ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<bitvector_shr_to>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %ecx,-0x2c(%rbp)
  cmpq   $0x0,-0x28(%rbp)
  je     L27
  cmpl   $0x0,-0x2c(%rbp)
  jne    L34
  mov    -0x28(%rbp),%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  jmp    L43
L34:
  mov    -0x2c(%rbp),%eax
  sar    $0x5,%eax
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  je     L2
  mov    -0x4(%rbp),%eax
  cltq
  sub    %rax,-0x28(%rbp)
  mov    -0x4(%rbp),%eax
  lea    0x0(,%rax,4),%edx
  mov    -0x28(%rbp),%rax
  lea    0x0(,%rax,4),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  andl   $0x1f,-0x2c(%rbp)
  mov    -0x4(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rax,-0x20(%rbp)
L2:
  movl   $0x0,-0x8(%rbp)
  jmp    L4
L35:
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x2c(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%ecx
  sar    %cl,%esi
  mov    -0x8(%rbp),%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    $0x20,%eax
  sub    -0x2c(%rbp),%eax
  mov    %eax,%ecx
  shl    %cl,%edx
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  or     %esi,%edx
  mov    %edx,(%rax)
  addl   $0x1,-0x8(%rbp)
L4:
  mov    -0x8(%rbp),%eax
  cltq
  mov    -0x28(%rbp),%rdx
  sub    $0x1,%rdx
  cmp    %rdx,%rax
  jb     L35
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%esi
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x2c(%rbp),%eax
  mov    %eax,%ecx
  sar    %cl,%esi
  mov    %esi,%eax
  mov    %eax,(%rdx)
  jmp    L43
L27:
  nop
L43:
  leave
  ret
```

assistant: ```c
void bitvector_shr_to(u_int32_t *dest, u_int32_t *b, size_t n, u_int32_t s) {
  u_int32_t i, j;
  if (n == 0)
    return;
  if (s == 0) {
    memcpy(dest, b, n * 4);
    return;
  }
  j = (s >> 5);
  if (j) {
    n -= j;
    memset(&dest[n], 0, j * 4);
    s &= 31;
    b = &b[j];
  }
  for (i = 0; i < n - 1; i++) {
    dest[i] = (b[i] >> s) | (b[i + 1] << (32 - s));
  }
  dest[i] = b[i] >> s;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<bitvector_shr_to>:
  endbr64
  test   %rdx,%rdx
  je     L60
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    %rsi,%rbx
  mov    %rdx,%r13
  mov    %ecx,%ebp
  test   %ecx,%ecx
  je     L48
  mov    %ecx,%edx
  sar    $0x5,%edx
  jne    L25
L11:
  mov    %r13,%r9
  sub    $0x1,%r9
  je     L30
  lea    -0x4(,%r13,4),%r8
  mov    $0x0,%eax
  mov    $0x20,%edi
  sub    %ebp,%edi
L8:
  mov    0x4(%rbx,%rax,1),%edx
  mov    %edi,%ecx
  shl    %cl,%edx
  mov    (%rbx,%rax,1),%esi
  mov    %ebp,%ecx
  sar    %cl,%esi
  or     %esi,%edx
  mov    %edx,(%r12,%rax,1)
  add    $0x4,%rax
  cmp    %r8,%rax
  jne    L8
L30:
  mov    (%rbx,%r9,4),%eax
  mov    %ebp,%ecx
  sar    %cl,%eax
  mov    %eax,(%r12,%r9,4)
L7:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L48:
  lea    0x0(,%rdx,4),%rdx
  call   <memcpy@plt>
  jmp    L7
L25:
  movslq %edx,%r14
  sub    %r14,%r13
  shl    $0x2,%edx
  lea    (%rdi,%r13,4),%rdi
  mov    $0x0,%esi
  call   <memset@plt>
  and    $0x1f,%ebp
  lea    (%rbx,%r14,4),%rbx
  jmp    L11
L60:
  ret
```

