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
<main_menu>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <title@plt>
  mov    $0x1b,%esi
  mov    $0x6,%edi
  lea    D56(%rip),%rdx
  call   <mvaddstr@plt>
  mov    $0x1b,%esi
  mov    $0x7,%edi
  lea    D4(%rip),%rdx
  call   <mvaddstr@plt>
  mov    $0x1b,%esi
  mov    $0x8,%edi
  lea    D27(%rip),%rdx
  call   <mvaddstr@plt>
  mov    $0x1b,%esi
  mov    $0x9,%edi
  lea    D40(%rip),%rdx
  call   <mvaddstr@plt>
  mov    $0x1b,%esi
  mov    $0xb,%edi
  lea    D26(%rip),%rdx
  call   <mvaddstr@plt>
  mov    $0x1b,%esi
  mov    $0xc,%edi
  lea    D53(%rip),%rdx
  call   <mvaddstr@plt>
  mov    $0x1b,%esi
  mov    $0xd,%edi
  lea    D3(%rip),%rdx
  call   <mvaddstr@plt>
  mov    $0x1b,%esi
  mov    $0xf,%edi
  lea    D41(%rip),%rdx
  call   <mvaddstr@plt>
  mov    $0x1b,%esi
  mov    $0x10,%edi
  lea    D70(%rip),%rdx
  call   <mvaddstr@plt>
  xor    %esi,%esi
  xor    %edi,%edi
  call   <move@plt>
  xor    %eax,%eax
  call   <refresh@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    10b0 <getch@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D56
type_guessed: string
name: extract_symbolic_data
arguments: data_label: D4
type_guessed: string
name: extract_symbolic_data
arguments: data_label: D27
type_guessed: string
name: extract_symbolic_data
arguments: data_label: D40
type_guessed: string
name: extract_symbolic_data
arguments: data_label: D26
type_guessed: string
name: extract_symbolic_data
arguments: data_label: D53
type_guessed: string
name: extract_symbolic_data
arguments: data_label: D3
type_guessed: string
name: extract_symbolic_data
arguments: data_label: D41
type_guessed: string
name: extract_symbolic_data
arguments: data_label: D70
type_guessed: string

tool: D56: [L]oad a file

tool: D4: [W]rite RAM to file

tool: D27: [D]isply RAM

tool: D40: [F]ill RAM with blank state (0xFFFF)

