ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<sin>:
  endbr64
  sub    $0x18,%rsp
  movsd  %xmm0,0x8(%rsp)
  divsd  D85(%rip),%xmm0
  call   <floor@plt>
  pxor   %xmm3,%xmm3
  movsd  D91(%rip),%xmm2
  movsd  0x8(%rsp),%xmm1
  add    %eax,%eax
  movsd  D33(%rip),%xmm6
  cvtsi2sd %eax,%xmm3
  mulsd  %xmm2,%xmm3
  subsd  %xmm3,%xmm1
  comisd %xmm2,%xmm1
  movapd %xmm1,%xmm0
  jbe    L0
  subsd  %xmm2,%xmm0
  movsd  D58(%rip),%xmm6
L0:
  comisd D19(%rip),%xmm0
  jbe    L14
  subsd  %xmm0,%xmm2
  movapd %xmm2,%xmm0
L14:
  movapd %xmm0,%xmm2
  movapd %xmm0,%xmm1
  mulsd  %xmm0,%xmm2
  mulsd  %xmm2,%xmm1
  movapd %xmm2,%xmm5
  movapd %xmm2,%xmm4
  movapd %xmm2,%xmm3
  movapd %xmm2,%xmm7
  mulsd  %xmm1,%xmm5
  divsd  D37(%rip),%xmm1
  movapd %xmm1,%xmm8
  movapd %xmm0,%xmm1
  subsd  %xmm8,%xmm1
  mulsd  %xmm5,%xmm4
  divsd  D97(%rip),%xmm5
  addsd  %xmm5,%xmm1
  movapd %xmm1,%xmm0
  mulsd  %xmm4,%xmm3
  divsd  D12(%rip),%xmm4
  subsd  %xmm4,%xmm0
  mulsd  %xmm3,%xmm7
  divsd  D11(%rip),%xmm3
  addsd  %xmm3,%xmm0
  mulsd  %xmm7,%xmm2
  movapd %xmm7,%xmm1
  divsd  D64(%rip),%xmm1
  subsd  %xmm1,%xmm0
  divsd  D53(%rip),%xmm2
  add    $0x18,%rsp
  addsd  %xmm2,%xmm0
  mulsd  %xmm6,%xmm0
  ret
```

tool_calls: name: retrieve_data
arguments: label: D85
type_guessed: byte[8]
name: retrieve_data
arguments: label: D91
type_guessed: f64
name: retrieve_data
arguments: label: D33
type_guessed: f64
name: retrieve_data
arguments: label: D58
type_guessed: byte[8]
name: retrieve_data
arguments: label: D19
type_guessed: f64[2]
name: retrieve_data
arguments: label: D37
type_guessed: byte[8]
name: retrieve_data
arguments: label: D97
type_guessed: byte[8]
name: retrieve_data
arguments: label: D12
type_guessed: byte[8]
name: retrieve_data
arguments: label: D11
type_guessed: byte[8]
name: retrieve_data
arguments: label: D64
type_guessed: byte[8]
name: retrieve_data
arguments: label: D53
type_guessed: byte[8]

tool: {"D85": [0x18,0x2d,0x44,0x54,0xfb,0x21,0x19,0x40]}

tool: D91: 3.141592653589793

tool: D33: 1

tool: {"D58": [0x00,0x00,0x00,0x00,0x00,0x00,0xf0,0xbf]}

tool: {"D19": [1.5707963267948966,6.0]}

tool: {"D37": [0x00,0x00,0x00,0x00,0x00,0x00,0x18,0x40]}

tool: {"D97": [0x00,0x00,0x00,0x00,0x00,0x00,0x5e,0x40]}

tool: {"D12": [0x00,0x00,0x00,0x00,0x00,0xb0,0xb3,0x40]}

tool: {"D11": [0x00,0x00,0x00,0x00,0x00,0x26,0x16,0x41]}

tool: {"D64": [0x00,0x00,0x00,0x00,0xa8,0x08,0x83,0x41]}

tool: {"D53": [0x00,0x00,0x00,0xc0,0x8c,0x32,0xf7,0x41]}

assistant: ```c
double sin(double x) {
  double sign;
  double x2, x3, x5, x7, x9, x11, x13;
  sign = 1.0;
  x = x - floor(x / (2 * 3.1415926535897932384626433832795)) * 2 * 3.1415926535897932384626433832795;
  if (x > 3.1415926535897932384626433832795) {
    x = x - 3.1415926535897932384626433832795;
    sign = -1;
  }
  if (x > 1.5707963267948966192313216916398) {
    x = 3.1415926535897932384626433832795 - x;
  }
  x2 = x * x;
  x3 = x * x2;
  x5 = x3 * x2;
  x7 = x5 * x2;
  x9 = x7 * x2;
  x11 = x9 * x2;
  x13 = x11 * x2;
  return sign * (x - x3 / 6 + x5 / 120 - x7 / 5040 + x9 / 362880 - x11 / 39916800 + x13 / 6227020800);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<read_until>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %edx,%eax
