ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What is the c source code of the assembly code below?

```asm
<read_table>:
  endbr64
  push   %r12
  mov    %rdi,%rsi
  push   %rbp
  mov    %rdi,%rbp
  lea    D3(%rip),%rdi
  push   %rbx
  mov    $0x5f5e100,%ebx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  call   <printf@plt>
  imul   $0x293b497,%rbp,%rdx
  lea    0x0(%rbp,%rbp,4),%rax
  lea    (%rax,%rax,4),%rax
  shl    $0x2,%rax
  xor    %rbp,%rax
  add    $0x527692e,%rdx
  xor    %rdx,%rax
  mov    %rax,(%rsp)
  nopl   0x0(%rax,%rax,1)
L16:
  mov    %r12,%rdi
  call   <nextrand@plt>
  mov    %rbp,%rsi
  movslq %eax,%rdx
  mov    %eax,%ecx
  imul   $0x57619f1,%rdx,%rdx
  sar    $0x1f,%ecx
  sar    $0x27,%rdx
  sub    %ecx,%edx
  imul   $0x1770,%edx,%edx
  sub    %edx,%eax
  mov    %eax,%edi
  call   <test_lookup@plt>
  sub    $0x1,%rbx
  jne    L16
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D3
data_type: string

tool: D3: Running with %ld


assistant: ```c
void *read_table(void *val) {
  size_t id = (size_t)val;
  printf("Running with %ld\n", id);
  uint64_t rng = id ^ (id * 100) ^ ((id + 2) * 43234455);
  for (size_t i = 0; i < (100 * 100 * 100 * 100); i++) {
    test_lookup(nextrand(&rng) % (3000 * 2), id);
  }
  return 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
<read_table>:
  endbr64
  push   %r12
  mov    %rdi,%rsi
  push   %rbp
  mov    %rdi,%rbp
  lea    D0(%rip),%rdi
  push   %rbx
  mov    $0x5f5e100,%ebx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  call   <printf@plt>
  imul   $0x293b497,%rbp,%rdx
  lea    0x0(%rbp,%rbp,4),%rax
  lea    (%rax,%rax,4),%rax
  shl    $0x2,%rax
  xor    %rbp,%rax
  add    $0x527692e,%rdx
  xor    %rdx,%rax
  mov    %rax,(%rsp)
  nopl   0x0(%rax,%rax,1)
L3:
  mov    %r12,%rdi
  call   <nextrand@plt>
  mov    %rbp,%rsi
  movslq %eax,%rdx
  mov    %eax,%ecx
  imul   $0x57619f1,%rdx,%rdx
  sar    $0x1f,%ecx
  sar    $0x27,%rdx
  sub    %ecx,%edx
  imul   $0x1770,%edx,%edx
  sub    %edx,%eax
  mov    %eax,%edi
  call   <test_lookup@plt>
  sub    $0x1,%rbx
  jne    L3
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D0
type_guessed: string

tool: D0: Running with %ld


assistant: ```c
void *read_table(void *val) {
  size_t id = (size_t)val;
  printf("Running with %ld\n", id);
  uint64_t rng = id ^ (id * 100) ^ ((id + 2) * 43234455);
  for (size_t i = 0; i < (100 * 100 * 100 * 100); i++) {
    test_lookup(nextrand(&rng) % (3000 * 2), id);
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<geta>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0xc(%rbp)
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L1
  call   <__stack_chk_fail@plt>
L1:
  leave
  ret
```

assistant: ```c
int *geta() {
  int a = 0;
  return &a;
}
```
user: Converte this assembly snippet into its equivalent C code:

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

assistant: ```c
int *geta() {
  int a = 0;
  return &a;
}
```
user: What would the C source look like for the assembly code provided here?