assistant: ```c
void bitvector_shr_to(u_int32_t *dest, u_int32_t *b, size_t n, u_int32_t s) {
  u_int32_t i, j;
  if (n == 0)
    return;
  if (s == 0) {
    memcpy(dest, b, n * 4);
    return;
  }
  j = (s >> 5);
  if (j) {
    n -= j;
    memset(&dest[n], 0, j * 4);
    s &= 31;
    b = &b[j];
  }
  for (i = 0; i < n - 1; i++) {
    dest[i] = (b[i] >> s) | (b[i + 1] << (32 - s));
  }
  dest[i] = b[i] >> s;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<bitvector_shr_to>:
  endbr64
  test   %rdx,%rdx
  je     L11
  push   %r14
  push   %r13
  mov    %rdx,%r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %ecx,%ebx
  test   %ecx,%ecx
  je     L30
  mov    %ecx,%edx
  sar    $0x5,%edx
  jne    L59
  mov    %r13,%r9
  sub    $0x1,%r9
  je     L13
L39:
  mov    $0x20,%edi
  lea    -0x4(,%r13,4),%r8
  xor    %eax,%eax
  sub    %ebx,%edi
  nopl   0x0(%rax)
L37:
  mov    0x0(%rbp,%rax,1),%edx
  mov    0x4(%rbp,%rax,1),%esi
  mov    %ebx,%ecx
  sar    %cl,%edx
  mov    %edi,%ecx
  shl    %cl,%esi
  or     %esi,%edx
  mov    %edx,(%r12,%rax,1)
  add    $0x4,%rax
  cmp    %rax,%r8
  jne    L37
  lea    0x0(,%r9,4),%rax
  add    %rax,%rbp
  add    %rax,%r12
L13:
  mov    0x0(%rbp),%eax
  mov    %ebx,%ecx
  sar    %cl,%eax
  mov    %eax,(%r12)
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
L11:
  ret
  nopl   0x0(%rax)
L59:
  movslq %edx,%r14
  xor    %esi,%esi
  shl    $0x2,%edx
  and    $0x1f,%ebx
  sub    %r14,%r13
  lea    0x0(%rbp,%r14,4),%rbp
  lea    (%rdi,%r13,4),%rdi
  call   <memset@plt>
  mov    %r13,%r9
  sub    $0x1,%r9
  jne    L39
  jmp    L13
  nopl   0x0(%rax,%rax,1)
L30:
  pop    %rbx
  lea    0x0(,%rdx,4),%rdx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  jmp    1070 <memcpy@plt>
```

assistant: ```c
void bitvector_shr_to(u_int32_t *dest, u_int32_t *b, size_t n, u_int32_t s) {
  u_int32_t i, j;
  if (n == 0)
    return;
  if (s == 0) {
    memcpy(dest, b, n * 4);
    return;
  }
  j = (s >> 5);
  if (j) {
    n -= j;
    memset(&dest[n], 0, j * 4);
    s &= 31;
    b = &b[j];
  }
  for (i = 0; i < n - 1; i++) {
    dest[i] = (b[i] >> s) | (b[i + 1] << (32 - s));
  }
  dest[i] = b[i] >> s;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<AppUpdaterCAN>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <ConfigureBridge@plt>
  mov    $0x0,%edi
  call   <ConfigureCANInterface@plt>
  mov    $0x0,%eax
  call   <UpdaterCAN@plt>
  nop
  pop    %rbp
  ret
```

assistant: ```c
void AppUpdaterCAN(void) {
  ConfigureBridge();
  ConfigureCANInterface(0);
  UpdaterCAN();
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<AppUpdaterCAN>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <ConfigureBridge@plt>
  mov    $0x0,%edi
  call   <ConfigureCANInterface@plt>
  mov    $0x0,%eax
  call   <UpdaterCAN@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void AppUpdaterCAN(void) {
  ConfigureBridge();
  ConfigureCANInterface(0);
  UpdaterCAN();
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<AppUpdaterCAN>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <ConfigureBridge@plt>
  xor    %edi,%edi
  call   <ConfigureCANInterface@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1070 <UpdaterCAN@plt>
```

assistant: ```c
void AppUpdaterCAN(void) {
  ConfigureBridge();
  ConfigureCANInterface(0);
  UpdaterCAN();
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<AppUpdaterCAN>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <ConfigureBridge@plt>
  xor    %edi,%edi
  call   <ConfigureCANInterface@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1070 <UpdaterCAN@plt>
```

