Thread: What's the difference between long and long int?

What's the difference between long and long int?
Similarly, what's the difference between short and short int?

When I use typeid on them (using visual c++) they retain their types as 'long' and 'short' instead of being converted implicitly to 'long int' or 'short int' as I thought they would have.

Code:

    long var = 5.5;
    std::cout << typeid(var).name();

output: long

long and short also occupy the same amount of space as long int and short int accoring to sizeof().

So really how are they different, and why are long and short valid types when they're supposed to be modifiers? It's juzt confusing. Because I remember seeing long in examples (now forgotten where I saw them) and always thought that it was implicitly long int.

I tried searching the web for a long time but didn't get any proper material about this.

long means long int. They are exactly the same. Most programmers prefer to say long.
short and short int are also exactly the same.
So are long long and long long int.
So are unsigned and unsigned int.

> But then why are we able to use long and short as data types at all?
History - because that's what archaic C allowed you to do.

It's allowed because there's no other reasonable interpretation of long or long long or unsigned or unsigned long or unsigned long long. Why force people to put int after all of those? It adds absolutely nothing.

Originally Posted by Nwb

Hmmm would you happen to know why C allowed that? But thanks Salem, that makes sense.

Because "long" was used before "long int" was supported in most C Compilers. It was likely done because the compiler was easier to write that way.

Tim S.

Originally Posted by Nwb

My point was that having two types (...) that basically are the same thing can cause confusion.

long and long int both denote the same type, so there is only one type. long is a simple type specifier for the type long int.

Originally Posted by Nwb

(as I found that 'long' wouldn't be implicitly converted to 'long int')

It is true that a long wouldn't be implicitly converted to long int, but that's because they are of the same type. It would be like trying to implicitly convert an int to an int.

Originally Posted by Nwb

Take the example of function overloading, how do you know which overload would be taken (keep in mind overloading is based on types not values)?

Good question. So, you're saying that you can compile this program without an error?

Code:

#include <iostream>

void foo(long int x)
{
    std::cout << "long int" << std::endl;
}

void foo(long x)
{
    std::cout << "long" << std::endl;
}

int main()
{
}

If so, you might want to change to a standard conforming compiler. You cannot overload with the corresponding parameter differing only by one being long and the other being long int because long and long int denote the exact same type.

Originally Posted by Nwb

And remember this was after I thought 'long' and 'long int' are different types, now I found that they're not.

Eh, you just wrote that they are two types, and now you say that they are the same type?

I was explaining my initial assumption and my conclusion ;p
I thought they were two different types. I found out that 'long int' is nothing but 'long'. And hence there's only one type.

I get that it wasn't that clear, but I was trying to tell john.c why it would cause confusion for having two types that did the same thing. But then again it's not two types. It's one type. So ya I messed up that post.

You see, I was about to post that when I decided to test whether 'long int' was 'long int' and then when I found out that it was 'long', I lost my train of thought and posted that saying two contradicting things.

Don't forget there are such things as 'long double' as well.

Originally Posted by flp1969

Depending on the platform, int and long int are completely different "types". On Windows, both have the same size (32 bits), but on SysV ABI for x86-64 int is 32 bits long, and long int, 64

Pure idiocy, responding to a question that wasn't asked.
Nobody said that int and long were the same.
Just that long and long int are the same.
Jesus!

> Depending on the platform, int and long int are completely different "types".
No, they ARE different types - period.
If they happen to have the same number of bits and representation on a platform doesn't make them suddenly the same type.
Equivalence in storage capacity doesn't make for equality of type.

Originally Posted by C99

6.3.1 Arithmetic operands
6.3.1.1 Boolean, characters, and integers
1 Every integer type has an integer conversion rank defined as follows:
— No two signed integer types shall have the same rank, even if they have the same
representation.
— The rank of a signed integer type shall be greater than the rank of any signed integer
type with less precision.
— The rank of long long int shall be greater than the rank of long int, which
shall be greater than the rank of int, which shall be greater than the rank of short
int, which shall be greater than the rank of signed char.

Code:

int a = 1;
long int b = 2L;
long int c = a + b;

So 'a' will always be promoted to long int before the addition (during semantic parsing), even if that eventually turns into a nop at code generation time.

By the way.. how is this function call not ambiguous?

Code:

#include <iostream>

void print(int) {
    std::cout << "Int\n";
}

void print(long long int) {
    std::cout << "Long\n";
}

int main()
{
    std::cout << "Type of literal: " << typeid(5l).name();
    std::cout << "\nFunction called: ";
    print(5l);
    std::cin.get();
}

Short also calls the 'int' overload. How does the compiler know that 'short int' must call 'int' instead of 'long' (I wrote another overload for long)?