ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    $0x1,%ebx
  mov    $0x86,%ebp
  lea    D1(%rip),%r12
L1:
  mov    %ebp,%eax
  mov    $0x0,%edx
  div    %ebx
  add    $0x1,%ebx
  mov    %ebp,%ecx
  mov    %eax,%edx
  mov    %ebx,%esi
  mov    %r12,%rdi
  call   <printf@plt>
  cmp    $0xd,%ebx
  jne    L1
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D1
data_type: string

tool: D1: %08x
%08x
%08x


assistant: ```c
void main() {
  unsigned x = 134;
  unsigned y = 1;
  unsigned q, r, p, i;
  for (i = 0; i < 12; i++) {
    q = x / y;
    r = x % y;
    p = q * y + r;
    y++;
    printf("%08x\n%08x\n%08x\n", y, q, p);
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r12
  lea    D2(%rip),%r12
  push   %rbp
  mov    $0x86,%ebp
  push   %rbx
  mov    $0x1,%ebx
  nopl   0x0(%rax)
L5:
  xor    %edx,%edx
  mov    %ebp,%eax
  mov    $0x86,%ecx
  mov    %r12,%rdi
  div    %ebx
  add    $0x1,%ebx
  mov    %ebx,%esi
  mov    %eax,%edx
  call   <printf@plt>
  cmp    $0xd,%ebx
  jne    L5
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D2
data_type: string

tool: D2: %08x
%08x
%08x


assistant: ```c
void main() {
  unsigned x = 134;
  unsigned y = 1;
  unsigned q, r, p, i;
  for (i = 0; i < 12; i++) {
    q = x / y;
    r = x % y;
    p = q * y + r;
    y++;
    printf("%08x\n%08x\n%08x\n", y, q, p);
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %r12
  lea    D2(%rip),%r12
  push   %rbp
  mov    $0x86,%ebp
  push   %rbx
  mov    $0x1,%ebx
  nopl   0x0(%rax)
L5:
  xor    %edx,%edx
  mov    %ebp,%eax
  mov    $0x86,%ecx
  mov    %r12,%rdi
  div    %ebx
  add    $0x1,%ebx
  mov    %ebx,%esi
  mov    %eax,%edx
  call   <printf@plt>
  cmp    $0xd,%ebx
  jne    L5
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: unpack_data
arguments: label: D2
type_infered: string

