ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 这段汇编代码用C语言表示是什么样的？

```asm
<Try>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %rcx,-0x30(%rbp)
  mov    %r8,-0x38(%rbp)
  mov    %r9,-0x40(%rbp)
  movl   $0x0,-0x4(%rbp)
  mov    -0x20(%rbp),%rax
  movl   $0x0,(%rax)
  jmp    L44
L39:
  addl   $0x1,-0x4(%rbp)
  mov    -0x20(%rbp),%rax
  movl   $0x0,(%rax)
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  test   %eax,%eax
  je     L44
  mov    -0x14(%rbp),%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  test   %eax,%eax
  je     L44
  mov    -0x14(%rbp),%eax
  sub    -0x4(%rbp),%eax
  cltq
  add    $0x7,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  test   %eax,%eax
  je     L44
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x40(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  mov    %eax,(%rdx)
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  movl   $0x0,(%rax)
  mov    -0x14(%rbp),%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movl   $0x0,(%rax)
  mov    -0x14(%rbp),%eax
  sub    -0x4(%rbp),%eax
  cltq
  add    $0x7,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  movl   $0x0,(%rax)
  cmpl   $0x7,-0x14(%rbp)
  jg     L47
  mov    -0x14(%rbp),%eax
  lea    0x1(%rax),%edi
  mov    -0x40(%rbp),%r8
  mov    -0x38(%rbp),%rsi
  mov    -0x30(%rbp),%rcx
  mov    -0x28(%rbp),%rdx
  mov    -0x20(%rbp),%rax
  mov    %r8,%r9
  mov    %rsi,%r8
  mov    %rax,%rsi
  mov    $0x0,%eax
  call   <Try>
  mov    -0x20(%rbp),%rax
  mov    (%rax),%eax
  test   %eax,%eax
  jne    L44
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  movl   $0x1,(%rax)
  mov    -0x14(%rbp),%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movl   $0x1,(%rax)
  mov    -0x14(%rbp),%eax
  sub    -0x4(%rbp),%eax
  cltq
  add    $0x7,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  movl   $0x1,(%rax)
  jmp    L44
L47:
  mov    -0x20(%rbp),%rax
  movl   $0x1,(%rax)
L44:
  mov    -0x20(%rbp),%rax
  mov    (%rax),%eax
  test   %eax,%eax
  jne    L85
  cmpl   $0x8,-0x4(%rbp)
  jne    L39
L85:
  nop
  leave
  ret
```

assistant: ```c
Try(i, q, a, b, c, x) int i, *q, a[], b[], c[], x[];
{
  int j;
  j = 0;
  *q = 0;
  while ((!*q) && (j != 8)) {
    j = j + 1;
    *q = 0;
    if (b[j] && a[i + j] && c[i - j + 7]) {
      x[i] = j;
      b[j] = 0;
      a[i + j] = 0;
      c[i - j + 7] = 0;
      if (i < 8) {
        Try(i + 1, q, a, b, c, x);
        if (!*q) {
          b[j] = 1;
          a[i + j] = 1;
          c[i - j + 7] = 1;
        }
      } else
        *q = 1;
    }
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<Try>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %edi,%r12d