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

assistant: ```c
int *geta() {
  int a = 0;
  return &a;
}
```
user: Translate the below assembly code into C programming language:

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

assistant: ```c
int *geta() {
  int a = 0;
  return &a;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movabs $0x622061206d612049,%rax
  mov    %rax,-0x2c(%rbp)
  movl   $0x2e796f,-0x24(%rbp)
  movl   $0x0,-0x30(%rbp)
  jmp    L37
L20:
  mov    -0x30(%rbp),%eax
  cltq
  lea    -0x2c(%rbp),%rdx
  add    %rdx,%rax
  mov    -0x30(%rbp),%edx
  movslq %edx,%rdx
  lea    -0x20(%rbp),%rcx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  addl   $0x1,-0x30(%rbp)
L37:
  mov    -0x30(%rbp),%eax
  cltq
  lea    -0x2c(%rbp),%rdx
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L20
  mov    -0x30(%rbp),%eax
  cltq
  lea    -0x20(%rbp),%rdx
  add    %rdx,%rax
  movb   $0x0,(%rax)
  lea    -0x2c(%rbp),%rax
  mov    %rax,%rsi
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D23(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x0,-0x30(%rbp)
  jmp    L32
L43:
  mov    -0x30(%rbp),%eax
  cltq
  movzbl -0x20(%rbp,%rax,1),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <putchar@plt>
  addl   $0x1,-0x30(%rbp)
L32:
  mov    -0x30(%rbp),%eax
  cltq
  movzbl -0x20(%rbp,%rax,1),%eax
  test   %al,%al
  jne    L43
  lea    -0x20(%rbp),%rax
  mov    %rax,%rsi
  lea    D22(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L36
  call   <__stack_chk_fail@plt>
L36:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D4
type_infered: string
name: disassemble_data
arguments: tag: D23
type_infered: string
name: disassemble_data
arguments: tag: D22
type_infered: string

tool: D4: string a is:%s


tool: D23: string b is:

tool: D22: 
%s

assistant: ```c
void main() {
  char a[] = "I am a boy.", b[20];
  int i;
  for (i = 0; *(a + i) != '\0'; i++) {
    *(b + i) = *(a + i);
  }
  *(b + i) = '\0';
  printf("string a is:%s\n", a);
  printf("string b is:");
  for (i = 0; b[i] != '\0'; i++) {
    printf("%c", b[i]);
  }
  printf("\n%s", b);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  movabs $0x622061206d612049,%rdx
  mov    %rdx,0x4(%rsp)
  movl   $0x2e796f,0xc(%rsp)
  test   %dl,%dl
  je     L27
L55:
  mov    %dl,0x10(%rsp,%rax,1)
  add    $0x1,%rax
  movzbl 0x4(%rsp,%rax,1),%edx
  test   %dl,%dl
  jne    L55
L20:
  movb   $0x0,0x10(%rsp,%rax,1)
  lea    0x4(%rsp),%rsi
  lea    D22(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D3(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movzbl 0x10(%rsp),%edi
  test   %dil,%dil
  je     L2
  lea    0x11(%rsp),%rbx
L51:
  movsbl %dil,%edi
  call   <putchar@plt>
  add    $0x1,%rbx
  movzbl -0x1(%rbx),%edi
  test   %dil,%dil
  jne    L51
L2:
  lea    0x10(%rsp),%rsi
  lea    D1(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L21
  add    $0x30,%rsp
  pop    %rbx
  ret
L27:
  mov    $0x0,%eax
  jmp    L20
L21:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D22
type_infered: string
