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

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D15(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x20(%rbp),%rax
  mov    %rax,%rsi
  lea    D14(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x20(%rbp),%edx
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  mov    %eax,-0x1c(%rbp)
  mov    -0x1c(%rbp),%ecx
  mov    %ecx,%eax
  shl    $0x2,%eax
  add    %ecx,%eax
  add    %eax,%eax
  sub    %eax,%edx
  mov    %edx,-0x1c(%rbp)
  mov    -0x20(%rbp),%edx
  movslq %edx,%rax
  imul   $0x51eb851f,%rax,%rax
  shr    $0x20,%rax
  sar    $0x5,%eax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  imul   $0x64,%eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x18(%rbp)
  mov    -0x20(%rbp),%edx
  movslq %edx,%rax
  imul   $0x10624dd3,%rax,%rax
  shr    $0x20,%rax
  sar    $0x6,%eax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  imul   $0x3e8,%eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  movslq %eax,%rdx
  imul   $0x51eb851f,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x5,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x20(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x10624dd3,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x6,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x1c(%rbp),%edx
  mov    -0x18(%rbp),%eax
  add    %eax,%edx
  mov    -0x14(%rbp),%eax
  add    %eax,%edx
  mov    -0x10(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x20(%rbp),%eax
  mov    -0xc(%rbp),%edx
  mov    %eax,%esi
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L6
  call   <__stack_chk_fail@plt>
L6:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D15
type_infered: string
name: disassemble_data
arguments: data_tag: D14
type_infered: string
name: disassemble_data
arguments: data_tag: D8
type_infered: string

tool: D15: Insira o numero inicial de 4 algarismos: 

tool: D14: %i

tool: D8: A soma dos algarismos %i eh %i

assistant: ```c
void main(void) {
  int numInicial, numSomado, num1, num2, num3, num4;
  printf("Insira o numero inicial de 4 algarismos: ");
  scanf("%i", &numInicial);
  num1 = numInicial % 10;
  num2 = numInicial % 100 / 10;
  num3 = numInicial % 1000 / 100;
  num4 = numInicial / 1000;
  numSomado = num1 + num2 + num3 + num4;
  printf("A soma dos algarismos %i eh %i", numInicial, numSomado);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D0(%rip),%rdi
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D18(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%esi
  movslq %esi,%rax
  imul   $0x51eb851f,%rax,%rdx
  sar    $0x25,%rdx
  mov    %esi,%edi
  sar    $0x1f,%edi
  sub    %edi,%edx
  imul   $0x64,%edx,%edx
  mov    %esi,%ecx
  sub    %edx,%ecx
  movslq %ecx,%rdx
  imul   $0x66666667,%rdx,%rdx
  sar    $0x22,%rdx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  imul   $0x66666667,%rax,%rcx
  sar    $0x22,%rcx
  sub    %edi,%ecx
  lea    (%rcx,%rcx,4),%ecx
  add    %ecx,%ecx
  mov    %esi,%r8d
  sub    %ecx,%r8d
  add    %r8d,%edx
  imul   $0x10624dd3,%rax,%rax
  sar    $0x26,%rax
  sub    %edi,%eax
  imul   $0x3e8,%eax,%ecx
  mov    %esi,%edi
  sub    %ecx,%edi
  movslq %edi,%rcx
  imul   $0x51eb851f,%rcx,%rcx
  sar    $0x25,%rcx
  sar    $0x1f,%edi
  sub    %edi,%ecx
  add    %ecx,%edx
  add    %eax,%edx
  lea    D27(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x18,%rsp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D0
datatype: string
name: parse_data
arguments: data_tag: D18
datatype: string
name: parse_data
arguments: data_tag: D27
datatype: string

tool: D0: Insira o numero inicial de 4 algarismos: 

