How to convert a type to a new type based on it?

Question content

I created a new type to add custom methods specific to my application needs

type content html.node

I know I can do this by deriving html.node from a content var of type <div class="code" style="position:relative; padding:0px; margin:0px;"><pre class='brush:php;toolbar:false;'>node := html.node(content)</pre><div class="contentsignin">Copy after login</div></div> However, there is a var of type

html.node

, how to convert it to content? I found out that doing the following...

ndoe, err := htmlquery.loadurl(page.url)

content := content(node)

...can not work. It gave me this error:

cannot convert doc (variable of type *html.Node) to type Content

Correct answer

https://www.php.cn/link/b14fba037d119d307e3b6ceb2a9fbb31

If the node is

html.node

use: <div class="code" style="position:relative; padding:0px; margin:0px;"><pre class='brush:php;toolbar:false;'>c := content(node) </pre><div class="contentsignin">Copy after login</div></div> If the node is

*html.node

use: <div class="code" style="position:relative; padding:0px; margin:0px;"><pre class='brush:php;toolbar:false;'>c := (*content)(node) </pre><div class="contentsignin">Copy after login</div></div> Comments on the specification linked above:

"If a type begins with the operator

*... it must be enclosed in parentheses if necessary to avoid ambiguity." If you want

content

instead of *content then you need to dereference the pointer before or after the conversion, for example:

c := Content(*node) // dereference before conversion
// or
c := *(*Content)(node) // dereference after conversion

https://www.php.cn/link/fb9d433088a21464e7d634c4e190b31a

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