ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<ti_ema>:
  endbr64
  push   %r15
  mov    $0x1,%eax
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  cvttsd2si (%rdx),%r14d
  test   %r14d,%r14d
  jle    L39
  mov    %edi,%ebx
  mov    %rdx,%rdi
  mov    (%rsi),%r13
  mov    (%rcx),%r12
  mov    %rdx,%rbp
  mov    %rcx,%r15
  call   <ti_ema_start@plt>
  mov    %eax,%r8d
  xor    %eax,%eax
  cmp    %ebx,%r8d
  jge    L39
  pxor   %xmm0,%xmm0
  movsd  0x0(%r13),%xmm1
  lea    0x8(%r12),%rcx
  movsd  D5(%rip),%xmm2
  cvtsi2sd %r14d,%xmm0
  addsd  D6(%rip),%xmm0
  movsd  %xmm1,(%r12)
  divsd  %xmm0,%xmm2
  cmp    $0x1,%ebx
  jle    L36
  mov    %ebx,%edx
  mov    $0x1,%eax
  nopl   0x0(%rax,%rax,1)
L31:
  movsd  0x0(%r13,%rax,8),%xmm0
  subsd  %xmm1,%xmm0
  mulsd  %xmm2,%xmm0
  addsd  %xmm0,%xmm1
  movsd  %xmm1,(%r12,%rax,8)
  add    $0x1,%rax
  cmp    %rax,%rdx
  jne    L31
  lea    -0x1(%rbx),%eax
  lea    (%rcx,%rax,8),%rcx
L36:
  sub    (%r15),%rcx
  mov    %rbp,%rdi
  sar    $0x3,%rcx
  mov    %rcx,%r12
  call   <ti_ema_start@plt>
  xor    %edi,%edi
  sub    %eax,%ebx
  movslq %ebx,%rbx
  cmp    %r12,%rbx
  sete   %dil
  call   <assert@plt>
  xor    %eax,%eax
L39:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: read_data
arguments: data_label: D5
type: f32[2]
name: read_data
arguments: data_label: D6
type: f32[2]

tool: {"D5": [0.0,2.0]}

tool: {"D6": [0.0,1.875]}

assistant: ```c
int ti_ema(int size, double const *const *inputs, double const *options, double *const *outputs) {
  const double *input = inputs[0];
  const int period = (int)options[0];
  double *output = outputs[0];
  if (period < 1)
    return 1;
  if (size <= ti_ema_start(options))
    return 0;
  const double per = 2 / ((double)period + 1);
  double val = input[0];
  *output++ = val;
  int i;
  for (i = 1; i < size; ++i) {
    val = (input[i] - val) * per + val;
    *output++ = val;
  }
  assert(output - outputs[0] == size - ti_ema_start(options));
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<syntaxerror>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  nop
  pop    %rbp
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D8
datatype: string

tool: D8: syntaxerror() should not be called in AMBER


assistant: ```c
syntaxerror() {
  printf("syntaxerror() should not be called in AMBER\n");
  exit(1);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<syntaxerror>:
  endbr64
  sub    $0x8,%rsp
  lea    D6(%rip),%rdi
  call   <printf@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: data_tag: D6
type: string

tool: D6: syntaxerror() should not be called in AMBER


assistant: ```c
syntaxerror() {
  printf("syntaxerror() should not be called in AMBER\n");
  exit(1);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<syntaxerror>:
  endbr64
  sub    $0x8,%rsp
  lea    D0(%rip),%rdi
  call   <printf@plt>
  mov    $0x1,%edi
  add    $0x8,%rsp
  jmp    1070 <exit@plt>
```

tool_calls: name: retrieve_data
arguments: label: D0
data_type: string

tool: D0: syntaxerror() should not be called in AMBER


assistant: ```c
syntaxerror() {
  printf("syntaxerror() should not be called in AMBER\n");
  exit(1);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<syntaxerror>:
  endbr64
  sub    $0x8,%rsp
  lea    D9(%rip),%rdi
  call   <printf@plt>
  mov    $0x1,%edi
  add    $0x8,%rsp
  jmp    1070 <exit@plt>
```

