ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What would the C source look like for the assembly code provided here?

```asm
<bottom_up_cut_rod>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  mov    %edx,-0x30(%rbp)
  mov    -0x30(%rbp),%eax
  add    $0x1,%eax
  cltq
  shl    $0x2,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movl   $0x0,(%rax)
  movl   $0x1,-0x18(%rbp)
  jmp    L6
L39:
  movl   $0xffffffff,-0x14(%rbp)
  movl   $0x1,-0x10(%rbp)
  jmp    L30
L14:
  mov    -0x10(%rbp),%eax
  cltq
  shl    $0x2,%rax
  lea    -0x4(%rax),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x18(%rbp),%eax
  sub    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x8(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%eax
  add    %eax,%edx
  mov    -0x14(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <max@plt>
  mov    %eax,-0x14(%rbp)
  addl   $0x1,-0x10(%rbp)
L30:
  mov    -0x10(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jle    L14
  mov    -0x18(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x8(%rbp),%rax
  add    %rax,%rdx
  mov    -0x14(%rbp),%eax
  mov    %eax,(%rdx)
  addl   $0x1,-0x18(%rbp)
L6:
  mov    -0x30(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jge    L39
  mov    -0x30(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    -0xc(%rbp),%eax
  leave
  ret
```

assistant: ```c
int bottom_up_cut_rod(int *p, int size_p, int rod_length) {
  int *result = malloc((rod_length + 1) * sizeof(int));
  result[0] = 0;
  for (int i = 1; i < rod_length + 1; i++) {
    int maxP = -1;
    for (int j = 1; j <= i; j++) {
      maxP = max(maxP, p[j - 1] + result[i - j]);
    }
    result[i] = maxP;
  }
  int r = result[rod_length];
  free(result);
  return r;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<bottom_up_cut_rod>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%r13
  mov    %edx,%r15d
  lea    0x4(,%rdx,4),%edi
  call   <malloc@plt>
  mov    %rax,0x8(%rsp)
  movl   $0x0,(%rax)
  test   %r15d,%r15d
  jle    L11
  lea    0x4(%rax),%r14
  mov    $0x1,%r12d
  jmp    L35
L45:
  mov    -0x4(%rbp),%esi
  add    -0x4(%r13,%rbx,4),%esi
  call   <max@plt>
  mov    %eax,%edi
  add    $0x1,%rbx
  sub    $0x4,%rbp
  cmp    %ebx,%r12d
  jge    L45
L6:
  mov    %edi,(%r14)
  add    $0x1,%r12d
  add    $0x4,%r14
  cmp    %r12d,%r15d
  jl     L11
L35:
  mov    %r14,%rbp
  mov    $0x1,%ebx
  mov    $0xffffffff,%edi
  test   %r12d,%r12d
  jg     L45
  jmp    L6
L11:
  movslq %r15d,%r15
  mov    0x8(%rsp),%rdi
  mov    (%rdi,%r15,4),%ebx
  call   <free@plt>
  mov    %ebx,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
int bottom_up_cut_rod(int *p, int size_p, int rod_length) {
  int *result = malloc((rod_length + 1) * sizeof(int));
  result[0] = 0;
  for (int i = 1; i < rod_length + 1; i++) {
    int maxP = -1;
    for (int j = 1; j <= i; j++) {
      maxP = max(maxP, p[j - 1] + result[i - j]);
    }
    result[i] = maxP;
  }
  int r = result[rod_length];
  free(result);
  return r;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<bottom_up_cut_rod>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  movslq %edx,%r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  lea    0x4(,%r13,4),%edi
  push   %rbx
  sub    $0x18,%rsp
  call   <malloc@plt>
  mov    %rax,0x8(%rsp)
  movl   $0x0,(%rax)
  test   %r13d,%r13d
  jle    L47
  lea    0x4(%rax),%r12
  xor    %esi,%esi
  mov    $0x1,%ebx
L6:
  mov    %r12,%r14
  mov    $0x2,%r15d
  mov    $0xffffffff,%edi
  jmp    L46
L4:
  mov    -0x4(%r14),%esi
  add    $0x1,%r15
L46:
  add    -0x8(%rbp,%r15,4),%esi
  sub    $0x4,%r14
  call   <max@plt>
  mov    %eax,%edi
  cmp    %r15d,%ebx
  jge    L4
  add    $0x1,%ebx
  mov    %eax,(%r12)
  cmp    %ebx,%r13d
  jl     L47
