C语言编译时，提示invalid conversion from ‘char’ to ‘char*’ [-fpermissive]错误，请问哪里问题?

Question

编译环境：
Armbian（arm嵌入式linux）64位系统，基于Ubuntu Bionic 18.04 LTS 4.19内核
错误提示：

tools.c:63:13: error: invalid conversion from ‘char’ to ‘char*’ [-fpermissive]
*pout = '\0';

^~~~

部分代码内容：
这是图片：

以下是文字代码：
int code_convert(char *from_charset, char *to_charset, char *inbuf, size_t inlen, char *outbuf, size_t outlen)
{

iconv_t cd;

char **pin = &inbuf;

char **pout = &outbuf;

cd = iconv_open(to_charset, from_charset);  
if (cd == 0)  
    return -1;  
memset(outbuf, 0, outlen);  
if (iconv(cd, pin, &inlen, pout, &outlen) == -1)  
    return -1;  
iconv_close(cd);  
*pout = '\0';  

return 0;  

}

Answer 1

pout是指向指针的指针，*pout是指向char的指针，**pout才是char；改成**pout = '\0'; 就对了