ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Converte this assembly snippet into its equivalent C code:

```asm
<reverse_string>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  test   %rdi,%rdi
  setne  %dil
  movzbl %dil,%edi
  call   <assert@plt>
  mov    %rbx,%rdi
  call   <strlen@plt>
  cmp    $0x1,%eax
  jle    L18
  mov    %eax,%ebp
  mov    $0x0,%r12d
  jmp    L74
L3:
  lea    (%rcx,%rdx,1),%eax
  mov    %rdx,%rdi
  add    $0x1,%rdx
  cmpb   $0x20,-0x1(%rdx)
  je     L3
L63:
  movslq %eax,%rcx
  testb  $0xdf,(%rbx,%rcx,1)
  je     L54
  add    $0x1,%eax
  cltq
L14:
  mov    %eax,%r12d
  movzbl (%rbx,%rax,1),%edx
  mov    %rax,%rcx
  add    $0x1,%rax
  test   $0xdf,%dl
  jne    L14
L8:
  lea    -0x1(%rbx,%rcx,1),%rsi
  call   <reverse_word@plt>
  cmp    %r12d,%ebp
  jle    L72
L74:
  movslq %r12d,%rax
  lea    (%rbx,%rax,1),%rdi
  lea    0x1(%rbx,%rax,1),%rax
  mov    %rax,%rdx
  lea    0x1(%r12),%ecx
  sub    %eax,%ecx
  mov    %r12d,%eax
  cmpb   $0x20,(%rdi)
  je     L3
  jmp    L63
L54:
  mov    %eax,%r12d
  jmp    L8
L72:
  movslq %ebp,%rbp
  lea    -0x1(%rbx,%rbp,1),%rsi
  mov    %rbx,%rdi
  call   <reverse_word@plt>
L18:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void reverse_string(char *str) {
  assert(str != NULL);
  int len = strlen(str);
  if (len < 2)
    return;
  int i = 0;
  char *p = NULL;
  char *q = NULL;
  while (i < len) {
    while (str[i] == ' ' && str[i] != '\0')
      i++;
    p = str + i;
    while (str[i] != ' ' && str[i] != '\0')
      i++;
    q = str + i;
    reverse_word(p, q - 1);
  }
  reverse_word(str, str + len - 1);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<reverse_string>:
  endbr64
  push   %r12
  test   %rdi,%rdi
  push   %rbp
  mov    %rdi,%rbp
  setne  %dil
  push   %rbx
  movzbl %dil,%edi
  xor    %ebx,%ebx
  call   <assert@plt>
  mov    %rbp,%rdi
  call   <strlen@plt>
  movslq %eax,%r12
  cmp    $0x1,%r12d
  jle    L44
  xchg   %ax,%ax
L26:
  movslq %ebx,%rcx
  lea    0x1(%rbx),%eax
  lea    0x0(%rbp,%rcx,1),%rdi
  cltq
  movzbl (%rdi),%edx
  cmp    $0x20,%dl
  jne    L14
  nopl   (%rax)
L48:
  mov    %eax,%ebx
  lea    0x0(%rbp,%rax,1),%rdi
  mov    %rax,%rcx
  add    $0x1,%rax
  movzbl -0x1(%rbp,%rax,1),%edx
  cmp    $0x20,%dl
  je     L48
L14:
  and    $0xdf,%edx
  je     L2
  add    $0x1,%ebx
  movslq %ebx,%rax
  xchg   %ax,%ax
L33:
  movzbl 0x0(%rbp,%rax,1),%edx
  mov    %eax,%ebx
  mov    %rax,%rcx
  add    $0x1,%rax
  and    $0xdf,%edx
  jne    L33
L2:
  lea    -0x1(%rbp,%rcx,1),%rsi
  call   <reverse_word@plt>
  cmp    %ebx,%r12d
  jg     L26
  lea    -0x1(%rbp,%r12,1),%rsi
  pop    %rbx
  mov    %rbp,%rdi
  pop    %rbp
  pop    %r12
  jmp    1090 <reverse_word@plt>
L44:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void reverse_string(char *str) {
  assert(str != NULL);
  int len = strlen(str);
  if (len < 2)
    return;
  int i = 0;
  char *p = NULL;
  char *q = NULL;
  while (i < len) {
    while (str[i] == ' ' && str[i] != '\0')
      i++;
    p = str + i;
    while (str[i] != ' ' && str[i] != '\0')
      i++;
    q = str + i;
    reverse_word(p, q - 1);
  }
  reverse_word(str, str + len - 1);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<reverse_string>:
  endbr64
  push   %r12
  test   %rdi,%rdi
  push   %rbp
  mov    %rdi,%rbp
  setne  %dil
  push   %rbx
  movzbl %dil,%edi
  xor    %ebx,%ebx
  call   <assert@plt>
  mov    %rbp,%rdi
  call   <strlen@plt>
  movslq %eax,%r12
  cmp    $0x1,%r12d
  jle    L39
  xchg   %ax,%ax
L41:
  movslq %ebx,%rcx
  lea    0x1(%rbx),%eax
  lea    0x0(%rbp,%rcx,1),%rdi
  cltq
  movzbl (%rdi),%edx
  cmp    $0x20,%dl
  jne    L14
  nopl   (%rax)
L28:
  mov    %eax,%ebx
  lea    0x0(%rbp,%rax,1),%rdi
  mov    %rax,%rcx
  add    $0x1,%rax
  movzbl -0x1(%rbp,%rax,1),%edx
  cmp    $0x20,%dl
  je     L28
L14:
  and    $0xdf,%edx
  je     L9
  add    $0x1,%ebx
  movslq %ebx,%rax
  xchg   %ax,%ax
L45:
  movzbl 0x0(%rbp,%rax,1),%edx
  mov    %eax,%ebx
  mov    %rax,%rcx
  add    $0x1,%rax
  and    $0xdf,%edx
  jne    L45
L9:
  lea    -0x1(%rbp,%rcx,1),%rsi
  call   <reverse_word@plt>
  cmp    %ebx,%r12d
  jg     L41
  lea    -0x1(%rbp,%r12,1),%rsi
  pop    %rbx
  mov    %rbp,%rdi
  pop    %rbp
  pop    %r12
  jmp    1090 <reverse_word@plt>
L39:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void reverse_string(char *str) {
  assert(str != NULL);
  int len = strlen(str);
  if (len < 2)
    return;
  int i = 0;
  char *p = NULL;
  char *q = NULL;
  while (i < len) {
    while (str[i] == ' ' && str[i] != '\0')
      i++;
    p = str + i;
    while (str[i] != ' ' && str[i] != '\0')
      i++;
    q = str + i;
    reverse_word(p, q - 1);
  }
  reverse_word(str, str + len - 1);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<vl_tgetflag>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x28(%rbp),%rdx
  lea    -0x12(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    %eax,%edi
  call   <tgetflag@plt>
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L2
  call   <__stack_chk_fail@plt>
L2:
  leave
  ret
```