  add    $0x4,%r12
  mov    %eax,%esi
  jmp    L6
  nopl   0x0(%rax)
L47:
  mov    0x8(%rsp),%rdi
  mov    (%rdi,%r13,4),%r12d
  call   <free@plt>
  add    $0x18,%rsp
  pop    %rbx
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
int bottom_up_cut_rod(int *p, int size_p, int rod_length) {
  int *result = malloc((rod_length + 1) * sizeof(int));
  result[0] = 0;
  for (int i = 1; i < rod_length + 1; i++) {
    int maxP = -1;
    for (int j = 1; j <= i; j++) {
      maxP = max(maxP, p[j - 1] + result[i - j]);
    }
    result[i] = maxP;
  }
  int r = result[rod_length];
  free(result);
  return r;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<bottom_up_cut_rod>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  movslq %edx,%r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  lea    0x4(,%r13,4),%edi
  push   %rbx
  sub    $0x18,%rsp
  call   <malloc@plt>
  mov    %rax,0x8(%rsp)
  movl   $0x0,(%rax)
  test   %r13d,%r13d
  jle    L22
  lea    0x4(%rax),%r12
  mov    $0x1,%ebx
  xchg   %ax,%ax
L2:
  mov    %r12,%r14
  mov    $0x1,%r15d
  mov    $0xffffffff,%edi
  xchg   %ax,%ax
L21:
  mov    -0x4(%r14),%esi
  add    -0x4(%rbp,%r15,4),%esi
  add    $0x1,%r15
  sub    $0x4,%r14
  call   <max@plt>
  mov    %eax,%edi
  cmp    %r15d,%ebx
  jge    L21
  add    $0x1,%ebx
  mov    %eax,(%r12)
  add    $0x4,%r12
  cmp    %ebx,%r13d
  jge    L2
L22:
  mov    0x8(%rsp),%rdi
  mov    (%rdi,%r13,4),%r12d
  call   <free@plt>
  add    $0x18,%rsp
  pop    %rbx
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
int bottom_up_cut_rod(int *p, int size_p, int rod_length) {
  int *result = malloc((rod_length + 1) * sizeof(int));
  result[0] = 0;
  for (int i = 1; i < rod_length + 1; i++) {
    int maxP = -1;
    for (int j = 1; j <= i; j++) {
      maxP = max(maxP, p[j - 1] + result[i - j]);
    }
    result[i] = maxP;
  }
  int r = result[rod_length];
  free(result);
  return r;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<space_relation>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %rdi,%rbx
  movq   %xmm0,%r12
  mov    %rcx,%rbp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
L45:
  movsd  (%rsi,%rax,1),%xmm0
  subsd  (%rdx,%rax,1),%xmm0
  movsd  %xmm0,(%rsp,%rax,1)
  add    $0x8,%rax
  cmp    $0x18,%rax
  jne    L45
  mov    %rsp,%r14
  mov    %r14,%rdi
  call   <norm@plt>
  movq   %xmm0,%r13
  mov    %rbp,%rsi
  mov    %r14,%rdi
  call   <dot_product@plt>
  movq   %xmm0,%rbp
  lea    D37(%rip),%rsi
  mov    %rbx,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jne    L40
  movq   %r12,%xmm1
  movq   %r13,%xmm2
  comisd %xmm2,%xmm1
  setae  %al
  movzbl %al,%eax
L39:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L42
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L40:
  lea    D40(%rip),%rsi
  mov    %rbx,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jne    L27
  movq   %r13,%xmm3
  movq   %r12,%xmm4
  comisd %xmm4,%xmm3
  seta   %al
  movzbl %al,%eax
  jmp    L39
L27:
  lea    D19(%rip),%rsi
  mov    %rbx,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jne    L9
  pxor   %xmm0,%xmm0
  movq   %rbp,%xmm5
  comisd %xmm5,%xmm0
  setae  %al
  movzbl %al,%eax
  jmp    L39
L9:
  lea    D16(%rip),%rsi
  mov    %rbx,%rdi
  call   <strcmp@plt>
  mov    %rax,%rdx
  mov    $0x0,%eax
  test   %rdx,%rdx
  jne    L39
  movq   %rbp,%xmm6
  comisd D0(%rip),%xmm6
  seta   %al
  movzbl %al,%eax
  jmp    L39
L42:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D37
type_infered: string
name: disassemble_data
arguments: tag: D40
type_infered: string
name: disassemble_data
arguments: tag: D19
type_infered: string
name: disassemble_data
arguments: tag: D16
type_infered: string
name: disassemble_data
arguments: tag: D0
type_infered: f64