name: parse_data
arguments: label: D3
type_infered: string
name: parse_data
arguments: label: D1
type_infered: string

tool: D22: string a is:%s


tool: D3: string b is:

tool: D1: 
%s

assistant: ```c
void main() {
  char a[] = "I am a boy.", b[20];
  int i;
  for (i = 0; *(a + i) != '\0'; i++) {
    *(b + i) = *(a + i);
  }
  *(b + i) = '\0';
  printf("string a is:%s\n", a);
  printf("string b is:");
  for (i = 0; b[i] != '\0'; i++) {
    printf("%c", b[i]);
  }
  printf("\n%s", b);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    $0x49,%edx
  push   %rbx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  movabs $0x622061206d612049,%rax
  movl   $0x2e796f,0xc(%rsp)
  lea    0x10(%rsp),%rbp
  lea    0x4(%rsp),%rsi
  mov    %rax,0x4(%rsp)
  xor    %eax,%eax
L11:
  mov    %dl,0x0(%rbp,%rax,1)
  add    $0x1,%rax
  movzbl (%rsi,%rax,1),%edx
  test   %dl,%dl
  jne    L11
  movb   $0x0,0x10(%rsp,%rax,1)
  lea    D5(%rip),%rdi
  xor    %eax,%eax
  lea    0x11(%rsp),%rbx
  call   <printf@plt>
  lea    D4(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  movsbl 0x10(%rsp),%edi
  test   %dil,%dil
  je     L22
L36:
  call   <putchar@plt>
  movsbl (%rbx),%edi
  add    $0x1,%rbx
  test   %dil,%dil
  jne    L36
L22:
  xor    %eax,%eax
  mov    %rbp,%rsi
  lea    D27(%rip),%rdi
  call   <printf@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D5
datatype: string
name: read_data
arguments: label: D4
datatype: string
name: read_data
arguments: label: D27
datatype: string

tool: D5: string a is:%s


tool: D4: string b is:

tool: D27: 
%s

assistant: ```c
void main() {
  char a[] = "I am a boy.", b[20];
  int i;
  for (i = 0; *(a + i) != '\0'; i++) {
    *(b + i) = *(a + i);
  }
  *(b + i) = '\0';
  printf("string a is:%s\n", a);
  printf("string b is:");
  for (i = 0; b[i] != '\0'; i++) {
    printf("%c", b[i]);
  }
  printf("\n%s", b);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  mov    $0x49,%edx
  push   %rbx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  movabs $0x622061206d612049,%rax
  movl   $0x2e796f,0xc(%rsp)
  lea    0x10(%rsp),%rbp
  lea    0x4(%rsp),%rsi
  mov    %rax,0x4(%rsp)
  xor    %eax,%eax
L15:
  mov    %dl,0x0(%rbp,%rax,1)
  add    $0x1,%rax
  movzbl (%rsi,%rax,1),%edx
  test   %dl,%dl
  jne    L15
  movb   $0x0,0x10(%rsp,%rax,1)
  lea    D28(%rip),%rdi
  xor    %eax,%eax
  lea    0x11(%rsp),%rbx
  call   <printf@plt>
  lea    D9(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  movsbl 0x10(%rsp),%edi
  test   %dil,%dil
  je     L33
L28:
  call   <putchar@plt>
  movsbl (%rbx),%edi
  add    $0x1,%rbx
  test   %dil,%dil
  jne    L28
L33:
  xor    %eax,%eax
  mov    %rbp,%rsi
  lea    D13(%rip),%rdi
  call   <printf@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L18
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  ret
L18:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D28
type_infered: string
name: parse_data
arguments: label: D9
type_infered: string
name: parse_data
arguments: label: D13
type_infered: string

