ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<kelvin_to_TTTa>:
  endbr64
  ucomisd D39(%rip),%xmm0
  push   %r12
  mov    %rdi,%r12
  jp     L2
  je     L8
L2:
  subsd  D17(%rip),%xmm0
  mulsd  D1(%rip),%xmm0
  mov    %r12,%rdi
  lea    D11(%rip),%rsi
  cvttsd2si %xmm0,%eax
  mov    %eax,%edx
  mov    %eax,%ecx
  neg    %edx
  and    $0xfffffffe,%ecx
  or     $0x1,%edx
  test   %eax,%eax
  cmovns %ecx,%edx
  xor    %eax,%eax
  call   <sprintf@plt>
  mov    %r12,%rax
  pop    %r12
  ret
  nopl   0x0(%rax)
L8:
  mov    %r12,%rax
  movl   $0x2f2f2f,(%rdi)
  pop    %r12
  ret
```

tool_calls: name: read_data
arguments: tag: D39
datatype: f64
name: read_data
arguments: tag: D17
datatype: f64
name: read_data
arguments: tag: D1
datatype: f64
name: read_data
arguments: tag: D11
datatype: string

tool: D39: 1.7e+38

tool: D17: 273.15

tool: D1: 10

tool: D11: %03d

assistant: ```c
char *kelvin_to_TTTa(char *target, double T) {
  int ix;
  if (T == (1.7e38)) {
    sprintf(target, "///");
    return target;
  }
  ix = (int)(10.0 * (T - 273.15));
  if (ix >= 0) {
    ix &= ~1;
  } else {
    ix = (-ix);
    ix |= 1;
  }
  sprintf(target, "%03d", ix);
  return target;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<kelvin_to_TTTa>:
  endbr64
  ucomisd D39(%rip),%xmm0
  push   %r12
  mov    %rdi,%r12
  jp     L3
  je     L2
L3:
  subsd  D31(%rip),%xmm0
  mulsd  D15(%rip),%xmm0
  mov    %r12,%rdi
  lea    D3(%rip),%rsi
  cvttsd2si %xmm0,%eax
  mov    %eax,%edx
  mov    %eax,%ecx
  neg    %edx
  and    $0xfffffffe,%ecx
  or     $0x1,%edx
  test   %eax,%eax
  cmovns %ecx,%edx
  xor    %eax,%eax
  call   <sprintf@plt>
  mov    %r12,%rax
  pop    %r12
  ret
  nopl   0x0(%rax)
L2:
  mov    %r12,%rax
  movl   $0x2f2f2f,(%rdi)
  pop    %r12
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D39
datatype: f64
name: retrieve_data
arguments: data_tag: D31
datatype: f64
name: retrieve_data
arguments: data_tag: D15
datatype: f64
name: retrieve_data
arguments: data_tag: D3
datatype: string

tool: D39: 1.7e+38

tool: D31: 273.15

tool: D15: 10

tool: D3: %03d

assistant: ```c
char *kelvin_to_TTTa(char *target, double T) {
  int ix;
  if (T == (1.7e38)) {
    sprintf(target, "///");
    return target;
  }
  ix = (int)(10.0 * (T - 273.15));
  if (ix >= 0) {
    ix &= ~1;
  } else {
    ix = (-ix);
    ix |= 1;
  }
  sprintf(target, "%03d", ix);
  return target;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<unlock_mem>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x20(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <munlock@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  je     L5
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <perror@plt>
  mov    $0xffffffff,%edi
  call   <exit@plt>
L5:
  nop
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D9
type: string

tool: D9: munlock

assistant: ```c
void unlock_mem(void *mmap_pointer, unsigned long memory) {
  int unlock;
  unlock = munlock(mmap_pointer, memory);
  if (unlock != 0) {
    perror("munlock");
    exit(-1);
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<unlock_mem>:
  endbr64
  sub    $0x8,%rsp
  call   <munlock@plt>
  test   %eax,%eax
  jne    L4
L2:
  add    $0x8,%rsp
  ret
L4:
  lea    D2(%rip),%rdi
  call   <perror@plt>
  mov    $0xffffffff,%edi
  call   <exit@plt>
  jmp    L2
```

