ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What would the C source look like for the assembly code provided here:

```asm
<create_givens>:
  endbr64
  push   %r12
  movapd %xmm0,%xmm2
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x10,%rsp
  ucomisd D2(%rip),%xmm1
  jp     L2
  jne    L2
  mov    D14(%rip),%rax
  mov    %rax,(%rdi)
  movq   $0x0,(%rsi)
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L2:
  movapd %xmm1,%xmm0
  movsd  %xmm1,0x8(%rsp)
  movsd  %xmm2,(%rsp)
  call   <fabs@plt>
  movsd  (%rsp),%xmm2
  mov    %rax,%r12
  movapd %xmm2,%xmm0
  call   <fabs@plt>
  movsd  (%rsp),%xmm2
  movsd  0x8(%rsp),%xmm1
  cmp    %rax,%r12
  jle    L8
  xorpd  D76(%rip),%xmm2
  divsd  %xmm1,%xmm2
  movapd %xmm2,%xmm0
  movsd  %xmm2,(%rsp)
  mulsd  %xmm2,%xmm0
  addsd  D14(%rip),%xmm0
  cvttsd2si %xmm0,%edi
  call   <sqrt@plt>
  movsd  (%rsp),%xmm2
  movapd %xmm0,%xmm1
  movsd  D14(%rip),%xmm0
  divsd  %xmm1,%xmm0
  mulsd  %xmm0,%xmm2
  movsd  %xmm0,(%rbx)
  movsd  %xmm2,0x0(%rbp)
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L8:
  xorpd  D76(%rip),%xmm1
  divsd  %xmm2,%xmm1
  movapd %xmm1,%xmm0
  movsd  %xmm1,(%rsp)
  mulsd  %xmm1,%xmm0
  addsd  D14(%rip),%xmm0
  cvttsd2si %xmm0,%edi
  call   <sqrt@plt>
  movsd  (%rsp),%xmm1
  movapd %xmm0,%xmm2
  movsd  D14(%rip),%xmm0
  divsd  %xmm2,%xmm0
  mulsd  %xmm0,%xmm1
  movsd  %xmm0,0x0(%rbp)
  movsd  %xmm1,(%rbx)
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D2
type_infered: byte[8]
name: extract_symbolic_data
arguments: label: D14
type_infered: f64
name: extract_symbolic_data
arguments: label: D76
type_infered: byte[16]

tool: {"D2": [0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00]}

tool: D14: 1

tool: {"D76": [0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x80,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00]}

assistant: ```c
void create_givens(const double a, const double b, double *c, double *s) {
  if (b == 0) {
    *c = 1;
    *s = 0;
  } else if (fabs(b) > fabs(a)) {
    double t = -a / b;
    double s1 = 1.0 / sqrt(1 + t * t);
    *s = s1;
    *c = s1 * t;
  } else {
    double t = -b / a;
    double c1 = 1.0 / sqrt(1 + t * t);
    *c = c1;
    *s = c1 * t;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<sal_strdup>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  jne    L29
  mov    $0x0,%eax
  jmp    L14
L29:
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strdup@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L6
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,%esi
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x1,%edi
  call   <exit@plt>
L6:
  mov    -0x8(%rbp),%rax
L14:
  leave
  ret
```

tool_calls: name: read_data
arguments: data_label: D8
type_guessed: string

tool: D8: Cannot alloc %d bytes.


assistant: ```c
char *sal_strdup(const char *s) {
  char *ptr;
  if (s == NULL)
    return NULL;
  ptr = strdup(s);
  if (ptr == NULL) {
    printf("Cannot alloc %d bytes.\n", (int)strlen(s));
    exit(1);
  }
  return ptr;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<sal_strdup>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  test   %rdi,%rdi
  je     L8
  call   <strdup@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  je     L39
L18:
  mov    %rbp,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L39:
  mov    %rbx,%rdi
  call   <strlen@plt>
  mov    %eax,%esi
  lea    D3(%rip),%rdi
  call   <printf@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  jmp    L18
L8:
  mov    %rdi,%rbp
  jmp    L18
```

tool_calls: name: disassemble_data
arguments: data_tag: D3
datatype: string

tool: D3: Cannot alloc %d bytes.


assistant: ```c
char *sal_strdup(const char *s) {
  char *ptr;
  if (s == NULL)
    return NULL;
  ptr = strdup(s);
  if (ptr == NULL) {
    printf("Cannot alloc %d bytes.\n", (int)strlen(s));
    exit(1);
  }
  return ptr;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<sal_strdup>:
  endbr64
  push   %r12
  push   %rbp
  sub    $0x8,%rsp
  test   %rdi,%rdi
  je     L14
  mov    %rdi,%rbp
  call   <strdup@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L12
L0:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L14:
  xor    %r12d,%r12d
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
L12:
  mov    %rbp,%rdi
  call   <strlen@plt>
  lea    D3(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  jmp    L0
```

