ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<cb_asprint>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r14
  mov    %rsi,%rbx
  mov    %edx,%ebp
  mov    %ecx,%r12d
  mov    %r8d,%r13d
  mov    $0x0,%edi
  call   <cb_sprint@plt>
  movslq %eax,%rdi
  add    $0x1,%rdi
  call   <malloc@plt>
  mov    %rax,(%r14)
  test   %rax,%rax
  je     L4
  mov    %rax,%rdi
  mov    %r13d,%r8d
  mov    %r12d,%ecx
  mov    %ebp,%edx
  mov    %rbx,%rsi
  call   <cb_sprint@plt>
L2:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L4:
  mov    $0xffffffff,%eax
  jmp    L2
```

assistant: ```c
int cb_asprint(char **ret, const char *buf, ssize_t length, ssize_t max_width, ssize_t max_lines) {
  size_t n = cb_sprint(NULL, buf, length, max_width, max_lines);
  *ret = (char *)malloc((n + 1) * sizeof(char));
  if (!*ret) {
    return -1;
  }
  return cb_sprint(*ret, buf, length, max_width, max_lines);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<cb_asprint>:
  endbr64
  push   %r14
  mov    %r8d,%r14d
  push   %r13
  mov    %ecx,%r13d
  push   %r12
  mov    %edx,%r12d
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  xor    %edi,%edi
  call   <cb_sprint@plt>
  movslq %eax,%rdi
  add    $0x1,%rdi
  call   <malloc@plt>
  mov    %rax,(%rbx)
  test   %rax,%rax
  je     L3
  pop    %rbx
  mov    %r14d,%r8d
  mov    %r13d,%ecx
  mov    %r12d,%edx
  mov    %rbp,%rsi
  mov    %rax,%rdi
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  jmp    1060 <cb_sprint@plt>
L3:
  pop    %rbx
  mov    $0xffffffff,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int cb_asprint(char **ret, const char *buf, ssize_t length, ssize_t max_width, ssize_t max_lines) {
  size_t n = cb_sprint(NULL, buf, length, max_width, max_lines);
  *ret = (char *)malloc((n + 1) * sizeof(char));
  if (!*ret) {
    return -1;
  }
  return cb_sprint(*ret, buf, length, max_width, max_lines);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<cb_asprint>:
  endbr64
  push   %r14
  mov    %r8d,%r14d
  push   %r13
  mov    %ecx,%r13d
  push   %r12
  mov    %edx,%r12d
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  xor    %edi,%edi
  call   <cb_sprint@plt>
  movslq %eax,%rdi
  add    $0x1,%rdi
  call   <malloc@plt>
  mov    %rax,(%rbx)
  test   %rax,%rax
  je     L1
  pop    %rbx
  mov    %r14d,%r8d
  mov    %r13d,%ecx
  mov    %r12d,%edx
  mov    %rbp,%rsi
  mov    %rax,%rdi
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  jmp    1060 <cb_sprint@plt>
L1:
  pop    %rbx
  mov    $0xffffffff,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int cb_asprint(char **ret, const char *buf, ssize_t length, ssize_t max_width, ssize_t max_lines) {
  size_t n = cb_sprint(NULL, buf, length, max_width, max_lines);
  *ret = (char *)malloc((n + 1) * sizeof(char));
  if (!*ret) {
    return -1;
  }
  return cb_sprint(*ret, buf, length, max_width, max_lines);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<kprintf>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xe0,%rsp
  mov    %rdi,-0xc8(%rbp)
  mov    %rsi,-0xd0(%rbp)
  mov    %rdx,-0xd8(%rbp)
