ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<symodd>:
  endbr64
  movslq %edi,%rcx
  mov    %rsi,%r8
  shl    $0x2,%rcx
  lea    (%rsi,%rcx,1),%rsi
  cmp    %rsi,%r8
  jae    L4
  lea    0x4(%r8),%rdi
  mov    %rdx,%rax
  lea    -0x1(%rcx),%r9
  sub    %rdi,%rax
  cmp    $0x8,%rax
  jbe    L24
  cmp    $0xb,%r9
  jbe    L24
  mov    %r9,%r10
  xor    %eax,%eax
  shr    $0x2,%r10
  add    $0x1,%r10
  mov    %r10,%rdi
  shr    $0x2,%rdi
  shl    $0x4,%rdi
  nopl   (%rax)
L106:
  movups (%r8,%rax,1),%xmm1
  movups %xmm1,(%rdx,%rax,1)
  add    $0x10,%rax
  cmp    %rax,%rdi
  jne    L106
  mov    %r10,%r11
  and    $0xfffffffffffffffc,%r11
  lea    0x0(,%r11,4),%rax
  lea    (%r8,%rax,1),%rdi
  add    %rdx,%rax
  cmp    %r11,%r10
  je     L112
  movss  (%rdi),%xmm0
  lea    0x4(%rdi),%r10
  movss  %xmm0,(%rax)
  cmp    %r10,%rsi
  jbe    L112
  movss  0x4(%rdi),%xmm0
  lea    0x8(%rdi),%r10
  movss  %xmm0,0x4(%rax)
  cmp    %r10,%rsi
  jbe    L112
  movss  0x8(%rdi),%xmm0
  movss  %xmm0,0x8(%rax)
L112:
  shr    $0x2,%r9
  lea    0x4(%rdx,%r9,4),%rdx
L4:
  sub    $0x8,%rsi
  cmp    %rsi,%r8
  jae    L127
  lea    -0x9(%rcx),%rdi
  mov    %rdi,%rax
  shr    $0x2,%rax
  lea    0x1(%rax),%r9
  neg    %rax
  lea    -0x8(%rcx,%rax,4),%rax
  lea    (%rdx,%r9,4),%r10
  add    %r8,%rax
  cmp    %rax,%r10
  lea    -0x4(%r8,%rcx,1),%r10
  setbe  %al
  cmp    %r10,%rdx
  setae  %r10b
  or     %r10b,%al
  je     L10
  cmp    $0xb,%rdi
  jbe    L10
  mov    %r9,%r10
  lea    -0x14(%r8,%rcx,1),%rax
  mov    %rdx,%rcx
  shr    $0x2,%r10
  mov    %rax,%rdi
  shl    $0x4,%r10
  sub    %r10,%rdi
  nopl   (%rax)
L73:
  movups (%rax),%xmm0
  sub    $0x10,%rax
  add    $0x10,%rcx
  shufps $0x1b,%xmm0,%xmm0
  movups %xmm0,-0x10(%rcx)
  cmp    %rdi,%rax
  jne    L73
  mov    %r9,%rcx
  and    $0xfffffffffffffffc,%rcx
  mov    %rcx,%rax
  neg    %rax
  shl    $0x2,%rax
  add    %rax,%rsi
  sub    %rax,%rdx
  cmp    %rcx,%r9
  je     L127
  movss  (%rsi),%xmm0
  lea    -0x4(%rsi),%rax
  movss  %xmm0,(%rdx)
  cmp    %rax,%r8
  jae    L127
  movss  -0x4(%rsi),%xmm0
  lea    -0x8(%rsi),%rax
  movss  %xmm0,0x4(%rdx)
  cmp    %rax,%r8
  jae    L127
  movss  -0x8(%rsi),%xmm0
  movss  %xmm0,0x8(%rdx)
  ret
  nopl   0x0(%rax,%rax,1)
L10:
  movss  (%rsi),%xmm0
  sub    $0x4,%rsi
  add    $0x4,%rdx
  movss  %xmm0,-0x4(%rdx)
  cmp    %rsi,%r8
  jb     L10
L127:
  ret
  nopw   0x0(%rax,%rax,1)
L24:
  mov    %rdx,%rdi
  mov    %r8,%rax
  cs nopw 0x0(%rax,%rax,1)
L91:
  movss  (%rax),%xmm0
  add    $0x4,%rax
  add    $0x4,%rdi
  movss  %xmm0,-0x4(%rdi)
  cmp    %rax,%rsi
  ja     L91
  jmp    L112
```

assistant: ```c
void symodd(n, v, u) int n;
float v[], u[];
{
  float *vp, *up, *vn = v + n;
  for (vp = v, up = u; vp < vn; vp++, up++)
    *up = *vp;
  for (vp = vn - 2; vp > v; vp--, up++)
    *up = *vp;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<map_key_RND_to_SP>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  and    $0x30,%eax
  test   %eax,%eax
  je     L100
  mov    -0x4(%rbp),%eax
  and    $0x4,%eax
  test   %eax,%eax
  jne    L103
  mov    -0x4(%rbp),%eax
  and    $0x1,%eax
  test   %eax,%eax
  jne    L20
  mov    -0x4(%rbp),%eax
  and    $0x8,%eax
  test   %eax,%eax
  jne    L140
  mov    -0x4(%rbp),%eax
  and    $0x2,%eax
  test   %eax,%eax
  je     L143
  mov    $0x8,%eax
  jmp    L43
L143:
  mov    $0x9,%eax
  jmp    L43
L140:
  mov    $0x7,%eax
  jmp    L43
L20:
  mov    $0x6,%eax
  jmp    L43
L103:
  mov    $0x5,%eax
  jmp    L43
L100:
  mov    -0x4(%rbp),%eax
  and    $0x4,%eax
  test   %eax,%eax
  jne    L144
  mov    -0x4(%rbp),%eax
  and    $0x1,%eax
  test   %eax,%eax
  jne    L71
  mov    -0x4(%rbp),%eax
  and    $0x8,%eax
  test   %eax,%eax
  jne    L146
  mov    -0x4(%rbp),%eax
  add    %eax,%eax
  and    $0x4,%eax
  jmp    L43
L146:
  mov    $0x3,%eax
  jmp    L43
L71:
  mov    $0x2,%eax
  jmp    L43
L144:
  mov    $0x1,%eax
L43:
  pop    %rbp
  ret
```