  mov    %rsi,%rbp
  mov    %rdx,0x8(%rsp)
  mov    %rcx,%r13
  mov    %r8,0x10(%rsp)
  mov    %r9,0x20(%rsp)
  movslq %edi,%rax
  lea    (%r9,%rax,4),%rax
  mov    %rax,0x18(%rsp)
  mov    $0x1,%ebx
  lea    0x1(%rdi),%eax
  mov    %eax,0x2c(%rsp)
  jmp    L52
L24:
  mov    0x20(%rsp),%r9
  mov    %rsi,%r8
  mov    %r13,%rcx
  mov    0x8(%rsp),%rdx
  mov    %rbp,%rsi
  mov    0x2c(%rsp),%edi
  mov    $0x0,%eax
  call   <Try>
  cmpl   $0x0,0x0(%rbp)
  jne    L16
  movl   $0x1,0x0(%r13,%rbx,4)
  movl   $0x1,(%r14)
  movl   $0x1,(%r15)
L30:
  cmpl   $0x0,0x0(%rbp)
  sete   %dl
  cmp    $0x8,%ebx
  setne  %al
  add    $0x1,%rbx
  test   %al,%dl
  je     L16
L52:
  movl   $0x0,0x0(%rbp)
  cmpl   $0x0,0x0(%r13,%rbx,4)
  je     L30
  lea    (%r12,%rbx,1),%eax
  cltq
  mov    0x8(%rsp),%rsi
  lea    (%rsi,%rax,4),%r14
  cmpl   $0x0,(%r14)
  je     L30
  mov    %r12d,%eax
  sub    %ebx,%eax
  cltq
  mov    0x10(%rsp),%rsi
  lea    0x1c(%rsi,%rax,4),%r15
  cmpl   $0x0,(%r15)
  je     L30
  mov    0x18(%rsp),%rax
  mov    %ebx,(%rax)
  movl   $0x0,0x0(%r13,%rbx,4)
  movl   $0x0,(%r14)
  movl   $0x0,(%r15)
  cmp    $0x7,%r12d
  jle    L24
  movl   $0x1,0x0(%rbp)
L16:
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
Try(i, q, a, b, c, x) int i, *q, a[], b[], c[], x[];
{
  int j;
  j = 0;
  *q = 0;
  while ((!*q) && (j != 8)) {
    j = j + 1;
    *q = 0;
    if (b[j] && a[i + j] && c[i - j + 7]) {
      x[i] = j;
      b[j] = 0;
      a[i + j] = 0;
      c[i - j + 7] = 0;
      if (i < 8) {
        Try(i + 1, q, a, b, c, x);
        if (!*q) {
          b[j] = 1;
          a[i + j] = 1;
          c[i - j + 7] = 1;
        }
      } else
        *q = 1;
    }
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<Try>:
  endbr64
  push   %r15
  movslq %edi,%rax
  push   %r14
  mov    %rdx,%r14
  push   %r13
  mov    %rax,%r13
  push   %r12
  push   %rbp
  lea    (%r9,%rax,4),%rbp
  push   %rbx
  mov    $0x1,%ebx
  sub    $0x28,%rsp
L46:
  movl   $0x0,(%rsi)
  mov    (%rcx,%rbx,4),%r11d
  test   %r11d,%r11d
  je     L44
  lea    0x0(%r13,%rbx,1),%eax
  cltq
  lea    (%r14,%rax,4),%r12
  mov    (%r12),%r10d
  test   %r10d,%r10d
  je     L44
  mov    %r13d,%eax
  sub    %ebx,%eax
  cltq
  lea    0x1c(%r8,%rax,4),%r15
  mov    (%r15),%edi
  test   %edi,%edi
  je     L44
  mov    %ebx,0x0(%rbp)
  movl   $0x0,(%rcx,%rbx,4)
  movl   $0x0,(%r12)
  movl   $0x0,(%r15)
  cmp    $0x7,%r13d
  jle    L27
  movl   $0x1,(%rsi)
L11:
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  xchg   %ax,%ax
L44:
  mov    $0x1,%edx
L9:
  cmp    $0x8,%ebx
  setne  %al
  add    $0x1,%rbx
  test   %dl,%al
  jne    L46
