public static function pagination( $pages, $paged, $range = 2, $show

syntax error, unexpected 'public' (T_PUBLIC), expecting end of file
のエラーが出ます原因が全く分かりません。
回答よろしくお願いします。

php
1public static function pagination( $pages, $paged, $range = 2, $show_only = false ) {
2  $pages = ( int ) $pages;    //float型で渡ってくるので明示的に int型 へ
3  $paged = $paged ?: 1;       //get_query_var('paged')をそのまま投げても大丈夫なように
4      $text_first   = "« 最初へ";
5      $text_before  = "‹ 前へ";
6      $text_next    = "次へ ›";
7      $text_last    = "最後へ »";
8    if ( $show_only && $pages === 1 ) {
9        echo '<div class="pagination"><span class="current pager">1</span></div>';
10        return;
11      }
12  
13    if ( $pages === 1 ) return;    // １ページのみで表示設定もない場合
14    if ( 1 !== $pages ) {
15        echo '<div class="pagination"><span class="page_num">Page ', $paged ,' of ', $pages ,'</span>';
16    if ( $paged > $range + 1 ) {
17        echo '<a href="', get_pagenum_link(1) ,'" class="first">', $text_first ,'</a>';
18        }
19    if ( $paged > 1 ) {
20        echo '<a href="', get_pagenum_link( $paged - 1 ) ,'" class="prev">', $text_before ,'</a>';
21    }
22    for ( $i = 1; $i <= $pages; $i++ ) {
23      if ( $i <= $paged + $range && $i >= $paged - $range ) {
24        if ( $paged === $i ) {
25          echo '<span class="current pager">', $i ,'</span>';
26        } else {
27          echo '<a href="', get_pagenum_link( $i ) ,'" class="pager">', $i ,'</a>';
28        }
29      }
30    }
31    if ( $paged < $pages ) {
32      echo '<a href="', get_pagenum_link( $paged + 1 ) ,'" class="next">', $text_next ,'</a>';
33    }
34    if ( $paged + $range < $pages ) {
35      echo '<a href="', get_pagenum_link( $pages ) ,'" class="last">', $text_last ,'</a>';
36    }
37      echo '</div>';
38  }
39}