tool_calls: name: parse_data
arguments: label: D9
data_type: string

tool: D9: syntaxerror() should not be called in AMBER


assistant: ```c
syntaxerror() {
  printf("syntaxerror() should not be called in AMBER\n");
  exit(1);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<get_char_utf8>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x28(%rbp)
  mov    -0x28(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,-0x8(%rbp)
  movl   $0x1,-0x10(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L25
L15:
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  movzbl (%rax),%eax
  mov    %al,-0x11(%rbp)
  movzbl -0x11(%rbp),%eax
  and    $0xfc,%eax
  cmp    $0xfc,%eax
  jne    L138
  movzbl -0x11(%rbp),%eax
  and    $0x1,%eax
  mov    %eax,-0xc(%rbp)
  movl   $0x5,-0x10(%rbp)
  jmp    L25
L138:
  movzbl -0x11(%rbp),%eax
  and    $0xf8,%eax
  cmp    $0xf8,%eax
  jne    L19
  movzbl -0x11(%rbp),%eax
  and    $0x3,%eax
  mov    %eax,-0xc(%rbp)
  movl   $0x4,-0x10(%rbp)
  jmp    L25
L19:
  movzbl -0x11(%rbp),%eax
  and    $0xf0,%eax
  cmp    $0xf0,%eax
  jne    L88
  movzbl -0x11(%rbp),%eax
  and    $0x7,%eax
  mov    %eax,-0xc(%rbp)
  movl   $0x3,-0x10(%rbp)
  jmp    L25
L88:
  movzbl -0x11(%rbp),%eax
  and    $0xe0,%eax
  cmp    $0xe0,%eax
  jne    L134
  movzbl -0x11(%rbp),%eax
  and    $0xf,%eax
  mov    %eax,-0xc(%rbp)
  movl   $0x2,-0x10(%rbp)
  jmp    L25
L134:
  movzbl -0x11(%rbp),%eax
  and    $0xc0,%eax
  cmp    $0xc0,%eax
  jne    L70
  movzbl -0x11(%rbp),%eax
  and    $0x1f,%eax
  mov    %eax,-0xc(%rbp)
  movl   $0x1,-0x10(%rbp)
  jmp    L25
L70:
  movzbl -0x11(%rbp),%eax
  test   %al,%al
  jns    L121
  shll   $0x6,-0xc(%rbp)
  movzbl -0x11(%rbp),%eax
  and    $0x3f,%eax
  or     %eax,-0xc(%rbp)
  jmp    L25
L121:
  movzbl -0x11(%rbp),%eax
  mov    %eax,-0xc(%rbp)
L25:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L40
  mov    -0x10(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0x10(%rbp)
  test   %eax,%eax
  jne    L15
L40:
  mov    -0x28(%rbp),%rax
  mov    -0x8(%rbp),%rdx
  mov    %rdx,(%rax)
  mov    -0xc(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int get_char_utf8(const char **_src) {
  const unsigned char *src = (const unsigned char *)*_src;
  int expect = 1;
  unsigned int n = 0;
  while (*src && expect--) {
    unsigned char ch = *src++;
    if ((ch & 0b11111100) == 0b11111100) {
      n = ch & 0b00000001;
      expect = 5;
    } else if ((ch & 0b11111000) == 0b11111000) {
      n = ch & 0b00000011;
      expect = 4;
    } else if ((ch & 0b11110000) == 0b11110000) {
      n = ch & 0b00000111;
      expect = 3;
    } else if ((ch & 0b11100000) == 0b11100000) {
      n = ch & 0b00001111;
      expect = 2;
    } else if ((ch & 0b11000000) == 0b11000000) {
      n = ch & 0b00011111;
      expect = 1;
    } else if ((ch & 0b10000000) == 0b10000000) {
      n <<= 6;
      n |= ch & 0b00111111;
    } else {
      n = ch;
    }
  }
  *_src = (const char *)src;
  return (int)n;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<get_char_utf8>:
  endbr64
  mov    %rdi,%r8
  mov    (%rdi),%rdx
  movzbl (%rdx),%eax
  test   %al,%al
  je     L60
  push   %rbp
  push   %rbx
  mov    $0x0,%esi
  mov    $0x0,%edi
  mov    $0x1,%ebp
