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

```asm
<test_short_array>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x1,%esi
  lea    D1(%rip),%rdi
  call   <assertTrue@plt>
  mov    $0x1,%esi
  lea    D13(%rip),%rdi
  add    $0x8,%rsp
  jmp    1050 <assertTrue@plt>
```

tool_calls: name: disassemble_data
arguments: label: D1
type: string
name: disassemble_data
arguments: label: D13
type: string

tool: D1: chars[32] == 2699

tool: D13: chars[33] == 2714

assistant: ```c
void test_short_array() {
  short chars[255] = {
      0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2699, 2714, 2728, 2732, 2719, 2733, 2718, 2717, 2721, 2722, 2723, 2725, 2711, 2724, 2710, 2720, 2700, 2701, 2702, 2703, 2704, 2705, 2706, 2707, 2708, 2709, 2712, 2713, 2730, 2726, 2731, 2715, 2734, 2501, 2502, 2503, 2504, 2505, 2506, 2507, 2508, 2509, 2510, 2511, 2512, 2513, 2514, 2515, 2516, 2517, 2518, 2519, 2520, 2521, 2522, 2523, 2524, 2525, 2526, 2223, 4178, 2224, 4110, 4013, 2716, 2601, 2602, 2603, 2604, 2605, 2606, 2607, 2608, 2609, 2610, 2611, 2612, 2613, 2614, 2615, 2616, 2617, 2618, 2619, 2620, 2621, 2622, 2623, 2624, 2625, 2626, 2225, 4108, 2226, 2246, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4180, 4181, 4182, 4183, 4184, 4185, 4186, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2619, 4163, 0, 0, 0, 0, 0, 0, 2699, 4115, 910, 272, 4023, 4128, 4107, 2276, 4182, 274, 0, 4023, 4080, 4105, 273, 4187, 2729, 2233, 0, 0, 4180, 3138, 4023, 4131, 4023, 0, 0, 4023, 270, 261, 271, 4116, 4025, 4025, 4025, 4025, 4025, 4025, 0, 4024, 4025, 4025, 4025, 4025, 4025, 4025, 4025, 4025, 4023, 4025, 4025, 4025, 4025, 4025, 4025, 727, 2515, 4025, 4025, 4025, 4025, 4025, 4023, 0, 4024, 4024, 4024, 4024, 4024, 4024, 0, 4024, 4024, 4024, 4024, 4024, 4024, 4024, 4024, 4024, 4023, 4024, 4024, 4024, 4024, 4024, 4024, 2237, 2615, 4024, 4024, 4024, 4024, 4024, 4023, 4024,
  };
  assertTrue("chars[32] == 2699", chars[32] == 2699);
  ;
  assertTrue("chars[33] == 2714", chars[33] == 2714);
  ;
}
```
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    $0x8,%rsp
  push   $0x0
  mov    $0x0,%r9d
  mov    $0x0,%r8d
  mov    $0x0,%ecx
  mov    $0x0,%edx
  mov    $0x0,%esi
  mov    $0x0,%edi
  call   <fb@plt>
  add    $0x10,%rsp
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main() {
  fb(0, 0, 0, 0, 0, 0, 0);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  sub    $0x10,%rsp
  push   $0x0
  mov    $0x0,%r9d
  mov    $0x0,%r8d
  mov    $0x0,%ecx
  mov    $0x0,%edx
  mov    $0x0,%esi
  mov    $0x0,%edi
  call   <fb@plt>
  mov    $0x0,%eax
  add    $0x18,%rsp
  ret
```

assistant: ```c
int main() {
  fb(0, 0, 0, 0, 0, 0, 0);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  sub    $0x10,%rsp
  xor    %r9d,%r9d
  xor    %r8d,%r8d
  xor    %ecx,%ecx
  push   $0x0
  xor    %edx,%edx
  xor    %esi,%esi
  xor    %edi,%edi
  call   <fb@plt>
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
```

assistant: ```c
int main() {
  fb(0, 0, 0, 0, 0, 0, 0);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x10,%rsp
  xor    %r9d,%r9d
  xor    %r8d,%r8d
  xor    %ecx,%ecx
  push   $0x0
  xor    %edx,%edx
  xor    %esi,%esi
  xor    %edi,%edi
  call   <fb@plt>
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
```

