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

```asm
<aprintf>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  movl   $0x0,-0x4(%rbp)
  jmp    L18
L12:
  mov    -0x4(%rbp),%eax
  cltq
  shl    $0x4,%rax
  mov    %rax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  fldt   (%rax)
  lea    -0x10(%rsp),%rsp
  fstpt  (%rsp)
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x10,%rsp
  mov    -0x4(%rbp),%eax
  add    $0x1,%eax
  cmp    %eax,-0x1c(%rbp)
  jle    L7
  mov    -0x4(%rbp),%eax
  cltq
  add    $0x1,%rax
  shl    $0x4,%rax
  mov    %rax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  fldt   (%rax)
  lea    -0x10(%rsp),%rsp
  fstpt  (%rsp)
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x10,%rsp
L7:
  mov    -0x4(%rbp),%eax
  add    $0x2,%eax
  cmp    %eax,-0x1c(%rbp)
  jle    L40
  mov    -0x4(%rbp),%eax
  cltq
  add    $0x2,%rax
  shl    $0x4,%rax
  mov    %rax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  fldt   (%rax)
  lea    -0x10(%rsp),%rsp
  fstpt  (%rsp)
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x10,%rsp
L40:
  mov    -0x4(%rbp),%eax
  add    $0x3,%eax
  cmp    %eax,-0x1c(%rbp)
  jle    L48
  mov    -0x4(%rbp),%eax
  cltq
  add    $0x3,%rax
  shl    $0x4,%rax
  mov    %rax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  fldt   (%rax)
  lea    -0x10(%rsp),%rsp
  fstpt  (%rsp)
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x10,%rsp
L48:
  mov    $0xa,%edi
  call   <putchar@plt>
  addl   $0x4,-0x4(%rbp)
L18:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L12
  nop
  nop
  leave
  ret
```

tool_calls: name: read_data
arguments: tag: D12
type_guessed: string

tool: D12: 	%10.15Lg

assistant: ```c
void aprintf(char *str, int n, long double a[n]) {
  printf("%s\n", str);
  int i;
  for (i = 0; i < n; i += 4) {
    printf("\t%10.15Lg", a[i]);
    if (i + 1 < n)
      printf("\t%10.15Lg", a[i + 1]);
    if (i + 2 < n)
      printf("\t%10.15Lg", a[i + 2]);
    if (i + 3 < n)
      printf("\t%10.15Lg", a[i + 3]);
    printf("\n");
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<aprintf>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %esi,%r12d
  mov    %rdx,%rbp
  call   <puts@plt>
  test   %r12d,%r12d
  jle    L76
  lea    -0x1(%r12),%r15d
  and    $0xfffffffc,%r15d
  add    $0x5,%r15d
  mov    $0x1,%ebx
  lea    D6(%rip),%r13
  jmp    L23
L73:
  push   0x18(%rbp)
  push   0x10(%rbp)
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x10,%rsp
  jmp    L78
L30:
  push   0x28(%r14)
  push   0x20(%r14)
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x10,%rsp
  jmp    L79
L25:
  mov    $0xa,%edi
  call   <putchar@plt>
  add    $0x40,%rbp
  add    $0x4,%ebx
  cmp    %r15d,%ebx
  je     L76
L23:
  mov    %rbp,%r14
  push   0x8(%rbp)
  push   0x0(%rbp)
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x10,%rsp
  cmp    %ebx,%r12d
  jg     L73
L78:
  lea    0x1(%rbx),%eax
  cmp    %eax,%r12d
  jg     L30
L79:
  lea    0x2(%rbx),%eax
  cmp    %eax,%r12d
  jle    L25
  push   0x38(%r14)
  push   0x30(%r14)
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x10,%rsp
  jmp    L25
L76:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D6
type: string