tool: D18: %i

tool: D27: A soma dos algarismos %i eh %i

assistant: ```c
void main(void) {
  int numInicial, numSomado, num1, num2, num3, num4;
  printf("Insira o numero inicial de 4 algarismos: ");
  scanf("%i", &numInicial);
  num1 = numInicial % 10;
  num2 = numInicial % 100 / 10;
  num3 = numInicial % 1000 / 100;
  num4 = numInicial / 1000;
  numSomado = num1 + num2 + num3 + num4;
  printf("A soma dos algarismos %i eh %i", numInicial, numSomado);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D2(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D6(%rip),%rdi
  call   <scanf@plt>
  movslq 0x4(%rsp),%rax
  imul   $0x51eb851f,%rax,%rdx
  mov    %eax,%edi
  mov    %eax,%ecx
  mov    %rax,%rsi
  sar    $0x1f,%edi
  mov    %eax,%r8d
  sar    $0x25,%rdx
  sub    %edi,%edx
  imul   $0x64,%edx,%edx
  sub    %edx,%ecx
  movslq %ecx,%rdx
  sar    $0x1f,%ecx
  imul   $0x66666667,%rdx,%rdx
  sar    $0x22,%rdx
  sub    %ecx,%edx
  imul   $0x66666667,%rax,%rcx
  imul   $0x10624dd3,%rax,%rax
  sar    $0x22,%rcx
  sub    %edi,%ecx
  sar    $0x26,%rax
  lea    (%rcx,%rcx,4),%ecx
  sub    %edi,%eax
  mov    %esi,%edi
  add    %ecx,%ecx
  sub    %ecx,%r8d
  imul   $0x3e8,%eax,%ecx
  add    %r8d,%edx
  sub    %ecx,%edi
  movslq %edi,%rcx
  sar    $0x1f,%edi
  imul   $0x51eb851f,%rcx,%rcx
  sar    $0x25,%rcx
  sub    %edi,%ecx
  lea    D27(%rip),%rdi
  add    %ecx,%edx
  add    %eax,%edx
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  add    $0x18,%rsp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D2
type_guessed: string
name: disassemble_data
arguments: data_label: D6
type_guessed: string
name: disassemble_data
arguments: data_label: D27
type_guessed: string

tool: D2: Insira o numero inicial de 4 algarismos: 

tool: D6: %i

tool: D27: A soma dos algarismos %i eh %i

assistant: ```c
void main(void) {
  int numInicial, numSomado, num1, num2, num3, num4;
  printf("Insira o numero inicial de 4 algarismos: ");
  scanf("%i", &numInicial);
  num1 = numInicial % 10;
  num2 = numInicial % 100 / 10;
  num3 = numInicial % 1000 / 100;
  num4 = numInicial / 1000;
  numSomado = num1 + num2 + num3 + num4;
  printf("A soma dos algarismos %i eh %i", numInicial, numSomado);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D7(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D25(%rip),%rdi
  call   <scanf@plt>
  movslq 0x4(%rsp),%rax
  imul   $0x51eb851f,%rax,%rdx
  mov    %eax,%edi
  mov    %eax,%ecx
  mov    %rax,%rsi
  sar    $0x1f,%edi
  mov    %eax,%r8d
  sar    $0x25,%rdx
  sub    %edi,%edx
  imul   $0x64,%edx,%edx
  sub    %edx,%ecx
  movslq %ecx,%rdx
  sar    $0x1f,%ecx
  imul   $0x66666667,%rdx,%rdx
  sar    $0x22,%rdx
  sub    %ecx,%edx
  imul   $0x66666667,%rax,%rcx
  imul   $0x10624dd3,%rax,%rax
  sar    $0x22,%rcx
  sub    %edi,%ecx
  sar    $0x26,%rax
  lea    (%rcx,%rcx,4),%ecx
  sub    %edi,%eax
  mov    %esi,%edi
  add    %ecx,%ecx
  sub    %ecx,%r8d
  imul   $0x3e8,%eax,%ecx
  add    %r8d,%edx
  sub    %ecx,%edi
  movslq %edi,%rcx
  sar    $0x1f,%edi
  imul   $0x51eb851f,%rcx,%rcx
  sar    $0x25,%rcx
  sub    %edi,%ecx
  lea    D2(%rip),%rdi
  add    %ecx,%edx
  add    %eax,%edx
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x18,%rsp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D7
type_infered: string
name: retrieve_data
arguments: data_tag: D25
type_infered: string
name: retrieve_data
arguments: data_tag: D2
type_infered: string

tool: D7: Insira o numero inicial de 4 algarismos: 

tool: D25: %i

tool: D2: A soma dos algarismos %i eh %i

assistant: ```c
void main(void) {
  int numInicial, numSomado, num1, num2, num3, num4;
  printf("Insira o numero inicial de 4 algarismos: ");
  scanf("%i", &numInicial);
  num1 = numInicial % 10;
  num2 = numInicial % 100 / 10;
  num3 = numInicial % 1000 / 100;
  num4 = numInicial / 1000;
  numSomado = num1 + num2 + num3 + num4;
  printf("A soma dos algarismos %i eh %i", numInicial, numSomado);
}
```
user: Translate the below assembly code into C programming language:

