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

```asm
<hex_to_int>:
  endbr64
  movzbl (%rdi),%eax
  mov    %rdi,%r9
  test   %al,%al
  je     L47
  test   %esi,%esi
  jle    L47
  mov    $0x1,%edx
  xor    %r8d,%r8d
  jmp    L77
  nopl   0x0(%rax)
L88:
  movsbl %cl,%ecx
  add    %ecx,%r8d
L19:
  movzbl (%r9,%rdx,1),%eax
  cmp    %edx,%esi
  setg   %dil
  test   %al,%al
  setne  %cl
  add    $0x1,%rdx
  test   %cl,%dil
  je     L84
L77:
  lea    -0x30(%rax),%ecx
  shl    $0x4,%r8d
  cmp    $0x9,%cl
  jbe    L88
  lea    -0x41(%rax),%ecx
  cmp    $0x5,%cl
  ja     L37
  sub    $0x37,%eax
  movsbl %al,%eax
  add    %eax,%r8d
  jmp    L19
  xchg   %ax,%ax
L37:
  lea    -0x61(%rax),%ecx
  cmp    $0x5,%cl
  ja     L19
  sub    $0x57,%eax
  movsbl %al,%eax
  add    %eax,%r8d
  jmp    L19
  nopl   0x0(%rax,%rax,1)
L47:
  xor    %r8d,%r8d
L84:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int hex_to_int(char *hex, int l) {
  int n = 0;
  for (int i = 0; hex[i] && i < l; i++) {
    n <<= 4;
    if (hex[i] >= '0' && hex[i] <= '9') {
      n += hex[i] - '0';
    } else if (hex[i] >= 'A' && hex[i] <= 'F') {
      n += hex[i] - 'A' + 0x0A;
    } else if (hex[i] >= 'a' && hex[i] <= 'f') {
      n += hex[i] - 'a' + 0x0A;
    }
  }
  return n;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<test>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x4,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
int *test() {
  int *a2 = (int *)malloc(sizeof(int));
  return a2;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<test>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x4,%edi
  call   <malloc@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
int *test() {
  int *a2 = (int *)malloc(sizeof(int));
  return a2;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<test>:
  endbr64
  mov    $0x4,%edi
  jmp    1050 <malloc@plt>
```

assistant: ```c
int *test() {
  int *a2 = (int *)malloc(sizeof(int));
  return a2;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<test>:
  endbr64
  mov    $0x4,%edi
  jmp    1050 <malloc@plt>
```

assistant: ```c
int *test() {
  int *a2 = (int *)malloc(sizeof(int));
  return a2;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<EsNumero>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  cmpq   $0x2f,-0x8(%rbp)
  jbe    L11
  cmpq   $0x39,-0x8(%rbp)
  ja     L11
  mov    $0x1,%eax
  jmp    L27
L11:
  mov    $0x0,%eax
L27:
  pop    %rbp
  ret
```

assistant: ```c
int EsNumero(char caracter2[]) {
  if (caracter2 >= '0' && caracter2 <= '9') {
    return 1;
  } else {
    return 0;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<EsNumero>:
  endbr64
  sub    $0x30,%rdi
  cmp    $0x9,%rdi
  setbe  %al
  movzbl %al,%eax
  ret
```

assistant: ```c
int EsNumero(char caracter2[]) {
  if (caracter2 >= '0' && caracter2 <= '9') {
    return 1;
  } else {
    return 0;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<EsNumero>:
  endbr64
  sub    $0x30,%rdi
  xor    %eax,%eax
  cmp    $0x9,%rdi
  setbe  %al
  ret
```

assistant: ```c
int EsNumero(char caracter2[]) {
  if (caracter2 >= '0' && caracter2 <= '9') {
    return 1;
  } else {
    return 0;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<EsNumero>:
  endbr64
  sub    $0x30,%rdi
  xor    %eax,%eax
  cmp    $0x9,%rdi
  setbe  %al
  ret
```

assistant: ```c
int EsNumero(char caracter2[]) {
  if (caracter2 >= '0' && caracter2 <= '9') {
    return 1;
  } else {
    return 0;
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<copy_to_user>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    $0x0,%eax
  call   <assert_context_process@plt>
  mov    $0x0,%eax
  call   <nondet_ulong@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
unsigned long copy_to_user(void *to, void const *from, unsigned long n) {
  unsigned long tmp;
  {
    assert_context_process();
    tmp = nondet_ulong();
    return (tmp);
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<copy_to_user>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <assert_context_process@plt>
  mov    $0x0,%eax
  call   <nondet_ulong@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
unsigned long copy_to_user(void *to, void const *from, unsigned long n) {
  unsigned long tmp;
  {
    assert_context_process();
    tmp = nondet_ulong();
    return (tmp);
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<copy_to_user>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <assert_context_process@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1060 <nondet_ulong@plt>
```