tool: D28: string a is:%s


tool: D9: string b is:

tool: D13: 
%s

assistant: ```c
void main() {
  char a[] = "I am a boy.", b[20];
  int i;
  for (i = 0; *(a + i) != '\0'; i++) {
    *(b + i) = *(a + i);
  }
  *(b + i) = '\0';
  printf("string a is:%s\n", a);
  printf("string b is:");
  for (i = 0; b[i] != '\0'; i++) {
    printf("%c", b[i]);
  }
  printf("\n%s", b);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<nutate>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x70,%rsp
  mov    %rdi,-0x58(%rbp)
  mov    %rsi,-0x60(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x58(%rbp),%rax
  mov    (%rax),%rax
  lea    -0x50(%rbp),%rcx
  lea    -0x48(%rbp),%rdx
  mov    %rcx,%rsi
  mov    %rdx,%rdi
  movq   %rax,%xmm0
  call   <SinCos@plt>
  mov    -0x60(%rbp),%rax
  mov    (%rax),%rax
  lea    -0x40(%rbp),%rcx
  lea    -0x38(%rbp),%rdx
  mov    %rcx,%rsi
  mov    %rdx,%rdi
  movq   %rax,%xmm0
  call   <SinCos@plt>
  mov    $0x0,%eax
  call   <obliquity@plt>
  movq   %xmm0,%rax
  lea    -0x30(%rbp),%rcx
  lea    -0x28(%rbp),%rdx
  mov    %rcx,%rsi
  mov    %rdx,%rdi
  movq   %rax,%xmm0
  call   <SinCos@plt>
  movsd  -0x38(%rbp),%xmm0
  movsd  -0x40(%rbp),%xmm1
  divsd  %xmm1,%xmm0
  movsd  %xmm0,-0x20(%rbp)
  mov    $0x0,%eax
  call   <dEcLong@plt>
  movsd  -0x28(%rbp),%xmm2
  movsd  -0x48(%rbp),%xmm1
  mulsd  %xmm2,%xmm1
  movapd %xmm1,%xmm2
  mulsd  -0x20(%rbp),%xmm2
  movsd  -0x30(%rbp),%xmm1
  addsd  %xmm2,%xmm1
  mulsd  %xmm1,%xmm0
  movsd  %xmm0,-0x68(%rbp)
  mov    $0x0,%eax
  call   <dObliq@plt>
  movsd  -0x50(%rbp),%xmm1
  mulsd  %xmm1,%xmm0
  movapd %xmm0,%xmm1
  mulsd  -0x20(%rbp),%xmm1
  movsd  -0x68(%rbp),%xmm0
  subsd  %xmm1,%xmm0
  movsd  %xmm0,-0x18(%rbp)
  mov    $0x0,%eax
  call   <dEcLong@plt>
  movsd  -0x28(%rbp),%xmm2
  movsd  -0x50(%rbp),%xmm1
  mulsd  %xmm2,%xmm1
  mulsd  %xmm1,%xmm0
  movsd  %xmm0,-0x68(%rbp)
  mov    $0x0,%eax
  call   <dObliq@plt>
  movsd  -0x48(%rbp),%xmm1
  mulsd  %xmm1,%xmm0
  addsd  -0x68(%rbp),%xmm0
  movsd  %xmm0,-0x10(%rbp)
  mov    -0x58(%rbp),%rax
  movsd  (%rax),%xmm0
  addsd  -0x18(%rbp),%xmm0
  mov    -0x58(%rbp),%rax
  movsd  %xmm0,(%rax)
  mov    -0x60(%rbp),%rax
  movsd  (%rax),%xmm0
  addsd  -0x10(%rbp),%xmm0
  mov    -0x60(%rbp),%rax
  movsd  %xmm0,(%rax)
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L9
  call   <__stack_chk_fail@plt>
L9:
  leave
  ret
```

assistant: ```c
void nutate(double *RA, double *Dec) {
  double cosRA, sinRA, cosDec, sinDec, tanDec;
  double dRA1, dDec1;
  double cosOb, sinOb;
  SinCos(*RA, &sinRA, &cosRA);
  SinCos(*Dec, &sinDec, &cosDec);
  SinCos(obliquity(), &sinOb, &cosOb);
  tanDec = sinDec / cosDec;
  dRA1 = (dEcLong() * (cosOb + sinOb * sinRA * tanDec) - dObliq() * cosRA * tanDec);
  dDec1 = (dEcLong() * (sinOb * cosRA) + dObliq() * sinRA);
  *RA = (*RA + dRA1);
  *Dec = (*Dec + dDec1);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<nutate>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x58,%rsp
  mov    %rdi,%rbp
