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

```asm
<adjoint33>:
  endbr64
  movsd  0x20(%rsi),%xmm8
  movsd  0x28(%rsi),%xmm6
  movsd  0x38(%rsi),%xmm5
  movsd  0x40(%rsi),%xmm3
  movapd %xmm6,%xmm10
  movapd %xmm8,%xmm9
  movsd  0x8(%rsi),%xmm4
  movsd  0x10(%rsi),%xmm2
  mulsd  %xmm5,%xmm10
  movsd  (%rsi),%xmm0
  movsd  0x18(%rsi),%xmm7
  mulsd  %xmm3,%xmm9
  movsd  0x30(%rsi),%xmm1
  subsd  %xmm10,%xmm9
  movapd %xmm4,%xmm10
  mulsd  %xmm3,%xmm10
  movsd  %xmm9,(%rdi)
  movapd %xmm2,%xmm9
  mulsd  %xmm5,%xmm9
  subsd  %xmm10,%xmm9
  movapd %xmm2,%xmm10
  mulsd  %xmm8,%xmm10
  movsd  %xmm9,0x8(%rdi)
  movapd %xmm4,%xmm9
  mulsd  %xmm6,%xmm9
  subsd  %xmm10,%xmm9
  movapd %xmm7,%xmm10
  mulsd  %xmm3,%xmm10
  mulsd  %xmm0,%xmm3
  movsd  %xmm9,0x10(%rdi)
  movapd %xmm6,%xmm9
  mulsd  %xmm1,%xmm9
  mulsd  %xmm0,%xmm6
  subsd  %xmm10,%xmm9
  movsd  %xmm9,0x18(%rdi)
  movapd %xmm2,%xmm9
  mulsd  %xmm1,%xmm9
  mulsd  %xmm7,%xmm2
  subsd  %xmm9,%xmm3
  subsd  %xmm6,%xmm2
  movsd  %xmm3,0x20(%rdi)
  movapd %xmm8,%xmm3
  movsd  %xmm2,0x28(%rdi)
  mulsd  %xmm1,%xmm3
  movapd %xmm7,%xmm2
  mulsd  %xmm5,%xmm2
  mulsd  %xmm4,%xmm1
  mulsd  %xmm0,%xmm5
  mulsd  %xmm7,%xmm4
  mulsd  %xmm8,%xmm0
  subsd  %xmm3,%xmm2
  subsd  %xmm5,%xmm1
  movsd  %xmm2,0x30(%rdi)
  subsd  %xmm4,%xmm0
  movsd  %xmm1,0x38(%rdi)
  movsd  %xmm0,0x40(%rdi)
  ret
```

assistant: ```c
void adjoint33(double madj[3][3], double m[3][3]) {
  double m00, m01, m02, m10, m11, m12, m20, m21, m22;
  m00 = m[0][0];
  m01 = m[0][1];
  m02 = m[0][2];
  m10 = m[1][0];
  m11 = m[1][1];
  m12 = m[1][2];
  m20 = m[2][0];
  m21 = m[2][1];
  m22 = m[2][2];
  madj[0][0] = m11 * m22 - m12 * m21;
  madj[0][1] = m02 * m21 - m01 * m22;
  madj[0][2] = m01 * m12 - m02 * m11;
  madj[1][0] = m12 * m20 - m10 * m22;
  madj[1][1] = m00 * m22 - m02 * m20;
  madj[1][2] = m02 * m10 - m00 * m12;
  madj[2][0] = m10 * m21 - m11 * m20;
  madj[2][1] = m01 * m20 - m00 * m21;
  madj[2][2] = m00 * m11 - m01 * m10;
}
```
user: Translate the below assembly code into C programming language:

