ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 这段汇编代码用C语言表示是什么样的？

```asm
<ZapOSGLUVars>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <InitDrives@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1070 <ZapWinStateVars@plt>
```

assistant: ```c
void ZapOSGLUVars(void) {
  InitDrives();
  ZapWinStateVars();
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<ZapOSGLUVars>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <InitDrives@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1070 <ZapWinStateVars@plt>
```

assistant: ```c
void ZapOSGLUVars(void) {
  InitDrives();
  ZapWinStateVars();
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<program_main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  jmp    L27
L22:
  lea    D28(%rip),%rax
  mov    %rax,%rsi
  lea    D57(%rip),%rax
  mov    %rax,%rdi
  call   <test_subsequence@plt>
  test   %rax,%rax
  je     L36
  lea    D18(%rip),%rax
  jmp    L38
L36:
  lea    D11(%rip),%rax
L38:
  mov    %rax,%rsi
  lea    D49(%rip),%rax
  mov    %rax,%rdi
  call   <naive_printf@plt>
L27:
  lea    D28(%rip),%rax
  mov    %rax,%rdx
  lea    D57(%rip),%rax
  mov    %rax,%rsi
  lea    D45(%rip),%rax
  mov    %rax,%rdi
  call   <naive_scanf@plt>
  cmp    $0x2,%eax
  je     L22
  mov    $0x0,%eax
  pop    %rbp
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D18
type: string
name: disassemble_data
arguments: data_tag: D11
type: string
name: disassemble_data
arguments: data_tag: D49
type: string
name: disassemble_data
arguments: data_tag: D45
type: string

tool: D18: Yes

tool: D11: No

tool: D49: %s


tool: D45: %s%s

assistant: ```c
int program_main() {
  static char s1[100000 + 1], s2[100000 + 1];
  while (naive_scanf("%s%s", s1, s2) == 2)
    naive_printf("%s\n", (test_subsequence(s1, s2) ? "Yes" : "No"));
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<program_main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  lea    D22(%rip),%rbp
  lea    D5(%rip),%rbx
  lea    D3(%rip),%r12
  lea    D59(%rip),%r13
  jmp    L9
L7:
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <test_subsequence@plt>
  test   %rax,%rax
  lea    D48(%rip),%rsi
  cmovne %r13,%rsi
  lea    D4(%rip),%rdi
  call   <naive_printf@plt>
L9:
  mov    %rbp,%rdx
  mov    %rbx,%rsi
  mov    %r12,%rdi
  call   <naive_scanf@plt>
  cmp    $0x2,%eax
  je     L7
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D3
type_guessed: string
name: unpack_data
arguments: data_tag: D59
type_guessed: string
name: unpack_data
arguments: data_tag: D48
type_guessed: string
name: unpack_data
arguments: data_tag: D4
type_guessed: string

tool: D3: %s%s

tool: D59: Yes

tool: D48: No

tool: D4: %s


assistant: ```c
int program_main() {
  static char s1[100000 + 1], s2[100000 + 1];
  while (naive_scanf("%s%s", s1, s2) == 2)
    naive_printf("%s\n", (test_subsequence(s1, s2) ? "Yes" : "No"));
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<program_main>:
  endbr64
  push   %r15
  lea    D1(%rip),%r15
  push   %r14
  lea    D33(%rip),%r14
  push   %r13
  lea    D13(%rip),%r13
  push   %r12
  lea    D45(%rip),%r12
  push   %rbp
  lea    D55(%rip),%rbp
  push   %rbx
  lea    D28(%rip),%rbx
  sub    $0x8,%rsp
  jmp    L17
  xchg   %ax,%ax
L5:
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <test_subsequence@plt>
  mov    %r14,%rsi
  mov    %r13,%rdi
  test   %rax,%rax
  cmovne %r15,%rsi
  call   <naive_printf@plt>
L17:
  mov    %rbp,%rdx
  mov    %rbx,%rsi
  mov    %r12,%rdi
  call   <naive_scanf@plt>
  cmp    $0x2,%eax
  je     L5
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: parse_data
arguments: label: D1
type: string
name: parse_data
arguments: label: D33
type: string
name: parse_data
arguments: label: D13
type: string
name: parse_data
arguments: label: D45
type: string

tool: D1: Yes

tool: D33: No

tool: D13: %s


tool: D45: %s%s

assistant: ```c
int program_main() {
  static char s1[100000 + 1], s2[100000 + 1];
  while (naive_scanf("%s%s", s1, s2) == 2)
    naive_printf("%s\n", (test_subsequence(s1, s2) ? "Yes" : "No"));
  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
<program_main>:
  endbr64
  push   %r15
  lea    D58(%rip),%r15
  push   %r14
  lea    D56(%rip),%r14
  push   %r13
  lea    D33(%rip),%r13
  push   %r12
  lea    D7(%rip),%r12
  push   %rbp
  lea    D51(%rip),%rbp
  push   %rbx
  lea    D1(%rip),%rbx
  sub    $0x8,%rsp
  jmp    L19
  xchg   %ax,%ax
L8:
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <test_subsequence@plt>
  mov    %r14,%rsi
  mov    %r13,%rdi
  test   %rax,%rax
  cmovne %r15,%rsi
  call   <naive_printf@plt>
L19:
  mov    %rbp,%rdx
  mov    %rbx,%rsi
  mov    %r12,%rdi
  call   <naive_scanf@plt>
  cmp    $0x2,%eax
  je     L8
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: parse_data
arguments: label: D58
type_guessed: string
name: parse_data
arguments: label: D56
type_guessed: string
name: parse_data
arguments: label: D33
type_guessed: string
name: parse_data
arguments: label: D7
type_guessed: string

tool: D58: Yes

tool: D56: No

tool: D33: %s


tool: D7: %s%s

assistant: ```c
int program_main() {
  static char s1[100000 + 1], s2[100000 + 1];
  while (naive_scanf("%s%s", s1, s2) == 2)
    naive_printf("%s\n", (test_subsequence(s1, s2) ? "Yes" : "No"));
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<GetDistExtraBits>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x4,-0x4(%rbp)
  jg     L0
  mov    $0x0,%eax
  jmp    L10
L0:
  mov    -0x4(%rbp),%eax
  sub    $0x1,%eax
  bsr    %eax,%eax
  xor    $0x1f,%eax
  xor    $0x1f,%eax
  sub    $0x1,%eax
L10:
  pop    %rbp
  ret
```