tool: D37: inside

tool: D40: outside

tool: D19: below

tool: D16: above

tool: D0: 0

assistant: ```c
int space_relation(char *operator, double center[3], double vcen[3], double vrad, double vaxis[3]) {
  int k, t;
  double r, d;
  double vector[3];
  for (k = 0; k < 3; k++)
    vector[k] = center[k] - vcen[k];
  r = norm(vector);
  d = dot_product(vector, vaxis);
  if (strcmp(operator, "inside") == 0)
    t = (r <= vrad);
  else if (strcmp(operator, "outside") == 0)
    t = (r > vrad);
  else if (strcmp(operator, "below") == 0)
    t = (d <= 0.0);
  else if (strcmp(operator, "above") == 0)
    t = (d > 0.0);
  else
    t = 0;
  return (t);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<space_relation>:
  endbr64
  push   %r13
  mov    %rcx,%r13
  push   %r12
  mov    %rdi,%r12
  mov    %rsi,%rdi
  push   %rbp
  sub    $0x40,%rsp
  movsd  %xmm0,0x10(%rsp)
  lea    0x20(%rsp),%rbp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
L62:
  movsd  (%rdi,%rax,1),%xmm0
  subsd  (%rdx,%rax,1),%xmm0
  movsd  %xmm0,0x0(%rbp,%rax,1)
  add    $0x8,%rax
  cmp    $0x18,%rax
  jne    L62
  mov    %rbp,%rdi
  call   <norm@plt>
  mov    %r13,%rsi
  mov    %rbp,%rdi
  movsd  %xmm0,0x8(%rsp)
  call   <dot_product@plt>
  lea    D1(%rip),%rsi
  mov    %r12,%rdi
  movsd  %xmm0,0x18(%rsp)
  call   <strcmp@plt>
  test   %rax,%rax
  jne    L2
  xor    %eax,%eax
  movsd  0x10(%rsp),%xmm2
  comisd 0x8(%rsp),%xmm2
  setae  %al
L67:
  mov    0x38(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L30
  add    $0x40,%rsp
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   (%rax)
L2:
  lea    D34(%rip),%rsi
  mov    %r12,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jne    L24
  xor    %eax,%eax
  movsd  0x8(%rsp),%xmm3
  comisd 0x10(%rsp),%xmm3
  seta   %al
  jmp    L67
  nopw   0x0(%rax,%rax,1)
L24:
  lea    D24(%rip),%rsi
  mov    %r12,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jne    L60
  xor    %eax,%eax
  pxor   %xmm0,%xmm0
  comisd 0x18(%rsp),%xmm0
  setae  %al
  jmp    L67
L60:
  lea    D36(%rip),%rsi
  mov    %r12,%rdi
  call   <strcmp@plt>
  mov    %rax,%r8
  xor    %eax,%eax
  test   %r8,%r8
  jne    L67
  xor    %eax,%eax
  movsd  0x18(%rsp),%xmm6
  comisd D11(%rip),%xmm6
  seta   %al
  jmp    L67
L30:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D1
datatype: string
name: unpack_data
arguments: data_label: D34
datatype: string
name: unpack_data
arguments: data_label: D24
datatype: string
name: unpack_data
arguments: data_label: D36
datatype: string
name: unpack_data
arguments: data_label: D11
datatype: f64

