Is Linux kernel splice() zero copy?

Question

I know splice() is designed for zero copy and used Linux kernel pipe buffer to achieve that. For example if I wanted to copy data from one file descriptor(fp1) to another file descriptor(fp2), it didn't need to copy data from "kernel space->user space->kernel space". Instead it just copy data in kernel space the flow will be like "fp1 -> pipe_read -> pipe_write -> fp2". And my question is that dose kernel need to copy data between "fp1 -> pipe_read" and "pipe_write -> fp2"?

The Wikipedia said that:

Ideally, splice and vmsplice work by remapping pages and do not actually copy any data,    
which may improve I/O performance. As linear addresses do not necessarily correspond to
contiguous physical addresses, this may not be possible in all cases and on all hardware 
combinations.

I have already traced kernel source(3.12) for my question and I found that the flow between "fp1->write_pipe", in the end it would called kernel_readv() in fs/splice.c and then called "do_readv_writev()" and finally called "aio_write()"

558 static ssize_t kernel_readv(struct file *file, const struct iovec *vec,
559                 unsigned long vlen, loff_t offset)
//*vec would point to struct page which belong to pipe

The flow between "read_pipe -> fp2" in the end would call "__kernel_write()" and then called "fp2->f_op->write()"

430 ssize_t __kernel_write(struct file *file, const char *buf, size_t count, loff_t *pos)
//*buf is the pipe buffer

And I thought both "aio_write()" and "file->f_op_write()" would perform really data copy, so does splice() really perform zero copy?

Answer 1

As I understand splice(), it will read pages of fd1 and the MMU will map these pages. The reference created by the mapping will be put into the pipe and handed over to fd2. No real data should be copied in the process, as long as every participant has DMA available. If no DMA is available you need to copy data.