assistant: ```c
void AppUpdaterCAN(void) {
  ConfigureBridge();
  ConfigureCANInterface(0);
  UpdaterCAN();
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<lps>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    %edx,-0x1c(%rbp)
  mov    %ecx,-0x20(%rbp)
  cmpl   $0x1,-0x14(%rbp)
  jne    L40
  mov    -0x18(%rbp),%eax
  mov    %eax,%esi
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L56
L40:
  mov    -0x18(%rbp),%eax
  mov    %eax,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L14
L60:
  addl   $0x1,-0x4(%rbp)
  mov    -0x1c(%rbp),%eax
  add    %eax,-0x8(%rbp)
  mov    -0x8(%rbp),%eax
  cmp    -0x20(%rbp),%eax
  jg     L4
  mov    -0x8(%rbp),%eax
  mov    %eax,%esi
  lea    D27(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x1c(%rbp),%eax
  add    %eax,-0x8(%rbp)
L14:
  mov    -0x8(%rbp),%edx
  mov    -0x1c(%rbp),%eax
  add    %edx,%eax
  cmp    %eax,-0x20(%rbp)
  jge    L60
  jmp    L64
L4:
  nop
L64:
  mov    -0x8(%rbp),%eax
  cmp    -0x20(%rbp),%eax
  jg     L2
  mov    -0x18(%rbp),%eax
  mov    %eax,%esi
  lea    D27(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x1c(%rbp),%eax
  add    %eax,%eax
  add    %eax,-0x18(%rbp)
  addl   $0x1,-0x4(%rbp)
L2:
  mov    -0x1c(%rbp),%eax
  lea    (%rax,%rax,1),%edi
  mov    -0x14(%rbp),%eax
  sub    -0x4(%rbp),%eax
  mov    -0x20(%rbp),%edx
  mov    -0x18(%rbp),%esi
  mov    %edx,%ecx
  mov    %edi,%edx
  mov    %eax,%edi
  call   <lps>
L56:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D12
type_infered: string
name: unpack_data
arguments: data_label: D27
type_infered: string

tool: D12: 
Last man standing is %d.