  mov    %rcx,-0x38(%rbp)
  mov    %al,-0x28(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0x0,-0x10(%rbp)
  jmp    L3
L15:
  mov    -0x30(%rbp),%rax
  sub    -0x10(%rbp),%rax
  mov    %rax,%rdx
  mov    -0x38(%rbp),%rcx
  mov    -0x10(%rbp),%rax
  lea    (%rcx,%rax,1),%rsi
  lea    -0x18(%rbp),%rcx
  mov    -0x24(%rbp),%eax
  mov    %eax,%edi
  call   <receive@plt>
  test   %rax,%rax
  jne    L55
  mov    -0x18(%rbp),%rax
  test   %rax,%rax
  jne    L4
L55:
  mov    $0xffffffff,%eax
  jmp    L5
L4:
  mov    -0x18(%rbp),%rdx
  movsbl -0x28(%rbp),%eax
  mov    -0x38(%rbp),%rsi
  mov    -0x10(%rbp),%rcx
  add    %rsi,%rcx
  mov    %eax,%esi
  mov    %rcx,%rdi
  call   <memchr@plt>
  test   %rax,%rax
  je     L49
  mov    -0x18(%rbp),%rdx
  movsbl -0x28(%rbp),%eax
  mov    -0x38(%rbp),%rsi
  mov    -0x10(%rbp),%rcx
  add    %rsi,%rcx
  mov    %eax,%esi
  mov    %rcx,%rdi
  call   <memchr@plt>
  movb   $0x0,(%rax)
  mov    -0x10(%rbp),%rax
  mov    %eax,%edx
  mov    -0x18(%rbp),%rax
  add    %edx,%eax
  jmp    L5
L49:
  mov    -0x18(%rbp),%rax
  add    %rax,-0x10(%rbp)
L3:
  mov    -0x10(%rbp),%rax
  cmp    -0x30(%rbp),%rax
  jb     L15
  mov    $0xffffffff,%eax
L5:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L73
  call   <__stack_chk_fail@plt>
L73:
  leave
  ret
```

assistant: ```c
int read_until(int fd, size_t n, char terminator, char *buf) {
  size_t cgc_read = 0;
  while (cgc_read < n) {
    size_t tmp_read;
    if (receive(fd, buf + cgc_read, n - cgc_read, &tmp_read) != 0 || tmp_read == 0)
      return -1;
    if (memchr(buf + cgc_read, terminator, tmp_read) != NULL) {
      *((char *)memchr(buf + cgc_read, terminator, tmp_read)) = '\0';
      ;
      return cgc_read + tmp_read;
    } else {
      cgc_read += tmp_read;
    }
  }
  return -1;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<read_until>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %edi,0x4(%rsp)
  mov    %rcx,0x8(%rsp)
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  test   %rsi,%rsi
  je     L92
  mov    %rsi,%r12
  mov    %edx,%r14d
  mov    $0x0,%ebx
  lea    0x10(%rsp),%r15
L89:
  mov    0x8(%rsp),%rax
  lea    (%rax,%rbx,1),%rbp
  mov    %r12,%rdx
  sub    %rbx,%rdx
  mov    %r15,%rcx
  mov    %rbp,%rsi
  mov    0x4(%rsp),%edi
  call   <receive@plt>
  test   %rax,%rax
  jne    L14
  mov    0x10(%rsp),%rdx
  test   %rdx,%rdx
  je     L19
  movsbl %r14b,%r13d
  mov    %r13d,%esi
  mov    %rbp,%rdi
  call   <memchr@plt>
  test   %rax,%rax
  jne    L23
  mov    %rbx,%rax
  add    0x10(%rsp),%rax
  mov    %rax,%rbx
  cmp    %rax,%r12
  ja     L89
  mov    $0xffffffff,%eax
  jmp    L39
L23:
  mov    0x10(%rsp),%rdx
  mov    %r13d,%esi
  mov    %rbp,%rdi
  call   <memchr@plt>
  movb   $0x0,(%rax)
  mov    %ebx,%eax
  add    0x10(%rsp),%eax
  jmp    L39
L14:
  mov    $0xffffffff,%eax
L39:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L88
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L19:
  mov    $0xffffffff,%eax
  jmp    L39
L92:
  mov    $0xffffffff,%eax
  jmp    L39
L88:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int read_until(int fd, size_t n, char terminator, char *buf) {
