A C++ question, I know int* foo (void) foo will return a pointer to int type how about int &foo (void) what does it return? Thank a lot!
It is a pointer to function that returns int* and accepts int* and pointer to function that returns int* (and accepts undefined number of parameters; see comments).
c - type of int * () (int * , int * () ()) - Stack Overflow
That second memory address, then, is expected to hold an int. Do note that, while you are declaring a pointer to an int, the actual int is not allocated. So it is valid to say int *i = 23, which is saying "I have a variable and I want it to point to memory address 23 which will contain an int."
Difference between the int * i and int** i - Stack Overflow
int* i, int * i, int*i, and int *i are all exactly equivalent. This stems from the C compiler (and it's compatible C like systems) ignoring white space in token stream generated during the process of parsing the source code.
c - difference between int* i and int *i - Stack Overflow
The following code can pass compiling and will print 0 on the console. I saw similar code in STL. Does type int in C++ have a constructor? Is int() a call of some defined function? int main() { ...
In C-like languages, whitespace mostly doesn't matter. All versions you listed parse as the same three tokens: int & b so they mean the same to the compiler. The only time whitespace matters is when it separates two alphanumeric tokens, and even then the amount and type of whitespace doesn't matter, as long as there is some. But any sort of punctuation always becomes a separate token from ...