  mov    %rcx,-0x98(%rbp)
  mov    %r8,-0x90(%rbp)
  mov    %r9,-0x88(%rbp)
  test   %al,%al
  je     L7
  movaps %xmm0,-0x80(%rbp)
  movaps %xmm1,-0x70(%rbp)
  movaps %xmm2,-0x60(%rbp)
  movaps %xmm3,-0x50(%rbp)
  movaps %xmm4,-0x40(%rbp)
  movaps %xmm5,-0x30(%rbp)
  movaps %xmm6,-0x20(%rbp)
  movaps %xmm7,-0x10(%rbp)
L7:
  mov    -0xd8(%rbp),%rdx
  mov    -0xb8(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <va_start@plt>
  mov    -0xb8(%rbp),%ecx
  mov    -0xd8(%rbp),%rdx
  mov    -0xd0(%rbp),%rsi
  mov    -0xc8(%rbp),%rax
  mov    %rax,%rdi
  call   <vfkprintf@plt>
  mov    %eax,-0xb4(%rbp)
  mov    -0xb8(%rbp),%eax
  mov    %eax,%edi
  call   <va_end@plt>
  mov    -0xb4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int kprintf(int (*writefn)(void *arg, const char *str, int len), void *p, const char *fmt, ...) {
  va_list args;
  va_start(args, fmt);
  int ret = vfkprintf(writefn, p, fmt, args);
  va_end(args);
  return ret;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<kprintf>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r12
  mov    %rsi,%r13
  mov    %rdx,%rbx
  mov    %rdx,%rsi
  mov    $0x0,%ebp
  mov    %ebp,%edi
  call   <va_start@plt>
  mov    %ebp,%ecx
  mov    %rbx,%rdx
  mov    %r13,%rsi
  mov    %r12,%rdi
  call   <vfkprintf@plt>
  mov    %eax,%ebx
  mov    %ebp,%edi
  call   <va_end@plt>
  mov    %ebx,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int kprintf(int (*writefn)(void *arg, const char *str, int len), void *p, const char *fmt, ...) {
  va_list args;
  va_start(args, fmt);
  int ret = vfkprintf(writefn, p, fmt, args);
  va_end(args);
  return ret;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<kprintf>:
  endbr64
  push   %r14
  mov    %rsi,%r14
  mov    %rdx,%rsi
  push   %r13
  mov    %rdi,%r13
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  xor    %ebp,%ebp
  mov    %ebp,%edi
  sub    $0x8,%rsp
  call   <va_start@plt>
  mov    %ebp,%ecx
  mov    %r12,%rdx
  mov    %r14,%rsi
  mov    %r13,%rdi
  call   <vfkprintf@plt>
  mov    %ebp,%edi
  mov    %eax,%r12d
  call   <va_end@plt>
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int kprintf(int (*writefn)(void *arg, const char *str, int len), void *p, const char *fmt, ...) {
  va_list args;
  va_start(args, fmt);
  int ret = vfkprintf(writefn, p, fmt, args);
  va_end(args);
  return ret;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<kprintf>:
  endbr64
  push   %r14
  mov    %rsi,%r14
  mov    %rdx,%rsi
  push   %r13
  mov    %rdi,%r13
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  xor    %ebp,%ebp
  mov    %ebp,%edi
  sub    $0x8,%rsp
  call   <va_start@plt>
  mov    %ebp,%ecx
  mov    %r12,%rdx
  mov    %r14,%rsi
  mov    %r13,%rdi
  call   <vfkprintf@plt>
  mov    %ebp,%edi
  mov    %eax,%r12d
  call   <va_end@plt>
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int kprintf(int (*writefn)(void *arg, const char *str, int len), void *p, const char *fmt, ...) {
  va_list args;
  va_start(args, fmt);
  int ret = vfkprintf(writefn, p, fmt, args);
  va_end(args);
