ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  push   %r14
  xor    %eax,%eax
  push   %r13
  push   %r12
  push   %rbp
  sub    $0x8,%rsp
  call   <method@plt>
  xor    %eax,%eax
  call   <getter@plt>
  mov    $0xa,%edi
  mov    %eax,%r14d
  call   <function@plt>
  mov    $0x7,%edi
  mov    %eax,%r13d
  call   <ten@plt>
  mov    %eax,%edi
  call   <ten@plt>
  mov    %eax,%edi
  call   <square@plt>
  mov    $0x2,%esi
  mov    %eax,%edi
  call   <mul@plt>
  mov    $0x3c,%r9d
  mov    $0x32,%r8d
  mov    $0x28,%ecx
  mov    $0x1e,%edx
  mov    $0x14,%esi
  mov    $0xa,%edi
  mov    %eax,%r12d
  call   <sum@plt>
  mov    $0xa,%esi
  mov    $0xa,%edi
  mov    %eax,%ebp
  call   <assertEq@plt>
  mov    %r14d,%edi
  mov    $0xc8,%esi
  call   <assertEq@plt>
  mov    %r13d,%edi
  mov    $0xc8,%esi
  call   <assertEq@plt>
  mov    %r12d,%edi
  mov    $0xef420,%esi
  call   <assertEq@plt>
  mov    %ebp,%edi
  mov    $0xd2,%esi
  call   <assertEq@plt>
  xor    %eax,%eax
  call   <endTest@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int main() {
  int a = 10;
  method();
  int b = getter();
  int c = function(a);
  int d = mul(square(ten(ten(7))), 2);
  int e = sum(10, 20, 30, 40, 50, 60);
  assertEq(a, 10);
  assertEq(b, 200);
  assertEq(c, 200);
  assertEq(d, 980000);
  assertEq(e, 210);
  endTest();
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<simplehttp_async_finish>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <simplehttp_async_check@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L7
  mov    -0x8(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <simplehttp_request_finish@plt>
L7:
  nop
  leave
  ret
```

assistant: ```c
void simplehttp_async_finish(struct evhttp_request *req) {
  struct simplehttp_request *entry;
  do {
    ;
  } while (0);
  if ((entry = simplehttp_async_check(req))) {
    do {
      ;
    } while (0);
    simplehttp_request_finish(req, entry);
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<simplehttp_async_finish>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <simplehttp_async_check@plt>
  test   %rax,%rax
  je     L0
  mov    %rax,%rsi
  mov    %rbx,%rdi
  call   <simplehttp_request_finish@plt>
L0:
  pop    %rbx
  ret
```

assistant: ```c
void simplehttp_async_finish(struct evhttp_request *req) {
  struct simplehttp_request *entry;
  do {
    ;
  } while (0);
  if ((entry = simplehttp_async_check(req))) {
    do {
      ;
    } while (0);
    simplehttp_request_finish(req, entry);
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<simplehttp_async_finish>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  call   <simplehttp_async_check@plt>
  test   %rax,%rax
  je     L6
  mov    %rbp,%rdi
  mov    %rax,%rsi
  pop    %rbp
  jmp    1060 <simplehttp_request_finish@plt>
  xchg   %ax,%ax
L6:
  pop    %rbp
  ret
```

assistant: ```c
void simplehttp_async_finish(struct evhttp_request *req) {
  struct simplehttp_request *entry;
  do {
    ;
  } while (0);
  if ((entry = simplehttp_async_check(req))) {
    do {
      ;
    } while (0);
    simplehttp_request_finish(req, entry);
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<simplehttp_async_finish>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  call   <simplehttp_async_check@plt>
  test   %rax,%rax
  je     L7
  mov    %rbp,%rdi
  mov    %rax,%rsi
  pop    %rbp
  jmp    1060 <simplehttp_request_finish@plt>
  xchg   %ax,%ax
L7:
  pop    %rbp
  ret
```