tool: D1: inside

tool: D34: outside

tool: D24: below

tool: D36: above

tool: D11: 0

assistant: ```c
int space_relation(char *operator, double center[3], double vcen[3], double vrad, double vaxis[3]) {
  int k, t;
  double r, d;
  double vector[3];
  for (k = 0; k < 3; k++)
    vector[k] = center[k] - vcen[k];
  r = norm(vector);
  d = dot_product(vector, vaxis);
  if (strcmp(operator, "inside") == 0)
    t = (r <= vrad);
  else if (strcmp(operator, "outside") == 0)
    t = (r > vrad);
  else if (strcmp(operator, "below") == 0)
    t = (d <= 0.0);
  else if (strcmp(operator, "above") == 0)
    t = (d > 0.0);
  else
    t = 0;
  return (t);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<space_relation>:
  endbr64
  push   %r13
  push   %r12
  mov    %rcx,%r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x40,%rsp
  movupd (%rsi),%xmm1
  movupd (%rdx),%xmm4
  movsd  %xmm0,0x10(%rsp)
  movsd  0x10(%rsi),%xmm0
  subsd  0x10(%rdx),%xmm0
  lea    0x20(%rsp),%r13
  mov    %fs:0x28,%rcx
  mov    %rcx,0x38(%rsp)
  xor    %ecx,%ecx
  subpd  %xmm4,%xmm1
  mov    %r13,%rdi
  movsd  %xmm0,0x30(%rsp)
  movaps %xmm1,0x20(%rsp)
  call   <norm@plt>
  mov    %r12,%rsi
  mov    %r13,%rdi
  movsd  %xmm0,0x8(%rsp)
  call   <dot_product@plt>
  lea    D25(%rip),%rsi
  mov    %rbp,%rdi
  movsd  %xmm0,0x18(%rsp)
  call   <strcmp@plt>
  test   %rax,%rax
  jne    L71
  xor    %eax,%eax
  movsd  0x10(%rsp),%xmm2
  comisd 0x8(%rsp),%xmm2
  setae  %al
L1:
  mov    0x38(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L76
  add    $0x40,%rsp
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L71:
  lea    D27(%rip),%rsi
  mov    %rbp,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jne    L33
  xor    %eax,%eax
  movsd  0x8(%rsp),%xmm5
  comisd 0x10(%rsp),%xmm5
  seta   %al
  jmp    L1
  nop
L33:
  lea    D46(%rip),%rsi
  mov    %rbp,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jne    L30
  xor    %eax,%eax
  pxor   %xmm0,%xmm0
  comisd 0x18(%rsp),%xmm0
  setae  %al
  jmp    L1
L30:
  lea    D38(%rip),%rsi
  mov    %rbp,%rdi
  call   <strcmp@plt>
  mov    %rax,%r8
  xor    %eax,%eax
  test   %r8,%r8
  jne    L1
  xor    %eax,%eax
  movsd  0x18(%rsp),%xmm7
  comisd D7(%rip),%xmm7
  seta   %al
  jmp    L1
L76:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D25
type_guessed: string
name: extract_symbolic_data
arguments: tag: D27
type_guessed: string
name: extract_symbolic_data
arguments: tag: D46
type_guessed: string
name: extract_symbolic_data
arguments: tag: D38
type_guessed: string
name: extract_symbolic_data
arguments: tag: D7
type_guessed: f64