  size_t cgc_read = 0;
  while (cgc_read < n) {
    size_t tmp_read;
    if (receive(fd, buf + cgc_read, n - cgc_read, &tmp_read) != 0 || tmp_read == 0)
      return -1;
    if (memchr(buf + cgc_read, terminator, tmp_read) != NULL) {
      *((char *)memchr(buf + cgc_read, terminator, tmp_read)) = '\0';
      ;
      return cgc_read + tmp_read;
    } else {
      cgc_read += tmp_read;
    }
  }
  return -1;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<read_until>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rcx,0x8(%rsp)
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  test   %rsi,%rsi
  je     L37
  mov    %edi,%r14d
  mov    %rsi,%rbp
  xor    %ebx,%ebx
  movsbl %dl,%r12d
  lea    0x10(%rsp),%r15
  jmp    L18
  nop
L8:
  mov    0x10(%rsp),%rdx
  test   %rdx,%rdx
  je     L37
  mov    %r12d,%esi
  mov    %r13,%rdi
  call   <memchr@plt>
  test   %rax,%rax
  jne    L79
  add    0x10(%rsp),%rbx
  cmp    %rbx,%rbp
  jbe    L37
L18:
  mov    0x8(%rsp),%rax
  mov    %rbp,%rdx
  mov    %r15,%rcx
  mov    %r14d,%edi
  sub    %rbx,%rdx
  lea    (%rax,%rbx,1),%r13
  mov    %r13,%rsi
  call   <receive@plt>
  test   %rax,%rax
  je     L8
L37:
  mov    $0xffffffff,%eax
L24:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L55
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L79:
  mov    0x10(%rsp),%rdx
  mov    %r12d,%esi
  mov    %r13,%rdi
  call   <memchr@plt>
  movb   $0x0,(%rax)
  mov    0x10(%rsp),%eax
  add    %ebx,%eax
  jmp    L24
L55:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int read_until(int fd, size_t n, char terminator, char *buf) {
  size_t cgc_read = 0;
  while (cgc_read < n) {
    size_t tmp_read;
    if (receive(fd, buf + cgc_read, n - cgc_read, &tmp_read) != 0 || tmp_read == 0)
      return -1;
    if (memchr(buf + cgc_read, terminator, tmp_read) != NULL) {
      *((char *)memchr(buf + cgc_read, terminator, tmp_read)) = '\0';
      ;
      return cgc_read + tmp_read;
    } else {
      cgc_read += tmp_read;
    }
  }
  return -1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<read_until>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rcx,0x8(%rsp)
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  test   %rsi,%rsi
  je     L14
  mov    %edi,%r14d
  mov    %rsi,%rbp
  xor    %ebx,%ebx
  movsbl %dl,%r12d
  lea    0x10(%rsp),%r15
  jmp    L42
  nop
L6:
  mov    0x10(%rsp),%rdx
  test   %rdx,%rdx
  je     L14
  mov    %r12d,%esi
  mov    %r13,%rdi
  call   <memchr@plt>
  test   %rax,%rax
  jne    L2
  add    0x10(%rsp),%rbx
  cmp    %rbx,%rbp
  jbe    L14
L42:
  mov    0x8(%rsp),%rax
  mov    %rbp,%rdx
  mov    %r15,%rcx
  mov    %r14d,%edi
  sub    %rbx,%rdx
  lea    (%rax,%rbx,1),%r13
  mov    %r13,%rsi
  call   <receive@plt>
  test   %rax,%rax
  je     L6
L14:
  mov    $0xffffffff,%eax
L8:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L31
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L2:
  mov    0x10(%rsp),%rdx
  mov    %r12d,%esi
  mov    %r13,%rdi
  call   <memchr@plt>
  movb   $0x0,(%rax)
  mov    0x10(%rsp),%eax
  add    %ebx,%eax
  jmp    L8
