Why Does Printing a `bytes.Buffer` in Go Produce Different Results Depending on Whether It\'s a Pointer or a Value?

Different Behavior of Printed Bytes Buffers in Go

In Go, printing a bytes.Buffer value can yield distinct outputs depending on whether it's a pointer or a regular value.

When creating a bytes.Buffer with new(bytes.Buffer), we obtain a pointer to a buffer. Accessing the value's String() method and printing it outputs the buffer's content: Hello World.

However, using var buf bytes.Buffer directly creates a value of type bytes.Buffer. This value doesn't have the String() method, so its default format is printed. This results in the verbose output: {[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 difference arises because Go checks for a String() method when printing values. If it exists, the method is invoked to obtain the value's representation. For pointer values, such as *bytes.Buffer, the String() method is available, but for regular values like bytes.Buffer, it is not.

In contrast to pointers, regular values have a different default format when printed: {field0 field1 ...}, displaying their fields. This explains the different outputs observed when printing bytes.Buffer values depending on whether they're pointers or regular values.

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