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是将内核放在 0x0000_0000 - 0x1000_0000 可以解决这个问题,因为当内存小的时候,只有这个范围上会有可用的内存,如果在一个地址上没有物理内存,那么指定的程序就没有办法被加载到这里,也就没办法找到任务可用指令,程序运行的时候会触发异常 FetchException,然后跳转到异常处理的程序中,但是因为系统没有初始化,异常处理地址为 0,所以会跳转到 0,地址 0 处没有内容。

而在 loongarch64 上没有 sbi 的存在,所以就没有办法输出 sbi 的图标,rustsbi 无法在 loongarch64 上使用,所以指定 sbi 会导致异常,尝试使用qemu-system-loongarch64 -kernel target/loongarch64-unknown-none/release/os -nographic -drive file=../user/target/loongarch64-unknown-none/release/fs.img,if=none,format=raw,id=x0 -m 1G -smp 1 -D qemu.log -d in_asm,int,pcall,cpu_reset,guest_errors

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