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

tool: D2: munlock

assistant: ```c
void unlock_mem(void *mmap_pointer, unsigned long memory) {
  int unlock;
  unlock = munlock(mmap_pointer, memory);
  if (unlock != 0) {
    perror("munlock");
    exit(-1);
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<unlock_mem>:
  endbr64
  sub    $0x8,%rsp
  call   <munlock@plt>
  test   %eax,%eax
  jne    L5
  add    $0x8,%rsp
  ret
  cs nopw 0x0(%rax,%rax,1)
L5:
  lea    D2(%rip),%rdi
  call   <perror@plt>
  mov    $0xffffffff,%edi
  add    $0x8,%rsp
  jmp    1090 <exit@plt>
```

tool_calls: name: parse_data
arguments: label: D2
type_guessed: string

tool: D2: munlock

assistant: ```c
void unlock_mem(void *mmap_pointer, unsigned long memory) {
  int unlock;
  unlock = munlock(mmap_pointer, memory);
  if (unlock != 0) {
    perror("munlock");
    exit(-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
<unlock_mem>:
  endbr64
  sub    $0x8,%rsp
  call   <munlock@plt>
  test   %eax,%eax
  jne    L9
  add    $0x8,%rsp
  ret
  cs nopw 0x0(%rax,%rax,1)
L9:
  lea    D6(%rip),%rdi
  call   <perror@plt>
  mov    $0xffffffff,%edi
  add    $0x8,%rsp
  jmp    1090 <exit@plt>
```

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

tool: D6: munlock

assistant: ```c
void unlock_mem(void *mmap_pointer, unsigned long memory) {
  int unlock;
  unlock = munlock(mmap_pointer, memory);
  if (unlock != 0) {
    perror("munlock");
    exit(-1);
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<printstring>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L2
L15:
  mov    -0x4(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x4(%rbp)
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <putchar@plt>
L2:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L15
  nop
  nop
  leave
  ret
```

assistant: ```c
void printstring(char *c) {
  int i = 0;
  while (c[i] != 0) {
    putchar(c[i++]);
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<printstring>:
  endbr64
  movzbl (%rdi),%eax
  test   %al,%al
  je     L16
  push   %rbx
  lea    0x1(%rdi),%rbx
L5:
  movsbl %al,%edi
  call   <putchar@plt>
  add    $0x1,%rbx
  movzbl -0x1(%rbx),%eax
  test   %al,%al
  jne    L5
  pop    %rbx
  ret
L16:
  ret
```

assistant: ```c
void printstring(char *c) {
  int i = 0;
  while (c[i] != 0) {
    putchar(c[i++]);
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<printstring>:
  endbr64
  movzbl (%rdi),%eax
  test   %al,%al
  je     L10
  push   %rbx
  lea    0x1(%rdi),%rbx
L3:
  movsbl %al,%edi
  add    $0x1,%rbx
  call   <putchar@plt>
  movzbl -0x1(%rbx),%eax
  test   %al,%al
  jne    L3
  pop    %rbx
  ret
  cs nopw 0x0(%rax,%rax,1)
L10:
  ret
```

assistant: ```c
void printstring(char *c) {
  int i = 0;
  while (c[i] != 0) {
    putchar(c[i++]);
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<printstring>:
  endbr64
  movzbl (%rdi),%eax
  test   %al,%al
  je     L1
  push   %rbx
  lea    0x1(%rdi),%rbx
L12:
  movsbl %al,%edi
  add    $0x1,%rbx
  call   <putchar@plt>
  movzbl -0x1(%rbx),%eax
  test   %al,%al
  jne    L12
  pop    %rbx
  ret
  cs nopw 0x0(%rax,%rax,1)
L1:
  ret
```