tool: D27: %d 

assistant: ```c
void lps(int n, int a, int d, int m) {
  if (n == 1) {
    printf("\nLast man standing is %d.", a);
    return;
  }
  int i = a, j = 0, l, k;
  while (i + d <= m) {
    j++;
    i = i + d;
    if (i > m)
      break;
    printf("%d ", i);
    i = i + d;
  }
  if (i <= m) {
    printf("%d ", a);
    a = a + 2 * d;
    j++;
  }
  lps(n - j, a, 2 * d, m);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<lps>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %edi,0xc(%rsp)
  mov    %esi,0x8(%rsp)
  cmp    $0x1,%edi
  je     L36
  mov    %edx,%r12d
  mov    %ecx,%r13d
  lea    (%rsi,%rdx,1),%r15d
  lea    (%rdx,%rdx,1),%eax
  mov    %eax,0x4(%rsp)
  mov    %r15d,%ebx
  mov    $0x0,%ebp
  lea    D10(%rip),%r14
  mov    %esi,%eax
  cmp    %r15d,%ecx
  jl     L38
L43:
  add    $0x1,%ebp
  mov    %ebx,%esi
  mov    %r14,%rdi
  call   <printf@plt>
  lea    (%rbx,%r12,1),%eax
  add    0x4(%rsp),%ebx
  cmp    %ebx,%r13d
  jge    L43
L38:
  cmp    %eax,%r13d
  jge    L59
L54:
  lea    (%r12,%r12,1),%edx
  mov    0xc(%rsp),%edi
  sub    %ebp,%edi
  mov    %r13d,%ecx
  mov    0x8(%rsp),%esi
  call   <lps>
L11:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L36:
  mov    0x8(%rsp),%esi
  lea    D14(%rip),%rdi
  call   <printf@plt>
  jmp    L11
L59:
  mov    0x8(%rsp),%esi
  lea    D10(%rip),%rdi
  call   <printf@plt>
  lea    (%r12,%r15,1),%eax
  mov    %eax,0x8(%rsp)
  add    $0x1,%ebp
  jmp    L54
```

tool_calls: name: extract_symbolic_data
arguments: tag: D10
type: string
name: extract_symbolic_data
arguments: tag: D14
type: string

tool: D10: %d 

tool: D14: 
Last man standing is %d.

assistant: ```c
void lps(int n, int a, int d, int m) {
  if (n == 1) {
    printf("\nLast man standing is %d.", a);
    return;
  }
  int i = a, j = 0, l, k;
  while (i + d <= m) {
    j++;
    i = i + d;
    if (i > m)
      break;
    printf("%d ", i);
    i = i + d;
  }
  if (i <= m) {
    printf("%d ", a);
    a = a + 2 * d;
    j++;
  }
  lps(n - j, a, 2 * d, m);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<lps>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %edi,0x4(%rsp)
  mov    %esi,0x8(%rsp)
  cmp    $0x1,%edi
  je     L26
  mov    %edx,%ebp
  mov    %ecx,%r13d
  lea    D14(%rip),%r14
  nop
L29:
  mov    0x8(%rsp),%eax
  lea    0x0(%rbp,%rbp,1),%r12d
  xor    %ebx,%ebx
  add    %ebp,%eax
  mov    %eax,0xc(%rsp)
  mov    %eax,%r15d
  cmp    %eax,%r13d
  jl     L21
  nopl   0x0(%rax)
L30:
  mov    %r15d,%esi
  mov    %r14,%rdi
  add    $0x1,%ebx
  call   <printf@plt>
  lea    0x0(%rbp,%r15,1),%eax
  add    %r12d,%r15d
  cmp    %r15d,%r13d
  jge    L30
  lea    0x0(%rbp,%rbp,1),%r12d
  cmp    %r13d,%eax
  jle    L1
  sub    %ebx,0x4(%rsp)
  mov    0x4(%rsp),%eax
  cmp    $0x1,%eax
  je     L26
L75:
  mov    %r12d,%ebp
  jmp    L29
  nopw   0x0(%rax,%rax,1)
L21:
  cmp    0x8(%rsp),%r13d
  jl     L75
  nopw   0x0(%rax,%rax,1)
L1:
  mov    0x8(%rsp),%esi
  mov    %r14,%rdi
  add    $0x1,%ebx
  call   <printf@plt>
  sub    %ebx,0x4(%rsp)
  mov    0x4(%rsp),%eax
  add    0xc(%rsp),%ebp
  mov    %ebp,0x8(%rsp)
  cmp    $0x1,%eax
  jne    L75
L26:
  mov    0x8(%rsp),%esi
  add    $0x18,%rsp
  lea    D7(%rip),%rdi
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  jmp    1050 <printf@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D14
type: string
name: parse_data
arguments: data_tag: D7
type: string

tool: D14: %d 

tool: D7: 
Last man standing is %d.

assistant: ```c
void lps(int n, int a, int d, int m) {
  if (n == 1) {
    printf("\nLast man standing is %d.", a);
    return;
  }
  int i = a, j = 0, l, k;
  while (i + d <= m) {
    j++;
    i = i + d;
    if (i > m)
      break;
    printf("%d ", i);
    i = i + d;
  }
  if (i <= m) {
    printf("%d ", a);
    a = a + 2 * d;
    j++;
  }
  lps(n - j, a, 2 * d, m);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<lps>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %edi,0x4(%rsp)
  mov    %esi,0x8(%rsp)
  cmp    $0x1,%edi
  je     L69
  mov    %edx,%ebp
  mov    %ecx,%r12d
  lea    D9(%rip),%r14
  nop
L27:
  mov    0x8(%rsp),%eax
  lea    0x0(%rbp,%rbp,1),%r13d
  xor    %ebx,%ebx
  add    %ebp,%eax
  mov    %eax,0xc(%rsp)
  mov    %eax,%r15d
  cmp    %eax,%r12d
  jl     L49
  nopl   (%rax)
L10:
  mov    %r15d,%esi
  mov    %r14,%rdi
  add    $0x1,%ebx
  call   <printf@plt>
  lea    0x0(%rbp,%r15,1),%eax
  add    %r13d,%r15d
  cmp    %r15d,%r12d
  jge    L10
L45:
  lea    0x0(%rbp,%rbp,1),%r13d
  cmp    %eax,%r12d
  jge    L4
  sub    %ebx,0x4(%rsp)
  mov    0x4(%rsp),%eax
  cmp    $0x1,%eax
  je     L69
L23:
  mov    %r13d,%ebp
  jmp    L27
  nopw   0x0(%rax,%rax,1)
L4:
  mov    0x8(%rsp),%esi
  mov    %r14,%rdi
  add    $0x1,%ebx
  call   <printf@plt>
  sub    %ebx,0x4(%rsp)
  mov    0x4(%rsp),%eax
  add    0xc(%rsp),%ebp
  mov    %ebp,0x8(%rsp)
  cmp    $0x1,%eax
  jne    L23
L69:
  mov    0x8(%rsp),%esi
  add    $0x18,%rsp
  lea    D10(%rip),%rdi
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  jmp    1050 <printf@plt>
  nopw   0x0(%rax,%rax,1)
L49:
  mov    0x8(%rsp),%eax
  jmp    L45
```