assistant: ```c
int map_key_RND_to_SP(int key) {
  if (key & ((1 << 4) | (1 << 5))) {
    return (key & (1 << 2) ? (5) : key & (1 << 0) ? (6) : key & (1 << 3) ? (7) : key & (1 << 1) ? (8) : (9));
  } else {
    return (key & (1 << 2) ? (1) : key & (1 << 0) ? (2) : key & (1 << 3) ? (3) : key & (1 << 1) ? (4) : (0));
  }
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<map_key_RND_to_SP>:
  endbr64
  test   $0x30,%dil
  je     L0
  mov    $0x5,%eax
  test   $0x4,%dil
  jne    L1
  mov    $0x6,%eax
  test   $0x1,%dil
  jne    L1
  mov    $0x7,%eax
  test   $0x8,%dil
  jne    L1
  test   $0x2,%dil
  sete   %al
  movzbl %al,%eax
  add    $0x8,%eax
  ret
L0:
  mov    $0x1,%eax
  test   $0x4,%dil
  jne    L1
  mov    $0x2,%eax
  test   $0x1,%dil
  jne    L1
  lea    (%rdi,%rdi,1),%eax
  and    $0x4,%eax
  test   $0x8,%dil
  mov    $0x3,%edx
  cmovne %edx,%eax
L1:
  ret
```

assistant: ```c
int map_key_RND_to_SP(int key) {
  if (key & ((1 << 4) | (1 << 5))) {
    return (key & (1 << 2) ? (5) : key & (1 << 0) ? (6) : key & (1 << 3) ? (7) : key & (1 << 1) ? (8) : (9));
  } else {
    return (key & (1 << 2) ? (1) : key & (1 << 0) ? (2) : key & (1 << 3) ? (3) : key & (1 << 1) ? (4) : (0));
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<map_key_RND_to_SP>:
  endbr64
  mov    %edi,%edx
  and    $0x4,%edx
  test   $0x30,%dil
  je     L14
  mov    $0x5,%eax
  test   %edx,%edx
  jne    L39
  mov    $0x6,%eax
  test   $0x1,%dil
  jne    L39
  mov    $0x7,%eax
  test   $0x8,%dil
  jne    L39
  xor    %eax,%eax
  and    $0x2,%edi
  sete   %al
  add    $0x8,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L14:
  mov    $0x1,%eax
  test   %edx,%edx
  je     L52
L39:
  ret
  nopw   0x0(%rax,%rax,1)
L52:
  mov    $0x2,%eax
  test   $0x1,%dil
  jne    L39
  lea    (%rdi,%rdi,1),%eax
  mov    $0x3,%edx
  and    $0x4,%eax
  and    $0x8,%edi
  cmovne %edx,%eax
  ret
```