tool: D25: inside

tool: D27: outside

tool: D46: below

tool: D38: above

tool: D7: 0

assistant: ```c
int space_relation(char *operator, double center[3], double vcen[3], double vrad, double vaxis[3]) {
  int k, t;
  double r, d;
  double vector[3];
  for (k = 0; k < 3; k++)
    vector[k] = center[k] - vcen[k];
  r = norm(vector);
  d = dot_product(vector, vaxis);
  if (strcmp(operator, "inside") == 0)
    t = (r <= vrad);
  else if (strcmp(operator, "outside") == 0)
    t = (r > vrad);
  else if (strcmp(operator, "below") == 0)
    t = (d <= 0.0);
  else if (strcmp(operator, "above") == 0)
    t = (d > 0.0);
  else
    t = 0;
  return (t);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<dateToDays>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    %edx,-0x1c(%rbp)
  mov    %ecx,-0x20(%rbp)
  movl   $0x0,-0x4(%rbp)
  cmpl   $0xc,-0x18(%rbp)
  ja     L60
  mov    -0x18(%rbp),%eax
  lea    0x0(,%rax,4),%rdx
  lea    D2(%rip),%rax
  mov    (%rdx,%rax,1),%eax
  cltq
  lea    D2(%rip),%rdx
  add    %rdx,%rax
  notrack jmp *%rax
  mov    -0x14(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  jmp    L60
  mov    -0x14(%rbp),%eax
  add    $0x1f,%eax
  mov    %eax,-0x4(%rbp)
  jmp    L60
  mov    -0x14(%rbp),%eax
  lea    0x3b(%rax),%edx
  mov    -0x20(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x4(%rbp)
  jmp    L60
  mov    -0x14(%rbp),%eax
  lea    0x5a(%rax),%edx
  mov    -0x20(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x4(%rbp)
  jmp    L60
  mov    -0x14(%rbp),%eax
  lea    0x78(%rax),%edx
  mov    -0x20(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x4(%rbp)
  jmp    L60
  mov    -0x14(%rbp),%eax
  lea    0x97(%rax),%edx
  mov    -0x20(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x4(%rbp)
  jmp    L60
  mov    -0x14(%rbp),%eax
  lea    0xb5(%rax),%edx
  mov    -0x20(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x4(%rbp)
  jmp    L60
  mov    -0x14(%rbp),%eax
  lea    0xd4(%rax),%edx
  mov    -0x20(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x4(%rbp)
  jmp    L60
  mov    -0x14(%rbp),%eax
  lea    0xf3(%rax),%edx
  mov    -0x20(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x4(%rbp)
  jmp    L60
  mov    -0x14(%rbp),%eax
  lea    0x111(%rax),%edx
  mov    -0x20(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x4(%rbp)
  jmp    L60
  mov    -0x14(%rbp),%eax
  lea    0x130(%rax),%edx
  mov    -0x20(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x4(%rbp)
  jmp    L60
  mov    -0x14(%rbp),%eax
  lea    0x14e(%rax),%edx
  mov    -0x20(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x4(%rbp)
  nop
L60:
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int dateToDays(int d, int m, int y, int lp) {
  int nod = 0;
  switch (m) {
  case 1:
    nod = d;
    break;
  case 2:
    nod = d + 31;
    break;
  case 3:
    nod = d + 59 + lp;
    break;
  case 4:
    nod = d + 90 + lp;
    break;
  case 5:
    nod = d + 120 + lp;
    break;
  case 6:
    nod = d + 151 + lp;
    break;
  case 7:
    nod = d + 181 + lp;
    break;
  case 8:
    nod = d + 212 + lp;
    break;
  case 9:
    nod = d + 243 + lp;
    break;
  case 10:
    nod = d + 273 + lp;
    break;
  case 11:
    nod = d + 304 + lp;
    break;
  case 12:
    nod = d + 334 + lp;
    break;
  }
  return nod;
}
```
user: Translate the below assembly code into C programming language:

```asm
<dateToDays>:
  endbr64
  cmp    $0xc,%esi
  ja     L5
  mov    %esi,%esi
  lea    D2(%rip),%rdx
  movslq (%rdx,%rsi,4),%rax
  add    %rdx,%rax
  notrack jmp *%rax
  mov    %edi,%eax
  ret
  lea    0x1f(%rdi),%eax
  ret
  lea    0x3b(%rdi,%rcx,1),%eax
  ret
  lea    0x5a(%rdi,%rcx,1),%eax
  ret
  lea    0x78(%rdi,%rcx,1),%eax
  ret
  lea    0x97(%rdi,%rcx,1),%eax
  ret
  lea    0xb5(%rdi,%rcx,1),%eax
  ret
  lea    0xd4(%rdi,%rcx,1),%eax
  ret
  lea    0xf3(%rdi,%rcx,1),%eax
  ret
  lea    0x111(%rdi,%rcx,1),%eax
  ret
  lea    0x130(%rdi,%rcx,1),%eax
  ret
  lea    0x14e(%rdi,%rcx,1),%eax
  ret
L5:
  mov    $0x0,%eax
  ret
```

assistant: ```c
int dateToDays(int d, int m, int y, int lp) {
  int nod = 0;
  switch (m) {
  case 1:
    nod = d;
    break;
  case 2:
    nod = d + 31;
    break;
  case 3:
    nod = d + 59 + lp;
    break;
  case 4:
    nod = d + 90 + lp;
    break;
  case 5:
    nod = d + 120 + lp;
    break;
  case 6:
    nod = d + 151 + lp;
    break;
  case 7:
    nod = d + 181 + lp;
    break;
  case 8:
    nod = d + 212 + lp;
    break;
  case 9:
    nod = d + 243 + lp;
    break;
  case 10:
    nod = d + 273 + lp;
    break;
  case 11:
    nod = d + 304 + lp;
    break;
  case 12:
    nod = d + 334 + lp;
    break;
  }
  return nod;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<dateToDays>:
  endbr64
  cmp    $0xc,%esi
  ja     L2
  lea    D4(%rip),%rdx
  mov    %esi,%esi
  movslq (%rdx,%rsi,4),%rax
  add    %rdx,%rax
  notrack jmp *%rax
  mov    %edi,%eax
  ret
  nopl   0x0(%rax,%rax,1)
  lea    0x1f(%rdi),%eax
  ret
  nopl   0x0(%rax)
  lea    0x3b(%rdi,%rcx,1),%eax
  ret
  nopl   (%rax)
  lea    0x5a(%rdi,%rcx,1),%eax
  ret
  nopl   (%rax)
  lea    0x78(%rdi,%rcx,1),%eax
  ret
  nopl   (%rax)
  lea    0x97(%rdi,%rcx,1),%eax
  ret
  lea    0xb5(%rdi,%rcx,1),%eax
  ret
  nopl   0x0(%rax,%rax,1)
  lea    0xd4(%rdi,%rcx,1),%eax
  ret
  nopl   0x0(%rax,%rax,1)
  lea    0xf3(%rdi,%rcx,1),%eax
  ret
  nopl   0x0(%rax,%rax,1)
  lea    0x111(%rdi,%rcx,1),%eax
  ret
  nopl   0x0(%rax,%rax,1)
  lea    0x130(%rdi,%rcx,1),%eax
  ret
  nopl   0x0(%rax,%rax,1)
  lea    0x14e(%rdi,%rcx,1),%eax
  ret
L2:
  xor    %eax,%eax
  ret
```

