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

```asm
<fmt_hex_bytes>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    -0x40(%rbp),%rax
  mov    %rax,%rdi
  call   <fmt_strlen@plt>
  mov    %rax,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  and    $0x1,%eax
  test   %rax,%rax
  je     L39
  movq   $0x0,-0x38(%rbp)
  mov    $0x0,%eax
  jmp    L34
L39:
  mov    -0x20(%rbp),%rax
  shr    %rax
  mov    %rax,-0x18(%rbp)
  movq   $0x0,-0x30(%rbp)
  movq   $0x0,-0x28(%rbp)
  jmp    L33
L21:
  mov    -0x40(%rbp),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <_hex_nib@plt>
  shl    $0x4,%eax
  mov    %eax,%ebx
  mov    -0x30(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    -0x40(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <_hex_nib@plt>
  mov    %eax,%edx
  mov    -0x28(%rbp),%rax
  lea    0x0(,%rax,4),%rcx
  mov    -0x38(%rbp),%rax
  add    %rcx,%rax
  or     %ebx,%edx
  mov    %edx,(%rax)
  addq   $0x2,-0x30(%rbp)
  addq   $0x1,-0x28(%rbp)
L33:
  mov    -0x28(%rbp),%rax
  cmp    -0x18(%rbp),%rax
  jb     L21
  mov    -0x18(%rbp),%rax
L34:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
size_t fmt_hex_bytes(uint8_t *out, const char *hex) {
  size_t len = fmt_strlen(hex);
  if (len & 1) {
    out = NULL;
    return 0;
  }
  size_t final_len = len >> 1;
  for (size_t i = 0, j = 0; j < final_len; i += 2, j++) {
    out[j] = (_hex_nib(hex[i]) << 4) | _hex_nib(hex[i + 1]);
  }
  return final_len;
}
```
user: Translate the below assembly code into C programming language:

```asm
<fmt_hex_bytes>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    %rsi,%rbp
  mov    %rsi,%rdi
  call   <fmt_strlen@plt>
  mov    $0x0,%r14d
  test   $0x1,%al
  jne    L13
  shr    %rax
  mov    %rax,%r14
  je     L13
  add    $0x1,%rbp
  lea    0x0(%rbp,%rax,2),%r13
L6:
  movsbl -0x1(%rbp),%edi
  call   <_hex_nib@plt>
  mov    %eax,%ebx
  movsbl 0x0(%rbp),%edi
  call   <_hex_nib@plt>
  shl    $0x4,%ebx
  or     %eax,%ebx
  mov    %ebx,(%r12)
  add    $0x4,%r12
  add    $0x2,%rbp
  cmp    %r13,%rbp
  jne    L6
L13:
  mov    %r14,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
size_t fmt_hex_bytes(uint8_t *out, const char *hex) {
  size_t len = fmt_strlen(hex);
  if (len & 1) {
    out = NULL;
    return 0;
  }
  size_t final_len = len >> 1;
  for (size_t i = 0, j = 0; j < final_len; i += 2, j++) {
    out[j] = (_hex_nib(hex[i]) << 4) | _hex_nib(hex[i + 1]);
  }
  return final_len;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<fmt_hex_bytes>:
  endbr64
  push   %r14
  mov    %rdi,%r14
  mov    %rsi,%rdi
  push   %r13
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  call   <fmt_strlen@plt>
  test   $0x1,%al
  jne    L13
  shr    %rax
  mov    %rax,%r12
  je     L13
  add    $0x1,%rbp
  lea    0x0(%rbp,%rax,2),%r13
  nopw   0x0(%rax,%rax,1)
L14:
  movsbl -0x1(%rbp),%edi
  add    $0x2,%rbp
  add    $0x4,%r14
  call   <_hex_nib@plt>
  movsbl -0x2(%rbp),%edi
  mov    %eax,%ebx
  call   <_hex_nib@plt>
  shl    $0x4,%ebx
  or     %eax,%ebx
  mov    %ebx,-0x4(%r14)
  cmp    %r13,%rbp
  jne    L14
L13:
  pop    %rbx
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
size_t fmt_hex_bytes(uint8_t *out, const char *hex) {
  size_t len = fmt_strlen(hex);
  if (len & 1) {
    out = NULL;
    return 0;
  }
  size_t final_len = len >> 1;
  for (size_t i = 0, j = 0; j < final_len; i += 2, j++) {
    out[j] = (_hex_nib(hex[i]) << 4) | _hex_nib(hex[i + 1]);
  }
  return final_len;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<fmt_hex_bytes>:
  endbr64
  push   %r14
  mov    %rdi,%r14
  mov    %rsi,%rdi
  push   %r13
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  call   <fmt_strlen@plt>
  test   $0x1,%al
  jne    L16
  shr    %rax
  mov    %rax,%r12
  je     L16
  add    $0x1,%rbp
  lea    0x0(%rbp,%rax,2),%r13
  nopw   0x0(%rax,%rax,1)
L3:
  movsbl -0x1(%rbp),%edi
  add    $0x2,%rbp
  add    $0x4,%r14
  call   <_hex_nib@plt>
  movsbl -0x2(%rbp),%edi
  mov    %eax,%ebx
  call   <_hex_nib@plt>
  shl    $0x4,%ebx
  or     %eax,%ebx
  mov    %ebx,-0x4(%r14)
  cmp    %r13,%rbp
