c++ - When a function has a specific-size array parameter, why is it replaced with a pointer? -

given following program,

#include <iostream>  using namespace std;  void foo( char a[100] ) {     cout << "foo() " << sizeof( ) << endl; }  int main() {     char bar[100] = { 0 };     cout << "main() " << sizeof( bar ) << endl;     foo( bar );     return 0; } 

outputs

main() 100 foo() 4 

1. why array passed pointer first element?
2. is heritage c?
3. what standard say?
4. why strict type-safety of c++ dropped?

yes it's inherited c. function:

void foo ( char a[100] ); 

will have parameter adjusted pointer, , becomes:

void foo ( char * ); 

if want array type preserved, should pass in reference array:

void foo ( char (&a)[100] ); 

c++ '03 8.3.5/3:

...the type of function determined using following rules. type of each parameter determined own decl-specifier-seq , declarator. after determining type of each parameter, parameter of type "array of t" or "function returning t" adjusted "pointer t" or "pointer function returning t," respectively....

to explain syntax:

check "right-left" rule in google; found 1 description of here.

it applied example approximately follows:

void foo (char (&a)[100]); 

start @ identifier 'a'

'a'

move right - find ) reverse direction looking (. move left pass &

'a' reference

after & reach opening ( reverse again , right. see [100]

'a' reference array of 100

and reverse direction again until reach char:

'a' reference array of 100 chars