int & 是什么意思

Question

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!

Answer 1

It returns a reference to an int. References are similar to pointers but with some important distinctions. I'd recommend you read up on the differences between pointers, references, objects and primitive data types.

"Effective C++" and "More Effective C++" (both by Scott Meyers) have some good descriptions of the differences and when to use pointers vs references.

EDIT: There are a number of answers saying things along the lines of "references are just syntactic sugar for easier handling of pointers". They most certainly are not .

Consider the following code:

int a = 3;
int b = 4;
int* pointerToA = &a;
int* pointerToB = &b;
int* p = pointerToA;
p = pointerToB;
printf("%d %d %d\n", a, b, *p); // Prints 3 4 4
int& referenceToA = a;
int& referenceToB = b;
int& r = referenceToA;
r = referenceToB;
printf("%d %d %d\n", a, b, r); // Prints 4 4 4

The line p = pointerToB changes the value of p , ie it now points to a different piece of memory.

r = referenceToB does something completely different: it assigns the value of b to where the value of a used to be. It does not change r at all. r is still a reference to the same piece of memory.

The difference is subtle but very important.

If you still think that references are just syntactic sugar for pointer handling then please read Scott Meyers' books. He can explain the difference much better than I can.

Answer 2

Be careful here... you're walking the C/C++ line. There's a quite clear distinction but it doesn't always appear that way:

C++ : this often means a reference. For example, consider:

void func(int &x)
{
   x = 4;
}

void callfunc()
{
    int x = 7;
    func(x);
}

As such, C++ can pass by value or pass by reference.

C however has no such pass by reference functionality. & means "addressof" and is a way to formulate a pointer from a variable. However, consider this:

void func(int* x)
{
   *x = 4;
}

void callfunc()
{
    int x = 7;
    func(&x);
}

Deceptively similar, yet fundamentally different. What you are doing in C is passing a copy of the pointer. Now these things still point to the same area of memory, so the effect is like a pass by reference in terms of the pointed-to memory, but it is not a reference being passed in. It is a reference to a point in memory.

Try this (Compile as C):

#include <stdio.h>

void addptr(int* x)
{
    printf("Add-ptr scope 1:\n");
    printf("Addr: %p\n", x);
    printf("Pointed-to-memory: %d\n", *x);
    *x = *x + 7;
    x++;
    printf("Add-ptr scope 2:\n");
    printf("Addr: %p\n", x);
    printf("Pointed-to-memory: %d\n", *x);
}

int main(int argc, char** argv)
{
    int a = 7;
    int *y = &a;
    printf("Main-Scope 2:\n");
    printf("Addr: %p\n", y);
    printf("Pointed-to-memory: %d\n", *y);
    addptr(y);
    printf("Main-Scope 2:\n");
    printf("Addr: %p\n", y);
    printf("Pointed-to-memory: %d\n", *y);
    return 0;

}

If C had pass by reference, the incoming pointer address, when changed by addptr should be reflected in main , but it isn't. Pointers are still values.

So, C does not have any pass by reference mechanism. In C++, this exists, and that is what & means in function arguments etc.

Edit: You might be wondering why I can't do this demonstration in C++ easily. It's because I can't change the address of the reference. At all. From t his quite good guide to references :

How can you reseat a reference to make it refer to a different object?

No way.

You can't separate the reference from the referent.

Unlike a pointer, once a reference is bound to an object, it can not be "reseated" to another object. The reference itself isn't an object (it has no identity; taking the address of a reference gives you the address of the referent; remember: the reference is its referent).

In that sense, a reference is similar to a const pointer such as int* const p (as opposed to a pointer to const such as int const* p). But please don't confuse references with pointers; they're very different from the programmer's standpoint.

By request, on returning references:

#include <iostream>

using namespace std;

int& function(int f)
{
   f=f+3;
   return f;
}

int main(int argc, char** argv)
{
    int x = 7;
    int y;
    y = function(x);
    cout << "Input: " << x << endl;
    cout << "Output:" << y << endl;
    return 0;
}

Any good compiler ought to give you this warning message in some form:

exp.cpp:7:11: warning: reference to stack memory associated with local variable 'f' returned

What does this mean? Well, we know function arguments are pushed onto the stack (note: not actually on x64, they go into registers then the stack, but they are on the stack literally on x86) and what this warning is saying is that creating a reference to such an object is not a good idea, because it's not guaranteed to be left in place. The fact it is is just luck.

So what gives? Try this modified version:

#include <iostream>

using namespace std;

int& function(int& f)
{
    f=f+3;
    return f;
}

int main(int argc, char** argv)
{
    int x = 7;
    int y;
    y = function(x);
    cout << "Input: " << x << endl;
    cout << "Output:" << y << endl;
    return 0;
}

Run this, and you'll see both values get updated. What? Well they both refer to the same thing and that thing is being edited.

Answer 3

它返回对 int 变量的引用。

Answer 4

This question isn't C/C++ at all, as C does not have references, only pointers. An int& is a reference to an int. Also, you don't need void , it can just be int& foo();

Answer 5

From Alfred's comments

This is what the document says, Texas instrument's TMS320C28x C/C++ compiler intrinsics , page 122, int&__byte(int, unsigned int), I guess it is different from PC – Alfred Zhong

From the manual:

int &__byte(int *array, unsigned int byte_index);

MOVB array[byte_index].LSB, src

The lowest adressable unit in C28x is 16 bits. Therefore, normally you cannot access 8-bit MOVB dst, array[byte_index]. LSB entities off a memory location. This intrinsic helps access an 8-bit quantity off a memory location, and can be invoked as follows:

__byte(array,5) = 10;
b = __byte(array,20);

This just means that the function returns a reference to an integer that acts like an 8 bit quantity. Because the value is a reference modifying will modify the object at the destination (just like the MOVB) instruction, while assigning to b will copy (just like MOVB) to the destination.

Answer 6

just playing with variables to show you the meaning

int i = 5;
int * pI = &i;
int & referenceToI = * pI;
referenceToI = 4; // now i == 4

EDIT : References are just a syntactic sugar for easier pointers handling. at the assembly level, the code generated by the compiler returns to a you an address-pointer

Answer 7

https://isocpp.org/wiki/faq/references

This page says it succinctly! A reference is an alias. It IS the object, just by another name. The previous name is still valid. The reference is not a stand-alone object. It IS the object by a different name. You cannot operate on the reference because it is not a thing unto itself. It is just an alias of the original object.

It is that simple. Now you can think of it easily rather than the round and round in the above posts.

Answer 8

int& is a reference. To be more precise, a reference variable is simply an alternative name of an existing variable.

Example: Consider below code

main(){
    int x = 4;
    int& ref_of_x = x;
    // Now, modifying x modifies ref_of_x too and vice-versa
    x = 5;
    cout<<ref_of_x; //This prints 5
    
    ref_of_x = 15;
    cout<<x;  // This prints 15
    }