assistant: ```c
int GetDistExtraBits(int dist) {
  if (dist < 5)
    return 0;
  return (31 ^ __builtin_clz(dist - 1)) - 1;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<GetDistExtraBits>:
  endbr64
  mov    $0x0,%eax
  cmp    $0x4,%edi
  jle    L3
  sub    $0x1,%edi
  bsr    %edi,%edi
  lea    -0x1(%rdi),%eax
L3:
  ret
```

assistant: ```c
int GetDistExtraBits(int dist) {
  if (dist < 5)
    return 0;
  return (31 ^ __builtin_clz(dist - 1)) - 1;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<GetDistExtraBits>:
  endbr64
  xor    %eax,%eax
  cmp    $0x4,%edi
  jle    L8
  sub    $0x1,%edi
  bsr    %edi,%edi
  lea    -0x1(%rdi),%eax
L8:
  ret
```

assistant: ```c
int GetDistExtraBits(int dist) {
  if (dist < 5)
    return 0;
  return (31 ^ __builtin_clz(dist - 1)) - 1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<GetDistExtraBits>:
  endbr64
  xor    %eax,%eax
  cmp    $0x4,%edi
  jle    L9
  sub    $0x1,%edi
  bsr    %edi,%edi
  lea    -0x1(%rdi),%eax
L9:
  ret
```

assistant: ```c
int GetDistExtraBits(int dist) {
  if (dist < 5)
    return 0;
  return (31 ^ __builtin_clz(dist - 1)) - 1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<rk29_sdmmc_gpio_open>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  cmpl   $0x2,-0x4(%rbp)
  je     L31
  cmpl   $0x2,-0x4(%rbp)
  jg     L2
  cmpl   $0x0,-0x4(%rbp)
  je     L38
  cmpl   $0x1,-0x4(%rbp)
  je     L77
  jmp    L2
L31:
  nop
  jmp    L66
L2:
  nop
  jmp    L66
L38:
  nop
  jmp    L66
L77:
  nop
L66:
  nop
  pop    %rbp
  ret
```

