ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Demonstrate the C code that generates the assembly output below:

```asm
<wcschr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movq   $0x0,-0x8(%rbp)
L8:
  mov    -0x8(%rbp),%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jne    L25
  mov    -0x8(%rbp),%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  jmp    L37
L25:
  cmpl   $0x0,-0xc(%rbp)
  jne    L38
  mov    $0x0,%eax
  jmp    L37
L38:
  addq   $0x1,-0x8(%rbp)
  jmp    L8
L37:
  pop    %rbp
  ret
```

assistant: ```c
wchar_t *wcschr(const wchar_t *s, wchar_t c) {
  size_t n = 0;
  for (;;) {
    wchar_t wc = s[n];
    if (wc == c)
      return (wchar_t *)s + n;
    if (wc == L'\0')
      return NULL;
    n++;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<wcschr>:
  endbr64
  mov    (%rdi),%eax
  cmp    %eax,%esi
  je     L26
  add    $0x4,%rdi
L21:
  test   %eax,%eax
  je     L12
  mov    %rdi,%rdx
  add    $0x4,%rdi
  mov    -0x4(%rdi),%eax
  cmp    %esi,%eax
  jne    L21
  jmp    L17
L26:
  mov    %rdi,%rdx
  jmp    L17
L12:
  mov    $0x0,%edx
L17:
  mov    %rdx,%rax
  ret
```

assistant: ```c
wchar_t *wcschr(const wchar_t *s, wchar_t c) {
  size_t n = 0;
  for (;;) {
    wchar_t wc = s[n];
    if (wc == c)
      return (wchar_t *)s + n;
    if (wc == L'\0')
      return NULL;
    n++;
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<wcschr>:
  endbr64
  mov    (%rdi),%eax
  cmp    %eax,%esi
  je     L13
  add    $0x4,%rdi
  jmp    L38
L12:
  mov    %rdi,%r8
  mov    (%rdi),%eax
  add    $0x4,%rdi
  cmp    %esi,%eax
  je     L19
L38:
  test   %eax,%eax
  jne    L12
  xor    %r8d,%r8d
L19:
  mov    %r8,%rax
  ret
L13:
  mov    %rdi,%r8
  jmp    L19
```

assistant: ```c
wchar_t *wcschr(const wchar_t *s, wchar_t c) {
  size_t n = 0;
  for (;;) {
    wchar_t wc = s[n];
    if (wc == c)
      return (wchar_t *)s + n;
    if (wc == L'\0')
      return NULL;
    n++;
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<wcschr>:
  endbr64
  mov    (%rdi),%eax
  cmp    %eax,%esi
  je     L25
  add    $0x4,%rdi
  jmp    L8
L5:
  mov    %rdi,%r8
  mov    (%rdi),%eax
  add    $0x4,%rdi
  cmp    %esi,%eax
  je     L19
L8:
  test   %eax,%eax
  jne    L5
  xor    %r8d,%r8d
L19:
  mov    %r8,%rax
  ret
L25:
  mov    %rdi,%r8
  jmp    L19
```

assistant: ```c
wchar_t *wcschr(const wchar_t *s, wchar_t c) {
  size_t n = 0;
  for (;;) {
    wchar_t wc = s[n];
    if (wc == c)
      return (wchar_t *)s + n;
    if (wc == L'\0')
      return NULL;
    n++;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<ppMemF>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  lea    D16(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  movl   $0x0,-0x4(%rbp)
  jmp    L13
L10:
  mov    -0x4(%rbp),%eax
  cltq
  add    $0x3,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movss  (%rax),%xmm0
  pxor   %xmm2,%xmm2
  cvtss2sd %xmm0,%xmm2
  mov    -0x4(%rbp),%eax
  cltq
  add    $0x2,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movss  (%rax),%xmm0
  pxor   %xmm1,%xmm1
  cvtss2sd %xmm0,%xmm1
  mov    -0x4(%rbp),%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movss  (%rax),%xmm0
  cvtss2sd %xmm0,%xmm0
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movss  (%rax),%xmm3
  pxor   %xmm4,%xmm4
  cvtss2sd %xmm3,%xmm4
  movq   %xmm4,%rax
  movapd %xmm2,%xmm3
  movapd %xmm1,%xmm2
  movapd %xmm0,%xmm1
  movq   %rax,%xmm0
  lea    D15(%rip),%rax
  mov    %rax,%rdi
  mov    $0x4,%eax
  call   <printf@plt>
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  pxor   %xmm0,%xmm0
  movss  %xmm0,(%rax)
  mov    -0x4(%rbp),%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  pxor   %xmm0,%xmm0
  movss  %xmm0,(%rax)
  mov    -0x4(%rbp),%eax
  cltq
  add    $0x2,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  pxor   %xmm0,%xmm0
  movss  %xmm0,(%rax)
  mov    -0x4(%rbp),%eax
  cltq
  add    $0x3,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  pxor   %xmm0,%xmm0
  movss  %xmm0,(%rax)
  addl   $0x4,-0x4(%rbp)
L13:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L10
  nop
  nop
  leave
  ret
```

