Difference between inline and separate line variable initializations in the context of closures?

<p>I was going through the Tour of Go when I noticed in the exercise where we&#39;re supposed to generate the fibonacci sequence with a lambda function I got two different results between this: </p> <p>func fibonacci() func() int {<br/> x := 0<br/> y := 1<br/> return func() int {<br/> x, y = y, x+y<br/> return x<br/> }<br/> } </p> <p>and this:</p> <p>func fibonacci() func() int {<br/> x := 0<br/> y := 1<br/> return func() int {<br/> x = y<br/> y = x+y<br/> return x<br/> }<br/> } </p> <p>I thought both versions of the code would function the same but it didn&#39;t seem like the case.</p> <hr/>**评论：**<br/><br/>gohacker: <pre><p>This is unrelated to closures. </p> <blockquote> <p>The assignment proceeds in two phases. First, the operands of index expressions and pointer indirections (including implicit pointer indirections in selectors) on the left and the expressions on the right are all evaluated in the usual order. Second, the assignments are carried out in left-to-right order. </p> </blockquote> <p><a href="https://golang.org/ref/spec#Assignments" rel="nofollow">https://golang.org/ref/spec#Assignments</a></p></pre>Tikiatua: <pre><p>Hi there, </p> <p>The difference between the two versions is as follows: In the version x,y = y, x+y both variables are assigned the value &#34;at the same time&#34;. Therefore x will not have the value of y for the assignment y = x+y. In the second version, x is first assigned the value of y, then y is assigned the new value of x plus y.</p> <p>Hope this clarifies things for you.</p></pre>tezeev: <pre><p>To make the point clear, statement like</p> <pre><code>x, y = y, x + y </code></pre> <p>is equivalent to</p> <pre><code>t := y y = x + y x = t </code></pre></pre>