assistant: ```c
void rk29_sdmmc_gpio_open(int device_id, int on) {
  switch (device_id) {
  case 0: {
  } break;
  case 1: {
  } break;
  case 2:
    break;
  default:
    break;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<rk29_sdmmc_gpio_open>:
  endbr64
  ret
```

assistant: ```c
void rk29_sdmmc_gpio_open(int device_id, int on) {
  switch (device_id) {
  case 0: {
  } break;
  case 1: {
  } break;
  case 2:
    break;
  default:
    break;
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<rk29_sdmmc_gpio_open>:
  endbr64
  ret
```

assistant: ```c
void rk29_sdmmc_gpio_open(int device_id, int on) {
  switch (device_id) {
  case 0: {
  } break;
  case 1: {
  } break;
  case 2:
    break;
  default:
    break;
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x0,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L32
L13:
  mov    -0x4(%rbp),%eax
  and    $0x1,%eax
  test   %eax,%eax
  jne    L35
  mov    -0x4(%rbp),%eax
  add    %eax,-0x8(%rbp)
  addl   $0x1,-0x4(%rbp)
L32:
  cmpl   $0x9,-0x4(%rbp)
  jle    L13
  jmp    L34
L35:
  nop
L34:
  mov    -0x8(%rbp),%eax
  mov    %eax,%esi
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D3
data_type: string

tool: D3: %d


assistant: ```c
int main() {
  int sum = 0;
  for (int i = 0; i < 10; i++) {
    if (i % 2)
      break;
    sum += i;
  }
  printf("%d\n", sum);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%esi
  lea    D4(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D4
type_guessed: string

tool: D4: %d


assistant: ```c
int main() {
  int sum = 0;
  for (int i = 0; i < 10; i++) {
    if (i % 2)
      break;
    sum += i;
  }
  printf("%d\n", sum);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %esi,%esi
  lea    D5(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

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

tool: D5: %d


assistant: ```c
int main() {
  int sum = 0;
  for (int i = 0; i < 10; i++) {
    if (i % 2)
      break;
    sum += i;
  }
  printf("%d\n", sum);
}
```
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
  sub    $0x8,%rsp
  xor    %esi,%esi
  lea    D7(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D7
data_type: string

tool: D7: %d


assistant: ```c
int main() {
  int sum = 0;
  for (int i = 0; i < 10; i++) {
    if (i % 2)
      break;
    sum += i;
  }
  printf("%d\n", sum);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x0,%eax
  call   <read@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <fib@plt>
  mov    %eax,%edi
  call   <printInt@plt>
  cmpl   $0x9,-0x4(%rbp)
  jg     L1
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <printInt@plt>
L1:
  mov    $0xa,%edi
  call   <printChar@plt>
  nop
  leave
  ret
```

assistant: ```c
void main() {
  int x = read();
  printInt(fib(x));
  if (x < 10)
    printInt(x);
  printChar('\n');
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  push   %rbx
  mov    $0x0,%eax
  call   <read@plt>
  mov    %eax,%ebx
  mov    %eax,%edi
  call   <fib@plt>
  mov    %eax,%edi
  call   <printInt@plt>
  cmp    $0x9,%ebx
  jle    L7
L3:
  mov    $0xa,%edi
  call   <printChar@plt>
  pop    %rbx
  ret
L7:
  mov    %ebx,%edi
  call   <printInt@plt>
  jmp    L3
```

assistant: ```c
void main() {
  int x = read();
  printInt(fib(x));
  if (x < 10)
    printInt(x);
  printChar('\n');
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  xor    %eax,%eax
  call   <read@plt>
  mov    %eax,%edi
  mov    %eax,%ebp
  call   <fib@plt>
  mov    %eax,%edi
  call   <printInt@plt>
  cmp    $0x9,%ebp
  jle    L18
L6:
  mov    $0xa,%edi
  pop    %rbp
  jmp    10a0 <printChar@plt>
L18:
  mov    %ebp,%edi
  call   <printInt@plt>
  jmp    L6
```

