ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将以下的汇编代码转换成C语言代码：

```asm
<get_word>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  movl   $0xff,-0x14(%rbp)
  nop
L21:
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <getc@plt>
  mov    %eax,%ebx
  cmp    $0x20,%ebx
  jg     L126
  mov    D18(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,%ebx
  je     L126
  mov    $0x1,%eax
  jmp    L104
L126:
  mov    $0x0,%eax
L104:
  test   %eax,%eax
  jne    L21
  mov    D18(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,%ebx
  je     L28
L66:
  mov    -0x30(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x30(%rbp)
  mov    %ebx,%edx
  mov    %dl,(%rax)
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <getc@plt>
  mov    %eax,%ebx
  cmp    $0x20,%ebx
  jle    L137
  mov    D18(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,%ebx
  je     L137
  subl   $0x1,-0x14(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  je     L137
  mov    $0x1,%eax
  jmp    L99
L137:
  mov    $0x0,%eax
L99:
  test   %eax,%eax
  jne    L66
L28:
  mov    -0x30(%rbp),%rax
  movb   $0x0,(%rax)
  mov    -0x28(%rbp),%rax
  mov    %rax,%rsi
  mov    %ebx,%edi
  call   <ungetc@plt>
  mov    D18(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,%ebx
  jne    L74
  mov    $0x0,%eax
  jmp    L130
L74:
  cmpl   $0x0,-0x14(%rbp)
  je     L133
  mov    $0x1,%eax
  jmp    L130
L133:
  mov    $0x3,%eax
L130:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int get_word(FILE *in, char *w) {
  int bufsize = 255;
  register int c;
  while (c = getc(in), c <= 32 && c != EOF)
    ;
  if (c != EOF) {
    do {
      *w++ = (char)(c);
    } while (c = getc(in), c > 32 && c != EOF && --bufsize);
  }
  *w = '\0';
  ungetc(c, in);
  if (c == EOF)
    return 0;
  if (bufsize)
    return 1;
  return 3;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<get_word>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r13
  mov    %rsi,%r12
L85:
  mov    %r13,%rdi
  call   <getc@plt>
  mov    %eax,%ebx
  cmp    $0x20,%eax
  jg     L30
  mov    D9(%rip),%rax
  cmp    %ebx,(%rax)
  jne    L85
  mov    $0xff,%ebp
L20:
  movb   $0x0,(%r12)
  mov    %r13,%rsi
  mov    %ebx,%edi
  call   <ungetc@plt>
  mov    $0x0,%eax
  mov    D9(%rip),%rdx
  cmp    %ebx,(%rdx)
  je     L66
  cmp    $0x1,%ebp
  sbb    %eax,%eax
  and    $0x2,%eax
  add    $0x1,%eax
L66:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L30:
  mov    D9(%rip),%rax
  cmp    %ebx,(%rax)
  je     L38
  mov    $0xff,%ebp
L28:
  add    $0x1,%r12
  mov    %bl,-0x1(%r12)
  mov    %r13,%rdi
  call   <getc@plt>
  mov    %eax,%ebx
  cmp    $0x20,%eax
  jle    L20
  mov    D9(%rip),%rax
  cmp    %ebx,(%rax)
  je     L20
  sub    $0x1,%ebp
  jne    L28
  jmp    L20
L38:
  mov    $0xff,%ebp
  jmp    L20
```

assistant: ```c
int get_word(FILE *in, char *w) {
  int bufsize = 255;
  register int c;
  while (c = getc(in), c <= 32 && c != EOF)
    ;
  if (c != EOF) {
    do {
      *w++ = (char)(c);
    } while (c = getc(in), c > 32 && c != EOF && --bufsize);
  }
  *w = '\0';
  ungetc(c, in);