  mov    %rsi,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  lea    0x18(%rsp),%rsi
  lea    0x20(%rsp),%rdi
  movsd  0x0(%rbp),%xmm0
  call   <SinCos@plt>
  lea    0x28(%rsp),%rsi
  lea    0x30(%rsp),%rdi
  movsd  (%rbx),%xmm0
  call   <SinCos@plt>
  mov    $0x0,%eax
  call   <obliquity@plt>
  lea    0x38(%rsp),%rsi
  lea    0x40(%rsp),%rdi
  call   <SinCos@plt>
  movsd  0x30(%rsp),%xmm0
  divsd  0x28(%rsp),%xmm0
  movsd  %xmm0,(%rsp)
  mov    $0x0,%eax
  call   <dEcLong@plt>
  movapd %xmm0,%xmm1
  movsd  0x40(%rsp),%xmm0
  mulsd  0x20(%rsp),%xmm0
  mulsd  (%rsp),%xmm0
  addsd  0x38(%rsp),%xmm0
  mulsd  %xmm1,%xmm0
  movsd  %xmm0,0x8(%rsp)
  mov    $0x0,%eax
  call   <dObliq@plt>
  mulsd  0x18(%rsp),%xmm0
  mulsd  (%rsp),%xmm0
  movsd  0x8(%rsp),%xmm2
  subsd  %xmm0,%xmm2
  movsd  %xmm2,0x8(%rsp)
  mov    $0x0,%eax
  call   <dEcLong@plt>
  movapd %xmm0,%xmm1
  movsd  0x40(%rsp),%xmm0
  mulsd  0x18(%rsp),%xmm0
  mulsd  %xmm1,%xmm0
  movsd  %xmm0,(%rsp)
  mov    $0x0,%eax
  call   <dObliq@plt>
  mulsd  0x20(%rsp),%xmm0
  addsd  (%rsp),%xmm0
  movsd  0x8(%rsp),%xmm1
  addsd  0x0(%rbp),%xmm1
  movsd  %xmm1,0x0(%rbp)
  addsd  (%rbx),%xmm0
  movsd  %xmm0,(%rbx)
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x58,%rsp
  pop    %rbx
  pop    %rbp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void nutate(double *RA, double *Dec) {
  double cosRA, sinRA, cosDec, sinDec, tanDec;
  double dRA1, dDec1;
  double cosOb, sinOb;
  SinCos(*RA, &sinRA, &cosRA);
  SinCos(*Dec, &sinDec, &cosDec);
  SinCos(obliquity(), &sinOb, &cosOb);
  tanDec = sinDec / cosDec;
  dRA1 = (dEcLong() * (cosOb + sinOb * sinRA * tanDec) - dObliq() * cosRA * tanDec);
  dDec1 = (dEcLong() * (sinOb * cosRA) + dObliq() * sinRA);
  *RA = (*RA + dRA1);
  *Dec = (*Dec + dDec1);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<nutate>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x58,%rsp
  movsd  0x0(%rbp),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  lea    0x18(%rsp),%rsi
  lea    0x20(%rsp),%rdi
  call   <SinCos@plt>
  movsd  (%rbx),%xmm0
  lea    0x28(%rsp),%rsi
  lea    0x30(%rsp),%rdi
  call   <SinCos@plt>
  xor    %eax,%eax
  call   <obliquity@plt>
  lea    0x38(%rsp),%rsi
  lea    0x40(%rsp),%rdi
  call   <SinCos@plt>
  xor    %eax,%eax
  movsd  0x30(%rsp),%xmm0
  divsd  0x28(%rsp),%xmm0
  movsd  %xmm0,(%rsp)
  call   <dEcLong@plt>
  xor    %eax,%eax
  movapd %xmm0,%xmm1
  movsd  0x40(%rsp),%xmm0
  mulsd  0x20(%rsp),%xmm0
  mulsd  (%rsp),%xmm0
  addsd  0x38(%rsp),%xmm0
  mulsd  %xmm0,%xmm1
  movsd  %xmm1,0x8(%rsp)
  call   <dObliq@plt>
  mulsd  0x18(%rsp),%xmm0
  movsd  0x8(%rsp),%xmm1
  xor    %eax,%eax
  mulsd  (%rsp),%xmm0
  subsd  %xmm0,%xmm1
  movsd  %xmm1,0x8(%rsp)
  call   <dEcLong@plt>
  xor    %eax,%eax
  movapd %xmm0,%xmm2
  movsd  0x40(%rsp),%xmm0
  mulsd  0x18(%rsp),%xmm0
  mulsd  %xmm2,%xmm0
  movsd  %xmm0,(%rsp)
  call   <dObliq@plt>
  movsd  0x8(%rsp),%xmm1
  addsd  0x0(%rbp),%xmm1
  mulsd  0x20(%rsp),%xmm0
  addsd  (%rsp),%xmm0
  movsd  %xmm1,0x0(%rbp)
  addsd  (%rbx),%xmm0
  movsd  %xmm0,(%rbx)
