C++ – the difference between “long”, “long long”, “long int”, and “long long int” in C++

clong-integer

I am transitioning from Java to C++ and have some questions about the long data type. In Java, to hold an integer greater than 2³², you would simply write long x;. However, in C++, it seems that long is both a data type and a modifier.

There seems to be several ways to use long:

long x;
long long x;
long int x;
long long int x;

Also, it seems there are things such as:

long double x;

and so on.

What is the difference between all of these various data types, and do they all have the same purpose?

Best Answer

long and long int are identical. So are long long and long long int. In both cases, the int is optional.

As to the difference between the two sets, the C++ standard mandates minimum ranges for each, and that long long is at least as wide as long.

The controlling parts of the standard (C++11, but this has been around for a long time) are, for one, 3.9.1 Fundamental types, section 2 (a later section gives similar rules for the unsigned integral types):

There are five standard signed integer types : signed char, short int, int, long int, and long long int. In this list, each type provides at least as much storage as those preceding it in the list.

There's also a table 9 in 7.1.6.2 Simple type specifiers, which shows the "mappings" of the specifiers to actual types (showing that the int is optional), a section of which is shown below:

Specifier(s)         Type
-------------    -------------
long long int    long long int
long long        long long int
long int         long int
long             long int

Note the distinction there between the specifier and the type. The specifier is how you tell the compiler what the type is but you can use different specifiers to end up at the same type.

Hence long on its own is neither a type nor a modifier as your question posits, it's simply a specifier for the long int type. Ditto for long long being a specifier for the long long int type.

Although the C++ standard itself doesn't specify the minimum ranges of integral types, it does cite C99, in 1.2 Normative references, as applying. Hence the minimal ranges as set out in C99 5.2.4.2.1 Sizes of integer types <limits.h> are applicable.

In terms of long double, that's actually a floating point value rather than an integer. Similarly to the integral types, it's required to have at least as much precision as a double and to provide a superset of values over that type (meaning at least those values, not necessarily more values).

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C++ – the difference between #include and #include “filename”

In practice, the difference is in the location where the preprocessor searches for the included file.

For #include <filename> the preprocessor searches in an implementation dependent manner, normally in search directories pre-designated by the compiler/IDE. This method is normally used to include standard library header files.

For #include "filename" the preprocessor searches first in the same directory as the file containing the directive, and then follows the search path used for the #include <filename> form. This method is normally used to include programmer-defined header files.

A more complete description is available in the GCC documentation on search paths.

C++ – What are the differences between a pointer variable and a reference variable in C++

A pointer can be re-assigned:

int x = 5;
int y = 6;
int *p;
p = &x;
p = &y;
*p = 10;
assert(x == 5);
assert(y == 10);

A reference cannot be re-bound, and must be bound at initialization:

int x = 5;
int y = 6;
int &q; // error
int &r = x;

A pointer variable has its own identity: a distinct, visible memory address that can be taken with the unary & operator and a certain amount of space that can be measured with the sizeof operator. Using those operators on a reference returns a value corresponding to whatever the reference is bound to; the reference’s own address and size are invisible. Since the reference assumes the identity of the original variable in this way, it is convenient to think of a reference as another name for the same variable.
```
int x = 0;
int &r = x;
int *p = &x;
int *p2 = &r;

assert(p == p2); // &x == &r
assert(&p != &p2);
```

You can have arbitrarily nested pointers to pointers offering extra levels of indirection. References only offer one level of indirection.

int x = 0;
int y = 0;
int *p = &x;
int *q = &y;
int **pp = &p;

**pp = 2;
pp = &q; // *pp is now q
**pp = 4;

assert(y == 4);
assert(x == 2);

A pointer can be assigned nullptr, whereas a reference must be bound to an existing object. If you try hard enough, you can bind a reference to nullptr, but this is undefined and will not behave consistently.

/* the code below is undefined; your compiler may optimise it
 * differently, emit warnings, or outright refuse to compile it */

int &r = *static_cast<int *>(nullptr);

// prints "null" under GCC 10
std::cout
    << (&r != nullptr
        ? "not null" : "null")
    << std::endl;

bool f(int &r) { return &r != nullptr; }

// prints "not null" under GCC 10
std::cout
    << (f(*static_cast<int *>(nullptr))
        ? "not null" : "null")
    << std::endl;

You can, however, have a reference to a pointer whose value is nullptr.

Pointers can iterate over an array; you can use ++ to go to the next item that a pointer is pointing to, and + 4 to go to the 5th element. This is no matter what size the object is that the pointer points to.
A pointer needs to be dereferenced with * to access the memory location it points to, whereas a reference can be used directly. A pointer to a class/struct uses -> to access its members whereas a reference uses a ..
References cannot be put into an array, whereas pointers can be (Mentioned by user @litb)
Const references can be bound to temporaries. Pointers cannot (not without some indirection):
```
const int &x = int(12); // legal C++
int *y = &int(12); // illegal to take the address of a temporary.
```
This makes const & more convenient to use in argument lists and so forth.

Best Answer

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