ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x17,-0x2c(%rbp)
  movl   $0x5,-0x28(%rbp)
  movl   $0x5,-0x20(%rbp)
  movl   $0xa,-0x1c(%rbp)
  movl   $0xf,-0x18(%rbp)
  movl   $0x14,-0x14(%rbp)
  movl   $0x17,-0x10(%rbp)
  lea    -0x20(%rbp),%rdx
  mov    -0x28(%rbp),%ecx
  mov    -0x2c(%rbp),%eax
  mov    %ecx,%esi
  mov    %eax,%edi
  call   <buscaO@plt>
  mov    %eax,-0x24(%rbp)
  mov    -0x24(%rbp),%eax
  mov    %eax,%esi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L0
  call   <__stack_chk_fail@plt>
L0:
  leave
  ret
```

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

tool: D7: Index: %d

assistant: ```c
int main(void) {
  int elem = 23;
  int tam = 5;
  int vetor[] = {5, 10, 15, 20, 23};
  int result = buscaO(elem, tam, vetor);
  printf("Index: %d", result);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movl   $0x5,(%rsp)
  movl   $0xa,0x4(%rsp)
  movl   $0xf,0x8(%rsp)
  movl   $0x14,0xc(%rsp)
  movl   $0x17,0x10(%rsp)
  mov    %rsp,%rdx
  mov    $0x5,%esi
  mov    $0x17,%edi
  call   <buscaO@plt>
  mov    %eax,%esi
  lea    D8(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  mov    $0x0,%eax
  add    $0x28,%rsp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D8
data_type: string

tool: D8: Index: %d

assistant: ```c
int main(void) {
  int elem = 23;
  int tam = 5;
  int vetor[] = {5, 10, 15, 20, 23};
  int result = buscaO(elem, tam, vetor);
  printf("Index: %d", result);
  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    $0x28,%rsp
  mov    $0x5,%esi
  mov    $0x17,%edi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  movabs $0xa00000005,%rax
  mov    %rsp,%rdx
  movl   $0x17,0x10(%rsp)
  mov    %rax,(%rsp)
  movabs $0x140000000f,%rax
  mov    %rax,0x8(%rsp)
  call   <buscaO@plt>
  lea    D9(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  xor    %eax,%eax
  add    $0x28,%rsp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D9
datatype: string

tool: D9: Index: %d

assistant: ```c
int main(void) {
  int elem = 23;
  int tam = 5;
  int vetor[] = {5, 10, 15, 20, 23};
  int result = buscaO(elem, tam, vetor);
  printf("Index: %d", result);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  mov    $0x5,%esi
  mov    $0x17,%edi
  movdqa D2(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdx
  movl   $0x17,0x10(%rsp)
  movaps %xmm0,(%rsp)
  call   <buscaO@plt>
  lea    D0(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  xor    %eax,%eax
  add    $0x28,%rsp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D2
type_infered: byte[16]
name: retrieve_data
arguments: data_label: D0
type_infered: string

tool: {"D2": [0x05,0x00,0x00,0x00,0x0a,0x00,0x00,0x00,0x0f,0x00,0x00,0x00,0x14,0x00,0x00,0x00]}

tool: D0: Index: %d

assistant: ```c
int main(void) {
  int elem = 23;
  int tam = 5;
  int vetor[] = {5, 10, 15, 20, 23};
  int result = buscaO(elem, tam, vetor);
  printf("Index: %d", result);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  movl   $0x0,-0x4(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L18
L9:
  mov    -0x20(%rbp),%rax
  mov    (%rax),%rdx
  mov    -0x4(%rbp),%eax
  cltq
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x4(%rbp)
L18:
  mov    -0x4(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L9
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D18
datatype: string
name: retrieve_data
arguments: data_label: D8
datatype: string

tool: D18: show argv:

tool: D8: argv + %d = [%s]


assistant: ```c
int main(int argc, char *argv[]) {
  printf("show argv:\n");