assistant: ```c
int vl_tgetflag(const char *name) {
  char temp[10];
  return tgetflag(strcpy(temp, name));
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<vl_tgetflag>:
  endbr64
  sub    $0x28,%rsp
  mov    %rdi,%rsi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0xe(%rsp),%rdi
  call   <strcpy@plt>
  mov    %eax,%edi
  call   <tgetflag@plt>
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L2
  add    $0x28,%rsp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int vl_tgetflag(const char *name) {
  char temp[10];
  return tgetflag(strcpy(temp, name));
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<vl_tgetflag>:
  endbr64
  sub    $0x28,%rsp
  mov    %rdi,%rsi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0xe(%rsp),%rdi
  call   <strcpy@plt>
  mov    %eax,%edi
  call   <tgetflag@plt>
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L3
  add    $0x28,%rsp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int vl_tgetflag(const char *name) {
  char temp[10];
  return tgetflag(strcpy(temp, name));
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<vl_tgetflag>:
  endbr64
  sub    $0x28,%rsp
  mov    %rdi,%rsi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0xe(%rsp),%rdi
  call   <strcpy@plt>
  mov    %eax,%edi
  call   <tgetflag@plt>
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L3
  add    $0x28,%rsp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int vl_tgetflag(const char *name) {
  char temp[10];
  return tgetflag(strcpy(temp, name));
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<setdec8>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x7,-0x4(%rbp)
  jmp    L4
L16:
  mov    -0x1c(%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
  mov    %edx,%ecx
  sub    %eax,%ecx
  mov    %ecx,%eax
  lea    0x30(%rax),%ecx
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    %ecx,%edx
  mov    %dl,(%rax)
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x1c(%rbp)
  subl   $0x1,-0x4(%rbp)
L4:
  cmpl   $0x0,-0x4(%rbp)
  jns    L16
  mov    -0x18(%rbp),%rax
  add    $0x8,%rax
  movb   $0x0,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void setdec8(char *s, int i) {
  int j;
  for (j = 7; j >= 0; j--) {
    s[j] = '0' + i % 10;
    i /= 10;
  }
  s[8] = 0;
  return;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<setdec8>:
  endbr64
  lea    0x7(%rdi),%rdx
L5:
  movslq %esi,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  mov    %esi,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  lea    (%rax,%rax,4),%ecx
  add    %ecx,%ecx
  sub    %ecx,%esi
  add    $0x30,%esi
  mov    %sil,(%rdx)
  mov    %eax,%esi
  mov    %rdx,%rax
  sub    $0x1,%rdx
  cmp    %rdi,%rax
  jne    L5
  movb   $0x0,0x8(%rdi)
  ret
```

assistant: ```c
void setdec8(char *s, int i) {
  int j;
  for (j = 7; j >= 0; j--) {
    s[j] = '0' + i % 10;
    i /= 10;
  }
  s[8] = 0;
  return;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<setdec8>:
  endbr64
  lea    0x7(%rdi),%rdx
  nopl   0x0(%rax,%rax,1)
L0:
  movslq %esi,%rax
  mov    %esi,%ecx
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%ecx
  sar    $0x22,%rax
  sub    %ecx,%eax
  lea    (%rax,%rax,4),%ecx
  add    %ecx,%ecx
  sub    %ecx,%esi
  add    $0x30,%esi
  mov    %sil,(%rdx)
  mov    %eax,%esi
  mov    %rdx,%rax
  sub    $0x1,%rdx
  cmp    %rdi,%rax
  jne    L0
  movb   $0x0,0x8(%rdi)
  ret
```