assistant: ```c
void simplehttp_async_finish(struct evhttp_request *req) {
  struct simplehttp_request *entry;
  do {
    ;
  } while (0);
  if ((entry = simplehttp_async_check(req))) {
    do {
      ;
    } while (0);
    simplehttp_request_finish(req, entry);
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  lea    0x28(%rsp),%rbp
  mov    %rbx,%r12
  lea    D14(%rip),%r13
L41:
  mov    %r12,%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
  add    $0x4,%r12
  cmp    %rbp,%r12
  jne    L41
  mov    (%rsp),%r12d
  lea    0x4(%rsp),%rax
L79:
  mov    (%rax),%edx
  cmp    %edx,%r12d
  cmovl  %edx,%r12d
  add    $0x4,%rax
  cmp    %rbp,%rax
  jne    L79
  mov    %rbx,%rax
  mov    $0x0,%r13d
L83:
  add    (%rax),%r13d
  add    $0x4,%rax
  cmp    %rbp,%rax
  jne    L83
  movslq %r13d,%r14
  imul   $0x66666667,%r14,%r14
  sar    $0x22,%r14
  mov    %r13d,%eax
  sar    $0x1f,%eax
  sub    %eax,%r14d
  mov    $0xa,%edi
  lea    0x4(%rbx),%r8
  jmp    L21
L76:
  add    $0x4,%rax
  cmp    %rsi,%rax
  je     L113
L93:
  mov    (%rax),%edx
  mov    0x4(%rax),%ecx
  cmp    %ecx,%edx
  jge    L76
  mov    %ecx,(%rax)
  mov    %edx,0x4(%rax)
  jmp    L76
L30:
  sub    $0x1,%edi
  jne    L21
  lea    D6(%rip),%r15
L84:
  mov    (%rbx),%esi
  mov    %r15,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x4,%rbx
  cmp    %rbp,%rbx
  jne    L84
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    %r14d,%ecx
  mov    %r13d,%edx
  mov    %r12d,%esi
  lea    D23(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L82
  mov    $0x0,%eax
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L113:
  sub    $0x1,%edi
L21:
  cmp    $0x1,%edi
  jle    L30
  lea    -0x2(%rdi),%eax
  lea    (%r8,%rax,4),%rsi
  mov    %rbx,%rax
  jmp    L93
L82:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D14
type_infered: string
name: read_data
arguments: data_label: D6
type_infered: string
name: read_data
arguments: data_label: D23
type_infered: string

tool: D14: %d

tool: D6: %d 

tool: D23: max=%d,sum=%d,net=%d


assistant: ```c
int main() {
  int i;
  int str[10];
  for (i = 0; i < 10; i++) {
    scanf("%d", &str[i]);
  }
  int max = str[0];
  for (i = 1; i < 10; i++) {
    if (max < str[i]) {
      max = str[i];
    }
  }
  int sum = 0, net;
  for (i = 0; i < 10; i++) {
    sum += str[i];
  }
  net = sum / 10;
  int j, temp;
  for (j = 0; j < 10; j++) {
    for (i = 1; i < 10 - j; i++) {
      if (str[i - 1] < str[i]) {
        temp = str[i - 1];
        str[i - 1] = str[i];
        str[i] = temp;
      }
    }
  }
  for (i = 0; i < 10; i++) {
    printf("%d ", str[i]);
  }
  printf("\n");
  printf("max=%d,sum=%d,net=%d\n", max, sum, net);
  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
  lea    D3(%rip),%r12
  push   %rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r14
  lea    0x28(%rsp),%rbx
  mov    %r14,%rbp
  nopl   0x0(%rax)
L47:
  mov    %rbp,%rsi
  mov    %r12,%rdi
  add    $0x4,%rbp
  call   <scanf@plt>
  cmp    %rbx,%rbp
  jne    L47
  mov    (%rsp),%esi
  lea    0x4(%rsp),%r15
  mov    %r15,%rax
  mov    %esi,%ebp
  nopl   0x0(%rax)
L22:
  mov    (%rax),%edx
  cmp    %edx,%ebp
  cmovl  %edx,%ebp
  add    $0x4,%rax
  cmp    %rbx,%rax
  jne    L22
  mov    %r15,%rax
  mov    %esi,%edx
  xor    %r12d,%r12d
  jmp    L110
  nopw   0x0(%rax,%rax,1)
L17:
  mov    (%rax),%edx
  add    $0x4,%rax
L110:
  add    %edx,%r12d
  cmp    %rbx,%rax
  jne    L17
  movslq %r12d,%r13
  mov    %r12d,%eax
  mov    $0xa,%edi
  add    $0x4,%r14
  imul   $0x66666667,%r13,%r13
  sar    $0x1f,%eax
  sar    $0x22,%r13
  sub    %eax,%r13d
  cmp    $0x1,%edi
  je     L85
  nopl   0x0(%rax,%rax,1)
L118:
  lea    -0x2(%rdi),%eax
  lea    (%r14,%rax,4),%rcx
  mov    %r15,%rax
  jmp    L89
  nopl   0x0(%rax)
L78:
  mov    (%rax),%esi
  add    $0x4,%rax
L89:
  mov    (%rax),%edx
  cmp    %esi,%edx
  jle    L29
  mov    %edx,-0x4(%rax)
  mov    %esi,(%rax)
L29:
  cmp    %rcx,%rax
  jne    L78
  sub    $0x1,%edi
  mov    (%rsp),%esi
  cmp    $0x1,%edi
  jne    L118
L85:
  lea    D17(%rip),%r14
  jmp    L21
  nopw   0x0(%rax,%rax,1)
L93:
  mov    (%r15),%esi
  add    $0x4,%r15
L21:
  xor    %eax,%eax
  mov    %r14,%rdi
  call   <printf@plt>
  cmp    %rbx,%r15
  jne    L93
  mov    $0xa,%edi
  call   <putchar@plt>
  xor    %eax,%eax
  mov    %r13d,%ecx
  mov    %r12d,%edx
  mov    %ebp,%esi
  lea    D8(%rip),%rdi
  call   <printf@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L114
  add    $0x38,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L114:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D3
type: string
name: unpack_data
arguments: tag: D17
type: string
name: unpack_data
arguments: tag: D8
type: string