  if (c == EOF)
    return 0;
  if (bufsize)
    return 1;
  return 3;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<get_word>:
  endbr64
  push   %r14
  mov    %rdi,%r14
  push   %r13
  mov    D8(%rip),%r13
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  jmp    L58
  nopl   0x0(%rax,%rax,1)
L70:
  cmp    %eax,0x0(%r13)
  je     L11
L58:
  mov    %r14,%rdi
  call   <getc@plt>
  mov    %eax,%ebx
  cmp    $0x20,%eax
  jle    L70
  mov    $0xff,%r12d
  cmp    %eax,0x0(%r13)
  jne    L55
  jmp    L23
  nopl   0x0(%rax,%rax,1)
L95:
  cmp    %eax,0x0(%r13)
  je     L23
  sub    $0x1,%r12d
  je     L23
L55:
  mov    %bl,0x0(%rbp)
  mov    %r14,%rdi
  add    $0x1,%rbp
  call   <getc@plt>
  mov    %eax,%ebx
  cmp    $0x20,%eax
  jg     L95
L23:
  movb   $0x0,0x0(%rbp)
  mov    %r14,%rsi
  mov    %ebx,%edi
  call   <ungetc@plt>
  xor    %eax,%eax
  cmp    %ebx,0x0(%r13)
  je     L92
  cmp    $0x1,%r12d
  sbb    %eax,%eax
  and    $0x2,%eax
  add    $0x1,%eax
L92:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   (%rax)
L11:
  mov    $0xff,%r12d
  jmp    L23
```

assistant: ```c
int get_word(FILE *in, char *w) {
  int bufsize = 255;
  register int c;
  while (c = getc(in), c <= 32 && c != EOF)
    ;
  if (c != EOF) {
    do {
      *w++ = (char)(c);
    } while (c = getc(in), c > 32 && c != EOF && --bufsize);
  }
  *w = '\0';
  ungetc(c, in);
  if (c == EOF)
    return 0;
  if (bufsize)
    return 1;
  return 3;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<get_word>:
  endbr64
  push   %r14
  mov    %rdi,%r14
  push   %r13
  mov    D8(%rip),%r13
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  jmp    L33
  nopl   0x0(%rax,%rax,1)
L53:
  cmp    %eax,0x0(%r13)
  je     L20
L33:
  mov    %r14,%rdi
  call   <getc@plt>
  mov    %eax,%ebx
  cmp    $0x20,%eax
  jle    L53
  mov    $0xff,%r12d
  cmp    %eax,0x0(%r13)
  jne    L11
  jmp    L95
  nopl   0x0(%rax,%rax,1)
L89:
  cmp    %eax,0x0(%r13)
  je     L95
  sub    $0x1,%r12d
  je     L95
L11:
  mov    %bl,0x0(%rbp)
  mov    %r14,%rdi
  add    $0x1,%rbp
  call   <getc@plt>
  mov    %eax,%ebx
  cmp    $0x20,%eax
  jg     L89
L95:
  movb   $0x0,0x0(%rbp)
  mov    %r14,%rsi
  mov    %ebx,%edi
  call   <ungetc@plt>
  xor    %eax,%eax
  cmp    %ebx,0x0(%r13)
  je     L73
  cmp    $0x1,%r12d
  sbb    %eax,%eax
  and    $0x2,%eax
  add    $0x1,%eax
L73:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   (%rax)
L20:
  mov    $0xff,%r12d
  jmp    L95
```

assistant: ```c
int get_word(FILE *in, char *w) {
  int bufsize = 255;
  register int c;
  while (c = getc(in), c <= 32 && c != EOF)
    ;
  if (c != EOF) {
    do {
      *w++ = (char)(c);
    } while (c = getc(in), c > 32 && c != EOF && --bufsize);
  }
  *w = '\0';
  ungetc(c, in);
  if (c == EOF)
    return 0;
  if (bufsize)
    return 1;
  return 3;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<grmhd_disp_df>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x8,%rsp
  movsd  %xmm0,-0x48(%rbp)
  movsd  %xmm1,-0x50(%rbp)
  movsd  %xmm2,-0x58(%rbp)
  movsd  %xmm3,-0x60(%rbp)
  movsd  %xmm4,-0x68(%rbp)
