ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Translate the below assembly code into C programming language:

```asm
<filein_stats_protein>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <open_infile@plt>
L4:
  mov    $0x0,%eax
  call   <read_infile@plt>
  mov    %eax,%ebx
  mov    $0x0,%eax
  call   <read_infile@plt>
  mov    %eax,%ebp
  mov    $0x0,%eax
  call   <read_infile@plt>
  movsbl %al,%edx
  movsbl %bpl,%esi
  movsbl %bl,%edi
  call   <bioseq_dna_codon_protein@plt>
  cmp    $0x3f,%al
  je     L8
  test   %al,%al
  jne    L4
L8:
  lea    D8(%rip),%rdi
  call   <puts@plt>
  mov    $0x0,%eax
  call   <close_infile@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D8
type_guessed: string

tool: 

assistant: ```c
void filein_stats_protein() {
  char in[3] = {'\0', '\0', '\0'};
  int cont = 1;
  open_infile();
  while (cont == 1) {
    in[0] = read_infile();
    in[1] = read_infile();
    in[2] = read_infile();
    char out = bioseq_dna_codon_protein(in[0], in[1], in[2]);
    if (out == '?' || out == '\0')
      cont = 0;
  }
  puts("");
  close_infile();
}
```
user: Translate the below assembly code into C programming language:

```asm
<filein_stats_protein>:
  endbr64
  push   %rbp
  xor    %eax,%eax
  push   %rbx
  sub    $0x8,%rsp
  call   <open_infile@plt>
  nopl   0x0(%rax)
L17:
  xor    %eax,%eax
  call   <read_infile@plt>
  mov    %eax,%ebx
  xor    %eax,%eax
  call   <read_infile@plt>
  mov    %eax,%ebp
  xor    %eax,%eax
  call   <read_infile@plt>
  movsbl %bpl,%esi
  movsbl %bl,%edi
  movsbl %al,%edx
  call   <bioseq_dna_codon_protein@plt>
  cmp    $0x3f,%al
  je     L1
  test   %al,%al
  jne    L17
L1:
  lea    D6(%rip),%rdi
  call   <puts@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  jmp    1090 <close_infile@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D6
type: string

tool: 

assistant: ```c
void filein_stats_protein() {
  char in[3] = {'\0', '\0', '\0'};
  int cont = 1;
  open_infile();
  while (cont == 1) {
    in[0] = read_infile();
    in[1] = read_infile();
    in[2] = read_infile();
    char out = bioseq_dna_codon_protein(in[0], in[1], in[2]);
    if (out == '?' || out == '\0')
      cont = 0;
  }
  puts("");
  close_infile();
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<filein_stats_protein>:
  endbr64
  push   %rbp
  xor    %eax,%eax
  push   %rbx
  sub    $0x8,%rsp
  call   <open_infile@plt>
  nopl   0x0(%rax)
L12:
  xor    %eax,%eax
  call   <read_infile@plt>
  mov    %eax,%ebx
  xor    %eax,%eax
  call   <read_infile@plt>
  mov    %eax,%ebp
  xor    %eax,%eax
  call   <read_infile@plt>
  movsbl %bpl,%esi
  movsbl %bl,%edi
  movsbl %al,%edx