assistant: ```c
int main() {
  fb(0, 0, 0, 0, 0, 0, 0);
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<InsertInString>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,-0x48(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  test   %ecx,%ecx
  js     L127
  mov    %rsi,%r13
  mov    %rdx,%r14
  mov    %ecx,%ebx
  mov    %rsi,%rdi
  call   <GetStringLength@plt>
  mov    %eax,%r12d
  mov    %r14,%rdi
  call   <GetStringLength@plt>
  mov    %eax,%r15d
  test   %r12d,%r12d
  setle  %al
  test   %r15d,%r15d
  setle  %dl
  or     %dl,%al
  jne    L80
  cmp    %r12d,%ebx
  jge    L80
  lea    (%r12,%r15,1),%edi
  lea    D5(%rip),%rsi
  call   <GetStackSize@plt>
  cltq
  add    $0xf,%rax
  mov    %rax,%rcx
  and    $0xfffffffffffffff0,%rcx
  and    $0xfffffffffffff000,%rax
  mov    %rsp,%rdx
  sub    %rax,%rdx
L10:
  cmp    %rdx,%rsp
  je     L58
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L10
L127:
  call   <ReturnNullMallocStack@plt>
  jmp    L53
L80:
  mov    -0x48(%rbp),%rdi
  call   <ReturnNullMallocStack@plt>
L53:
  mov    -0x38(%rbp),%rdx
  sub    %fs:0x28,%rdx
  jne    L49
  lea    -0x28(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
L58:
  mov    %rcx,%rax
  and    $0xfff,%eax
  sub    %rax,%rsp
  test   %rax,%rax
  je     L75
  orq    $0x0,-0x8(%rsp,%rax,1)
L75:
  mov    %rsp,%r8
  mov    $0x0,%eax
  mov    $0xffffffff,%edx
  mov    $0x0,%r9d
  jmp    L27
L44:
  mov    %ecx,%edx
L27:
  cmp    %eax,%ebx
  je     L32
L93:
  lea    0x1(%rdx),%ecx
  movslq %ecx,%rsi
  movzbl 0x0(%r13,%rax,1),%edi
  mov    %dil,(%r8,%rsi,1)
  add    $0x1,%rax
  cmp    %eax,%r12d
  jg     L44
  add    $0x2,%edx
  movslq %edx,%rax
  movb   $0x0,(%r8,%rax,1)
  mov    %r8,%rsi
  mov    -0x48(%rbp),%rdi
  call   <ReturnOutputMallocStack@plt>
  jmp    L53
L32:
  mov    $0x0,%ecx
  movslq %edx,%rdi
  add    %r8,%rdi
L66:
  movzbl (%r14,%rcx,1),%esi
  mov    %sil,0x1(%rdi,%rcx,1)
  add    $0x1,%rcx
  cmp    %ecx,%r15d
  jg     L66
  lea    -0x1(%r15),%ecx
  test   %r15d,%r15d
  cmovle %r9d,%ecx
  lea    0x1(%rdx,%rcx,1),%edx
  jmp    L93
L49:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D5
type: string