  jne    L3
L16:
  pop    %rbx
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
size_t fmt_hex_bytes(uint8_t *out, const char *hex) {
  size_t len = fmt_strlen(hex);
  if (len & 1) {
    out = NULL;
    return 0;
  }
  size_t final_len = len >> 1;
  for (size_t i = 0, j = 0; j < final_len; i += 2, j++) {
    out[j] = (_hex_nib(hex[i]) << 4) | _hex_nib(hex[i + 1]);
  }
  return final_len;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x10(%rbp)
  movl   $0x0,-0xc(%rbp)
  lea    D36(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
L5:
  lea    D31(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x14(%rbp),%rax
  mov    %rax,%rsi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x14(%rbp),%eax
  test   %eax,%eax
  jne    L60
  mov    -0x10(%rbp),%edx
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D24(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L49
  jmp    L19
L60:
  mov    -0x14(%rbp),%eax
  cmp    %eax,-0xc(%rbp)
  jge    L28
  mov    -0x14(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  movl   $0x1,-0x10(%rbp)
  jmp    L5
L28:
  mov    -0x14(%rbp),%eax
  cmp    %eax,-0xc(%rbp)
  jne    L5
  addl   $0x1,-0x10(%rbp)
  jmp    L5
L19:
  call   <__stack_chk_fail@plt>
L49:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: label: D36
type: string
name: disassemble_data
arguments: label: D31
type: string
name: disassemble_data
arguments: label: D7
type: string
name: disassemble_data
arguments: label: D24
type: string

tool: D36: Ao finalizar, digite '00', sem aspas, para obter o resultado.


tool: D31: Insira um numero: 

tool: D7: %d

tool: D24: Maior numero: %d - quantidade de vezes: %d


assistant: ```c
int main() {
  int quantidade = 0, maior = 0;
  printf("Ao finalizar, digite '00', sem aspas, para obter o resultado.\n\n");
  while (1) {
    printf("Insira um numero: ");
    int num;
    scanf("%d", &num);
    if (!num)
      break;
    if (num > maior) {
      maior = num;
      quantidade = 1;
    } else if (num == maior)
      quantidade++;
  }
  printf("Maior numero: %d - quantidade de vezes: %d\n", maior, quantidade);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D14(%rip),%rdi
  call   <puts@plt>
  mov    $0x0,%ebx
  mov    $0x0,%ebp
  lea    D12(%rip),%r14
  lea    0x4(%rsp),%r13
  lea    D6(%rip),%r12
  jmp    L7
L20:
  mov    %eax,%ebx
  mov    $0x1,%ebp
L7:
  mov    %r14,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %r13,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  test   %eax,%eax
  je     L31
  cmp    %ebx,%eax
  jg     L20
  sete   %al
  movzbl %al,%eax
  add    %eax,%ebp
  jmp    L7
L31:
  mov    %ebp,%edx
  mov    %ebx,%esi
  lea    D13(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L47
  mov    $0x0,%eax
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L47:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D14
type_guessed: string
name: extract_symbolic_data
arguments: data_label: D12
type_guessed: string
name: extract_symbolic_data
arguments: data_label: D6
type_guessed: string
name: extract_symbolic_data
arguments: data_label: D13
type_guessed: string

tool: D14: Ao finalizar, digite '00', sem aspas, para obter o resultado.


tool: D12: Insira um numero: 

tool: D6: %d

tool: D13: Maior numero: %d - quantidade de vezes: %d


assistant: ```c
int main() {
  int quantidade = 0, maior = 0;
  printf("Ao finalizar, digite '00', sem aspas, para obter o resultado.\n\n");
  while (1) {
    printf("Insira um numero: ");
    int num;
    scanf("%d", &num);
    if (!num)
      break;
    if (num > maior) {
      maior = num;
      quantidade = 1;
    } else if (num == maior)
      quantidade++;
  }
  printf("Maior numero: %d - quantidade de vezes: %d\n", maior, quantidade);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %r14
  lea    D16(%rip),%rdi
  lea    D38(%rip),%r14
  push   %r13
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  lea    D15(%rip),%rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%r13
  call   <puts@plt>
  nopl   0x0(%rax)
L48:
  mov    %r14,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r13,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  test   %eax,%eax
  je     L35
  cmp    %ebp,%eax
  jg     L13
  sete   %al
  movzbl %al,%eax
  add    %eax,%r12d
  jmp    L48
  nopl   0x0(%rax)
L13:
  mov    %eax,%ebp
  mov    $0x1,%r12d
  jmp    L48
  nopw   0x0(%rax,%rax,1)
L35:
  xor    %eax,%eax
  mov    %r12d,%edx
  lea    D37(%rip),%rdi
  mov    %ebp,%esi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L17
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L17:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D16
type_guessed: string
name: retrieve_data
arguments: data_tag: D38
type_guessed: string
name: retrieve_data
arguments: data_tag: D15
type_guessed: string
name: retrieve_data
arguments: data_tag: D37
type_guessed: string

tool: D16: Ao finalizar, digite '00', sem aspas, para obter o resultado.


tool: D38: Insira um numero: 

tool: D15: %d

tool: D37: Maior numero: %d - quantidade de vezes: %d


assistant: ```c
int main() {
  int quantidade = 0, maior = 0;
  printf("Ao finalizar, digite '00', sem aspas, para obter o resultado.\n\n");
  while (1) {
    printf("Insira um numero: ");
    int num;
    scanf("%d", &num);
    if (!num)
      break;
    if (num > maior) {
      maior = num;
      quantidade = 1;
    } else if (num == maior)
      quantidade++;
  }
  printf("Maior numero: %d - quantidade de vezes: %d\n", maior, quantidade);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r14
  lea    D21(%rip),%rdi
  lea    D7(%rip),%r14
  push   %r13
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  lea    D0(%rip),%rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%r13
  call   <puts@plt>
  nopl   0x0(%rax)
L34:
  mov    %r14,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r13,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  test   %eax,%eax
  je     L18
  cmp    %ebp,%eax
  jg     L33
  sete   %al
  movzbl %al,%eax
  add    %eax,%r12d
  jmp    L34
  nopl   0x0(%rax)
L33:
  mov    %eax,%ebp
  mov    $0x1,%r12d
  jmp    L34
  nopw   0x0(%rax,%rax,1)
L18:
  xor    %eax,%eax
  mov    %r12d,%edx
  lea    D3(%rip),%rdi
  mov    %ebp,%esi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D21
type_guessed: string
name: retrieve_data
arguments: data_label: D7
type_guessed: string
name: retrieve_data
arguments: data_label: D0
type_guessed: string
name: retrieve_data
arguments: data_label: D3
type_guessed: string

tool: D21: Ao finalizar, digite '00', sem aspas, para obter o resultado.


tool: D7: Insira um numero: 

tool: D0: %d

tool: D3: Maior numero: %d - quantidade de vezes: %d


assistant: ```c
int main() {
  int quantidade = 0, maior = 0;
  printf("Ao finalizar, digite '00', sem aspas, para obter o resultado.\n\n");
  while (1) {
    printf("Insira um numero: ");
    int num;
    scanf("%d", &num);
    if (!num)
      break;
    if (num > maior) {
      maior = num;
      quantidade = 1;
    } else if (num == maior)
      quantidade++;
  }
  printf("Maior numero: %d - quantidade de vezes: %d\n", maior, quantidade);
}
```
user: Translate the below assembly code into C programming language:

```asm
<clean_hdr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  movl   $0x0,-0x14(%rbp)
  mov    -0x28(%rbp),%rax
  mov    $0x3a,%esi
  mov    %rax,%rdi
  call   <strchr@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L41
  mov    -0x8(%rbp),%rax
  add    $0x1,%rax
  mov    %rax,-0x10(%rbp)
  jmp    L4
L5:
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x20,%al
  jne    L48
  addl   $0x1,-0x14(%rbp)
  addq   $0x1,-0x10(%rbp)
L4:
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L5
  jmp    L47
L48:
  nop
L47:
  cmpl   $0x1,-0x14(%rbp)
  jle    L44
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,%edx
  mov    -0x8(%rbp),%rax
  sub    -0x28(%rbp),%rax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  sub    $0x2,%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%rax
  lea    0x2(%rax),%rcx
  mov    -0x10(%rbp),%rax
  mov    %rax,%rsi
  mov    %rcx,%rdi
  call   <memmove@plt>
  jmp    L44
L41:
  nop
L44:
  leave
  ret
```