tool: D2: %08x
%08x
%08x


assistant: ```c
void main() {
  unsigned x = 134;
  unsigned y = 1;
  unsigned q, r, p, i;
  for (i = 0; i < 12; i++) {
    q = x / y;
    r = x % y;
    p = q * y + r;
    y++;
    printf("%08x\n%08x\n%08x\n", y, q, p);
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  movl   $0x40000040,-0x8(%rbp)
  movl   $0x0,-0x14(%rbp)
  jmp    L83
L58:
  cmpl   $0x9,-0x14(%rbp)
  jg     L24
  mov    -0x14(%rbp),%eax
  add    $0x30,%eax
  movsbl %al,%eax
  jmp    L80
L24:
  mov    -0x14(%rbp),%eax
  add    $0x57,%eax
  movsbl %al,%eax
L80:
  mov    %eax,%edi
  call   <to_stdout@plt>
  mov    $0xa,%edi
  call   <to_stdout@plt>
  mov    -0x14(%rbp),%eax
  shl    $0x3,%eax
  add    $0x40000040,%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x8(%rbp),%eax
  mov    $0x0,%esi
  mov    %eax,%edi
  call   <startCounter@plt>
  movl   $0x1,-0x10(%rbp)
  movl   $0x0,-0xc(%rbp)
L36:
  mov    $0x0,%eax
  call   <readCounter@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jle    L53
  andl   $0x1,-0x10(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  jmp    L7
L53:
  movl   $0x0,-0x10(%rbp)
L7:
  mov    $0x0,%eax
  call   <readCounter@plt>
  and    $0x3fffffff,%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%eax
  and    $0x3fffffff,%eax
  cmp    %eax,%edx
  jl     L36
  cmpl   $0x0,-0x10(%rbp)
  je     L15
  mov    $0x6f,%edi
  call   <to_stdout@plt>
  mov    $0x6b,%edi
  call   <to_stdout@plt>
  jmp    L40
L15:
  mov    $0x65,%edi
  call   <to_stdout@plt>
  mov    $0x72,%edi
  call   <to_stdout@plt>
  mov    $0x72,%edi
  call   <to_stdout@plt>
L40:
  mov    $0xa,%edi
  call   <to_stdout@plt>
  addl   $0x1,-0x14(%rbp)
L83:
  cmpl   $0x3,-0x14(%rbp)
  jle    L58
  nop
  nop
  leave
  ret
```

assistant: ```c
void main(void) {
  int i, increased, new, old, newValue;
  newValue = 0x40000040;
  for (i = 0; i < 4; i++) {
    to_stdout((((i) < 10) ? ((i) + '0') : (((i) + 'a') - 10)));
    to_stdout('\n');
    newValue = 0x40000040 + (i << 3);
    startCounter(newValue, 0);
    increased = !((0 == 1));
    old = 0;
    do {
      if ((new = readCounter()) > old) {
        increased = increased & !((0 == 1));
        old = new;
      } else {
        increased = (0 == 1);
      }
    } while ((readCounter() & 0x3fffffff) < (newValue & 0x3fffffff));
    if (increased) {
      to_stdout('o');
      to_stdout('k');
    } else {
      to_stdout('e');
      to_stdout('r');
      to_stdout('r');
    }
    to_stdout('\n');
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    $0x40,%ebp
  mov    $0x0,%r14d
  mov    $0x1,%r15d
  jmp    L44
L21:
  lea    0x57(%r14),%edi
  jmp    L50
L99:
  mov    %r13d,%r12d
L80:
  mov    $0x0,%eax
  call   <readCounter@plt>
  and    $0x3fffffff,%eax
  cmp    %ebp,%eax
  jge    L86
L36:
  mov    $0x0,%eax
  call   <readCounter@plt>
  cmp    %eax,%ebx
  jge    L99
  mov    %eax,%ebx
  jmp    L80
L86:
  test   %r12d,%r12d
  je     L2
  mov    $0x6f,%edi
  call   <to_stdout@plt>
  mov    $0x6b,%edi
  call   <to_stdout@plt>
L64:
  mov    $0xa,%edi
  call   <to_stdout@plt>
  add    $0x1,%r14d
  add    $0x8,%ebp
  cmp    $0x4,%r14d
  je     L45
L44:
  cmp    $0x9,%r14d
  jg     L21
  lea    0x30(%r14),%edi
L50:
  call   <to_stdout@plt>
  mov    $0xa,%edi
  call   <to_stdout@plt>
  lea    0x40000000(%rbp),%edi
  mov    $0x0,%esi
  call   <startCounter@plt>
  mov    $0x0,%ebx
  mov    %r15d,%r12d
  mov    $0x0,%r13d
  jmp    L36
L2:
  mov    $0x65,%edi
  call   <to_stdout@plt>
  mov    $0x72,%edi
  call   <to_stdout@plt>
  mov    $0x72,%edi
  call   <to_stdout@plt>
  jmp    L64
L45:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void main(void) {
  int i, increased, new, old, newValue;
  newValue = 0x40000040;
  for (i = 0; i < 4; i++) {