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L27:
  mov    %r14,%rdx
  lea    0x1(%r13),%edi
  xor    %eax,%eax
  mov    %rsi,(%rsp)
  mov    %r9,0x18(%rsp)
  mov    %r8,0x10(%rsp)
  mov    %rcx,0x8(%rsp)
  call   <Try>
  mov    (%rsp),%rsi
  mov    (%rsi),%edx
  test   %edx,%edx
  jne    L11
  mov    0x8(%rsp),%rcx
  mov    0x10(%rsp),%r8
  mov    0x18(%rsp),%r9
  movl   $0x1,(%rcx,%rbx,4)
  movl   $0x1,(%r12)
  movl   $0x1,(%r15)
  mov    (%rsi),%eax
  test   %eax,%eax
  sete   %dl
  jmp    L9
```

assistant: ```c
Try(i, q, a, b, c, x) int i, *q, a[], b[], c[], x[];
{
  int j;
  j = 0;
  *q = 0;
  while ((!*q) && (j != 8)) {
    j = j + 1;
    *q = 0;
    if (b[j] && a[i + j] && c[i - j + 7]) {
      x[i] = j;
      b[j] = 0;
      a[i + j] = 0;
      c[i - j + 7] = 0;
      if (i < 8) {
        Try(i + 1, q, a, b, c, x);
        if (!*q) {
          b[j] = 1;
          a[i + j] = 1;
          c[i - j + 7] = 1;
        }
      } else
        *q = 1;
    }
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<ustrtoul>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  mov    -0x24(%rbp),%edx
  mov    -0x20(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <simple_strtoul@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x20(%rbp),%rax
  mov    (%rax),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  cmp    $0x6b,%eax
  je     L83
  cmp    $0x6b,%eax
  jg     L75
  cmp    $0x4d,%eax
  je     L89
  cmp    $0x4d,%eax
  jg     L75
  cmp    $0x47,%eax
  je     L66
  cmp    $0x4b,%eax
  je     L83
  jmp    L75
L66:
  shlq   $0xa,-0x8(%rbp)
L89:
  shlq   $0xa,-0x8(%rbp)
L83:
  shlq   $0xa,-0x8(%rbp)
  mov    -0x20(%rbp),%rax
  mov    (%rax),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  cmp    $0x69,%al
  jne    L75
  mov    -0x20(%rbp),%rax
  mov    (%rax),%rax
  add    $0x2,%rax
  movzbl (%rax),%eax
  cmp    $0x42,%al
  jne    L73
  mov    -0x20(%rbp),%rax
  mov    (%rax),%rax
  lea    0x3(%rax),%rdx
  mov    -0x20(%rbp),%rax
  mov    %rdx,(%rax)
  jmp    L75
L73:
  mov    -0x20(%rbp),%rax
  mov    (%rax),%rax
  lea    0x2(%rax),%rdx
  mov    -0x20(%rbp),%rax
  mov    %rdx,(%rax)
L75:
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
int ustrtoul(const char *cp, char **endp, unsigned int base) {
  unsigned long result = simple_strtoul(cp, endp, base);
  switch (**endp) {
  case 'G':
    result *= 1024;
  case 'M':
    result *= 1024;
  case 'K':
  case 'k':
    result *= 1024;
    if ((*endp)[1] == 'i') {
      if ((*endp)[2] == 'B')
        (*endp) += 3;
      else
        (*endp) += 2;
    }
  }
  return result;
}
```
user: What is the c source code of the assembly code below?