tool: D3: %d

tool: D17: %d 

tool: D8: max=%d,sum=%d,net=%d


assistant: ```c
int main() {
  int i;
  int str[10];
  for (i = 0; i < 10; i++) {
    scanf("%d", &str[i]);
  }
  int max = str[0];
  for (i = 1; i < 10; i++) {
    if (max < str[i]) {
      max = str[i];
    }
  }
  int sum = 0, net;
  for (i = 0; i < 10; i++) {
    sum += str[i];
  }
  net = sum / 10;
  int j, temp;
  for (j = 0; j < 10; j++) {
    for (i = 1; i < 10 - j; i++) {
      if (str[i - 1] < str[i]) {
        temp = str[i - 1];
        str[i - 1] = str[i];
        str[i] = temp;
      }
    }
  }
  for (i = 0; i < 10; i++) {
    printf("%d ", str[i]);
  }
  printf("\n");
  printf("max=%d,sum=%d,net=%d\n", max, sum, net);
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<solve_LatLonAlt>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  movsd  %xmm0,0x8(%rsp)
  movsd  %xmm1,0x18(%rsp)
  movq   %xmm2,%r14
  mov    %rdi,%rbp
  mov    %rsi,%r12
  mov    %rdx,%rbx
  movapd %xmm0,%xmm7
  mulsd  %xmm0,%xmm7
  movapd %xmm7,%xmm0
  movapd %xmm1,%xmm2
  mulsd  %xmm1,%xmm2
  addsd  %xmm2,%xmm0
  pxor   %xmm1,%xmm1
  ucomisd %xmm0,%xmm1
  ja     L42
  sqrtsd %xmm0,%xmm0
  movsd  %xmm0,0x10(%rsp)
L20:
  movsd  0x8(%rsp),%xmm2
  addsd  0x10(%rsp),%xmm2
  movapd %xmm2,%xmm1
  movsd  0x18(%rsp),%xmm0
  call   <atan2@plt>
  addsd  %xmm0,%xmm0
  movsd  %xmm0,(%r12)
  movsd  0x10(%rsp),%xmm1
  mulsd  D27(%rip),%xmm1
  movq   %r14,%xmm0
  divsd  %xmm1,%xmm0
  call   <atan@plt>
  movsd  %xmm0,0x0(%rbp)
  movq   $0x0,(%rbx)
  mov    $0x64,%r12d
L35:
  mov    (%rbx),%r13
  movsd  0x0(%rbp),%xmm0
  call   <sin@plt>
  mov    %eax,%edi
  mov    $0x2,%esi
  call   <pow@plt>
  mulsd  D25(%rip),%xmm0
  movsd  D54(%rip),%xmm1
  subsd  %xmm0,%xmm1
  pxor   %xmm5,%xmm5
  ucomisd %xmm1,%xmm5
  ja     L43
  sqrtsd %xmm1,%xmm1
L2:
  movsd  D38(%rip),%xmm3
  divsd  %xmm1,%xmm3
  movsd  %xmm3,0x8(%rsp)
  movsd  0x0(%rbp),%xmm0
  call   <cos@plt>
  movapd %xmm0,%xmm1
  movsd  0x10(%rsp),%xmm4
  movapd %xmm4,%xmm0
  divsd  %xmm1,%xmm0
  movsd  0x8(%rsp),%xmm3
  subsd  %xmm3,%xmm0
  movsd  %xmm0,(%rbx)
  movapd %xmm3,%xmm1
  mulsd  D25(%rip),%xmm1
  addsd  %xmm3,%xmm0
  divsd  %xmm0,%xmm1
  movsd  D54(%rip),%xmm0
  subsd  %xmm1,%xmm0
  mulsd  %xmm4,%xmm0
  movq   %r14,%xmm1
  divsd  %xmm0,%xmm1
  movapd %xmm1,%xmm0
  call   <atan@plt>
  movsd  %xmm0,0x0(%rbp)
  movsd  (%rbx),%xmm0
  movq   %r13,%xmm6
  subsd  %xmm6,%xmm0
  call   <fabs@plt>
  test   %eax,%eax
  jle    L15
  sub    $0x1,%r12d
  jne    L35
L15:
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L42:
  call   <sqrt@plt>
  movsd  %xmm0,0x10(%rsp)
  jmp    L20
L43:
  movapd %xmm1,%xmm0
  call   <sqrt@plt>
  movapd %xmm0,%xmm1
  jmp    L2
```

tool_calls: name: read_data
arguments: data_tag: D27
type: byte[8]
name: read_data
arguments: data_tag: D25
type: f64
name: read_data
arguments: data_tag: D54
type: f64