```asm
<adjoint33>:
  endbr64
  movupd (%rsi),%xmm0
  movsd  0x28(%rsi),%xmm9
  movupd 0x20(%rsi),%xmm1
  movsd  0x30(%rsi),%xmm3
  movapd %xmm0,%xmm4
  movapd %xmm9,%xmm5
  movsd  0x40(%rsi),%xmm7
  movupd 0x10(%rsi),%xmm2
  unpcklpd %xmm3,%xmm5
  unpckhpd %xmm1,%xmm4
  movsd  0x38(%rsi),%xmm6
  mulpd  %xmm5,%xmm4
  movsd  0x20(%rsi),%xmm5
  movapd %xmm7,%xmm8
  unpcklpd %xmm6,%xmm8
  movapd %xmm6,%xmm10
  unpcklpd %xmm7,%xmm5
  unpcklpd %xmm7,%xmm10
  mulpd  %xmm2,%xmm5
  subpd  %xmm5,%xmm4
  movapd %xmm1,%xmm5
  unpcklpd %xmm2,%xmm5
  mulpd  %xmm8,%xmm5
  movapd %xmm1,%xmm8
  unpckhpd %xmm0,%xmm8
  movups %xmm4,0x10(%rdi)
  movapd %xmm2,%xmm4
  mulpd  %xmm10,%xmm8
  movsd  %xmm7,%xmm4
  movapd %xmm3,%xmm7
  unpcklpd %xmm9,%xmm7
  subpd  %xmm8,%xmm5
  movups %xmm5,(%rdi)
  movapd %xmm0,%xmm5
  unpcklpd %xmm2,%xmm5
  mulpd  %xmm5,%xmm4
  movapd %xmm2,%xmm5
  unpcklpd %xmm0,%xmm5
  mulpd  %xmm7,%xmm5
  subpd  %xmm5,%xmm4
  movapd %xmm6,%xmm5
  unpcklpd %xmm3,%xmm5
  unpcklpd %xmm6,%xmm3
  movups %xmm4,0x20(%rdi)
  movapd %xmm2,%xmm4
  unpckhpd %xmm2,%xmm2
  unpckhpd %xmm0,%xmm4
  mulpd  %xmm5,%xmm4
  movapd %xmm1,%xmm5
  mulsd  %xmm0,%xmm1
  unpcklpd %xmm0,%xmm5
  unpckhpd %xmm0,%xmm0
  mulpd  %xmm3,%xmm5
  mulsd  %xmm2,%xmm0
  movapd %xmm4,%xmm3
  subpd  %xmm5,%xmm3
  subsd  %xmm0,%xmm1
  movups %xmm3,0x30(%rdi)
  movsd  %xmm1,0x40(%rdi)
  ret
```

assistant: ```c
void adjoint33(double madj[3][3], double m[3][3]) {
  double m00, m01, m02, m10, m11, m12, m20, m21, m22;
  m00 = m[0][0];
  m01 = m[0][1];
  m02 = m[0][2];
  m10 = m[1][0];
  m11 = m[1][1];
  m12 = m[1][2];
  m20 = m[2][0];
  m21 = m[2][1];
  m22 = m[2][2];
  madj[0][0] = m11 * m22 - m12 * m21;
  madj[0][1] = m02 * m21 - m01 * m22;
  madj[0][2] = m01 * m12 - m02 * m11;
  madj[1][0] = m12 * m20 - m10 * m22;
  madj[1][1] = m00 * m22 - m02 * m20;
  madj[1][2] = m02 * m10 - m00 * m12;
  madj[2][0] = m10 * m21 - m11 * m20;
  madj[2][1] = m01 * m20 - m00 * m21;
  madj[2][2] = m00 * m11 - m01 * m10;
}
```
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   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  movl   $0x64,-0x10(%rbp)
  jmp    L6
L9:
  mov    -0x10(%rbp),%edx
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %edx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%ecx
  mov    %ecx,%eax
  shl    $0x2,%eax
  add    %ecx,%eax
  add    %eax,%eax
  sub    %eax,%edx
  mov    %edx,-0xc(%rbp)
  mov    -0x10(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  sub    %eax,%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,-0x8(%rbp)
  mov    -0x8(%rbp),%ecx
  mov    %ecx,%eax
  shl    $0x2,%eax
  add    %ecx,%eax
  add    %eax,%eax
  sub    %eax,%edx
  mov    %edx,-0x8(%rbp)
  mov    -0x10(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x51eb851f,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x5,%edx
  sar    $0x1f,%eax
  sub    %eax,%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,-0x4(%rbp)
  mov    -0x4(%rbp),%ecx
  mov    %ecx,%eax
  shl    $0x2,%eax
  add    %ecx,%eax
  add    %eax,%eax
  sub    %eax,%edx
  mov    %edx,-0x4(%rbp)
  mov    -0xc(%rbp),%eax
  imul   %eax,%eax
  imul   -0xc(%rbp),%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%eax
  imul   %eax,%eax
  imul   -0x8(%rbp),%eax
  add    %eax,%edx
  mov    -0x4(%rbp),%eax
  imul   %eax,%eax
  imul   -0x4(%rbp),%eax
  add    %edx,%eax
  cmp    %eax,-0x10(%rbp)
  jne    L24
  mov    -0x10(%rbp),%eax
  mov    %eax,%esi
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L24:
  addl   $0x1,-0x10(%rbp)
L6:
  cmpl   $0x3e7,-0x10(%rbp)
  jle    L9
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D9
data_type: string
name: extract_symbolic_data
arguments: tag: D3
data_type: string

