Show return symbol on listings line breaks

When answering another answer, I found out that the tikzmark library for TikZ provides some features for use with the listings package. The following still requires you to manually place the return marks, so it is not done fully automatically. But it may still be better than no return marks at all ...

The macro takes two arguments, the first for the line number in question and the second for the number of line breaks. Note that I assumed the line spacing of the listings environment to be 1em, but it may be different in other circumstances.

documentclass[a4paper,12pt]{article}
usepackage{xcolor}
usepackage{listings}

usepackage{tikz}
usetikzlibrary{tikzmark}
usetikzmarklibrary{listings}

newcommand{continuingline}[2]{%
    begin{tikzpicture}[remember picture, overlay]
        foreach i in {1,...,#2} {
            node[anchor=base, align=center] at ([xshift=-1.25em, yshift={-i*1em}]pic cs:line-code-#1-start) {$hookrightarrow$};
        }
    end{tikzpicture}%
}

lstset{
    basicstyle=footnotesizettfamily,
    breakatwhitespace=true,  
    breaklines=true,         
    firstnumber=1,
    frame=leftline,       
    keepspaces=true,         
    numbers=left,            
    showspaces=false,        
    showstringspaces=false,  
    showtabs=false,          
    stepnumber=1,
    language=java,
    numberstyle=color{darkgray},
    commentstyle=color{gray},     
    stringstyle=color{red},
    keywordstyle=color{blue},
    rulecolor=color{black},
}

begin{document}

begin{lstlisting}[caption={You can easily recognize line breaks}, name=code]
public int nextInt(int n) {
 if (n<=0)
  throw new IllegalArgumentException("n is not positive");


 // this is a very very very very very very very very very very very very very very very very very very very very very very very very very very very long line
 if ((n & -n) == n)  // i.e., n is a power of 2
  return (int)((n * (long)next(31)) >> 31);

 int bits, val;
 do {
  bits = next(31);
  val = bits % n;
 } while(bits - val + (n-1) < 0);
 // this is another although not very very very very long but still quite long line
 return val;
}
end{lstlisting}

continuingline{6}{2}
continuingline{15}{1}

end{document}

---

Update: Better solution that also works for more than only one lstlisting environment.

The macro now takes three mandatory arguments, the first being the name of the listing, which has to be set via name=... as option to the relevant lstlisting environment. The second and third argument is again the relevant line number and the number of line breaks respectively.

Furthermore, the macro takes one optional argument, in case that the lineskip is something other than 1em. It defaults to 1em.

continuingline[<lineskip>]{<listings name>}{<line number>}{<line breaks>}

documentclass[a4paper,12pt]{article}
usepackage{xcolor}
usepackage{listings}

usepackage{tikz}
usetikzlibrary{tikzmark}
usetikzmarklibrary{listings}

newcommand{continuingline}[4][1em]{%
    begin{tikzpicture}[remember picture, overlay]
        foreach i in {1,...,#4} {
            node[anchor=base, align=center] at ([xshift=-1.25em, yshift={-i*#1}]pic cs:line-#2-#3-start) {$hookrightarrow$};
        }
    end{tikzpicture}%
}

lstset{
    basicstyle=footnotesizettfamily,
    breakatwhitespace=true,  
    breaklines=true,         
    firstnumber=1,
    frame=leftline,       
    keepspaces=true,         
    numbers=left,            
    showspaces=false,        
    showstringspaces=false,  
    showtabs=false,          
    stepnumber=1,
    language=java,
    numberstyle=color{darkgray},
    commentstyle=color{gray},     
    stringstyle=color{red},
    keywordstyle=color{blue},
    rulecolor=color{black},
}

begin{document}

begin{lstlisting}[caption={You can easily recognize line breaks}, name=codea]
public int nextInt(int n) {
 if (n<=0)
  throw new IllegalArgumentException("n is not positive");


 // this is a very very very very very very very very very very very very very very very very very very very very very very very very very very very long line
 if ((n & -n) == n)  // i.e., n is a power of 2
  return (int)((n * (long)next(31)) >> 31);

 int bits, val;
 do {
  bits = next(31);
  val = bits % n;
 } while(bits - val + (n-1) < 0);
 // this is another although not very very very very long but still quite long line
 return val;
}
end{lstlisting}

continuingline{codea}{6}{2}
continuingline{codea}{15}{1}

begin{lstlisting}[caption={You can again easily recognize line breaks}, name=codeb]
public int nextInt(int n) {
 if (n<=0)
  throw new IllegalArgumentException("n is not positive");
  // this is another unbelievably long and probably under normal circumstances totally unnesessary line
 return val;
}
end{lstlisting}

continuingline{codeb}{4}{1}

begin{lstlisting}[caption={You can again easily recognize line breaks}, name=codec, basicstyle=footnotesizettfamilylinespread{1.2}selectfont]
public int nextInt(int n) {
 if (n<=0)
  throw new IllegalArgumentException("n is not positive");
  // this is yet another unbelievably long and probably under normal circumstances totally unnesessary line
 return val;
}
end{lstlisting}

continuingline[1.2em]{codec}{4}{1}

end{document}