assistant: ```c
void clean_hdr(char *ln) {
  int spaces = 0;
  char *colon, *i;
  colon = strchr(ln, ':');
  if (colon == NULL)
    return;
  for (i = colon + 1; *i != '\0'; i++) {
    if (' ' == *i)
      spaces++;
    else
      break;
  }
  if (spaces > 1)
    memmove(colon + 2, i, strlen(ln) - (colon - ln) - 2);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<clean_hdr>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    $0x3a,%esi
  call   <strchr@plt>
  test   %rax,%rax
  je     L46
  mov    %rax,%rbp
  lea    0x1(%rax),%rdx
  cmpb   $0x20,0x1(%rax)
  jne    L46
  mov    %rdx,%rbx
  mov    $0x1,%eax
  sub    %edx,%eax
L42:
  lea    (%rax,%rbx,1),%edx
  add    $0x1,%rbx
  cmpb   $0x20,(%rbx)
  je     L42
  cmp    $0x1,%edx
  jg     L38
L46:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L38:
  mov    %r12,%rdi
  call   <strlen@plt>
  sub    %ebp,%r12d
  lea    -0x2(%r12,%rax,1),%edx
  lea    0x2(%rbp),%rdi
  mov    %rbx,%rsi
  call   <memmove@plt>
  jmp    L46
```

assistant: ```c
void clean_hdr(char *ln) {
  int spaces = 0;
  char *colon, *i;
  colon = strchr(ln, ':');
  if (colon == NULL)
    return;
  for (i = colon + 1; *i != '\0'; i++) {
    if (' ' == *i)
      spaces++;
    else
      break;
  }
  if (spaces > 1)
    memmove(colon + 2, i, strlen(ln) - (colon - ln) - 2);
}
```
user: Translate the below assembly code into C programming language:

```asm
<clean_hdr>:
  endbr64
  push   %r12
  mov    $0x3a,%esi
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  call   <strchr@plt>
  test   %rax,%rax
  je     L16
  cmpb   $0x20,0x1(%rax)
  mov    %rax,%rbx
  lea    0x1(%rax),%rbp
  jne    L16
  mov    $0x1,%eax
  sub    %ebp,%eax
  xchg   %ax,%ax
L7:
  lea    (%rax,%rbp,1),%edx
  add    $0x1,%rbp
  cmpb   $0x20,0x0(%rbp)
  je     L7
  cmp    $0x1,%edx
  jne    L6
L16:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L6:
  mov    %r12,%rdi
  sub    %ebx,%r12d
  call   <strlen@plt>
  lea    0x2(%rbx),%rdi
  mov    %rbp,%rsi
  pop    %rbx
  lea    -0x2(%r12,%rax,1),%edx
  pop    %rbp
  pop    %r12
  jmp    1090 <memmove@plt>
```