  mov    $0x2,%ebx
  mov    $0x3,%r11d
  mov    $0x4,%r10d
  mov    $0x5,%r9d
  jmp    L147
L42:
  mov    %eax,%edi
  and    $0x1,%edi
  mov    %r9d,%ecx
L97:
  movzbl (%rdx),%eax
  test   %al,%al
  je     L156
  lea    -0x1(%rcx),%esi
L147:
  add    $0x1,%rdx
  mov    %eax,%ecx
  and    $0xfffffffc,%ecx
  cmp    $0xfc,%cl
  je     L42
  mov    %eax,%ecx
  and    $0xfffffff8,%ecx
  cmp    $0xf8,%cl
  je     L11
  mov    %eax,%ecx
  and    $0xfffffff0,%ecx
  cmp    $0xf0,%cl
  je     L80
  mov    %eax,%ecx
  and    $0xffffffe0,%ecx
  cmp    $0xe0,%cl
  je     L61
  mov    %eax,%ecx
  and    $0xffffffc0,%ecx
  cmp    $0xc0,%cl
  je     L123
  test   %al,%al
  js     L98
  movzbl %al,%edi
L121:
  movzbl (%rdx),%eax
  test   %al,%al
  je     L156
  test   %esi,%esi
  je     L156
  sub    $0x1,%esi
  jmp    L147
L11:
  mov    %eax,%edi
  and    $0x3,%edi
  mov    %r10d,%ecx
  jmp    L97
L80:
  mov    %eax,%edi
  and    $0x7,%edi
  mov    %r11d,%ecx
  jmp    L97
L61:
  mov    %eax,%edi
  and    $0xf,%edi
  mov    %ebx,%ecx
  jmp    L97
L123:
  mov    %eax,%edi
  and    $0x1f,%edi
  mov    %ebp,%ecx
  jmp    L97
L98:
  shl    $0x6,%edi
  and    $0x3f,%eax
  or     %eax,%edi
  jmp    L121
L60:
  mov    $0x0,%edi
  mov    %rdx,(%r8)
  mov    %edi,%eax
  ret
L156:
  mov    %rdx,(%r8)
  mov    %edi,%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int get_char_utf8(const char **_src) {
  const unsigned char *src = (const unsigned char *)*_src;
  int expect = 1;
  unsigned int n = 0;
  while (*src && expect--) {
    unsigned char ch = *src++;
    if ((ch & 0b11111100) == 0b11111100) {
      n = ch & 0b00000001;
      expect = 5;
    } else if ((ch & 0b11111000) == 0b11111000) {
      n = ch & 0b00000011;
      expect = 4;
    } else if ((ch & 0b11110000) == 0b11110000) {
      n = ch & 0b00000111;
      expect = 3;
    } else if ((ch & 0b11100000) == 0b11100000) {
      n = ch & 0b00001111;
      expect = 2;
    } else if ((ch & 0b11000000) == 0b11000000) {
      n = ch & 0b00011111;
      expect = 1;
    } else if ((ch & 0b10000000) == 0b10000000) {
      n <<= 6;
      n |= ch & 0b00111111;
    } else {
      n = ch;
    }
  }
  *_src = (const char *)src;
  return (int)n;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<get_char_utf8>:
  endbr64
  mov    (%rdi),%rcx
  xor    %esi,%esi
  xor    %eax,%eax
  movzbl (%rcx),%edx
  test   %dl,%dl
  je     L122
L100:
  mov    %esi,%r8d
  nopl   (%rax)
L182:
  mov    %edx,%esi
  add    $0x1,%rcx
  and    $0xfffffffc,%esi
  cmp    $0xfc,%sil
  je     L166
  mov    %edx,%esi
  and    $0xfffffff8,%esi
  cmp    $0xf8,%sil
  je     L90
  mov    %edx,%esi
  and    $0xfffffff0,%esi
  cmp    $0xf0,%sil
  je     L33
  mov    %edx,%esi
  and    $0xffffffe0,%esi
  cmp    $0xe0,%sil
  je     L51
  mov    %edx,%esi
  and    $0xffffffc0,%esi
  cmp    $0xc0,%sil
  je     L92
  test   %dl,%dl
  js     L137
  movzbl %dl,%eax
L72:
  movzbl (%rcx),%edx
  test   %dl,%dl
  je     L122
  lea    -0x1(%r8),%esi
  test   %r8d,%r8d
  jne    L100
L122:
  mov    %rcx,(%rdi)
  ret
  nopw   0x0(%rax,%rax,1)
L166:
  mov    %edx,%eax
  movzbl (%rcx),%edx
  and    $0x1,%eax
  test   %dl,%dl
  je     L122
  mov    $0x4,%r8d
  jmp    L182
  nopl   0x0(%rax)
L90:
  mov    %edx,%eax
  movzbl (%rcx),%edx
  and    $0x3,%eax
  test   %dl,%dl
  je     L122
  mov    $0x3,%r8d
  jmp    L182
  nop
L33:
  mov    %edx,%eax
  movzbl (%rcx),%edx
  and    $0x7,%eax
  test   %dl,%dl
  je     L122
  mov    $0x2,%r8d
  jmp    L182
  nopw   0x0(%rax,%rax,1)
L51:
  mov    %edx,%eax
  movzbl (%rcx),%edx
  and    $0xf,%eax
  test   %dl,%dl
  je     L122
  mov    $0x1,%r8d
  jmp    L182
  nopw   0x0(%rax,%rax,1)
L92:
  mov    %edx,%eax
  movzbl (%rcx),%edx
  and    $0x1f,%eax
  test   %dl,%dl
  je     L122
  xor    %r8d,%r8d
  jmp    L182
  nopl   0x0(%rax,%rax,1)
L137:
  shl    $0x6,%eax
  and    $0x3f,%edx
  or     %edx,%eax
  jmp    L72
```

assistant: ```c
int get_char_utf8(const char **_src) {
  const unsigned char *src = (const unsigned char *)*_src;
  int expect = 1;
  unsigned int n = 0;
  while (*src && expect--) {
    unsigned char ch = *src++;
    if ((ch & 0b11111100) == 0b11111100) {
      n = ch & 0b00000001;
      expect = 5;
    } else if ((ch & 0b11111000) == 0b11111000) {
      n = ch & 0b00000011;
      expect = 4;
    } else if ((ch & 0b11110000) == 0b11110000) {
      n = ch & 0b00000111;
      expect = 3;
    } else if ((ch & 0b11100000) == 0b11100000) {
      n = ch & 0b00001111;
      expect = 2;
    } else if ((ch & 0b11000000) == 0b11000000) {
      n = ch & 0b00011111;
      expect = 1;
    } else if ((ch & 0b10000000) == 0b10000000) {
      n <<= 6;
      n |= ch & 0b00111111;
    } else {
      n = ch;
    }
  }
  *_src = (const char *)src;
  return (int)n;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<get_char_utf8>:
  endbr64
  mov    (%rdi),%rcx
  xor    %esi,%esi
  xor    %eax,%eax
  movzbl (%rcx),%edx
  test   %dl,%dl
  je     L103
L132:
  mov    %esi,%r8d
  nopl   (%rax)
L153:
  mov    %edx,%esi
  add    $0x1,%rcx
  and    $0xfffffffc,%esi
  cmp    $0xfc,%sil
  je     L72
  mov    %edx,%esi
  and    $0xfffffff8,%esi
  cmp    $0xf8,%sil
  je     L13
  mov    %edx,%esi
  and    $0xfffffff0,%esi
  cmp    $0xf0,%sil
  je     L118
  mov    %edx,%esi
  and    $0xffffffe0,%esi
  cmp    $0xe0,%sil
  je     L61
  mov    %edx,%esi
  and    $0xffffffc0,%esi
  cmp    $0xc0,%sil
  je     L64
  test   %dl,%dl
  js     L110
  movzbl %dl,%eax
L16:
  movzbl (%rcx),%edx
  test   %dl,%dl
  je     L103
  lea    -0x1(%r8),%esi
  test   %r8d,%r8d
  jne    L132
L103:
  mov    %rcx,(%rdi)
  ret
  nopw   0x0(%rax,%rax,1)
L72:
  mov    %edx,%eax
  movzbl (%rcx),%edx
  and    $0x1,%eax
  test   %dl,%dl
  je     L103
  mov    $0x4,%r8d
  jmp    L153
  nopl   0x0(%rax)
L13:
  mov    %edx,%eax
  movzbl (%rcx),%edx
  and    $0x3,%eax
  test   %dl,%dl
  je     L103
  mov    $0x3,%r8d
  jmp    L153
  nop
L118:
  mov    %edx,%eax
  movzbl (%rcx),%edx
  and    $0x7,%eax
  test   %dl,%dl
  je     L103
  mov    $0x2,%r8d
  jmp    L153
  nopw   0x0(%rax,%rax,1)
L61:
  mov    %edx,%eax
  movzbl (%rcx),%edx
  and    $0xf,%eax
  test   %dl,%dl
  je     L103
  mov    $0x1,%r8d
  jmp    L153
  nopw   0x0(%rax,%rax,1)
L64:
  mov    %edx,%eax
  movzbl (%rcx),%edx
  and    $0x1f,%eax
  test   %dl,%dl
  je     L103
  xor    %r8d,%r8d
  jmp    L153
  nopl   0x0(%rax,%rax,1)
L110:
  shl    $0x6,%eax
  and    $0x3f,%edx
  or     %edx,%eax
  jmp    L16
```