assistant: ```c
void setdec8(char *s, int i) {
  int j;
  for (j = 7; j >= 0; j--) {
    s[j] = '0' + i % 10;
    i /= 10;
  }
  s[8] = 0;
  return;
}
```
user: Translate the below assembly code into C programming language:

```asm
<setdec8>:
  endbr64
  movq   D3(%rip),%xmm0
  movslq %esi,%r8
  push   %rbp
  mov    %rdi,%rdx
  mov    %r8,%rax
  imul   $0x66666667,%r8,%r8
  push   %rbx
  mov    %eax,%ecx
  movb   $0x0,0x8(%rdx)
  sar    $0x1f,%ecx
  sar    $0x22,%r8
  sub    %ecx,%r8d
  movslq %r8d,%r10
  mov    %r8d,%ecx
  imul   $0x66666667,%r10,%r10
  sar    $0x1f,%ecx
  sar    $0x22,%r10
  sub    %ecx,%r10d
  movslq %r10d,%rbx
  mov    %r10d,%ecx
  imul   $0x66666667,%rbx,%rbx
  sar    $0x1f,%ecx
  sar    $0x22,%rbx
  sub    %ecx,%ebx
  movslq %ebx,%r11
  mov    %ebx,%ecx
  imul   $0x66666667,%r11,%r11
  sar    $0x1f,%ecx
  sar    $0x22,%r11
  sub    %ecx,%r11d
  movslq %r11d,%r9
  mov    %r11d,%ecx
  imul   $0x66666667,%r9,%r9
  sar    $0x1f,%ecx
  sar    $0x22,%r9
  sub    %ecx,%r9d
  movslq %r9d,%rdi
  mov    %r9d,%ecx
  imul   $0x66666667,%rdi,%rdi
  sar    $0x1f,%ecx
  sar    $0x22,%rdi
  sub    %ecx,%edi
  movslq %edi,%rsi
  mov    %edi,%ecx
  imul   $0x66666667,%rsi,%rsi
  sar    $0x1f,%ecx
  sar    $0x22,%rsi
  sub    %ecx,%esi
  movslq %esi,%rcx
  mov    %esi,%ebp
  imul   $0x66666667,%rcx,%rcx
  sar    $0x1f,%ebp
  sar    $0x22,%rcx
  sub    %ebp,%ecx
  lea    (%rcx,%rcx,4),%ebp
  mov    %esi,%ecx
  add    %ebp,%ebp
  sub    %ebp,%ecx
  lea    (%rsi,%rsi,4),%ebp
  mov    %edi,%esi
  add    %ebp,%ebp
  movzbl %cl,%ecx
  sub    %ebp,%esi
  lea    (%rdi,%rdi,4),%ebp
  mov    %r9d,%edi
  add    %ebp,%ebp
  movzbl %sil,%esi
  sub    %ebp,%edi
  lea    (%r9,%r9,4),%ebp
  mov    %r11d,%r9d
  add    %ebp,%ebp
  movzbl %dil,%edi
  sub    %ebp,%r9d
  lea    (%r11,%r11,4),%ebp
  mov    %ebx,%r11d
  add    %ebp,%ebp
  movzbl %r9b,%r9d
  sub    %ebp,%r11d
  lea    (%rbx,%rbx,4),%ebp
  mov    %r10d,%ebx
  add    %ebp,%ebp
  movzbl %r11b,%r11d
  sub    %ebp,%ebx
  lea    (%r10,%r10,4),%ebp
  mov    %r8d,%r10d
  lea    (%r8,%r8,4),%r8d
  add    %ebp,%ebp
  add    %r8d,%r8d
  sub    %ebp,%r10d
  sub    %r8d,%eax
  movzbl %r10b,%r8d
  movzbl %al,%eax
  shl    $0x8,%rax
  or     %r8,%rax
  movzbl %bl,%r8d
  pop    %rbx
  pop    %rbp
  shl    $0x8,%rax
  or     %r8,%rax
  shl    $0x8,%rax
  or     %r11,%rax
  shl    $0x8,%rax
  or     %r9,%rax
  shl    $0x8,%rax
  or     %rdi,%rax
  shl    $0x8,%rax
  or     %rsi,%rax
  shl    $0x8,%rax
  or     %rcx,%rax
  movq   %rax,%xmm1
  paddb  %xmm1,%xmm0
  movq   %xmm0,(%rdx)
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D3
datatype: byte[8]