  int i = 0;
  for (i = 0; i < argc; i++) {
    printf("argv + %d = [%s]\n", i, argv[0] + i);
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebp
  mov    %rsi,%r12
  lea    D16(%rip),%rdi
  call   <puts@plt>
  test   %ebp,%ebp
  jle    L19
  mov    %ebp,%ebp
  mov    $0x0,%ebx
  lea    D15(%rip),%r13
L16:
  mov    %rbx,%rdx
  add    (%r12),%rdx
  mov    %ebx,%esi
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x1,%rbx
  cmp    %rbp,%rbx
  jne    L16
L19:
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: disassemble_data
arguments: label: D16
type_infered: string
name: disassemble_data
arguments: label: D15
type_infered: string

tool: D16: show argv:

tool: D15: argv + %d = [%s]


assistant: ```c
int main(int argc, char *argv[]) {
  printf("show argv:\n");
  int i = 0;
  for (i = 0; i < argc; i++) {
    printf("argv + %d = [%s]\n", i, argv[0] + i);
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  movslq %edi,%rbp
  lea    D4(%rip),%rdi
  push   %rbx
  sub    $0x8,%rsp
  call   <puts@plt>
  test   %ebp,%ebp
  jle    L18
  xor    %ebx,%ebx
  lea    D19(%rip),%r13
  nopl   (%rax)
L14:
  mov    (%r12),%rdx
  mov    %ebx,%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  add    %rbx,%rdx
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    %rbp,%rbx
  jne    L14
L18:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: parse_data
arguments: label: D4
datatype: string
name: parse_data
arguments: label: D19
datatype: string

tool: D4: show argv:

tool: D19: argv + %d = [%s]


assistant: ```c
int main(int argc, char *argv[]) {
  printf("show argv:\n");
  int i = 0;
  for (i = 0; i < argc; i++) {
    printf("argv + %d = [%s]\n", i, argv[0] + i);
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  movslq %edi,%rbp
  lea    D19(%rip),%rdi
  push   %rbx
  sub    $0x8,%rsp
  call   <puts@plt>
  test   %ebp,%ebp
  jle    L1
  xor    %ebx,%ebx
  lea    D2(%rip),%r13
  nopl   (%rax)
L15:
  mov    (%r12),%rdx
  mov    %ebx,%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  add    %rbx,%rdx
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    %rbp,%rbx
  jne    L15
L1:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D19
datatype: string
name: retrieve_data
arguments: tag: D2
datatype: string

tool: D19: show argv:

tool: D2: argv + %d = [%s]


assistant: ```c
int main(int argc, char *argv[]) {
  printf("show argv:\n");
  int i = 0;
  for (i = 0; i < argc; i++) {
    printf("argv + %d = [%s]\n", i, argv[0] + i);
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ascii_to_integer>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L20
L38:
  mov    -0x4(%rbp),%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  sub    $0x30,%eax
  add    %eax,-0x4(%rbp)
L20:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x2f,%al
  jle    L4
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x39,%al
  jle    L38
L4:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L35
  movl   $0x0,-0x4(%rbp)
L35:
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int ascii_to_integer(char *string) {
  int value;
  int digit;
  value = 0;
  while (*string >= '0' && *string <= '9') {
    value *= 10;
    value += *string - '0';
  }
  if (*string != '\0')
    value = 0;
  return value;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ascii_to_integer>:
  endbr64
  movzbl (%rdi),%eax
  sub    $0x30,%eax
  cmp    $0x9,%al
  ja     L8
L7:
  jmp    L7
L8:
  mov    $0x0,%eax
  ret
```

assistant: ```c
int ascii_to_integer(char *string) {