  movsd  %xmm5,-0x70(%rbp)
  movsd  %xmm6,-0x78(%rbp)
  movsd  %xmm7,-0x80(%rbp)
  movsd  -0x48(%rbp),%xmm0
  subsd  -0x78(%rbp),%xmm0
  movsd  %xmm0,-0x40(%rbp)
  movsd  -0x80(%rbp),%xmm0
  mulsd  -0x48(%rbp),%xmm0
  subsd  0x10(%rbp),%xmm0
  movsd  %xmm0,-0x38(%rbp)
  movsd  -0x48(%rbp),%xmm0
  addsd  -0x70(%rbp),%xmm0
  movsd  %xmm0,-0x30(%rbp)
  movsd  -0x50(%rbp),%xmm1
  movsd  D32(%rip),%xmm0
  mulsd  %xmm1,%xmm0
  mulsd  -0x40(%rbp),%xmm0
  mulsd  -0x40(%rbp),%xmm0
  movsd  -0x40(%rbp),%xmm1
  mulsd  %xmm1,%xmm0
  movsd  %xmm0,-0x28(%rbp)
  movsd  -0x30(%rbp),%xmm0
  movapd %xmm0,%xmm1
  mulsd  %xmm0,%xmm1
  movsd  -0x58(%rbp),%xmm0
  subsd  %xmm1,%xmm0
  movsd  %xmm0,-0x20(%rbp)
  movsd  -0x30(%rbp),%xmm1
  movsd  D35(%rip),%xmm0
  mulsd  %xmm1,%xmm0
  movsd  %xmm0,-0x18(%rbp)
  movsd  -0x60(%rbp),%xmm0
  movq   D24(%rip),%xmm1
  xorpd  %xmm1,%xmm0
  mulsd  -0x40(%rbp),%xmm0
  movapd %xmm0,%xmm1
  mulsd  -0x40(%rbp),%xmm1
  movsd  -0x68(%rbp),%xmm0
  mulsd  -0x38(%rbp),%xmm0
  mulsd  -0x38(%rbp),%xmm0
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x10(%rbp)
  movsd  -0x60(%rbp),%xmm1
  movsd  D35(%rip),%xmm0
  mulsd  %xmm1,%xmm0
  movapd %xmm0,%xmm1
  mulsd  -0x40(%rbp),%xmm1
  movsd  -0x68(%rbp),%xmm0
  addsd  %xmm0,%xmm0
  mulsd  -0x80(%rbp),%xmm0
  mulsd  -0x38(%rbp),%xmm0
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movsd  -0x18(%rbp),%xmm0
  mulsd  -0x10(%rbp),%xmm0
  movapd %xmm0,%xmm1
  addsd  -0x28(%rbp),%xmm1
  movsd  -0x20(%rbp),%xmm0
  mulsd  -0x8(%rbp),%xmm0
  addsd  %xmm1,%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D32
type_infered: f64
name: unpack_data
arguments: data_tag: D35
type_infered: byte[8]
name: unpack_data
arguments: data_tag: D24
type_infered: byte[8]

tool: D32: 4

tool: {"D35": [0x00,0x00,0x00,0x00,0x00,0x00,0x00,0xc0]}

tool: {"D24": [0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x80]}

assistant: ```c
double grmhd_disp_df(double v, double C1, double C2, double C3, double C4, double bei, double vi, double b0, double bi) {
  double a = v - vi;
  double B = b0 * v - bi;
  double n = v + bei;
  double df1 = 4 * C1 * a * a * a;
  double f2 = C2 - n * n;
  double df2 = -2 * n;
  double f3 = -C3 * a * a + C4 * B * B;
  double df3 = -2 * C3 * a + 2 * C4 * b0 * B;
  return df1 + df2 * f3 + f2 * df3;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<grmhd_disp_df>:
  endbr64
  movapd %xmm0,%xmm8
  subsd  %xmm6,%xmm8
  movapd %xmm0,%xmm6
  mulsd  %xmm7,%xmm6
  subsd  0x8(%rsp),%xmm6
  addsd  %xmm0,%xmm5
  movapd %xmm3,%xmm0
  xorpd  D6(%rip),%xmm0
  mulsd  %xmm8,%xmm0
  mulsd  %xmm8,%xmm0
  movapd %xmm6,%xmm9
  mulsd  %xmm4,%xmm9
  mulsd  %xmm6,%xmm9
  addsd  %xmm9,%xmm0
  movsd  D25(%rip),%xmm9
  movapd %xmm5,%xmm10
  mulsd  %xmm9,%xmm10
  mulsd  %xmm10,%xmm0
  mulsd  D22(%rip),%xmm1
  mulsd  %xmm8,%xmm1
  mulsd  %xmm8,%xmm1
  mulsd  %xmm8,%xmm1
  addsd  %xmm1,%xmm0
  addsd  %xmm4,%xmm4
  mulsd  %xmm7,%xmm4
  mulsd  %xmm6,%xmm4
  mulsd  %xmm9,%xmm3
  mulsd  %xmm8,%xmm3
  addsd  %xmm3,%xmm4
  mulsd  %xmm5,%xmm5
  subsd  %xmm5,%xmm2
  mulsd  %xmm2,%xmm4