    to_stdout((((i) < 10) ? ((i) + '0') : (((i) + 'a') - 10)));
    to_stdout('\n');
    newValue = 0x40000040 + (i << 3);
    startCounter(newValue, 0);
    increased = !((0 == 1));
    old = 0;
    do {
      if ((new = readCounter()) > old) {
        increased = increased & !((0 == 1));
        old = new;
      } else {
        increased = (0 == 1);
      }
    } while ((readCounter() & 0x3fffffff) < (newValue & 0x3fffffff));
    if (increased) {
      to_stdout('o');
      to_stdout('k');
    } else {
      to_stdout('e');
      to_stdout('r');
      to_stdout('r');
    }
    to_stdout('\n');
  }
}
```
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   %r13
  mov    $0x30,%r13d
  push   %r12
  push   %rbp
  mov    $0x40,%ebp
  push   %rbx
  sub    $0x8,%rsp
L0:
  mov    %r13d,%edi
  xor    %ebx,%ebx
  mov    $0x1,%r12d
  call   <to_stdout@plt>
  mov    $0xa,%edi
  call   <to_stdout@plt>
  lea    0x40000000(%rbp),%edi
  xor    %esi,%esi
  call   <startCounter@plt>
  jmp    L9
  nopw   0x0(%rax,%rax,1)
L14:
  mov    %eax,%ebx
  xor    %eax,%eax
  call   <readCounter@plt>
  and    $0x3fffffff,%eax
  cmp    %ebp,%eax
  jge    L16
L9:
  xor    %eax,%eax
  call   <readCounter@plt>
  cmp    %eax,%ebx
  jl     L14
  xor    %eax,%eax
  xor    %r12d,%r12d
  call   <readCounter@plt>
  and    $0x3fffffff,%eax
  cmp    %ebp,%eax
  jl     L9
L16:
  test   %r12d,%r12d
  je     L20
  mov    $0x6f,%edi
  call   <to_stdout@plt>
  mov    $0x6b,%edi
  call   <to_stdout@plt>
L65:
  mov    $0xa,%edi
  add    $0x8,%ebp
  add    $0x1,%r13d
  call   <to_stdout@plt>
  cmp    $0x60,%ebp
  jne    L0
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  cs nopw 0x0(%rax,%rax,1)
L20:
  mov    $0x65,%edi
  call   <to_stdout@plt>
  mov    $0x72,%edi
  call   <to_stdout@plt>
  mov    $0x72,%edi
  call   <to_stdout@plt>
  jmp    L65
```

assistant: ```c
void main(void) {
  int i, increased, new, old, newValue;
  newValue = 0x40000040;
  for (i = 0; i < 4; i++) {
    to_stdout((((i) < 10) ? ((i) + '0') : (((i) + 'a') - 10)));
    to_stdout('\n');
    newValue = 0x40000040 + (i << 3);
    startCounter(newValue, 0);
    increased = !((0 == 1));
    old = 0;
    do {
      if ((new = readCounter()) > old) {
        increased = increased & !((0 == 1));
        old = new;
      } else {
        increased = (0 == 1);
      }
    } while ((readCounter() & 0x3fffffff) < (newValue & 0x3fffffff));
    if (increased) {
      to_stdout('o');
      to_stdout('k');
    } else {
      to_stdout('e');
      to_stdout('r');
      to_stdout('r');
    }
    to_stdout('\n');
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %r13
  mov    $0x30,%r13d
  push   %r12
  push   %rbp
  mov    $0x40,%ebp
  push   %rbx
  sub    $0x8,%rsp
L50:
  mov    %r13d,%edi
  xor    %ebx,%ebx
  mov    $0x1,%r12d
  call   <to_stdout@plt>
  mov    $0xa,%edi
  call   <to_stdout@plt>
  lea    0x40000000(%rbp),%edi
  xor    %esi,%esi
  call   <startCounter@plt>
  jmp    L47
  nopw   0x0(%rax,%rax,1)
L45:
  mov    %eax,%ebx
  xor    %eax,%eax
  call   <readCounter@plt>
  and    $0x3fffffff,%eax
  cmp    %ebp,%eax
  jge    L14
L47:
  xor    %eax,%eax
  call   <readCounter@plt>
  cmp    %eax,%ebx
  jl     L45
  xor    %eax,%eax
  xor    %r12d,%r12d
  call   <readCounter@plt>
  and    $0x3fffffff,%eax
  cmp    %ebp,%eax
  jl     L47
L14:
  test   %r12d,%r12d
  je     L67
  mov    $0x6f,%edi
  call   <to_stdout@plt>
  mov    $0x6b,%edi
L69:
  call   <to_stdout@plt>
  add    $0x8,%ebp
  mov    $0xa,%edi
  add    $0x1,%r13d
  call   <to_stdout@plt>
  cmp    $0x60,%ebp
  jne    L50
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  cs nopw 0x0(%rax,%rax,1)
L67:
  mov    $0x65,%edi
  call   <to_stdout@plt>
  mov    $0x72,%edi
  call   <to_stdout@plt>
  mov    $0x72,%edi
  jmp    L69
```