L31:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int read_until(int fd, size_t n, char terminator, char *buf) {
  size_t cgc_read = 0;
  while (cgc_read < n) {
    size_t tmp_read;
    if (receive(fd, buf + cgc_read, n - cgc_read, &tmp_read) != 0 || tmp_read == 0)
      return -1;
    if (memchr(buf + cgc_read, terminator, tmp_read) != NULL) {
      *((char *)memchr(buf + cgc_read, terminator, tmp_read)) = '\0';
      ;
      return cgc_read + tmp_read;
    } else {
      cgc_read += tmp_read;
    }
  }
  return -1;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<strrindex>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  movl   $0x0,-0x18(%rbp)
  movl   $0xffffffff,-0x14(%rbp)
  jmp    L32
L64:
  mov    -0x18(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x30(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L60
L61:
  addq   $0x1,-0x10(%rbp)
  addq   $0x1,-0x8(%rbp)
L60:
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L56
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L56
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%edx
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    %al,%dl
  je     L61
L56:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L12
  mov    -0x18(%rbp),%eax
  mov    %eax,-0x14(%rbp)
L12:
  addl   $0x1,-0x18(%rbp)
L32:
  mov    -0x18(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L64
  mov    -0x14(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int strrindex(char *s, char *t) {
  int i = 0;
  int res = -1;
  while (s[i]) {
    char *tmp = &s[i];
    char *t2 = t;
    while (*tmp && *t2 && *tmp == *t2)
      tmp++, t2++;
    if (*t2 == 0)
      res = i;
    i++;
  }
  return res;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<strrindex>:
  endbr64
  mov    $0xffffffff,%r10d
  mov    $0x0,%r9d
  cmpb   $0x0,(%rdi)
  jne    L6
L31:
  mov    %r10d,%eax
  ret
L52:
  cmpb   $0x0,(%rax)
  cmove  %r9d,%r10d
  add    $0x1,%r9d
  add    $0x1,%rdi
  cmpb   $0x0,(%rdi)
  je     L31
L6:
  movzbl (%rdi),%edx
  mov    %rsi,%rax
  mov    %rdi,%r8
  test   %dl,%dl
  je     L52
L24:
  movzbl (%rax),%ecx
  cmp    %dl,%cl
  jne    L52
  test   %cl,%cl
  je     L52
  add    $0x1,%r8
  add    $0x1,%rax
  movzbl (%r8),%edx
  test   %dl,%dl
  jne    L24
  jmp    L52
```

assistant: ```c
int strrindex(char *s, char *t) {
  int i = 0;
  int res = -1;
  while (s[i]) {
    char *tmp = &s[i];
    char *t2 = t;
    while (*tmp && *t2 && *tmp == *t2)
      tmp++, t2++;
    if (*t2 == 0)
      res = i;
    i++;
  }
  return res;
}
```
user: What is the c source code of the assembly code below:

```asm
<strrindex>:
  endbr64
  movzbl (%rdi),%edx
  mov    $0xffffffff,%r10d
  xor    %r9d,%r9d
  test   %dl,%dl
  je     L41
  nopl   0x0(%rax)
L37:
  mov    %rsi,%rcx
  mov    %rdi,%r8
  jmp    L16
L47:
  test   %al,%al
  je     L9
  movzbl 0x1(%r8),%edx
  add    $0x1,%r8
  lea    0x1(%rcx),%rax
  test   %dl,%dl
  je     L10
  mov    %rax,%rcx
L16:
  movzbl (%rcx),%eax
  cmp    %dl,%al
  je     L47
L9:
  test   %al,%al
  cmove  %r9d,%r10d
  movzbl 0x1(%rdi),%edx
  add    $0x1,%rdi
  add    $0x1,%r9d
  test   %dl,%dl
  jne    L37
L41:
  mov    %r10d,%eax
  ret
  nopl   0x0(%rax)
L10:
  movzbl 0x1(%rcx),%eax