assistant: ```c
void clean_hdr(char *ln) {
  int spaces = 0;
  char *colon, *i;
  colon = strchr(ln, ':');
  if (colon == NULL)
    return;
  for (i = colon + 1; *i != '\0'; i++) {
    if (' ' == *i)
      spaces++;
    else
      break;
  }
  if (spaces > 1)
    memmove(colon + 2, i, strlen(ln) - (colon - ln) - 2);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<clean_hdr>:
  endbr64
  push   %r12
  mov    $0x3a,%esi
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  call   <strchr@plt>
  test   %rax,%rax
  je     L37
  cmpb   $0x20,0x1(%rax)
  mov    %rax,%rbx
  lea    0x1(%rax),%rbp
  jne    L37
  mov    $0x1,%eax
  sub    %ebp,%eax
  xchg   %ax,%ax
L21:
  lea    (%rax,%rbp,1),%edx
  add    $0x1,%rbp
  cmpb   $0x20,0x0(%rbp)
  je     L21
  cmp    $0x1,%edx
  jne    L0
L37:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L0:
  mov    %r12,%rdi
  sub    %ebx,%r12d
  call   <strlen@plt>
  lea    0x2(%rbx),%rdi
  mov    %rbp,%rsi
  pop    %rbx
  lea    -0x2(%r12,%rax,1),%edx
  pop    %rbp
  pop    %r12
  jmp    1090 <memmove@plt>
```

assistant: ```c
void clean_hdr(char *ln) {
  int spaces = 0;
  char *colon, *i;
  colon = strchr(ln, ':');
  if (colon == NULL)
    return;
  for (i = colon + 1; *i != '\0'; i++) {
    if (' ' == *i)
      spaces++;
    else
      break;
  }
  if (spaces > 1)
    memmove(colon + 2, i, strlen(ln) - (colon - ln) - 2);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<gmtime>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  lea    D6(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <gmtime_r@plt>
  leave
  ret
```

