Why Does a Pointer Receiver in String Method Lead to a Dead Loop in Go?

Pointer Receiver vs Value Receiver in String Method

In Go, the String() method is used to control the string representation of a custom type. When defining the String() method, it's crucial to understand the difference between a pointer receiver and a value receiver.

Consider the following code:

type TT struct {
    a int
    b float32
    c string
}

func (t *TT) String() string {
    return fmt.Sprintf("%+v", *t)
}

In this code, the String() method has a pointer receiver, which means it operates on a pointer to a TT value. Calling tt.String() effectively calls (*tt).String().

Now, consider changing the String() method to the following:

func (t *TT) String() string {
    return fmt.Sprintf("%+v", t)
}

This change removes the pointer receiver, making the String() method operate on a value of type TT.

Why does this result in a dead loop?

The fmt package checks if the value being printed implements the Stringer interface (or has a String() method). If so, it will call that method to obtain the string representation. However, in this case, we have a pointer receiver for the String() method.

When we call tt.String(), the fmt package passes a value of type *TT. However, the String() method expects a value of type TT. This mismatch causes the fmt package to call the String() method again, which in turn will be called again, and so on, resulting in an infinite loop.

Prevention/Protection

To prevent this dead loop, ensure that the receiver type of the String() method matches the type of the value passed to the fmt package. If for some reason you need to use the same receiver type, you can create a new type using the type keyword and convert the value to the new type before calling fmt.Sprintf:

func (t TT) String() string {
    type TT2 TT
    return fmt.Sprintf("%+v", TT2(t))
}

By creating a new type, we remove all methods from the underlying type, effectively breaking the cycle and allowing us to safely call fmt.Sprintf.

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