tool_calls: name: parse_data
arguments: data_tag: D3
data_type: string

tool: D3: Cannot alloc %d bytes.


assistant: ```c
char *sal_strdup(const char *s) {
  char *ptr;
  if (s == NULL)
    return NULL;
  ptr = strdup(s);
  if (ptr == NULL) {
    printf("Cannot alloc %d bytes.\n", (int)strlen(s));
    exit(1);
  }
  return ptr;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<sal_strdup>:
  endbr64
  push   %r12
  push   %rbp
  sub    $0x8,%rsp
  test   %rdi,%rdi
  je     L11
  mov    %rdi,%rbp
  call   <strdup@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L8
L0:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L11:
  xor    %r12d,%r12d
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
L8:
  mov    %rbp,%rdi
  call   <strlen@plt>
  lea    D8(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  jmp    L0
```

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

tool: D8: Cannot alloc %d bytes.


assistant: ```c
char *sal_strdup(const char *s) {
  char *ptr;
  if (s == NULL)
    return NULL;
  ptr = strdup(s);
  if (ptr == NULL) {
    printf("Cannot alloc %d bytes.\n", (int)strlen(s));
    exit(1);
  }
  return ptr;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<dn_skipname>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L5
L31:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L17
  mov    -0x8(%rbp),%rax
  sub    -0x18(%rbp),%rax
  add    $0x1,%eax
  jmp    L59
L17:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0xbf,%al
  jbe    L69
  mov    -0x8(%rbp),%rax
  add    $0x1,%rax
  cmp    %rax,-0x20(%rbp)
  jbe    L14
  mov    -0x8(%rbp),%rax
  sub    -0x18(%rbp),%rax
  add    $0x2,%eax
  jmp    L59
L69:
  addq   $0x1,-0x8(%rbp)
L5:
  mov    -0x8(%rbp),%rax
  cmp    -0x20(%rbp),%rax
  jb     L31
  jmp    L7
L14:
  nop
L7:
  mov    $0xffffffff,%eax
L59:
  pop    %rbp
  ret
```