  addsd  %xmm4,%xmm0
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D6
type_infered: byte[16]
name: extract_symbolic_data
arguments: data_tag: D25
type_infered: byte[8]
name: extract_symbolic_data
arguments: data_tag: D22
type_infered: f64

tool: {"D6": [0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x80,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00]}

tool: {"D25": [0x00,0x00,0x00,0x00,0x00,0x00,0x00,0xc0]}

tool: D22: 4

assistant: ```c
double grmhd_disp_df(double v, double C1, double C2, double C3, double C4, double bei, double vi, double b0, double bi) {
  double a = v - vi;
  double B = b0 * v - bi;
  double n = v + bei;
  double df1 = 4 * C1 * a * a * a;
  double f2 = C2 - n * n;
  double df2 = -2 * n;
  double f3 = -C3 * a * a + C4 * B * B;
  double df3 = -2 * C3 * a + 2 * C4 * b0 * B;
  return df1 + df2 * f3 + f2 * df3;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<grmhd_disp_df>:
  endbr64
  movapd %xmm0,%xmm8
  addsd  %xmm0,%xmm5
  subsd  %xmm6,%xmm8
  movapd %xmm0,%xmm6
  movapd %xmm3,%xmm0
  mulsd  %xmm7,%xmm6
  xorpd  D5(%rip),%xmm0
  subsd  0x8(%rsp),%xmm6
  mulsd  D19(%rip),%xmm1
  movapd %xmm5,%xmm10
  mulsd  %xmm8,%xmm0
  mulsd  %xmm5,%xmm5
  movapd %xmm6,%xmm9
  mulsd  %xmm4,%xmm9
  addsd  %xmm4,%xmm4
  mulsd  %xmm8,%xmm1
  mulsd  %xmm8,%xmm0
  mulsd  %xmm7,%xmm4
  subsd  %xmm5,%xmm2
  mulsd  %xmm6,%xmm9
  mulsd  %xmm8,%xmm1
  mulsd  %xmm6,%xmm4
  addsd  %xmm9,%xmm0
  movsd  D28(%rip),%xmm9
  mulsd  %xmm8,%xmm1
  mulsd  %xmm9,%xmm3
  mulsd  %xmm9,%xmm10
  mulsd  %xmm8,%xmm3
  mulsd  %xmm10,%xmm0
  addsd  %xmm3,%xmm4
  addsd  %xmm1,%xmm0
  mulsd  %xmm2,%xmm4
  addsd  %xmm4,%xmm0
  ret
```

tool_calls: name: parse_data
arguments: data_label: D5
data_type: byte[16]
name: parse_data
arguments: data_label: D19
data_type: f64
name: parse_data
arguments: data_label: D28
data_type: byte[8]

tool: {"D5": [0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x80,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00]}

tool: D19: 4

tool: {"D28": [0x00,0x00,0x00,0x00,0x00,0x00,0x00,0xc0]}

assistant: ```c
double grmhd_disp_df(double v, double C1, double C2, double C3, double C4, double bei, double vi, double b0, double bi) {
  double a = v - vi;
  double B = b0 * v - bi;
  double n = v + bei;
  double df1 = 4 * C1 * a * a * a;
  double f2 = C2 - n * n;
  double df2 = -2 * n;
  double f3 = -C3 * a * a + C4 * B * B;
  double df3 = -2 * C3 * a + 2 * C4 * b0 * B;
  return df1 + df2 * f3 + f2 * df3;
}
```
user: What is the c source code of the assembly code below:

```asm
<grmhd_disp_df>:
  endbr64
  movapd %xmm0,%xmm8
  addsd  %xmm0,%xmm5
  subsd  %xmm6,%xmm8
  movapd %xmm0,%xmm6
  movapd %xmm3,%xmm0
  mulsd  %xmm7,%xmm6
  xorpd  D19(%rip),%xmm0
  subsd  0x8(%rsp),%xmm6
  mulsd  D2(%rip),%xmm1
  movapd %xmm5,%xmm10
  mulsd  %xmm8,%xmm0
  mulsd  %xmm5,%xmm5
  movapd %xmm6,%xmm9
  mulsd  %xmm4,%xmm9
  addsd  %xmm4,%xmm4
  mulsd  %xmm8,%xmm1
  mulsd  %xmm8,%xmm0
  mulsd  %xmm7,%xmm4
  subsd  %xmm5,%xmm2
  mulsd  %xmm6,%xmm9
  mulsd  %xmm8,%xmm1
  mulsd  %xmm6,%xmm4
  addsd  %xmm9,%xmm0