assistant: ```c
void main() {
  int x = read();
  printInt(fib(x));
  if (x < 10)
    printInt(x);
  printChar('\n');
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %rbp
  xor    %eax,%eax
  call   <read@plt>
  mov    %eax,%edi
  mov    %eax,%ebp
  call   <fib@plt>
  mov    %eax,%edi
  call   <printInt@plt>
  cmp    $0x9,%ebp
  jle    L8
L3:
  mov    $0xa,%edi
  pop    %rbp
  jmp    10a0 <printChar@plt>
L8:
  mov    %ebp,%edi
  call   <printInt@plt>
  jmp    L3
```

assistant: ```c
void main() {
  int x = read();
  printInt(fib(x));
  if (x < 10)
    printInt(x);
  printChar('\n');
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<get_m>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0x0,-0x18(%rbp)
  movq   $0x0,-0x10(%rbp)
  lea    -0x10(%rbp),%rdi
  lea    -0x18(%rbp),%rcx
  mov    -0x38(%rbp),%rdx
  mov    -0x30(%rbp),%rsi
  mov    -0x28(%rbp),%rax
  mov    %rdi,%r8
  mov    %rax,%rdi
  call   <integer_solve_quadratic@plt>
  test   %rax,%rax
  je     L5
  mov    -0x10(%rbp),%rax
  jmp    L25
L5:
  mov    $0xffffffff,%eax
L25:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L27
  call   <__stack_chk_fail@plt>
L27:
  leave
  ret
```