  return ret;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<cdflib_dt1>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %rdi,0x18(%rsp)
  mov    %rdx,%r14
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  call   <cdflib_dinvnr@plt>
  mov    %eax,%edi
  call   <fabs@plt>
  movapd %xmm0,%xmm5
  movsd  %xmm0,0x10(%rsp)
  mulsd  %xmm0,%xmm0
  movsd  %xmm0,0x20(%rsp)
  lea    D33(%rip),%rbx
  lea    D22(%rip),%r13
  lea    D10(%rip),%r12
  lea    0xa0(%rbx),%r15
  movsd  %xmm5,(%rsp)
  movsd  D24(%rip),%xmm6
  movsd  %xmm6,0x8(%rsp)
  lea    0x20(%rsp),%rbp
L3:
  mov    %rbp,%rdx
  mov    %r13,%rsi
  mov    %rbx,%rdi
  call   <cdflib_eval_pol@plt>
  movsd  0x8(%rsp),%xmm2
  mulsd  (%r14),%xmm2
  movsd  %xmm2,0x8(%rsp)
  mulsd  0x10(%rsp),%xmm0
  movapd %xmm2,%xmm1
  mulsd  (%r12),%xmm1
  divsd  %xmm1,%xmm0
  addsd  (%rsp),%xmm0
  movsd  %xmm0,(%rsp)
  add    $0x28,%rbx
  add    $0x4,%r13
  add    $0x8,%r12
  cmp    %r15,%rbx
  jne    L3
  mov    0x18(%rsp),%rax
  movsd  (%rax),%xmm0
  comisd D29(%rip),%xmm0
  jae    L10
  movsd  (%rsp),%xmm7
  xorpd  D3(%rip),%xmm7
  movsd  %xmm7,(%rsp)
L10:
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L24
  movsd  (%rsp),%xmm0
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L24:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D33
data_type: f64
name: disassemble_data
arguments: tag: D22
data_type: string
name: disassemble_data
arguments: tag: D10
data_type: f64
name: disassemble_data
arguments: tag: D24
data_type: f64
name: disassemble_data
arguments: tag: D29
data_type: f64[2]
name: disassemble_data
arguments: tag: D3
data_type: f32[4]

tool: D33: 1

tool: D22: 

tool: D10: 4

tool: D24: 1

tool: {"D29": [0.5,-0.0]}

tool: {"D3": [0.0,-0.0,0.0,0.0]}

assistant: ```c
double cdflib_dt1(double *p, double *q, double *df) {
  static const double coef[4][5] = {{1.0e0, 1.0e0, 0.0e0, 0.0e0, 0.0e0}, {3.0e0, 16.0e0, 5.0e0, 0.0e0, 0.0e0}, {-15.0e0, 17.0e0, 19.0e0, 3.0e0, 0.0e0}, {-945.0e0, -1920.0e0, 1482.0e0, 776.0e0, 79.0e0}};
  static const double denom[4] = {4.0e0, 96.0e0, 384.0e0, 92160.0e0};
  static const int ideg[4] = {2, 3, 4, 5};
  double dt1, denpow, sum, term, x, xp, xx;
  int i;
  x = fabs(cdflib_dinvnr(p, q));
  xx = x * x;
  sum = x;
  denpow = 1.0e0;
  for (i = 0; i < 4; i++) {
    term = cdflib_eval_pol(&coef[i][0], &ideg[i], &xx) * x;
    denpow *= *df;
    sum += (term / (denpow * denom[i]));
  }
  if (!(*p >= 0.5e0)) {
    goto S20;
  }
  xp = sum;
  goto S30;
S20:
  xp = -sum;
S30:
  dt1 = xp;
  return dt1;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<cdflib_dt1>:
  endbr64
  push   %r15
  lea    D39(%rip),%r15
  push   %r14
  lea    D15(%rip),%r14
  push   %r13
  lea    0xa0(%r14),%r13
  push   %r12
  push   %rbp
  lea    D40(%rip),%rbp
  push   %rbx
  mov    %rdx,%rbx
  sub    $0x38,%rsp
  mov    %rdi,0x18(%rsp)
  lea    0x20(%rsp),%r12
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  call   <cdflib_dinvnr@plt>
  mov    %eax,%edi
  call   <fabs@plt>
  movsd  D33(%rip),%xmm1
  movapd %xmm0,%xmm2
  movsd  %xmm0,0x10(%rsp)
  mulsd  %xmm0,%xmm0
  movsd  %xmm0,0x20(%rsp)