tool_calls: name: unpack_data
arguments: tag: D16
type_infered: string
name: unpack_data
arguments: tag: D15
type_infered: string

tool: D16: MEM1:

tool: D15: %lf, %lf, %lf, %lf


assistant: ```c
void ppMemF(float *mem, int len) {
  int i;
  printf("MEM1:\n");
  for (i = 0; i < len; i = i + 4) {
    printf("%lf, %lf, %lf, %lf\n", mem[i], mem[i + 1], mem[i + 2], mem[i + 3]);
    mem[i] = 0;
    mem[i + 1] = 0;
    mem[i + 2] = 0;
    mem[i + 3] = 0;
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<ppMemF>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    %esi,%ebp
  lea    D12(%rip),%rdi
  call   <puts@plt>
  test   %ebp,%ebp
  jle    L17
  mov    %r12,%rbx
  sub    $0x1,%ebp
  shr    $0x2,%ebp
  mov    %ebp,%ebp
  shl    $0x4,%rbp
  lea    0x10(%r12,%rbp,1),%r12
  lea    D6(%rip),%rbp
L6:
  pxor   %xmm0,%xmm0
  cvtss2sd (%rbx),%xmm0
  pxor   %xmm3,%xmm3
  cvtss2sd 0xc(%rbx),%xmm3
  pxor   %xmm2,%xmm2
  cvtss2sd 0x8(%rbx),%xmm2
  pxor   %xmm1,%xmm1
  cvtss2sd 0x4(%rbx),%xmm1
  mov    %rbp,%rdi
  mov    $0x4,%eax
  call   <printf@plt>
  movl   $0x0,(%rbx)
  movl   $0x0,0x4(%rbx)
  movl   $0x0,0x8(%rbx)
  movl   $0x0,0xc(%rbx)
  add    $0x10,%rbx
  cmp    %r12,%rbx
  jne    L6
L17:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: read_data
arguments: data_label: D12
type: string
name: read_data
arguments: data_label: D6
type: string

tool: D12: MEM1:

tool: D6: %lf, %lf, %lf, %lf


assistant: ```c
void ppMemF(float *mem, int len) {
  int i;
  printf("MEM1:\n");
  for (i = 0; i < len; i = i + 4) {
    printf("%lf, %lf, %lf, %lf\n", mem[i], mem[i + 1], mem[i + 2], mem[i + 3]);
    mem[i] = 0;
    mem[i + 1] = 0;
    mem[i + 2] = 0;
    mem[i + 3] = 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
<ppMemF>:
  endbr64
  push   %r12
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %rdi,%rbx
  lea    D9(%rip),%rdi
  call   <puts@plt>
  test   %ebp,%ebp
  jle    L13
  sub    $0x1,%ebp
  shr    $0x2,%ebp
  shl    $0x4,%rbp
  lea    0x10(%rbx,%rbp,1),%r12
  lea    D2(%rip),%rbp
  nopl   0x0(%rax,%rax,1)
L18:
  pxor   %xmm0,%xmm0
  pxor   %xmm3,%xmm3
  pxor   %xmm2,%xmm2
  mov    %rbp,%rdi
  pxor   %xmm1,%xmm1
  mov    $0x4,%eax
  cvtss2sd (%rbx),%xmm0
  cvtss2sd 0xc(%rbx),%xmm3
  cvtss2sd 0x8(%rbx),%xmm2
  cvtss2sd 0x4(%rbx),%xmm1
  add    $0x10,%rbx
  call   <printf@plt>
  movl   $0x0,-0x10(%rbx)
  movl   $0x0,-0xc(%rbx)
  movl   $0x0,-0x8(%rbx)
  movl   $0x0,-0x4(%rbx)
  cmp    %rbx,%r12
  jne    L18
L13:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: read_data
arguments: data_tag: D9
type: string
name: read_data
arguments: data_tag: D2
type: string