assistant: ```c
int dn_skipname(const unsigned char *s, const unsigned char *end) {
  const unsigned char *p;
  for (p = s; p < end; p++)
    if (!*p)
      return p - s + 1;
    else if (*p >= 192) {
      if (p + 1 < end)
        return p - s + 2;
      else
        break;
    }
  return -1;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<dn_skipname>:
  endbr64
  cmp    %rsi,%rdi
  jae    L3
  mov    %rdi,%rax
L31:
  movzbl (%rax),%edx
  test   %dl,%dl
  je     L10
  cmp    $0xbf,%dl
  ja     L35
  add    $0x1,%rax
  cmp    %rax,%rsi
  jne    L31
  mov    $0xffffffff,%eax
  ret
L10:
  sub    %rdi,%rax
  add    $0x1,%eax
  ret
L35:
  lea    0x1(%rax),%rdx
  sub    %rdi,%rax
  add    $0x2,%eax
  cmp    %rdx,%rsi
  mov    $0xffffffff,%edx
  cmovbe %edx,%eax
  ret
L3:
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int dn_skipname(const unsigned char *s, const unsigned char *end) {
  const unsigned char *p;
  for (p = s; p < end; p++)
    if (!*p)
      return p - s + 1;
    else if (*p >= 192) {
      if (p + 1 < end)
        return p - s + 2;
      else
        break;
    }
  return -1;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<dn_skipname>:
  endbr64
  cmp    %rsi,%rdi
  jae    L15
  mov    %rdi,%rax
  jmp    L42
  xchg   %ax,%ax
L37:
  lea    0x1(%rax),%rcx
  cmp    $0xbf,%dl
  ja     L3
  cmp    %rcx,%rsi
  je     L15
  mov    %rcx,%rax
L42:
  movzbl (%rax),%edx
  test   %dl,%dl
  jne    L37
  sub    %rdi,%rax
  add    $0x1,%eax
  ret
  nop
L15:
  mov    $0xffffffff,%eax
  ret
  cs nopw 0x0(%rax,%rax,1)
L3:
  sub    %rdi,%rax
  mov    $0xffffffff,%edx
  add    $0x2,%eax
  cmp    %rcx,%rsi
  cmovbe %edx,%eax
  ret
```

assistant: ```c
int dn_skipname(const unsigned char *s, const unsigned char *end) {
  const unsigned char *p;
  for (p = s; p < end; p++)
    if (!*p)
      return p - s + 1;
    else if (*p >= 192) {
      if (p + 1 < end)
        return p - s + 2;
      else
        break;
    }
  return -1;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<dn_skipname>:
  endbr64
  cmp    %rsi,%rdi
  jae    L21
  mov    %rdi,%rax
  jmp    L37
  xchg   %ax,%ax
L25:
  lea    0x1(%rax),%rcx
  cmp    $0xbf,%dl
  ja     L30
  cmp    %rcx,%rsi
  je     L21
  mov    %rcx,%rax
L37:
  movzbl (%rax),%edx
  test   %dl,%dl
  jne    L25
  sub    %rdi,%rax
  add    $0x1,%eax
  ret
  nop
L21:
  mov    $0xffffffff,%eax
  ret
  cs nopw 0x0(%rax,%rax,1)
L30:
  sub    %rdi,%rax
  mov    $0xffffffff,%edx
  add    $0x2,%eax
  cmp    %rcx,%rsi
  cmovbe %edx,%eax
  ret
```

assistant: ```c
int dn_skipname(const unsigned char *s, const unsigned char *end) {
  const unsigned char *p;
  for (p = s; p < end; p++)
    if (!*p)
      return p - s + 1;
    else if (*p >= 192) {
      if (p + 1 < end)
        return p - s + 2;
      else
        break;
    }
  return -1;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<line2argv>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  cmpq   $0x0,-0x28(%rbp)
  jne    L9
  mov    -0x18(%rbp),%rax
  movl   $0x0,(%rax)
  mov    -0x20(%rbp),%rax
  mov    (%rax),%rax
  movq   $0x0,(%rax)
  mov    $0x0,%eax
  jmp    L13
L9:
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  add    $0x2,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x28(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    -0x28(%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
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <l2a@plt>
  mov    -0x18(%rbp),%rdx
  mov    %eax,(%rdx)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  add    $0x1,%eax
  cltq
  shl    $0x3,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,%rdx
  mov    -0x20(%rbp),%rax
  mov    %rdx,(%rax)
  mov    -0x20(%rbp),%rax
  mov    (%rax),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <l2a@plt>
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
L13:
  leave
  ret
```

assistant: ```c
int line2argv(int *pargc, char ***pargv, char *line) {
  char *lincp;
  if (line == NULL) {
    *pargc = 0;
    **pargv = NULL;
    return (0);
  }
  lincp = (char *)malloc((strlen(line) + 2) * sizeof(char));
  strcpy(lincp, line);
  lincp[strlen(line) + 1] = '\0';
  *pargc = l2a(lincp, NULL);
  *pargv = (char **)malloc(((*pargc) + 1) * sizeof(char *));
  l2a(lincp, *pargv);
  free(lincp);
  return (*pargc);
}
```
user: What is the c source code of the assembly code below?

```asm
<line2argv>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r12
  mov    %rsi,%r13
  test   %rdx,%rdx
  je     L4
  mov    %rdx,%rbx
  mov    %rdx,%rdi
  call   <strlen@plt>
  lea    0x2(%rax),%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  mov    %rbx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    %rbx,%rdi
  call   <strlen@plt>
  cltq
  movb   $0x0,0x1(%rbp,%rax,1)
  mov    $0x0,%esi
  mov    %rbp,%rdi
  call   <l2a@plt>
  mov    %eax,(%r12)
  lea    0x8(,%rax,8),%edi
  call   <malloc@plt>
  mov    %rax,%rsi
  mov    %rax,0x0(%r13)
  mov    %rbp,%rdi
  call   <l2a@plt>
  mov    %rbp,%rdi
  call   <free@plt>
  mov    (%r12),%eax
L16:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L4:
  movl   $0x0,(%rdi)
  mov    (%rsi),%rax
  movq   $0x0,(%rax)
  mov    $0x0,%eax
  jmp    L16
```