tool_calls: name: read_data
arguments: data_label: D9
type_guessed: string
name: read_data
arguments: data_label: D10
type_guessed: string

tool: D9: %d 

tool: D10: 
Last man standing is %d.

assistant: ```c
void lps(int n, int a, int d, int m) {
  if (n == 1) {
    printf("\nLast man standing is %d.", a);
    return;
  }
  int i = a, j = 0, l, k;
  while (i + d <= m) {
    j++;
    i = i + d;
    if (i > m)
      break;
    printf("%d ", i);
    i = i + d;
  }
  if (i <= m) {
    printf("%d ", a);
    a = a + 2 * d;
    j++;
  }
  lps(n - j, a, 2 * d, m);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<factor>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %edi,-0x14(%rbp)
  cmpl   $0x1,-0x14(%rbp)
  jne    L3
  mov    $0x1,%eax
  jmp    L2
L3:
  mov    -0x14(%rbp),%eax
  movslq %eax,%rbx
  mov    -0x14(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,%edi
  call   <factor>
  imul   %rbx,%rax
L2:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
long factor(int n) {
  if (n == 1)
    return 1;
  return n * factor(n - 1);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<factor>:
  endbr64
  mov    $0x1,%eax
  cmp    $0x1,%edi
  je     L5
  push   %rbx
  mov    %edi,%ebx
  lea    -0x1(%rdi),%edi
  call   <factor>
  mov    %rax,%rdx
  movslq %ebx,%rax
  imul   %rdx,%rax
  pop    %rbx
  ret
L5:
  ret
```

assistant: ```c
long factor(int n) {
  if (n == 1)
    return 1;
  return n * factor(n - 1);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<factor>:
  endbr64
  mov    $0x1,%r8d
  cmp    $0x1,%edi
  je     L1
  movslq %edi,%rdx
  lea    -0x2(%rdi),%esi
  mov    $0x1,%r8d
  lea    -0x1(%rdx),%rax
  mov    %rax,%rcx
  sub    %rsi,%rcx
  jmp    L24
  nopw   0x0(%rax,%rax,1)
L20:
  sub    $0x1,%rax
L24:
  imul   %rdx,%r8
  mov    %rax,%rdx
  cmp    %rcx,%rax
  jne    L20
L1:
  mov    %r8,%rax
  ret
```