tool: D6: 	%10.15Lg

assistant: ```c
void aprintf(char *str, int n, long double a[n]) {
  printf("%s\n", str);
  int i;
  for (i = 0; i < n; i += 4) {
    printf("\t%10.15Lg", a[i]);
    if (i + 1 < n)
      printf("\t%10.15Lg", a[i + 1]);
    if (i + 2 < n)
      printf("\t%10.15Lg", a[i + 2]);
    if (i + 3 < n)
      printf("\t%10.15Lg", a[i + 3]);
    printf("\n");
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<aprintf>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  call   <puts@plt>
  test   %r12d,%r12d
  jle    L31
  lea    -0x1(%r12),%r14d
  mov    $0x1,%ebx
  lea    D7(%rip),%r13
  and    $0xfffffffc,%r14d
  add    $0x5,%r14d
  nopl   0x0(%rax)
L11:
  push   0x8(%rbp)
  mov    %r13,%rdi
  xor    %eax,%eax
  push   0x0(%rbp)
  call   <printf@plt>
  pop    %r9
  pop    %r10
  cmp    %ebx,%r12d
  jle    L18
  push   0x18(%rbp)
  mov    %r13,%rdi
  xor    %eax,%eax
  push   0x10(%rbp)
  call   <printf@plt>
  pop    %rdi
  pop    %r8
L18:
  lea    0x1(%rbx),%eax
  cmp    %eax,%r12d
  jle    L29
  push   0x28(%rbp)
  mov    %r13,%rdi
  xor    %eax,%eax
  push   0x20(%rbp)
  call   <printf@plt>
  pop    %rcx
  pop    %rsi
L29:
  lea    0x2(%rbx),%eax
  cmp    %eax,%r12d
  jle    L19
  push   0x38(%rbp)
  mov    %r13,%rdi
  xor    %eax,%eax
  push   0x30(%rbp)
  call   <printf@plt>
  pop    %rax
  pop    %rdx
L19:
  mov    $0xa,%edi
  add    $0x4,%ebx
  add    $0x40,%rbp
  call   <putchar@plt>
  cmp    %r14d,%ebx
  jne    L11
L31:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

tool_calls: name: parse_data
arguments: tag: D7
datatype: string

tool: D7: 	%10.15Lg

assistant: ```c
void aprintf(char *str, int n, long double a[n]) {
  printf("%s\n", str);
  int i;
  for (i = 0; i < n; i += 4) {
    printf("\t%10.15Lg", a[i]);
    if (i + 1 < n)
      printf("\t%10.15Lg", a[i + 1]);
    if (i + 2 < n)
      printf("\t%10.15Lg", a[i + 2]);
    if (i + 3 < n)
      printf("\t%10.15Lg", a[i + 3]);
    printf("\n");
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<aprintf>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  call   <puts@plt>
  test   %r12d,%r12d
  jle    L34
  lea    -0x1(%r12),%r14d
  mov    $0x1,%ebx
  lea    D7(%rip),%r13
  and    $0xfffffffc,%r14d
  add    $0x5,%r14d
  nopl   0x0(%rax)
L16:
  push   0x8(%rbp)
  mov    %r13,%rdi
  xor    %eax,%eax
  push   0x0(%rbp)
  call   <printf@plt>
  pop    %r9
  pop    %r10
  cmp    %r12d,%ebx
  jge    L6
  push   0x18(%rbp)
  mov    %r13,%rdi
  xor    %eax,%eax
  push   0x10(%rbp)
  call   <printf@plt>
  pop    %rdi
  pop    %r8
L6:
  lea    0x1(%rbx),%eax
  cmp    %eax,%r12d
  jle    L40
  push   0x28(%rbp)
  mov    %r13,%rdi
  xor    %eax,%eax
  push   0x20(%rbp)
  call   <printf@plt>
  pop    %rcx
  pop    %rsi
L40:
  lea    0x2(%rbx),%eax
  cmp    %eax,%r12d
  jg     L49
  mov    $0xa,%edi
  add    $0x4,%ebx
  add    $0x40,%rbp
  call   <putchar@plt>
  cmp    %r14d,%ebx
  jne    L16
L34:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   (%rax)
L49:
  push   0x38(%rbp)
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x4,%ebx
  push   0x30(%rbp)
  add    $0x40,%rbp
  call   <printf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  pop    %rax
  pop    %rdx
  cmp    %r14d,%ebx
  jne    L16
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