  movsd  D14(%rip),%xmm9
  mulsd  %xmm8,%xmm1
  mulsd  %xmm9,%xmm3
  mulsd  %xmm9,%xmm10
  mulsd  %xmm8,%xmm3
  mulsd  %xmm10,%xmm0
  addsd  %xmm3,%xmm4
  addsd  %xmm1,%xmm0
  mulsd  %xmm2,%xmm4
  addsd  %xmm4,%xmm0
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D19
type_guessed: byte[16]
name: disassemble_data
arguments: data_label: D2
type_guessed: f64
name: disassemble_data
arguments: data_label: D14
type_guessed: byte[8]

tool: {"D19": [0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x80,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00]}

tool: D2: 4

tool: {"D14": [0x00,0x00,0x00,0x00,0x00,0x00,0x00,0xc0]}

assistant: ```c
double grmhd_disp_df(double v, double C1, double C2, double C3, double C4, double bei, double vi, double b0, double bi) {
  double a = v - vi;
  double B = b0 * v - bi;
  double n = v + bei;
  double df1 = 4 * C1 * a * a * a;
  double f2 = C2 - n * n;
  double df2 = -2 * n;
  double f3 = -C3 * a * a + C4 * B * B;
  double df3 = -2 * C3 * a + 2 * C4 * b0 * B;
  return df1 + df2 * f3 + f2 * df3;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D29(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  jmp    L16
L15:
  mov    -0x10(%rbp),%edx
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%ecx
  mov    %ecx,%eax
  shl    $0x2,%eax
  add    %ecx,%eax
  add    %eax,%eax
  sub    %eax,%edx
  mov    %edx,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x10(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x10(%rbp)
L16:
  mov    -0x10(%rbp),%eax
  test   %eax,%eax
  jne    L15
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L19
  call   <__stack_chk_fail@plt>
L19:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D29
type_guessed: string
name: retrieve_data
arguments: data_label: D5
type_guessed: string

tool: D29: Enter your number

tool: D5: %i

assistant: ```c
int main(void) {
  int number, right_digit;
  printf("Enter your number\n");
  scanf("%i", &number);
  while (number != 0) {
    right_digit = number % 10;
    printf("%i", right_digit);
    number = number / 10;
  }
  printf("\n");
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D18(%rip),%rdi
  call   <puts@plt>
  lea    0x4(%rsp),%rsi
  lea    D24(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%esi
  test   %esi,%esi
  je     L0
  lea    D24(%rip),%rbx
L10:
  movslq %esi,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  mov    %esi,%edx
  sar    $0x1f,%edx
  sub    %edx,%eax
  lea    (%rax,%rax,4),%eax
  add    %eax,%eax
  sub    %eax,%esi
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x4(%rsp),%eax
  movslq %eax,%rsi
  imul   $0x66666667,%rsi,%rsi
  sar    $0x22,%rsi
  sar    $0x1f,%eax
  sub    %eax,%esi
  mov    %esi,0x4(%rsp)
  jne    L10
L0:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  mov    $0x0,%eax
  add    $0x10,%rsp
  pop    %rbx
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D18
type_guessed: string
name: disassemble_data
arguments: label: D24
type_guessed: string

tool: D18: Enter your number

tool: D24: %i

assistant: ```c
int main(void) {
  int number, right_digit;