```asm
<ustrtoul>:
  endbr64
  push   %rbx
  mov    %rsi,%rbx
  call   <simple_strtoul@plt>
  mov    (%rbx),%rcx
  movzbl (%rcx),%edx
  cmp    $0x4d,%dl
  je     L31
  jg     L53
  cmp    $0x47,%dl
  je     L28
  cmp    $0x4b,%dl
  jne    L75
L18:
  shl    $0xa,%rax
  cmpb   $0x69,0x1(%rcx)
  jne    L75
  cmpb   $0x42,0x2(%rcx)
  je     L58
  add    $0x2,%rcx
  mov    %rcx,(%rbx)
  jmp    L75
L53:
  cmp    $0x6b,%dl
  je     L18
L75:
  pop    %rbx
  ret
L28:
  shl    $0xa,%rax
L31:
  shl    $0xa,%rax
  jmp    L18
L58:
  add    $0x3,%rcx
  mov    %rcx,(%rbx)
  jmp    L75
```

assistant: ```c
int ustrtoul(const char *cp, char **endp, unsigned int base) {
  unsigned long result = simple_strtoul(cp, endp, base);
  switch (**endp) {
  case 'G':
    result *= 1024;
  case 'M':
    result *= 1024;
  case 'K':
  case 'k':
    result *= 1024;
    if ((*endp)[1] == 'i') {
      if ((*endp)[2] == 'B')
        (*endp) += 3;
      else
        (*endp) += 2;
    }
  }
  return result;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<ustrtoul>:
  endbr64
  push   %rbx
  mov    %rsi,%rbx
  call   <simple_strtoul@plt>
  mov    (%rbx),%rcx
  movzbl (%rcx),%edx
  cmp    $0x4d,%dl
  je     L5
  jg     L43
  cmp    $0x47,%dl
  jne    L38
  shl    $0xa,%rax
L5:
  shl    $0xa,%rax
L30:
  shl    $0xa,%rax
  cmpb   $0x69,0x1(%rcx)
  je     L39
L76:
  pop    %rbx
  ret
  nopl   0x0(%rax,%rax,1)
L38:
  cmp    $0x4b,%dl
  jne    L76
  shl    $0xa,%rax
  cmpb   $0x69,0x1(%rcx)
  jne    L76
L39:
  cmpb   $0x42,0x2(%rcx)
  je     L57
  add    $0x2,%rcx
  mov    %rcx,(%rbx)
  pop    %rbx
  ret
  cs nopw 0x0(%rax,%rax,1)
L43:
  cmp    $0x6b,%dl
  je     L30
  pop    %rbx
  ret
  nopw   0x0(%rax,%rax,1)
L57:
  add    $0x3,%rcx
  mov    %rcx,(%rbx)
  pop    %rbx
  ret
```

assistant: ```c
int ustrtoul(const char *cp, char **endp, unsigned int base) {
  unsigned long result = simple_strtoul(cp, endp, base);
  switch (**endp) {
  case 'G':
    result *= 1024;
  case 'M':
    result *= 1024;
  case 'K':
  case 'k':
    result *= 1024;
    if ((*endp)[1] == 'i') {
      if ((*endp)[2] == 'B')
        (*endp) += 3;
      else
        (*endp) += 2;
    }
  }
  return result;
}
```
user: What is the c source code of the assembly code below:

```asm
<ustrtoul>:
  endbr64
  push   %rbx
  mov    %rsi,%rbx
  call   <simple_strtoul@plt>
  mov    (%rbx),%rcx
  movzbl (%rcx),%edx
  cmp    $0x4d,%dl
  je     L72
  jg     L45
  cmp    $0x47,%dl
  jne    L7
  shl    $0xa,%rax
L72:
  shl    $0xa,%rax
L33:
  shl    $0xa,%rax
  cmpb   $0x69,0x1(%rcx)
  je     L64
L68:
  pop    %rbx
  ret
  nopl   0x0(%rax,%rax,1)
L7:
  cmp    $0x4b,%dl
  jne    L68
  shl    $0xa,%rax
  cmpb   $0x69,0x1(%rcx)
  jne    L68
L64:
  cmpb   $0x42,0x2(%rcx)
  je     L77
  add    $0x2,%rcx
  mov    %rcx,(%rbx)
  pop    %rbx
  ret
  cs nopw 0x0(%rax,%rax,1)
L45:
  cmp    $0x6b,%dl
  je     L33
  pop    %rbx
  ret
  nopw   0x0(%rax,%rax,1)
L77:
  add    $0x3,%rcx
  mov    %rcx,(%rbx)
  pop    %rbx
  ret
```

assistant: ```c
int ustrtoul(const char *cp, char **endp, unsigned int base) {
  unsigned long result = simple_strtoul(cp, endp, base);
  switch (**endp) {
  case 'G':
    result *= 1024;
  case 'M':
    result *= 1024;
  case 'K':
  case 'k':
    result *= 1024;
    if ((*endp)[1] == 'i') {
      if ((*endp)[2] == 'B')
        (*endp) += 3;