  call   <bioseq_dna_codon_protein@plt>
  cmp    $0x3f,%al
  je     L16
  test   %al,%al
  jne    L12
L16:
  lea    D0(%rip),%rdi
  call   <puts@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  jmp    1090 <close_infile@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D0
type_guessed: string

tool: 

assistant: ```c
void filein_stats_protein() {
  char in[3] = {'\0', '\0', '\0'};
  int cont = 1;
  open_infile();
  while (cont == 1) {
    in[0] = read_infile();
    in[1] = read_infile();
    in[2] = read_infile();
    char out = bioseq_dna_codon_protein(in[0], in[1], in[2]);
    if (out == '?' || out == '\0')
      cont = 0;
  }
  puts("");
  close_infile();
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<cc_format_pos_rank_new>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  js     L1
  cmpl   $0x19,-0x14(%rbp)
  jle    L26
L1:
  mov    $0x0,%eax
  jmp    L10
L26:
  mov    $0x4,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x14(%rbp),%eax
  add    $0x1,%eax
  movzbl %al,%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,%ecx
  lea    D8(%rip),%rdx
  mov    $0x3,%esi
  mov    %rax,%rdi
  call   <snprintf@plt>
  mov    -0x8(%rbp),%rax
L10:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D8
data_type: string

tool: D8: %-hhu

assistant: ```c
char *cc_format_pos_rank_new(int const j) {
  if ((j < (0)) || ((25) < j))
    return NULL;
  char *new = (char *)malloc(4);
  snprintf(new, 3, "%-hhu", (unsigned char)(j + 1));
  return new;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<cc_format_pos_rank_new>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    $0x0,%ebp
  cmp    $0x19,%edi
  ja     L8
  mov    %edi,%ebx
  mov    $0x4,%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  lea    0x1(%rbx),%ecx
  movzbl %cl,%ecx
  lea    D0(%rip),%rdx
  mov    $0x3,%esi
  mov    %rax,%rdi
  call   <snprintf@plt>
L8:
  mov    %rbp,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

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

tool: D0: %-hhu

assistant: ```c
char *cc_format_pos_rank_new(int const j) {
  if ((j < (0)) || ((25) < j))
    return NULL;
  char *new = (char *)malloc(4);
  snprintf(new, 3, "%-hhu", (unsigned char)(j + 1));
  return new;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<cc_format_pos_rank_new>:
  endbr64
  push   %r12
  xor    %r12d,%r12d
  push   %rbx
  sub    $0x8,%rsp
  cmp    $0x19,%edi
  ja     L9
  mov    %edi,%ebx
  mov    $0x4,%edi
  call   <malloc@plt>
  lea    0x1(%rbx),%ecx
  mov    $0x3,%esi
  lea    D6(%rip),%rdx
  movzbl %cl,%ecx
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <snprintf@plt>
L9:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  ret
```