assistant: ```c
long factor(int n) {
  if (n == 1)
    return 1;
  return n * factor(n - 1);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<factor>:
  endbr64
  mov    $0x1,%r8d
  cmp    $0x1,%edi
  je     L15
  movslq %edi,%rdx
  lea    -0x2(%rdi),%esi
  mov    $0x1,%r8d
  lea    -0x1(%rdx),%rax
  mov    %rax,%rcx
  sub    %rsi,%rcx
  jmp    L27
  nopw   0x0(%rax,%rax,1)
L23:
  sub    $0x1,%rax
L27:
  imul   %rdx,%r8
  mov    %rax,%rdx
  cmp    %rcx,%rax
  jne    L23
L15:
  mov    %r8,%rax
  ret
```

assistant: ```c
long factor(int n) {
  if (n == 1)
    return 1;
  return n * factor(n - 1);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<strlen_constant1>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  lea    D9(%rip),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cmp    $0x3,%eax
  nop
  leave
  ret
```

tool_calls: name: read_data
arguments: data_tag: D9
type_infered: string

tool: D9: 123

assistant: ```c
void strlen_constant1() {
  const char *a = "123";
  if (strlen(a) != 3)
    (void)*(char *)0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<strlen_constant1>:
  endbr64
  sub    $0x8,%rsp
  lea    D1(%rip),%rdi
  call   <strlen@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D1
datatype: string

tool: D1: 123

assistant: ```c
void strlen_constant1() {
  const char *a = "123";
  if (strlen(a) != 3)
    (void)*(char *)0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<strlen_constant1>:
  endbr64
  lea    D3(%rip),%rdi
  jmp    1050 <strlen@plt>
```

tool_calls: name: read_data
arguments: label: D3
data_type: string

tool: D3: 123

assistant: ```c
void strlen_constant1() {
  const char *a = "123";
  if (strlen(a) != 3)
    (void)*(char *)0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<strlen_constant1>:
  endbr64
  lea    D5(%rip),%rdi
  jmp    1050 <strlen@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D5
data_type: string

tool: D5: 123

assistant: ```c
void strlen_constant1() {
  const char *a = "123";
  if (strlen(a) != 3)
    (void)*(char *)0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<f91>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x64,-0x4(%rbp)
  jg     L12
  mov    -0x4(%rbp),%eax
  add    $0xb,%eax
  mov    %eax,%edi
  call   <f91>
  mov    %eax,%edi
  call   <f91>
  jmp    L1
L12:
  mov    -0x4(%rbp),%eax
  sub    $0xa,%eax
L1:
  leave
  ret
```