name: read_data
arguments: data_tag: D38
type: f64

tool: {"D27": [0xd6,0x02,0x1c,0xde,0x28,0xc9,0xef,0x3f]}

tool: D25: 0.00669437999014132

tool: D54: 1

tool: D38: 6378137

assistant: ```c
void solve_LatLonAlt(double x_n, double y_n, double z_n, double *lat, double *lon, double *alt) {
  int iterations = 100;
  const double a = (6378137.0);
  const double e2 = (0.00669437999014132);
  const double p = sqrt(x_n * x_n + y_n * y_n);
  *lon = 2.0 * atan2(y_n, x_n + p);
  *lat = atan(z_n / (p * (1.0 - e2)));
  *alt = 0.0;
  while (iterations > 0) {
    double tmp = *alt;
    double N = a / sqrt(1.0 - e2 * pow(sin(*lat), 2));
    *alt = p / cos(*lat) - N;
    *lat = atan(z_n / (p * (1.0 - e2 * N / (N + *alt))));
    if (fabs(*alt - tmp) < 1e-3)
      break;
    iterations--;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<solve_LatLonAlt>:
  endbr64
  movapd %xmm0,%xmm4
  movapd %xmm1,%xmm3
  push   %r14
  pxor   %xmm7,%xmm7
  mulsd  %xmm0,%xmm0
  push   %r12
  movq   %xmm2,%r14
  mov    %rsi,%r12
  mulsd  %xmm1,%xmm1
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x28,%rsp
  addsd  %xmm1,%xmm0
  ucomisd %xmm0,%xmm7
  ja     L36
  sqrtsd %xmm0,%xmm0
  movsd  %xmm0,0x18(%rsp)
L10:
  addsd  0x18(%rsp),%xmm4
  movapd %xmm3,%xmm0
  movapd %xmm4,%xmm1
  call   <atan2@plt>
  movsd  0x18(%rsp),%xmm1
  mulsd  D41(%rip),%xmm1
  addsd  %xmm0,%xmm0
  movsd  %xmm0,(%r12)
  movq   %r14,%xmm0
  mov    $0x64,%r12d
  divsd  %xmm1,%xmm0
  call   <atan@plt>
  movq   $0x0,0x10(%rsp)
  movsd  %xmm0,(%rbx)
  movq   $0x0,0x0(%rbp)
  jmp    L68
  xchg   %ax,%ax
L7:
  sub    $0x1,%r12d
  je     L9
  movsd  0x0(%rbp),%xmm3
  movsd  %xmm3,0x10(%rsp)
L68:
  movsd  (%rbx),%xmm0
  call   <sin@plt>
  mov    $0x2,%esi
  mov    %eax,%edi
  call   <pow@plt>
  mulsd  D2(%rip),%xmm0
  pxor   %xmm6,%xmm6
  movsd  D19(%rip),%xmm1
  subsd  %xmm0,%xmm1
  ucomisd %xmm1,%xmm6
  ja     L42
  sqrtsd %xmm1,%xmm1
L19:
  movsd  D51(%rip),%xmm2
  movsd  (%rbx),%xmm0
  divsd  %xmm1,%xmm2
  movsd  %xmm2,0x8(%rsp)
  call   <cos@plt>
  movsd  0x18(%rsp),%xmm5
  movsd  0x8(%rsp),%xmm2
  movapd %xmm0,%xmm1
  movapd %xmm5,%xmm0
  divsd  %xmm1,%xmm0
  movsd  D2(%rip),%xmm1
  mulsd  %xmm2,%xmm1
  subsd  %xmm2,%xmm0
  movsd  %xmm0,0x0(%rbp)
  addsd  %xmm2,%xmm0
  divsd  %xmm0,%xmm1
  movsd  D19(%rip),%xmm0
  subsd  %xmm1,%xmm0
  movq   %r14,%xmm1
  mulsd  %xmm5,%xmm0
  divsd  %xmm0,%xmm1
  movapd %xmm1,%xmm0
  call   <atan@plt>
  movsd  %xmm0,(%rbx)
  movsd  0x0(%rbp),%xmm0
  subsd  0x10(%rsp),%xmm0
  call   <fabs@plt>
  test   %eax,%eax
  jg     L7
L9:
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r14
  ret
L42:
  movapd %xmm1,%xmm0
  call   <sqrt@plt>
  movapd %xmm0,%xmm1
  jmp    L19
L36:
  movsd  %xmm3,0x10(%rsp)
  movsd  %xmm4,0x8(%rsp)
  call   <sqrt@plt>
  movsd  0x10(%rsp),%xmm3
  movsd  0x8(%rsp),%xmm4
  movsd  %xmm0,0x18(%rsp)
  jmp    L10
```