tool: {"D3": [0x30,0x30,0x30,0x30,0x30,0x30,0x30,0x30]}

assistant: ```c
void setdec8(char *s, int i) {
  int j;
  for (j = 7; j >= 0; j--) {
    s[j] = '0' + i % 10;
    i /= 10;
  }
  s[8] = 0;
  return;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<greetingsMsg>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D87(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D58(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D21(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D73(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D89(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D82(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D100(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D58(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D42(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
  pop    %rbp
  ret
```

tool_calls: name: read_data
arguments: data_tag: D0
type: string
name: read_data
arguments: data_tag: D87
type: string
name: read_data
arguments: data_tag: D58
type: string
name: read_data
arguments: data_tag: D21
type: string
name: read_data
arguments: data_tag: D73
type: string
name: read_data
arguments: data_tag: D89
type: string
name: read_data
arguments: data_tag: D82
type: string
name: read_data
arguments: data_tag: D8
type: string
name: read_data
arguments: data_tag: D100
type: string
name: read_data
arguments: data_tag: D42
type: string

tool: D0: This is the program to give out solution for a cubic function in Latex form.


tool: D87: The program will accept integer or rational number(x/y with x,y are integers).


tool: D58: ________________________________________________________________________________


tool: D21: The program use integer as a base starage type for operation.


tool: D73: It can reduce fractions to its irreducible form.


tool: D89: Root reduction is not an available feature.


tool: D82: The result is in Latex format.


tool: D8: Please copy the result in this program to https ://www.codecogs.com/latex/eqneditor.php 
in order to get a mathematical expression of that result.


tool: D100: Accepted input methods: a, a/b with a, b are integers.


tool: D42: Equation to solve: A*X^3 + B*X^2 + C*X + D = 0


assistant: ```c
void greetingsMsg() {
  printf("This is the program to give out solution for a cubic function in Latex form.\n");
  printf("The program will accept integer or rational number(x/y with x,y are integers).\n");
  printf("________________________________________________________________________________\n");
  printf("The program use integer as a base starage type for operation.\n");
  printf("It can reduce fractions to its irreducible form.\n");
  printf("Root reduction is not an available feature.\n");
  printf("The result is in Latex format.\n");
  printf("Please copy the result in this program to https ://www.codecogs.com/latex/eqneditor.php \nin order to get a mathematical expression of that result.\n");
  printf("Accepted input methods: a, a/b with a, b are integers.\n");
  printf("________________________________________________________________________________\n");
  printf("Equation to solve: A*X^3 + B*X^2 + C*X + D = 0\n");
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<greetingsMsg>:
  endbr64
  push   %rbx
  lea    D93(%rip),%rdi
  call   <printf@plt>
  lea    D21(%rip),%rdi
  call   <printf@plt>
  lea    D94(%rip),%rbx
  mov    %rbx,%rdi
  call   <printf@plt>
  lea    D25(%rip),%rdi
  call   <printf@plt>
  lea    D43(%rip),%rdi
  call   <printf@plt>
  lea    D53(%rip),%rdi
  call   <printf@plt>
  lea    D12(%rip),%rdi
  call   <printf@plt>
  lea    D91(%rip),%rdi
  call   <printf@plt>
  lea    D52(%rip),%rdi
  call   <printf@plt>
  mov    %rbx,%rdi
  call   <printf@plt>
  lea    D1(%rip),%rdi
  call   <printf@plt>
  pop    %rbx
  ret
```