assistant: ```c
int f91(int N) {
  if (N <= 100)
    return (f91(f91(N + 11)));
  else
    return (N - 10);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<f91>:
  endbr64
  lea    -0xa(%rdi),%eax
  cmp    $0x64,%edi
  jle    L9
  ret
L9:
  sub    $0x8,%rsp
  add    $0xb,%edi
  call   <f91>
  mov    %eax,%edi
  call   <f91>
  add    $0x8,%rsp
  ret
```

assistant: ```c
int f91(int N) {
  if (N <= 100)
    return (f91(f91(N + 11)));
  else
    return (N - 10);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<f91>:
  endbr64
  mov    %edi,%eax
  cmp    $0x64,%edi
  jg     L36
L99:
  add    $0xb,%eax
  cmp    $0x64,%eax
  jg     L354
L259:
  add    $0xb,%eax
  cmp    $0x64,%eax
  jg     L254
L178:
  add    $0xb,%eax
  cmp    $0x64,%eax
  jg     L368
L169:
  add    $0xb,%eax
  cmp    $0x64,%eax
  jg     L328
L351:
  add    $0xb,%eax
  cmp    $0x64,%eax
  jg     L395
L173:
  add    $0xb,%eax
  cmp    $0x64,%eax
  jg     L234
L348:
  add    $0xb,%eax
  cmp    $0x64,%eax
  jg     L226
L411:
  add    $0xb,%eax
  cmp    $0x64,%eax
  jg     L260
  sub    $0x8,%rsp
  nopl   0x0(%rax,%rax,1)
L372:
  lea    0xb(%rax),%edi
  call   <f91>
  cmp    $0x64,%eax
  jle    L372
  sub    $0xa,%eax
  cmp    $0x64,%eax
  jg     L31
L70:
  add    $0xb,%eax
  cmp    $0x64,%eax
  jle    L372
  sub    $0xa,%eax
  cmp    $0x64,%eax
  jle    L70
L31:
  sub    $0xa,%eax
  cmp    $0x64,%eax
  jg     L35
L37:
  add    $0xb,%eax
  cmp    $0x64,%eax
  jle    L70
  sub    $0xa,%eax
  cmp    $0x64,%eax
  jle    L37
L35:
  sub    $0xa,%eax
  cmp    $0x64,%eax
  jg     L187
L392:
  add    $0xb,%eax
  cmp    $0x64,%eax
  jle    L37
  sub    $0xa,%eax
  cmp    $0x64,%eax
  jle    L392
L187:
  sub    $0xa,%eax
  cmp    $0x64,%eax
  jg     L44
L397:
  add    $0xb,%eax
  cmp    $0x64,%eax
  jle    L392
  sub    $0xa,%eax
  cmp    $0x64,%eax
  jle    L397
L44:
  sub    $0xa,%eax
  cmp    $0x64,%eax
  jg     L308
L299:
  add    $0xb,%eax
  cmp    $0x64,%eax
  jle    L397
  sub    $0xa,%eax
  cmp    $0x64,%eax
  jle    L299
L308:
  sub    $0xa,%eax
  cmp    $0x64,%eax
  jg     L330
L180:
  add    $0xb,%eax
  cmp    $0x64,%eax
  jle    L299
  sub    $0xa,%eax
  cmp    $0x64,%eax
  jle    L180
L330:
  sub    $0xa,%eax
  cmp    $0x64,%eax
  jg     L298
L214:
  add    $0xb,%eax
  cmp    $0x64,%eax
  jle    L180
  sub    $0xa,%eax
  cmp    $0x64,%eax
  jle    L214
L298:
  sub    $0xa,%eax
  cmp    $0x64,%eax
  jg     L154
L108:
  add    $0xb,%eax
  cmp    $0x64,%eax
  jle    L214
  sub    $0xa,%eax
  cmp    $0x64,%eax
  jle    L108
L154:
  sub    $0xa,%eax
  add    $0x8,%rsp
  ret
  nopl   (%rax)
L260:
  sub    $0xa,%eax
  cmp    $0x64,%eax
  jle    L411
L226:
  sub    $0xa,%eax
  cmp    $0x64,%eax
  jle    L348
L234:
  sub    $0xa,%eax
  cmp    $0x64,%eax
  jle    L173
L395:
  sub    $0xa,%eax
  cmp    $0x64,%eax
  jle    L351
L328:
  sub    $0xa,%eax
  cmp    $0x64,%eax
  jle    L169
L368:
  sub    $0xa,%eax
  cmp    $0x64,%eax
  jle    L178
L254:
  sub    $0xa,%eax
  cmp    $0x64,%eax
  jle    L259
L354:
  sub    $0xa,%eax
  cmp    $0x64,%eax
  jle    L99
L36:
  sub    $0xa,%eax
  ret
```

assistant: ```c
int f91(int N) {
  if (N <= 100)
    return (f91(f91(N + 11)));
  else
    return (N - 10);
}
```