assistant: ```c
int map_key_RND_to_SP(int key) {
  if (key & ((1 << 4) | (1 << 5))) {
    return (key & (1 << 2) ? (5) : key & (1 << 0) ? (6) : key & (1 << 3) ? (7) : key & (1 << 1) ? (8) : (9));
  } else {
    return (key & (1 << 2) ? (1) : key & (1 << 0) ? (2) : key & (1 << 3) ? (3) : key & (1 << 1) ? (4) : (0));
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<map_key_RND_to_SP>:
  endbr64
  mov    %edi,%edx
  and    $0x4,%edx
  test   $0x30,%dil
  je     L44
  mov    $0x5,%eax
  test   %edx,%edx
  jne    L50
  mov    $0x6,%eax
  test   $0x1,%dil
  jne    L50
  mov    $0x7,%eax
  test   $0x8,%dil
  jne    L50
  xor    %eax,%eax
  and    $0x2,%edi
  sete   %al
  add    $0x8,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L44:
  mov    $0x1,%eax
  test   %edx,%edx
  je     L23
L50:
  ret
  nopw   0x0(%rax,%rax,1)
L23:
  mov    $0x2,%eax
  test   $0x1,%dil
  jne    L50
  lea    (%rdi,%rdi,1),%eax
  mov    $0x3,%edx
  and    $0x4,%eax
  and    $0x8,%edi
  cmovne %edx,%eax
  ret
```

assistant: ```c
int map_key_RND_to_SP(int key) {
  if (key & ((1 << 4) | (1 << 5))) {
    return (key & (1 << 2) ? (5) : key & (1 << 0) ? (6) : key & (1 << 3) ? (7) : key & (1 << 1) ? (8) : (9));
  } else {
    return (key & (1 << 2) ? (1) : key & (1 << 0) ? (2) : key & (1 << 3) ? (3) : key & (1 << 1) ? (4) : (0));
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %edi,-0x44(%rbp)
  mov    %rsi,-0x50(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x30(%rbp),%rax
  mov    $0x28,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  movl   $0x0,-0x34(%rbp)
  jmp    L5
L1:
  lea    -0x30(%rbp),%rax
  mov    %rax,%rdi
  call   <foo1@plt>
  lea    -0x30(%rbp),%rax
  mov    %rax,%rdi
  call   <foo2@plt>
  addl   $0x1,-0x34(%rbp)
L5:
  cmpl   $0x9,-0x34(%rbp)
  jle    L1
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L19
  call   <__stack_chk_fail@plt>
L19:
  leave
  ret
```