assistant: ```c
int line2argv(int *pargc, char ***pargv, char *line) {
  char *lincp;
  if (line == NULL) {
    *pargc = 0;
    **pargv = NULL;
    return (0);
  }
  lincp = (char *)malloc((strlen(line) + 2) * sizeof(char));
  strcpy(lincp, line);
  lincp[strlen(line) + 1] = '\0';
  *pargc = l2a(lincp, NULL);
  *pargv = (char **)malloc(((*pargc) + 1) * sizeof(char *));
  l2a(lincp, *pargv);
  free(lincp);
  return (*pargc);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<line2argv>:
  endbr64
  push   %r13
  mov    %rsi,%r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  test   %rdx,%rdx
  je     L4
  mov    %rdx,%rdi
  mov    %rdx,%rbp
  call   <strlen@plt>
  lea    0x2(%rax),%edi
  call   <malloc@plt>
  mov    %rbp,%rsi
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <strcpy@plt>
  mov    %rbp,%rdi
  call   <strlen@plt>
  xor    %esi,%esi
  mov    %r12,%rdi
  cltq
  movb   $0x0,0x1(%r12,%rax,1)
  call   <l2a@plt>
  mov    %eax,(%rbx)
  lea    0x8(,%rax,8),%edi
  call   <malloc@plt>
  mov    %r12,%rdi
  mov    %rax,0x0(%r13)
  mov    %rax,%rsi
  call   <l2a@plt>
  mov    %r12,%rdi
  call   <free@plt>
  mov    (%rbx),%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  cs nopw 0x0(%rax,%rax,1)
L4:
  mov    (%rsi),%rax
  movl   $0x0,(%rdi)
  movq   $0x0,(%rax)
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int line2argv(int *pargc, char ***pargv, char *line) {
  char *lincp;
  if (line == NULL) {
    *pargc = 0;
    **pargv = NULL;
    return (0);
  }
  lincp = (char *)malloc((strlen(line) + 2) * sizeof(char));
  strcpy(lincp, line);
  lincp[strlen(line) + 1] = '\0';
  *pargc = l2a(lincp, NULL);
  *pargv = (char **)malloc(((*pargc) + 1) * sizeof(char *));
  l2a(lincp, *pargv);
  free(lincp);
  return (*pargc);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<line2argv>:
  endbr64
  push   %r13
  mov    %rsi,%r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  test   %rdx,%rdx
  je     L3
  mov    %rdx,%rdi
  mov    %rdx,%rbp
  call   <strlen@plt>
  lea    0x2(%rax),%edi
  call   <malloc@plt>
  mov    %rbp,%rsi
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <strcpy@plt>
  mov    %rbp,%rdi
  call   <strlen@plt>
  xor    %esi,%esi
  mov    %r12,%rdi
  cltq
  movb   $0x0,0x1(%r12,%rax,1)
  call   <l2a@plt>
  mov    %eax,(%rbx)
  lea    0x8(,%rax,8),%edi
  call   <malloc@plt>
  mov    %r12,%rdi
  mov    %rax,0x0(%r13)
  mov    %rax,%rsi
  call   <l2a@plt>
  mov    %r12,%rdi
  call   <free@plt>
  mov    (%rbx),%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  cs nopw 0x0(%rax,%rax,1)
L3:
  mov    (%rsi),%rax
  movl   $0x0,(%rdi)
  movq   $0x0,(%rax)
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int line2argv(int *pargc, char ***pargv, char *line) {
  char *lincp;
  if (line == NULL) {
    *pargc = 0;
    **pargv = NULL;
    return (0);
  }
  lincp = (char *)malloc((strlen(line) + 2) * sizeof(char));
  strcpy(lincp, line);
  lincp[strlen(line) + 1] = '\0';
  *pargc = l2a(lincp, NULL);
  *pargv = (char **)malloc(((*pargc) + 1) * sizeof(char *));
  l2a(lincp, *pargv);
  free(lincp);
  return (*pargc);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<frequency>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,%eax
  mov    %esi,-0x18(%rbp)
  mov    %edx,-0x1c(%rbp)
  mov    %rcx,-0x28(%rbp)
  mov    %al,-0x14(%rbp)
  movl   $0x0,-0x4(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L4
L8:
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x18(%rbp),%eax
  cltq
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    %al,-0x14(%rbp)
  sete   %al
  movzbl %al,%eax
  add    %eax,-0x4(%rbp)
  addl   $0x1,-0x8(%rbp)
L4:
  mov    -0x8(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L8
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int frequency(char allele, int site, int nsam, char **list) {
  int i, count = 0;
  for (i = 0; i < nsam; i++)
    count += (list[i][site] == allele ? 1 : 0);
  return (count);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<frequency>:
  endbr64
  test   %edx,%edx
  jle    L4
  mov    %rcx,%rax
  lea    -0x1(%rdx),%edx
  lea    0x8(%rcx,%rdx,8),%r8
  mov    $0x0,%edx
  movslq %esi,%rsi
L14:
  mov    (%rax),%rcx
  cmp    %dil,(%rcx,%rsi,1)
  sete   %cl
  movzbl %cl,%ecx
  add    %ecx,%edx
  add    $0x8,%rax
  cmp    %r8,%rax
  jne    L14
L9:
  mov    %edx,%eax
  ret
L4:
  mov    $0x0,%edx
  jmp    L9
```

assistant: ```c
int frequency(char allele, int site, int nsam, char **list) {
  int i, count = 0;
  for (i = 0; i < nsam; i++)
    count += (list[i][site] == allele ? 1 : 0);
  return (count);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<frequency>:
  endbr64
  test   %edx,%edx
  jle    L10
  lea    -0x1(%rdx),%eax
  movslq %esi,%rsi
  lea    0x8(%rcx,%rax,8),%r8
  xor    %eax,%eax
  nopl   (%rax)
L8:
  mov    (%rcx),%rdx
  cmp    %dil,(%rdx,%rsi,1)
  sete   %dl
  add    $0x8,%rcx
  movzbl %dl,%edx
  add    %edx,%eax
  cmp    %rcx,%r8
  jne    L8
  ret
  nopl   0x0(%rax)
L10:
  xor    %eax,%eax
  ret
```