tool_calls: name: unpack_data
arguments: tag: D7
type_guessed: string

tool: D7: 	%10.15Lg

assistant: ```c
void aprintf(char *str, int n, long double a[n]) {
  printf("%s\n", str);
  int i;
  for (i = 0; i < n; i += 4) {
    printf("\t%10.15Lg", a[i]);
    if (i + 1 < n)
      printf("\t%10.15Lg", a[i + 1]);
    if (i + 2 < n)
      printf("\t%10.15Lg", a[i + 2]);
    if (i + 3 < n)
      printf("\t%10.15Lg", a[i + 3]);
    printf("\n");
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<valve_open>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x1,%esi
  mov    $0x11,%edi
  call   <GPIODirection@plt>
  mov    $0x1,%esi
  mov    $0x1b,%edi
  call   <GPIODirection@plt>
  mov    $0x1,%esi
  mov    $0x16,%edi
  call   <GPIODirection@plt>
  mov    $0x0,%esi
  mov    $0x11,%edi
  call   <GPIOWrite@plt>
  cmp    $0xffffffff,%eax
  je     L54
  mov    $0x0,%esi
  mov    $0x1b,%edi
  call   <GPIOWrite@plt>
  cmp    $0xffffffff,%eax
  jne    L34
L54:
  mov    $0xffffffff,%eax
  jmp    L14
L34:
  mov    $0x1,%esi
  mov    $0x16,%edi
  call   <GPIOWrite@plt>
  mov    $0x1,%esi
  mov    $0x11,%edi
  call   <GPIOWrite@plt>
  cmp    $0xffffffff,%eax
  jne    L52
  mov    $0xffffffff,%eax
  jmp    L14
L52:
  mov    $0xa,%edi
  call   <sleep@plt>
  mov    $0x0,%esi
  mov    $0x11,%edi
  call   <GPIOWrite@plt>
  cmp    $0xffffffff,%eax
  jne    L68
  mov    $0xffffffff,%eax
  jmp    L14
L68:
  mov    $0x0,%eax
L14:
  pop    %rbp
  ret
```

assistant: ```c
int valve_open() {
  GPIODirection(17, 1);
  GPIODirection(27, 1);
  GPIODirection(22, 1);
  if ((-1 == GPIOWrite(17, 0)) || (-1 == GPIOWrite(27, 0)))
    return (-1);
  GPIOWrite(22, 1);
  if (-1 == GPIOWrite(17, 1))
    return (-1);
  sleep(10);
  if (-1 == GPIOWrite(17, 0))
    return (-1);
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<valve_open>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x1,%esi
  mov    $0x11,%edi
  call   <GPIODirection@plt>
  mov    $0x1,%esi
  mov    $0x1b,%edi
  call   <GPIODirection@plt>
  mov    $0x1,%esi
  mov    $0x16,%edi
  call   <GPIODirection@plt>
  mov    $0x0,%esi
  mov    $0x11,%edi
  call   <GPIOWrite@plt>
  cmp    $0xffffffff,%eax
  je     L27
  mov    $0x0,%esi
  mov    $0x1b,%edi
  call   <GPIOWrite@plt>
  cmp    $0xffffffff,%eax
  je     L27
  mov    $0x1,%esi
  mov    $0x16,%edi
  call   <GPIOWrite@plt>
  mov    $0x1,%esi
  mov    $0x11,%edi
  call   <GPIOWrite@plt>
  cmp    $0xffffffff,%eax
  je     L27
  mov    $0xa,%edi
  call   <sleep@plt>
  mov    $0x0,%esi
  mov    $0x11,%edi
  call   <GPIOWrite@plt>
  cmp    $0xffffffff,%eax
  sete   %al
  movzbl %al,%eax
  neg    %eax
L27:
  add    $0x8,%rsp
  ret
```