tool_calls: name: unpack_data
arguments: label: D41
type_infered: byte[8]
name: unpack_data
arguments: label: D2
type_infered: f64
name: unpack_data
arguments: label: D19
type_infered: f64
name: unpack_data
arguments: label: D51
type_infered: f64

tool: {"D41": [0xd6,0x02,0x1c,0xde,0x28,0xc9,0xef,0x3f]}

tool: D2: 0.00669437999014132

tool: D19: 1

tool: D51: 6378137

assistant: ```c
void solve_LatLonAlt(double x_n, double y_n, double z_n, double *lat, double *lon, double *alt) {
  int iterations = 100;
  const double a = (6378137.0);
  const double e2 = (0.00669437999014132);
  const double p = sqrt(x_n * x_n + y_n * y_n);
  *lon = 2.0 * atan2(y_n, x_n + p);
  *lat = atan(z_n / (p * (1.0 - e2)));
  *alt = 0.0;
  while (iterations > 0) {
    double tmp = *alt;
    double N = a / sqrt(1.0 - e2 * pow(sin(*lat), 2));
    *alt = p / cos(*lat) - N;
    *lat = atan(z_n / (p * (1.0 - e2 * N / (N + *alt))));
    if (fabs(*alt - tmp) < 1e-3)
      break;
    iterations--;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<solve_LatLonAlt>:
  endbr64
  movapd %xmm0,%xmm4
  movapd %xmm1,%xmm3
  push   %r14
  pxor   %xmm7,%xmm7
  mulsd  %xmm0,%xmm0
  push   %r12
  movq   %xmm2,%r14
  mov    %rsi,%r12
  mulsd  %xmm1,%xmm1
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x28,%rsp
  addsd  %xmm1,%xmm0
  ucomisd %xmm0,%xmm7
  ja     L52
  sqrtsd %xmm0,%xmm0
  movsd  %xmm0,0x18(%rsp)
L31:
  addsd  0x18(%rsp),%xmm4
  movapd %xmm3,%xmm0
  movapd %xmm4,%xmm1
  call   <atan2@plt>
  movsd  0x18(%rsp),%xmm1
  mulsd  D19(%rip),%xmm1
  addsd  %xmm0,%xmm0
  movsd  %xmm0,(%r12)
  movq   %r14,%xmm0
  mov    $0x64,%r12d
  divsd  %xmm1,%xmm0
  call   <atan@plt>
  movq   $0x0,0x10(%rsp)
  movsd  %xmm0,(%rbx)
  movq   $0x0,0x0(%rbp)
  jmp    L4
  xchg   %ax,%ax
L64:
  sub    $0x1,%r12d
  je     L26
  movsd  0x0(%rbp),%xmm3
  movsd  %xmm3,0x10(%rsp)
L4:
  movsd  (%rbx),%xmm0
  call   <sin@plt>
  mov    $0x2,%esi
  mov    %eax,%edi
  call   <pow@plt>
  mulsd  D21(%rip),%xmm0
  pxor   %xmm6,%xmm6
  movsd  D18(%rip),%xmm1
  subsd  %xmm0,%xmm1
  ucomisd %xmm1,%xmm6
  ja     L23
  sqrtsd %xmm1,%xmm1
L9:
  movsd  D31(%rip),%xmm2
  movsd  (%rbx),%xmm0
