为什么用cpu_to_le32来表示DMA地址而不用长度？(书上的一个例子代码)

Question

我在读这本书 Essential Linux设备驱动程序 (Sreekrishnan Venkateswaran著)和第10章。清单 10.5. 设置 DMA 描述符和缓冲区我明白了

/* Device-specific data structure for the Ethernet Function */
struct device_data {
  struct pci_dev *pdev; /* The PCI Device structure */
  struct net_device *ndev; /* The Net Device structure */
  void *dma_buffer_rx; /* Kernel virtual address of the receive descriptor */
  dma_addr_t dma_bus_rx; /* Bus address of the receive descriptor */
  void *dma_buffer_tx; /* Kernel virtual address of the transmit descriptor */
  dma_addr_t dma_bus_tx; /* Bus address of the transmit descriptor */
  /* ... */
  spin_lock_t device_lock; /* Serialize */
} *mydev_data;

/* On-card registers related to DMA */
#define DMA_RX_REGISTER_OFFSET 0x0 /* Offset of the register holding the bus address of the RX descriptor */
#define DMA_TX_REGISTER_OFFSET 0x4 /* Offset of the register holding the bus address of the TX descriptor */
#define CONTROL_REGISTER 0x8 /* Offset of the control register */
/* Control Register Defines */
#define INITIATE_XMIT 0x1
/* Descriptor control word definitions */
#define FREE_FLAG 0x1 /* Free Descriptor */
#define INTERRUPT_FLAG 0x2 /* Assert interrupt after DMA */

/* Invoked from Listing 10.3 */
static void dma_descriptor_setup(struct pci_dev *pdev)
{
  /* Allocate receive DMA descriptors and buffers */
  mydev_data->dma_buffer_rx = pci_alloc_consistent(pdev, 3096, &mydev_data->dma_bus_rx);
  /* Fill the two receive descriptors as shown in Figure 10.2 */
  /* RX descriptor 1 */
  mydev_data->dma_buffer_rx[0] = cpu_to_le32((unsigned long)(mydev_data->dma_bus_rx + 24)); /* Buffer address */
  mydev_data->dma_buffer_rx[1] = 1536; /* Buffer length */
  mydev_data->dma_buffer_rx[2] = FREE_FLAG; /* Descriptor is free */
  /* RX descriptor 2 */
  mydev_data->dma_buffer_rx[3] = cpu_to_le32((unsigned long)(mydev_data->dma_bus_rx + 1560)); /* Buffer address */
  mydev_data->dma_buffer_rx[4] = 1536; /* Buffer length */
  mydev_data->dma_buffer_rx[5] = FREE_FLAG; /* Descriptor is free */
  wmb(); /* Write Memory Barrier */
  /* Write the address of the receive descriptor to the appropriate register in the card. The I/O base address, ioaddr, was populated in Listing 10.3 */
  outl(cpu_to_le32((unsigned long)mydev_data->dma_bus_rx), ioaddr + DMA_RX_REGISTER_OFFSET);
  /* Allocate transmit DMA descriptors and buffers */
  mydev_data->dma_buffer_tx = pci_alloc_consistent(pdev, 3096, &mydev_data->dma_bus_tx);
  /* Fill the two transmit descriptors as shown in Figure 10.2 */
  /* TX descriptor 1 */
  mydev_data->dma_buffer_tx[0] = cpu_to_le32((unsigned long)(mydev_data->dma_bus_tx + 24)); /* Buffer address */  <---- line A
  mydev_data->dma_buffer_tx[1] = 1536; /* Buffer length */    <---- line B
  /* Valid descriptor. Generate an interrupt after completing the DMA */
  mydev_data->dma_buffer_tx[2] = (FREE_FLAG | INTERRUPT_FLAG);
  /* TX descriptor 2 */
  mydev_data->dma_buffer_tx[3] = cpu_to_le32((unsigned long)(mydev_data->dma_bus_tx + 1560)); /* Buffer address */
  mydev_data->dma_buffer_tx[4] = 1536; /* Buffer length */
  mydev_data->dma_buffer_tx[5] = (FREE_FLAG | INTERRUPT_FLAG);
  wmb(); /* Write Memory Barrier */
  /* Write the address of the transmit descriptor to the appropriate register in the card. The I/O base, ioaddr, was populated in Listing 10.3 */
  outl(cpu_to_le32((unsigned long)mydev_data->dma_bus_tx), ioaddr + DMA_TX_REGISTER_OFFSET);
}

/* Invoked from Listing 10.3 */
static void dma_descriptor_release(struct pci_dev *pdev)
{
  pci_free_consistent(pdev, 3096, mydev_data->dma_bus_tx);
  pci_free_consistent(pdev, 3096, mydev_data->dma_bus_rx);
}

在代码中，驱动程序为DMA描述符和DMA缓冲区准备了一个缓冲区，利用 pci_alloc_consistent() 并将它们设置好，并将缓冲区地址（总线地址）传递给硬件，确保它是以小恩迪安格式，用 cpu_to_le32(). 所以我理解HW看到的是缓冲区描述符。但是，在描述符中，为什么要用 cpu_to_le32() 描述符地址（上面A行），而不是下面的缓冲区长度（上面B行）？难道HW只看到缓冲区地址而看不到大小？还是书上的错误？对了，这是针对一个虚构的PCI以太网芯片驱动的。

Answer 1

在实践中，这样的方法的确像是一个错误。但理论上，有可能一个字段(address)无论如何都是小恩迪安，而另一种(length)是在原生模式下。如今，在FPGA的帮助下，我想人们甚至可以尝试实现这种理论上的情况。

因此，在给定的背景下，特别是考虑到你说的这是为 "我 "准备的。虚构的PCI以太网芯片虽说作者在这里的用意是什么，但有点难说。