assistant: ```c
int dateToDays(int d, int m, int y, int lp) {
  int nod = 0;
  switch (m) {
  case 1:
    nod = d;
    break;
  case 2:
    nod = d + 31;
    break;
  case 3:
    nod = d + 59 + lp;
    break;
  case 4:
    nod = d + 90 + lp;
    break;
  case 5:
    nod = d + 120 + lp;
    break;
  case 6:
    nod = d + 151 + lp;
    break;
  case 7:
    nod = d + 181 + lp;
    break;
  case 8:
    nod = d + 212 + lp;
    break;
  case 9:
    nod = d + 243 + lp;
    break;
  case 10:
    nod = d + 273 + lp;
    break;
  case 11:
    nod = d + 304 + lp;
    break;
  case 12:
    nod = d + 334 + lp;
    break;
  }
  return nod;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<dateToDays>:
  endbr64
  cmp    $0xc,%esi
  ja     L0
  lea    D7(%rip),%rdx
  mov    %esi,%esi
  movslq (%rdx,%rsi,4),%rax
  add    %rdx,%rax
  notrack jmp *%rax
  mov    %edi,%eax
  ret
  nopl   0x0(%rax,%rax,1)
  lea    0x1f(%rdi),%eax
  ret
  nopl   0x0(%rax)
  lea    0x3b(%rdi,%rcx,1),%eax
  ret
  nopl   (%rax)
  lea    0x5a(%rdi,%rcx,1),%eax
  ret
  nopl   (%rax)
  lea    0x78(%rdi,%rcx,1),%eax
  ret
  nopl   (%rax)
  lea    0x97(%rdi,%rcx,1),%eax
  ret
  lea    0xb5(%rdi,%rcx,1),%eax
  ret
  nopl   0x0(%rax,%rax,1)
  lea    0xd4(%rdi,%rcx,1),%eax
  ret
  nopl   0x0(%rax,%rax,1)
  lea    0xf3(%rdi,%rcx,1),%eax
  ret
  nopl   0x0(%rax,%rax,1)
  lea    0x111(%rdi,%rcx,1),%eax
  ret
  nopl   0x0(%rax,%rax,1)
  lea    0x130(%rdi,%rcx,1),%eax
  ret
  nopl   0x0(%rax,%rax,1)
  lea    0x14e(%rdi,%rcx,1),%eax
  ret
L0:
  xor    %eax,%eax
  ret
```

assistant: ```c
int dateToDays(int d, int m, int y, int lp) {
  int nod = 0;
  switch (m) {
  case 1:
    nod = d;
    break;
  case 2:
    nod = d + 31;
    break;
  case 3:
    nod = d + 59 + lp;
    break;
  case 4:
    nod = d + 90 + lp;
    break;
  case 5:
    nod = d + 120 + lp;
    break;
  case 6:
    nod = d + 151 + lp;
    break;
  case 7:
    nod = d + 181 + lp;
    break;
  case 8:
    nod = d + 212 + lp;
    break;
  case 9:
    nod = d + 243 + lp;
    break;
  case 10:
    nod = d + 273 + lp;
    break;
  case 11:
    nod = d + 304 + lp;
    break;
  case 12:
    nod = d + 334 + lp;
    break;
  }
  return nod;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<make_multi_box2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    %rcx,-0x40(%rbp)
  mov    %r8d,-0x44(%rbp)
  mov    %r9d,-0x48(%rbp)
  movsd  %xmm0,-0x50(%rbp)
  mov    -0x44(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0xc(%rbp)
  mov    0x10(%rbp),%eax
  sub    $0x1,%eax
  imul   0x18(%rbp),%eax
  mov    %eax,%eax
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0x14(%rbp)
  jmp    L18
L33:
  movl   $0x0,-0x10(%rbp)
  jmp    L3
L56:
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x10(%rbp),%eax
  cltq
  shl    $0x3,%rax
  add    %rdx,%rax
  movq   $0xffffffffffffffff,(%rax)
  addl   $0x1,-0x10(%rbp)
L3:
  mov    -0x10(%rbp),%eax
  cmp    %eax,-0x44(%rbp)
  ja     L56
  addl   $0x1,-0x14(%rbp)
L18:
  mov    -0x14(%rbp),%eax
  cmp    %eax,-0x44(%rbp)
  ja     L33
  mov    -0x8(%rbp),%rax
  mov    %eax,-0x18(%rbp)
  jmp    L55
L31:
  mov    -0x28(%rbp),%rax
  mov    (%rax),%rdx
  mov    -0x18(%rbp),%eax
  cltq
  shl    $0x3,%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  divsd  -0x50(%rbp),%xmm0
  cvttsd2si %xmm0,%eax
  and    -0xc(%rbp),%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x48(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,%eax
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x18(%rbp),%eax
  cltq
  sub    -0x8(%rbp),%rax
  shl    $0x3,%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  divsd  -0x50(%rbp),%xmm0
  cvttsd2si %xmm0,%eax
  and    -0xc(%rbp),%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x10(%rbp),%eax
  cltq
  shl    $0x3,%rax
  add    %rdx,%rax
  mov    -0x18(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,8),%rcx
  mov    -0x38(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rax),%rax
  mov    %rax,(%rdx)
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x10(%rbp),%eax
  cltq
  shl    $0x3,%rax
  add    %rax,%rdx
  mov    -0x18(%rbp),%eax
  cltq
  mov    %rax,(%rdx)
  addl   $0x1,-0x18(%rbp)
L55:
  mov    -0x18(%rbp),%eax
  cltq
  cmp    %rax,-0x40(%rbp)
  ja     L31
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void make_multi_box2(double **ser, long **box, long *list, unsigned long l, unsigned int bs, unsigned int dim, unsigned int emb, unsigned int del, double eps) {
  int i, x, y;
  int ib = bs - 1;
  long back = (emb - 1) * del;
  for (x = 0; x < bs; x++)
    for (y = 0; y < bs; y++)
      box[x][y] = -1;
  for (i = back; i < l; i++) {
    x = (int)(ser[0][i] / eps) & ib;
    y = (int)(ser[dim - 1][i - back] / eps) & ib;
    list[i] = box[x][y];
    box[x][y] = i;
  }
}
```