assistant: ```c
void printstring(char *c) {
  int i = 0;
  while (c[i] != 0) {
    putchar(c[i++]);
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<double_reduc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  pxor   %xmm0,%xmm0
  movss  %xmm0,-0xc(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L17
L3:
  movl   $0x0,-0x4(%rbp)
  jmp    L36
L12:
  mov    -0x4(%rbp),%eax
  shl    $0x2,%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movss  (%rax),%xmm0
  movss  -0xc(%rbp),%xmm1
  addss  %xmm1,%xmm0
  movss  %xmm0,-0xc(%rbp)
  mov    -0x4(%rbp),%eax
  shl    $0x2,%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movss  (%rax),%xmm0
  movss  -0xc(%rbp),%xmm1
  addss  %xmm1,%xmm0
  movss  %xmm0,-0xc(%rbp)
  mov    -0x4(%rbp),%eax
  shl    $0x2,%eax
  cltq
  add    $0x2,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movss  (%rax),%xmm0
  movss  -0xc(%rbp),%xmm1
  addss  %xmm1,%xmm0
  movss  %xmm0,-0xc(%rbp)
  mov    -0x4(%rbp),%eax
  shl    $0x2,%eax
  cltq
  add    $0x3,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movss  (%rax),%xmm0
  movss  -0xc(%rbp),%xmm1
  addss  %xmm1,%xmm0
  movss  %xmm0,-0xc(%rbp)
  addl   $0x1,-0x4(%rbp)
L36:
  cmpl   $0xf,-0x4(%rbp)
  jle    L12
  addl   $0x1,-0x8(%rbp)
L17:
  cmpl   $0x7,-0x8(%rbp)
  jle    L3
  movss  -0xc(%rbp),%xmm0
  pop    %rbp
  ret
```

assistant: ```c
float double_reduc(float *i) {
  float l = 0;
  for (int a = 0; a < 8; a++)
    for (int b = 0; b < 16; b++) {
      l += i[b * 4];
      l += i[b * 4 + 1];
      l += i[b * 4 + 2];
      l += i[b * 4 + 3];
    }
  return l;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<double_reduc>:
  endbr64
  mov    $0x8,%ecx
  pxor   %xmm0,%xmm0
  lea    0x100(%rdi),%rdx
L9:
  mov    %rdi,%rax
L8:
  addss  (%rax),%xmm0
  addss  0x4(%rax),%xmm0
  addss  0x8(%rax),%xmm0
  addss  0xc(%rax),%xmm0
  add    $0x10,%rax
  cmp    %rdx,%rax
  jne    L8
  sub    $0x1,%ecx
  jne    L9
  ret
```

assistant: ```c
float double_reduc(float *i) {
  float l = 0;
  for (int a = 0; a < 8; a++)
    for (int b = 0; b < 16; b++) {
      l += i[b * 4];
      l += i[b * 4 + 1];
      l += i[b * 4 + 2];
      l += i[b * 4 + 3];
    }
  return l;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<double_reduc>:
  endbr64
  mov    $0x8,%ecx
  pxor   %xmm0,%xmm0
  lea    0x100(%rdi),%rdx
  nopl   0x0(%rax)
L15:
  mov    %rdi,%rax
  nopl   0x0(%rax,%rax,1)
L8:
  addss  (%rax),%xmm0
  add    $0x10,%rax
  addss  -0xc(%rax),%xmm0
  addss  -0x8(%rax),%xmm0
  addss  -0x4(%rax),%xmm0
  cmp    %rax,%rdx
  jne    L8
  sub    $0x1,%ecx
  jne    L15
  ret
```

assistant: ```c
float double_reduc(float *i) {
  float l = 0;
  for (int a = 0; a < 8; a++)
    for (int b = 0; b < 16; b++) {
      l += i[b * 4];
      l += i[b * 4 + 1];
      l += i[b * 4 + 2];
      l += i[b * 4 + 3];
    }
  return l;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<mmap_file>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    D2(%rip),%rax
  mov    (%rax),%edx
  mov    -0x18(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <open@plt>
  mov    %eax,-0xc(%rbp)
  cmpl   $0x0,-0xc(%rbp)
  jns    L13
  mov    -0xc(%rbp),%eax
  jmp    L35
L13:
  lea    -0x10(%rbp),%rdx
  mov    -0xc(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <fstat@plt>
  cmp    $0xffffffff,%eax
  jne    L49
  mov    $0x0,%eax
  jmp    L35
L49:
  mov    D30(%rip),%rax
  mov    (%rax),%ecx
  mov    D11(%rip),%rax
  mov    (%rax),%edx
  mov    -0x10(%rbp),%eax
  mov    -0xc(%rbp),%esi
  mov    $0x0,%r9d
  mov    %esi,%r8d
  mov    %eax,%esi
  mov    $0x0,%edi
  call   <mmap@plt>
  mov    -0x20(%rbp),%rdx
  mov    %rax,(%rdx)
  mov    -0x20(%rbp),%rax
  mov    (%rax),%rdx
  mov    D33(%rip),%rax
  mov    (%rax),%rax
  cmp    %rax,%rdx
  jne    L7
  mov    $0x0,%eax
  jmp    L35
L7:
  mov    -0xc(%rbp),%eax
  mov    %eax,%edi
  call   <close@plt>
  mov    $0x1,%eax
L35:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L68
  call   <__stack_chk_fail@plt>
L68:
  leave
  ret
```

