Why Does Printing a `bytes.Buffer` in Go Produce Different Output Depending on Whether It\'s a Pointer or Not?

Different Behavior in Printing bytes.Buffer in Go

In Go, the behavior of printing a value of type bytes.Buffer varies depending on whether it is a pointer or a non-pointer value.

Consider the following code example:

buf := new(bytes.Buffer)
buf.WriteString("Hello world")
fmt.Println(buf)

In this case, using the * operator to dereference the buffer pointer invokes the String() method of the bytes.Buffer type. This method returns a string representation of the buffer's content, resulting in "Hello World" being printed.

Now, let's modify the code as follows:

var buf bytes.Buffer
buf.WriteString("Hello world")
fmt.Println(buf)

Without using the * operator, we are passing a non-pointer value of type bytes.Buffer. Unlike the previous example, the fmt package does not call the String() method in this case. Instead, it prints the buffer as a struct value, using the default formatting:

{[72 101 108 108 111 32 119 111 114 108 100] 0 [72 101 108 108 111 32 119 111 114 108 100 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0] 0}

This output represents the internal structure of the bytes.Buffer, including the slice backing the buffer and the length and capacity fields.

The key difference between these two examples lies in the String() method. When printing a value of type *bytes.Buffer, the fmt package utilizes the String() method to obtain a string representation of the buffer's content. Since the non-pointer value of type bytes.Buffer does not have this method, the default struct formatting is employed.

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