Possible mistake in constructor initialization chapter

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1 
class DependencyExample {
private:
    std::string name;           // Declared first
    int nameLength;             // Declared second - depends on name
    
public:
    // WRONG: nameLength depends on name, but name isn't initialized yet
    DependencyExample(const std::string& n)
        : nameLength(n.length()), name(n) {
        // nameLength will be garbage because name isn't initialized yet!
    }
    
    // CORRECT: Declare members in dependency order
    DependencyExample(const std::string& n)
        : name(n), nameLength(name.length()) {
        // name is initialized first, then nameLength
    }
};

// nameLength will be garbage because name isn’t initialized yet!
is not true because compiler will initialize it in order of which members are declared. It was mentioned just before this.

2

You’re right. nameLength will not be garbage.

The order of initialization would be name = n and then namelength = n.size() in the first constructor as well. Since n is always defined, the answer, namelength will still be defined. Let me think of a better example.

3 
class DependencyExample {
private:
    int nameLength;             // Declared first - but depends on name!
    std::string name;           // Declared second!
    
public:
    // WRONG: nameLength depends on name, but name isn't initialized yet
    DependencyExample(const std::string& n)
        : name(n), nameLength(name.size()) {
        // nameLength will be garbage because name isn't initialized yet!
    }
    
    // CORRECT: Declare members in dependency order
    DependencyExample(const std::string& n)
        : nameLength(n.size()), name(n) {
        // name is initialized first, then nameLength
    }
};

This is the example I wanted to bring up here. nameLength is always going to be undefined because it is dependent on name.size() which is not initialized at the time of being called.

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