```asm
<memcmp>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  mov    %rdi,-0x10(%rbp)
  mov    %rsi,-0x18(%rbp)
  mov    %edx,-0x1c(%rbp)
  mov    $0x0,%eax
  mov    -0x10(%rbp),%rdx
  mov    -0x18(%rbp),%rsi
  mov    -0x1c(%rbp),%ecx
  mov    %rdx,%rdi
  cld
  repz cmpsb %es:(%rdi),%ds:(%rsi)
  je     L0
  mov    $0x1,%eax
  jl     L0
  neg    %eax
L0:
  mov    %eax,%ebx
  mov    %ebx,%eax
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int memcmp(const void *cs, const void *ct, int count) {
  register int __res;
  __asm__("cld\n\t"
          "repe\n\t"
          "cmpsb\n\t"
          "je 1f\n\t"
          "movl $1,%%eax\n\t"
          "jl 1f\n\t"
          "negl %%eax\n"
          "1:"
          : "=a"(__res)
          : "0"(0), "D"(cs), "S"(ct), "c"(count));
  return __res;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<memcmp>:
  endbr64
  mov    $0x0,%eax
  mov    %edx,%ecx
  cld
  repz cmpsb %es:(%rdi),%ds:(%rsi)
  je     L9
  mov    $0x1,%eax
  jl     L9
  neg    %eax
L9:
  ret
```

assistant: ```c
int memcmp(const void *cs, const void *ct, int count) {
  register int __res;
  __asm__("cld\n\t"
          "repe\n\t"
          "cmpsb\n\t"
          "je 1f\n\t"
          "movl $1,%%eax\n\t"
          "jl 1f\n\t"
          "negl %%eax\n"
          "1:"
          : "=a"(__res)
          : "0"(0), "D"(cs), "S"(ct), "c"(count));
  return __res;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<memcmp>:
  endbr64
  xor    %eax,%eax
  mov    %edx,%ecx
  cld
  repz cmpsb %es:(%rdi),%ds:(%rsi)
  je     L15
  mov    $0x1,%eax
  jl     L15
  neg    %eax
L15:
  ret
```

assistant: ```c
int memcmp(const void *cs, const void *ct, int count) {
  register int __res;
  __asm__("cld\n\t"
          "repe\n\t"
          "cmpsb\n\t"
          "je 1f\n\t"
          "movl $1,%%eax\n\t"
          "jl 1f\n\t"
          "negl %%eax\n"
          "1:"
          : "=a"(__res)
          : "0"(0), "D"(cs), "S"(ct), "c"(count));
  return __res;
}
```
user: What is the c source code of the assembly code below:

```asm
<memcmp>:
  endbr64
  xor    %eax,%eax
  mov    %edx,%ecx
  cld
  repz cmpsb %es:(%rdi),%ds:(%rsi)
  je     L0
  mov    $0x1,%eax
  jl     L0
  neg    %eax
L0:
  ret
```

assistant: ```c
int memcmp(const void *cs, const void *ct, int count) {
  register int __res;
  __asm__("cld\n\t"
          "repe\n\t"
          "cmpsb\n\t"
          "je 1f\n\t"
          "movl $1,%%eax\n\t"
          "jl 1f\n\t"
          "negl %%eax\n"
          "1:"
          : "=a"(__res)
          : "0"(0), "D"(cs), "S"(ct), "c"(count));
  return __res;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<precedGEQ>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,%edi
  call   <preced@plt>
  mov    %eax,-0x8(%rbp)
  mov    -0x18(%rbp),%eax
  mov    %eax,%edi
  call   <preced@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0x3,-0x8(%rbp)
  jne    L22
  cmpl   $0x2,-0x4(%rbp)
  jle    L22
  movl   $0x0,-0x8(%rbp)
L22:
  cmpl   $0x2,-0x8(%rbp)
  jne    L8
  cmpl   $0x1,-0x4(%rbp)
  jle    L8
  movl   $0x0,-0x8(%rbp)
L8:
  mov    -0x8(%rbp),%eax
  cmp    -0x4(%rbp),%eax
  setge  %al
  movzbl %al,%eax
  leave
  ret
```

assistant: ```c
int precedGEQ(int lfttok, int rhttok) {
  int pl = preced(lfttok);
  int pr = preced(rhttok);
  if (pl == 3 && pr >= 3)
    pl = 0;
  if (pl == 2 && pr >= 2)
    pl = 0;
  return pl >= pr;
}
```
user: What is the c source code of the assembly code below?