  int value;
  int digit;
  value = 0;
  while (*string >= '0' && *string <= '9') {
    value *= 10;
    value += *string - '0';
  }
  if (*string != '\0')
    value = 0;
  return value;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<ascii_to_integer>:
  endbr64
  movzbl (%rdi),%eax
  sub    $0x30,%eax
  cmp    $0x9,%al
  ja     L9
L3:
  jmp    L3
L9:
  xor    %eax,%eax
  ret
```

assistant: ```c
int ascii_to_integer(char *string) {
  int value;
  int digit;
  value = 0;
  while (*string >= '0' && *string <= '9') {
    value *= 10;
    value += *string - '0';
  }
  if (*string != '\0')
    value = 0;
  return value;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<ascii_to_integer>:
  endbr64
  movzbl (%rdi),%eax
  sub    $0x30,%eax
  cmp    $0x9,%al
  ja     L18
L19:
  jmp    L19
L18:
  xor    %eax,%eax
  ret
```

assistant: ```c
int ascii_to_integer(char *string) {
  int value;
  int digit;
  value = 0;
  while (*string >= '0' && *string <= '9') {
    value *= 10;
    value += *string - '0';
  }
  if (*string != '\0')
    value = 0;
  return value;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<strcpy>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  jmp    L17
L0:
  mov    -0x10(%rbp),%rdx
  lea    0x1(%rdx),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rcx
  mov    %rcx,-0x8(%rbp)
  movzbl (%rdx),%edx
  mov    %dl,(%rax)
L17:
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L0
  mov    -0x8(%rbp),%rax
  movb   $0x0,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void strcpy(char *dst, const char *src) {
  while (*src != '\0')
    *dst++ = *src++;
  *dst = '\0';
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<strcpy>:
  endbr64
  movzbl (%rsi),%eax
  test   %al,%al
  je     L17
L1:
  add    $0x1,%rsi
  add    $0x1,%rdi
  mov    %al,-0x1(%rdi)
  movzbl (%rsi),%eax
  test   %al,%al
  jne    L1
L17:
  movb   $0x0,(%rdi)
  ret
```

assistant: ```c
void strcpy(char *dst, const char *src) {
  while (*src != '\0')
    *dst++ = *src++;
  *dst = '\0';
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<strcpy>:
  endbr64
  jmp    L14
  cs nopw 0x0(%rax,%rax,1)
L4:
  mov    %al,(%rdi)
  add    $0x1,%rdi
  add    $0x1,%rsi
L14:
  movzbl (%rsi),%eax
  test   %al,%al
  jne    L4
  movb   $0x0,(%rdi)
  ret
```

assistant: ```c
void strcpy(char *dst, const char *src) {
  while (*src != '\0')
    *dst++ = *src++;
  *dst = '\0';
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<strcpy>:
  endbr64
  jmp    L10
  cs nopw 0x0(%rax,%rax,1)
L8:
  mov    %al,(%rdi)
  add    $0x1,%rdi
  add    $0x1,%rsi
L10:
  movzbl (%rsi),%eax
  test   %al,%al
  jne    L8
  movb   $0x0,(%rdi)
  ret
```

assistant: ```c
void strcpy(char *dst, const char *src) {
  while (*src != '\0')
    *dst++ = *src++;
  *dst = '\0';
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<quicksort>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    -0x20(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jle    L7
  mov    -0x20(%rbp),%edx
  mov    -0x1c(%rbp),%ecx
  mov    -0x18(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <partition@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    -0x1c(%rbp),%ecx
  mov    -0x18(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <quicksort>
  mov    -0x4(%rbp),%eax
  lea    0x1(%rax),%ecx
  mov    -0x20(%rbp),%edx
  mov    -0x18(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <quicksort>
L7:
  nop
  leave
  ret
```

assistant: ```c
void quicksort(long int x[], int low, int high) {
  int pivot, i;
