使用不带Unicode字节顺序标记的iconv进行字符转换

Question

如何在不添加字节顺序标记的情况下使iconv转换为UTF-*编码？

这段代码显示了这个问题，它是在64位Linux下用G++4.8编译的：

#include <iostream>
#include <iconv.h>
#include <cstdint>
#include <iomanip>
using namespace std;
int main() {
        const wchar_t *ws = L"Hello, world.n";
        for(size_t ii = 0; ii < 15; ii++) {
                cout << setw(2) << hex << uint32_t(ws[ii]) << " ";
        }
        cout << endl;
        iconv_t conv = iconv_open("UTF-16", "UTF-32");
        char *outbuf = new char[14 * 4];
        char *inptr = const_cast<char*>(reinterpret_cast<const char *>(ws));
        char *outptr = outbuf;
        size_t in_len = 14 * 4;
        size_t out_len = 14 * 4;
        size_t result = iconv(conv, &inptr, &in_len, &outptr, &out_len);
        uint16_t *encoded = reinterpret_cast<uint16_t*>(outbuf);
        for(size_t ii = 0; ii < 15; ii++) {
                cout << setw(2) << hex << encoded[ii] << " ";
        }
        cout << endl;
}

该输出：

48 65 6c 6c 6f 2c 20 77 6f 72 6c 64 2e  a 0
feff 48 65 6c 6c 6f 2c 20 77 6f 72 6c 64 2e a

在中不存在的结果字符串的开头清楚地显示BOM

Answer 1

我发现这对我有用：

iconv_t conv = iconv_open("UTF16LE", "UTF32");

尽管我认为这取决于实现。

还有一个类似的问题：使用iconv在没有BOM 的情况下从UTF-16BE转换为UTF-8