  printf("Enter your number\n");
  scanf("%i", &number);
  while (number != 0) {
    right_digit = number % 10;
    printf("%i", right_digit);
    number = number / 10;
  }
  printf("\n");
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbx
  lea    D15(%rip),%rdi
  lea    D2(%rip),%rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  lea    0x4(%rsp),%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%esi
  test   %esi,%esi
  je     L10
  nopl   0x0(%rax)
L16:
  movslq %esi,%rax
  mov    %esi,%edx
  mov    %rbx,%rdi
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%edx
  sar    $0x22,%rax
  sub    %edx,%eax
  lea    (%rax,%rax,4),%eax
  add    %eax,%eax
  sub    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  movslq 0x4(%rsp),%rsi
  mov    %rsi,%rax
  imul   $0x66666667,%rsi,%rsi
  sar    $0x1f,%eax
  sar    $0x22,%rsi
  sub    %eax,%esi
  mov    %esi,0x4(%rsp)
  jne    L16
L10:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L20
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  ret
L20:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D15
type_guessed: string
name: disassemble_data
arguments: tag: D2
type_guessed: string

tool: D15: Enter your number

tool: D2: %i

assistant: ```c
int main(void) {
  int number, right_digit;
  printf("Enter your number\n");
  scanf("%i", &number);
  while (number != 0) {
    right_digit = number % 10;
    printf("%i", right_digit);
    number = number / 10;
  }
  printf("\n");
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  push   %rbx
  lea    D17(%rip),%rdi
  lea    D18(%rip),%rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  lea    0x4(%rsp),%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%esi
  test   %esi,%esi
  je     L5
  nopl   0x0(%rax)
L17:
  movslq %esi,%rax
  mov    %esi,%edx
  mov    %rbx,%rdi
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%edx
  sar    $0x22,%rax
  sub    %edx,%eax
  lea    (%rax,%rax,4),%eax
  add    %eax,%eax
  sub    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  movslq 0x4(%rsp),%rsi
  mov    %rsi,%rax
  imul   $0x66666667,%rsi,%rsi
  sar    $0x1f,%eax
  sar    $0x22,%rsi
  sub    %eax,%esi
  mov    %esi,0x4(%rsp)
  jne    L17
L5:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L16
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  ret
L16:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D17
type: string
name: disassemble_data
arguments: label: D18
type: string

tool: D17: Enter your number

tool: D18: %i

assistant: ```c
int main(void) {
  int number, right_digit;
  printf("Enter your number\n");
  scanf("%i", &number);
  while (number != 0) {
    right_digit = number % 10;
    printf("%i", right_digit);
    number = number / 10;
  }
  printf("\n");
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<test_negative>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0xfffffffd,-0x30(%rbp)
  movl   $0xfffffffe,-0x2c(%rbp)
  movl   $0xffffffff,-0x28(%rbp)
  movl   $0x0,-0x24(%rbp)
  movl   $0x1,-0x20(%rbp)
  movl   $0x2,-0x1c(%rbp)
  movl   $0x3,-0x18(%rbp)
  movl   $0x4,-0x14(%rbp)