  if (high > low) {
    pivot = partition(x, low, high);
    quicksort(x, low, pivot - 1);
    quicksort(x, pivot + 1, high);
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<quicksort>:
  endbr64
  cmp    %esi,%edx
  jg     L9
  ret
L9:
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r12
  mov    %esi,%r13d
  mov    %edx,%ebx
  call   <partition@plt>
  mov    %eax,%ebp
  lea    -0x1(%rax),%edx
  mov    %r13d,%esi
  mov    %r12,%rdi
  call   <quicksort>
  lea    0x1(%rbp),%esi
  mov    %ebx,%edx
  mov    %r12,%rdi
  call   <quicksort>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
void quicksort(long int x[], int low, int high) {
  int pivot, i;
  if (high > low) {
    pivot = partition(x, low, high);
    quicksort(x, low, pivot - 1);
    quicksort(x, pivot + 1, high);
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<ui2hex>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,%eax
  mov    %al,-0x4(%rbp)
  movzbl -0x4(%rbp),%eax
  cmp    $0xf,%eax
  ja     L19
  mov    %eax,%eax
  lea    0x0(,%rax,4),%rdx
  lea    D9(%rip),%rax
  mov    (%rdx,%rax,1),%eax
  cltq
  lea    D9(%rip),%rdx
  add    %rdx,%rax
  notrack jmp *%rax
  mov    $0x30,%eax
  jmp    L29
  mov    $0x31,%eax
  jmp    L29
  mov    $0x32,%eax
  jmp    L29
  mov    $0x33,%eax
  jmp    L29
  mov    $0x34,%eax
  jmp    L29
  mov    $0x35,%eax
  jmp    L29
  mov    $0x36,%eax
  jmp    L29
  mov    $0x37,%eax
  jmp    L29
  mov    $0x38,%eax
  jmp    L29
  mov    $0x39,%eax
  jmp    L29
  mov    $0x41,%eax
  jmp    L29
  mov    $0x42,%eax
  jmp    L29
  mov    $0x43,%eax
  jmp    L29
  mov    $0x44,%eax
  jmp    L29
  mov    $0x45,%eax
  jmp    L29
  mov    $0x46,%eax
  jmp    L29
L19:
  mov    $0x20,%eax
L29:
  pop    %rbp
  ret
```

assistant: ```c
char ui2hex(unsigned char ui) {
  switch (ui) {
  case 0x0: {
    return ('0');
  }
  case 0x1: {
    return ('1');
  }
  case 0x2: {
    return ('2');
  }
  case 0x3: {
    return ('3');
  }
  case 0x4: {
    return ('4');
  }
  case 0x5: {
    return ('5');
  }
  case 0x6: {
    return ('6');
  }
  case 0x7: {
    return ('7');
  }
  case 0x8: {
    return ('8');
  }
  case 0x9: {
    return ('9');
  }
  case 0xa: {
    return ('A');
  }
  case 0xb: {
    return ('B');
  }
  case 0xc: {
    return ('C');
  }
  case 0xd: {
    return ('D');
  }
  case 0xe: {
    return ('E');
  }
  case 0xf: {
    return ('F');
  }
  }
  return (' ');
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<ui2hex>:
  endbr64
  cmp    $0xf,%dil
  ja     L0
  movzbl %dil,%edi
  lea    D8(%rip),%rdx
  movslq (%rdx,%rdi,4),%rax
  add    %rdx,%rax
  notrack jmp *%rax
  mov    $0x31,%eax
  ret
  mov    $0x32,%eax
  ret
  mov    $0x33,%eax
  ret
  mov    $0x34,%eax
  ret
  mov    $0x35,%eax
  ret
  mov    $0x36,%eax
  ret
  mov    $0x37,%eax
  ret
  mov    $0x38,%eax
  ret
  mov    $0x39,%eax
  ret
  mov    $0x41,%eax
  ret
  mov    $0x42,%eax
  ret
  mov    $0x43,%eax
  ret
  mov    $0x44,%eax
  ret
  mov    $0x45,%eax
  ret
  mov    $0x46,%eax
  ret
L0:
  mov    $0x20,%eax
  ret
  mov    $0x30,%eax
  ret
```

assistant: ```c
char ui2hex(unsigned char ui) {
  switch (ui) {
  case 0x0: {
    return ('0');
  }
  case 0x1: {
    return ('1');
  }
  case 0x2: {
    return ('2');
  }
  case 0x3: {
    return ('3');
  }
  case 0x4: {
    return ('4');
  }
  case 0x5: {
    return ('5');
  }
  case 0x6: {
    return ('6');
  }
  case 0x7: {
    return ('7');
  }
  case 0x8: {
    return ('8');
  }
  case 0x9: {
    return ('9');
  }
  case 0xa: {
    return ('A');