assistant: ```c
int main(int argc, char *argv) {
  int v[10];
  memset(v, 0, sizeof(v));
  for (int j = 0; j < 10; j++) {
    foo1(&v[0]);
    foo2(v);
  }
#pragma omp taskwait
  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   %rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  mov    $0x28,%edx
  mov    $0x0,%esi
  call   <memset@plt>
  mov    $0xa,%ebx
  mov    %rsp,%rbp
L7:
  mov    %rbp,%rdi
  call   <foo1@plt>
  mov    %rbp,%rdi
  call   <foo2@plt>
  sub    $0x1,%ebx
  jne    L7
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  mov    $0x0,%eax
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(int argc, char *argv) {
  int v[10];
  memset(v, 0, sizeof(v));
  for (int j = 0; j < 10; j++) {
    foo1(&v[0]);
    foo2(v);
  }
#pragma omp taskwait
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    $0x28,%edx
  xor    %esi,%esi
  push   %rbx
  mov    $0xa,%ebx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%rdi
  call   <memset@plt>
  nopl   0x0(%rax)
L7:
  mov    %rbp,%rdi
  call   <foo1@plt>
  mov    %rbp,%rdi
  call   <foo2@plt>
  sub    $0x1,%ebx
  jne    L7
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L17
  add    $0x38,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L17:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(int argc, char *argv) {
  int v[10];
  memset(v, 0, sizeof(v));
  for (int j = 0; j < 10; j++) {
    foo1(&v[0]);
    foo2(v);
  }
#pragma omp taskwait
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    $0x28,%edx
  xor    %esi,%esi
  push   %rbx
  mov    $0xa,%ebx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%rdi
  call   <memset@plt>
  nopl   0x0(%rax)
L10:
  mov    %rbp,%rdi
  call   <foo1@plt>
  mov    %rbp,%rdi
  call   <foo2@plt>
  sub    $0x1,%ebx
  jne    L10
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L12
  add    $0x38,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L12:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(int argc, char *argv) {
  int v[10];
  memset(v, 0, sizeof(v));
  for (int j = 0; j < 10; j++) {
    foo1(&v[0]);
    foo2(v);
  }
#pragma omp taskwait
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %edi,-0x34(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x64,-0x28(%rbp)
  movss  D15(%rip),%xmm0
  movss  %xmm0,-0x24(%rbp)
  movsd  D42(%rip),%xmm0
  movsd  %xmm0,-0x20(%rbp)
  movb   $0x41,-0x29(%rbp)
  movl   $0x6e686f4a,-0x14(%rbp)
  movb   $0x0,-0x10(%rbp)
  movl   $0x766c6953,-0xf(%rbp)
  movw   $0x7265,-0xb(%rbp)
  movb   $0x0,-0x9(%rbp)
  mov    -0x28(%rbp),%eax
  mov    %eax,%esi
  lea    D36(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  pxor   %xmm1,%xmm1
  cvtss2sd -0x24(%rbp),%xmm1
  movq   %xmm1,%rax
  movq   %rax,%xmm0
  lea    D44(%rip),%rax
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    -0x20(%rbp),%rax
  movq   %rax,%xmm0
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  movsbl -0x29(%rbp),%eax
  mov    %eax,%esi
  lea    D50(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x14(%rbp),%rax
  mov    %rax,%rsi
  lea    D54(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0xf(%rbp),%rax
  mov    %rax,%rsi
  lea    D62(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movsbl -0x29(%rbp),%edx
  lea    -0xf(%rbp),%rcx
  lea    -0x14(%rbp),%rax
  mov    %rax,%rsi
  lea    D79(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L7
  call   <__stack_chk_fail@plt>
L7:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D15
type: byte[4]
name: unpack_data
arguments: data_label: D42
type: f64
name: unpack_data
arguments: data_label: D36
type: string
name: unpack_data
arguments: data_label: D44
type: string
name: unpack_data
arguments: data_label: D3
type: string
name: unpack_data
arguments: data_label: D50
type: string
name: unpack_data
arguments: data_label: D54
type: string
name: unpack_data
arguments: data_label: D62
type: string
name: unpack_data
arguments: data_label: D79
type: string

tool: {"D15": [0xb6,0xf3,0x9d,0x3f]}

tool: D42: 5467.7894

tool: D36: %d miles away


tool: D44: %f power left


tool: D3: %f amount of super powers


tool: D50: %c is an initial


tool: D54: %s is my first name


tool: D62: %s is my last name 


tool: D79: My whole name is %s %c %s.


