Printing array element memory adresses C and C++, why different output? [duplicate]

Answer 1

You must cast &array[i] to void*

for(int i =0; i<10;i++){
    std::cout<<(void*)&array[i]<<std::endl;
}

This is because C++ streams work differently for different types. For example when you pass char* to it, your data is treated as a C-string - thus it is printed as a list of characters.

You must explicitly tell C++ that you want to print an address by casting.

By the way (void*) is not the best way to do that as you should avoid C-like casting. Always use C++-style casting (static_cast, dynamic_cast, reinterpret_cast). In this case static_cast would do the job.

Answer 2

Cast the address to void* before printing, in C++ the operator<< of ostream is overloaded for (const) char* so that it thinks it's a c-style string:

char array[10];
for(int i =0; i<10;i++){
    std::cout << static_cast<void*>(&array[i]) << std::endl;
}

Also see this answer of mine.

Answer 3

std::cout does not necessarily treat pointers as pointers. A char* pointer could be a string, for instance. Taking the address of an element in a char array basically outputs the substring from that point.

Answer 4

The type of &array[i] is char*, and so cout<< thinks that you want to print a string.