  }
  case 0xb: {
    return ('B');
  }
  case 0xc: {
    return ('C');
  }
  case 0xd: {
    return ('D');
  }
  case 0xe: {
    return ('E');
  }
  case 0xf: {
    return ('F');
  }
  }
  return (' ');
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<ui2hex>:
  endbr64
  mov    $0x20,%eax
  cmp    $0xf,%dil
  ja     L9
  movzbl %dil,%edi
  lea    D0(%rip),%rax
  movzbl (%rax,%rdi,1),%eax
L9:
  ret
```

assistant: ```c
char ui2hex(unsigned char ui) {
  switch (ui) {
  case 0x0: {
    return ('0');
  }
  case 0x1: {
    return ('1');
  }
  case 0x2: {
    return ('2');
  }
  case 0x3: {
    return ('3');
  }
  case 0x4: {
    return ('4');
  }
  case 0x5: {
    return ('5');
  }
  case 0x6: {
    return ('6');
  }
  case 0x7: {
    return ('7');
  }
  case 0x8: {
    return ('8');
  }
  case 0x9: {
    return ('9');
  }
  case 0xa: {
    return ('A');
  }
  case 0xb: {
    return ('B');
  }
  case 0xc: {
    return ('C');
  }
  case 0xd: {
    return ('D');
  }
  case 0xe: {
    return ('E');
  }
  case 0xf: {
    return ('F');
  }
  }
  return (' ');
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<ui2hex>:
  endbr64
  mov    $0x20,%eax
  cmp    $0xf,%dil
  ja     L2
  movzbl %dil,%edi
  lea    D6(%rip),%rax
  movzbl (%rax,%rdi,1),%eax
L2:
  ret
```

assistant: ```c
char ui2hex(unsigned char ui) {
  switch (ui) {
  case 0x0: {
    return ('0');
  }
  case 0x1: {
    return ('1');
  }
  case 0x2: {
    return ('2');
  }
  case 0x3: {
    return ('3');
  }
  case 0x4: {
    return ('4');
  }
  case 0x5: {
    return ('5');
  }
  case 0x6: {
    return ('6');
  }
  case 0x7: {
    return ('7');
  }
  case 0x8: {
    return ('8');
  }
  case 0x9: {
    return ('9');
  }
  case 0xa: {
    return ('A');
  }
  case 0xb: {
    return ('B');
  }
  case 0xc: {
    return ('C');
  }
  case 0xd: {
    return ('D');
  }
  case 0xe: {
    return ('E');
  }
  case 0xf: {
    return ('F');
  }
  }
  return (' ');
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<P>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0x6,-0x30(%rbp)
  movq   $0x2,-0x28(%rbp)
  movq   $0x4,-0x20(%rbp)
  movq   $0x8,-0x18(%rbp)
  cmpq   $0x0,-0x38(%rbp)
  jg     L12
  mov    $0x1,%eax
  jmp    L11
L12:
  mov    -0x38(%rbp),%rax
  cqto
  shr    $0x3e,%rdx
  add    %rdx,%rax
  and    $0x3,%eax
  sub    %rdx,%rax
  mov    -0x30(%rbp,%rax,8),%rax
L11:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L1
  call   <__stack_chk_fail@plt>
L1:
  leave
  ret
```

assistant: ```c
long long int P(long long int k) {
  long long int arr[] = {6, 2, 4, 8};
  if (k < 1)
    return 1;
  else
    return arr[k % 4];
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<P>:
  endbr64
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  movq   $0x6,(%rsp)
  movq   $0x2,0x8(%rsp)
  movq   $0x4,0x10(%rsp)
  movq   $0x8,0x18(%rsp)
  mov    $0x1,%eax
  test   %rdi,%rdi
  jle    L18
  mov    %rdi,%rax
  sar    $0x3f,%rax
  shr    $0x3e,%rax
  add    %rax,%rdi
  and    $0x3,%edi
  sub    %rax,%rdi
  mov    (%rsp,%rdi,8),%rax
L18:
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L15
  add    $0x38,%rsp
  ret
L15:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
long long int P(long long int k) {
  long long int arr[] = {6, 2, 4, 8};
  if (k < 1)
    return 1;
  else
    return arr[k % 4];
}
```
user: What is the c source code of the assembly code below:

```asm
<P>:
  endbr64
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  movq   $0x6,(%rsp)
  mov    $0x1,%eax
  movq   $0x2,0x8(%rsp)