L2:
  mov    %rbp,%rsi
  mov    %r14,%rdi
  mov    %r12,%rdx
  movsd  %xmm2,0x8(%rsp)
  movsd  %xmm1,(%rsp)
  add    $0x28,%r14
  add    $0x4,%rbp
  add    $0x8,%r15
  call   <cdflib_eval_pol@plt>
  movsd  (%rsp),%xmm1
  mulsd  (%rbx),%xmm1
  movsd  -0x8(%r15),%xmm3
  mulsd  0x10(%rsp),%xmm0
  movsd  0x8(%rsp),%xmm2
  mulsd  %xmm1,%xmm3
  divsd  %xmm3,%xmm0
  addsd  %xmm0,%xmm2
  cmp    %r13,%r14
  jne    L2
  mov    0x18(%rsp),%rax
  movsd  (%rax),%xmm0
  comisd D1(%rip),%xmm0
  jae    L3
  xorpd  D9(%rip),%xmm2
L3:
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x38,%rsp
  movapd %xmm2,%xmm0
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D39
type_guessed: f64
name: disassemble_data
arguments: tag: D15
type_guessed: f64
name: disassemble_data
arguments: tag: D40
type_guessed: string
name: disassemble_data
arguments: tag: D33
type_guessed: f64
name: disassemble_data
arguments: tag: D1
type_guessed: f64[2]
name: disassemble_data
arguments: tag: D9
type_guessed: f32[4]

tool: D39: 4

tool: D15: 1

tool: D40: 

tool: D33: 1

tool: {"D1": [0.5,-0.0]}

tool: {"D9": [0.0,-0.0,0.0,0.0]}

assistant: ```c
double cdflib_dt1(double *p, double *q, double *df) {
  static const double coef[4][5] = {{1.0e0, 1.0e0, 0.0e0, 0.0e0, 0.0e0}, {3.0e0, 16.0e0, 5.0e0, 0.0e0, 0.0e0}, {-15.0e0, 17.0e0, 19.0e0, 3.0e0, 0.0e0}, {-945.0e0, -1920.0e0, 1482.0e0, 776.0e0, 79.0e0}};
  static const double denom[4] = {4.0e0, 96.0e0, 384.0e0, 92160.0e0};
  static const int ideg[4] = {2, 3, 4, 5};
  double dt1, denpow, sum, term, x, xp, xx;
  int i;
  x = fabs(cdflib_dinvnr(p, q));
  xx = x * x;
  sum = x;
  denpow = 1.0e0;
  for (i = 0; i < 4; i++) {
    term = cdflib_eval_pol(&coef[i][0], &ideg[i], &xx) * x;
    denpow *= *df;
    sum += (term / (denpow * denom[i]));
  }
  if (!(*p >= 0.5e0)) {
    goto S20;
  }
  xp = sum;
  goto S30;
S20:
  xp = -sum;
S30:
  dt1 = xp;
  return dt1;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<cdflib_dt1>:
  endbr64
  push   %r15
  lea    D59(%rip),%r15
  push   %r14
  lea    D30(%rip),%r14
  push   %r13
  lea    0xa0(%r14),%r13
  push   %r12
  push   %rbp
  lea    D23(%rip),%rbp
  push   %rbx
  mov    %rdx,%rbx
  sub    $0x38,%rsp
  mov    %rdi,0x18(%rsp)
  lea    0x20(%rsp),%r12
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  call   <cdflib_dinvnr@plt>
  mov    %eax,%edi
  call   <fabs@plt>
  movsd  D38(%rip),%xmm1
  movapd %xmm0,%xmm2
  movsd  %xmm0,0x10(%rsp)
  mulsd  %xmm0,%xmm0
  movsd  %xmm0,0x20(%rsp)
L9:
  mov    %rbp,%rsi
  mov    %r14,%rdi
  mov    %r12,%rdx
  movsd  %xmm2,0x8(%rsp)
  movsd  %xmm1,(%rsp)
  add    $0x28,%r14
  add    $0x4,%rbp
  add    $0x8,%r15
  call   <cdflib_eval_pol@plt>
  movsd  (%rsp),%xmm1
  mulsd  (%rbx),%xmm1
  movsd  -0x8(%r15),%xmm3
  mulsd  0x10(%rsp),%xmm0
  movsd  0x8(%rsp),%xmm2
  mulsd  %xmm1,%xmm3
  divsd  %xmm3,%xmm0
  addsd  %xmm0,%xmm2
  cmp    %r13,%r14
  jne    L9
  mov    0x18(%rsp),%rax
  movsd  (%rax),%xmm0
  comisd D22(%rip),%xmm0
  jae    L25
  xorpd  D32(%rip),%xmm2
L25:
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x38,%rsp
