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MTD设备的逻辑擦除块大小可以增加吗?

[英]Can the logical erase block size of an MTD device be increased?

The minimum erase block size for jffs2 (mtd-utils version 1.5.0, mkfs.jffs2 revision 1.60) seems to be 8KiB: jffs2(mtd-utils版本1.5.0, mkfs.jffs2修订版1.60)的最小擦除块大小似乎是8KiB:

Erase size 0x1000 too small. Increasing to 8KiB minimum

However I am running Linux 3.10 with an at25df321a, 但是我正在使用at25df321a运行Linux 3.10,

m25p80 spi32766.0: at25df321a (4096 Kbytes),

and the erase block size is only 4KiB: 并且擦除块大小仅为4KiB:

mtd5
Name:                           spi32766.0
Type:                           nor
Eraseblock size:                4096 bytes, 4.0 KiB
Amount of eraseblocks:          1024 (4194304 bytes, 4.0 MiB)
Minimum input/output unit size: 1 byte
Sub-page size:                  1 byte
Character device major/minor:   90:10
Bad blocks are allowed:         false
Device is writable:             true

Is there a way to make the mtd system treat multiple erase blocks as one? 有没有办法使mtd系统将多个擦除块视为一个? Maybe some ioctl or module parameter? 也许一些ioctl或模块参数?

If I flash a jffs2 image with larger erase block size, I get lots of kernel error messages, missing files and sometimes panic. 如果以较大的擦除块大小刷新jffs2图像,则会收到许多内核错误消息,文件丢失甚至有时出现紧急情况。

workaround 解决方法

I found that flasherase --jffs2 results in a working filesystem inspite of the 4KiB erase block size. 我发现flasherase --jffs2 4KiB擦除块大小, flasherase --jffs2使文件系统正常工作。 So I hacked the mkfs.jfss2.c file and the resulting image seems to work fine. 因此,我入侵了mkfs.jfss2.c文件,生成的图像似乎工作正常。 I'll give it some testing. 我会做一些测试。

diff -rupN orig/mkfs.jffs2.c new/mkfs.jffs2.c
--- orig/mkfs.jffs2.c   2014-10-20 15:43:31.751696500 +0200
+++ new/mkfs.jffs2.c    2014-10-20 15:43:12.623431400 +0200
@@ -1659,11 +1659,11 @@ int main(int argc, char **argv)
                                                  }
                                                  erase_block_size *= units;

-                                                 /* If it's less than 8KiB, they're not allowed */
-                                                 if (erase_block_size < 0x2000) {
-                                                         fprintf(stderr, "Erase size 0x%x too small. Increasing to 8KiB minimum\n",
+                                                 /* If it's less than 4KiB, they're not allowed */
+                                                 if (erase_block_size < 0x1000) {
+                                                         fprintf(stderr, "Erase size 0x%x too small. Increasing to 4KiB minimum\n",
                                                                          erase_block_size);
-                                                         erase_block_size = 0x2000;
+                                                         erase_block_size = 0x1000;
                                                  }
                                                  break;
                                          }

http://lists.infradead.org/pipermail/linux-mtd/2010-September/031876.html http://lists.infradead.org/pipermail/linux-mtd/2010-September/031876.html

JFFS2 should be able to fit at least one node to eraseblock. JFFS2应该能够容纳至少一个节点以擦除块。 The maximum node size is 4KiB+few bytes. 最大节点大小为4KiB +几个字节。 This is why the minimum eraseblocks size is 8KiB. 这就是为什么最小擦除块大小为8KiB的原因。

But in practice, even 8KiB is bad because you and up with wasting a lot of space at the end of eraseblocks. 但是实际上,即使8KiB也很糟糕,因为您和其他人在擦除块末尾浪费了大量空间。

You should join several erasblock into one virtual eraseblock of 64 or 128 KiB and use it - this will be more optimal. 您应该将多个erasblock合并为一个64或128 KiB的虚拟擦除块,然后使用它—这样做会更好。

Some drivers have already implemented this. 一些驱动程序已经实现了这一点。 I know about 我知道

MTD_SPI_NOR_USE_4K_SECTORS

Linux configuration option. Linux配置选项。 It have to be set to "n" to enable large erase sectors of size 0x00010000. 必须将其设置为“ n”以启用大小为0x00010000的较大擦除扇区。

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