tool: D9: MEM1:

tool: D2: %lf, %lf, %lf, %lf


assistant: ```c
void ppMemF(float *mem, int len) {
  int i;
  printf("MEM1:\n");
  for (i = 0; i < len; i = i + 4) {
    printf("%lf, %lf, %lf, %lf\n", mem[i], mem[i + 1], mem[i + 2], mem[i + 3]);
    mem[i] = 0;
    mem[i + 1] = 0;
    mem[i + 2] = 0;
    mem[i + 3] = 0;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ppMemF>:
  endbr64
  push   %r12
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %rdi,%rbx
  lea    D0(%rip),%rdi
  call   <puts@plt>
  test   %ebp,%ebp
  jle    L10
  sub    $0x1,%ebp
  shr    $0x2,%ebp
  shl    $0x4,%rbp
  lea    0x10(%rbx,%rbp,1),%r12
  lea    D3(%rip),%rbp
  nopl   0x0(%rax,%rax,1)
L16:
  pxor   %xmm0,%xmm0
  pxor   %xmm3,%xmm3
  pxor   %xmm2,%xmm2
  mov    %rbp,%rdi
  pxor   %xmm1,%xmm1
  mov    $0x4,%eax
  cvtss2sd (%rbx),%xmm0
  cvtss2sd 0xc(%rbx),%xmm3
  cvtss2sd 0x8(%rbx),%xmm2
  cvtss2sd 0x4(%rbx),%xmm1
  call   <printf@plt>
  pxor   %xmm4,%xmm4
  movups %xmm4,(%rbx)
  add    $0x10,%rbx
  cmp    %r12,%rbx
  jne    L16
L10:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D0
type: string
name: unpack_data
arguments: data_label: D3
type: string

tool: D0: MEM1:

tool: D3: %lf, %lf, %lf, %lf


assistant: ```c
void ppMemF(float *mem, int len) {
  int i;
  printf("MEM1:\n");
  for (i = 0; i < len; i = i + 4) {
    printf("%lf, %lf, %lf, %lf\n", mem[i], mem[i + 1], mem[i + 2], mem[i + 3]);
    mem[i] = 0;
    mem[i + 1] = 0;
    mem[i + 2] = 0;
    mem[i + 3] = 0;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<parse_host>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <gethostbyname@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L6
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <usage@plt>
L6:
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rax
  leave
  ret
```

tool_calls: name: read_data
arguments: label: D5
type_infered: string

tool: D5: Unknown host address

assistant: ```c
char *parse_host(const char *str) {
  struct hostent *host = gethostbyname(str);
  if (!host)
    usage("Unknown host address");
  return host->h_addr;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<parse_host>:
  endbr64
  push   %rbx
  call   <gethostbyname@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L0
L15:
  mov    (%rbx),%rax
  pop    %rbx
  ret
L0:
  lea    D3(%rip),%rdi
  call   <usage@plt>
  jmp    L15
```

tool_calls: name: retrieve_data
arguments: data_label: D3
datatype: string

tool: D3: Unknown host address

assistant: ```c
char *parse_host(const char *str) {
  struct hostent *host = gethostbyname(str);
  if (!host)
    usage("Unknown host address");
  return host->h_addr;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<parse_host>:
  endbr64
  push   %rbx
  call   <gethostbyname@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L8
  mov    (%rbx),%rax
  pop    %rbx
  ret
  nopw   0x0(%rax,%rax,1)
L8:
  lea    D6(%rip),%rdi
  call   <usage@plt>
  mov    (%rbx),%rax
  pop    %rbx
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D6
data_type: string

tool: D6: Unknown host address

assistant: ```c
char *parse_host(const char *str) {
  struct hostent *host = gethostbyname(str);
  if (!host)
    usage("Unknown host address");
  return host->h_addr;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<parse_host>:
  endbr64
  push   %rbx
  call   <gethostbyname@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L6
  mov    (%rbx),%rax
  pop    %rbx
  ret
  nopw   0x0(%rax,%rax,1)
L6:
  lea    D8(%rip),%rdi
  call   <usage@plt>
  mov    (%rbx),%rax
  pop    %rbx
  ret
```

tool_calls: name: read_data
arguments: data_tag: D8
datatype: string

tool: D8: Unknown host address

assistant: ```c
char *parse_host(const char *str) {
  struct hostent *host = gethostbyname(str);
  if (!host)
    usage("Unknown host address");
  return host->h_addr;
}
```
user: What is the c source code of the assembly code below:

```asm
<get_hbm_index>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x10(%rbp)
  movl   $0x4e20,-0xc(%rbp)
  jmp    L4
L42:
  mov    -0x10(%rbp),%eax
  cltq
  mov    -0xc(%rbp,%rax,4),%eax
  cmp    %eax,-0x14(%rbp)
  jl     L8
  addl   $0x1,-0x10(%rbp)
L4:
  lea    -0xc(%rbp),%rax
  mov    %rax,%rdi
  call   <ARRAY_SIZE@plt>
  cmp    %eax,-0x10(%rbp)
  jl     L42
  jmp    L35
L8:
  nop
L35:
  mov    -0x10(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L31
  call   <__stack_chk_fail@plt>
L31:
  leave
  ret
```