  jmp    L9
```

assistant: ```c
int strrindex(char *s, char *t) {
  int i = 0;
  int res = -1;
  while (s[i]) {
    char *tmp = &s[i];
    char *t2 = t;
    while (*tmp && *t2 && *tmp == *t2)
      tmp++, t2++;
    if (*t2 == 0)
      res = i;
    i++;
  }
  return res;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<strrindex>:
  endbr64
  xor    %r9d,%r9d
  cmpb   $0x0,(%rdi)
  mov    $0xffffffff,%r10d
  je     L2
  nopw   0x0(%rax,%rax,1)
L85:
  movzbl (%rdi),%edx
  mov    %rsi,%rcx
  mov    %rdi,%r8
  test   %dl,%dl
  jne    L7
  jmp    L44
  nopw   0x0(%rax,%rax,1)
L66:
  test   %al,%al
  je     L35
  movzbl 0x1(%r8),%edx
  add    $0x1,%r8
  lea    0x1(%rcx),%rax
  test   %dl,%dl
  je     L58
  mov    %rax,%rcx
L7:
  movzbl (%rcx),%eax
  cmp    %dl,%al
  je     L66
L35:
  test   %al,%al
  cmove  %r9d,%r10d
  add    $0x1,%rdi
  add    $0x1,%r9d
  cmpb   $0x0,(%rdi)
  jne    L85
L2:
  mov    %r10d,%eax
  ret
  cs nopw 0x0(%rax,%rax,1)
L58:
  movzbl 0x1(%rcx),%eax
  jmp    L35
L44:
  movzbl (%rsi),%eax
  jmp    L35
```

assistant: ```c
int strrindex(char *s, char *t) {
  int i = 0;
  int res = -1;
  while (s[i]) {
    char *tmp = &s[i];
    char *t2 = t;
    while (*tmp && *t2 && *tmp == *t2)
      tmp++, t2++;
    if (*t2 == 0)
      res = i;
    i++;
  }
  return res;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0xea0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
L17:
  movl   $0x0,-0x3e98(%rbp)
  lea    -0x3e9c(%rbp),%rax
  mov    %rax,%rsi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x3e9c(%rbp),%eax
  test   %eax,%eax
  je     L60
L36:
  mov    -0x3e9c(%rbp),%edx
  mov    -0x3e98(%rbp),%eax
  cltq
  mov    %edx,-0x3e90(%rbp,%rax,4)
  addl   $0x1,-0x3e98(%rbp)
  lea    -0x3e9c(%rbp),%rax
  mov    %rax,%rsi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x3e9c(%rbp),%eax
  test   %eax,%eax
  jne    L36
  movl   $0x1,-0x3e94(%rbp)
  jmp    L22
L56:
  mov    -0x3e94(%rbp),%eax
  cltq
  mov    -0x3e90(%rbp,%rax,4),%eax
  mov    -0x3e90(%rbp),%esi
  mov    -0x3e94(%rbp),%edx
  lea    -0x1(%rdx),%ecx
  sub    %esi,%eax
  mov    %eax,%edx
  movslq %ecx,%rax
  mov    %edx,-0x1f50(%rbp,%rax,4)
  mov    -0x3e94(%rbp),%eax
  sub    $0x1,%eax
  cltq
  mov    -0x1f50(%rbp,%rax,4),%eax
  test   %eax,%eax
  jns    L54
  mov    -0x3e94(%rbp),%eax
  sub    $0x1,%eax
  cltq
  mov    -0x1f50(%rbp,%rax,4),%eax
  mov    -0x3e94(%rbp),%edx
  lea    -0x1(%rdx),%ecx
  neg    %eax
  mov    %eax,%edx
  movslq %ecx,%rax
  mov    %edx,-0x1f50(%rbp,%rax,4)
L54:
  addl   $0x1,-0x3e94(%rbp)
L22:
  mov    -0x3e94(%rbp),%eax
  cmp    -0x3e98(%rbp),%eax
  jl     L56
  mov    -0x3e98(%rbp),%eax
  lea    -0x1(%rax),%edx
  lea    -0x1f50(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <gcdOfArr@plt>
  mov    %eax,%esi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L17
L60:
  nop
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L33
  call   <__stack_chk_fail@plt>
L33:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: tag: D7
type_guessed: string
name: unpack_data
arguments: tag: D9
type_guessed: string

tool: D7: %d

tool: D9: %d


assistant: ```c
int main(void) {
  int arr[2000], diffArr[2000], num, mSize, i;
