Re: An array is just a pointer

SG <>
Tue, 22 Mar 2011 08:55:41 -0700 (PDT)
On 22 Mrz., 15:32, Paul wrote:

Leigh wrote:

No the correct terminology is "array-to-pointer conversion".

No it's not the array doesn't change at all, its not converted to anyt=


In this instance, you demonstrated that you don't understand the
meaning of the word "conversion" as it is used in the C++ ISO standard
and several posts that mentioned array-to-pointer conversion.
Conversion does NOT imply that the source changes. Conversion is only
about creating something new based on something old. Example:

  struct arraylike {
    int data[99];
    operator int*() {return &(data[0]);}

  int main() {
    arraylike arr;
    arr[0] = 1729; // [] works ...
    int* ptr = arr; // initialization works ...
    // ... because arr is implicitly convertible to an int*
    // arr still exists and has not changed at all, though.

I have to agree with Alf Steinbach. You are basically wrong about
everything at every conceivable scale. You're clueless and
_not_even_trying_ to understand what people write.


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