ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 下面的汇编指令对应的C源代码是什么：

```asm
<memmove>:
  endbr64
  mov    %rdi,%rax
  test   %edx,%edx
  jle    L45
  movslq %edx,%rcx
  lea    (%rdi,%rcx,1),%rdi
  cmp    %rsi,%rdi
  jbe    L182
  add    %rsi,%rcx
  cmp    %rax,%rcx
  jbe    L182
  cmp    %rsi,%rax
  jae    L125
  lea    0x1(%rsi),%r8
  mov    %rax,%rcx
  lea    -0x1(%rdx),%edi
  mov    %edx,%r9d
  sub    %r8,%rcx
  cmp    $0xe,%rcx
  jbe    L157
  cmp    $0x6,%edi
  jbe    L157
  cmp    $0xe,%edi
  jbe    L102
  mov    %edx,%r8d
  xor    %ecx,%ecx
  shr    $0x4,%r8d
  shl    $0x4,%r8
  nopl   (%rax)
L106:
  movdqu (%rsi,%rcx,1),%xmm0
  movups %xmm0,(%rax,%rcx,1)
  add    $0x10,%rcx
  cmp    %rcx,%r8
  jne    L106
  mov    %edx,%r11d
  and    $0xfffffff0,%r11d
  mov    %r11d,%r10d
  sub    %r11d,%edi
  lea    (%rax,%r10,1),%rcx
  lea    (%rsi,%r10,1),%r8
  cmp    %r11d,%edx
  je     L45
  sub    %r11d,%edx
  mov    %edx,%r9d
  lea    -0x1(%rdx),%edx
  cmp    $0x6,%edx
  jbe    L30
L101:
  mov    (%rsi,%r10,1),%rdx
  mov    %rdx,(%rax,%r10,1)
  mov    %r9d,%edx
  and    $0xfffffff8,%edx
  mov    %edx,%esi
  sub    %edx,%edi
  add    %rsi,%rcx
  add    %rsi,%r8
  cmp    %edx,%r9d
  je     L45
L30:
  movzbl (%r8),%edx
  mov    %dl,(%rcx)
  test   %edi,%edi
  je     L45
  movzbl 0x1(%r8),%edx
  mov    %dl,0x1(%rcx)
  cmp    $0x1,%edi
  je     L45
  movzbl 0x2(%r8),%edx
  mov    %dl,0x2(%rcx)
  cmp    $0x2,%edi
  je     L45
  movzbl 0x3(%r8),%edx
  mov    %dl,0x3(%rcx)
  cmp    $0x3,%edi
  je     L45
  movzbl 0x4(%r8),%edx
  mov    %dl,0x4(%rcx)
  cmp    $0x4,%edi
  je     L45
  movzbl 0x5(%r8),%edx
  mov    %dl,0x5(%rcx)
  cmp    $0x5,%edi
  je     L45
  movzbl 0x6(%r8),%edx
  mov    %dl,0x6(%rcx)
  ret
  nopl   0x0(%rax,%rax,1)
L125:
  jbe    L45
  sub    $0x1,%edx
  mov    $0xfffffffffffffffe,%r8
  sub    %rdx,%r8
  mov    $0xffffffffffffffff,%rdx
  xchg   %ax,%ax
L82:
  movzbl (%rcx,%rdx,1),%esi
  mov    %sil,(%rdi,%rdx,1)
  sub    $0x1,%rdx
  cmp    %r8,%rdx
  jne    L82
L45:
  ret
  nopw   0x0(%rax,%rax,1)
L182:
  mov    %rax,%rdi
  jmp    1050 <memcpy@plt>
L157:
  mov    %edx,%edx
  xor    %ecx,%ecx
  nopl   0x0(%rax)
L74:
  movzbl (%rsi,%rcx,1),%edi
  mov    %dil,(%rax,%rcx,1)
  add    $0x1,%rcx
  cmp    %rdx,%rcx
  jne    L74
  ret
L102:
  mov    %rsi,%r8
  mov    %rax,%rcx
  xor    %r10d,%r10d
  jmp    L101
```

assistant: ```c
char *memmove(char *dst, char *src, int n) {
  char *realdst = dst;
  if (n <= 0)
    return dst;
  if (src >= dst + n || dst >= src + n)
    return memcpy(dst, src, n);
  if (src > dst) {
    while (--n >= 0)
      *dst++ = *src++;
  } else if (src < dst) {
    src += n;
    dst += n;
    while (--n >= 0)
      *--dst = *--src;
  }
  return realdst;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<divideby>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  jmp    L25
L14:
  mov    -0x4(%rbp),%eax
  cltd
  idivl  -0x8(%rbp)
  mov    %eax,-0x4(%rbp)
L25:
  mov    -0x4(%rbp),%eax
  cltd
  idivl  -0x8(%rbp)
  mov    %edx,%eax
  test   %eax,%eax
  je     L14
  cmpl   $0x1,-0x4(%rbp)
  jne    L39
  mov    $0x1,%eax
  jmp    L2
L39:
  mov    $0x0,%eax
L2:
  pop    %rbp
  ret
```

assistant: ```c
int divideby(int M, int d) {
  while (M % d == 0) {
    M = M / d;
  }
  if (M == 1) {
    return 1;
  }