  divsd  %xmm1,%xmm2
  movsd  %xmm2,0x8(%rsp)
  call   <cos@plt>
  movsd  0x18(%rsp),%xmm5
  movsd  0x8(%rsp),%xmm2
  movapd %xmm0,%xmm1
  movapd %xmm5,%xmm0
  divsd  %xmm1,%xmm0
  movsd  D21(%rip),%xmm1
  mulsd  %xmm2,%xmm1
  subsd  %xmm2,%xmm0
  movsd  %xmm0,0x0(%rbp)
  addsd  %xmm2,%xmm0
  divsd  %xmm0,%xmm1
  movsd  D18(%rip),%xmm0
  subsd  %xmm1,%xmm0
  movq   %r14,%xmm1
  mulsd  %xmm5,%xmm0
  divsd  %xmm0,%xmm1
  movapd %xmm1,%xmm0
  call   <atan@plt>
  movsd  %xmm0,(%rbx)
  movsd  0x0(%rbp),%xmm0
  subsd  0x10(%rsp),%xmm0
  call   <fabs@plt>
  test   %eax,%eax
  jg     L64
L26:
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r14
  ret
L23:
  movapd %xmm1,%xmm0
  call   <sqrt@plt>
  movapd %xmm0,%xmm1
  jmp    L9
L52:
  movsd  %xmm3,0x10(%rsp)
  movsd  %xmm4,0x8(%rsp)
  call   <sqrt@plt>
  movsd  0x10(%rsp),%xmm3
  movsd  0x8(%rsp),%xmm4
  movsd  %xmm0,0x18(%rsp)
  jmp    L31
```

tool_calls: name: parse_data
arguments: data_label: D19
type: byte[8]
name: parse_data
arguments: data_label: D21
type: f64
name: parse_data
arguments: data_label: D18
type: f64
name: parse_data
arguments: data_label: D31
type: f64

tool: {"D19": [0xd6,0x02,0x1c,0xde,0x28,0xc9,0xef,0x3f]}

tool: D21: 0.00669437999014132

tool: D18: 1

tool: D31: 6378137

assistant: ```c
void solve_LatLonAlt(double x_n, double y_n, double z_n, double *lat, double *lon, double *alt) {
  int iterations = 100;
  const double a = (6378137.0);
  const double e2 = (0.00669437999014132);
  const double p = sqrt(x_n * x_n + y_n * y_n);
  *lon = 2.0 * atan2(y_n, x_n + p);
  *lat = atan(z_n / (p * (1.0 - e2)));
  *alt = 0.0;
  while (iterations > 0) {
    double tmp = *alt;
    double N = a / sqrt(1.0 - e2 * pow(sin(*lat), 2));
    *alt = p / cos(*lat) - N;
    *lat = atan(z_n / (p * (1.0 - e2 * N / (N + *alt))));
    if (fabs(*alt - tmp) < 1e-3)
      break;
    iterations--;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<__copy_to_user>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    -0x18(%rbp),%rdx
  mov    -0x10(%rbp),%rcx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <__copy_to_user_inatomic@plt>
  leave
  ret
```

