Is there a way to check that a variable is an int?

Question

Had a look at the <typeinfo> and <type_traits> header files, but the only way i could see to do it was to use a rouge integer to compare against...

#include <iostream>
#include <typeinfo>
using namespace std;

int main()
{
int a = 7; //rogue number for use in typeid to test against..is there some other way?
int myint = 40;

if(typeid(myint).name() == typeid(a).name()){
cout << "It is an integer";
}

return 0;
}
Is there some other way?

Answer 1

x
if( (int)(x) == x)
cout<<"yes";
else
cout<<"no";

Answer 2

I have once stumbled over this and found it helpful. I hope you'll find what you need.

https://stackoverflow.com/questions/1986418/typeid-versus-typeof-in-c

Answer 3

Honfu
Had a scan of your link..but can't find anything that does what I was after. In python you have isinstance function. in c++ you have 'is_class' and pass a class name to it and it will return true or false, but is_intergral only takes a type name e.g 'int' rather than on object of that type e.g '5'.
Thanks anyway.

Answer 4

Use decltype to get the type.
if( std::is_integral<decltype(5)>::value )
{
std::cout << "integral\n";
}

But you really do not need this because 5 will always be an int, 5.f will always be a float and 5.0 will always be a double.
I've never needed this other than with templates.

Answer 5

To be honest, I can remember why I need to do this, it was a small project I was working on (for myself) a few months back.
But.... can you pass a variable of type 'int' into it rather than a constant, '5' in you example.
Thanks for the decltype suggestion, I'll look more into it.

Answer 6

int x;
if( std::is_integral<decltype( x )>::value )

FYI, integral and int are not the same, the above also returns true for long long or char, for example.
std::is_same<int, decltype( x )>::value would return true just for an int type.

Answer 7

Why would you even need to?
A variable with the type int will always be an int.
This isn't python or js.

The only case I can think of is in combination with templates where you want the template to accept only integrals, for example.
In that case you can combine std::enable_if and std::is_integral for all integral types or std::enable_if and std::is_same<int,T> just for ints.