只需一步,快速开始
root@(none):/# cat /proc/mtd 5 {; S: o$ R" o6 d# T3 |dev: size erasesize name + k! ]3 o, U. V7 I! }2 V# B) G4 j& `mtd0: 00020000 00020000 "boot" . {# \8 c( k: U! ]9 n: emtd1: 00100000 00020000 "enva" * F; t' C% M) {3 zmtd2: 00100000 00020000 "envb"3 S; d5 e/ j9 C% e! W mtd3: 00400000 00020000 "KernelA"/ }- S A. ]' k) @( B mtd4: 01900000 00020000 "AppA"& K& S2 V9 x/ }7 [% z T mtd5: 02800000 00020000 "RootfsA"8 X8 a# m. g. }/ |. x7 W mtd6: 00400000 00020000 "KernelB"; B1 l# i* i( ]6 s, J! j2 } mtd7: 01900000 00020000 "AppB"" f& t0 z3 F1 U( m" { F mtd8: 02800000 00020000 "RootfsB"! L6 O) S9 {7 ^2 L" r mtd9: 00200000 00020000 "ConfigurationA"( _3 t. {: m9 ?1 o( C# T& q- P mtd10: 00a00000 00020000 "ConfigurationB"& M- p9 F' o' ~) D1 @ mtd11: 00400000 00020000 "UserLocalCT" - F# A, e( V8 k$ K2 xmtd12: 00e00000 00020000 "Userdata" % B2 A0 Z# e( Jmtd13: 00800000 00020000 "Framework1") t' ~1 p) R9 u# V# S mtd14: 00800000 00020000 "Framework2"* k9 W* R) o0 M' K mtd15: 04200000 00020000 "Apps" 8 W9 C- g9 Z! n3 `6 vmtd16: 003e0000 00020000 "other" 1 Y; f. M! b. k# r9 v. _mtd17: 02416000 0001f000 "rootfs"
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tm5880891 发表于 2023-12-8 13:55# h f2 Z- ]- l# q 我滴可以,发来我弄
pensz 发表于 2023-12-8 15:41/ F2 t' k, u) V& _' x& s 方便说下怎么弄嘛,主要想学习下相关技术。谢谢
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