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<divideby>:
  endbr64
  mov    %edi,%eax
  cltd
  idiv   %esi
  test   %edx,%edx
  jne    L2
L6:
  mov    %edi,%eax
  cltd
  idiv   %esi
  mov    %eax,%edi
  cltd
  idiv   %esi
  test   %edx,%edx
  je     L6
L2:
  cmp    $0x1,%edi
  sete   %al
  movzbl %al,%eax
  ret
```

assistant: ```c
int divideby(int M, int d) {
  while (M % d == 0) {
    M = M / d;
  }
  if (M == 1) {
    return 1;
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<divideby>:
  endbr64
  mov    %edi,%eax
  cltd
  idiv   %esi
  test   %edx,%edx
  jne    L11
  nopl   (%rax)
L1:
  mov    %edi,%eax
  cltd
  idiv   %esi
  cltd
  mov    %eax,%edi
  idiv   %esi
  test   %edx,%edx
  je     L1
L11:
  xor    %eax,%eax
  cmp    $0x1,%edi
  sete   %al
  ret
```

assistant: ```c
int divideby(int M, int d) {
  while (M % d == 0) {
    M = M / d;
  }
  if (M == 1) {
    return 1;
  }
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<divideby>:
  endbr64
  mov    %edi,%eax
  cltd
  idiv   %esi
  test   %edx,%edx
  jne    L12
  nopl   (%rax)
L0:
  mov    %edi,%eax
  cltd
  idiv   %esi
  cltd
  mov    %eax,%edi
  idiv   %esi
  test   %edx,%edx
  je     L0
L12:
  xor    %eax,%eax
  cmp    $0x1,%edi
  sete   %al
  ret
```

assistant: ```c
int divideby(int M, int d) {
  while (M % d == 0) {
    M = M / d;
  }
  if (M == 1) {
    return 1;
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<strv_merge_concat>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  cmpq   $0x0,-0x30(%rbp)
  jne    L115
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <strv_copy@plt>
  jmp    L33
L115:
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <strv_length@plt>
  mov    %rax,%rbx
  mov    -0x30(%rbp),%rax
  mov    %rax,%rdi
  call   <strv_length@plt>
  add    %rbx,%rax
  add    $0x1,%rax
  mov    %rax,%rsi
  mov    $0x8,%edi
  call   <malloc_multiply@plt>
  mov    %rax,-0x18(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  jne    L119
  mov    $0x0,%eax
  jmp    L33
L119:
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x20(%rbp)
  cmpq   $0x0,-0x28(%rbp)
  je     L10
  jmp    L106
L90:
  mov    -0x28(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <strdup@plt>
  mov    %rax,%rdx
  mov    -0x20(%rbp),%rax
  mov    %rdx,(%rax)
  mov    -0x20(%rbp),%rax
  mov    (%rax),%rax
  test   %rax,%rax
  je     L56
  addq   $0x8,-0x20(%rbp)
  addq   $0x8,-0x28(%rbp)
L106:
  mov    -0x28(%rbp),%rax
  mov    (%rax),%rax
  test   %rax,%rax
  jne    L90
  jmp    L10
L24:
  mov    -0x30(%rbp),%rax
  mov    (%rax),%rax
  mov    -0x38(%rbp),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strappend@plt>
  mov    -0x20(%rbp),%rdx
  mov    %rax,(%rdx)
  mov    -0x20(%rbp),%rax
  mov    (%rax),%rax
  test   %rax,%rax
  je     L31
  addq   $0x8,-0x20(%rbp)
  addq   $0x8,-0x30(%rbp)
L10:
  mov    -0x30(%rbp),%rax
  mov    (%rax),%rax
  test   %rax,%rax
  jne    L24
  mov    -0x20(%rbp),%rax
  movq   $0x0,(%rax)
  mov    -0x18(%rbp),%rax
  jmp    L33
L56:
  nop
  jmp    L75
L31:
  nop
L75:
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strv_free@plt>
  mov    $0x0,%eax
L33:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
char **strv_merge_concat(char **a, char **b, const char *suffix) {
  char **r, **k;
  if (!b)
    return strv_copy(a);