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x58,%rsp
  pop    %rbx
  pop    %rbp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void nutate(double *RA, double *Dec) {
  double cosRA, sinRA, cosDec, sinDec, tanDec;
  double dRA1, dDec1;
  double cosOb, sinOb;
  SinCos(*RA, &sinRA, &cosRA);
  SinCos(*Dec, &sinDec, &cosDec);
  SinCos(obliquity(), &sinOb, &cosOb);
  tanDec = sinDec / cosDec;
  dRA1 = (dEcLong() * (cosOb + sinOb * sinRA * tanDec) - dObliq() * cosRA * tanDec);
  dDec1 = (dEcLong() * (sinOb * cosRA) + dObliq() * sinRA);
  *RA = (*RA + dRA1);
  *Dec = (*Dec + dDec1);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<nutate>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x58,%rsp
  movsd  0x0(%rbp),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  lea    0x18(%rsp),%rsi
  lea    0x20(%rsp),%rdi
  call   <SinCos@plt>
  movsd  (%rbx),%xmm0
  lea    0x28(%rsp),%rsi
  lea    0x30(%rsp),%rdi
  call   <SinCos@plt>
  xor    %eax,%eax
  call   <obliquity@plt>
  lea    0x38(%rsp),%rsi
  lea    0x40(%rsp),%rdi
  call   <SinCos@plt>
  xor    %eax,%eax
  movsd  0x30(%rsp),%xmm0
  divsd  0x28(%rsp),%xmm0
  movsd  %xmm0,(%rsp)
  call   <dEcLong@plt>
  xor    %eax,%eax
  movapd %xmm0,%xmm1
  movsd  0x40(%rsp),%xmm0
  mulsd  0x20(%rsp),%xmm0
  mulsd  (%rsp),%xmm0
  addsd  0x38(%rsp),%xmm0
  mulsd  %xmm0,%xmm1
  movsd  %xmm1,0x8(%rsp)
  call   <dObliq@plt>
  mulsd  0x18(%rsp),%xmm0
  movsd  0x8(%rsp),%xmm1
  xor    %eax,%eax
  mulsd  (%rsp),%xmm0
  subsd  %xmm0,%xmm1
  movsd  %xmm1,0x8(%rsp)
  call   <dEcLong@plt>
  xor    %eax,%eax
  movapd %xmm0,%xmm2
  movsd  0x40(%rsp),%xmm0
  mulsd  0x18(%rsp),%xmm0
  mulsd  %xmm2,%xmm0
  movsd  %xmm0,(%rsp)
  call   <dObliq@plt>
  movsd  0x8(%rsp),%xmm1
  addsd  0x0(%rbp),%xmm1
  mulsd  0x20(%rsp),%xmm0
  addsd  (%rsp),%xmm0
  movsd  %xmm1,0x0(%rbp)
  addsd  (%rbx),%xmm0
  movsd  %xmm0,(%rbx)
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0x58,%rsp
  pop    %rbx
  pop    %rbp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void nutate(double *RA, double *Dec) {
  double cosRA, sinRA, cosDec, sinDec, tanDec;
  double dRA1, dDec1;
  double cosOb, sinOb;
  SinCos(*RA, &sinRA, &cosRA);
  SinCos(*Dec, &sinDec, &cosDec);
  SinCos(obliquity(), &sinOb, &cosOb);
  tanDec = sinDec / cosDec;
  dRA1 = (dEcLong() * (cosOb + sinOb * sinRA * tanDec) - dObliq() * cosRA * tanDec);
  dDec1 = (dEcLong() * (sinOb * cosRA) + dObliq() * sinRA);
  *RA = (*RA + dRA1);
  *Dec = (*Dec + dDec1);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<usage>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0xffffffff,%edi
  call   <exit@plt>
  nop
  pop    %rbp
  ret
```

tool_calls: name: read_data
arguments: data_tag: D2
datatype: string

tool: D2: Usage:
flocksplit path len


assistant: ```c
void usage() {
  printf("Usage:\n"
         "flocksplit path len\n");
  exit(-1);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<usage>:
  endbr64
  sub    $0x8,%rsp
  lea    D9(%rip),%rdi
  call   <printf@plt>
  mov    $0xffffffff,%edi