assistant: ```c
int main(int argc, char *argv[]) {
  int distance = 100;
  float power = 1.234f;
  double superPower = 5467.7894;
  char initial = 'A';
  char firstName[] = "John";
  char lastName[] = "Silver";
  printf("%d miles away\n", distance);
  printf("%f power left\n", power);
  printf("%f amount of super powers\n", superPower);
  printf("%c is an initial\n", initial);
  printf("%s is my first name\n", firstName);
  printf("%s is my last name \n", lastName);
  printf("My whole name is %s %c %s.\n", firstName, initial, lastName);
  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   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movl   $0x6e686f4a,0xc(%rsp)
  movb   $0x0,0x10(%rsp)
  movl   $0x766c6953,0x11(%rsp)
  movw   $0x7265,0x15(%rsp)
  movb   $0x0,0x17(%rsp)
  mov    $0x64,%esi
  lea    D36(%rip),%rdi
  call   <printf@plt>
  movsd  D72(%rip),%xmm0
  lea    D69(%rip),%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  movsd  D35(%rip),%xmm0
  lea    D85(%rip),%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    $0x41,%esi
  lea    D14(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0xc(%rsp),%rbx
  mov    %rbx,%rsi
  lea    D46(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0x11(%rsp),%rbp
  mov    %rbp,%rsi
  lea    D23(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %rbp,%rcx
  mov    $0x41,%edx
  mov    %rbx,%rsi
  lea    D79(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  mov    $0x0,%eax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D36
type_infered: string
name: disassemble_data
arguments: label: D72
type_infered: byte[8]
name: disassemble_data
arguments: label: D69
type_infered: string
name: disassemble_data
arguments: label: D35
type_infered: f64
name: disassemble_data
arguments: label: D85
type_infered: string
name: disassemble_data
arguments: label: D14
type_infered: string
name: disassemble_data
arguments: label: D46
type_infered: string
name: disassemble_data
arguments: label: D23
type_infered: string
name: disassemble_data
arguments: label: D79
type_infered: string

tool: D36: %d miles away


tool: {"D72": [0x00,0x00,0x00,0xc0,0x76,0xbe,0xf3,0x3f]}

tool: D69: %f power left


tool: D35: 5467.7894

tool: D85: %f amount of super powers


tool: D14: %c is an initial


tool: D46: %s is my first name


tool: D23: %s is my last name 


tool: D79: My whole name is %s %c %s.