assistant: ```c
int get_hbm_index(int idx) {
  int i = 0;
  int idx_list[] = {20000};
  while (i < ARRAY_SIZE(idx_list)) {
    if (idx < idx_list[i])
      break;
    i++;
  }
  return i;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<get_hbm_index>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %edi,%ebx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0x4e20,0x4(%rsp)
  lea    0x4(%rsp),%rdi
  call   <ARRAY_SIZE@plt>
  test   %eax,%eax
  jle    L1
  cmp    0x4(%rsp),%ebx
  jl     L16
  lea    0x4(%rsp),%rdi
  call   <ARRAY_SIZE@plt>
  mov    $0x1,%eax
L29:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L43
  add    $0x10,%rsp
  pop    %rbx
  ret
L1:
  mov    $0x0,%eax
  jmp    L29
L16:
  mov    $0x0,%eax
  jmp    L29
L43:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int get_hbm_index(int idx) {
  int i = 0;
  int idx_list[] = {20000};
  while (i < ARRAY_SIZE(idx_list)) {
    if (idx < idx_list[i])
      break;
    i++;
  }
  return i;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<get_hbm_index>:
  endbr64
  push   %rbp
  push   %rbx
  mov    %edi,%ebx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rbp
  movl   $0x4e20,0x4(%rsp)
  mov    %rbp,%rdi
  call   <ARRAY_SIZE@plt>
  test   %eax,%eax
  jle    L11
  cmp    0x4(%rsp),%ebx
  jl     L11
  mov    %rbp,%rdi
  call   <ARRAY_SIZE@plt>
  mov    $0x1,%eax
L33:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L2
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nop
L11:
  xor    %eax,%eax
  jmp    L33
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int get_hbm_index(int idx) {
  int i = 0;
  int idx_list[] = {20000};
  while (i < ARRAY_SIZE(idx_list)) {
    if (idx < idx_list[i])
      break;
    i++;
  }
  return i;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<get_hbm_index>:
  endbr64
  push   %rbp
  push   %rbx
  mov    %edi,%ebx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rbp
  movl   $0x4e20,0x4(%rsp)
  mov    %rbp,%rdi
  call   <ARRAY_SIZE@plt>
  test   %eax,%eax
  jle    L9
  cmp    0x4(%rsp),%ebx
  jl     L9
  mov    %rbp,%rdi
  call   <ARRAY_SIZE@plt>
  mov    $0x1,%eax
L27:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L25
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nop
L9:
  xor    %eax,%eax
  jmp    L27
L25:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int get_hbm_index(int idx) {
  int i = 0;
  int idx_list[] = {20000};
  while (i < ARRAY_SIZE(idx_list)) {
    if (idx < idx_list[i])
      break;
    i++;
  }
  return i;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<fmt_ulong>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r12
  push   %rbx
  mov    %rdi,%rcx
  mov    $0x1,%ebx
  mov    %rsi,%r12
  jmp    L25
L15:
  add    $0x1,%ebx
  movabs $0xcccccccccccccccd,%rdx
  mov    %r12,%rax
  mul    %rdx
  mov    %rdx,%rax
  shr    $0x3,%rax
  mov    %rax,%r12
L25:
  cmp    $0x9,%r12
  ja     L15
  test   %rcx,%rcx
  je     L7
  mov    %ebx,%eax
  add    %rax,%rcx
L34:
  movabs $0xcccccccccccccccd,%rdx
  mov    %rsi,%rax
  mul    %rdx
  shr    $0x3,%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  add    %rax,%rax
  mov    %rsi,%rdx
  sub    %rax,%rdx
  mov    %edx,%eax
  add    $0x30,%eax
  sub    $0x1,%rcx
  mov    %al,(%rcx)
  movabs $0xcccccccccccccccd,%rdx
  mov    %rsi,%rax
  mul    %rdx
  mov    %rdx,%rax
  shr    $0x3,%rax
  mov    %rax,%rsi
  test   %rsi,%rsi
  jne    L34
L7:
  mov    %ebx,%eax
  pop    %rbx
  pop    %r12
  pop    %rbp
  ret
```

assistant: ```c
unsigned int fmt_ulong(s, u)
register char *s;
register unsigned long u;
{
  register unsigned int len;
  register unsigned long q;
  len = 1;
  q = u;
  while (q > 9) {
    ++len;
    q /= 10;
  }
  if (s) {
    s += len;
    do {
      *--s = '0' + (u % 10);
      u /= 10;
    } while (u);
  }
  return len;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<fmt_ulong>:
  endbr64
  mov    %rsi,%rcx
  cmp    $0x9,%rsi
  jbe    L21
  mov    %rsi,%rdx
  mov    $0x1,%r8d
  movabs $0xcccccccccccccccd,%r9
L35:
  add    $0x1,%r8d
  mov    %rdx,%rsi
  mov    %rdx,%rax
  mul    %r9
  shr    $0x3,%rdx
  cmp    $0x63,%rsi
  ja     L35
L28:
  test   %rdi,%rdi
  je     L1
  mov    %r8d,%eax
  add    %rax,%rdi
  movabs $0xcccccccccccccccd,%r9
L27:
  sub    $0x1,%rdi
  mov    %rcx,%rax
  mul    %r9
  shr    $0x3,%rdx
  lea    (%rdx,%rdx,4),%rsi
  add    %rsi,%rsi
  mov    %rcx,%rax
  sub    %rsi,%rax
  add    $0x30,%eax
  mov    %al,(%rdi)
  mov    %rcx,%rax
  mov    %rdx,%rcx
  cmp    $0x9,%rax
  ja     L27
L1:
  mov    %r8d,%eax
  ret
L21:
  mov    $0x1,%r8d
  jmp    L28
```