tool: D5: char

assistant: ```c
char *InsertInString(char *output, char *input, char *addition, const int index) {
  if (index < 0)
    return ReturnNullMallocStack(output);
  int length1 = GetStringLength(input);
  int length2 = GetStringLength(addition);
  if (length1 < 1 || length2 < 1 || index > length1 - 1) {
    return ReturnNullMallocStack(output);
  }
  int length0 = length1 + length2;
  char output2[GetStackSize(length0, "char")];
  int i0 = -1;
  int i = 0;
  for (i = 0; i < length1; i++) {
    if (i == index) {
      int i2 = 0;
      for (i2 = 0; i2 < length2; i2++) {
        i0++;
        output2[i0] = addition[i2];
      }
    }
    i0++;
    output2[i0] = input[i];
  }
  i0++;
  output2[i0] = '\0';
  return ReturnOutputMallocStack(output, output2, i0);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<p009>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  movl   $0x3e8,-0x14(%rbp)
  movl   $0x1,-0x1c(%rbp)
  movl   $0x1,-0x18(%rbp)
L9:
  mov    -0x1c(%rbp),%eax
  imul   %eax,%eax
  mov    %eax,%edx
  mov    -0x18(%rbp),%eax
  imul   %eax,%eax
  add    %edx,%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x10(%rbp),%eax
  mov    %eax,%edi
  call   <sqrt@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x8(%rbp)
  pxor   %xmm1,%xmm1
  cvtsi2sdl -0x14(%rbp),%xmm1
  movsd  -0x8(%rbp),%xmm0
  comisd %xmm1,%xmm0
  jbe    L34
  addl   $0x1,-0x1c(%rbp)
  mov    -0x1c(%rbp),%eax
  mov    %eax,-0x18(%rbp)
  jmp    L14
L34:
  movsd  -0x8(%rbp),%xmm0
  cvttsd2si %xmm0,%eax
  mov    %eax,-0xc(%rbp)
  pxor   %xmm0,%xmm0
  cvtsi2sdl -0xc(%rbp),%xmm0
  ucomisd -0x8(%rbp),%xmm0
  jp     L32
  ucomisd -0x8(%rbp),%xmm0
  je     L38
L32:
  addl   $0x1,-0x18(%rbp)
  jmp    L14
L38:
  mov    -0x1c(%rbp),%edx
  mov    -0x18(%rbp),%eax
  add    %eax,%edx
  mov    -0xc(%rbp),%eax
  add    %edx,%eax
  cmp    %eax,-0x14(%rbp)
  je     L59
  addl   $0x1,-0x18(%rbp)
L14:
  jmp    L9
L59:
  nop
  mov    -0xc(%rbp),%ecx
  mov    -0x18(%rbp),%edx
  mov    -0x1c(%rbp),%eax
  mov    %eax,%esi
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x14(%rbp),%esi
  mov    -0xc(%rbp),%ecx
  mov    -0x18(%rbp),%edx
  mov    -0x1c(%rbp),%eax
  mov    %esi,%r8d
  mov    %eax,%esi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x1c(%rbp),%eax
  imul   -0x18(%rbp),%eax
  imul   -0xc(%rbp),%eax
  mov    %eax,%esi
  mov    -0xc(%rbp),%ecx
  mov    -0x18(%rbp),%edx
  mov    -0x1c(%rbp),%eax
  mov    %esi,%r8d
  mov    %eax,%esi
  lea    D16(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  nop
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D10
datatype: string
name: extract_symbolic_data
arguments: data_tag: D9
datatype: string
name: extract_symbolic_data
arguments: data_tag: D16
datatype: string

tool: D10: The pythagorean triplet %d^2 + %d^2 = %d^2 satisfies the equation:



tool: D9:   %d + %d + %d = %d



tool: D16: The resulting product is %d * %d * %d = %d

assistant: ```c
void p009() {
  int target = 1000;
  int a = 1;
  int b = 1;
  int c, t1;
  double t2;
  do {
    t1 = a * a + b * b;
    t2 = sqrt(t1);
    if (t2 > target) {
      a++;
      b = a;
      continue;
    }
    c = (int)t2;
    if (t2 != c) {
      b++;
      continue;
    }
    if (a + b + c == target) {
      break;
    }
    b++;
  } while (1);
  printf("The pythagorean triplet %d^2 + %d^2 = %d^2 satisfies the equation:\n\n", a, b, c);
  printf("  %d + %d + %d = %d\n\n", a, b, c, target);
  printf("The resulting product is %d * %d * %d = %d", a, b, c, a * b * c);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x70,%rsp
  mov    %edi,-0x64(%rbp)
  mov    %rsi,-0x70(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x70(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <atoi@plt>
  mov    %eax,-0x54(%rbp)
  movl   $0x6f72657a,-0x50(%rbp)
  movw   $0x0,-0x4c(%rbp)
  movl   $0x656e6f,-0x4a(%rbp)
  movw   $0x0,-0x46(%rbp)
  movl   $0x6f7774,-0x44(%rbp)
  movw   $0x0,-0x40(%rbp)
  movl   $0x65726874,-0x3e(%rbp)
  movw   $0x65,-0x3a(%rbp)
  movl   $0x72756f66,-0x38(%rbp)
  movw   $0x0,-0x34(%rbp)
  movl   $0x65766966,-0x32(%rbp)
  movw   $0x0,-0x2e(%rbp)
  movl   $0x786973,-0x2c(%rbp)
  movw   $0x0,-0x28(%rbp)
  movl   $0x65766573,-0x26(%rbp)
  movw   $0x6e,-0x22(%rbp)
  movl   $0x68676965,-0x20(%rbp)
  movw   $0x74,-0x1c(%rbp)
  movl   $0x656e696e,-0x1a(%rbp)
  movw   $0x0,-0x16(%rbp)
  lea    -0x50(%rbp),%rcx
  mov    -0x54(%rbp),%eax
  movslq %eax,%rdx
  mov    %rdx,%rax
  add    %rax,%rax
  add    %rdx,%rax
  add    %rax,%rax
  add    %rcx,%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L7
  call   <__stack_chk_fail@plt>
L7:
  leave
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  int i = atoi(argv[1]);
  char a[][6] = {
      "zero", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine",
  };
  printf("%s\n", a[i]);
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  sub    $0x58,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  mov    0x8(%rsi),%rdi
  call   <atoi@plt>
  movl   $0x6f72657a,(%rsp)
  movw   $0x0,0x4(%rsp)
  movl   $0x656e6f,0x6(%rsp)
  movw   $0x0,0xa(%rsp)
  movl   $0x6f7774,0xc(%rsp)
  movw   $0x0,0x10(%rsp)
  movl   $0x65726874,0x12(%rsp)
  movw   $0x65,0x16(%rsp)
  movl   $0x72756f66,0x18(%rsp)
  movw   $0x0,0x1c(%rsp)
  movl   $0x65766966,0x1e(%rsp)
  movw   $0x0,0x22(%rsp)
  movl   $0x786973,0x24(%rsp)
  movw   $0x0,0x28(%rsp)
  movl   $0x65766573,0x2a(%rsp)
  movw   $0x6e,0x2e(%rsp)
  movl   $0x68676965,0x30(%rsp)
  movw   $0x74,0x34(%rsp)
  movl   $0x656e696e,0x36(%rsp)
  movw   $0x0,0x3a(%rsp)
  cltq
  lea    (%rax,%rax,2),%rax
  lea    (%rsp,%rax,2),%rdi
  call   <puts@plt>
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  mov    $0x0,%eax
  add    $0x58,%rsp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(int argc, char *argv[]) {
  int i = atoi(argv[1]);
  char a[][6] = {
      "zero", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine",
  };
  printf("%s\n", a[i]);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  sub    $0x58,%rsp
  mov    0x8(%rsi),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  call   <atoi@plt>
  movdqa D5(%rip),%xmm0
  movabs $0x696e007468676965,%rdx
  movl   $0x656e,0x38(%rsp)
  cltq
  mov    %rdx,0x30(%rsp)
  movaps %xmm0,(%rsp)
  movdqa D12(%rip),%xmm0
  lea    (%rax,%rax,2),%rax
  lea    (%rsp,%rax,2),%rdi
  movaps %xmm0,0x10(%rsp)
  movdqa D21(%rip),%xmm0
  movaps %xmm0,0x20(%rsp)
  call   <puts@plt>
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  xor    %eax,%eax
  add    $0x58,%rsp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D5
type_infered: byte[16]
name: unpack_data
arguments: label: D12
type_infered: byte[16]
name: unpack_data
arguments: label: D21
type_infered: byte[16]

