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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Os Dev的PCI IDE教程中的函数insl是做什么的?
Run Code Online (Sandbox Code Playgroud)void ide_read_buffer(... unsigned int buffer ...)
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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