Re: [Dev-C++] Number of elements of a multidimensional

[Per W. <pw...@ia...>]

Not true that sizeof never returns size of array. It depends on what
declaration the compiler sees when processing the sizeof operator.

void func(int *array) {
    // sizeof(array) =3D sizeof(int*)
}

void func(int array[]) {
    // sizeof(array) =3D sizeof(int*)
}

void func() {
    int array[4];
    // sizeof(array) =3D 4*sizeof(int)
}

Big gotcha about this.
    char buf[100];
    write(fd,buf,sizeof(buf));
converted to:
    char *buf =3D malloc(100);
    write(fd,buf,sizeof(buf))
would fail miserably, because the second version would only write 4 bytes
in a 32-bit machine...

/Per W

On Thu, 28 Nov 2002, Carlos Garc=EDa del Monte wrote:

> The sizeof applied to a array will give you the size of a pointer, never
> the size of the array.
> Say char *array; or char array[];
> Then sizeof(array) =3D 4. (at least in my intelIII x86, with win2k 4 is
> the size for any kind of pointer).
>=20
> The only solution is to have some special value for the end of the
> array. Which for strings is '\0', and you can try to set another
> depending on which kind of data you're working with and which range of
> values you can find, so you can then choose on of the values which are
> out of the range to set as that special value. So you can reach for it
> and you can then now, with the approppriate functions you can write the
> total size of the array and the size of each row of the multidimensional
> array.
>=20
> Samuel Keding escribi=F3:
>=20
> > Isn't there a keyword sizeof in C/C++? It seems like I
> > remember one but I've never had to use it. I think
> > there is one and I think it returns the number of
> > bytes in the array of objects. All you would have to
> > do is divide by the number of bytes that is in just
> > one object. I may be off on this. Like I said, I
> > haven't really had to use this before.
> >
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