assistant: ```c
__attribute__((always_)) unsigned long __copy_to_user(void *to, const void *from, unsigned long n) {
  do {
    do {
    } while (0);
  } while (0);
  return __copy_to_user_inatomic(to, from, n);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<__copy_to_user>:
  endbr64
  sub    $0x8,%rsp
  call   <__copy_to_user_inatomic@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
__attribute__((always_)) unsigned long __copy_to_user(void *to, const void *from, unsigned long n) {
  do {
    do {
    } while (0);
  } while (0);
  return __copy_to_user_inatomic(to, from, n);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<__copy_to_user>:
  endbr64
  jmp    1050 <__copy_to_user_inatomic@plt>
```

assistant: ```c
__attribute__((always_)) unsigned long __copy_to_user(void *to, const void *from, unsigned long n) {
  do {
    do {
    } while (0);
  } while (0);
  return __copy_to_user_inatomic(to, from, n);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<__copy_to_user>:
  endbr64
  jmp    1050 <__copy_to_user_inatomic@plt>
```

assistant: ```c
__attribute__((always_)) unsigned long __copy_to_user(void *to, const void *from, unsigned long n) {
  do {
    do {
    } while (0);
  } while (0);
  return __copy_to_user_inatomic(to, from, n);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<handle_intel>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %edi,%ebx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  cmp    $0x1,%esi
  seta   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  movl   $0x0,(%rsp)
  movl   $0x0,0x4(%rsp)
  mov    $0x1,%r12d
  mov    $0x1,%r14d
  mov    %rsp,%rbp
  jmp    L12
L58:
  lea    0x4(%rsp),%rdx
  mov    %rbp,%rcx
  mov    %r13d,%esi
  mov    %ebx,%edi
  call   <intel_check_word@plt>
  test   %rax,%rax
  jne    L94
  lea    0x4(%rsp),%rdx
  mov    %rbp,%rcx
  mov    $0x0,%esi
  mov    %ebx,%edi
  call   <intel_check_word@plt>
  test   %rax,%rax
  jne    L94
  lea    0x4(%rsp),%rdx
  mov    %rbp,%rcx
  mov    $0x0,%esi
  mov    %ebx,%edi
  call   <intel_check_word@plt>
  test   %rax,%rax
  jne    L94
  lea    0x4(%rsp),%rdx
  mov    %rbp,%rcx
  mov    $0x0,%esi
  mov    %ebx,%edi
  call   <intel_check_word@plt>
  test   %rax,%rax
  jne    L94
  lea    0x1(%r12),%edx
  cmp    %r12d,%r14d
  jbe    L87
  mov    %edx,%r12d
L12:
  mov    $0x0,%r8d
  mov    $0x0,%esi
  mov    %esi,%ecx
  mov    %esi,%edx
  mov    %r13d,%esi
  mov    $0x2,%edi
  call   <__cpuid@plt>
  cmp    $0x1,%r12d
  jne    L58
  movzbl %r13b,%r14d
  mov    $0x0,%r13b
  jmp    L58
L87:
  mov    D8(%rip),%rdx
  cmp    %ebx,(%rdx)
  jg     L94
  mov    D0(%rip),%rdx
  cmp    %ebx,(%rdx)
  jl     L94
  mov    (%rsp),%eax
  neg    %eax
  sbb    %rax,%rax
L94:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L104
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L104:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
long int __attribute__((no)) handle_intel(int name, unsigned int maxidx) {
  assert(maxidx >= 2);
  unsigned int cnt = 0;
  unsigned int max = 1;
  long int result = 0;
  bool no_level_2_or_3 = false;
  bool has_level_2 = false;
  while (cnt++ < max) {
    unsigned int eax;
    unsigned int ebx;
    unsigned int ecx;
    unsigned int edx;
    __cpuid(2, eax, ebx, ecx, edx);
    if (cnt == 1) {
      max = eax & 0xff;
      eax &= 0xffffff00;
    }
    result = intel_check_word(name, eax, &has_level_2, &no_level_2_or_3);
    if (result != 0)
      return result;
    result = intel_check_word(name, ebx, &has_level_2, &no_level_2_or_3);
    if (result != 0)
      return result;
    result = intel_check_word(name, ecx, &has_level_2, &no_level_2_or_3);
    if (result != 0)
      return result;
    result = intel_check_word(name, edx, &has_level_2, &no_level_2_or_3);
    if (result != 0)
      return result;
  }
  if (name >= _SC_LEVEL2_CACHE_SIZE && name <= _SC_LEVEL3_CACHE_LINESIZE && no_level_2_or_3)
    return -1;
  return 0;
}
```