To loop over a 3x3 array called "a" in C++ i used the following code.
int a[3][3] {};
for(auto &b: a) {
for(auto &c: b) {
std::cout << c << std::endl;
}
}
If I had to replace the "auto", I would intuitively try
int a[3][3] {};
for(int &(b[3]): a) {
for(int &c: b) {
std::cout << c << std::endl;
}
}
But this does not work. Instead I figured out that the following works.
int a[3][3] {};
for(int (&b)[3]: a) {
for(int &c: b) {
std::cout << c << std::endl;
}
}
So the question is: Why does the latter example work?
I thought I needed a reference to an array of length 3, but instead I need an array of length 3 containing references.
If you have an array like this
T a[N1][N2][N3];
where T
is some type specifier and N1
, N2
, and N3
the sizes of the array then a reference to this array will look like
T ( &ra )[N1][N2][N3] = a;
If you need to declare a reference to the element of the array that (the element) has the type T[N2][N3]
then you can write for example
T ( &e )[N2][N3] = a[0];
Returning to your example you declared an array
int a[3][3] {};
elements of this array have the type int[3]
. So to declare a reference to elements of the array you have to write
for ( int ( %row )[3] : a )
As for this declaration
int &(b[3])
(where the parentheses are redundant) then it declares an array of three elements with the type int &
. However according to the C++ Standard you may not declare an array of references.
From the C++ Standard (8.3.4 Arrays)
1 In a declaration T D where D has the form
D1 [ constant-expressionopt] attribute-specifier-seqopt
and the type of the identifier in the declaration T D1 is “derived-declarator-type-list T”, then the type of the identifier of D is an array type; if the type of the identifier of D contains the auto type-specifier, the program is ill-formed. T is called the array element type; this type shall not be a reference type, the (possibly cvqualified) type void, a function type or an abstract class type....