assistant: ```c
int main(int argc, char *argv[]) {
  int distance = 100;
  float power = 1.234f;
  double superPower = 5467.7894;
  char initial = 'A';
  char firstName[] = "John";
  char lastName[] = "Silver";
  printf("%d miles away\n", distance);
  printf("%f power left\n", power);
  printf("%f amount of super powers\n", superPower);
  printf("%c is an initial\n", initial);
  printf("%s is my first name\n", firstName);
  printf("%s is my last name \n", lastName);
  printf("My whole name is %s %c %s.\n", firstName, initial, lastName);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %r12
  mov    $0x7265,%edx
  mov    $0x64,%esi
  lea    D53(%rip),%rdi
  push   %rbp
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %dx,0x15(%rsp)
  lea    0xc(%rsp),%rbp
  lea    0x11(%rsp),%r12
  movl   $0x6e686f4a,0xc(%rsp)
  movb   $0x0,0x10(%rsp)
  movl   $0x766c6953,0x11(%rsp)
  movb   $0x0,0x17(%rsp)
  call   <printf@plt>
  movsd  D56(%rip),%xmm0
  mov    $0x1,%eax
  lea    D20(%rip),%rdi
  call   <printf@plt>
  movsd  D18(%rip),%xmm0
  mov    $0x1,%eax
  lea    D64(%rip),%rdi
  call   <printf@plt>
  mov    $0x41,%esi
  lea    D68(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rsi
  lea    D51(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r12,%rsi
  lea    D2(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  mov    %r12,%rcx
  mov    $0x41,%edx
  mov    %rbp,%rsi
  lea    D16(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D53
type_guessed: string
name: read_data
arguments: data_tag: D56
type_guessed: byte[8]
name: read_data
arguments: data_tag: D20
type_guessed: string
name: read_data
arguments: data_tag: D18
type_guessed: f64
name: read_data
arguments: data_tag: D64
type_guessed: string
name: read_data
arguments: data_tag: D68
type_guessed: string
name: read_data
arguments: data_tag: D51
type_guessed: string
name: read_data
arguments: data_tag: D2
type_guessed: string
name: read_data
arguments: data_tag: D16
type_guessed: string

tool: D53: %d miles away


tool: {"D56": [0x00,0x00,0x00,0xc0,0x76,0xbe,0xf3,0x3f]}

tool: D20: %f power left


tool: D18: 5467.7894

tool: D64: %f amount of super powers


tool: D68: %c is an initial


tool: D51: %s is my first name


tool: D2: %s is my last name 


tool: D16: My whole name is %s %c %s.


assistant: ```c
int main(int argc, char *argv[]) {
  int distance = 100;
  float power = 1.234f;
  double superPower = 5467.7894;
  char initial = 'A';
  char firstName[] = "John";
  char lastName[] = "Silver";
  printf("%d miles away\n", distance);
  printf("%f power left\n", power);
  printf("%f amount of super powers\n", superPower);
  printf("%c is an initial\n", initial);
  printf("%s is my first name\n", firstName);
  printf("%s is my last name \n", lastName);
  printf("My whole name is %s %c %s.\n", firstName, initial, lastName);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  push   %r12
  mov    $0x7265,%edx
  mov    $0x64,%esi
  lea    D81(%rip),%rdi
  push   %rbp
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %dx,0x15(%rsp)
  lea    0xc(%rsp),%rbp
  lea    0x11(%rsp),%r12
  movl   $0x6e686f4a,0xc(%rsp)
  movb   $0x0,0x10(%rsp)
  movl   $0x766c6953,0x11(%rsp)
  movb   $0x0,0x17(%rsp)
  call   <printf@plt>
  movsd  D47(%rip),%xmm0
  mov    $0x1,%eax
  lea    D0(%rip),%rdi
  call   <printf@plt>
  movsd  D71(%rip),%xmm0
  mov    $0x1,%eax
  lea    D84(%rip),%rdi
  call   <printf@plt>
  mov    $0x41,%esi
  lea    D23(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rsi
  lea    D69(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r12,%rsi
  lea    D44(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  mov    %r12,%rcx
  mov    $0x41,%edx
  mov    %rbp,%rsi
  lea    D86(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D81
type_infered: string
name: parse_data
arguments: tag: D47
type_infered: byte[8]
name: parse_data
arguments: tag: D0
type_infered: string
name: parse_data
arguments: tag: D71
type_infered: f64
name: parse_data
arguments: tag: D84
type_infered: string
name: parse_data
arguments: tag: D23
type_infered: string
name: parse_data
arguments: tag: D69
type_infered: string
name: parse_data
arguments: tag: D44
type_infered: string
name: parse_data
arguments: tag: D86
type_infered: string

tool: D81: %d miles away


tool: {"D47": [0x00,0x00,0x00,0xc0,0x76,0xbe,0xf3,0x3f]}

tool: D0: %f power left


tool: D71: 5467.7894

tool: D84: %f amount of super powers


tool: D23: %c is an initial


tool: D69: %s is my first name


tool: D44: %s is my last name 


tool: D86: My whole name is %s %c %s.