  r = ((char **)malloc_multiply(sizeof(char *), (strv_length(a) + strv_length(b) + 1)));
  if (!r)
    return NULL;
  k = r;
  if (a)
    for (; *a; k++, a++) {
      *k = strdup(*a);
      if (!*k)
        goto fail;
    }
  for (; *b; k++, b++) {
    *k = strappend(*b, suffix);
    if (!*k)
      goto fail;
  }
  *k = NULL;
  return r;
fail:
  strv_free(r);
  return NULL;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<strv_merge_concat>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r12
  test   %rsi,%rsi
  je     L23
  mov    %rsi,%rbp
  mov    %rdx,%r13
  call   <strv_length@plt>
  mov    %rax,%rbx
  mov    %rbp,%rdi
  call   <strv_length@plt>
  lea    0x1(%rbx,%rax,1),%rsi
  mov    $0x8,%edi
  call   <malloc_multiply@plt>
  mov    %rax,%r14
  mov    %rax,%r15
  test   %rax,%rax
  je     L45
  test   %r12,%r12
  je     L47
  mov    (%r12),%rdi
  mov    %rax,%rbx
  sub    %rax,%r12
  test   %rdi,%rdi
  je     L98
L56:
  call   <strdup@plt>
  mov    %rax,(%rbx)
  test   %rax,%rax
  je     L38
  add    $0x8,%rbx
  mov    (%rbx,%r12,1),%rdi
  test   %rdi,%rdi
  jne    L56
L98:
  mov    0x0(%rbp),%rdi
  test   %rdi,%rdi
  je     L51
L37:
  mov    %r13,%rsi
  call   <strappend@plt>
  mov    %rax,(%rbx)
  test   %rax,%rax
  je     L38
  add    $0x8,%rbx
  add    $0x8,%rbp
  mov    0x0(%rbp),%rdi
  test   %rdi,%rdi
  jne    L37
L51:
  movq   $0x0,(%rbx)
L73:
  mov    %r15,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L23:
  call   <strv_copy@plt>
  mov    %rax,%r15
  jmp    L73
L47:
  mov    %rax,%rbx
  jmp    L98
L38:
  mov    %r14,%rdi
  call   <strv_free@plt>
  mov    $0x0,%r15d
  jmp    L73
L45:
  mov    $0x0,%r15d
  jmp    L73
```

assistant: ```c
char **strv_merge_concat(char **a, char **b, const char *suffix) {
  char **r, **k;
  if (!b)
    return strv_copy(a);
  r = ((char **)malloc_multiply(sizeof(char *), (strv_length(a) + strv_length(b) + 1)));
  if (!r)
    return NULL;
  k = r;
  if (a)
    for (; *a; k++, a++) {
      *k = strdup(*a);
      if (!*k)
        goto fail;
    }
  for (; *b; k++, b++) {
    *k = strappend(*b, suffix);
    if (!*k)
      goto fail;
  }
  *k = NULL;
  return r;
fail:
  strv_free(r);
  return NULL;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<strv_merge_concat>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  test   %rsi,%rsi
  je     L60
  mov    %rsi,%rbp
  mov    %rdx,%r13
  call   <strv_length@plt>
  mov    %rbp,%rdi
  mov    %rax,%rbx
  call   <strv_length@plt>
  mov    $0x8,%edi
  lea    0x1(%rbx,%rax,1),%rsi
  call   <malloc_multiply@plt>
  mov    %rax,%r14
  test   %rax,%rax
  je     L82
  test   %r12,%r12
  je     L38
  mov    (%r12),%rdi
  mov    %rax,%rbx
  sub    %rax,%r12
  test   %rdi,%rdi
  je     L99
  nopl   0x0(%rax)
L37:
  call   <strdup@plt>
  mov    %rax,(%rbx)
  test   %rax,%rax
  je     L73
  mov    0x8(%rbx,%r12,1),%rdi
  add    $0x8,%rbx
  test   %rdi,%rdi
  jne    L37
  mov    0x0(%rbp),%rdi
  test   %rdi,%rdi
  je     L12
  nopl   0x0(%rax)
L31:
  mov    %r13,%rsi
  call   <strappend@plt>
  mov    %rax,(%rbx)
  test   %rax,%rax
  je     L73
  add    $0x8,%rbx
  add    $0x8,%rbp
L99:
  mov    0x0(%rbp),%rdi
  test   %rdi,%rdi
  jne    L31
L12:
  movq   $0x0,(%rbx)
L82:
  pop    %rbx
  mov    %r14,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
L38:
  mov    %rax,%rbx
  jmp    L99