      else
        (*endp) += 2;
    }
  }
  return result;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<xstrndup>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  cmp    -0x20(%rbp),%rax
  jbe    L7
  mov    -0x20(%rbp),%rax
  mov    %rax,-0x10(%rbp)
L7:
  mov    -0x10(%rbp),%rax
  mov    $0x1,%esi
  mov    %rax,%rdi
  call   <xmalloc_inc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L11
  mov    $0x0,%eax
  jmp    L5
L11:
  mov    -0x10(%rbp),%rdx
  mov    -0x18(%rbp),%rcx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0x8(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x8(%rbp),%rax
L5:
  leave
  ret
```

assistant: ```c
char *xstrndup(const char *s, size_t len) {
  char *new;
  size_t n = strlen(s);
  if (n > len)
    n = len;
  new = xmalloc_inc(n, 1);
  if (new == NULL)
    return NULL;
  memcpy(new, s, n);
  new[n] = '\0';
  return new;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<xstrndup>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    %rsi,%rbp
  call   <strlen@plt>
  cmp    %rbp,%rax
  cmovbe %rax,%rbp
  mov    $0x1,%esi
  mov    %rbp,%rdi
  call   <xmalloc_inc@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L1
  mov    %rbp,%rdx
  mov    %r12,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  movb   $0x0,(%rbx,%rbp,1)
L1:
  mov    %rbx,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *xstrndup(const char *s, size_t len) {
  char *new;
  size_t n = strlen(s);
  if (n > len)
    n = len;
  new = xmalloc_inc(n, 1);
  if (new == NULL)
    return NULL;
  memcpy(new, s, n);
  new[n] = '\0';
  return new;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<xstrndup>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rsi,%rbx
  call   <strlen@plt>
  mov    $0x1,%esi
  cmp    %rbx,%rax
  cmovbe %rax,%rbx
  mov    %rbx,%rdi
  call   <xmalloc_inc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L4
  mov    %rbx,%rdx
  mov    %rbp,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  movb   $0x0,(%r12,%rbx,1)
L4:
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *xstrndup(const char *s, size_t len) {
  char *new;
  size_t n = strlen(s);
  if (n > len)
    n = len;
  new = xmalloc_inc(n, 1);
  if (new == NULL)
    return NULL;
  memcpy(new, s, n);
  new[n] = '\0';
  return new;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<xstrndup>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rsi,%rbx
  call   <strlen@plt>
  mov    $0x1,%esi
  cmp    %rbx,%rax
  cmovbe %rax,%rbx
  mov    %rbx,%rdi
  call   <xmalloc_inc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L1
  mov    %rbx,%rdx
  mov    %rbp,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  movb   $0x0,(%r12,%rbx,1)
L1:
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *xstrndup(const char *s, size_t len) {
  char *new;
  size_t n = strlen(s);
  if (n > len)
    n = len;
  new = xmalloc_inc(n, 1);
  if (new == NULL)
    return NULL;
  memcpy(new, s, n);
  new[n] = '\0';
  return new;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<readline>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    D1(%rip),%rax
  mov    (%rax),%edx
  mov    -0x10(%rbp),%rax
  lea    -0x1(%rax),%rcx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <fgets@plt>
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cltq