assistant: ```c
int get_char_utf8(const char **_src) {
  const unsigned char *src = (const unsigned char *)*_src;
  int expect = 1;
  unsigned int n = 0;
  while (*src && expect--) {
    unsigned char ch = *src++;
    if ((ch & 0b11111100) == 0b11111100) {
      n = ch & 0b00000001;
      expect = 5;
    } else if ((ch & 0b11111000) == 0b11111000) {
      n = ch & 0b00000011;
      expect = 4;
    } else if ((ch & 0b11110000) == 0b11110000) {
      n = ch & 0b00000111;
      expect = 3;
    } else if ((ch & 0b11100000) == 0b11100000) {
      n = ch & 0b00001111;
      expect = 2;
    } else if ((ch & 0b11000000) == 0b11000000) {
      n = ch & 0b00011111;
      expect = 1;
    } else if ((ch & 0b10000000) == 0b10000000) {
      n <<= 6;
      n |= ch & 0b00111111;
    } else {
      n = ch;
    }
  }
  *_src = (const char *)src;
  return (int)n;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<process>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  mov    -0x8(%rbp),%edx
  mov    -0x4(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <add@plt>
  mov    %eax,%esi
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x8(%rbp),%edx
  mov    -0x4(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <sub@plt>
  mov    %eax,%esi
  lea    D25(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x8(%rbp),%edx
  mov    -0x4(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <mul@plt>
  mov    %eax,%esi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_label: D10
datatype: string
name: parse_data
arguments: data_label: D25
datatype: string
name: parse_data
arguments: data_label: D7
datatype: string

