C++ Core Guidelines: Declarations and Initialisations

Let’s continue our tour through the rules for expressions and statements in the C++ core guidelines. This post will be about declarations and initializations.

To be honest, most of the rules in this post are quite obvious but they often provide one or the other very interesting insight; therefore, I will mainly write in this post about these special aspects. Here are the rules for today:

Here are the details.

ES.11: Use `auto` to avoid redundant repetition of type names

The example from the guidelines is not promising to me. So, let me give you another one. Changing your code may become a piece of cake if you use auto.

The following example is based on auto. You do have not to think about the types; therefore, you can not make an error. This means the type of res will be int at the end.

auto a = 5;
auto b = 10;
auto sum =  a * b * 3;
auto res = sum + 10; 
std::cout << typeid(res).name();         // i

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If you decide to change the literal b from int to double (2) or use in (3) a float literal instead of the int literal. No problem. It will be automatically handled for you.

auto a = 5;
auto b = 10.5;                           // (1)
auto sum = a * b * 3;
auto res = sum * 10;  
std::cout << typeid(res).name();         // d
  
auto a = 5;
auto b = 10;
auto sum = a * b * 3.1f;                // (2)
auto res = sum * 10;  
std::cout << typeid(res).name();        // f

ES.12: Do not reuse names in nested scopes

This is one of these quite apparent rules. You should not reuse names in nested scopes for readability and maintenance reasons.

// shadow.cpp

#include <iostream>

int shadow(bool cond){
  int d = 0;
  if (cond){
    d = 1;
  }
  else {
    int d = 2;
    d = 3;
  }
  return d;
}

int main(){

    std::cout << std::endl;
    
    std::cout << shadow(true) << std::endl;      
    std::cout << shadow(false) << std::endl;     

    std::cout << std::endl;
    
}

What will be the output of the program? Confused by the d’s? Here is the result.

This was easy! Right? But the same phenomenon is quite surprising in class hierarchies.

// shadowClass.cpp

#include <iostream>
#include <string>

struct Base{
    void shadow(std::string){                           // 2
        std::cout << "Base::shadow" << std::endl;       
    }
};

struct Derived: Base{
    void shadow(int){                                   // 3
        std::cout << "Derived::shadow" << std::endl;    
    }
};

int main(){
    
    std::cout << std::endl;
    
    Derived derived;
    
    derived.shadow(std::string{});                      // 1
    derived.shadow(int{});                              
    
    std::cout << std::endl;
    
}

Both structs Base and Derived have a method shadow. The one in the base accepts a std::string (2) and the other one an int (3). When I invoke the object derived with a default-constructed std::string (1), I may assume that the base version will be called. Wrong! Because the method shadow is implemented in the class Derived, the methods of the base class will not be considered during name resolution. Here is the output of my Gcc.

To fix this issue, shadow must be known to Derived.

struct Derived: Base{
    using Base::shadow;              // 1
    void shadow(int){
        std::cout << "Derived::shadow" << std::endl;   
    }
};

You have to put a Base::shadow (1) into Derived. Now the program behaves as expected.

ES.20: Always initialize an object

The rules on which object will be initialized or not are pretty tricky to get right in C++. Here is a simple example.

struct T1 {};
class T2{
 public:
    T2() {} 
};

int n;                 // OK

int main(){
  int n2;              // ERROR
  std::string s;       // OK
  T1 t1;               // OK
  T2 t2;               // OK                  
}

n is a global variable, which will be initialized to 0. This will not hold for n2 because it is a local variable and will not be initialized. But they will be initialized if you use a user-defined type such as std::string, T1, or T2 in a local scope.

If that is too difficult for you, I have a simple fix. Use auto. Now, you can not forget to initialize the variable. The compiler will check this.

struct T1 {};
class T2{
 public:
    T2() {}
};

auto n = 0;

int main(){
  auto n2 = 0;
  auto s = ""s;      
  auto t1 = T1();               
  auto t2 = T2();
}

ES.21: Don’t introduce a variable (or constant) before you need to use it

I think this is trivial. We program C++, not C.

ES.22: Don’t declare a variable until you have a value to initialize it with

You may have a so-called used-before-set error if you don’t follow this rule. Have a look at the guidelines.

int var;  

if (cond)      // some non-trivial condition
    Set(&var);
else if (cond2 || !cond3) {
    var = Set2(3.14);
}

// use var

Do you know if one of the conditions holds? If not, var as a local built-in variable is not initialized.

ES.23: Prefer the `{}`-initializer syntax

There are a lot of reasons for using {}-initialization:

always applicable
overcomes the most vexing parse
prevents narrowing

You have to keep a particular rule in mind. If you combine auto with an {}-initialization, you will get an std::initializer_list in C++14 but not in C++17.

For all the details, read my previous post to {}-Initialisation.

ES.24: Use a `unique_ptr<T>` to hold pointers

I will make it short. A std::unique_ptr<T> is as efficient as a raw pointer by design but has a great added value: it takes care of its resource. This means: don’t use a raw pointer. If you are curious about the details of std::unique_ptr, read my two posts to std::unqiue_ptr.

What’s next?

We are not done with the rules for declarations in C++. The remaining will follow in the next post.

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