assistant: ```c
void main(void) {
  int i, increased, new, old, newValue;
  newValue = 0x40000040;
  for (i = 0; i < 4; i++) {
    to_stdout((((i) < 10) ? ((i) + '0') : (((i) + 'a') - 10)));
    to_stdout('\n');
    newValue = 0x40000040 + (i << 3);
    startCounter(newValue, 0);
    increased = !((0 == 1));
    old = 0;
    do {
      if ((new = readCounter()) > old) {
        increased = increased & !((0 == 1));
        old = new;
      } else {
        increased = (0 == 1);
      }
    } while ((readCounter() & 0x3fffffff) < (newValue & 0x3fffffff));
    if (increased) {
      to_stdout('o');
      to_stdout('k');
    } else {
      to_stdout('e');
      to_stdout('r');
      to_stdout('r');
    }
    to_stdout('\n');
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<vspen>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L16
  mov    $0x0,%eax
  jmp    L22
L16:
  mov    -0x8(%rbp),%rax
  mov    %eax,%edi
  call   <point@plt>
  test   %eax,%eax
  jne    L38
  mov    $0x0,%eax
  jmp    L22
L38:
  mov    -0x8(%rbp),%rax
  lea    D8(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <match@plt>
L22:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D8
type_guessed: string

tool: D8: \^

assistant: ```c
int vspen(char *s) {
  if (s == 0)
    return (0);
  if (!point((int)s))
    return (0);
  return (match(s, "\\^"));
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<vspen>:
  endbr64
  mov    $0x0,%eax
  test   %rdi,%rdi
  je     L17
  push   %rbx
  mov    %rdi,%rbx
  call   <point@plt>
  test   %eax,%eax
  jne    L29
L4:
  pop    %rbx
  ret
L29:
  lea    D6(%rip),%rsi
  mov    %rbx,%rdi
  call   <match@plt>
  jmp    L4
L17:
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D6
type_infered: string

tool: D6: \^

assistant: ```c
int vspen(char *s) {
  if (s == 0)
    return (0);
  if (!point((int)s))
    return (0);
  return (match(s, "\\^"));
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<vspen>:
  endbr64
  test   %rdi,%rdi
  je     L19
  push   %rbp
  mov    %rdi,%rbp
  call   <point@plt>
  test   %eax,%eax
  jne    L6
  xor    %eax,%eax
  pop    %rbp
  ret
  nopw   0x0(%rax,%rax,1)
L6:
  mov    %rbp,%rdi
  lea    D2(%rip),%rsi
  pop    %rbp
  jmp    1070 <match@plt>
L19:
  xor    %eax,%eax
  ret
```

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

tool: D2: \^

assistant: ```c
int vspen(char *s) {
  if (s == 0)
    return (0);
  if (!point((int)s))
    return (0);
  return (match(s, "\\^"));
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<vspen>:
  endbr64
  test   %rdi,%rdi
  je     L15
  push   %rbp
  mov    %rdi,%rbp
  call   <point@plt>
  test   %eax,%eax
  jne    L17
  xor    %eax,%eax
  pop    %rbp
  ret
  nopw   0x0(%rax,%rax,1)
L17:
  mov    %rbp,%rdi
  lea    D7(%rip),%rsi
  pop    %rbp
  jmp    1070 <match@plt>
L15:
  xor    %eax,%eax
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D7
datatype: string

tool: D7: \^

assistant: ```c
int vspen(char *s) {
  if (s == 0)
    return (0);
  if (!point((int)s))
    return (0);
  return (match(s, "\\^"));
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<utime>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    D7(%rip),%rax
  mov    (%rax),%edx
  mov    D18(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xffffffff,%eax
  pop    %rbp
  ret
```

assistant: ```c
int utime(const char *filename, const struct utimbuf *buf) {
  errno = ENOSYS;
  return -1;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<utime>:
  endbr64
  mov    D7(%rip),%rax
  mov    (%rax),%edx
  mov    D9(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int utime(const char *filename, const struct utimbuf *buf) {
  errno = ENOSYS;
  return -1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<utime>:
  endbr64
  mov    D19(%rip),%rax
  mov    (%rax),%edx
  mov    D3(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int utime(const char *filename, const struct utimbuf *buf) {
  errno = ENOSYS;
  return -1;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<utime>:
  endbr64
  mov    D9(%rip),%rax
  mov    (%rax),%edx
  mov    D19(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int utime(const char *filename, const struct utimbuf *buf) {
  errno = ENOSYS;
  return -1;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<crypto_cbc_encrypt>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    %rcx,-0x20(%rbp)
  mov    -0xc(%rbp),%eax
  sub    $0xa,%eax
  cmp    $0x2,%eax
  ja     L1
  mov    -0x20(%rbp),%rdx
  mov    -0x18(%rbp),%rcx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <aes_cbc_encrypt@plt>
  nop
L1:
  nop
  leave
  ret
```

assistant: ```c
void crypto_cbc_encrypt(void *ctx, int cipher, unsigned char *plaintxt, unsigned char *ciphertxt) {
  switch (cipher) {
  case 10:
  case 11:
  case 12:
    aes_cbc_encrypt(ctx, plaintxt, ciphertxt);
    break;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<crypto_cbc_encrypt>:
  endbr64
  sub    $0xa,%esi
  cmp    $0x2,%esi
  jbe    L7
  ret
L7:
  sub    $0x8,%rsp
  mov    %rdx,%rax
  mov    %rcx,%rdx
  mov    %rax,%rsi
  call   <aes_cbc_encrypt@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void crypto_cbc_encrypt(void *ctx, int cipher, unsigned char *plaintxt, unsigned char *ciphertxt) {
  switch (cipher) {
  case 10:
  case 11:
  case 12:
    aes_cbc_encrypt(ctx, plaintxt, ciphertxt);
    break;
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<crypto_cbc_encrypt>:
  endbr64
  sub    $0xa,%esi
  mov    %rdx,%r8
  cmp    $0x2,%esi
  jbe    L8
  ret
L8:
  mov    %rcx,%rdx
  mov    %r8,%rsi
  jmp    1050 <aes_cbc_encrypt@plt>
```