assistant: ```c
int valve_open() {
  GPIODirection(17, 1);
  GPIODirection(27, 1);
  GPIODirection(22, 1);
  if ((-1 == GPIOWrite(17, 0)) || (-1 == GPIOWrite(27, 0)))
    return (-1);
  GPIOWrite(22, 1);
  if (-1 == GPIOWrite(17, 1))
    return (-1);
  sleep(10);
  if (-1 == GPIOWrite(17, 0))
    return (-1);
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<valve_open>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x1,%esi
  mov    $0x11,%edi
  call   <GPIODirection@plt>
  mov    $0x1,%esi
  mov    $0x1b,%edi
  call   <GPIODirection@plt>
  mov    $0x1,%esi
  mov    $0x16,%edi
  call   <GPIODirection@plt>
  xor    %esi,%esi
  mov    $0x11,%edi
  call   <GPIOWrite@plt>
  cmp    $0xffffffff,%eax
  je     L11
  xor    %esi,%esi
  mov    $0x1b,%edi
  call   <GPIOWrite@plt>
  cmp    $0xffffffff,%eax
  je     L11
  mov    $0x1,%esi
  mov    $0x16,%edi
  call   <GPIOWrite@plt>
  mov    $0x1,%esi
  mov    $0x11,%edi
  call   <GPIOWrite@plt>
  cmp    $0xffffffff,%eax
  je     L11
  mov    $0xa,%edi
  call   <sleep@plt>
  xor    %esi,%esi
  mov    $0x11,%edi
  call   <GPIOWrite@plt>
  cmp    $0xffffffff,%eax
  sete   %al
  movzbl %al,%eax
  neg    %eax
L7:
  add    $0x8,%rsp
  ret
L11:
  mov    $0xffffffff,%eax
  jmp    L7
```

assistant: ```c
int valve_open() {
  GPIODirection(17, 1);
  GPIODirection(27, 1);
  GPIODirection(22, 1);
  if ((-1 == GPIOWrite(17, 0)) || (-1 == GPIOWrite(27, 0)))
    return (-1);
  GPIOWrite(22, 1);
  if (-1 == GPIOWrite(17, 1))
    return (-1);
  sleep(10);
  if (-1 == GPIOWrite(17, 0))
    return (-1);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<valve_open>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x1,%esi
  mov    $0x11,%edi
  call   <GPIODirection@plt>
  mov    $0x1,%esi
  mov    $0x1b,%edi
  call   <GPIODirection@plt>
  mov    $0x1,%esi
  mov    $0x16,%edi
  call   <GPIODirection@plt>
  xor    %esi,%esi
  mov    $0x11,%edi
  call   <GPIOWrite@plt>
  cmp    $0xffffffff,%eax
  je     L19
  xor    %esi,%esi
  mov    $0x1b,%edi
  call   <GPIOWrite@plt>
  cmp    $0xffffffff,%eax
  je     L19
  mov    $0x1,%esi
  mov    $0x16,%edi
  call   <GPIOWrite@plt>
  mov    $0x1,%esi
  mov    $0x11,%edi
  call   <GPIOWrite@plt>
  cmp    $0xffffffff,%eax
  je     L19
  mov    $0xa,%edi
  call   <sleep@plt>
  xor    %esi,%esi
  mov    $0x11,%edi
  call   <GPIOWrite@plt>
  cmp    $0xffffffff,%eax
  sete   %al
  movzbl %al,%eax
  neg    %eax
L24:
  add    $0x8,%rsp
  ret
L19:
  mov    $0xffffffff,%eax
  jmp    L24
```

assistant: ```c
int valve_open() {
  GPIODirection(17, 1);
  GPIODirection(27, 1);
  GPIODirection(22, 1);
  if ((-1 == GPIOWrite(17, 0)) || (-1 == GPIOWrite(27, 0)))
    return (-1);
  GPIOWrite(22, 1);
  if (-1 == GPIOWrite(17, 1))
    return (-1);
  sleep(10);
  if (-1 == GPIOWrite(17, 0))
    return (-1);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<strapp>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movq   $0x0,-0x8(%rbp)
  jmp    L9
L4:
  mov    -0x18(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  add    %rax,%rdx
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  addq   $0x1,-0x8(%rbp)
  addq   $0x1,-0x20(%rbp)
L9:
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L4
  mov    -0x8(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
size_t strapp(char *target, const char *app) {
  size_t delta = 0;
  while (*app != '\0') {
    target[delta] = *app;
    delta++;
    app++;
  }
  return delta;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<strapp>:
  endbr64
  movzbl (%rsi),%edx
  test   %dl,%dl
  je     L15
  mov    $0x0,%eax
L1:
  mov    %dl,(%rdi,%rax,1)
  add    $0x1,%rax
  movzbl (%rsi,%rax,1),%edx
  test   %dl,%dl
  jne    L1
  ret
L15:
  mov    $0x0,%eax
  ret
```

assistant: ```c
size_t strapp(char *target, const char *app) {
  size_t delta = 0;
  while (*app != '\0') {
    target[delta] = *app;
    delta++;
    app++;
  }
  return delta;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<strapp>:
  endbr64
  movzbl (%rsi),%edx
  xor    %eax,%eax
  test   %dl,%dl
  je     L9
  nopl   (%rax)
L14:
  mov    %dl,(%rdi,%rax,1)
  add    $0x1,%rax
  movzbl (%rsi,%rax,1),%edx
  test   %dl,%dl
  jne    L14
  ret
L9:
  ret
```