assistant: ```c
struct tm *gmtime(const time_t *timep) {
  static struct tm shared;
  return gmtime_r(timep, &shared);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<gmtime>:
  endbr64
  sub    $0x8,%rsp
  lea    D9(%rip),%rsi
  call   <gmtime_r@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
struct tm *gmtime(const time_t *timep) {
  static struct tm shared;
  return gmtime_r(timep, &shared);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<gmtime>:
  endbr64
  lea    D4(%rip),%rsi
  jmp    1050 <gmtime_r@plt>
```

assistant: ```c
struct tm *gmtime(const time_t *timep) {
  static struct tm shared;
  return gmtime_r(timep, &shared);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<gmtime>:
  endbr64
  lea    D4(%rip),%rsi
  jmp    1050 <gmtime_r@plt>
```

assistant: ```c
struct tm *gmtime(const time_t *timep) {
  static struct tm shared;
  return gmtime_r(timep, &shared);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<muld>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  movsd  %xmm0,-0x18(%rbp)
  movsd  %xmm1,-0x20(%rbp)
  movsd  %xmm2,-0x28(%rbp)
  movsd  %xmm3,-0x30(%rbp)
  mov    %edi,-0x34(%rbp)
  movsd  -0x18(%rbp),%xmm0
  movsd  D3(%rip),%xmm1
  comisd %xmm1,%xmm0
  jb     L93
  movsd  D4(%rip),%xmm0
  comisd -0x18(%rbp),%xmm0
  jae    L33
L93:
  movsd  -0x20(%rbp),%xmm0
  movsd  D3(%rip),%xmm1
  comisd %xmm1,%xmm0
  jb     L18
  movsd  D4(%rip),%xmm0
  comisd -0x20(%rbp),%xmm0
  jae    L33
L18:
  pxor   %xmm0,%xmm0
  comisd -0x18(%rbp),%xmm0
  jbe    L186
  pxor   %xmm0,%xmm0
  comisd -0x20(%rbp),%xmm0
  jbe    L186
  movsd  -0x18(%rbp),%xmm0
  movq   D33(%rip),%xmm1
  xorpd  %xmm0,%xmm1
  movsd  -0x30(%rbp),%xmm0
  movq   D33(%rip),%xmm2
  xorpd  %xmm2,%xmm0
  divsd  -0x20(%rbp),%xmm0
  comisd %xmm1,%xmm0
  jae    L33
L186:
  pxor   %xmm0,%xmm0
  comisd -0x18(%rbp),%xmm0
  jbe    L109
  movsd  -0x20(%rbp),%xmm0
  pxor   %xmm1,%xmm1
  comisd %xmm1,%xmm0
  jbe    L109
  movsd  -0x28(%rbp),%xmm0
  movapd %xmm0,%xmm1
  divsd  -0x20(%rbp),%xmm1
  movsd  -0x18(%rbp),%xmm0
  comisd %xmm1,%xmm0
  jae    L33
L109:
  movsd  -0x18(%rbp),%xmm0
  pxor   %xmm1,%xmm1
  comisd %xmm1,%xmm0
  jbe    L174
  pxor   %xmm0,%xmm0
  comisd -0x20(%rbp),%xmm0
  jbe    L174
  movsd  -0x28(%rbp),%xmm0
  movapd %xmm0,%xmm1
  divsd  -0x18(%rbp),%xmm1
  movsd  -0x20(%rbp),%xmm0
  comisd %xmm1,%xmm0
  jae    L33
L174:
  movsd  -0x18(%rbp),%xmm0
  pxor   %xmm1,%xmm1
  comisd %xmm1,%xmm0
  jbe    L181
  movsd  -0x20(%rbp),%xmm0
  pxor   %xmm1,%xmm1
  comisd %xmm1,%xmm0
  jbe    L181
  movsd  -0x30(%rbp),%xmm0
  divsd  -0x20(%rbp),%xmm0
  comisd -0x18(%rbp),%xmm0
  jb     L181
L33:
  mov    $0x1,%eax
  jmp    L95
L181:
  mov    $0x0,%eax
L95:
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jne    L32
  cmpl   $0x0,-0x34(%rbp)
  je     L32
  lea    D25(%rip),%rax
  mov    %rax,%rdi
  call   <warning@plt>
  movl   $0x1,-0x4(%rbp)
L32:
  mov    -0x4(%rbp),%eax
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D3
type: byte[8]
name: unpack_data
arguments: data_tag: D4
type: byte[8]
name: unpack_data
arguments: data_tag: D33
type: byte[8]
name: unpack_data
arguments: data_tag: D25
type: string