assistant: ```c
unsigned long copy_to_user(void *to, void const *from, unsigned long n) {
  unsigned long tmp;
  {
    assert_context_process();
    tmp = nondet_ulong();
    return (tmp);
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<copy_to_user>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <assert_context_process@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1060 <nondet_ulong@plt>
```

assistant: ```c
unsigned long copy_to_user(void *to, void const *from, unsigned long n) {
  unsigned long tmp;
  {
    assert_context_process();
    tmp = nondet_ulong();
    return (tmp);
  }
}
```
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    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movl   $0x64655a,0xf(%rsp)
  movl   $0x77616853,0x13(%rsp)
  movb   $0x0,0x17(%rsp)
  mov    $0x64,%esi
  lea    D33(%rip),%rdi
  call   <printf@plt>
  movsd  D12(%rip),%xmm0
  lea    D89(%rip),%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  movsd  D49(%rip),%xmm0
  lea    D69(%rip),%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    $0x41,%esi
  lea    D67(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0xf(%rsp),%rbx
  mov    %rbx,%rsi
  lea    D42(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0x13(%rsp),%rbp
  mov    %rbp,%rsi
  lea    D11(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %rbp,%rcx
  mov    $0x41,%edx
  mov    %rbx,%rsi
  lea    D21(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  mov    $0x0,%eax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  ret
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D33
datatype: string
name: retrieve_data
arguments: tag: D12
datatype: byte[8]
name: retrieve_data
arguments: tag: D89
datatype: string
name: retrieve_data
arguments: tag: D49
datatype: f64
name: retrieve_data
arguments: tag: D69
datatype: string
name: retrieve_data
arguments: tag: D67
datatype: string
name: retrieve_data
arguments: tag: D42
datatype: string
name: retrieve_data
arguments: tag: D11
datatype: string
name: retrieve_data
arguments: tag: D21
datatype: string

tool: D33: You are %d kilometres away.


tool: {"D12": [0x00,0x00,0x00,0x60,0x8f,0xc2,0x02,0x40]}

tool: D89: You have %f levels of power.


tool: D49: 56789.4532

tool: D69: You have %f awesome super powers.


tool: D67: I have an initial %c.


tool: D42: My first name is %s.


tool: D11: My last name is %s.


tool: D21: My whole name is %s %c. %s.


assistant: ```c
int main(int argc, char *argv[]) {
  int distance = 100;
  float power = 2.345f;
  double super_power = 56789.4532;
  char initial = 'A';
  char first_name[] = "Zed";
  char last_name[] = "Shaw";
  printf("You are %d kilometres away.\n", distance);
  printf("You have %f levels of power.\n", power);
  printf("You have %f awesome super powers.\n", super_power);
  printf("I have an initial %c.\n", initial);
  printf("My first name is %s.\n", first_name);
  printf("My last name is %s.\n", last_name);
  printf("My whole name is %s %c. %s.\n", first_name, initial, last_name);
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %r12
  mov    $0x64,%esi
  lea    D52(%rip),%rdi
  push   %rbp
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movl   $0x64655a,0xf(%rsp)
  lea    0xf(%rsp),%rbp
  lea    0x13(%rsp),%r12
  movl   $0x77616853,0x13(%rsp)
  movb   $0x0,0x17(%rsp)
  call   <printf@plt>
  movsd  D58(%rip),%xmm0
  mov    $0x1,%eax
  lea    D60(%rip),%rdi
  call   <printf@plt>
  movsd  D64(%rip),%xmm0
  mov    $0x1,%eax
  lea    D72(%rip),%rdi
  call   <printf@plt>
  mov    $0x41,%esi
  lea    D49(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rsi
  lea    D46(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r12,%rsi
  lea    D68(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  mov    %r12,%rcx
  mov    $0x41,%edx
  mov    %rbp,%rsi
  lea    D69(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D52
type: string
name: unpack_data
arguments: label: D58
type: byte[8]
name: unpack_data
arguments: label: D60
type: string
name: unpack_data
arguments: label: D64
type: f64
name: unpack_data
arguments: label: D72
type: string
name: unpack_data
arguments: label: D49
type: string
name: unpack_data
arguments: label: D46
type: string
name: unpack_data
arguments: label: D68
type: string
name: unpack_data
arguments: label: D69
type: string

tool: D52: You are %d kilometres away.