  while (1) {
    mSize = 0;
    scanf("%d", &num);
    if (num == 0)
      break;
    do {
      arr[mSize] = num;
      mSize++;
      scanf("%d", &num);
    } while (num != 0);
    for (i = 1; i < mSize; i++) {
      diffArr[i - 1] = arr[i] - arr[0];
      if (diffArr[i - 1] < 0)
        diffArr[i - 1] = -diffArr[i - 1];
    }
    printf("%d\n", gcdOfArr(diffArr, mSize - 1));
  }
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0xea8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x3e98(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%r13
  lea    D9(%rip),%r12
  lea    D4(%rip),%r14
  jmp    L85
L1:
  mov    %ecx,%esi
  sub    %edx,%esi
  mov    %esi,0x1f50(%rsp,%rax,4)
L3:
  add    $0x1,%rax
  cmp    %eax,%ebp
  jle    L26
L36:
  mov    0x14(%rsp,%rax,4),%edx
  mov    %edx,%esi
  sub    %ecx,%esi
  js     L1
  mov    %esi,0x1f50(%rsp,%rax,4)
  jmp    L3
L26:
  lea    0x1f50(%rsp),%rdi
  mov    %ebp,%esi
  call   <gcdOfArr@plt>
  mov    %eax,%esi
  mov    %r14,%rdi
  call   <printf@plt>
L85:
  mov    %r13,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  cmpl   $0x0,0xc(%rsp)
  je     L28
  lea    0x10(%rsp),%r15
  mov    $0x0,%ebx
L86:
  mov    0xc(%rsp),%eax
  mov    %eax,(%r15)
  mov    %ebx,%ebp
  add    $0x1,%ebx
  mov    %r13,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  add    $0x4,%r15
  cmpl   $0x0,0xc(%rsp)
  jne    L86
  cmp    $0x1,%ebx
  jle    L26
  mov    0x10(%rsp),%ecx
  mov    $0x0,%eax
  jmp    L36
L28:
  mov    0x3e98(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L22
  mov    $0x0,%eax
  add    $0x3ea8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L22:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D9
data_type: string
name: retrieve_data
arguments: tag: D4
data_type: string

tool: D9: %d

tool: D4: %d


assistant: ```c
int main(void) {
  int arr[2000], diffArr[2000], num, mSize, i;
  while (1) {
    mSize = 0;
    scanf("%d", &num);
    if (num == 0)
      break;
    do {
      arr[mSize] = num;
      mSize++;
      scanf("%d", &num);
    } while (num != 0);
    for (i = 1; i < mSize; i++) {
      diffArr[i - 1] = arr[i] - arr[0];
      if (diffArr[i - 1] < 0)
        diffArr[i - 1] = -diffArr[i - 1];
    }
    printf("%d\n", gcdOfArr(diffArr, mSize - 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
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0xea8,%rsp
  lea    D18(%rip),%rbx
  lea    D0(%rip),%r12
  mov    %fs:0x28,%rax
  mov    %rax,0x3e98(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rbp
  nop
L59:
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%eax
  test   %eax,%eax
  je     L16
  xor    %r13d,%r13d
  lea    0x10(%rsp),%r15
  nopl   0x0(%rax,%rax,1)
L34:
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  mov    %eax,(%r15,%r13,4)
  mov    %r13d,%r14d
  call   <scanf@plt>
  mov    0xc(%rsp),%eax
  add    $0x1,%r13
  test   %eax,%eax
  jne    L34
  test   %r14d,%r14d
  je     L23
  mov    0x10(%rsp),%ecx
  xor    %edx,%edx
  lea    0x1f50(%rsp),%rdi
  nopl   0x0(%rax)
L11:
  mov    0x4(%r15,%rdx,4),%eax
  mov    %ecx,%esi
  sub    %eax,%esi
  sub    %ecx,%eax
  cmovs  %esi,%eax
