Usage of dollar "$" in xsl - xslt

Question

In the following piece of code, I want to understand what is "$type" here and how it is being used.
How this if condition is being applied using "$type".
<xsl:template name="CodValue">
<xsl:param name="type"/>
<xsl:param name="nodeNM">category</xsl:param>
<xsl:element name="{$nodeNM}">
<xsl:if test="$type">
<xsl:attribute name="xsi:type">
<xsl:value-of select="$type"/>
</xsl:attribute>
</xsl:if>
</xsl:element>
</xsl:template>

Answer 1

For this template, you need a parameter, another one is optional nodeNM. You can call the parameter like this:
<xsl:call-template name="CodValue">
<xsl:with-param name="type" select="123" />
</xsl:call-template>
or
<xsl:call-template name="CodValue">
<xsl:with-param name="type">123</xsl:with-param>
</xsl:call-template>
type is a variable in CodValue, so you can print it via
<xsl:value-of select="$type" />
or via {$type} in attributes.
Suggestions:
$nodeNM looks like the name of a tag (html-tags if you produce HTML-Code).
$type (if returns true from xpath, ie. if not empty) will create a xsi:type-Tag-Attribute.
So if you call
<xsl:call-template name="CodValue">
<xsl:with-param name="type" select="123" />
</xsl:call-template>
Your XML will be transformed into
<category xsi:type="123" />

Answer 2

$ is used to reference a variable inside an XPath expression.
In this particular case, $type is declared earlier by <xsl:param name="type"/>. However, it has not been given a value so you will need to use <xsl:with-param> when calling the template to so that you can provide a value.
n.b. The variable $nodeNM was given a default value, so you don't need to specify that when calling the template.