assistant: ```c
void crypto_cbc_encrypt(void *ctx, int cipher, unsigned char *plaintxt, unsigned char *ciphertxt) {
  switch (cipher) {
  case 10:
  case 11:
  case 12:
    aes_cbc_encrypt(ctx, plaintxt, ciphertxt);
    break;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<crypto_cbc_encrypt>:
  endbr64
  sub    $0xa,%esi
  mov    %rdx,%r8
  cmp    $0x2,%esi
  jbe    L8
  ret
L8:
  mov    %rcx,%rdx
  mov    %r8,%rsi
  jmp    1050 <aes_cbc_encrypt@plt>
```

assistant: ```c
void crypto_cbc_encrypt(void *ctx, int cipher, unsigned char *plaintxt, unsigned char *ciphertxt) {
  switch (cipher) {
  case 10:
  case 11:
  case 12:
    aes_cbc_encrypt(ctx, plaintxt, ciphertxt);
    break;
  }
}
```
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    $0x10,%rsp
  movl   $0x0,-0x4(%rbp)
  jmp    L18
L4:
  mov    -0x4(%rbp),%eax
  add    $0x1,%eax
  pxor   %xmm1,%xmm1
  cvtsi2sd %eax,%xmm1
  movsd  D30(%rip),%xmm0
  mulsd  %xmm1,%xmm0
  cvttsd2si %xmm0,%edx
  movsd  D14(%rip),%xmm0
  mov    D22(%rip),%rax
  mov    %edx,%edi
  movapd %xmm0,%xmm1
  movq   %rax,%xmm0
  call   <beta@plt>
  movq   %xmm0,%rdx
  mov    -0x4(%rbp),%eax
  add    $0x1,%eax
  pxor   %xmm1,%xmm1
  cvtsi2sd %eax,%xmm1
  movsd  D30(%rip),%xmm0
  mulsd  %xmm0,%xmm1
  movq   %xmm1,%rax
  movq   %rdx,%xmm1
  movq   %rax,%xmm0
  lea    D31(%rip),%rax
  mov    %rax,%rdi
  mov    $0x2,%eax
  call   <printf@plt>
  addl   $0x1,-0x4(%rbp)
L18:
  cmpl   $0x3,-0x4(%rbp)
  jle    L4
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0x0,%eax
  call   <getch@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: read_data
arguments: data_label: D30
type_guessed: f64
name: read_data
arguments: data_label: D14
type_guessed: f64
name: read_data
arguments: data_label: D22
type_guessed: f64
name: read_data
arguments: data_label: D31
type_guessed: string