```asm
<precedGEQ>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %esi,%ebp
  call   <preced@plt>
  mov    %eax,%ebx
  mov    %ebp,%edi
  call   <preced@plt>
  cmp    $0x3,%ebx
  jne    L37
  cmp    $0x2,%eax
  jle    L37
  mov    $0x0,%ebx
  jmp    L10
L37:
  cmp    $0x2,%ebx
  jne    L10
  cmp    $0x1,%eax
  mov    $0x0,%edx
  cmovg  %edx,%ebx
L10:
  cmp    %eax,%ebx
  setge  %al
  movzbl %al,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int precedGEQ(int lfttok, int rhttok) {
  int pl = preced(lfttok);
  int pr = preced(rhttok);
  if (pl == 3 && pr >= 3)
    pl = 0;
  if (pl == 2 && pr >= 2)
    pl = 0;
  return pl >= pr;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<precedGEQ>:
  endbr64
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  sub    $0x8,%rsp
  call   <preced@plt>
  mov    %ebp,%edi
  mov    %eax,%ebx
  call   <preced@plt>
  cmp    $0x3,%ebx
  jne    L7
  cmp    $0x2,%eax
  jle    L7
  xor    %ebx,%ebx
L9:
  cmp    %eax,%ebx
  setge  %al
  add    $0x8,%rsp
  movzbl %al,%eax
  pop    %rbx
  pop    %rbp
  ret
  nopl   (%rax)
L7:
  cmp    $0x2,%ebx
  jne    L9
  xor    %edx,%edx
  cmp    $0x1,%eax
  cmovg  %edx,%ebx
  cmp    %eax,%ebx
  setge  %al
  add    $0x8,%rsp
  movzbl %al,%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int precedGEQ(int lfttok, int rhttok) {
  int pl = preced(lfttok);
  int pr = preced(rhttok);
  if (pl == 3 && pr >= 3)
    pl = 0;
  if (pl == 2 && pr >= 2)
    pl = 0;
  return pl >= pr;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<precedGEQ>:
  endbr64
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  sub    $0x8,%rsp
  call   <preced@plt>
  mov    %ebp,%edi
  mov    %eax,%ebx
  call   <preced@plt>
  cmp    $0x3,%ebx
  jne    L3
  cmp    $0x2,%eax
  jle    L3
  xor    %ebx,%ebx
L21:
  cmp    %eax,%ebx
  setge  %al
  add    $0x8,%rsp
  movzbl %al,%eax
  pop    %rbx
  pop    %rbp
  ret
  nopl   (%rax)
L3:
  cmp    $0x2,%ebx
  jne    L21
  xor    %edx,%edx
  cmp    $0x1,%eax
  cmovg  %edx,%ebx
  cmp    %eax,%ebx
  setge  %al
  add    $0x8,%rsp
  movzbl %al,%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int precedGEQ(int lfttok, int rhttok) {
  int pl = preced(lfttok);
  int pr = preced(rhttok);
  if (pl == 3 && pr >= 3)
    pl = 0;
  if (pl == 2 && pr >= 2)
    pl = 0;
  return pl >= pr;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<run_hooks>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  jmp    L1
L15:
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rax
  call   *%rax
L1:
  addq   $0x8,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rax
  test   %rax,%rax
  jne    L15
  nop
  nop
  leave
  ret
```

assistant: ```c
void run_hooks(void (*const list[])(void)) {
  while (*++list)
    (**list)();
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<run_hooks>:
  endbr64
  push   %rbx
  lea    0x8(%rdi),%rbx
  mov    0x8(%rdi),%rax
  test   %rax,%rax
  je     L19
L8:
  call   *%rax
  add    $0x8,%rbx
  mov    (%rbx),%rax
  test   %rax,%rax
  jne    L8
L19:
  pop    %rbx
  ret
```