tool_calls: name: parse_data
arguments: label: D6
data_type: string

tool: D6: %-hhu

assistant: ```c
char *cc_format_pos_rank_new(int const j) {
  if ((j < (0)) || ((25) < j))
    return NULL;
  char *new = (char *)malloc(4);
  snprintf(new, 3, "%-hhu", (unsigned char)(j + 1));
  return new;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<cc_format_pos_rank_new>:
  endbr64
  push   %r12
  xor    %r12d,%r12d
  push   %rbx
  sub    $0x8,%rsp
  cmp    $0x19,%edi
  ja     L5
  mov    %edi,%ebx
  mov    $0x4,%edi
  call   <malloc@plt>
  lea    0x1(%rbx),%ecx
  mov    $0x3,%esi
  lea    D3(%rip),%rdx
  movzbl %cl,%ecx
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <snprintf@plt>
L5:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D3
type_guessed: string

tool: D3: %-hhu

assistant: ```c
char *cc_format_pos_rank_new(int const j) {
  if ((j < (0)) || ((25) < j))
    return NULL;
  char *new = (char *)malloc(4);
  snprintf(new, 3, "%-hhu", (unsigned char)(j + 1));
  return new;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <fred@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main() {
  fred();
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <fred@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  fred();
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <fred@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  fred();
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <fred@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  fred();
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<strtok_csv>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  cmpq   $0x0,-0x20(%rbp)
  je     L127
  cmpq   $0x0,-0x28(%rbp)
  jne    L85
L127:
  mov    $0x0,%eax
  jmp    L26
L85:
  cmpq   $0x0,-0x18(%rbp)
  jne    L74
  mov    -0x28(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,-0x10(%rbp)
  cmpq   $0x0,-0x10(%rbp)
  jne    L7
  mov    $0x0,%eax
  jmp    L26
L74:
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x10(%rbp)
L7:
  mov    -0x20(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strstr@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L79
  mov    -0x28(%rbp),%rax
  movq   $0x0,(%rax)
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  movslq %eax,%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x8(%rbp)
  jmp    L138
L79:
  mov    -0x8(%rbp),%rax
  movb   $0x0,(%rax)
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    -0x28(%rbp),%rax
  mov    %rdx,(%rax)
  jmp    L138
L46:
  addq   $0x1,-0x10(%rbp)
L138:
  mov    -0x10(%rbp),%rax
  cmp    -0x8(%rbp),%rax
  jae    L36
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L46
L36:
  subq   $0x1,-0x8(%rbp)
  jmp    L97
L89:
  mov    -0x8(%rbp),%rax
  movb   $0x0,(%rax)
  subq   $0x1,-0x8(%rbp)
L97:
  mov    -0x8(%rbp),%rax
  cmp    -0x10(%rbp),%rax
  jbe    L133
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L89
L133:
  mov    -0x10(%rbp),%rax
L26:
  leave
  ret
```

assistant: ```c
char *strtok_csv(char *string, const char *separators, char **savedptr) {
  char *p, *s;
  if (separators == NULL || savedptr == NULL) {
    return NULL;
  }
  if (string == NULL) {
    p = *savedptr;
    if (p == NULL) {
      return NULL;
    }
  } else {
    p = string;
  }
  s = strstr(p, separators);
  if (s == NULL) {
    *savedptr = NULL;
    s = p + strlen(p);
  } else {
    *s = 0;
    *savedptr = s + 1;
  }
  while (p < s && isspace(*p)) {
    ++p;
  }
  --s;
  while (s > p && isspace(*s)) {
    *s = 0;
    --s;
  }
  return p;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<strtok_csv>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  test   %rsi,%rsi
  je     L39
  mov    %rdi,%rbp
  mov    %rdx,%r12
  test   %rdx,%rdx
  je     L39
  test   %rdi,%rdi
  je     L63
L86:
  mov    %rbp,%rdi
  call   <strstr@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L34
  movb   $0x0,(%rax)
  lea    0x1(%rax),%rax
  mov    %rax,(%r12)
L106:
  cmp    %rbx,%rbp
  jae    L99
L38:
  movsbl 0x0(%rbp),%edi
  call   <isspace@plt>
  test   %rax,%rax
  je     L99
  add    $0x1,%rbp
  cmp    %rbx,%rbp
  jne    L38
  mov    %rbx,%rbp
L99:
  sub    $0x1,%rbx
  cmp    %rbp,%rbx
  jbe    L28
L128:
  movsbl (%rbx),%edi
  call   <isspace@plt>
  test   %rax,%rax
  je     L28
  movb   $0x0,(%rbx)
  sub    $0x1,%rbx
  cmp    %rbp,%rbx
  jne    L128
L28:
  mov    %rbp,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L63:
  mov    (%rdx),%rbp
  test   %rbp,%rbp
  jne    L86
  jmp    L28
L34:
  movq   $0x0,(%r12)
  mov    %rbp,%rdi
  call   <strlen@plt>
  movslq %eax,%rbx
  add    %rbp,%rbx
  jmp    L106
L39:
  mov    $0x0,%ebp
  jmp    L28
```

assistant: ```c
char *strtok_csv(char *string, const char *separators, char **savedptr) {
  char *p, *s;
  if (separators == NULL || savedptr == NULL) {
    return NULL;
  }
  if (string == NULL) {
    p = *savedptr;
    if (p == NULL) {
      return NULL;
    }
  } else {
    p = string;
  }
  s = strstr(p, separators);
  if (s == NULL) {
    *savedptr = NULL;
    s = p + strlen(p);
  } else {
    *s = 0;
    *savedptr = s + 1;
  }
  while (p < s && isspace(*p)) {
    ++p;
  }
  --s;
  while (s > p && isspace(*s)) {
    *s = 0;
    --s;
  }
  return p;
}
```
user: What is the c source code of the assembly code below?

