Hi
1) When I have
I get a pointer to the whole array. But since it is an expression, I believe I should get a pointer to pointer to char, i.e. char ** (cf. answer #2 in Question 1.32). Why is it a pointer to the whole array, then?Code:&"test"
2) When I have the initialization
then str is an array, right?Code:char str[] = "test";
Best,
Niles.
Well a pointer to an array (well that one in particular) would be
It would be the same with your 2nd example as wellCode:char (*ptr)[5] = &"test";
Code:char (*ptr)[5] = &str;
If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
If at first you don't succeed, try writing your phone number on the exam paper.
Recall what I quoted from the C standard back in your Two-dimensional arrays with strings thread.Originally Posted by Niels_M
Yes.
Look up a C++ Reference and learn How To Ask Questions The Smart Way
Ahh, yes - it is an exception. I should have known that.
Thanks to both of you.