assistant: ```c
int mmap_file(const char *filename, unsigned char **map) {
  int fd;
  struct stat st;
  fd = open(filename, O_RDONLY);
  if (fd < 0)
    return fd;
  if (fstat(fd, &st) == -1)
    return 0;
  *map = mmap(NULL, st.st_size, PROT_READ, MAP_PRIVATE, fd, 0);
  if (*map == MAP_FAILED)
    return 0;
  close(fd);
  return 1;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<mmap_file>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rsi,%rbp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    D20(%rip),%rax
  mov    (%rax),%esi
  call   <open@plt>
  mov    %eax,%ebx
  test   %eax,%eax
  js     L27
  lea    0x4(%rsp),%rsi
  mov    %eax,%edi
  call   <fstat@plt>
  cmp    $0xffffffff,%eax
  je     L17
  mov    $0x0,%r9d
  mov    %ebx,%r8d
  mov    D30(%rip),%rax
  mov    (%rax),%ecx
  mov    D13(%rip),%rax
  mov    (%rax),%edx
  mov    0x4(%rsp),%esi
  mov    $0x0,%edi
  call   <mmap@plt>
  mov    %rax,0x0(%rbp)
  mov    D37(%rip),%rdx
  cmp    (%rdx),%rax
  je     L20
  mov    %ebx,%edi
  call   <close@plt>
  mov    $0x1,%ebx
L27:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  mov    %ebx,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L17:
  mov    $0x0,%ebx
  jmp    L27
L20:
  mov    $0x0,%ebx
  jmp    L27
L4:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int mmap_file(const char *filename, unsigned char **map) {
  int fd;
  struct stat st;
  fd = open(filename, O_RDONLY);
  if (fd < 0)
    return fd;
  if (fstat(fd, &st) == -1)
    return 0;
  *map = mmap(NULL, st.st_size, PROT_READ, MAP_PRIVATE, fd, 0);
  if (*map == MAP_FAILED)
    return 0;
  close(fd);
  return 1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<mmap_file>:
  endbr64
  push   %r12
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  mov    D37(%rip),%rax
  mov    (%rax),%esi
  call   <open@plt>
  mov    %eax,%r12d
  test   %eax,%eax
  js     L22
  lea    0x4(%rsp),%rsi
  mov    %eax,%edi
  call   <fstat@plt>
  cmp    $0xffffffff,%eax
  je     L1
  mov    D3(%rip),%rax
  mov    0x4(%rsp),%esi
  xor    %r9d,%r9d
  xor    %edi,%edi
  mov    %r12d,%r8d
  mov    (%rax),%ecx
  mov    D22(%rip),%rax
  mov    (%rax),%edx
  call   <mmap@plt>
  mov    D14(%rip),%rdx
  mov    %rax,(%rbx)
  cmp    (%rdx),%rax
  je     L1
  mov    %r12d,%edi
  mov    $0x1,%r12d
  call   <close@plt>
L22:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L32
  add    $0x18,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %r12
  ret
  nopl   (%rax)
L1:
  xor    %r12d,%r12d
  jmp    L22
L32:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int mmap_file(const char *filename, unsigned char **map) {
  int fd;
  struct stat st;
  fd = open(filename, O_RDONLY);
  if (fd < 0)
    return fd;
  if (fstat(fd, &st) == -1)
    return 0;
  *map = mmap(NULL, st.st_size, PROT_READ, MAP_PRIVATE, fd, 0);
  if (*map == MAP_FAILED)
    return 0;
  close(fd);
  return 1;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<mmap_file>:
  endbr64
  push   %r12
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  mov    D15(%rip),%rax
  mov    (%rax),%esi
  call   <open@plt>
  mov    %eax,%r12d
  test   %eax,%eax
  js     L7
  lea    0x4(%rsp),%rsi
  mov    %eax,%edi
  call   <fstat@plt>