```asm
<strtok_csv>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  test   %rsi,%rsi
  je     L50
  mov    %rdx,%r12
  test   %rdx,%rdx
  je     L50
  mov    %rdi,%rbp
  test   %rdi,%rdi
  je     L41
L82:
  mov    %rbp,%rdi
  call   <strstr@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L132
  movb   $0x0,(%rax)
  lea    0x1(%rax),%rax
  mov    %rax,(%r12)
L33:
  cmp    %rbx,%rbp
  jb     L71
  jmp    L9
  nop
L97:
  add    $0x1,%rbp
  cmp    %rbx,%rbp
  je     L93
L71:
  movsbl 0x0(%rbp),%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L97
L9:
  mov    %rbp,%r12
L68:
  sub    $0x1,%rbx
  cmp    %r12,%rbx
  ja     L95
  jmp    L140
  nopl   (%rax)
L20:
  movb   $0x0,(%rbx)
  sub    $0x1,%rbx
  cmp    %r12,%rbx
  je     L140
L95:
  movsbl (%rbx),%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L20
L140:
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L93:
  mov    %rbx,%r12
  jmp    L68
  nopl   (%rax)
L41:
  mov    (%rdx),%rbp
  test   %rbp,%rbp
  jne    L82
L50:
  xor    %r12d,%r12d
  pop    %rbx
  pop    %rbp
  mov    %r12,%rax
  pop    %r12
  ret
  nop
L132:
  movq   $0x0,(%r12)
  mov    %rbp,%rdi
  call   <strlen@plt>
  movslq %eax,%rbx
  add    %rbp,%rbx
  jmp    L33
```

assistant: ```c
char *strtok_csv(char *string, const char *separators, char **savedptr) {
  char *p, *s;
  if (separators == NULL || savedptr == NULL) {
    return NULL;
  }
  if (string == NULL) {
    p = *savedptr;
    if (p == NULL) {
      return NULL;
    }
  } else {
    p = string;
  }
  s = strstr(p, separators);
  if (s == NULL) {
    *savedptr = NULL;
    s = p + strlen(p);
  } else {
    *s = 0;
    *savedptr = s + 1;
  }
  while (p < s && isspace(*p)) {
    ++p;
  }
  --s;
  while (s > p && isspace(*s)) {
    *s = 0;
    --s;
  }
  return p;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<strtok_csv>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  test   %rsi,%rsi
  je     L40
  mov    %rdx,%r12
  test   %rdx,%rdx
  je     L40
  mov    %rdi,%rbp
  test   %rdi,%rdi
  je     L37
L126:
  mov    %rbp,%rdi
  call   <strstr@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L100
  movb   $0x0,(%rax)
  lea    0x1(%rax),%rax
  mov    %rax,(%r12)
L84:
  cmp    %rbx,%rbp
  jb     L83
  jmp    L2
  nop
L73:
  add    $0x1,%rbp
  cmp    %rbx,%rbp
  je     L124
L83:
  movsbl 0x0(%rbp),%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L73
L2:
  mov    %rbp,%r12
L32:
  sub    $0x1,%rbx
  cmp    %r12,%rbx
  ja     L63
  jmp    L107
  nopl   (%rax)
L77:
  movb   $0x0,(%rbx)
  sub    $0x1,%rbx
  cmp    %r12,%rbx
  je     L107
L63:
  movsbl (%rbx),%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L77
L107:
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L124:
  mov    %rbx,%r12
  jmp    L32
  nopl   (%rax)
L37:
  mov    (%rdx),%rbp
  test   %rbp,%rbp
  jne    L126
L40:
  xor    %r12d,%r12d
  pop    %rbx
  pop    %rbp
  mov    %r12,%rax
  pop    %r12
  ret
  nop
L100:
  movq   $0x0,(%r12)
  mov    %rbp,%rdi
  call   <strlen@plt>
  movslq %eax,%rbx
  add    %rbp,%rbx
  jmp    L84
```