  call   <exit@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D9
type: string

tool: D9: Usage:
flocksplit path len


assistant: ```c
void usage() {
  printf("Usage:\n"
         "flocksplit path len\n");
  exit(-1);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<usage>:
  endbr64
  sub    $0x8,%rsp
  lea    D8(%rip),%rdi
  call   <printf@plt>
  mov    $0xffffffff,%edi
  add    $0x8,%rsp
  jmp    1070 <exit@plt>
```

tool_calls: name: parse_data
arguments: tag: D8
data_type: string

tool: D8: Usage:
flocksplit path len


assistant: ```c
void usage() {
  printf("Usage:\n"
         "flocksplit path len\n");
  exit(-1);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<usage>:
  endbr64
  sub    $0x8,%rsp
  lea    D1(%rip),%rdi
  call   <printf@plt>
  mov    $0xffffffff,%edi
  add    $0x8,%rsp
  jmp    1070 <exit@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D1
type: string

tool: D1: Usage:
flocksplit path len


assistant: ```c
void usage() {
  printf("Usage:\n"
         "flocksplit path len\n");
  exit(-1);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x62626161,-0xf(%rbp)
  movw   $0x6363,-0xb(%rbp)
  movb   $0x0,-0x9(%rbp)
  lea    -0xf(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x14(%rbp)
  mov    -0x14(%rbp),%ecx
  lea    -0xf(%rbp),%rax
  mov    $0x1,%edx
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <rearrange@plt>
  lea    -0xf(%rbp),%rax
  mov    %rax,%rsi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L3
  call   <__stack_chk_fail@plt>
L3:
  leave
  ret
```

tool_calls: name: parse_data
arguments: label: D5
data_type: string

tool: D5: %s


assistant: ```c
void main() {
  char str[] = "aabbcc";
  int len = strlen(str);
  rearrange(str, len, 1);
  printf("%s\n", str);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0x62626161,0x1(%rsp)
  movw   $0x6363,0x5(%rsp)
  movb   $0x0,0x7(%rsp)
  lea    0x1(%rsp),%rbx
  mov    %rbx,%rdi
  call   <strlen@plt>
  mov    %eax,%esi
  mov    $0x1,%edx
  mov    %rbx,%rdi
  call   <rearrange@plt>
  mov    %rbx,%rsi
  lea    D2(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  add    $0x10,%rsp
  pop    %rbx
  ret
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D2
datatype: string

tool: D2: %s


assistant: ```c
void main() {
  char str[] = "aabbcc";
  int len = strlen(str);
  rearrange(str, len, 1);
  printf("%s\n", str);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x1(%rsp),%rbp
  mov    $0x6363,%eax
  movl   $0x62626161,0x1(%rsp)
  mov    %rbp,%rdi
  mov    %ax,0x5(%rsp)
  movb   $0x0,0x7(%rsp)
  call   <strlen@plt>
  mov    %rbp,%rdi
  mov    $0x1,%edx
  mov    %eax,%esi
  call   <rearrange@plt>
  mov    %rbp,%rsi
  lea    D9(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x10,%rsp
  pop    %rbp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

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

tool: D9: %s


assistant: ```c
void main() {
  char str[] = "aabbcc";
  int len = strlen(str);
  rearrange(str, len, 1);
  printf("%s\n", str);
}
```