assistant: ```c
size_t strapp(char *target, const char *app) {
  size_t delta = 0;
  while (*app != '\0') {
    target[delta] = *app;
    delta++;
    app++;
  }
  return delta;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<strapp>:
  endbr64
  movzbl (%rsi),%edx
  xor    %eax,%eax
  test   %dl,%dl
  je     L16
  nopl   (%rax)
L10:
  mov    %dl,(%rdi,%rax,1)
  add    $0x1,%rax
  movzbl (%rsi,%rax,1),%edx
  test   %dl,%dl
  jne    L10
  ret
L16:
  ret
```

assistant: ```c
size_t strapp(char *target, const char *app) {
  size_t delta = 0;
  while (*app != '\0') {
    target[delta] = *app;
    delta++;
    app++;
  }
  return delta;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<half>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  movl   $0x0,-0x8(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,%esi
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L32
L24:
  movl   $0x1,-0x4(%rbp)
  jmp    L27
L39:
  addl   $0x1,-0x4(%rbp)
L27:
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  mov    $0x2,%edi
  call   <pow@plt>
  cmp    %eax,-0x14(%rbp)
  jge    L39
  mov    -0x4(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,%esi
  mov    $0x2,%edi
  call   <pow@plt>
  mov    %eax,%edx
  mov    -0x4(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,%esi
  lea    D14(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x4(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,%esi
  mov    $0x2,%edi
  call   <pow@plt>
  sub    %eax,-0x14(%rbp)
  mov    -0x4(%rbp),%eax
  add    %eax,-0x8(%rbp)
L32:
  cmpl   $0x1,-0x14(%rbp)
  jg     L24
  mov    -0x14(%rbp),%edx
  mov    -0x8(%rbp),%eax
  mov    %eax,%esi
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  nop
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D12
type_infered: string
name: retrieve_data
arguments: data_label: D14
type_infered: string
name: retrieve_data
arguments: data_label: D10
type_infered: string

tool: D12: %d/2


tool: D14: i = %d, d = %d


tool: D10: quotient = %d, remainder = %d


assistant: ```c
void half(int n) {
  int q = 0, i;
  printf("%d/2\n", n);
  while (n >= 2) {
    for (i = 1; ((int)pow(2, i)) <= n; i++)
      ;
    printf("i = %d, d = %d\n", i - 1, (int)pow(2, i - 1));
    n -= (int)pow(2, i - 1);
    q += i;
  }
  printf("quotient = %d, remainder = %d\n", q, n);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<half>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%ebp
  mov    %edi,%esi
  lea    D20(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%r13d
  lea    D2(%rip),%r14
  cmp    $0x1,%ebp
  jg     L16
L21:
  mov    %ebp,%edx
  mov    %r13d,%esi
  lea    D4(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L20:
  add    $0x1,%ebx
L18:
  mov    %ebx,%esi
  mov    $0x2,%edi
  call   <pow@plt>
  cmp    %eax,%ebp
  jge    L20
  lea    -0x1(%rbx),%r12d
  mov    %r12d,%esi
  mov    $0x2,%edi
  call   <pow@plt>
  mov    %eax,%edx
  mov    %r12d,%esi
  mov    %r14,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %r12d,%esi
  mov    $0x2,%edi
  call   <pow@plt>
  sub    %eax,%ebp
  add    %ebx,%r13d
  cmp    $0x1,%ebp
  jle    L21
L16:
  mov    $0x1,%ebx
  jmp    L18
```

tool_calls: name: retrieve_data
arguments: tag: D20
type_guessed: string
name: retrieve_data
arguments: tag: D2
type_guessed: string
name: retrieve_data
arguments: tag: D4
type_guessed: string