tool: {"D58": [0x00,0x00,0x00,0x60,0x8f,0xc2,0x02,0x40]}

tool: D60: You have %f levels of power.


tool: D64: 56789.4532

tool: D72: You have %f awesome super powers.


tool: D49: I have an initial %c.


tool: D46: My first name is %s.


tool: D68: My last name is %s.


tool: D69: My whole name is %s %c. %s.


assistant: ```c
int main(int argc, char *argv[]) {
  int distance = 100;
  float power = 2.345f;
  double super_power = 56789.4532;
  char initial = 'A';
  char first_name[] = "Zed";
  char last_name[] = "Shaw";
  printf("You are %d kilometres away.\n", distance);
  printf("You have %f levels of power.\n", power);
  printf("You have %f awesome super powers.\n", super_power);
  printf("I have an initial %c.\n", initial);
  printf("My first name is %s.\n", first_name);
  printf("My last name is %s.\n", last_name);
  printf("My whole name is %s %c. %s.\n", first_name, initial, last_name);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r12
  mov    $0x64,%esi
  lea    D53(%rip),%rdi
  push   %rbp
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movl   $0x64655a,0xf(%rsp)
  lea    0xf(%rsp),%rbp
  lea    0x13(%rsp),%r12
  movl   $0x77616853,0x13(%rsp)
  movb   $0x0,0x17(%rsp)
  call   <printf@plt>
  movsd  D56(%rip),%xmm0
  mov    $0x1,%eax
  lea    D31(%rip),%rdi
  call   <printf@plt>
  movsd  D24(%rip),%xmm0
  mov    $0x1,%eax
  lea    D2(%rip),%rdi
  call   <printf@plt>
  mov    $0x41,%esi
  lea    D11(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rsi
  lea    D62(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r12,%rsi
  lea    D72(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  mov    %r12,%rcx
  mov    $0x41,%edx
  mov    %rbp,%rsi
  lea    D34(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D53
type: string
name: read_data
arguments: data_tag: D56
type: byte[8]
name: read_data
arguments: data_tag: D31
type: string
name: read_data
arguments: data_tag: D24
type: f64
name: read_data
arguments: data_tag: D2
type: string
name: read_data
arguments: data_tag: D11
type: string
name: read_data
arguments: data_tag: D62
type: string
name: read_data
arguments: data_tag: D72
type: string
name: read_data
arguments: data_tag: D34
type: string

tool: D53: You are %d kilometres away.


tool: {"D56": [0x00,0x00,0x00,0x60,0x8f,0xc2,0x02,0x40]}

tool: D31: You have %f levels of power.


tool: D24: 56789.4532

tool: D2: You have %f awesome super powers.


tool: D11: I have an initial %c.


tool: D62: My first name is %s.


tool: D72: My last name is %s.


tool: D34: My whole name is %s %c. %s.