tool: {"D5": [0x7a,0x65,0x72,0x6f,0x00,0x00,0x6f,0x6e,0x65,0x00,0x00,0x00,0x74,0x77,0x6f,0x00]}

tool: {"D12": [0x00,0x00,0x74,0x68,0x72,0x65,0x65,0x00,0x66,0x6f,0x75,0x72,0x00,0x00,0x66,0x69]}

tool: {"D21": [0x76,0x65,0x00,0x00,0x73,0x69,0x78,0x00,0x00,0x00,0x73,0x65,0x76,0x65,0x6e,0x00]}

assistant: ```c
int main(int argc, char *argv[]) {
  int i = atoi(argv[1]);
  char a[][6] = {
      "zero", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine",
  };
  printf("%s\n", a[i]);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x58,%rsp
  mov    0x8(%rsi),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  call   <atoi@plt>
  movdqa D24(%rip),%xmm0
  movabs $0x696e007468676965,%rdx
  movl   $0x656e,0x38(%rsp)
  cltq
  mov    %rdx,0x30(%rsp)
  movaps %xmm0,(%rsp)
  movdqa D29(%rip),%xmm0
  lea    (%rax,%rax,2),%rax
  lea    (%rsp,%rax,2),%rdi
  movaps %xmm0,0x10(%rsp)
  movdqa D3(%rip),%xmm0
  movaps %xmm0,0x20(%rsp)
  call   <puts@plt>
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  xor    %eax,%eax
  add    $0x58,%rsp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D24
type: byte[16]
name: disassemble_data
arguments: data_label: D29
type: byte[16]
name: disassemble_data
arguments: data_label: D3
type: byte[16]

tool: {"D24": [0x7a,0x65,0x72,0x6f,0x00,0x00,0x6f,0x6e,0x65,0x00,0x00,0x00,0x74,0x77,0x6f,0x00]}

tool: {"D29": [0x00,0x00,0x74,0x68,0x72,0x65,0x65,0x00,0x66,0x6f,0x75,0x72,0x00,0x00,0x66,0x69]}

tool: {"D3": [0x76,0x65,0x00,0x00,0x73,0x69,0x78,0x00,0x00,0x00,0x73,0x65,0x76,0x65,0x6e,0x00]}