  mov    %eax,(%rdi,%rdx,4)
  add    $0x1,%rdx
  cmp    %edx,%r14d
  jg     L11
L24:
  mov    %r14d,%esi
  call   <gcdOfArr@plt>
  mov    %r12,%rdi
  mov    %eax,%esi
  call   <printf@plt>
  jmp    L59
  nopl   0x0(%rax)
L16:
  mov    0x3e98(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L48
  add    $0x3ea8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L23:
  lea    0x1f50(%rsp),%rdi
  jmp    L24
L48:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D18
data_type: string
name: parse_data
arguments: label: D0
data_type: string

tool: D18: %d

tool: D0: %d


assistant: ```c
int main(void) {
  int arr[2000], diffArr[2000], num, mSize, i;
  while (1) {
    mSize = 0;
    scanf("%d", &num);
    if (num == 0)
      break;
    do {
      arr[mSize] = num;
      mSize++;
      scanf("%d", &num);
    } while (num != 0);
    for (i = 1; i < mSize; i++) {
      diffArr[i - 1] = arr[i] - arr[0];
      if (diffArr[i - 1] < 0)
        diffArr[i - 1] = -diffArr[i - 1];
    }
    printf("%d\n", gcdOfArr(diffArr, mSize - 1));
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<decryption>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  jmp    L6
L2:
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <rot13@plt>
L6:
  mov    -0x8(%rbp),%rax
  mov    %rax,%rsi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  test   %rax,%rax
  jne    L2
  nop
  nop
  leave
  ret
```

tool_calls: name: unpack_data
arguments: tag: D9
type_guessed: string

tool: D9: %1000s

assistant: ```c
void decryption(char dec[1000]) {
  while (scanf("%1000s", dec)) {
    rot13(dec);
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<decryption>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  lea    D8(%rip),%rbp
  jmp    L12
L13:
  mov    %rbx,%rdi
  call   <rot13@plt>
L12:
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  test   %rax,%rax
  jne    L13
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: read_data
arguments: data_tag: D8
data_type: string

tool: D8: %1000s

assistant: ```c
void decryption(char dec[1000]) {
  while (scanf("%1000s", dec)) {
    rot13(dec);
  }
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<decryption>:
  endbr64
  push   %rbp
  lea    D1(%rip),%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  jmp    L10
  cs nopw 0x0(%rax,%rax,1)
L15:
  mov    %rbx,%rdi
  call   <rot13@plt>
L10:
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  test   %rax,%rax
  jne    L15
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D1
data_type: string

tool: D1: %1000s

assistant: ```c
void decryption(char dec[1000]) {
  while (scanf("%1000s", dec)) {
    rot13(dec);
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<decryption>:
  endbr64
  push   %rbp
  lea    D5(%rip),%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  jmp    L2
  cs nopw 0x0(%rax,%rax,1)
L1:
  mov    %rbx,%rdi
  call   <rot13@plt>
L2:
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  test   %rax,%rax
  jne    L1
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D5
type_infered: string

tool: D5: %1000s

assistant: ```c
void decryption(char dec[1000]) {
  while (scanf("%1000s", dec)) {
    rot13(dec);
  }
}
```