使用iconv时要注意入参的涵义和类型

64位linux

#include <stdio.h>
#include <iconv.h>
#include <string.h>

int code_conv(const char* fromcode,const char* tocode,const char* str_in,int inlen,char* str_out,int outsize)
{
    if (NULL==str_in || inlen<=0 || NULL==str_out || outsize<=0)
    {
        return -1; //入参错误
    }

    iconv_t i_cv = iconv_open(tocode, fromcode);
printf(" i_cv   :%ld\n", i_cv);
    if ((iconv_t)-1 == i_cv)
    {
        return -2; //转换描述符申请失败
    }

printf("sizeof(iconv_t):%d\n", sizeof(iconv_t));
    char** pp_in = (char**)&str_in;
    char** pp_out = (char**)&str_out;
int b=1;
    int inleft = inlen;
    int a=1;
    size_t outleft = outsize;

    if ((size_t)-1 == iconv(i_cv, pp_in, (size_t*)&inleft, pp_out, (size_t*)&outleft))
    {
        perror("iconv");
printf(" i_cv   :%ld\n", i_cv);
        iconv_close(i_cv);
        return -3; //转码失败
    }

    iconv_close(i_cv);
    return outsize-outleft; //转码后的长度
}

main()
{
    char *in_utf8 = "姝ｅ?ㄥ??瑁?";
    char *in_gb2312 = "正在安装";
    char str_out[512] = {0}; //据chenx说,此处out数组小可能导致转码失败
    int in_len = strlen(in_gb2312);
    int out_size = sizeof(str_out);

    int n = code_conv("gb2312", "utf-8", in_gb2312, in_len, str_out, out_size);
    if(n > 0)
    {
        printf("out:%s\n", str_out);
        printf("out len:%d\n", n);
    }
    else
    {
        perror("code_conv");
    }
            printf("out:%s\n", str_out);
        printf("out len:%d\n", n);
}

i_cv :324074912
sizeof(iconv_t):8
iconv: Invalid or incomplete multibyte or wide character
 i_cv :324074911
段错误 (core dumped)

(size_t*)&inleft的用法显然是不对的。i_cv的值减少了1！如果在iconv_close尝试给i_cv加1,则报错：code_conv.cpp:30: 错误：‘void *’ 型指针用在了算术表达式
这是怎么发生的呢？

（未完待续）