Arrays and pointers in C + +

The first thing to make clear is that the array type and pointer type are really different types!

Char *p; Type is char*

Char str[6];//type is char [6]

Type information can be output using the following statement (requires #include<typeinfo>)

Std::cout << typeid (P). Name () << Std::endl;
Std::cout << typeid (str). Name () << Std::endl;

But sometimes we always define a function like this:

void foo (char *p) {

cout << p << Endl;

cout << sizeof (p)/sizeof (char); Output is 4

}

We found that we dropped Str in and it worked. So to be taken for granted, I define a function like this:

void foo (char str[6]) {

cout << str << Endl;

cout << sizeof (str)/sizeof (char) << endl;//output is also 4

}

We found that the input p is not a problem, there is a problem, and the second output is confusing, according to the type information it should output 6.

The problem here is that when you use a non-reference type as a formal parameter, the array type is automatically dropped to the pointer type to handle , that is, Foo (char str[6]) is exactly the same as Foo (char *p)!

To illustrate this point of knowledge, we changed the function prototype to the following:

void foo (char (&STR) [6]) {

cout << str << Endl;

cout << sizeof (str)/sizeof (char) << endl;//output to 6

}

At this point, we find that only the input char[6] type can be compiled, and the second output is 6, that is, the array length information is normally preserved during the parameter transfer.

When a reference type is a parameter, the array type does not fall to the pointer type!

Finally, based on this, we can write a function that takes an array of any dimension as a parameter, of course, using a template, because arrays of different lengths belong to different types:

Template<int N, int m>
int foo (int const (&A) [n][m], int const (&B) [n][m]) {
Std::cout << typeID (a). Name () << Std::endl; Output INT[N][M]
return 0;
}

Arrays and pointers in C + +