  lea    -0x20(%rbp),%rdx
  lea    -0x30(%rbp),%rax
  mov    $0x4,%ecx
  mov    $0x4,%esi
  mov    %rax,%rdi
  call   <median@plt>
  movq   %xmm0,%rax
  movq   %rax,%xmm1
  ucomisd D11(%rip),%xmm1
  jp     L39
  movq   %rax,%xmm2
  ucomisd D11(%rip),%xmm2
  jne    L39
  lea    -0x30(%rbp),%rdx
  lea    -0x20(%rbp),%rax
  mov    $0x4,%ecx
  mov    $0x4,%esi
  mov    %rax,%rdi
  call   <median@plt>
  movq   %xmm0,%rax
  movq   %rax,%xmm3
  ucomisd D11(%rip),%xmm3
  jp     L39
  movq   %rax,%xmm4
  ucomisd D11(%rip),%xmm4
  jne    L39
  lea    -0x20(%rbp),%rcx
  lea    -0x30(%rbp),%rax
  mov    $0x4,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <median_opt@plt>
  movq   %xmm0,%rax
  movq   %rax,%xmm5
  ucomisd D11(%rip),%xmm5
  jp     L39
  movq   %rax,%xmm6
  ucomisd D11(%rip),%xmm6
  jne    L39
  lea    -0x30(%rbp),%rcx
  lea    -0x20(%rbp),%rax
  mov    $0x4,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <median_opt@plt>
  movq   %xmm0,%rax
  movq   %rax,%xmm7
  ucomisd D11(%rip),%xmm7
  jp     L39
  movq   %rax,%xmm1
  ucomisd D11(%rip),%xmm1
  jne    L39
  mov    $0x1,%eax
  jmp    L33
L39:
  mov    $0x0,%eax
L33:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L86
  call   <__stack_chk_fail@plt>
L86:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: label: D11
type_guessed: f64

tool: D11: 0.5

assistant: ```c
int test_negative() {
  int a[] = {-3, -2, -1, 0};
  int b[] = {1, 2, 3, 4};
  return 0.5 == median(a, 4, b, 4) && 0.5 == median(b, 4, a, 4) && 0.5 == median_opt(a, b, 4) && 0.5 == median_opt(b, a, 4);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<test_negative>:
  endbr64
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  movl   $0xfffffffd,(%rsp)
  movl   $0xfffffffe,0x4(%rsp)
  movl   $0xffffffff,0x8(%rsp)
  movl   $0x0,0xc(%rsp)
  movl   $0x1,0x10(%rsp)
  movl   $0x2,0x14(%rsp)
  movl   $0x3,0x18(%rsp)
  movl   $0x4,0x1c(%rsp)
  lea    0x10(%rsp),%rdx
  mov    %rsp,%rdi
  mov    $0x4,%ecx
  mov    $0x4,%esi
  call   <median@plt>
  ucomisd D6(%rip),%xmm0
  jp     L96
  je     L98
L96:
  mov    $0x0,%eax
L17:
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L72
  add    $0x38,%rsp
  ret
L98:
  mov    %rsp,%rdx
  lea    0x10(%rsp),%rdi
  mov    $0x4,%ecx
  mov    $0x4,%esi
  call   <median@plt>
  ucomisd D6(%rip),%xmm0
  jp     L21
  je     L90
L21:
  mov    $0x0,%eax
  jmp    L17
L90:
  lea    0x10(%rsp),%rsi
  mov    %rsp,%rdi
  mov    $0x4,%edx
  call   <median_opt@plt>
  ucomisd D6(%rip),%xmm0
  jp     L64
  je     L25
L64:
  mov    $0x0,%eax
  jmp    L17
L25:
  mov    %rsp,%rsi
  lea    0x10(%rsp),%rdi
  mov    $0x4,%edx
  call   <median_opt@plt>
  ucomisd D6(%rip),%xmm0
  setnp  %al
  movzbl %al,%eax
  mov    $0x0,%edx
  cmovne %edx,%eax
  jmp    L17
L72:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D6
type_infered: f64

tool: D6: 0.5

assistant: ```c
int test_negative() {
  int a[] = {-3, -2, -1, 0};
  int b[] = {1, 2, 3, 4};
  return 0.5 == median(a, 4, b, 4) && 0.5 == median(b, 4, a, 4) && 0.5 == median_opt(a, b, 4) && 0.5 == median_opt(b, a, 4);
}
```