assistant: ```c
int main(int argc, char *argv[]) {
  int distance = 100;
  float power = 2.345f;
  double super_power = 56789.4532;
  char initial = 'A';
  char first_name[] = "Zed";
  char last_name[] = "Shaw";
  printf("You are %d kilometres away.\n", distance);
  printf("You have %f levels of power.\n", power);
  printf("You have %f awesome super powers.\n", super_power);
  printf("I have an initial %c.\n", initial);
  printf("My first name is %s.\n", first_name);
  printf("My last name is %s.\n", last_name);
  printf("My whole name is %s %c. %s.\n", first_name, initial, last_name);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<str_to_upper_case>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,-0x28(%rbp)
  movl   $0x0,-0x1c(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  add    $0x1,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x18(%rbp)
L1:
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    -0x1c(%rbp),%edx
  movslq %edx,%rcx
  mov    -0x18(%rbp),%rdx
  lea    (%rcx,%rdx,1),%rbx
  mov    %eax,%edi
  call   <toupper@plt>
  mov    %al,(%rbx)
  addl   $0x1,-0x1c(%rbp)
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L1
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x18(%rbp),%rax
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
char *str_to_upper_case(char str[]) {
  int i = 0;
  char *buffer;
  buffer = malloc(sizeof(char) * strlen(str) + 1);
  do {
    buffer[i] = toupper(str[i]);
    ++i;
  } while (str[i] != '\0');
  buffer[i] = '\0';
  return buffer;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<str_to_upper_case>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  call   <strlen@plt>
  lea    0x1(%rax),%edi
  call   <malloc@plt>
  mov    %rax,%r12
  mov    $0x1,%ebx
L4:
  movsbl -0x1(%rbp,%rbx,1),%edi
  call   <toupper@plt>
  mov    %al,-0x1(%r12,%rbx,1)
  mov    %rbx,%rax
  add    $0x1,%rbx
  cmpb   $0x0,-0x1(%rbp,%rbx,1)
  jne    L4
  movb   $0x0,(%r12,%rax,1)
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *str_to_upper_case(char str[]) {
  int i = 0;
  char *buffer;
  buffer = malloc(sizeof(char) * strlen(str) + 1);
  do {
    buffer[i] = toupper(str[i]);
    ++i;
  } while (str[i] != '\0');
  buffer[i] = '\0';
  return buffer;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<str_to_upper_case>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    $0x1,%ebx
  call   <strlen@plt>
  lea    0x1(%rax),%edi
  call   <malloc@plt>
  movsbl 0x0(%rbp),%edi
  mov    %rax,%r12
  nopl   0x0(%rax)
L8:
  call   <toupper@plt>
  mov    %al,-0x1(%r12,%rbx,1)
  mov    %rbx,%rax
  add    $0x1,%rbx
  movsbl -0x1(%rbp,%rbx,1),%edi
  test   %dil,%dil
  jne    L8
  movb   $0x0,(%r12,%rax,1)
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *str_to_upper_case(char str[]) {
  int i = 0;
  char *buffer;
  buffer = malloc(sizeof(char) * strlen(str) + 1);
  do {
    buffer[i] = toupper(str[i]);
    ++i;
  } while (str[i] != '\0');
  buffer[i] = '\0';
  return buffer;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<str_to_upper_case>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    $0x1,%ebx
  call   <strlen@plt>
  lea    0x1(%rax),%edi
  call   <malloc@plt>
  mov    %rax,%r12
L6:
  movsbl -0x1(%rbp,%rbx,1),%edi
  call   <toupper@plt>
  mov    %al,-0x1(%r12,%rbx,1)
  mov    %rbx,%rax
  add    $0x1,%rbx
  cmpb   $0x0,-0x1(%rbp,%rbx,1)
  jne    L6
  movb   $0x0,(%r12,%rax,1)
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *str_to_upper_case(char str[]) {
  int i = 0;
  char *buffer;
  buffer = malloc(sizeof(char) * strlen(str) + 1);
  do {
    buffer[i] = toupper(str[i]);
    ++i;
  } while (str[i] != '\0');
  buffer[i] = '\0';
  return buffer;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<displayHexRecord>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L36
L29:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  movzbl %al,%eax
  mov    %eax,%esi
  lea    D20(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x4(%rbp)
L36:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L29
  jmp    L39
L13:
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x4(%rbp)
L39:
  mov    -0x4(%rbp),%eax
  cmp    -0x20(%rbp),%eax
  jl     L13
  lea    D22(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x0,-0x4(%rbp)
  jmp    L66
L9:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isgraph@plt>
  test   %rax,%rax
  je     L62
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <putchar@plt>
  jmp    L59
L62:
  mov    $0x2e,%edi
  call   <putchar@plt>
L59:
  addl   $0x1,-0x4(%rbp)
L66:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L9
  mov    $0xa,%edi
  call   <putchar@plt>
  nop
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D20
type: string
name: retrieve_data
arguments: data_tag: D5
type: string
name: retrieve_data
arguments: data_tag: D22
type: string

tool: D20: %02x 

tool: D5:    

tool: D22: : 