  movapd %xmm2,%xmm0
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D59
type: f64
name: extract_symbolic_data
arguments: tag: D30
type: f64
name: extract_symbolic_data
arguments: tag: D23
type: string
name: extract_symbolic_data
arguments: tag: D38
type: f64
name: extract_symbolic_data
arguments: tag: D22
type: f64[2]
name: extract_symbolic_data
arguments: tag: D32
type: f32[4]

tool: D59: 4

tool: D30: 1

tool: D23: 

tool: D38: 1

tool: {"D22": [0.5,-0.0]}

tool: {"D32": [0.0,-0.0,0.0,0.0]}

assistant: ```c
double cdflib_dt1(double *p, double *q, double *df) {
  static const double coef[4][5] = {{1.0e0, 1.0e0, 0.0e0, 0.0e0, 0.0e0}, {3.0e0, 16.0e0, 5.0e0, 0.0e0, 0.0e0}, {-15.0e0, 17.0e0, 19.0e0, 3.0e0, 0.0e0}, {-945.0e0, -1920.0e0, 1482.0e0, 776.0e0, 79.0e0}};
  static const double denom[4] = {4.0e0, 96.0e0, 384.0e0, 92160.0e0};
  static const int ideg[4] = {2, 3, 4, 5};
  double dt1, denpow, sum, term, x, xp, xx;
  int i;
  x = fabs(cdflib_dinvnr(p, q));
  xx = x * x;
  sum = x;
  denpow = 1.0e0;
  for (i = 0; i < 4; i++) {
    term = cdflib_eval_pol(&coef[i][0], &ideg[i], &xx) * x;
    denpow *= *df;
    sum += (term / (denpow * denom[i]));
  }
  if (!(*p >= 0.5e0)) {
    goto S20;
  }
  xp = sum;
  goto S30;
S20:
  xp = -sum;
S30:
  dt1 = xp;
  return dt1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ft_is_sort>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x0,-0x8(%rbp)
  mov    -0x18(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  movsbl %al,%edx
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    -0x20(%rbp),%rcx
  mov    %edx,%esi
  mov    %eax,%edi
  call   *%rcx
  mov    %eax,-0x4(%rbp)
  jmp    L36
L2:
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x1(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%edx
  mov    -0x8(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    -0x20(%rbp),%rcx
  mov    %edx,%esi
  mov    %eax,%edi
  call   *%rcx
  cmp    %eax,-0x4(%rbp)
  je     L34
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x1(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%edx
  mov    -0x8(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    -0x20(%rbp),%rcx
  mov    %edx,%esi
  mov    %eax,%edi
  call   *%rcx
  test   %eax,%eax
  je     L34
  mov    $0x0,%eax
  jmp    L45
L34:
  addl   $0x1,-0x8(%rbp)
L36:
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x1(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L2
  mov    $0x1,%eax
L45:
  leave
  ret
```

assistant: ```c
int ft_is_sort(char *tab, int (*f)(int, int)) {
  int i;
  int order;
  i = 0;
  order = f(tab[0], tab[1]);
  while (tab[i + 1]) {
    if (f(tab[i], tab[i + 1]) != order && f(tab[i], tab[i + 1]) != 0)
      return (0);
    i++;
  }
  return (1);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<ft_is_sort>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r13
  mov    %rsi,%rbp
  movsbl 0x1(%rdi),%esi
  movsbl (%rdi),%edi
  call   *%rbp
  mov    %eax,%r12d
  lea    0x1(%r13),%rbx
  movzbl 0x1(%r13),%esi
  test   %sil,%sil
  jne    L33
  mov    $0x1,%eax
  jmp    L32
L8:
  add    $0x1,%rbx
  movzbl (%rbx),%esi
  test   %sil,%sil
  je     L49
L33:
  movsbl %sil,%esi
  movsbl -0x1(%rbx),%edi
  call   *%rbp
  cmp    %r12d,%eax
  je     L8
  movsbl (%rbx),%esi
  movsbl -0x1(%rbx),%edi
  call   *%rbp
  test   %eax,%eax
  je     L8
  mov    $0x0,%eax
  jmp    L32
L49:
  mov    $0x1,%eax
L32:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int ft_is_sort(char *tab, int (*f)(int, int)) {
  int i;
  int order;
  i = 0;
  order = f(tab[0], tab[1]);
  while (tab[i + 1]) {
    if (f(tab[i], tab[i + 1]) != order && f(tab[i], tab[i + 1]) != 0)
      return (0);
    i++;
  }
  return (1);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<ft_is_sort>:
  endbr64
  push   %r13
  mov    %rdi,%r13
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  movsbl 0x1(%rdi),%esi
  movsbl (%rdi),%edi
  call   *%rbp
  movsbl 0x1(%r13),%esi
  test   %sil,%sil
  je     L20
  mov    %eax,%r12d
  lea    0x1(%r13),%rbx
  xchg   %ax,%ax
L9:
  movsbl -0x1(%rbx),%edi
  call   *%rbp
  cmp    %r12d,%eax
  je     L35
  movsbl (%rbx),%esi
  movsbl -0x1(%rbx),%edi
  call   *%rbp
  test   %eax,%eax
  jne    L3
L35:
  movsbl 0x1(%rbx),%esi
  add    $0x1,%rbx
  test   %sil,%sil
  jne    L9
L20:
  add    $0x8,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   (%rax)
L3:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int ft_is_sort(char *tab, int (*f)(int, int)) {
  int i;
  int order;
  i = 0;
  order = f(tab[0], tab[1]);
  while (tab[i + 1]) {
    if (f(tab[i], tab[i + 1]) != order && f(tab[i], tab[i + 1]) != 0)
      return (0);
    i++;
  }
  return (1);
}
```
user: What is the c source code of the assembly code below?

```asm
<ft_is_sort>:
  endbr64
  push   %r13
  mov    %rdi,%r13
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  movsbl 0x1(%rdi),%esi
  movsbl (%rdi),%edi
  call   *%rbp
  movsbl 0x1(%r13),%esi
  test   %sil,%sil
  je     L19
  mov    %eax,%r12d
  lea    0x1(%r13),%rbx
  xchg   %ax,%ax
L15:
  movsbl -0x1(%rbx),%edi
  call   *%rbp
  cmp    %r12d,%eax
  je     L3
  movsbl (%rbx),%esi
  movsbl -0x1(%rbx),%edi
  call   *%rbp
  test   %eax,%eax
  jne    L34
L3:
  movsbl 0x1(%rbx),%esi
  add    $0x1,%rbx
  test   %sil,%sil
  jne    L15
L19:
  add    $0x8,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   (%rax)
L34:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int ft_is_sort(char *tab, int (*f)(int, int)) {
  int i;
  int order;
  i = 0;
  order = f(tab[0], tab[1]);
  while (tab[i + 1]) {
    if (f(tab[i], tab[i + 1]) != order && f(tab[i], tab[i + 1]) != 0)
      return (0);
    i++;
  }
  return (1);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x20(%rbp),%rax
  mov    %rax,%rsi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  movl   $0x1,-0x1c(%rbp)
  jmp    L33
L34:
  lea    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D16(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    -0x18(%rbp),%rax
  shl    $0x2,%rax
  test   %rax,%rax
  js     L56
  pxor   %xmm1,%xmm1
  cvtsi2sd %rax,%xmm1
  jmp    L15
L56:
  mov    %rax,%rdx
  shr    %rdx
  and    $0x1,%eax