assistant: ```c
unsigned int fmt_ulong(s, u)
register char *s;
register unsigned long u;
{
  register unsigned int len;
  register unsigned long q;
  len = 1;
  q = u;
  while (q > 9) {
    ++len;
    q /= 10;
  }
  if (s) {
    s += len;
    do {
      *--s = '0' + (u % 10);
      u /= 10;
    } while (u);
  }
  return len;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<fmt_ulong>:
  endbr64
  mov    %rsi,%rcx
  cmp    $0x9,%rsi
  jbe    L2
  mov    %rsi,%rdx
  mov    $0x1,%r8d
  movabs $0xcccccccccccccccd,%r9
L9:
  mov    %rdx,%rax
  mov    %rdx,%rsi
  add    $0x1,%r8d
  mul    %r9
  shr    $0x3,%rdx
  cmp    $0x63,%rsi
  ja     L9
L39:
  test   %rdi,%rdi
  je     L28
  movabs $0xcccccccccccccccd,%r9
  mov    %r8d,%eax
  add    %rax,%rdi
  nopl   0x0(%rax)
L17:
  mov    %rcx,%rax
  sub    $0x1,%rdi
  mul    %r9
  mov    %rcx,%rax
  shr    $0x3,%rdx
  lea    (%rdx,%rdx,4),%rsi
  add    %rsi,%rsi
  sub    %rsi,%rax
  add    $0x30,%eax
  mov    %al,(%rdi)
  mov    %rcx,%rax
  mov    %rdx,%rcx
  cmp    $0x9,%rax
  ja     L17
L28:
  mov    %r8d,%eax
  ret
L2:
  mov    $0x1,%r8d
  jmp    L39
```

assistant: ```c
unsigned int fmt_ulong(s, u)
register char *s;
register unsigned long u;
{
  register unsigned int len;
  register unsigned long q;
  len = 1;
  q = u;
  while (q > 9) {
    ++len;
    q /= 10;
  }
  if (s) {
    s += len;
    do {
      *--s = '0' + (u % 10);
      u /= 10;
    } while (u);
  }
  return len;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<fmt_ulong>:
  endbr64
  mov    %rsi,%rcx
  cmp    $0x9,%rsi
  jbe    L4
  mov    %rsi,%rdx
  mov    $0x1,%r8d
  movabs $0xcccccccccccccccd,%r9
L25:
  mov    %rdx,%rax
  mov    %rdx,%rsi
  add    $0x1,%r8d
  mul    %r9
  shr    $0x3,%rdx
  cmp    $0x63,%rsi
  ja     L25
L45:
  test   %rdi,%rdi
  je     L7
  movabs $0xcccccccccccccccd,%r9
  mov    %r8d,%eax
  add    %rax,%rdi
  nopl   0x0(%rax)
L10:
  mov    %rcx,%rax
  sub    $0x1,%rdi
  mul    %r9
  mov    %rcx,%rax
  shr    $0x3,%rdx
  lea    (%rdx,%rdx,4),%rsi
  add    %rsi,%rsi
  sub    %rsi,%rax
  add    $0x30,%eax
  mov    %al,(%rdi)
  mov    %rcx,%rax
  mov    %rdx,%rcx
  cmp    $0x9,%rax
  ja     L10
L7:
  mov    %r8d,%eax
  ret
L4:
  mov    $0x1,%r8d
  jmp    L45
```