assistant: ```c
void displayHexRecord(char *data, int count, int record_length) {
  int i;
  for (i = 0; i < count; i++)
    printf("%02x ", data[i] & 0xff);
  for (; i < record_length; i++)
    printf("   ");
  printf(": ");
  for (i = 0; i < count; i++) {
    if (isgraph(data[i]))
      putchar(data[i]);
    else
      putchar('.');
  }
  putchar('\n');
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<displayHexRecord>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    %esi,%r12d
  mov    %edx,%r13d
  test   %esi,%esi
  jle    L83
  mov    %rdi,%rbx
  lea    -0x1(%rsi),%eax
  lea    0x1(%rdi,%rax,1),%r15
  lea    D2(%rip),%r14
L9:
  movzbl (%rbx),%esi
  mov    %r14,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x1,%rbx
  cmp    %r15,%rbx
  jne    L9
  mov    %r12d,%ebx
  cmp    %r13d,%r12d
  jge    L85
L98:
  lea    D24(%rip),%r14
L56:
  mov    %r14,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x1,%ebx
  cmp    %ebx,%r13d
  jg     L56
  lea    D14(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  test   %r12d,%r12d
  jle    L13
L81:
  mov    $0x0,%ebx
  jmp    L105
L106:
  mov    $0x2e,%edi
  call   <putchar@plt>
L97:
  add    $0x1,%rbx
  cmp    %ebx,%r12d
  jle    L13
L105:
  movsbl 0x0(%rbp,%rbx,1),%edi
  call   <isgraph@plt>
  test   %rax,%rax
  je     L106
  movsbl 0x0(%rbp,%rbx,1),%edi
  call   <putchar@plt>
  jmp    L97
L85:
  lea    D14(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L81
L83:
  mov    $0x0,%ebx
  test   %edx,%edx
  jg     L98
  lea    D14(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L13:
  mov    $0xa,%edi
  call   <putchar@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D2
type: string
name: retrieve_data
arguments: tag: D24
type: string
name: retrieve_data
arguments: tag: D14
type: string

tool: D2: %02x 

tool: D24:    

tool: D14: : 

assistant: ```c
void displayHexRecord(char *data, int count, int record_length) {
  int i;
  for (i = 0; i < count; i++)
    printf("%02x ", data[i] & 0xff);
  for (; i < record_length; i++)
    printf("   ");
  printf(": ");
  for (i = 0; i < count; i++) {
    if (isgraph(data[i]))
      putchar(data[i]);
    else
      putchar('.');
  }
  putchar('\n');
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<displayHexRecord>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  mov    %edx,%r13d
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  test   %esi,%esi
  jle    L84
  lea    -0x1(%rsi),%eax
  mov    %rdi,%rbx
  lea    D1(%rip),%r14
  lea    0x1(%rdi,%rax,1),%r15
  nopl   (%rax)
L78:
  movzbl (%rbx),%esi
  mov    %r14,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    %r15,%rbx
  jne    L78
  mov    %r12d,%ebx
  cmp    %r13d,%r12d
  jge    L11
L103:
  lea    D24(%rip),%r14
  nopl   0x0(%rax)
L41:
  mov    %r14,%rdi
  xor    %eax,%eax
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    %ebx,%r13d
  jg     L41
  xor    %eax,%eax
  lea    D29(%rip),%rdi
  call   <printf@plt>
  test   %r12d,%r12d
  jle    L56
L45:
  xor    %ebx,%ebx
  jmp    L48
  nopl   0x0(%rax)
L114:
  movsbl 0x0(%rbp,%rbx,1),%edi
  add    $0x1,%rbx
  call   <putchar@plt>
  cmp    %ebx,%r12d
  jle    L56
L48:
  movsbl 0x0(%rbp,%rbx,1),%edi
  call   <isgraph@plt>
  test   %rax,%rax
  jne    L114
  mov    $0x2e,%edi
  add    $0x1,%rbx
  call   <putchar@plt>
  cmp    %ebx,%r12d
  jg     L48
L56:
  add    $0x8,%rsp
  mov    $0xa,%edi
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  jmp    1070 <putchar@plt>
  nopl   (%rax)
L84:
  xor    %ebx,%ebx
  test   %edx,%edx
  jg     L103
  lea    D29(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L56
  nopw   0x0(%rax,%rax,1)
L11:
  lea    D29(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L45
```

tool_calls: name: parse_data
arguments: data_label: D1
type_infered: string
name: parse_data
arguments: data_label: D24
type_infered: string
name: parse_data
arguments: data_label: D29
type_infered: string

tool: D1: %02x 

tool: D24:    

tool: D29: : 

assistant: ```c
void displayHexRecord(char *data, int count, int record_length) {
  int i;
  for (i = 0; i < count; i++)
    printf("%02x ", data[i] & 0xff);
  for (; i < record_length; i++)
    printf("   ");
  printf(": ");
  for (i = 0; i < count; i++) {
    if (isgraph(data[i]))
      putchar(data[i]);
    else
      putchar('.');
  }
  putchar('\n');
}
```