Variable Declarations

Like most compiled languages, C++ is statically typed . All variables must be declared to be of a specific type before they can be used. A variable’s type cannot be changed once it is declared.

C++ is (nearly) strongly typed. Mixed-mode expressions are limited and many conversions must be explicit.

Variables are declared by indicating the type followed by a comma-separated list of variables followed by a semicolon.

int i, j, k;
float  x, y;

Initializing at Compile Time

Variables can be declared and initialized at the same time:

float x=1.e.8, y=42.;
int i, j, k, counter=0;

C++, but not C, permits two additional formats for initialization. Constructor initialization encloses the initial value in parentheses. Uniform initialization uses curly braces.

float x(0.);
int   i{0};
int   j=2;


Start Geany or your preferred editor or IDE. Type

#include <iostream>

int main() {
/* Variables
   Author:  Your Name
    float x,y;
    int i,j=11;

    std::cout<<"Reals are "<<x<<" "<<y<<"\n";
    std::cout<<"Integers are "<<i<<" "<<j<<"\n";
    return 0;

Variables can be declared and initialized anywhere in the body of the code as long as they are declared before they are used. There are some implications for variable scope, which we will discuss later.

#include <cmath>
#include <iostream>

int main() {
/* More about variables
   Author:  Your Name
    float x;
    int i,j=11;


    float y=sqrt(x);

    int k;

    std::cout<<"Reals are "<<x<<" "<<y<<"\n";
    std::cout<<"Integers are "<<i<<" "<<j<<" "<<k<<"\n";

    return 0;


A recent introduction to the C++ (since C++11) standard is the auto declaration. We have learned that the compiler is able to infer the type of literals. Thus it can deduce the type of a variable that is initialized to a literal value. Moreover, if a variable is initialized to a variable previously declared to be a named type, the compiler can assign the same type to the new variable.

float x=12;
auto  y=x;

auto i=12;

Any auto variable must be initialized.

auto z;
error: declaration of ‘auto z’ has no initializer

If initializing an auto variable with a literal, be sure the type of the literal is what you intend. Keep in mind that 3 and 3.0 are distinct types to the compiler. The typeid built-in will return the type.

#include <iostream>

int main() {

    float w=11.;

    auto x=w;
    auto y=3.;
    auto z=3;
    std::cout<<typeid(x).name()<<" "<<typeid(y).name()<<" "<<typeid(z).name()<<"\n";
    return 0;

On Unix with g++ this results in

f d i

Recall that the literal 3.0 is a double.

If multiple assignments are made with auto they must be the same.

auto i=0, f=3.14159;


auto i=0;
auto f=3.14159;


Similar to auto, we can use the C++11 declaration decltype to declare that a variable is of the same type as another, previously-declared variable.

float x;
decltype(x) y;

Variable y will take the same type as x. This declaration was intended primarily for more advanced applications such as lambda functions and templates, where it may be difficult to determine the type at compile time.



Programmers can declare a variable to have a fixed value that cannot be changed. In C++ we add the qualifier const to these variables.

const double pi=3.141592653589793;

Variables declared const must be initialized when they are declared. Any attempt to change them will result in a fatal compiler error.


A volatile variable is one that the compiler may not change and that may be beyond the control of the program itself. The value of such a variable must always be read directly from main memory. An example might be a program that obtains data from an instrument. We do not want the compiler to modify the instrument readings so we would declare those variables volatile.

volatile float voltage;