assistant: ```c
void run_hooks(void (*const list[])(void)) {
  while (*++list)
    (**list)();
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<run_hooks>:
  endbr64
  mov    0x8(%rdi),%rax
  test   %rax,%rax
  je     L6
  push   %rbx
  lea    0x8(%rdi),%rbx
  nopw   0x0(%rax,%rax,1)
L9:
  call   *%rax
  mov    0x8(%rbx),%rax
  add    $0x8,%rbx
  test   %rax,%rax
  jne    L9
  pop    %rbx
  ret
  nopl   0x0(%rax)
L6:
  ret
```

assistant: ```c
void run_hooks(void (*const list[])(void)) {
  while (*++list)
    (**list)();
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<run_hooks>:
  endbr64
  mov    0x8(%rdi),%rax
  test   %rax,%rax
  je     L14
  push   %rbx
  lea    0x8(%rdi),%rbx
  nopw   0x0(%rax,%rax,1)
L11:
  call   *%rax
  mov    0x8(%rbx),%rax
  add    $0x8,%rbx
  test   %rax,%rax
  jne    L11
  pop    %rbx
  ret
  nopl   0x0(%rax)
L14:
  ret
```

assistant: ```c
void run_hooks(void (*const list[])(void)) {
  while (*++list)
    (**list)();
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<setCode>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  shr    $0xc,%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%rax
  add    $0x7,%rax
  mov    %dl,(%rax)
  mov    -0xc(%rbp),%eax
  shr    $0x4,%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%rax
  add    $0x8,%rax
  mov    %dl,(%rax)
  mov    -0x8(%rbp),%rax
  add    $0x9,%rax
  movzbl (%rax),%edx
  mov    -0x8(%rbp),%rax
  add    $0x9,%rax
  and    $0xf,%edx
  mov    %dl,(%rax)
  mov    -0x8(%rbp),%rax
  add    $0x9,%rax
  movzbl (%rax),%edx
  mov    -0xc(%rbp),%eax
  shl    $0x4,%eax
  mov    %eax,%ecx
  mov    -0x8(%rbp),%rax
  add    $0x9,%rax
  or     %ecx,%edx
  mov    %dl,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void setCode(unsigned char *packet, unsigned int code) {
  packet[7] = (code >> 12) & 0xff;
  packet[8] = (code >> 4) & 0xff;
  packet[9] &= 0xf;
  ;
  packet[9] |= ((code)&0xf) << 4;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<setCode>:
  endbr64
  mov    %esi,%eax
  shr    $0xc,%eax
  mov    %al,0x7(%rdi)
  mov    %esi,%eax
  shr    $0x4,%eax
  mov    %al,0x8(%rdi)
  movzbl 0x9(%rdi),%eax
  and    $0xf,%eax
  shl    $0x4,%esi
  or     %esi,%eax
  mov    %al,0x9(%rdi)
  ret
```

assistant: ```c
void setCode(unsigned char *packet, unsigned int code) {
  packet[7] = (code >> 12) & 0xff;
  packet[8] = (code >> 4) & 0xff;
  packet[9] &= 0xf;
  ;
  packet[9] |= ((code)&0xf) << 4;
}
```
user: Translate the below assembly code into C programming language:

```asm
<setCode>:
  endbr64
  mov    %esi,%eax
  shr    $0xc,%eax
  mov    %al,0x7(%rdi)
  mov    %esi,%eax
  shl    $0x4,%esi
  shr    $0x4,%eax
  mov    %al,0x8(%rdi)
  movzbl 0x9(%rdi),%eax
  and    $0xf,%eax
  or     %esi,%eax
  mov    %al,0x9(%rdi)
  ret
```