tool: D9: water flower number (100<=n<1000) are :

tool: D3: %d 

assistant: ```c
int main(int argc, char **argv) {
  int i, j, k, n;
  printf("water flower number (100<=n<1000) are :\n");
  for (n = 100; n < 1000; ++n) {
    i = n % 10;
    j = (n / 10) % 10;
    k = ((n / 10) / 10) % 10;
    if ((i * i * i) + (j * j * j) + (k * k * k) == n) {
      printf("%d ", n);
    }
  }
  printf("\n");
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  lea    D7(%rip),%rdi
  call   <puts@plt>
  mov    $0x64,%ebx
  lea    D0(%rip),%rbp
L21:
  add    $0x1,%ebx
  cmp    $0x3e8,%ebx
  je     L16
  movslq %ebx,%rcx
  imul   $0x66666667,%rcx,%rsi
  sar    $0x22,%rsi
  mov    %ebx,%r8d
  sar    $0x1f,%r8d
  sub    %r8d,%esi
  mov    %esi,%edi
  lea    (%rsi,%rsi,4),%eax
  add    %eax,%eax
  mov    %ebx,%esi
  sub    %eax,%esi
  movslq %edi,%rdx
  imul   $0x66666667,%rdx,%rdx
  sar    $0x22,%rdx
  mov    %edi,%eax
  sar    $0x1f,%eax
  sub    %eax,%edx
  lea    (%rdx,%rdx,4),%eax
  add    %eax,%eax
  sub    %eax,%edi
  mov    %edi,%edx
  imul   $0x51eb851f,%rcx,%rcx
  sar    $0x25,%rcx
  sub    %r8d,%ecx
  movslq %ecx,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  mov    %ecx,%edi
  sar    $0x1f,%edi
  sub    %edi,%eax
  lea    (%rax,%rax,4),%edi
  add    %edi,%edi
  mov    %ecx,%eax
  sub    %edi,%eax
  mov    %esi,%ecx
  imul   %esi,%ecx
  imul   %esi,%ecx
  mov    %edx,%esi
  imul   %edx,%esi
  imul   %esi,%edx
  add    %ecx,%edx
  mov    %eax,%ecx
  imul   %eax,%ecx
  imul   %ecx,%eax
  add    %edx,%eax
  cmp    %ebx,%eax
  jne    L21
  mov    %ebx,%esi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L21
L16:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D7
type_infered: string
name: extract_symbolic_data
arguments: label: D0
type_infered: string

tool: D7: water flower number (100<=n<1000) are :

tool: D0: %d 

assistant: ```c
int main(int argc, char **argv) {
  int i, j, k, n;
  printf("water flower number (100<=n<1000) are :\n");
  for (n = 100; n < 1000; ++n) {
    i = n % 10;
    j = (n / 10) % 10;
    k = ((n / 10) / 10) % 10;
    if ((i * i * i) + (j * j * j) + (k * k * k) == n) {
      printf("%d ", n);
    }
  }
  printf("\n");
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %r12
  lea    D12(%rip),%rdi
  mov    $0x64,%r12d
  push   %rbp
  mov    $0xcccccccd,%ebp
  push   %rbx
  lea    D1(%rip),%rbx
  call   <puts@plt>
  cs nopw 0x0(%rax,%rax,1)
L6:
  add    $0x1,%r12d
  cmp    $0x3e8,%r12d
  je     L15
L25:
  mov    %r12d,%edx
  mov    %rdx,%rsi
  imul   $0x51eb851f,%rdx,%rdx
  imul   %rbp,%rsi
  shr    $0x25,%rdx
  shr    $0x23,%rsi
  lea    (%rsi,%rsi,4),%eax
  mov    %esi,%ecx
  mov    %r12d,%esi
  add    %eax,%eax
  sub    %eax,%esi
  mov    %ecx,%eax
  imul   %rbp,%rax
  shr    $0x23,%rax
  lea    (%rax,%rax,4),%edi
  mov    %ecx,%eax
  mov    %esi,%ecx
  imul   %esi,%ecx
  add    %edi,%edi
  sub    %edi,%eax
  imul   %esi,%ecx
  mov    %eax,%esi
  imul   %eax,%esi
  imul   %esi,%eax
  add    %ecx,%eax
  mov    %edx,%ecx
  imul   %edx,%ecx
  imul   %ecx,%edx
  add    %edx,%eax
  cmp    %r12d,%eax
  jne    L6
  mov    %r12d,%esi
  mov    %rbx,%rdi
  xor    %eax,%eax
  add    $0x1,%r12d
  call   <printf@plt>
  cmp    $0x3e8,%r12d
  jne    L25
L15:
  mov    $0xa,%edi
  call   <putchar@plt>
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: read_data
arguments: tag: D12
type_guessed: string
name: read_data
arguments: tag: D1
type_guessed: string

