sed：match＆＃39; var = var＆＃39;并删除

Question

我想使用sed从包含模式myvar = myvar的文件中删除行。因此，例如，我编辑的汇编文件包含如下行：

wmemcopy = wmemcopy

我尝试了sed命令：

sed '/^\([a-z]*\) = \1$/d'

但是该命令无法找到并删除这些行。是否有用于完成此操作的sed命令？

编辑：

如果我使用sed -i '/^\([a-z]*\) = \1$/d' wtent.s尝试命令，则不会删除这些行。这是我尝试此命令的文件：

    .file   1 "wtent.c"
    .section .mdebug.abi32
    .previous
    .gnu_attribute 4, 1
    .text
    .align  2
    .globl  updwtmp
    .hidden updwtmp
    .set    nomips16
    .ent    updwtmp
    .type   updwtmp, @function
updwtmp:
    .frame  $sp,32,$31      # vars= 0, regs= 3/0, args= 16, gp= 0
    .mask   0x80030000,-4
    .fmask  0x00000000,0
    addiu   $sp,$sp,-32
    move    $6,$0
    sw  $17,24($sp)
    sw  $31,28($sp)
    move    $17,$5
    sw  $16,20($sp)
    li  $5,9            # 0x9
#APP
 # 40 "libc/misc/utmp/wtent.c" 1
    .set    noreorder
    li  $2, 4005            # open
    syscall
    .set    reorder
 # 0 "" 2
#NO_APP
    move    $16,$2
    beq $7,$0,$L2
    sw  $2,errno
    j   $L1
$L2:
    bltz    $2,$L1
    move    $4,$2
    li  $5,1            # 0x1
    jal lockf
    bne $2,$0,$L1
    move    $4,$16
    move    $5,$17
    li  $6,384          # 0x180
#APP
 # 43 "libc/misc/utmp/wtent.c" 1
    .set    noreorder
    li  $2, 4004            # write
    syscall
    .set    reorder
 # 0 "" 2
#NO_APP
    beq $7,$0,$L6
    sw  $2,errno
$L6:
    move    $4,$16
    move    $5,$0
    move    $6,$0
    jal lockf
    move    $4,$16
#APP
 # 45 "libc/misc/utmp/wtent.c" 1
    .set    noreorder
    li  $2, 4006            # close
    syscall
    .set reorder
 # 0 "" 2
#NO_APP
$L1:
    lw  $31,28($sp)
    lw  $17,24($sp)
    lw  $16,20($sp)
    addiu   $sp,$sp,32
    j   $31
    .end    updwtmp
    .size   updwtmp, .-updwtmp
    .globl  updwtmp
    updwtmp = updwtmp
    .ident  "GCC: (GNU) 4.7.0"