tool: D10: a + b = %d


tool: D25: a - b = %d


tool: D7: a * b = %d


assistant: ```c
void process(int a, int b) {
  printf("a + b = %d\n", add(a, b));
  printf("a - b = %d\n", sub(a, b));
  printf("a * b = %d\n", mul(a, b));
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<process>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebx
  mov    %esi,%ebp
  call   <add@plt>
  mov    %eax,%esi
  lea    D19(%rip),%rdi
  call   <printf@plt>
  mov    %ebp,%esi
  mov    %ebx,%edi
  call   <sub@plt>
  mov    %eax,%esi
  lea    D11(%rip),%rdi
  call   <printf@plt>
  mov    %ebp,%esi
  mov    %ebx,%edi
  call   <mul@plt>
  mov    %eax,%esi
  lea    D0(%rip),%rdi
  call   <printf@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: read_data
arguments: tag: D19
datatype: string
name: read_data
arguments: tag: D11
datatype: string
name: read_data
arguments: tag: D0
datatype: string

tool: D19: a + b = %d


tool: D11: a - b = %d


tool: D0: a * b = %d


assistant: ```c
void process(int a, int b) {
  printf("a + b = %d\n", add(a, b));
  printf("a - b = %d\n", sub(a, b));
  printf("a * b = %d\n", mul(a, b));
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<process>:
  endbr64
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  mov    %edi,%ebp
  sub    $0x8,%rsp
  call   <add@plt>
  lea    D19(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    %r12d,%esi
  mov    %ebp,%edi
  call   <sub@plt>
  lea    D29(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    %r12d,%esi
  mov    %ebp,%edi
  call   <mul@plt>
  add    $0x8,%rsp
  lea    D24(%rip),%rdi
  pop    %rbp
  mov    %eax,%esi
  pop    %r12
  jmp    1090 <printf@plt>
```

tool_calls: name: read_data
arguments: data_tag: D19
datatype: string
name: read_data
arguments: data_tag: D29
datatype: string
name: read_data
arguments: data_tag: D24
datatype: string

tool: D19: a + b = %d


tool: D29: a - b = %d


tool: D24: a * b = %d


assistant: ```c
void process(int a, int b) {
  printf("a + b = %d\n", add(a, b));
  printf("a - b = %d\n", sub(a, b));
  printf("a * b = %d\n", mul(a, b));
}
```
user: What is the c source code of the assembly code below:

```asm
<process>:
  endbr64
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  mov    %edi,%ebp
  sub    $0x8,%rsp
  call   <add@plt>
  lea    D25(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    %r12d,%esi
  mov    %ebp,%edi
  call   <sub@plt>
  lea    D29(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    %r12d,%esi
  mov    %ebp,%edi
  call   <mul@plt>
  add    $0x8,%rsp
  lea    D22(%rip),%rdi
  pop    %rbp
  mov    %eax,%esi
  pop    %r12
  jmp    1090 <printf@plt>
```

tool_calls: name: unpack_data
arguments: tag: D25
datatype: string
name: unpack_data
arguments: tag: D29
datatype: string
name: unpack_data
arguments: tag: D22
datatype: string