tool: D12: water flower number (100<=n<1000) are :

tool: D1: %d 

assistant: ```c
int main(int argc, char **argv) {
  int i, j, k, n;
  printf("water flower number (100<=n<1000) are :\n");
  for (n = 100; n < 1000; ++n) {
    i = n % 10;
    j = (n / 10) % 10;
    k = ((n / 10) / 10) % 10;
    if ((i * i * i) + (j * j * j) + (k * k * k) == n) {
      printf("%d ", n);
    }
  }
  printf("\n");
  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
<main>:
  endbr64
  push   %r12
  lea    D11(%rip),%rdi
  mov    $0x65,%r12d
  push   %rbp
  mov    $0xcccccccd,%ebp
  push   %rbx
  lea    D7(%rip),%rbx
  call   <puts@plt>
  cs nopw 0x0(%rax,%rax,1)
L26:
  mov    %r12d,%edx
  mov    %rdx,%rsi
  imul   $0x51eb851f,%rdx,%rdx
  imul   %rbp,%rsi
  shr    $0x25,%rdx
  shr    $0x23,%rsi
  lea    (%rsi,%rsi,4),%eax
  mov    %esi,%ecx
  mov    %r12d,%esi
  add    %eax,%eax
  sub    %eax,%esi
  mov    %ecx,%eax
  imul   %rbp,%rax
  shr    $0x23,%rax
  lea    (%rax,%rax,4),%edi
  mov    %ecx,%eax
  mov    %esi,%ecx
  imul   %esi,%ecx
  add    %edi,%edi
  sub    %edi,%eax
  imul   %esi,%ecx
  mov    %eax,%esi
  imul   %eax,%esi
  imul   %esi,%eax
  add    %ecx,%eax
  mov    %edx,%ecx
  imul   %edx,%ecx
  imul   %ecx,%edx
  add    %edx,%eax
  cmp    %r12d,%eax
  je     L4
  add    $0x1,%r12d
  cmp    $0x3e8,%r12d
  jne    L26
L11:
  mov    $0xa,%edi
  call   <putchar@plt>
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L4:
  mov    %r12d,%esi
  mov    %rbx,%rdi
  xor    %eax,%eax
  add    $0x1,%r12d
  call   <printf@plt>
  cmp    $0x3e8,%r12d
  jne    L26
  jmp    L11
```

tool_calls: name: disassemble_data
arguments: data_tag: D11
datatype: string
name: disassemble_data
arguments: data_tag: D7
datatype: string

tool: D11: water flower number (100<=n<1000) are :

tool: D7: %d 

assistant: ```c
int main(int argc, char **argv) {
  int i, j, k, n;
  printf("water flower number (100<=n<1000) are :\n");
  for (n = 100; n < 1000; ++n) {
    i = n % 10;
    j = (n / 10) % 10;
    k = ((n / 10) / 10) % 10;
    if ((i * i * i) + (j * j * j) + (k * k * k) == n) {
      printf("%d ", n);
    }
  }
  printf("\n");
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<rand>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0xc(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x0,%edi
  call   <llib_rnd_u32@plt>
  mov    -0xc(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L5
  call   <__stack_chk_fail@plt>
L5:
  leave
  ret
```