assistant: ```c
int frequency(char allele, int site, int nsam, char **list) {
  int i, count = 0;
  for (i = 0; i < nsam; i++)
    count += (list[i][site] == allele ? 1 : 0);
  return (count);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<frequency>:
  endbr64
  test   %edx,%edx
  jle    L11
  lea    -0x1(%rdx),%eax
  movslq %esi,%rsi
  lea    0x8(%rcx,%rax,8),%r8
  xor    %eax,%eax
  nopl   (%rax)
L18:
  mov    (%rcx),%rdx
  cmp    %dil,(%rdx,%rsi,1)
  sete   %dl
  add    $0x8,%rcx
  movzbl %dl,%edx
  add    %edx,%eax
  cmp    %rcx,%r8
  jne    L18
  ret
  nopl   0x0(%rax)
L11:
  xor    %eax,%eax
  ret
```

assistant: ```c
int frequency(char allele, int site, int nsam, char **list) {
  int i, count = 0;
  for (i = 0; i < nsam; i++)
    count += (list[i][site] == allele ? 1 : 0);
  return (count);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<fprimes>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %rcx,-0x30(%rbp)
  mov    %r8,-0x38(%rbp)
  pxor   %xmm1,%xmm1
  cvtsi2sdl -0x14(%rbp),%xmm1
  movsd  D28(%rip),%xmm0
  divsd  %xmm1,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L0
L7:
  movsd  -0x8(%rbp),%xmm0
  movq   D15(%rip),%xmm1
  xorpd  %xmm1,%xmm0
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm2
  movsd  D28(%rip),%xmm1
  addsd  %xmm2,%xmm1
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  divsd  %xmm1,%xmm0
  movsd  %xmm0,(%rax)
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm1
  movsd  D36(%rip),%xmm0
  addsd  %xmm0,%xmm1
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  addsd  %xmm0,%xmm1
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  movsd  -0x8(%rbp),%xmm0
  divsd  %xmm1,%xmm0
  movsd  %xmm0,(%rax)
  addl   $0x1,-0xc(%rbp)
L0:
  mov    -0xc(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L7
  nop
  nop
  pop    %rbp
  ret
```