  movq   $0x4,0x10(%rsp)
  movq   $0x8,0x18(%rsp)
  test   %rdi,%rdi
  jle    L5
  and    $0x3,%edi
  mov    (%rsp,%rdi,8),%rax
L5:
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L8
  add    $0x38,%rsp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
long long int P(long long int k) {
  long long int arr[] = {6, 2, 4, 8};
  if (k < 1)
    return 1;
  else
    return arr[k % 4];
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<P>:
  endbr64
  sub    $0x38,%rsp
  movdqa D15(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    $0x1,%eax
  movaps %xmm0,(%rsp)
  movdqa D11(%rip),%xmm0
  movaps %xmm0,0x10(%rsp)
  test   %rdi,%rdi
  jle    L17
  and    $0x3,%edi
  mov    (%rsp,%rdi,8),%rax
L17:
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L11
  add    $0x38,%rsp
  ret
L11:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D15
type_infered: byte[16]
name: read_data
arguments: tag: D11
type_infered: byte[16]

tool: {"D15": [0x06,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x02,0x00,0x00,0x00,0x00,0x00,0x00,0x00]}

tool: {"D11": [0x04,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x08,0x00,0x00,0x00,0x00,0x00,0x00,0x00]}

assistant: ```c
long long int P(long long int k) {
  long long int arr[] = {6, 2, 4, 8};
  if (k < 1)
    return 1;
  else
    return arr[k % 4];
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<_vRcSLE4442_Write>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,%eax
  mov    %esi,-0x18(%rbp)
  mov    %rdx,-0x20(%rbp)
  mov    %al,-0x14(%rbp)
  mov    $0x200,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  movzbl -0x14(%rbp),%edx
  mov    -0x18(%rbp),%esi
  mov    -0x18(%rbp),%ecx
  mov    -0x8(%rbp),%rax
  mov    %esi,%r9d
  mov    %ecx,%r8d
  mov    %edx,%ecx
  mov    $0x23f4,%edx
  lea    D4(%rip),%rsi
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <sprintf@plt>
  mov    %eax,-0x10(%rbp)
  mov    -0x10(%rbp),%edx
  mov    -0x8(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0x18(%rbp),%edx
  mov    -0x20(%rbp),%rax
  mov    %rax,%rsi
  mov    %rcx,%rdi
  call   <memcpy@plt>
  mov    -0x18(%rbp),%eax
  add    %eax,-0x10(%rbp)
  mov    -0x10(%rbp),%edx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  lea    D7(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <sprintf@plt>
  add    %eax,-0x10(%rbp)
  mov    -0x10(%rbp),%ecx
  mov    -0x8(%rbp),%rax
  mov    $0x5,%edx
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <iDoRemoteFunc@plt>
  mov    %eax,-0xc(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  cmpl   $0x0,-0xc(%rbp)
  je     L2
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  call   <vWriteLogTxt@plt>
L2:
  nop
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D4
type: string
name: extract_symbolic_data
arguments: data_label: D7
type: string
name: extract_symbolic_data
arguments: data_label: D8
type: string

tool: D4: %u(%u,%u,b%03u

tool: D7: )

tool: D8: _SLE4442_Write iDoRemoteFunc ret = %d

assistant: ```c
void _vRcSLE4442_Write(unsigned char Addr, int DataLen, unsigned char *DataBuff) {
  unsigned int uiDataLen, uiRet;
  char *psFunCall = malloc(512);
  uiDataLen = sprintf(psFunCall, "%u(%u,%u,b%03u", 9204, Addr, DataLen, DataLen);
  memcpy(psFunCall + uiDataLen, DataBuff, DataLen);
  uiDataLen += DataLen;
  uiDataLen += sprintf(psFunCall + uiDataLen, ")");
  uiRet = iDoRemoteFunc((unsigned char *)psFunCall, uiDataLen, 5);
  free(psFunCall);
  if (uiRet) {
    vWriteLogTxt("_SLE4442_Write iDoRemoteFunc ret = %d", uiRet);
  }
}
```