使用scatter-gather处理短读/写的技术？

Scatter-gather – readv()/ writev()/ preadv()/ pwritev() – 在单个系统调用中读/写可变数量的iovec结构.基本上它从第0个iovec到第N个顺序读/写每个缓冲区.但是根据文档,它在readv / writev调用上的返回次数也少于请求的次数.我想知道是否有一种标准/最佳实践/优雅方式来处理这种情况.

如果我们只是处理一堆字符缓冲区或类似的东西,这不是什么大问题.但其中一个细节是使用分散 – 聚集结构和/或离散变量作为单独的iovec项目.你如何处理readv / writev只读取/写入一个结构的一部分或一半长或类似的情况.

下面是我得到的一些人为的代码：

int fd;

struct iovec iov[3];

long aLong = 74775767;
int  aInt  = 949;
char aBuff[100];  //filled from where ever

ssize_t bytesWritten = 0;
ssize_t bytesToWrite = 0;

iov[0].iov_base = &aLong;
iov[0].iov_len = sizeof(aLong);
bytesToWrite += iov[0].iov_len;

iov[1].iov_base = &aInt;
iov[1].iov_len = sizeof(aint);
bytesToWrite += iov[1].iov_len;

iov[2].iov_base = &aBuff;
iov[2].iov_len = sizeof(aBuff);
bytesToWrite += iov[2].iov_len;

bytesWritten = writev(fd,iov,3);

if (bytesWritten == -1)
{
    //handle error
}

if (bytesWritten < bytesToWritE)
    //how to gracefully conTinue?.........

for (;;) {
    written = writev(fd,iov+cur,count-cur);
    if (written < 0) goto error;
    while (cur < count && written >= iov[cur].iov_len)
        written -= iov[cur++].iov_len;
    if (cur == count) break;
    iov[cur].iov_base = (char *)iov[cur].iov_base + written;
    iov[cur].iov_len -= written;
}