Address of a temporary in Go?

pointers return go temporary rvalue
Matt Joiner picture Matt Joiner · May 10, 2012 · Viewed 20.1k times · Source

What's the cleanest way to handle a case such as this:

func a() string {
    /* doesn't matter */
}

b *string = &a()

This generates the error:

cannot take the address of a()

My understanding is that Go automatically promotes a local variable to the heap if its address is taken. Here it's clear that the address of the return value is to be taken. What's an idiomatic way to handle this?

Answer

zzzz picture zzzz · May 10, 2012

The address operator returns a pointer to something having a "home", e.g. a variable. The value of the expression in your code is "homeless". if you really need a *string, you'll have to do it in 2 steps:

tmp := a(); b := &tmp

Note that while there are completely valid use cases for *string, many times it's a mistake to use them. In Go string is a value type, but a cheap one to pass around (a pointer and an int). String's value is immutable, changing a *string changes where the "home" points to, not the string value, so in most cases *string is not needed at all.

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