tool: {"D3": [0x00,0x00,0x00,0x00,0x00,0x00,0xf0,0xbf]}

tool: {"D4": [0x00,0x00,0x00,0x00,0x00,0x00,0xf0,0x3f]}

tool: {"D33": [0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x80]}

tool: D25: overflow in constant expression


assistant: ```c
int muld(double x, double y, double min, double max, int needconst) {
  int cond = ((x >= -1 && x <= 1) || (y >= -1 && y <= 1) || (x < 0 && y < 0 && -x <= max / -y) || (x < 0 && y > 0 && x >= min / y) || (x > 0 && y < 0 && y >= min / x) || (x > 0 && y > 0 && x <= max / y));
  if (!cond && needconst) {
    warning("overflow in constant expression\n");
    cond = 1;
  }
  return cond;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<muld>:
  endbr64
  movsd  D37(%rip),%xmm4
  comisd %xmm4,%xmm0
  jb     L0
  movsd  D12(%rip),%xmm5
  mov    $0x1,%eax
  comisd %xmm0,%xmm5
  jae    L91
L0:
  comisd %xmm4,%xmm1
  jb     L123
  movsd  D12(%rip),%xmm4
  mov    $0x1,%eax
  comisd %xmm1,%xmm4
  jae    L91
L123:
  pxor   %xmm4,%xmm4
  comisd %xmm0,%xmm4
  ja     L28
L48:
  comisd %xmm4,%xmm0
  jbe    L116
  comisd %xmm1,%xmm4
  ja     L62
L47:
  comisd %xmm4,%xmm1
  jbe    L116
  divsd  %xmm1,%xmm3
  mov    $0x1,%eax
  comisd %xmm0,%xmm3
  jae    L91
  nopl   (%rax)
L116:
  test   %edi,%edi
  je     L112
  sub    $0x8,%rsp
  lea    D23(%rip),%rdi
  call   <warning@plt>
  mov    $0x1,%eax
  add    $0x8,%rsp
  ret
  xchg   %ax,%ax
L28:
  comisd %xmm1,%xmm4
  ja     L86
L97:
  comisd %xmm4,%xmm1
  jbe    L48
  movapd %xmm2,%xmm5
  mov    $0x1,%eax
  divsd  %xmm1,%xmm5
  comisd %xmm5,%xmm0
  jb     L48
L91:
  ret
L62:
  divsd  %xmm0,%xmm2
  mov    $0x1,%eax
  comisd %xmm2,%xmm1
  jb     L47
  ret
L86:
  movq   D46(%rip),%xmm5
  movapd %xmm0,%xmm6
  mov    $0x1,%eax
  xorpd  %xmm5,%xmm6
  xorpd  %xmm3,%xmm5
  divsd  %xmm1,%xmm5
  comisd %xmm6,%xmm5
  jb     L97
  ret
  nopl   0x0(%rax)
L112:
  xor    %eax,%eax
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D37
type_infered: byte[8]
name: disassemble_data
arguments: tag: D12
type_infered: byte[8]
name: disassemble_data
arguments: tag: D23
type_infered: string
name: disassemble_data
arguments: tag: D46
type_infered: byte[8]

tool: {"D37": [0x00,0x00,0x00,0x00,0x00,0x00,0xf0,0xbf]}

tool: {"D12": [0x00,0x00,0x00,0x00,0x00,0x00,0xf0,0x3f]}

tool: D23: overflow in constant expression


tool: {"D46": [0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x80]}

assistant: ```c
int muld(double x, double y, double min, double max, int needconst) {
  int cond = ((x >= -1 && x <= 1) || (y >= -1 && y <= 1) || (x < 0 && y < 0 && -x <= max / -y) || (x < 0 && y > 0 && x >= min / y) || (x > 0 && y < 0 && y >= min / x) || (x > 0 && y > 0 && x <= max / y));
  if (!cond && needconst) {
    warning("overflow in constant expression\n");
    cond = 1;
  }
  return cond;
}
```