  cmp    $0xffffffff,%eax
  je     L22
  mov    D3(%rip),%rax
  mov    0x4(%rsp),%esi
  xor    %r9d,%r9d
  xor    %edi,%edi
  mov    %r12d,%r8d
  mov    (%rax),%ecx
  mov    D20(%rip),%rax
  mov    (%rax),%edx
  call   <mmap@plt>
  mov    D21(%rip),%rdx
  mov    %rax,(%rbx)
  cmp    (%rdx),%rax
  je     L22
  mov    %r12d,%edi
  mov    $0x1,%r12d
  call   <close@plt>
L7:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L17
  add    $0x18,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %r12
  ret
  nopl   (%rax)
L22:
  xor    %r12d,%r12d
  jmp    L7
L17:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int mmap_file(const char *filename, unsigned char **map) {
  int fd;
  struct stat st;
  fd = open(filename, O_RDONLY);
  if (fd < 0)
    return fd;
  if (fstat(fd, &st) == -1)
    return 0;
  *map = mmap(NULL, st.st_size, PROT_READ, MAP_PRIVATE, fd, 0);
  if (*map == MAP_FAILED)
    return 0;
  close(fd);
  return 1;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<e_line_read>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x18(%rbp)
  movl   $0x0,-0x4(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L77
L79:
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0x14(%rbp),%eax
  mov    $0x1,%edx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <read@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0x1,-0x4(%rbp)
  jne    L37
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0xa,%al
  je     L37
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L37
  addl   $0x1,-0x8(%rbp)
L77:
  mov    -0x18(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0x8(%rbp)
  jl     L79
L37:
  cmpl   $0x1,-0x4(%rbp)
  je     L73
  cmpl   $0x0,-0x8(%rbp)
  jne    L73
  mov    $0xffffffff,%eax
  jmp    L15
L73:
  mov    -0x18(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0x8(%rbp)
  jne    L58
  subl   $0x1,-0x8(%rbp)
L58:
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x1(%rax),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    $0x0,%eax
L15:
  leave
  ret
```

assistant: ```c
int e_line_read(int n, char *s, int max) {
  int i, ret = 0;
  for (i = 0; i < max - 1; i++)
    if ((ret = read(n, s + i, 1)) != 1 || s[i] == '\n' || s[i] == '\0')
      break;
  if (ret != 1 && i == 0)
    return (-1);
  if (i == max - 1)
    i--;
  s[i + 1] = '\0';
  return (0);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<e_line_read>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  lea    -0x1(%rdx),%r15d
  test   %r15d,%r15d
  jle    L25
  mov    %edi,%r13d
  mov    %rsi,%r14
  mov    %rsi,%rbx
  mov    %r15d,%r12d
  mov    $0x0,%ebp
L35:
  mov    $0x1,%edx
  mov    %rbx,%rsi
  mov    %r13d,%edi
  call   <read@plt>
  cmp    $0x1,%eax
  jne    L50
  movzbl (%rbx),%eax
  test   %al,%al
  je     L36
  cmp    $0xa,%al
  je     L36
  add    $0x1,%ebp
  add    $0x1,%rbx
  cmp    %r12d,%ebp
  jne    L35
  mov    %r12d,%ebp
  jmp    L36
L50:
  test   %ebp,%ebp
  je     L47
L36:
  cmp    %ebp,%r15d
  sete   %al
  movzbl %al,%eax
  sub    %eax,%ebp
  movslq %ebp,%rbp
  movb   $0x0,0x1(%r14,%rbp,1)
  mov    $0x0,%eax
L44:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L25:
  mov    $0xffffffff,%eax
  jmp    L44
L47:
  mov    $0xffffffff,%eax
  jmp    L44
```