assistant: ```c
int get_m(long a, long b, long c) {
  long x1 = 0;
  long x2 = 0;
  if (integer_solve_quadratic(a, b, c, &x1, &x2))
    return x2;
  else
    return -1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<get_m>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movq   $0x0,0x8(%rsp)
  movq   $0x0,0x10(%rsp)
  lea    0x8(%rsp),%rcx
  lea    0x10(%rsp),%r8
  call   <integer_solve_quadratic@plt>
  test   %rax,%rax
  je     L5
  mov    0x10(%rsp),%eax
L17:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L3
  add    $0x28,%rsp
  ret
L5:
  mov    $0xffffffff,%eax
  jmp    L17
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int get_m(long a, long b, long c) {
  long x1 = 0;
  long x2 = 0;
  if (integer_solve_quadratic(a, b, c, &x1, &x2))
    return x2;
  else
    return -1;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<get_m>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x8(%rsp),%rcx
  lea    0x10(%rsp),%r8
  movq   $0x0,0x8(%rsp)
  movq   $0x0,0x10(%rsp)
  call   <integer_solve_quadratic@plt>
  test   %rax,%rax
  je     L8
  mov    0x10(%rsp),%eax
L11:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L5
  add    $0x28,%rsp
  ret
  nopw   0x0(%rax,%rax,1)
L8:
  mov    $0xffffffff,%eax
  jmp    L11
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int get_m(long a, long b, long c) {
  long x1 = 0;
  long x2 = 0;
  if (integer_solve_quadratic(a, b, c, &x1, &x2))
    return x2;
  else
    return -1;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<get_m>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x8(%rsp),%rcx
  lea    0x10(%rsp),%r8
  movq   $0x0,0x8(%rsp)
  movq   $0x0,0x10(%rsp)
  call   <integer_solve_quadratic@plt>
  test   %rax,%rax
  je     L16
  mov    0x10(%rsp),%eax
L22:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L6
  add    $0x28,%rsp
  ret
  nopw   0x0(%rax,%rax,1)
L16:
  mov    $0xffffffff,%eax
  jmp    L22
L6:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int get_m(long a, long b, long c) {
  long x1 = 0;
  long x2 = 0;
  if (integer_solve_quadratic(a, b, c, &x1, &x2))
    return x2;
  else
    return -1;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<llrintl>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0x0,-0x10(%rbp)
  fldt   0x10(%rbp)
  fistpll -0x10(%rbp)
  mov    -0x10(%rbp),%rax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L0
  call   <__stack_chk_fail@plt>
L0:
  leave
  ret
```

assistant: ```c
long long llrintl(long double x) {
  long long retval = 0ll;
  __asm__ __volatile__("fistpll %0" : "=m"(retval) : "t"(x) : "st");
  return retval;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<llrintl>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  fldt   0x20(%rsp)
  fistpll (%rsp)
  mov    (%rsp),%rax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L3
  add    $0x18,%rsp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
long long llrintl(long double x) {
  long long retval = 0ll;
  __asm__ __volatile__("fistpll %0" : "=m"(retval) : "t"(x) : "st");
  return retval;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<llrintl>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  fldt   0x20(%rsp)
  fistpll (%rsp)
  mov    (%rsp),%rax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L2
  add    $0x18,%rsp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
long long llrintl(long double x) {
  long long retval = 0ll;
  __asm__ __volatile__("fistpll %0" : "=m"(retval) : "t"(x) : "st");
  return retval;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<llrintl>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  fldt   0x20(%rsp)
  fistpll (%rsp)
  mov    (%rsp),%rax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L5
  add    $0x18,%rsp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
long long llrintl(long double x) {
  long long retval = 0ll;
  __asm__ __volatile__("fistpll %0" : "=m"(retval) : "t"(x) : "st");
  return retval;
}
```
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   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x0,-0x4(%rbp)
  jmp    L9
L13:
  mov    -0x4(%rbp),%eax
  and    $0x1,%eax
  test   %eax,%eax
  je     L41
  mov    -0x4(%rbp),%edx
  movslq %edx,%rax
  imul   $0x55555556,%rax,%rax
  shr    $0x20,%rax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  mov    %eax,%ecx
  add    %ecx,%ecx
  add    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  test   %eax,%eax
  jne    L41
  mov    -0x4(%rbp),%edx
  movslq %edx,%rax
  imul   $0xffffffff92492493,%rax,%rax
  shr    $0x20,%rax
  add    %edx,%eax
  sar    $0x2,%eax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  mov    %eax,%ecx
  shl    $0x3,%ecx
  sub    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  test   %eax,%eax
  jne    L41
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L41:
  addl   $0x1,-0x4(%rbp)
L9:
  cmpl   $0x7530,-0x4(%rbp)
  jle    L13
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D9
type: string

tool: D9: %d


assistant: ```c
int main() {
  int c;
  for (c = 0; c <= 30000; c++) {
    if (c % 2 != 0) {
      if ((c % 3 == 0) && (c % 7 == 0)) {
        printf("%d\n", c);
      }
    }
  }
  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   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    $0x0,%ebx
  lea    D8(%rip),%rbp
L3:
  add    $0x1,%ebx
  cmp    $0x7531,%ebx
  je     L9
  test   $0x1,%bl
  je     L3
  movslq %ebx,%rax
  imul   $0x55555556,%rax,%rdx
  shr    $0x20,%rdx
  mov    %ebx,%esi
  sar    $0x1f,%esi
  sub    %esi,%edx
  lea    (%rdx,%rdx,2),%edx
  mov    %ebx,%ecx
  sub    %edx,%ecx
  imul   $0xffffffff92492493,%rax,%rax
  shr    $0x20,%rax
  add    %ebx,%eax
  sar    $0x2,%eax
  sub    %esi,%eax
  lea    0x0(,%rax,8),%edx
  sub    %eax,%edx
  mov    %ebx,%eax
  sub    %edx,%eax
  or     %eax,%ecx
  jne    L3
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  jmp    L3
L9:
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D8
type: string