assistant: ```c
int main(int argc, char *argv[]) {
  int distance = 100;
  float power = 1.234f;
  double superPower = 5467.7894;
  char initial = 'A';
  char firstName[] = "John";
  char lastName[] = "Silver";
  printf("%d miles away\n", distance);
  printf("%f power left\n", power);
  printf("%f amount of super powers\n", superPower);
  printf("%c is an initial\n", initial);
  printf("%s is my first name\n", firstName);
  printf("%s is my last name \n", lastName);
  printf("My whole name is %s %c %s.\n", firstName, initial, lastName);
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<getbaud>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0xffffffffffffffff,-0x10(%rbp)
  lea    -0x18(%rbp),%rdx
  mov    -0x24(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <tcgetattr@plt>
  lea    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <cfgetispeed@plt>
  mov    %eax,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  add    $0xffffff80,%eax
  cmp    $0x10,%eax
  ja     L101
  mov    %eax,%eax
  lea    0x0(,%rax,4),%rdx
  lea    D0(%rip),%rax
  mov    (%rdx,%rax,1),%eax
  cltq
  lea    D0(%rip),%rdx
  add    %rdx,%rax
  notrack jmp *%rax
  movq   $0x0,-0x10(%rbp)
  jmp    L101
  movq   $0x32,-0x10(%rbp)
  jmp    L101
  movq   $0x6e,-0x10(%rbp)
  jmp    L101
  movq   $0x86,-0x10(%rbp)
  jmp    L101
  movq   $0x96,-0x10(%rbp)
  jmp    L101
  movq   $0xc8,-0x10(%rbp)
  jmp    L101
  movq   $0x12c,-0x10(%rbp)
  jmp    L101
  movq   $0x258,-0x10(%rbp)
  jmp    L101
  movq   $0x4b0,-0x10(%rbp)
  jmp    L101
  movq   $0x708,-0x10(%rbp)
  jmp    L101
  movq   $0x960,-0x10(%rbp)
  jmp    L101
  movq   $0x12c0,-0x10(%rbp)
  jmp    L101
  movq   $0x2580,-0x10(%rbp)
  jmp    L101
  movq   $0x4b00,-0x10(%rbp)
  jmp    L101
  movq   $0x9600,-0x10(%rbp)
  jmp    L101
  movq   $0xe100,-0x10(%rbp)
  jmp    L101
  movq   $0x1c200,-0x10(%rbp)
  nop
L101:
  mov    -0x10(%rbp),%rax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L63
  call   <__stack_chk_fail@plt>
L63:
  leave
  ret
```

assistant: ```c
long int getbaud(int fd) {
  struct termios termAttr;
  long int inputSpeed = -1;
  speed_t baudRate;
  tcgetattr(fd, &termAttr);
  baudRate = cfgetispeed(&termAttr);
  switch (baudRate) {
  case B0:
    inputSpeed = 0;
    break;
  case B50:
    inputSpeed = 50;
    break;
  case B110:
    inputSpeed = 110;
    break;
  case B134:
    inputSpeed = 134;
    break;
  case B150:
    inputSpeed = 150;
    break;
  case B200:
    inputSpeed = 200;
    break;
  case B300:
    inputSpeed = 300;
    break;
  case B600:
    inputSpeed = 600;
    break;
  case B1200:
    inputSpeed = 1200;
    break;
  case B1800:
    inputSpeed = 1800;
    break;
  case B2400:
    inputSpeed = 2400;
    break;
  case B4800:
    inputSpeed = 4800;
    break;
  case B9600:
    inputSpeed = 9600;
    break;
  case B19200:
    inputSpeed = 19200;
    break;
  case B38400:
    inputSpeed = 38400;
    break;
  case B57600:
    inputSpeed = 57600;
    break;
  case B115200:
    inputSpeed = 115200;
    break;
  }
  return inputSpeed;
}
```