Os Dev的PCI IDE教程中的函数insl是做什么的?

mik*_*ike 2 c x86 assembly osdev ata

这是调用 insl 的函数。

void ide_read_buffer(unsigned char channel, unsigned char reg, unsigned int buffer,
                     unsigned int quads) {
   /* WARNING: This code contains a serious bug. The inline assembly trashes ES and
    *           ESP for all of the code the compiler generates between the inline
    *           assembly blocks.
    */
   if (reg > 0x07 && reg < 0x0C)
      ide_write(channel, ATA_REG_CONTROL, 0x80 | channels[channel].nIEN);
   asm("pushw %es; movw %ds, %ax; movw %ax, %es");
   if (reg < 0x08)
      insl(channels[channel].base  + reg - 0x00, buffer, quads);
   else if (reg < 0x0C)
      insl(channels[channel].base  + reg - 0x06, buffer, quads);
   else if (reg < 0x0E)
      insl(channels[channel].ctrl  + reg - 0x0A, buffer, quads);
   else if (reg < 0x16)
      insl(channels[channel].bmide + reg - 0x0E, buffer, quads);
   asm("popw %es;");
   if (reg > 0x07 && reg < 0x0C)
      ide_write(channel, ATA_REG_CONTROL, channels[channel].nIEN);
}
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链接在这里,https://wiki.osdev.org/PCI_IDE_Controller#Commands

Mar*_*nau 6

Os Dev的PCI IDE教程中的函数insl是做什么的?

void ide_read_buffer(... unsigned int buffer ...)
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很明显为什么你不理解代码:

unsigned int buffer显然是教程代码中的一个错误。应该是unsigned int * buffer

现在很清楚了insl:它quad从第一个参数中给出的端口读取时间并将结果写入数组buffer

该函数的黑盒行为可以用以下方式解释:

void insl(unsigned reg, unsigned int *buffer, int quads)
{
    int index;
    for(index = 0; index < quads; index++)
    {
        buffer[index] = inl(reg);
    }
}
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