assistant: ```c
char *strtok_csv(char *string, const char *separators, char **savedptr) {
  char *p, *s;
  if (separators == NULL || savedptr == NULL) {
    return NULL;
  }
  if (string == NULL) {
    p = *savedptr;
    if (p == NULL) {
      return NULL;
    }
  } else {
    p = string;
  }
  s = strstr(p, separators);
  if (s == NULL) {
    *savedptr = NULL;
    s = p + strlen(p);
  } else {
    *s = 0;
    *savedptr = s + 1;
  }
  while (p < s && isspace(*p)) {
    ++p;
  }
  --s;
  while (s > p && isspace(*s)) {
    *s = 0;
    --s;
  }
  return p;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<stb_duplower>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strdup@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  jmp    L19
L9:
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <tolower@plt>
  mov    -0x10(%rbp),%rdx
  mov    %al,(%rdx)
  addq   $0x1,-0x10(%rbp)
L19:
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L9
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *stb_duplower(char *s) {
  char *p = strdup(s), *q = p;
  while (*q) {
    *q = tolower(*q);
    ++q;
  }
  return p;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<stb_duplower>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <strdup@plt>
  mov    %rax,%rbp
  movzbl (%rax),%edi
  test   %dil,%dil
  je     L11
  mov    %rax,%rbx
L7:
  movsbl %dil,%edi
  call   <tolower@plt>
  mov    %al,(%rbx)
  add    $0x1,%rbx
  movzbl (%rbx),%edi
  test   %dil,%dil
  jne    L7
L11:
  mov    %rbp,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
char *stb_duplower(char *s) {
  char *p = strdup(s), *q = p;
  while (*q) {
    *q = tolower(*q);
    ++q;
  }
  return p;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<stb_duplower>:
  endbr64
  push   %r12
  push   %rbx
  sub    $0x8,%rsp
  call   <strdup@plt>
  movsbl (%rax),%edi
  mov    %rax,%r12
  test   %dil,%dil
  je     L12
  mov    %rax,%rbx
  xchg   %ax,%ax
L2:
  call   <tolower@plt>
  add    $0x1,%rbx
  mov    %al,-0x1(%rbx)
  movsbl (%rbx),%edi
  test   %dil,%dil
  jne    L2
L12:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  ret
```

assistant: ```c
char *stb_duplower(char *s) {
  char *p = strdup(s), *q = p;
  while (*q) {
    *q = tolower(*q);
    ++q;
  }
  return p;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<stb_duplower>:
  endbr64
  push   %r12
  push   %rbx
  sub    $0x8,%rsp
  call   <strdup@plt>
  movsbl (%rax),%edi
  mov    %rax,%r12
  test   %dil,%dil
  je     L9
  mov    %rax,%rbx
  xchg   %ax,%ax
L2:
  call   <tolower@plt>
  add    $0x1,%rbx
  mov    %al,-0x1(%rbx)
  movsbl (%rbx),%edi
  test   %dil,%dil
  jne    L2
L9:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  ret
```

assistant: ```c
char *stb_duplower(char *s) {
  char *p = strdup(s), *q = p;
  while (*q) {
    *q = tolower(*q);
    ++q;
  }
  return p;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0x0,-0x10(%rbp)
  movl   $0x0,-0x14(%rbp)
  movl   $0x0,-0x24(%rbp)
  movl   $0x0,-0x20(%rbp)
  movl   $0x0,-0x1c(%rbp)
  pxor   %xmm0,%xmm0
  movss  %xmm0,-0x18(%rbp)
  mov    $0x190,%edi
  call   <malloc@plt>
  mov    %rax,-0x10(%rbp)
  lea    -0x18(%rbp),%rcx
  lea    -0x20(%rbp),%rdx
  lea    -0x24(%rbp),%rax
  mov    %rax,%rsi
  lea    D16(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  jmp    L29
L24:
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x10(%rbp),%rax
  add    %rax,%rdx
  mov    -0x1c(%rbp),%eax
  mov    %eax,(%rdx)
  addl   $0x1,-0x14(%rbp)
L29:
  lea    -0x1c(%rbp),%rax
  mov    %rax,%rsi
  lea    D20(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  cmp    $0x1,%eax
  je     L24
  mov    -0x14(%rbp),%edx
  mov    -0x10(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <mostra@plt>
  mov    -0x20(%rbp),%edx
  mov    -0x24(%rbp),%ecx
  mov    -0x10(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <polinomio@plt>
  pxor   %xmm0,%xmm0
  cvtsi2sdl -0x14(%rbp),%xmm0
  mov    -0x18(%rbp),%ecx
  mov    -0x20(%rbp),%edx
  mov    -0x24(%rbp),%eax
  movapd %xmm0,%xmm1
  movd   %ecx,%xmm0
  mov    %eax,%esi
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x18(%rbp),%esi
  mov    -0x20(%rbp),%edx
  mov    -0x24(%rbp),%ecx
  mov    -0x10(%rbp),%rax
  movd   %esi,%xmm0
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <pcd@plt>
  movq   %xmm0,%rsi
  mov    -0x18(%rbp),%ecx
  mov    -0x20(%rbp),%edx
  mov    -0x24(%rbp),%eax
  movq   %rsi,%xmm1
  movd   %ecx,%xmm0
  mov    %eax,%esi
  lea    D35(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L8
  call   <__stack_chk_fail@plt>
L8:
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D16
type: string
name: parse_data
arguments: data_tag: D20
type: string
name: parse_data
arguments: data_tag: D10
type: string
name: parse_data
arguments: data_tag: D35
type: string