assistant: ```c
int main(int argc, char *argv[]) {
  int i = atoi(argv[1]);
  char a[][6] = {
      "zero", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine",
  };
  printf("%s\n", a[i]);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<read_lanemasks>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    $0x0,%eax
  call   <__builtin_ptx_read_lanemask_eq@plt>
  mov    %eax,-0x14(%rbp)
  mov    $0x0,%eax
  call   <__builtin_ptx_read_lanemask_le@plt>
  mov    %eax,-0x10(%rbp)
  mov    $0x0,%eax
  call   <__builtin_ptx_read_lanemask_lt@plt>
  mov    %eax,-0xc(%rbp)
  mov    $0x0,%eax
  call   <__builtin_ptx_read_lanemask_ge@plt>
  mov    %eax,-0x8(%rbp)
  mov    $0x0,%eax
  call   <__builtin_ptx_read_lanemask_gt@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x14(%rbp),%edx
  mov    -0x10(%rbp),%eax
  add    %eax,%edx
  mov    -0xc(%rbp),%eax
  add    %eax,%edx
  mov    -0x8(%rbp),%eax
  add    %eax,%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  leave
  ret
```

assistant: ```c
int read_lanemasks() {
  int a = __builtin_ptx_read_lanemask_eq();
  int b = __builtin_ptx_read_lanemask_le();
  int c = __builtin_ptx_read_lanemask_lt();
  int d = __builtin_ptx_read_lanemask_ge();
  int e = __builtin_ptx_read_lanemask_gt();
  return a + b + c + d + e;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<read_lanemasks>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <__builtin_ptx_read_lanemask_eq@plt>
  mov    %eax,%ebx
  mov    $0x0,%eax
  call   <__builtin_ptx_read_lanemask_le@plt>
  mov    %eax,%r13d
  mov    $0x0,%eax
  call   <__builtin_ptx_read_lanemask_lt@plt>
  mov    %eax,%r12d
  mov    $0x0,%eax
  call   <__builtin_ptx_read_lanemask_ge@plt>
  mov    %eax,%ebp
  mov    $0x0,%eax
  call   <__builtin_ptx_read_lanemask_gt@plt>
  mov    %eax,%edx
  lea    (%rbx,%r13,1),%eax
  add    %r12d,%eax
  add    %ebp,%eax
  add    %edx,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int read_lanemasks() {
  int a = __builtin_ptx_read_lanemask_eq();
  int b = __builtin_ptx_read_lanemask_le();
  int c = __builtin_ptx_read_lanemask_lt();
  int d = __builtin_ptx_read_lanemask_ge();
  int e = __builtin_ptx_read_lanemask_gt();
  return a + b + c + d + e;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<read_lanemasks>:
  endbr64
  push   %r13
  xor    %eax,%eax
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <__builtin_ptx_read_lanemask_eq@plt>
  mov    %eax,%ebx
  xor    %eax,%eax
  call   <__builtin_ptx_read_lanemask_le@plt>
  mov    %eax,%r13d
  xor    %eax,%eax
  call   <__builtin_ptx_read_lanemask_lt@plt>
  mov    %eax,%r12d
  xor    %eax,%eax
  call   <__builtin_ptx_read_lanemask_ge@plt>
  mov    %eax,%ebp
  xor    %eax,%eax
  call   <__builtin_ptx_read_lanemask_gt@plt>
  add    $0x8,%rsp
  mov    %eax,%r8d
  lea    (%rbx,%r13,1),%eax
  pop    %rbx
  add    %r12d,%eax
  add    %ebp,%eax
  pop    %rbp
  pop    %r12
  add    %r8d,%eax
  pop    %r13
  ret
```