tool_calls: name: retrieve_data
arguments: label: D93
data_type: string
name: retrieve_data
arguments: label: D21
data_type: string
name: retrieve_data
arguments: label: D94
data_type: string
name: retrieve_data
arguments: label: D25
data_type: string
name: retrieve_data
arguments: label: D43
data_type: string
name: retrieve_data
arguments: label: D53
data_type: string
name: retrieve_data
arguments: label: D12
data_type: string
name: retrieve_data
arguments: label: D91
data_type: string
name: retrieve_data
arguments: label: D52
data_type: string
name: retrieve_data
arguments: label: D1
data_type: string

tool: D93: This is the program to give out solution for a cubic function in Latex form.


tool: D21: The program will accept integer or rational number(x/y with x,y are integers).


tool: D94: ________________________________________________________________________________


tool: D25: The program use integer as a base starage type for operation.


tool: D43: It can reduce fractions to its irreducible form.


tool: D53: Root reduction is not an available feature.


tool: D12: The result is in Latex format.


tool: D91: Please copy the result in this program to https ://www.codecogs.com/latex/eqneditor.php 
in order to get a mathematical expression of that result.


tool: D52: Accepted input methods: a, a/b with a, b are integers.


tool: D1: Equation to solve: A*X^3 + B*X^2 + C*X + D = 0


assistant: ```c
void greetingsMsg() {
  printf("This is the program to give out solution for a cubic function in Latex form.\n");
  printf("The program will accept integer or rational number(x/y with x,y are integers).\n");
  printf("________________________________________________________________________________\n");
  printf("The program use integer as a base starage type for operation.\n");
  printf("It can reduce fractions to its irreducible form.\n");
  printf("Root reduction is not an available feature.\n");
  printf("The result is in Latex format.\n");
  printf("Please copy the result in this program to https ://www.codecogs.com/latex/eqneditor.php \nin order to get a mathematical expression of that result.\n");
  printf("Accepted input methods: a, a/b with a, b are integers.\n");
  printf("________________________________________________________________________________\n");
  printf("Equation to solve: A*X^3 + B*X^2 + C*X + D = 0\n");
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<greetingsMsg>:
  endbr64
  push   %rbp
  lea    D18(%rip),%rdi
  lea    D42(%rip),%rbp
  call   <printf@plt>
  lea    D85(%rip),%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <printf@plt>
  lea    D71(%rip),%rdi
  call   <printf@plt>
  lea    D26(%rip),%rdi
  call   <printf@plt>
  lea    D48(%rip),%rdi
  call   <printf@plt>
  lea    D11(%rip),%rdi
  call   <printf@plt>
  lea    D97(%rip),%rdi
  call   <printf@plt>
  lea    D66(%rip),%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <printf@plt>
  lea    D9(%rip),%rdi
  pop    %rbp
  jmp    1050 <printf@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D18
type: string
name: disassemble_data
arguments: data_tag: D42
type: string
name: disassemble_data
arguments: data_tag: D85
type: string
name: disassemble_data
arguments: data_tag: D71
type: string
name: disassemble_data
arguments: data_tag: D26
type: string
name: disassemble_data
arguments: data_tag: D48
type: string
name: disassemble_data
arguments: data_tag: D11
type: string
name: disassemble_data
arguments: data_tag: D97
type: string
name: disassemble_data
arguments: data_tag: D66
type: string
name: disassemble_data
arguments: data_tag: D9
type: string

tool: D18: This is the program to give out solution for a cubic function in Latex form.