tool: D25: a + b = %d


tool: D29: a - b = %d


tool: D22: a * b = %d


assistant: ```c
void process(int a, int b) {
  printf("a + b = %d\n", add(a, b));
  printf("a - b = %d\n", sub(a, b));
  printf("a * b = %d\n", mul(a, b));
}
```
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    $0x60,%rsp
  mov    %edi,-0x54(%rbp)
  mov    %rsi,-0x60(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x48(%rbp)
  movl   $0x1,-0x48(%rbp)
  jmp    L1
L24:
  mov    -0x48(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x60(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x48(%rbp),%eax
  mov    %eax,%esi
  lea    D11(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x48(%rbp)
L1:
  mov    -0x48(%rbp),%eax
  cmp    -0x54(%rbp),%eax
  jl     L24
  lea    D34(%rip),%rax
  mov    %rax,-0x40(%rbp)
  lea    D23(%rip),%rax
  mov    %rax,-0x38(%rbp)
  lea    D7(%rip),%rax
  mov    %rax,-0x30(%rbp)
  lea    D36(%rip),%rax
  mov    %rax,-0x28(%rbp)
  movq   $0x0,-0x20(%rbp)
  movq   $0x0,-0x18(%rbp)
  movl   $0x6,-0x44(%rbp)
  movl   $0x0,-0x48(%rbp)
  jmp    L35
L2:
  lea    -0x40(%rbp),%rdx
  mov    -0x48(%rbp),%eax
  cltq
  shl    $0x3,%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0x48(%rbp),%eax
  cltq
  mov    -0x40(%rbp,%rax,8),%rdx
  mov    -0x48(%rbp),%eax
  mov    %eax,%esi
  lea    D13(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x48(%rbp)
L35:
  mov    -0x48(%rbp),%eax
  cmp    -0x44(%rbp),%eax
  jl     L2
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L28
  call   <__stack_chk_fail@plt>
L28:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D11
data_type: string
name: retrieve_data
arguments: data_label: D34
data_type: string
name: retrieve_data
arguments: data_label: D23
data_type: string
name: retrieve_data
arguments: data_label: D7
data_type: string
name: retrieve_data
arguments: data_label: D36
data_type: string
name: retrieve_data
arguments: data_label: D13
data_type: string

tool: D11: arg %d: %s


tool: D34: California

tool: D23: Oregon

tool: D7: Washington

tool: D36: Texas

tool: D13: state %d: %s, %p


assistant: ```c
int main(int argc, char *argv[]) {
  int i = 0;
  for (i = 1; i < argc; i++) {
    printf("arg %d: %s\n", i, argv[i]);
  }
  char *states[] = {"California", "Oregon", "Washington", "Texas", NULL, NULL};
  int num_states = 6;
  for (i = 0; i < num_states; i++) {
    printf("state %d: %s, %p\n", i, states[i], &(states[i]));
  };
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x48,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  cmp    $0x1,%edi
  jle    L25
  mov    %rsi,%r12
  mov    %edi,%ebp
  mov    $0x1,%ebx
  lea    D14(%rip),%r13
L24:
  mov    (%r12,%rbx,8),%rdx
  mov    %ebx,%esi
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x1,%rbx
  cmp    %rbp,%rbx
  jne    L24
L25:
  lea    D47(%rip),%rax
  mov    %rax,(%rsp)
  lea    D56(%rip),%rax
  mov    %rax,0x8(%rsp)
  lea    D4(%rip),%rax
  mov    %rax,0x10(%rsp)
  lea    D37(%rip),%rax
  mov    %rax,0x18(%rsp)
  movq   $0x0,0x20(%rsp)
  movq   $0x0,0x28(%rsp)
  mov    %rsp,%rbp
  mov    $0x0,%ebx
  lea    D27(%rip),%r12
L34:
  mov    %rbp,%rcx
  mov    0x0(%rbp),%rdx
  mov    %ebx,%esi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x1,%ebx
  add    $0x8,%rbp
  cmp    $0x6,%ebx
  jne    L34
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L30
  mov    $0x0,%eax
  add    $0x48,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L30:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D14
datatype: string
name: retrieve_data
arguments: data_label: D47
datatype: string
name: retrieve_data
arguments: data_label: D56
datatype: string
name: retrieve_data
arguments: data_label: D4
datatype: string
name: retrieve_data
arguments: data_label: D37
datatype: string
name: retrieve_data
arguments: data_label: D27
datatype: string

tool: D14: arg %d: %s


tool: D47: California

tool: D56: Oregon

tool: D4: Washington

tool: D37: Texas

tool: D27: state %d: %s, %p


assistant: ```c
int main(int argc, char *argv[]) {
  int i = 0;
  for (i = 1; i < argc; i++) {
    printf("arg %d: %s\n", i, argv[i]);
  }
  char *states[] = {"California", "Oregon", "Washington", "Texas", NULL, NULL};
  int num_states = 6;
  for (i = 0; i < num_states; i++) {
    printf("state %d: %s, %p\n", i, states[i], &(states[i]));
  };
  return 0;
}
```