tool: D16: %d %d %f


tool: D20: %d 

tool: D10: Dados> a: %d, b: %d, x: %0.2f, tam: %d


tool: D35: p_%d,%d(%0.2f) = %0.2f 


assistant: ```c
int main() {
  int *v = NULL, tam = 0, a = 0, b = 0;
  int lido = 0;
  float x = 0;
  v = malloc(sizeof(int) * 100);
  scanf("%d %d %f\n", &a, &b, &x);
  while (scanf("%d ", &lido) == 1) {
    v[tam] = lido;
    tam++;
  }
  if (1) {
    mostra(v, tam);
    polinomio(v, a, b);
    printf("Dados> a: %d, b: %d, x: %0.2f, tam: %d\n", a, b, x, tam);
  }
  printf("p_%d,%d(%0.2f) = %0.2f \n", a, b, x, pcd(v, a, b, x));
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movl   $0x0,0x8(%rsp)
  movl   $0x0,0xc(%rsp)
  movl   $0x0,0x10(%rsp)
  movl   $0x0,0x14(%rsp)
  mov    $0x190,%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  lea    0x14(%rsp),%rcx
  lea    0xc(%rsp),%rdx
  lea    0x8(%rsp),%rsi
  lea    D9(%rip),%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    $0x0,%ebx
  lea    D36(%rip),%r12
  jmp    L6
L3:
  mov    0x10(%rsp),%eax
  mov    %eax,0x0(%rbp,%rbx,4)
  add    $0x1,%rbx
L6:
  lea    0x10(%rsp),%rsi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  cmp    $0x1,%eax
  je     L3
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <mostra@plt>
  mov    0xc(%rsp),%edx
  mov    0x8(%rsp),%esi
  mov    %rbp,%rdi
  call   <polinomio@plt>
  pxor   %xmm1,%xmm1
  cvtsi2sd %ebx,%xmm1
  movss  0x14(%rsp),%xmm0
  mov    0xc(%rsp),%edx
  mov    0x8(%rsp),%esi
  lea    D15(%rip),%rdi
  call   <printf@plt>
  movss  0x14(%rsp),%xmm0
  mov    0xc(%rsp),%edx
  mov    0x8(%rsp),%esi
  mov    %rbp,%rdi
  call   <pcd@plt>
  movapd %xmm0,%xmm1
  movss  0x14(%rsp),%xmm0
  mov    0xc(%rsp),%edx
  mov    0x8(%rsp),%esi
  lea    D8(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L17
  mov    $0x0,%eax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L17:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D9
datatype: string
name: retrieve_data
arguments: tag: D36
datatype: string
name: retrieve_data
arguments: tag: D15
datatype: string
name: retrieve_data
arguments: tag: D8
datatype: string

tool: D9: %d %d %f


tool: D36: %d 

tool: D15: Dados> a: %d, b: %d, x: %0.2f, tam: %d


tool: D8: p_%d,%d(%0.2f) = %0.2f 


assistant: ```c
int main() {
  int *v = NULL, tam = 0, a = 0, b = 0;
  int lido = 0;
  float x = 0;
  v = malloc(sizeof(int) * 100);
  scanf("%d %d %f\n", &a, &b, &x);
  while (scanf("%d ", &lido) == 1) {
    v[tam] = lido;
    tam++;
  }
  if (1) {
    mostra(v, tam);
    polinomio(v, a, b);
    printf("Dados> a: %d, b: %d, x: %0.2f, tam: %d\n", a, b, x, tam);
  }
  printf("p_%d,%d(%0.2f) = %0.2f \n", a, b, x, pcd(v, a, b, x));
  return 0;
}
```