assistant: ```c
__attribute__((unused)) int rand() {
  unsigned int ui;
  llib_rnd_u32(0, &ui);
  return (int)ui;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<rand>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  mov    $0x0,%edi
  call   <llib_rnd_u32@plt>
  mov    0x4(%rsp),%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L5
  add    $0x18,%rsp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
__attribute__((unused)) int rand() {
  unsigned int ui;
  llib_rnd_u32(0, &ui);
  return (int)ui;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<rand>:
  endbr64
  sub    $0x18,%rsp
  xor    %edi,%edi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <llib_rnd_u32@plt>
  mov    0x4(%rsp),%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L5
  add    $0x18,%rsp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
__attribute__((unused)) int rand() {
  unsigned int ui;
  llib_rnd_u32(0, &ui);
  return (int)ui;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<rand>:
  endbr64
  sub    $0x18,%rsp
  xor    %edi,%edi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <llib_rnd_u32@plt>
  mov    0x4(%rsp),%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L7
  add    $0x18,%rsp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
__attribute__((unused)) int rand() {
  unsigned int ui;
  llib_rnd_u32(0, &ui);
  return (int)ui;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x1,-0x18(%rbp)
  jmp    L36
L2:
  mov    -0x18(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x18(%rbp)
  mov    %eax,%esi
  lea    D38(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movzbl -0x10(%rbp),%eax
  cmp    $0x23,%al
  jne    L39
  movzbl -0x11(%rbp),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <next@plt>
  mov    %eax,%ecx
  lea    -0xe(%rbp),%rax
  mov    %rax,%rdx
  mov    %ecx,%esi
  lea    D46(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L36
L39:
  movzbl -0x10(%rbp),%eax
  cmp    $0x62,%al
  jne    L45
  movzbl -0x11(%rbp),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <prev@plt>
  mov    %eax,%ecx
  lea    -0xe(%rbp),%rax
  mov    %rax,%rdx
  mov    %ecx,%esi
  lea    D48(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L36
L45:
  lea    D42(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
L36:
  lea    -0xe(%rbp),%rdx
  lea    -0x11(%rbp),%rax
  mov    %rax,%rsi
  lea    D24(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  cmp    $0x2,%eax
  je     L2
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L46
  call   <__stack_chk_fail@plt>
L46:
  leave
  ret
```

tool_calls: name: parse_data
arguments: tag: D38
type_guessed: string
name: parse_data
arguments: tag: D46
type_guessed: string
name: parse_data
arguments: tag: D48
type_guessed: string
name: parse_data
arguments: tag: D42
type_guessed: string
name: parse_data
arguments: tag: D24
type_guessed: string

tool: D38: Case %d: 

tool: D46: %cb %s


tool: D48: %c# %s


tool: D42: UNIQUE

tool: D24: %s %s

assistant: ```c
int main() {
  char note[3], tonality[6];
  int case_number = 1;
  while (scanf("%s %s", note, tonality) == 2) {
    printf("Case %d: ", case_number++);
    if (note[1] == '#') {
      printf("%cb %s\n", next(note[0]), tonality);
    } else if (note[1] == 'b') {
      printf("%c# %s\n", prev(note[0]), tonality);
    } else {
      printf("UNIQUE\n");
    }
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    $0x1,%ebx
  lea    D2(%rip),%r12
  lea    D48(%rip),%r13
  lea    D25(%rip),%r14
  lea    D16(%rip),%r15
  jmp    L18
L42:
  movsbl 0xf(%rsp),%edi
  call   <next@plt>
  mov    %eax,%esi
  lea    0x12(%rsp),%rdx
  lea    D14(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L16
L54:
  movsbl 0xf(%rsp),%edi
  call   <prev@plt>
  mov    %eax,%esi
  lea    0x12(%rsp),%rdx
  mov    %r15,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L16:
  mov    %ebp,%ebx
L18:
  lea    0x12(%rsp),%rdx
  lea    0xf(%rsp),%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  cmp    $0x2,%eax
  jne    L56
  lea    0x1(%rbx),%ebp
  mov    %ebx,%esi
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movzbl 0x10(%rsp),%eax
  cmp    $0x23,%al
  je     L42
  cmp    $0x62,%al
  je     L54
  mov    %r14,%rdi
  call   <puts@plt>
  jmp    L16
L56:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  mov    $0x0,%eax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D2
type: string
name: retrieve_data
arguments: tag: D48
type: string
name: retrieve_data