  lea    -0x1(%rax),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *readline(char *s, size_t max) {
  fgets(s, max - 1, stdin);
  s[strlen(s) - 1] = '\0';
  return s;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<readline>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x1,%rsi
  mov    D4(%rip),%rax
  mov    (%rax),%edx
  call   <fgets@plt>
  mov    %rbx,%rdi
  call   <strlen@plt>
  cltq
  movb   $0x0,-0x1(%rbx,%rax,1)
  mov    %rbx,%rax
  pop    %rbx
  ret
```

assistant: ```c
char *readline(char *s, size_t max) {
  fgets(s, max - 1, stdin);
  s[strlen(s) - 1] = '\0';
  return s;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<readline>:
  endbr64
  mov    D5(%rip),%rax
  push   %r12
  sub    $0x1,%rsi
  mov    %rdi,%r12
  mov    (%rax),%edx
  call   <fgets@plt>
  mov    %r12,%rdi
  call   <strlen@plt>
  cltq
  movb   $0x0,-0x1(%r12,%rax,1)
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
char *readline(char *s, size_t max) {
  fgets(s, max - 1, stdin);
  s[strlen(s) - 1] = '\0';
  return s;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<readline>:
  endbr64
  mov    D2(%rip),%rax
  push   %r12
  sub    $0x1,%rsi
  mov    %rdi,%r12
  mov    (%rax),%edx
  call   <fgets@plt>
  mov    %r12,%rdi
  call   <strlen@plt>
  cltq
  movb   $0x0,-0x1(%r12,%rax,1)
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
char *readline(char *s, size_t max) {
  fgets(s, max - 1, stdin);
  s[strlen(s) - 1] = '\0';
  return s;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<llspeech_free_string_array>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  je     L39
  movl   $0x0,-0x4(%rbp)
  jmp    L35
L28:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  test   %rax,%rax
  je     L15
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <free@plt>
L15:
  addl   $0x1,-0x4(%rbp)
L35:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L28
  cmpq   $0x0,-0x18(%rbp)
  je     L14
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  jmp    L14
L39:
  nop
L14:
  leave
  ret
```

assistant: ```c
void llspeech_free_string_array(char **array, int len) {
  int i;
  if (!array)
    return;
  for (i = 0; i < len; i++) {
    if (array[i])
      free(array[i]);
  }
  if (array)
    free(array);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<llspeech_free_string_array>:
  endbr64
  test   %rdi,%rdi
  je     L11
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  test   %esi,%esi
  jle    L55
  mov    %rdi,%rbx
  lea    -0x1(%rsi),%eax
  lea    0x8(%rdi,%rax,8),%rbp
  jmp    L34
L5:
  add    $0x8,%rbx
  cmp    %rbp,%rbx
  je     L55
L34:
  mov    (%rbx),%rdi
  test   %rdi,%rdi
  je     L5
  call   <free@plt>
  jmp    L5
L55:
  mov    %r12,%rdi
  call   <free@plt>
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L11:
  ret
```

assistant: ```c
void llspeech_free_string_array(char **array, int len) {
  int i;
  if (!array)
    return;
  for (i = 0; i < len; i++) {
    if (array[i])
      free(array[i]);
  }
  if (array)
    free(array);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<llspeech_free_string_array>:
  endbr64
  test   %rdi,%rdi
  je     L24
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  test   %esi,%esi
  jle    L32
  lea    -0x1(%rsi),%eax
  mov    %rdi,%rbx
  lea    0x8(%rdi,%rax,8),%rbp
  nop
L30:
  mov    (%rbx),%rdi
  test   %rdi,%rdi