tool: D20: %d/2


tool: D2: i = %d, d = %d


tool: D4: quotient = %d, remainder = %d


assistant: ```c
void half(int n) {
  int q = 0, i;
  printf("%d/2\n", n);
  while (n >= 2) {
    for (i = 1; ((int)pow(2, i)) <= n; i++)
      ;
    printf("i = %d, d = %d\n", i - 1, (int)pow(2, i - 1));
    n -= (int)pow(2, i - 1);
    q += i;
  }
  printf("quotient = %d, remainder = %d\n", q, n);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<half>:
  endbr64
  push   %r14
  mov    %edi,%esi
  xor    %eax,%eax
  lea    D3(%rip),%r14
  push   %r13
  xor    %r13d,%r13d
  push   %r12
  mov    %edi,%r12d
  lea    D26(%rip),%rdi
  push   %rbp
  push   %rbx
  call   <printf@plt>
  cmp    $0x1,%r12d
  jle    L1
  nop
L25:
  mov    $0x1,%ebx
  jmp    L39
  nopw   0x0(%rax,%rax,1)
L13:
  add    $0x1,%ebx
L39:
  mov    %ebx,%esi
  mov    $0x2,%edi
  call   <pow@plt>
  cmp    %eax,%r12d
  jge    L13
  lea    -0x1(%rbx),%ebp
  mov    $0x2,%edi
  add    %ebx,%r13d
  mov    %ebp,%esi
  call   <pow@plt>
  mov    %ebp,%esi
  mov    %r14,%rdi
  mov    %eax,%edx
  xor    %eax,%eax
  call   <printf@plt>
  mov    %ebp,%esi
  mov    $0x2,%edi
  call   <pow@plt>
  sub    %eax,%r12d
  cmp    $0x1,%r12d
  jg     L25
L1:
  pop    %rbx
  mov    %r12d,%edx
  pop    %rbp
  mov    %r13d,%esi
  pop    %r12
  lea    D22(%rip),%rdi
  pop    %r13
  xor    %eax,%eax
  pop    %r14
  jmp    1060 <printf@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D3
type_guessed: string
name: disassemble_data
arguments: tag: D26
type_guessed: string
name: disassemble_data
arguments: tag: D22
type_guessed: string

tool: D3: i = %d, d = %d


tool: D26: %d/2


tool: D22: quotient = %d, remainder = %d


assistant: ```c
void half(int n) {
  int q = 0, i;
  printf("%d/2\n", n);
  while (n >= 2) {
    for (i = 1; ((int)pow(2, i)) <= n; i++)
      ;
    printf("i = %d, d = %d\n", i - 1, (int)pow(2, i - 1));
    n -= (int)pow(2, i - 1);
    q += i;
  }
  printf("quotient = %d, remainder = %d\n", q, n);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<half>:
  endbr64
  push   %r14
  mov    %edi,%esi
  xor    %eax,%eax
  lea    D7(%rip),%r14
  push   %r13
  xor    %r13d,%r13d
  push   %r12
  mov    %edi,%r12d
  lea    D26(%rip),%rdi
  push   %rbp
  push   %rbx
  call   <printf@plt>
  cmp    $0x1,%r12d
  jle    L22
  nop
L26:
  mov    $0x1,%ebx
  jmp    L24
  nopw   0x0(%rax,%rax,1)
L3:
  add    $0x1,%ebx
L24:
  mov    %ebx,%esi
  mov    $0x2,%edi
  call   <pow@plt>
  cmp    %eax,%r12d
  jge    L3
  lea    -0x1(%rbx),%ebp
  mov    $0x2,%edi
  add    %ebx,%r13d
  mov    %ebp,%esi
  call   <pow@plt>
  mov    %ebp,%esi
  mov    %r14,%rdi
  mov    %eax,%edx
  xor    %eax,%eax
  call   <printf@plt>
  mov    %ebp,%esi
  mov    $0x2,%edi
  call   <pow@plt>
  sub    %eax,%r12d
  cmp    $0x1,%r12d
  jg     L26
L22:
  pop    %rbx
  mov    %r12d,%edx
  pop    %rbp
  mov    %r13d,%esi
  pop    %r12
  lea    D9(%rip),%rdi
  pop    %r13
  xor    %eax,%eax
  pop    %r14
  jmp    1060 <printf@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D7
type_guessed: string
name: extract_symbolic_data
arguments: label: D26
type_guessed: string
name: extract_symbolic_data
arguments: label: D9
type_guessed: string