arguments: tag: D25
type: string
name: retrieve_data
arguments: tag: D16
type: string
name: retrieve_data
arguments: tag: D14
type: string

tool: D2: %s %s

tool: D48: Case %d: 

tool: D25: UNIQUE

tool: D16: %c# %s


tool: D14: %cb %s


assistant: ```c
int main() {
  char note[3], tonality[6];
  int case_number = 1;
  while (scanf("%s %s", note, tonality) == 2) {
    printf("Case %d: ", case_number++);
    if (note[1] == '#') {
      printf("%cb %s\n", next(note[0]), tonality);
    } else if (note[1] == 'b') {
      printf("%c# %s\n", prev(note[0]), tonality);
    } else {
      printf("UNIQUE\n");
    }
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r15
  mov    $0x1,%r15d
  push   %r14
  lea    D33(%rip),%r14
  push   %r13
  push   %r12
  lea    D38(%rip),%r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x12(%rsp),%rbp
  lea    0xf(%rsp),%r13
  jmp    L65
  nopw   0x0(%rax,%rax,1)
L63:
  lea    D2(%rip),%rdi
  call   <puts@plt>
L35:
  mov    %ebx,%r15d
L65:
  mov    %rbp,%rdx
  mov    %r13,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  cmp    $0x2,%eax
  jne    L21
  mov    %r15d,%esi
  mov    %r14,%rdi
  lea    0x1(%r15),%ebx
  xor    %eax,%eax
  call   <printf@plt>
  movzbl 0x10(%rsp),%eax
  cmp    $0x23,%al
  je     L44
  cmp    $0x62,%al
  jne    L63
  movsbl 0xf(%rsp),%edi
  call   <prev@plt>
  mov    %rbp,%rdx
  lea    D9(%rip),%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L35
  nopw   0x0(%rax,%rax,1)
L44:
  movsbl 0xf(%rsp),%edi
  call   <next@plt>
  mov    %rbp,%rdx
  lea    D24(%rip),%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L35
  nop
L21:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L17
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L17:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D33
type_infered: string
name: read_data
arguments: data_label: D38
type_infered: string
name: read_data
arguments: data_label: D2
type_infered: string
name: read_data
arguments: data_label: D9
type_infered: string
name: read_data
arguments: data_label: D24
type_infered: string

tool: D33: Case %d: 

tool: D38: %s %s

tool: D2: UNIQUE

tool: D9: %c# %s


tool: D24: %cb %s


assistant: ```c
int main() {
  char note[3], tonality[6];
  int case_number = 1;
  while (scanf("%s %s", note, tonality) == 2) {
    printf("Case %d: ", case_number++);
    if (note[1] == '#') {
      printf("%cb %s\n", next(note[0]), tonality);
    } else if (note[1] == 'b') {
      printf("%c# %s\n", prev(note[0]), tonality);
    } else {
      printf("UNIQUE\n");
    }
  }
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %r15
  mov    $0x1,%r15d
  push   %r14
  lea    D42(%rip),%r14
  push   %r13
  push   %r12
  lea    D21(%rip),%r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x12(%rsp),%rbp
  lea    0xf(%rsp),%r13
  jmp    L1
  nopw   0x0(%rax,%rax,1)
L15:
  lea    D4(%rip),%rdi
  call   <puts@plt>
L6:
  mov    %ebx,%r15d
L1:
  mov    %rbp,%rdx
  mov    %r13,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  cmp    $0x2,%eax
  jne    L53
  mov    %r15d,%esi
  mov    %r14,%rdi
  lea    0x1(%r15),%ebx
  xor    %eax,%eax
  call   <printf@plt>
  movzbl 0x10(%rsp),%eax
  cmp    $0x23,%al
  je     L21
  cmp    $0x62,%al
  jne    L15
  movsbl 0xf(%rsp),%edi
  call   <prev@plt>
  mov    %rbp,%rdx
  lea    D38(%rip),%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L6
  nopw   0x0(%rax,%rax,1)
L21:
  movsbl 0xf(%rsp),%edi
  call   <next@plt>
  mov    %rbp,%rdx
  lea    D19(%rip),%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L6
  nop
L53:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L43
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L43:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D42
type_infered: string
name: retrieve_data
arguments: tag: D21
type_infered: string
name: retrieve_data
arguments: tag: D4
type_infered: string
name: retrieve_data
arguments: tag: D38
type_infered: string
name: retrieve_data
arguments: tag: D19
type_infered: string

tool: D42: Case %d: 

tool: D21: %s %s

tool: D4: UNIQUE

tool: D38: %c# %s


tool: D19: %cb %s


assistant: ```c
int main() {
  char note[3], tonality[6];
  int case_number = 1;
  while (scanf("%s %s", note, tonality) == 2) {
    printf("Case %d: ", case_number++);
    if (note[1] == '#') {
      printf("%cb %s\n", next(note[0]), tonality);
    } else if (note[1] == 'b') {
      printf("%c# %s\n", prev(note[0]), tonality);
    } else {
      printf("UNIQUE\n");
    }
  }
  return 0;
}
```