tool: D42: ________________________________________________________________________________


tool: D85: The program will accept integer or rational number(x/y with x,y are integers).


tool: D71: The program use integer as a base starage type for operation.


tool: D26: It can reduce fractions to its irreducible form.


tool: D48: Root reduction is not an available feature.


tool: D11: The result is in Latex format.


tool: D97: Please copy the result in this program to https ://www.codecogs.com/latex/eqneditor.php 
in order to get a mathematical expression of that result.


tool: D66: Accepted input methods: a, a/b with a, b are integers.


tool: D9: Equation to solve: A*X^3 + B*X^2 + C*X + D = 0


assistant: ```c
void greetingsMsg() {
  printf("This is the program to give out solution for a cubic function in Latex form.\n");
  printf("The program will accept integer or rational number(x/y with x,y are integers).\n");
  printf("________________________________________________________________________________\n");
  printf("The program use integer as a base starage type for operation.\n");
  printf("It can reduce fractions to its irreducible form.\n");
  printf("Root reduction is not an available feature.\n");
  printf("The result is in Latex format.\n");
  printf("Please copy the result in this program to https ://www.codecogs.com/latex/eqneditor.php \nin order to get a mathematical expression of that result.\n");
  printf("Accepted input methods: a, a/b with a, b are integers.\n");
  printf("________________________________________________________________________________\n");
  printf("Equation to solve: A*X^3 + B*X^2 + C*X + D = 0\n");
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<greetingsMsg>:
  endbr64
  push   %rbp
  lea    D30(%rip),%rdi
  lea    D2(%rip),%rbp
  call   <printf@plt>
  lea    D93(%rip),%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <printf@plt>
  lea    D50(%rip),%rdi
  call   <printf@plt>
  lea    D55(%rip),%rdi
  call   <printf@plt>
  lea    D89(%rip),%rdi
  call   <printf@plt>
  lea    D63(%rip),%rdi
  call   <printf@plt>
  lea    D24(%rip),%rdi
  call   <printf@plt>
  lea    D0(%rip),%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <printf@plt>
  lea    D26(%rip),%rdi
  pop    %rbp
  jmp    1050 <printf@plt>
```

tool_calls: name: unpack_data
arguments: tag: D30
type_guessed: string
name: unpack_data
arguments: tag: D2
type_guessed: string
name: unpack_data
arguments: tag: D93
type_guessed: string
name: unpack_data
arguments: tag: D50
type_guessed: string
name: unpack_data
arguments: tag: D55
type_guessed: string
name: unpack_data
arguments: tag: D89
type_guessed: string
name: unpack_data
arguments: tag: D63
type_guessed: string
name: unpack_data
arguments: tag: D24
type_guessed: string
name: unpack_data
arguments: tag: D0
type_guessed: string
name: unpack_data
arguments: tag: D26
type_guessed: string

tool: D30: This is the program to give out solution for a cubic function in Latex form.


tool: D2: ________________________________________________________________________________


tool: D93: The program will accept integer or rational number(x/y with x,y are integers).


tool: D50: The program use integer as a base starage type for operation.


tool: D55: It can reduce fractions to its irreducible form.


tool: D89: Root reduction is not an available feature.


tool: D63: The result is in Latex format.


tool: D24: Please copy the result in this program to https ://www.codecogs.com/latex/eqneditor.php 
in order to get a mathematical expression of that result.


tool: D0: Accepted input methods: a, a/b with a, b are integers.


tool: D26: Equation to solve: A*X^3 + B*X^2 + C*X + D = 0


assistant: ```c
void greetingsMsg() {
  printf("This is the program to give out solution for a cubic function in Latex form.\n");
  printf("The program will accept integer or rational number(x/y with x,y are integers).\n");
  printf("________________________________________________________________________________\n");
  printf("The program use integer as a base starage type for operation.\n");
  printf("It can reduce fractions to its irreducible form.\n");
  printf("Root reduction is not an available feature.\n");
  printf("The result is in Latex format.\n");
  printf("Please copy the result in this program to https ://www.codecogs.com/latex/eqneditor.php \nin order to get a mathematical expression of that result.\n");
  printf("Accepted input methods: a, a/b with a, b are integers.\n");
  printf("________________________________________________________________________________\n");
  printf("Equation to solve: A*X^3 + B*X^2 + C*X + D = 0\n");
}
```