeev_ is a function and you can find its documentation here: http://www.netlib.org/clapack/old/double/dgeev.c

dgeev_ is declared in header in the following way, so we can use C function in C++ codes.

extern "C" {
void dgeev_(char*, char*, int*, double*, int*, double*, double*, double*,
 int*, double*, int*, double*, int*, int*);
}

As we can see, a is static array, and dgeev_ need a variable of type doublereal * for the 4th parameter.

What does (double*)a do here?

Answer 1

It's either a mistake or the author is playing a trick on you.

He's used a C-style cast to "hack" the value a into the double* type, so that a two-dimensional array may be treated as a single-dimensional array. ( doublereal is just an alias for double .)

Problem is, a isn't a double* , nor is it actually convertible to a double* . I don't know what it is convertible to, since VLAs are only available in C++ as a GCC extension; there is no standard wording to consult. I'm going to just conclude that this program has undefined behaviour (though it probably appears to work, due to how these things are practically laid out in memory).

If the author wants to be able to safely read this array using single-dimension indices, he should have declared it that way to begin with (optionally providing the illusion of 2D indexes on top of the data store).