  nopl   (%rax)
L73:
  mov    %r14,%rdi
  xor    %r14d,%r14d
  call   <strv_free@plt>
  pop    %rbx
  mov    %r14,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nop
L60:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  jmp    10d0 <strv_copy@plt>
```

assistant: ```c
char **strv_merge_concat(char **a, char **b, const char *suffix) {
  char **r, **k;
  if (!b)
    return strv_copy(a);
  r = ((char **)malloc_multiply(sizeof(char *), (strv_length(a) + strv_length(b) + 1)));
  if (!r)
    return NULL;
  k = r;
  if (a)
    for (; *a; k++, a++) {
      *k = strdup(*a);
      if (!*k)
        goto fail;
    }
  for (; *b; k++, b++) {
    *k = strappend(*b, suffix);
    if (!*k)
      goto fail;
  }
  *k = NULL;
  return r;
fail:
  strv_free(r);
  return NULL;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<strv_merge_concat>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  test   %rsi,%rsi
  je     L51
  mov    %rsi,%rbp
  mov    %rdx,%r13
  call   <strv_length@plt>
  mov    %rbp,%rdi
  mov    %rax,%rbx
  call   <strv_length@plt>
  mov    $0x8,%edi
  lea    0x1(%rbx,%rax,1),%rsi
  call   <malloc_multiply@plt>
  mov    %rax,%r14
  test   %rax,%rax
  je     L11
  test   %r12,%r12
  je     L32
  mov    (%r12),%rdi
  mov    %rax,%rbx
  sub    %rax,%r12
  test   %rdi,%rdi
  je     L10
  nopl   0x0(%rax)
L39:
  call   <strdup@plt>
  mov    %rax,(%rbx)
  test   %rax,%rax
  je     L40
  mov    0x8(%rbx,%r12,1),%rdi
  add    $0x8,%rbx
  test   %rdi,%rdi
  jne    L39
  mov    0x0(%rbp),%rdi
  test   %rdi,%rdi
  je     L43
  nopl   0x0(%rax)
L71:
  mov    %r13,%rsi
  call   <strappend@plt>
  mov    %rax,(%rbx)
  test   %rax,%rax
  je     L40
  add    $0x8,%rbx
  add    $0x8,%rbp
L10:
  mov    0x0(%rbp),%rdi
  test   %rdi,%rdi
  jne    L71
L43:
  movq   $0x0,(%rbx)
L11:
  pop    %rbx
  mov    %r14,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
L32:
  mov    %rax,%rbx
  jmp    L10
  nopl   (%rax)
L40:
  mov    %r14,%rdi
  xor    %r14d,%r14d
  call   <strv_free@plt>
  pop    %rbx
  mov    %r14,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nop
L51:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  jmp    10d0 <strv_copy@plt>
```

assistant: ```c
char **strv_merge_concat(char **a, char **b, const char *suffix) {
  char **r, **k;
  if (!b)
    return strv_copy(a);
  r = ((char **)malloc_multiply(sizeof(char *), (strv_length(a) + strv_length(b) + 1)));
  if (!r)
    return NULL;
  k = r;
  if (a)
    for (; *a; k++, a++) {
      *k = strdup(*a);
      if (!*k)
        goto fail;
    }
  for (; *b; k++, b++) {
    *k = strappend(*b, suffix);
    if (!*k)
      goto fail;
  }
  *k = NULL;
  return r;
fail:
  strv_free(r);
  return NULL;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<bitcount>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L18
L14:
  mov    -0x14(%rbp),%eax
  sub    $0x1,%eax
  and    %eax,-0x14(%rbp)
  addl   $0x1,-0x4(%rbp)
L18:
  cmpl   $0x0,-0x14(%rbp)
  jne    L14
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int bitcount(unsigned x) {
  int b = 0;
  while (x > 0) {
    x &= (x - 1);
    b++;
  }
  return b;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<bitcount>:
  endbr64
  test   %edi,%edi
  je     L7
  mov    $0x0,%edx
L9:
  lea    -0x1(%rdi),%eax
  add    $0x1,%edx
  and    %eax,%edi
  jne    L9