tool: D7: i = %d, d = %d


tool: D26: %d/2


tool: D9: quotient = %d, remainder = %d


assistant: ```c
void half(int n) {
  int q = 0, i;
  printf("%d/2\n", n);
  while (n >= 2) {
    for (i = 1; ((int)pow(2, i)) <= n; i++)
      ;
    printf("i = %d, d = %d\n", i - 1, (int)pow(2, i - 1));
    n -= (int)pow(2, i - 1);
    q += i;
  }
  printf("quotient = %d, remainder = %d\n", q, n);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<next_segment>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movq   $0x0,-0x8(%rbp)
  jmp    L3
L11:
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rax
  lea    0x1(%rax),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,(%rax)
  addq   $0x1,-0x8(%rbp)
L3:
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L11
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rax
  lea    0x1(%rax),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,(%rax)
  addq   $0x1,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
size_t next_segment(char **segment) {
  size_t used = 0;
  while (**segment != '\0') {
    ++*segment, ++used;
  }
  ++*segment, ++used;
  return (used);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<next_segment>:
  endbr64
  mov    (%rdi),%rcx
  cmpb   $0x0,(%rcx)
  je     L4
  lea    0x1(%rcx),%rdx
  mov    $0x0,%eax
L16:
  mov    %rdx,%rcx
  mov    %rdx,(%rdi)
  add    $0x1,%rax
  add    $0x1,%rdx
  cmpb   $0x0,-0x1(%rdx)
  jne    L16
L26:
  add    $0x1,%rcx
  mov    %rcx,(%rdi)
  add    $0x1,%rax
  ret
L4:
  mov    $0x0,%eax
  jmp    L26
```

assistant: ```c
size_t next_segment(char **segment) {
  size_t used = 0;
  while (**segment != '\0') {
    ++*segment, ++used;
  }
  ++*segment, ++used;
  return (used);
}
```
user: What is the c source code of the assembly code below?

```asm
<next_segment>:
  endbr64
  mov    (%rdi),%rdx
  cmpb   $0x0,(%rdx)
  je     L2
  lea    0x1(%rdx),%rsi
  xor    %ecx,%ecx
  nopw   0x0(%rax,%rax,1)
L10:
  mov    %rsi,%rdx
  mov    %rsi,(%rdi)
  mov    %rcx,%rax
  add    $0x1,%rcx
  cmpb   $0x0,(%rdx)
  lea    0x1(%rsi),%rsi
  jne    L10
  add    $0x1,%rdx
  add    $0x2,%rax
  mov    %rdx,(%rdi)
  ret
  nopw   0x0(%rax,%rax,1)
L2:
  add    $0x1,%rdx
  mov    $0x1,%eax
  mov    %rdx,(%rdi)
  ret
```

assistant: ```c
size_t next_segment(char **segment) {
  size_t used = 0;
  while (**segment != '\0') {
    ++*segment, ++used;
  }
  ++*segment, ++used;
  return (used);
}
```
user: What is the c source code of the assembly code below:

```asm
<next_segment>:
  endbr64
  mov    (%rdi),%rdx
  cmpb   $0x0,(%rdx)
  je     L1
  lea    0x1(%rdx),%rsi
  xor    %ecx,%ecx
  nopw   0x0(%rax,%rax,1)
L7:
  mov    %rsi,%rdx
  mov    %rsi,(%rdi)
  mov    %rcx,%rax
  add    $0x1,%rcx
  cmpb   $0x0,(%rdx)
  lea    0x1(%rsi),%rsi
  jne    L7
  add    $0x1,%rdx
  add    $0x2,%rax
  mov    %rdx,(%rdi)
  ret
  nopw   0x0(%rax,%rax,1)
L1:
  add    $0x1,%rdx
  mov    $0x1,%eax
  mov    %rdx,(%rdi)
  ret
```

assistant: ```c
size_t next_segment(char **segment) {
  size_t used = 0;
  while (**segment != '\0') {
    ++*segment, ++used;
  }
  ++*segment, ++used;
  return (used);
}
```