  or     %rax,%rdx
  pxor   %xmm0,%xmm0
  cvtsi2sd %rdx,%xmm0
  movapd %xmm0,%xmm1
  addsd  %xmm0,%xmm1
L15:
  mov    -0x18(%rbp),%rax
  test   %rax,%rax
  js     L3
  pxor   %xmm0,%xmm0
  cvtsi2sd %rax,%xmm0
  jmp    L45
L3:
  mov    %rax,%rdx
  shr    %rdx
  and    $0x1,%eax
  or     %rax,%rdx
  pxor   %xmm0,%xmm0
  cvtsi2sd %rdx,%xmm0
  addsd  %xmm0,%xmm0
L45:
  mulsd  %xmm0,%xmm1
  movsd  D30(%rip),%xmm0
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x10(%rbp)
  mov    -0x10(%rbp),%rdx
  mov    -0x1c(%rbp),%eax
  movq   %rdx,%xmm0
  mov    %eax,%esi
  lea    D36(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x1c(%rbp)
L33:
  mov    -0x20(%rbp),%eax
  cmp    %eax,-0x1c(%rbp)
  jle    L34
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L10
  call   <__stack_chk_fail@plt>
L10:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: label: D5
type_infered: string
name: disassemble_data
arguments: label: D16
type_infered: string
name: disassemble_data
arguments: label: D30
type_infered: f64
name: disassemble_data
arguments: label: D36
type_infered: string

tool: D5: %d

tool: D16: %llu

tool: D30: 0.25

tool: D36: Case %d: %.2f


assistant: ```c
int main() {
  int i, j, c, len, num, temp, t;
  unsigned long long int r;
  double res;
  scanf("%d", &t);
  i = 1;
  while (i <= t) {
    scanf("%llu", &r);
    res = 4 * r * (double)r + 0.25;
    printf("Case %d: %.2f\n", i, res);
    i++;
  }
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rsi
  lea    D3(%rip),%rdi
  call   <scanf@plt>
  cmpl   $0x0,0xc(%rsp)
  jle    L69
  mov    $0x1,%ebx
  lea    0x10(%rsp),%r13
  lea    D24(%rip),%r12
  lea    D13(%rip),%rbp
  jmp    L26
L64:
  shr    %rdx
  pxor   %xmm0,%xmm0
  cvtsi2sd %rdx,%xmm0
  addsd  %xmm0,%xmm0
  jmp    L39
L48:
  mov    %rax,%rdx
  shr    %rdx
  and    $0x1,%eax
  or     %rax,%rdx
  pxor   %xmm1,%xmm1
  cvtsi2sd %rdx,%xmm1
  addsd  %xmm1,%xmm1
L56:
  mulsd  %xmm1,%xmm0
  addsd  D28(%rip),%xmm0
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  add    $0x1,%ebx
  cmp    %ebx,0xc(%rsp)
  jl     L69
L26:
  mov    %r13,%rsi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    0x10(%rsp),%rax
  mov    %rax,%rdx
  shl    $0x2,%rdx
  js     L64
  pxor   %xmm0,%xmm0
  cvtsi2sd %rdx,%xmm0
L39:
  test   %rax,%rax
  js     L48
  pxor   %xmm1,%xmm1
  cvtsi2sd %rax,%xmm1
  jmp    L56
L69:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L22
  mov    $0x0,%eax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L22:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D3
datatype: string
name: parse_data
arguments: data_tag: D24
datatype: string
name: parse_data
arguments: data_tag: D13
datatype: string
name: parse_data
arguments: data_tag: D28
datatype: f64

tool: D3: %d

tool: D24: %llu

tool: D13: Case %d: %.2f


tool: D28: 0.25

assistant: ```c
int main() {
  int i, j, c, len, num, temp, t;
  unsigned long long int r;
  double res;
  scanf("%d", &t);
  i = 1;
  while (i <= t) {
    scanf("%llu", &r);
    res = 4 * r * (double)r + 0.25;
    printf("Case %d: %.2f\n", i, res);
    i++;
  }
  return 0;
}
```