L11:
  mov    %edx,%eax
  ret
L7:
  mov    $0x0,%edx
  jmp    L11
```

assistant: ```c
int bitcount(unsigned x) {
  int b = 0;
  while (x > 0) {
    x &= (x - 1);
    b++;
  }
  return b;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<bitcount>:
  endbr64
  xor    %r8d,%r8d
  test   %edi,%edi
  je     L1
  nopl   0x0(%rax,%rax,1)
L14:
  lea    -0x1(%rdi),%eax
  add    $0x1,%r8d
  and    %eax,%edi
  jne    L14
L1:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int bitcount(unsigned x) {
  int b = 0;
  while (x > 0) {
    x &= (x - 1);
    b++;
  }
  return b;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<bitcount>:
  endbr64
  xor    %r8d,%r8d
  test   %edi,%edi
  je     L1
  nopl   0x0(%rax,%rax,1)
L10:
  lea    -0x1(%rdi),%eax
  add    $0x1,%r8d
  and    %eax,%edi
  jne    L10
L1:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int bitcount(unsigned x) {
  int b = 0;
  while (x > 0) {
    x &= (x - 1);
    b++;
  }
  return b;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<Teclado>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,%eax
  mov    %esi,-0x8(%rbp)
  mov    %edx,-0xc(%rbp)
  mov    %al,-0x4(%rbp)
  cmpb   $0x1b,-0x4(%rbp)
  jne    L5
  mov    $0x0,%edi
  call   <exit@plt>
L5:
  nop
  leave
  ret
```

assistant: ```c
void Teclado(unsigned char key, int x, int y) {
  if (key == 27)
    exit(0);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<Teclado>:
  endbr64
  cmp    $0x1b,%dil
  je     L3
  ret
L3:
  sub    $0x8,%rsp
  mov    $0x0,%edi
  call   <exit@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void Teclado(unsigned char key, int x, int y) {
  if (key == 27)
    exit(0);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<Teclado>:
  endbr64
  cmp    $0x1b,%dil
  je     L1
  ret
  nopl   0x0(%rax,%rax,1)
L1:
  xor    %edi,%edi
  jmp    1050 <exit@plt>
```

assistant: ```c
void Teclado(unsigned char key, int x, int y) {
  if (key == 27)
    exit(0);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<Teclado>:
  endbr64
  cmp    $0x1b,%dil
  je     L2
  ret
  nopl   0x0(%rax,%rax,1)
L2:
  xor    %edi,%edi
  jmp    1050 <exit@plt>
```

assistant: ```c
void Teclado(unsigned char key, int x, int y) {
  if (key == 27)
    exit(0);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<csum>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movq   $0x0,-0x8(%rbp)
  jmp    L11
L2:
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  movzwl (%rax),%eax
  movzwl %ax,%eax
  add    %rax,-0x8(%rbp)
  subl   $0x2,-0x1c(%rbp)
L11:
  cmpl   $0x1,-0x1c(%rbp)
  jg     L2
  cmpl   $0x0,-0x1c(%rbp)
  jle    L15
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  add    %rax,-0x8(%rbp)
  jmp    L15
L42:
  mov    -0x8(%rbp),%rax
  movzwl %ax,%edx
  mov    -0x8(%rbp),%rax
  shr    $0x10,%rax
  add    %rdx,%rax
  mov    %rax,-0x8(%rbp)
L15:
  mov    -0x8(%rbp),%rax
  shr    $0x10,%rax
  test   %rax,%rax
  jne    L42
  mov    -0x8(%rbp),%rax
  not    %eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned short csum(unsigned char *addr, int count) {
  unsigned long sum = 0;
  while (count > 1) {
    sum += *(unsigned short *)addr++;
    count -= 2;
  }
  if (count > 0)
    sum += *(unsigned char *)addr;
  while (sum >> 16)
    sum = (sum & 0xffff) + (sum >> 16);
  return ~sum;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<csum>:
  endbr64
  cmp    $0x1,%esi
  jle    L27
  lea    -0x2(%rsi),%r8d
  shr    %r8d
  lea    0x1(%r8),%ecx
  add    %rdi,%rcx
  mov    $0x0,%eax
L50:
  add    $0x1,%rdi
  movzwl -0x1(%rdi),%edx
  add    %rdx,%rax
  cmp    %rcx,%rdi
  jne    L50
  neg    %r8d
  lea    -0x2(%rsi,%r8,2),%edx
  test   %edx,%edx
  jle    L69
L12:
  movzbl (%rcx),%edx
  add    %rdx,%rax
L69:
  mov    %rax,%rdx
  shr    $0x10,%rdx
  je     L67
L43:
  movzwl %ax,%eax
  add    %rdx,%rax
  mov    %rax,%rdx
  shr    $0x10,%rdx
  jne    L43
L67:
  not    %eax
  ret
L27:
  mov    %rdi,%rcx
  mov    $0x0,%eax
  je     L12
  jmp    L67
```

assistant: ```c
unsigned short csum(unsigned char *addr, int count) {
  unsigned long sum = 0;
  while (count > 1) {
    sum += *(unsigned short *)addr++;
    count -= 2;
  }
  if (count > 0)
    sum += *(unsigned char *)addr;
  while (sum >> 16)
    sum = (sum & 0xffff) + (sum >> 16);
  return ~sum;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<csum>:
  endbr64
  cmp    $0x1,%esi
  jle    L5
  sub    $0x2,%esi
  xor    %eax,%eax
  mov    %esi,%r8d
  shr    %r8d
  lea    0x1(%r8),%ecx
  add    %rdi,%rcx
  nopl   0x0(%rax,%rax,1)
L57:
  movzwl (%rdi),%edx
  add    $0x1,%rdi
  add    %rdx,%rax
  cmp    %rdi,%rcx
  jne    L57
  neg    %r8d
  lea    (%rsi,%r8,2),%edx
  cmp    $0x1,%edx
  jne    L30
  jmp    L62
  nopl   (%rax)
L69:
  movzwl %ax,%eax
  add    %rdx,%rax
L30:
  mov    %rax,%rdx
  shr    $0x10,%rdx
  jne    L69
  not    %eax
  ret
L63:
  mov    %rdi,%rcx
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L62:
  movzbl (%rcx),%edx
  add    %rdx,%rax
  jmp    L30
  nopl   0x0(%rax,%rax,1)
L5:
  je     L63
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
unsigned short csum(unsigned char *addr, int count) {
  unsigned long sum = 0;
  while (count > 1) {
    sum += *(unsigned short *)addr++;
    count -= 2;
  }
  if (count > 0)
    sum += *(unsigned char *)addr;
  while (sum >> 16)
    sum = (sum & 0xffff) + (sum >> 16);
  return ~sum;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<csum>:
  endbr64
  cmp    $0x1,%esi
  jle    L39
  sub    $0x2,%esi
  xor    %eax,%eax
  mov    %esi,%r8d
  shr    %r8d
  lea    0x1(%r8),%ecx
  add    %rdi,%rcx
  nopl   0x0(%rax,%rax,1)
L65:
  movzwl (%rdi),%edx
  add    $0x1,%rdi
  add    %rdx,%rax
  cmp    %rcx,%rdi
  jne    L65
  neg    %r8d
  lea    (%rsi,%r8,2),%edx
  cmp    $0x1,%edx
  jne    L10
  jmp    L68
  nopl   (%rax)
L34:
  movzwl %ax,%eax
  add    %rdx,%rax
L10:
  mov    %rax,%rdx
  shr    $0x10,%rdx
  jne    L34
  not    %eax
  ret
L53:
  mov    %rdi,%rcx
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L68:
  movzbl (%rcx),%edx
  add    %rdx,%rax
  jmp    L10
  nopl   0x0(%rax,%rax,1)
L39:
  je     L53
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
unsigned short csum(unsigned char *addr, int count) {
  unsigned long sum = 0;
  while (count > 1) {
    sum += *(unsigned short *)addr++;
    count -= 2;
  }
  if (count > 0)
    sum += *(unsigned char *)addr;
  while (sum >> 16)
    sum = (sum & 0xffff) + (sum >> 16);
  return ~sum;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <test_length@plt>
  mov    $0x0,%eax
  call   <test_get@plt>
  mov    $0x0,%eax
  call   <test_append@plt>
  mov    $0x0,%eax
  call   <test_prepend@plt>
  mov    $0x0,%eax