assistant: ```c
int e_line_read(int n, char *s, int max) {
  int i, ret = 0;
  for (i = 0; i < max - 1; i++)
    if ((ret = read(n, s + i, 1)) != 1 || s[i] == '\n' || s[i] == '\0')
      break;
  if (ret != 1 && i == 0)
    return (-1);
  if (i == max - 1)
    i--;
  s[i + 1] = '\0';
  return (0);
}
```
user: What is the c source code of the assembly code below:

```asm
<e_line_read>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  lea    -0x1(%rdx),%r12d
  push   %rbp
  push   %rbx
  test   %r12d,%r12d
  jle    L52
  mov    %edi,%r14d
  mov    %rsi,%r13
  mov    %rsi,%rbx
  xor    %ebp,%ebp
  jmp    L14
  nopw   0x0(%rax,%rax,1)
L38:
  movzbl (%rbx),%eax
  add    $0x1,%ebp
  cmp    $0xa,%al
  je     L16
  test   %al,%al
  je     L16
  add    $0x1,%rbx
  cmp    %r12d,%ebp
  je     L16
L14:
  mov    $0x1,%edx
  mov    %rbx,%rsi
  mov    %r14d,%edi
  call   <read@plt>
  cmp    $0x1,%eax
  je     L38
  test   %ebp,%ebp
  je     L52
  add    $0x1,%ebp
L16:
  movslq %ebp,%rbp
  xor    %eax,%eax
  movb   $0x0,0x0(%r13,%rbp,1)
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nop
L52:
  pop    %rbx
  mov    $0xffffffff,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int e_line_read(int n, char *s, int max) {
  int i, ret = 0;
  for (i = 0; i < max - 1; i++)
    if ((ret = read(n, s + i, 1)) != 1 || s[i] == '\n' || s[i] == '\0')
      break;
  if (ret != 1 && i == 0)
    return (-1);
  if (i == max - 1)
    i--;
  s[i + 1] = '\0';
  return (0);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<e_line_read>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  lea    -0x1(%rdx),%r12d
  push   %rbp
  push   %rbx
  test   %r12d,%r12d
  jle    L37
  mov    %edi,%r14d
  mov    %rsi,%r13
  mov    %rsi,%rbx
  xor    %ebp,%ebp
  jmp    L0
  nopw   0x0(%rax,%rax,1)
L17:
  movzbl (%rbx),%eax
  add    $0x1,%ebp
  cmp    $0xa,%al
  je     L22
  test   %al,%al
  je     L22
  add    $0x1,%rbx
  cmp    %r12d,%ebp
  je     L22
L0:
  mov    $0x1,%edx
  mov    %rbx,%rsi
  mov    %r14d,%edi
  call   <read@plt>
  cmp    $0x1,%eax
  je     L17
  test   %ebp,%ebp
  je     L37
  add    $0x1,%ebp
L22:
  movslq %ebp,%rbp
  xor    %eax,%eax
  movb   $0x0,0x0(%r13,%rbp,1)
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nop
L37:
  pop    %rbx
  mov    $0xffffffff,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int e_line_read(int n, char *s, int max) {
  int i, ret = 0;
  for (i = 0; i < max - 1; i++)
    if ((ret = read(n, s + i, 1)) != 1 || s[i] == '\n' || s[i] == '\0')
      break;
  if (ret != 1 && i == 0)
    return (-1);
  if (i == max - 1)
    i--;
  s[i + 1] = '\0';
  return (0);
}
```
user: What is the c source code of the assembly code below:

```asm
<gain>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  movsd  %xmm0,-0x18(%rbp)
  movsd  %xmm1,-0x20(%rbp)
  movsd  -0x18(%rbp),%xmm1
  movsd  D6(%rip),%xmm0
  addsd  %xmm0,%xmm1
  movq   %xmm1,%rax
  movq   %rax,%xmm0
  call   <floor@plt>
  addsd  %xmm0,%xmm0
  movsd  %xmm0,-0x10(%rbp)
  movsd  -0x18(%rbp),%xmm0
  addsd  %xmm0,%xmm0
  subsd  -0x10(%rbp),%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movsd  -0x8(%rbp),%xmm0
  pxor   %xmm1,%xmm1
  comisd %xmm1,%xmm0
  jbe    L11
  movsd  -0x20(%rbp),%xmm0
  mov    -0x8(%rbp),%rax
  movapd %xmm0,%xmm1
  movq   %rax,%xmm0
  call   <bias@plt>
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  addsd  -0x10(%rbp),%xmm0
  jmp    L13
L11:
  movsd  -0x8(%rbp),%xmm0
  movq   D29(%rip),%xmm1
  xorpd  %xmm1,%xmm0
  movq   %xmm0,%rax
  movsd  -0x20(%rbp),%xmm0
  movapd %xmm0,%xmm1
  movq   %rax,%xmm0
  call   <bias@plt>
  movapd %xmm0,%xmm1
  movsd  -0x10(%rbp),%xmm0
  subsd  %xmm1,%xmm0
L13:
  movsd  D6(%rip),%xmm1
  mulsd  %xmm1,%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  leave
  ret
```

tool_calls: name: read_data
arguments: data_label: D6
data_type: f64
name: read_data
arguments: data_label: D29
data_type: f32[2]

tool: D6: 0.5

tool: {"D29": [0.0,-0.0]}

assistant: ```c
double gain(double x, double p) {
  double i, px;
  i = 2.0 * floor(x + 0.5);
  px = 2.0 * x - i;
  return 0.5 * (px > 0.0 ? i + bias(px, p) : i - bias(-px, p));
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<gain>:
  endbr64
  sub    $0x18,%rsp
  movapd %xmm0,%xmm3
  addsd  D60(%rip),%xmm3
  movapd %xmm3,%xmm4
  movsd  D39(%rip),%xmm5
  movapd %xmm3,%xmm2
  andpd  %xmm5,%xmm2
  movsd  D18(%rip),%xmm6
  ucomisd %xmm2,%xmm6
  jbe    L17
  cvttsd2si %xmm3,%rax
  pxor   %xmm2,%xmm2
  cvtsi2sd %rax,%xmm2
  movapd %xmm2,%xmm4
  cmpnlesd %xmm3,%xmm4
  movsd  D10(%rip),%xmm6
  andpd  %xmm6,%xmm4
  subsd  %xmm4,%xmm2
  andnpd %xmm3,%xmm5
  movapd %xmm2,%xmm4
  orpd   %xmm5,%xmm4
L17:
  movapd %xmm4,%xmm7
  addsd  %xmm4,%xmm7
  addsd  %xmm0,%xmm0
  movsd  %xmm7,0x8(%rsp)
  subsd  %xmm7,%xmm0
  comisd D28(%rip),%xmm0
  jbe    L12
  call   <bias@plt>
  addsd  0x8(%rsp),%xmm0
L0:
  mulsd  D60(%rip),%xmm0
  add    $0x18,%rsp
  ret
L12:
  xorpd  D36(%rip),%xmm0
  call   <bias@plt>
  movapd %xmm0,%xmm1
  movsd  0x8(%rsp),%xmm0
  subsd  %xmm1,%xmm0
  jmp    L0
```

tool_calls: name: retrieve_data
arguments: label: D60
datatype: f64
name: retrieve_data
arguments: label: D39
datatype: byte[8]
name: retrieve_data
arguments: label: D18
datatype: f32[2]
name: retrieve_data
arguments: label: D10
datatype: f32[2]
name: retrieve_data
arguments: label: D28
datatype: f32[4]
name: retrieve_data
arguments: label: D36
datatype: f32[4]

tool: D60: 0.5

tool: {"D39": [0xff,0xff,0xff,0xff,0xff,0xff,0xff,0x7f]}

tool: {"D18": [0.0,176.0]}

tool: {"D10": [0.0,1.875]}

tool: {"D28": [0.0,0.0,0.0,-0.0]}

tool: {"D36": [0.0,-0.0,0.0,0.0]}

assistant: ```c
double gain(double x, double p) {
  double i, px;
  i = 2.0 * floor(x + 0.5);
  px = 2.0 * x - i;
  return 0.5 * (px > 0.0 ? i + bias(px, p) : i - bias(-px, p));
}
```