assistant: ```c
unsigned int fmt_ulong(s, u)
register char *s;
register unsigned long u;
{
  register unsigned int len;
  register unsigned long q;
  len = 1;
  q = u;
  while (q > 9) {
    ++len;
    q /= 10;
  }
  if (s) {
    s += len;
    do {
      *--s = '0' + (u % 10);
      u /= 10;
    } while (u);
  }
  return len;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movl   $0x0,-0x4(%rbp)
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main(void) {
  loff_t offset = 0;
  (void)offset;
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  mov    $0x0,%eax
  ret
```

assistant: ```c
int main(void) {
  loff_t offset = 0;
  (void)offset;
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int main(void) {
  loff_t offset = 0;
  (void)offset;
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int main(void) {
  loff_t offset = 0;
  (void)offset;
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<get_next_operand>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    %edx,%ebx
  movslq %edx,%rdx
  movzbl (%rsi,%rdx,1),%eax
  cmp    $0x2c,%al
  je     L83
  mov    %rsi,%rbp
  cmp    $0xa,%al
  je     L21
  mov    D5(%rip),%rcx
  cmp    %al,(%rcx)
  je     L21
  test   %al,%al
  je     L21
  mov    %rdx,%rcx
  mov    $0x0,%esi
  cmp    $0x20,%al
  je     L23
  mov    D5(%rip),%r9
  mov    %rdi,%r10
  sub    %rdx,%r10
  movabs $0x100100000400,%r11
  jmp    L65
L83:
  movb   $0x2c,(%rdi)
  movb   $0x0,0x1(%rdi)
  jmp    L1
L23:
  mov    $0x0,%esi
  jmp    L50
L21:
  movb   $0x0,(%r12)
L1:
  mov    %ebx,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L61:
  add    $0x1,%rcx
L65:
  mov    %ecx,%r8d
  mov    %ecx,%ebx
  cmp    %al,(%r9)
  setne  %dil
  test   %al,%al
  setne  %dl
  test   %dl,%dil
  je     L50
  cmp    $0x9,%al
  je     L50
  mov    %al,(%r10,%rcx,1)
  add    $0x1,%esi
  lea    0x1(%r8),%ebx
  movzbl 0x1(%rbp,%rcx,1),%eax
  cmp    $0x2c,%al
  ja     L61
  add    $0x1,%rcx
  bt     %rax,%r11
  jae    L65
L50:
  movslq %esi,%rsi
  movb   $0x0,(%r12,%rsi,1)
  mov    %r12,%rdi
  call   <is_current_word_empty@plt>
  test   %rax,%rax
  je     L1
  lea    0x1(%rbx),%edx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <get_next_operand>
  mov    %eax,%ebx
  jmp    L1
```

assistant: ```c
int get_next_operand(char *current_word, char *line, int last_position) {
  int position = last_position;
  int i = 0;
  if (line[position] == ',') {
    current_word[0] = ',';
    current_word[1] = '\0';
    return position;
  }
  if (line[position] == '\n' || line[position] == EOF || line[position] == '\0') {
    current_word[i] = '\0';
    return position;
  }
  while (line[position] != ',' && line[position] != '\n' && line[position] != ' ' && line[position] != EOF && line[position] != '\0' && line[position] != '\t') {
    current_word[i] = line[position];
    i++;
    position++;
  }
  current_word[i] = '\0';
  if (is_current_word_empty(current_word))
    position = get_next_operand(current_word, line, position + 1);
  return position;
}
```