tool: D30: 0.2

tool: D14: 2

tool: D22: 0.5

tool: D31: 
beta(0.5,2.0,%4.2f)=%2.5f

assistant: ```c
main() {
  int n;
  for (n = 0; n < 4; n++) {
    printf("\nbeta(0.5,2.0,%4.2f)=%2.5f", (n + 1) * 0.2, beta(0.5, 2.0, (n + 1) * 0.2));
  }
  printf("\n");
  getch();
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    $0x0,%ebx
  lea    D1(%rip),%rbp
L9:
  add    $0x1,%ebx
  pxor   %xmm0,%xmm0
  cvtsi2sd %ebx,%xmm0
  mulsd  D14(%rip),%xmm0
  movsd  %xmm0,0x8(%rsp)
  cvttsd2si %xmm0,%edi
  mov    D24(%rip),%rax
  movq   %rax,%xmm1
  mov    D2(%rip),%rax
  movq   %rax,%xmm0
  call   <beta@plt>
  movapd %xmm0,%xmm1
  movsd  0x8(%rsp),%xmm0
  mov    %rbp,%rdi
  mov    $0x2,%eax
  call   <printf@plt>
  cmp    $0x4,%ebx
  jne    L9
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0x0,%eax
  call   <getch@plt>
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D1
data_type: string
name: retrieve_data
arguments: data_tag: D14
data_type: f64
name: retrieve_data
arguments: data_tag: D24
data_type: f64
name: retrieve_data
arguments: data_tag: D2
data_type: f64

tool: D1: 
beta(0.5,2.0,%4.2f)=%2.5f

tool: D14: 0.2

tool: D24: 2

tool: D2: 0.5

assistant: ```c
main() {
  int n;
  for (n = 0; n < 4; n++) {
    printf("\nbeta(0.5,2.0,%4.2f)=%2.5f", (n + 1) * 0.2, beta(0.5, 2.0, (n + 1) * 0.2));
  }
  printf("\n");
  getch();
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D24(%rip),%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x18,%rsp
L9:
  add    $0x1,%ebx
  pxor   %xmm2,%xmm2
  mov    D28(%rip),%rax
  cvtsi2sd %ebx,%xmm2
  mulsd  D9(%rip),%xmm2
  movq   %rax,%xmm1
  mov    D34(%rip),%rax
  movq   %rax,%xmm0
  cvttsd2si %xmm2,%edi
  movsd  %xmm2,0x8(%rsp)
  call   <beta@plt>
  movsd  0x8(%rsp),%xmm2
  mov    %rbp,%rdi
  mov    $0x2,%eax
  movapd %xmm0,%xmm1
  movapd %xmm2,%xmm0
  call   <printf@plt>
  cmp    $0x4,%ebx
  jne    L9
  mov    $0xa,%edi
  call   <putchar@plt>
  xor    %eax,%eax
  call   <getch@plt>
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: read_data
arguments: data_tag: D24
type: string
name: read_data
arguments: data_tag: D28
type: f64
name: read_data
arguments: data_tag: D9
type: f64
name: read_data
arguments: data_tag: D34
type: f64

tool: D24: 
beta(0.5,2.0,%4.2f)=%2.5f

tool: D28: 2

tool: D9: 0.2

tool: D34: 0.5

assistant: ```c
main() {
  int n;
  for (n = 0; n < 4; n++) {
    printf("\nbeta(0.5,2.0,%4.2f)=%2.5f", (n + 1) * 0.2, beta(0.5, 2.0, (n + 1) * 0.2));
  }
  printf("\n");
  getch();
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D14(%rip),%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x18,%rsp
L8:
  add    $0x1,%ebx
  pxor   %xmm2,%xmm2
  mov    D31(%rip),%rax
  cvtsi2sd %ebx,%xmm2
  mulsd  D17(%rip),%xmm2
  movq   %rax,%xmm1
  mov    D15(%rip),%rax
  movq   %rax,%xmm0
  cvttsd2si %xmm2,%edi
  movsd  %xmm2,0x8(%rsp)
  call   <beta@plt>
  movsd  0x8(%rsp),%xmm2
  mov    %rbp,%rdi
  mov    $0x2,%eax
  movapd %xmm0,%xmm1
  movapd %xmm2,%xmm0
  call   <printf@plt>
  cmp    $0x4,%ebx
  jne    L8
  mov    $0xa,%edi
  call   <putchar@plt>
  xor    %eax,%eax
  call   <getch@plt>
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D14
type: string
name: unpack_data
arguments: data_tag: D31
type: f64
name: unpack_data
arguments: data_tag: D17
type: f64
name: unpack_data
arguments: data_tag: D15
type: f64

tool: D14: 
beta(0.5,2.0,%4.2f)=%2.5f

tool: D31: 2

tool: D17: 0.2

tool: D15: 0.5

assistant: ```c
main() {
  int n;
  for (n = 0; n < 4; n++) {
    printf("\nbeta(0.5,2.0,%4.2f)=%2.5f", (n + 1) * 0.2, beta(0.5, 2.0, (n + 1) * 0.2));
  }
  printf("\n");
  getch();
}
```