  call   <test_list@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main(void) {
  test_length();
  test_get();
  test_append();
  test_prepend();
  test_list();
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <test_length@plt>
  mov    $0x0,%eax
  call   <test_get@plt>
  mov    $0x0,%eax
  call   <test_append@plt>
  mov    $0x0,%eax
  call   <test_prepend@plt>
  mov    $0x0,%eax
  call   <test_list@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  test_length();
  test_get();
  test_append();
  test_prepend();
  test_list();
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <test_length@plt>
  xor    %eax,%eax
  call   <test_get@plt>
  xor    %eax,%eax
  call   <test_append@plt>
  xor    %eax,%eax
  call   <test_prepend@plt>
  xor    %eax,%eax
  call   <test_list@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  test_length();
  test_get();
  test_append();
  test_prepend();
  test_list();
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <test_length@plt>
  xor    %eax,%eax
  call   <test_get@plt>
  xor    %eax,%eax
  call   <test_append@plt>
  xor    %eax,%eax
  call   <test_prepend@plt>
  xor    %eax,%eax
  call   <test_list@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  test_length();
  test_get();
  test_append();
  test_prepend();
  test_list();
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<Bits_popCountx64>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  movl   $0x0,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L7
L13:
  mov    -0x4(%rbp),%eax
  mov    -0x14(%rbp),%edx
  mov    %eax,%ecx
  sar    %cl,%edx
  mov    %edx,%eax
  and    $0x1,%eax
  add    %eax,-0x8(%rbp)
  addl   $0x1,-0x4(%rbp)
L7:
  cmpl   $0x3f,-0x4(%rbp)
  jle    L13
  mov    -0x8(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int Bits_popCountx64(uint64_t number) {
  int out = 0;
  for (int i = 0; i < 64; i++) {
    out += ((number >> i) & 1);
  }
  return out;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<Bits_popCountx64>:
  endbr64
  mov    $0x0,%ecx
  mov    $0x0,%edx
L2:
  mov    %edi,%eax
  sar    %cl,%eax
  and    $0x1,%eax
  add    %eax,%edx
  add    $0x1,%ecx
  cmp    $0x40,%ecx
  jne    L2
  mov    %edx,%eax
  ret
```

assistant: ```c
int Bits_popCountx64(uint64_t number) {
  int out = 0;
  for (int i = 0; i < 64; i++) {
    out += ((number >> i) & 1);
  }
  return out;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<Bits_popCountx64>:
  endbr64
  xor    %ecx,%ecx
  xor    %r8d,%r8d
  nopl   0x0(%rax)
L0:
  mov    %edi,%eax
  sar    %cl,%eax
  add    $0x1,%ecx
  and    $0x1,%eax
  add    %eax,%r8d
  cmp    $0x40,%ecx
  jne    L0
  mov    %r8d,%eax
  ret
```

assistant: ```c
int Bits_popCountx64(uint64_t number) {
  int out = 0;
  for (int i = 0; i < 64; i++) {
    out += ((number >> i) & 1);
  }
  return out;
}
```
user: Translate the below assembly code into C programming language:

```asm
<Bits_popCountx64>:
  endbr64
  xor    %ecx,%ecx
  xor    %r8d,%r8d
  nopl   0x0(%rax)
L0:
  mov    %edi,%eax
  sar    %cl,%eax
  add    $0x1,%ecx
  and    $0x1,%eax
  add    %eax,%r8d
  cmp    $0x40,%ecx
  jne    L0
  mov    %r8d,%eax
  ret
```

assistant: ```c
int Bits_popCountx64(uint64_t number) {
  int out = 0;
  for (int i = 0; i < 64; i++) {
    out += ((number >> i) & 1);
  }
  return out;
}
```