assistant: ```c
void setCode(unsigned char *packet, unsigned int code) {
  packet[7] = (code >> 12) & 0xff;
  packet[8] = (code >> 4) & 0xff;
  packet[9] &= 0xf;
  ;
  packet[9] |= ((code)&0xf) << 4;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<cond_timedwait_and_unlock_if_cancelled>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    D9(%rip),%rax
  mov    (%rax),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x20(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <pthread_cleanup_push@plt>
  mov    -0x28(%rbp),%rdx
  mov    -0x20(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <pthread_cond_timedwait@plt>
  mov    %eax,-0xc(%rbp)
  mov    $0x0,%edi
  call   <pthread_cleanup_pop@plt>
  mov    -0xc(%rbp),%eax
  leave
  ret
```

assistant: ```c
int cond_timedwait_and_unlock_if_cancelled(pthread_cond_t *cond, pthread_mutex_t *mutex, struct timespec *abstime) {
  typedef void (*routine)(void *);
  routine r = (routine)pthread_mutex_unlock;
  int err;
  pthread_cleanup_push(r, (void *)mutex);
  err = pthread_cond_timedwait(cond, mutex, abstime);
  pthread_cleanup_pop(0);
  return err;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<cond_timedwait_and_unlock_if_cancelled>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    %rsi,%rbx
  mov    %rdx,%r12
  mov    D7(%rip),%rax
  mov    (%rax),%rdi
  call   <pthread_cleanup_push@plt>
  mov    %r12,%rdx
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <pthread_cond_timedwait@plt>
  mov    %eax,%ebx
  mov    $0x0,%edi
  call   <pthread_cleanup_pop@plt>
  mov    %ebx,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int cond_timedwait_and_unlock_if_cancelled(pthread_cond_t *cond, pthread_mutex_t *mutex, struct timespec *abstime) {
  typedef void (*routine)(void *);
  routine r = (routine)pthread_mutex_unlock;
  int err;
  pthread_cleanup_push(r, (void *)mutex);
  err = pthread_cond_timedwait(cond, mutex, abstime);
  pthread_cleanup_pop(0);
  return err;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<cond_timedwait_and_unlock_if_cancelled>:
  endbr64
  mov    D6(%rip),%rax
  push   %r13
  mov    %rdx,%r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    (%rax),%rdi
  mov    %rsi,%rbp
  call   <pthread_cleanup_push@plt>
  mov    %r13,%rdx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <pthread_cond_timedwait@plt>
  xor    %edi,%edi
  mov    %eax,%r12d
  call   <pthread_cleanup_pop@plt>
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int cond_timedwait_and_unlock_if_cancelled(pthread_cond_t *cond, pthread_mutex_t *mutex, struct timespec *abstime) {
  typedef void (*routine)(void *);
  routine r = (routine)pthread_mutex_unlock;
  int err;
  pthread_cleanup_push(r, (void *)mutex);
  err = pthread_cond_timedwait(cond, mutex, abstime);
  pthread_cleanup_pop(0);
  return err;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<cond_timedwait_and_unlock_if_cancelled>:
  endbr64
  mov    D2(%rip),%rax
  push   %r13
  mov    %rdx,%r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    (%rax),%rdi
  mov    %rsi,%rbp
  call   <pthread_cleanup_push@plt>
  mov    %r13,%rdx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <pthread_cond_timedwait@plt>
  xor    %edi,%edi
  mov    %eax,%r12d
  call   <pthread_cleanup_pop@plt>
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int cond_timedwait_and_unlock_if_cancelled(pthread_cond_t *cond, pthread_mutex_t *mutex, struct timespec *abstime) {
  typedef void (*routine)(void *);
  routine r = (routine)pthread_mutex_unlock;
  int err;
  pthread_cleanup_push(r, (void *)mutex);
  err = pthread_cond_timedwait(cond, mutex, abstime);
  pthread_cleanup_pop(0);
  return err;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main02>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D6(%rip),%rax
  mov    %rax,%rsi
  lea    D13(%rip),%rax
  mov    %rax,%rdi
  call   <fopen@plt>
  mov    %rax,-0x10(%rbp)
  movl   $0x12345678,-0x14(%rbp)
  mov    -0x10(%rbp),%rdx
  lea    -0x14(%rbp),%rax
  mov    %rdx,%rcx
  mov    $0x4,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <fwrite@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <fclose@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L5
  call   <__stack_chk_fail@plt>
L5:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D6
type: string
name: extract_symbolic_data
arguments: tag: D13
type: string

tool: D6: wb

tool: D13: D:\temp\a.dat

assistant: ```c
int main02() {
  FILE *p = fopen("D:\\temp\\a.dat", "wb");
  int a = 0x12345678;
  fwrite(&a, 0, 4, p);
  fclose(p);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main02>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D15(%rip),%rsi
  lea    D16(%rip),%rdi
  call   <fopen@plt>
  mov    %rax,%rbx
  movl   $0x12345678,0x4(%rsp)
  lea    0x4(%rsp),%rdi
  mov    %rax,%rcx
  mov    $0x4,%edx
  mov    $0x0,%esi
  call   <fwrite@plt>
  mov    %rbx,%rdi
  call   <fclose@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  mov    $0x0,%eax
  add    $0x10,%rsp
  pop    %rbx
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D15
type_guessed: string
name: parse_data
arguments: data_label: D16
type_guessed: string

tool: D15: wb

tool: D16: D:\temp\a.dat

assistant: ```c
int main02() {
  FILE *p = fopen("D:\\temp\\a.dat", "wb");
  int a = 0x12345678;
  fwrite(&a, 0, 4, p);
  fclose(p);
  return 0;
}
```