assistant: ```c
int read_lanemasks() {
  int a = __builtin_ptx_read_lanemask_eq();
  int b = __builtin_ptx_read_lanemask_le();
  int c = __builtin_ptx_read_lanemask_lt();
  int d = __builtin_ptx_read_lanemask_ge();
  int e = __builtin_ptx_read_lanemask_gt();
  return a + b + c + d + e;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<read_lanemasks>:
  endbr64
  push   %r13
  xor    %eax,%eax
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <__builtin_ptx_read_lanemask_eq@plt>
  mov    %eax,%ebx
  xor    %eax,%eax
  call   <__builtin_ptx_read_lanemask_le@plt>
  mov    %eax,%r13d
  xor    %eax,%eax
  call   <__builtin_ptx_read_lanemask_lt@plt>
  mov    %eax,%r12d
  xor    %eax,%eax
  call   <__builtin_ptx_read_lanemask_ge@plt>
  mov    %eax,%ebp
  xor    %eax,%eax
  call   <__builtin_ptx_read_lanemask_gt@plt>
  add    $0x8,%rsp
  mov    %eax,%r8d
  lea    (%rbx,%r13,1),%eax
  pop    %rbx
  add    %r12d,%eax
  add    %ebp,%eax
  pop    %rbp
  pop    %r12
  add    %r8d,%eax
  pop    %r13
  ret
```

assistant: ```c
int read_lanemasks() {
  int a = __builtin_ptx_read_lanemask_eq();
  int b = __builtin_ptx_read_lanemask_le();
  int c = __builtin_ptx_read_lanemask_lt();
  int d = __builtin_ptx_read_lanemask_ge();
  int e = __builtin_ptx_read_lanemask_gt();
  return a + b + c + d + e;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<get_dynamical_diff_other>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movsd  %xmm0,-0x18(%rbp)
  movsd  -0x18(%rbp),%xmm0
  movsd  D27(%rip),%xmm1
  subsd  %xmm1,%xmm0
  movsd  %xmm0,-0x10(%rbp)
  movsd  -0x10(%rbp),%xmm0
  mulsd  %xmm0,%xmm0
  movsd  %xmm0,-0x10(%rbp)
  movsd  -0x10(%rbp),%xmm0
  movsd  D25(%rip),%xmm1
  divsd  %xmm1,%xmm0
  movsd  D10(%rip),%xmm1
  subsd  %xmm1,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movsd  -0x8(%rbp),%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: data_label: D27
type_infered: byte[8]
name: parse_data
arguments: data_label: D25
type_infered: byte[8]
name: parse_data
arguments: data_label: D10
type_infered: byte[8]

tool: {"D27": [0x00,0x00,0x00,0x00,0xa2,0x2c,0x42,0x41]}

tool: {"D25": [0x00,0x00,0x00,0x00,0xcd,0x92,0x83,0x41]}

tool: {"D10": [0x00,0x00,0x00,0x00,0x00,0x00,0x2e,0x40]}

assistant: ```c
double get_dynamical_diff_other(double JD) {
  double TD;
  double a;
  a = (JD - 2382148);
  a *= a;
  TD = -15 + a / 41048480;
  return (TD);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<get_dynamical_diff_other>:
  endbr64
  subsd  D23(%rip),%xmm0
  mulsd  %xmm0,%xmm0
  divsd  D14(%rip),%xmm0
  subsd  D26(%rip),%xmm0
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D23
type_guessed: byte[8]
name: extract_symbolic_data
arguments: tag: D14
type_guessed: byte[8]
name: extract_symbolic_data
arguments: tag: D26
type_guessed: byte[8]

tool: {"D23": [0x00,0x00,0x00,0x00,0xa2,0x2c,0x42,0x41]}

tool: {"D14": [0x00,0x00,0x00,0x00,0xcd,0x92,0x83,0x41]}

tool: {"D26": [0x00,0x00,0x00,0x00,0x00,0x00,0x2e,0x40]}

assistant: ```c
double get_dynamical_diff_other(double JD) {
  double TD;
  double a;
  a = (JD - 2382148);
  a *= a;
  TD = -15 + a / 41048480;
  return (TD);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<get_dynamical_diff_other>:
  endbr64
  subsd  D18(%rip),%xmm0
  mulsd  %xmm0,%xmm0
  divsd  D19(%rip),%xmm0
  subsd  D16(%rip),%xmm0
  ret
```

tool_calls: name: read_data
arguments: label: D18
data_type: byte[8]
name: read_data
arguments: label: D19
data_type: byte[8]
name: read_data
arguments: label: D16
data_type: byte[8]

tool: {"D18": [0x00,0x00,0x00,0x00,0xa2,0x2c,0x42,0x41]}

tool: {"D19": [0x00,0x00,0x00,0x00,0xcd,0x92,0x83,0x41]}

tool: {"D16": [0x00,0x00,0x00,0x00,0x00,0x00,0x2e,0x40]}

assistant: ```c
double get_dynamical_diff_other(double JD) {
  double TD;
  double a;
  a = (JD - 2382148);
  a *= a;
  TD = -15 + a / 41048480;
  return (TD);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<get_dynamical_diff_other>:
  endbr64
  subsd  D18(%rip),%xmm0
  mulsd  %xmm0,%xmm0
  divsd  D17(%rip),%xmm0
  subsd  D24(%rip),%xmm0
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D18
data_type: byte[8]
name: extract_symbolic_data
arguments: label: D17
data_type: byte[8]
name: extract_symbolic_data
arguments: label: D24
data_type: byte[8]

tool: {"D18": [0x00,0x00,0x00,0x00,0xa2,0x2c,0x42,0x41]}

tool: {"D17": [0x00,0x00,0x00,0x00,0xcd,0x92,0x83,0x41]}

tool: {"D24": [0x00,0x00,0x00,0x00,0x00,0x00,0x2e,0x40]}

assistant: ```c
double get_dynamical_diff_other(double JD) {
  double TD;
  double a;
  a = (JD - 2382148);
  a *= a;
  TD = -15 + a / 41048480;
  return (TD);
}
```