tool: D8: %d


assistant: ```c
int main() {
  int c;
  for (c = 0; c <= 30000; c++) {
    if (c % 2 != 0) {
      if ((c % 3 == 0) && (c % 7 == 0)) {
        printf("%d\n", c);
      }
    }
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %r12
  mov    $0xaaaaaaab,%r12d
  push   %rbp
  lea    D0(%rip),%rbp
  push   %rbx
  xor    %ebx,%ebx
  nopw   0x0(%rax,%rax,1)
L12:
  add    $0x1,%ebx
  cmp    $0x7531,%ebx
  je     L17
L31:
  test   $0x1,%bl
  je     L12
  mov    %ebx,%edx
  mov    %ebx,%ecx
  mov    %rdx,%rax
  imul   $0x24924925,%rdx,%rdx
  imul   %r12,%rax
  shr    $0x20,%rdx
  shr    $0x21,%rax
  lea    (%rax,%rax,2),%eax
  sub    %eax,%ecx
  mov    %ebx,%eax
  sub    %edx,%eax
  shr    %eax
  add    %edx,%eax
  shr    $0x2,%eax
  lea    0x0(,%rax,8),%edx
  sub    %eax,%edx
  mov    %ebx,%eax
  sub    %edx,%eax
  or     %eax,%ecx
  jne    L12
  mov    %ebx,%esi
  mov    %rbp,%rdi
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    $0x7531,%ebx
  jne    L31
L17:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D0
datatype: string

tool: D0: %d


assistant: ```c
int main() {
  int c;
  for (c = 0; c <= 30000; c++) {
    if (c % 2 != 0) {
      if ((c % 3 == 0) && (c % 7 == 0)) {
        printf("%d\n", c);
      }
    }
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %r12
  mov    $0xaaaaaaab,%r12d
  push   %rbp
  lea    D2(%rip),%rbp
  push   %rbx
  mov    $0x1,%ebx
  jmp    L44
  nopl   0x0(%rax)
L31:
  add    $0x1,%ebx
  cmp    $0x7531,%ebx
  je     L2
L44:
  test   $0x1,%bl
  je     L31
  mov    %ebx,%edx
  mov    %ebx,%ecx
  mov    %rdx,%rax
  imul   $0x24924925,%rdx,%rdx
  imul   %r12,%rax
  shr    $0x20,%rdx
  shr    $0x21,%rax
  lea    (%rax,%rax,2),%eax
  sub    %eax,%ecx
  mov    %ebx,%eax
  sub    %edx,%eax
  shr    %eax
  add    %edx,%eax
  shr    $0x2,%eax
  lea    0x0(,%rax,8),%edx
  sub    %eax,%edx
  mov    %ebx,%eax
  sub    %edx,%eax
  or     %eax,%ecx
  jne    L31
  mov    %ebx,%esi
  mov    %rbp,%rdi
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    $0x7531,%ebx
  jne    L44
L2:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D2
type_infered: string

tool: D2: %d


assistant: ```c
int main() {
  int c;
  for (c = 0; c <= 30000; c++) {
    if (c % 2 != 0) {
      if ((c % 3 == 0) && (c % 7 == 0)) {
        printf("%d\n", c);
      }
    }
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<append_to_string_with_size>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    -0x18(%rbp),%rdx
  add    %rdx,%rax
  lea    0x1(%rax),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <realloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L2
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x1,%edi
  call   <exit@plt>
L2:
  mov    -0x10(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcat@plt>
  mov    -0x8(%rbp),%rax
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D6
type_guessed: string

tool: D6: Out of Memory!! Fuuuuuuuu

assistant: ```c
char *append_to_string_with_size(char *destination, char *append_string, long long size) {
  destination = realloc(destination, strlen(destination) + size + 1);
  if (destination == NULL) {
    printf("Out of Memory!! Fuuuuuuuu");
    exit(1);
  }
  strcat(destination, append_string);
  return destination;
}
```