  je     L19
  call   <free@plt>
L19:
  add    $0x8,%rbx
  cmp    %rbp,%rbx
  jne    L30
L32:
  pop    %rbx
  mov    %r12,%rdi
  pop    %rbp
  pop    %r12
  jmp    1050 <free@plt>
  nopw   0x0(%rax,%rax,1)
L24:
  ret
```

assistant: ```c
void llspeech_free_string_array(char **array, int len) {
  int i;
  if (!array)
    return;
  for (i = 0; i < len; i++) {
    if (array[i])
      free(array[i]);
  }
  if (array)
    free(array);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<llspeech_free_string_array>:
  endbr64
  test   %rdi,%rdi
  je     L39
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  test   %esi,%esi
  jle    L38
  lea    -0x1(%rsi),%eax
  mov    %rdi,%rbx
  lea    0x8(%rdi,%rax,8),%rbp
  nop
L32:
  mov    (%rbx),%rdi
  test   %rdi,%rdi
  je     L29
  call   <free@plt>
L29:
  add    $0x8,%rbx
  cmp    %rbp,%rbx
  jne    L32
L38:
  pop    %rbx
  mov    %r12,%rdi
  pop    %rbp
  pop    %r12
  jmp    1050 <free@plt>
  nopw   0x0(%rax,%rax,1)
L39:
  ret
```

assistant: ```c
void llspeech_free_string_array(char **array, int len) {
  int i;
  if (!array)
    return;
  for (i = 0; i < len; i++) {
    if (array[i])
      free(array[i]);
  }
  if (array)
    free(array);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<ast_args>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  movsd  %xmm0,(%rsp)
  movapd %xmm0,%xmm1
  mulsd  %xmm0,%xmm1
  movsd  %xmm1,0x8(%rsp)
  mulsd  %xmm0,%xmm1
  movsd  %xmm1,0x10(%rsp)
  mulsd  %xmm1,%xmm0
  movsd  %xmm0,0x18(%rsp)
  lea    D10(%rip),%rbx
  mov    %rdi,%rbp
  lea    0xc8(%rbx),%r14
  mov    %rsp,%r12
  jmp    L47
L5:
  mulsd  D16(%rip),%xmm0
  mov    D21(%rip),%rax
  movq   %rax,%xmm1
  call   <fmod@plt>
  movsd  %xmm0,0x0(%r13)
  add    $0x28,%rbx
  add    $0x8,%rbp
  cmp    %r14,%rbx
  je     L28
L47:
  mov    %rbx,%rdx
  movsd  D34(%rip),%xmm0
  mulsd  (%rbx),%xmm0
  mov    %rbp,%r13
  movsd  %xmm0,0x0(%rbp)
  mov    $0x0,%eax
L9:
  movsd  0x8(%rdx,%rax,1),%xmm1
  mulsd  (%rax,%r12,1),%xmm1
  addsd  %xmm1,%xmm0
  add    $0x8,%rax
  cmp    $0x20,%rax
  jne    L9
  jmp    L5
L28:
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L49
  add    $0x30,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L49:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D10
type: f64
name: unpack_data
arguments: data_tag: D16
type: byte[8]
name: unpack_data
arguments: data_tag: D21
type: byte[8]
name: unpack_data
arguments: data_tag: D34
type: f64

tool: D10: 134.96340251

tool: {"D16": [0x9d,0x8f,0xff,0xb2,0xa5,0x55,0xd4,0x3e]}

tool: {"D21": [0x18,0x2d,0x44,0x54,0xfb,0x21,0x19,0x40]}

tool: D34: 3600

assistant: ```c
void ast_args(double t, double *f) {
  static const double fc[][5] = {{134.96340251, 1717915923.2178, 31.8792, 0.051635, -0.00024470}, {357.52910918, 129596581.0481, -0.5532, 0.000136, -0.00001149}, {93.27209062, 1739527262.8478, -12.7512, -0.001037, 0.00000417}, {297.85019547, 1602961601.2090, -6.3706, 0.006593, -0.00003169}, {125.04455501, -6962890.2665, 7.4722, 0.007702 - 0.00005939}};
  double tt[4];
  int i, j;
  for (tt[0] = t, i = 1; i < 4; i++)
    tt[i] = tt[i - 1] * t;
  for (i = 0; i < 5; i++) {
    f[i] = fc[i][0] * 3600.0;
    for (j = 0; j < 4; j++)
      f[i] += fc[i][j + 1] * tt[j];
    f[i] = fmod(f[i] * ((3.1415926535897932 / 180.0) / 3600.0), 2.0 * 3.1415926535897932);
  }
}
```