tool_calls: name: disassemble_data
arguments: label: D28
type: f64
name: disassemble_data
arguments: label: D15
type: byte[8]
name: disassemble_data
arguments: label: D36
type: f64

tool: D28: 1

tool: {"D15": [0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x80]}

tool: D36: 2

assistant: ```c
void fprimes(const int n, const double *expz, const double *expmz, double *f1prime, double *f2prime) {
  int i;
  double ninv;
  ninv = 1.0 / n;
  for (i = 0; i < n; i++) {
    f1prime[i] = -ninv / (1.0 + expz[i]);
    f2prime[i] = ninv / (2.0 + expz[i] + expmz[i]);
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<fprimes>:
  endbr64
  pxor   %xmm0,%xmm0
  cvtsi2sd %edi,%xmm0
  movsd  D34(%rip),%xmm2
  divsd  %xmm0,%xmm2
  test   %edi,%edi
  jle    L12
  mov    %edi,%edi
  mov    $0x0,%eax
  movapd %xmm2,%xmm5
  xorpd  D2(%rip),%xmm5
  movsd  D34(%rip),%xmm4
  movsd  D27(%rip),%xmm3
L3:
  movapd %xmm4,%xmm1
  addsd  (%rsi,%rax,8),%xmm1
  movapd %xmm5,%xmm0
  divsd  %xmm1,%xmm0
  movsd  %xmm0,(%rcx,%rax,8)
  movapd %xmm3,%xmm0
  addsd  (%rsi,%rax,8),%xmm0
  addsd  (%rdx,%rax,8),%xmm0
  movapd %xmm2,%xmm1
  divsd  %xmm0,%xmm1
  movsd  %xmm1,(%r8,%rax,8)
  add    $0x1,%rax
  cmp    %rdi,%rax
  jne    L3
L12:
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D34
type_infered: f64
name: retrieve_data
arguments: tag: D2
type_infered: byte[16]
name: retrieve_data
arguments: tag: D27
type_infered: f64

tool: D34: 1

tool: {"D2": [0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x80,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00]}

tool: D27: 2

assistant: ```c
void fprimes(const int n, const double *expz, const double *expmz, double *f1prime, double *f2prime) {
  int i;
  double ninv;
  ninv = 1.0 / n;
  for (i = 0; i < n; i++) {
    f1prime[i] = -ninv / (1.0 + expz[i]);
    f2prime[i] = ninv / (2.0 + expz[i] + expmz[i]);
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<fprimes>:
  endbr64
  pxor   %xmm0,%xmm0
  movsd  D4(%rip),%xmm2
  cvtsi2sd %edi,%xmm0
  movapd %xmm2,%xmm3
  divsd  %xmm0,%xmm3
  test   %edi,%edi
  jle    L6
  movapd %xmm3,%xmm5
  movsd  D2(%rip),%xmm4
  movslq %edi,%rdi
  xor    %eax,%eax
  xorpd  D25(%rip),%xmm5
  nopl   0x0(%rax)
L11:
  movsd  (%rsi,%rax,8),%xmm1
  movapd %xmm5,%xmm0
  addsd  %xmm2,%xmm1
  divsd  %xmm1,%xmm0
  movapd %xmm3,%xmm1
  movsd  %xmm0,(%rcx,%rax,8)
  movsd  (%rsi,%rax,8),%xmm0
  addsd  %xmm4,%xmm0
  addsd  (%rdx,%rax,8),%xmm0
  divsd  %xmm0,%xmm1
  movsd  %xmm1,(%r8,%rax,8)
  add    $0x1,%rax
  cmp    %rax,%rdi
  jne    L11
L6:
  ret
```

tool_calls: name: read_data
arguments: tag: D4
type_guessed: f64
name: read_data
arguments: tag: D2
type_guessed: f64
name: read_data
arguments: tag: D25
type_guessed: byte[16]

tool: D4: 1

tool: D2: 2

tool: {"D25": [0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x80,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00]}

assistant: ```c
void fprimes(const int n, const double *expz, const double *expmz, double *f1prime, double *f2prime) {
  int i;
  double ninv;
  ninv = 1.0 / n;
  for (i = 0; i < n; i++) {
    f1prime[i] = -ninv / (1.0 + expz[i]);
    f2prime[i] = ninv / (2.0 + expz[i] + expmz[i]);
  }
}
```