tool: D26: [R]ead contents of PIC into RAM

tool: D53: [P]rogram PIC from RAM

tool: D3: [V]erify PIC with RAM

tool: D41: [C]onfigure PICPRG

tool: D70: [Q]uit

assistant: ```c
int main_menu() {
  title();
  mvaddstr(6, 27, "[L]oad a file");
  mvaddstr(7, 27, "[W]rite RAM to file");
  mvaddstr(8, 27, "[D]isply RAM");
  mvaddstr(9, 27, "[F]ill RAM with blank state (0xFFFF)");
  mvaddstr(11, 27, "[R]ead contents of PIC into RAM");
  mvaddstr(12, 27, "[P]rogram PIC from RAM");
  mvaddstr(13, 27, "[V]erify PIC with RAM");
  mvaddstr(15, 27, "[C]onfigure PICPRG");
  mvaddstr(16, 27, "[Q]uit");
  move(0, 0);
  refresh();
  return getch();
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<num_to_size_str>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %edx,-0x14(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  jne    L71
  mov    -0x10(%rbp),%rcx
  mov    -0x8(%rbp),%rax
  lea    D45(%rip),%rdx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <snprintf@plt>
  jmp    L69
L71:
  mov    -0x14(%rbp),%eax
  and    $0x3fffffff,%eax
  test   %eax,%eax
  jne    L61
  mov    -0x14(%rbp),%eax
  lea    0x3fffffff(%rax),%edx
  test   %eax,%eax
  cmovs  %edx,%eax
  sar    $0x1e,%eax
  mov    %eax,%edx
  mov    -0x10(%rbp),%rsi
  mov    -0x8(%rbp),%rax
  mov    %edx,%ecx
  lea    D30(%rip),%rdx
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <snprintf@plt>
  jmp    L69
L61:
  mov    -0x14(%rbp),%eax
  and    $0xfffff,%eax
  test   %eax,%eax
  jne    L11
  mov    -0x14(%rbp),%eax
  lea    0xfffff(%rax),%edx
  test   %eax,%eax
  cmovs  %edx,%eax
  sar    $0x14,%eax
  mov    %eax,%edx
  mov    -0x10(%rbp),%rsi
  mov    -0x8(%rbp),%rax
  mov    %edx,%ecx
  lea    D22(%rip),%rdx
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <snprintf@plt>
  jmp    L69
L11:
  mov    -0x14(%rbp),%eax
  and    $0x3ff,%eax
  test   %eax,%eax
  jne    L46
  mov    -0x14(%rbp),%eax
  lea    0x3ff(%rax),%edx
  test   %eax,%eax
  cmovs  %edx,%eax
  sar    $0xa,%eax
  mov    %eax,%edx
  mov    -0x10(%rbp),%rsi
  mov    -0x8(%rbp),%rax
  mov    %edx,%ecx
  lea    D11(%rip),%rdx
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <snprintf@plt>
  jmp    L69
L46:
  mov    -0x14(%rbp),%edx
  mov    -0x10(%rbp),%rsi
  mov    -0x8(%rbp),%rax
  mov    %edx,%ecx
  lea    D31(%rip),%rdx
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <snprintf@plt>
L69:
  mov    -0x8(%rbp),%rax
  leave
  ret
```

tool_calls: name: parse_data
arguments: tag: D45
datatype: string
name: parse_data
arguments: tag: D30
datatype: string
name: parse_data
arguments: tag: D22
datatype: string
name: parse_data
arguments: tag: D11
datatype: string
name: parse_data
arguments: tag: D31
datatype: string

tool: D45: 0

tool: D30: %uG

tool: D22: %uM

tool: D11: %uK

tool: D31: %u

assistant: ```c
unsigned char *num_to_size_str(unsigned char *buf, size_t buf_size, int num) {
  if (!num)
    snprintf(buf, buf_size, "0");
  else if (!(num % (1024 * 1024 * 1024)))
    snprintf(buf, buf_size, "%uG", num / (1024 * 1024 * 1024));
  else if (!(num % (1024 * 1024)))
    snprintf(buf, buf_size, "%uM", num / (1024 * 1024));
  else if (!(num % (1024)))
    snprintf(buf, buf_size, "%uK", num / (1024));
  else
    snprintf(buf, buf_size, "%u", num);
  return buf;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<num_to_size_str>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  test   %edx,%edx
  je     L0
  test   $0x3fffffff,%edx
  je     L24
  test   $0xfffff,%edx
  je     L61
  test   $0x3ff,%edx
  jne    L63
  lea    0x3ff(%rdx),%ecx
  test   %edx,%edx
  cmovns %edx,%ecx
  sar    $0xa,%ecx
  lea    D18(%rip),%rdx
  mov    $0x0,%eax
  call   <snprintf@plt>
  jmp    L15
L0:
  lea    D37(%rip),%rdx
  mov    $0x0,%eax
  call   <snprintf@plt>
L15:
  mov    %rbx,%rax
  pop    %rbx
  ret
L24:
  lea    0x3fffffff(%rdx),%ecx
  test   %edx,%edx
  cmovns %edx,%ecx
  sar    $0x1e,%ecx
  lea    D3(%rip),%rdx
  mov    $0x0,%eax
  call   <snprintf@plt>
  jmp    L15
L61:
  lea    0xfffff(%rdx),%ecx
  test   %edx,%edx
  cmovns %edx,%ecx
  sar    $0x14,%ecx
  lea    D12(%rip),%rdx
  mov    $0x0,%eax
  call   <snprintf@plt>
  jmp    L15
L63:
  mov    %edx,%ecx
  lea    D27(%rip),%rdx
  mov    $0x0,%eax
  call   <snprintf@plt>
  jmp    L15
```

tool_calls: name: extract_symbolic_data
arguments: tag: D18
datatype: string
name: extract_symbolic_data
arguments: tag: D37
datatype: string
name: extract_symbolic_data
arguments: tag: D3
datatype: string
name: extract_symbolic_data
arguments: tag: D12
datatype: string
name: extract_symbolic_data
arguments: tag: D27
datatype: string

tool: D18: %uK

tool: D37: 0

tool: D3: %uG

tool: D12: %uM

tool: D27: %u

assistant: ```c
unsigned char *num_to_size_str(unsigned char *buf, size_t buf_size, int num) {
  if (!num)
    snprintf(buf, buf_size, "0");
  else if (!(num % (1024 * 1024 * 1024)))
    snprintf(buf, buf_size, "%uG", num / (1024 * 1024 * 1024));
  else if (!(num % (1024 * 1024)))
    snprintf(buf, buf_size, "%uM", num / (1024 * 1024));
  else if (!(num % (1024)))
    snprintf(buf, buf_size, "%uK", num / (1024));
  else
    snprintf(buf, buf_size, "%u", num);
  return buf;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<num_to_size_str>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  test   %edx,%edx
  je     L32
  test   $0x3fffffff,%edx
  je     L12
  test   $0xfffff,%edx
  je     L36
  test   $0x3ff,%edx
  jne    L1
  test   %edx,%edx
  lea    0x3ff(%rdx),%eax
  mov    %edx,%ecx
  cmovs  %eax,%ecx
  lea    D24(%rip),%rdx
  xor    %eax,%eax
  sar    $0xa,%ecx
  call   <snprintf@plt>
  mov    %r12,%rax
  pop    %r12
  ret
  nopl   (%rax)
L12:
  test   %edx,%edx
  lea    0x3fffffff(%rdx),%eax
  mov    %edx,%ecx
  cmovs  %eax,%ecx
  lea    D16(%rip),%rdx
  xor    %eax,%eax
  sar    $0x1e,%ecx
  call   <snprintf@plt>
  mov    %r12,%rax
  pop    %r12
  ret
  nopl   0x0(%rax)
L32:
  lea    D35(%rip),%rdx
  xor    %eax,%eax
  call   <snprintf@plt>
  mov    %r12,%rax
  pop    %r12
  ret
  nopl   0x0(%rax)
L1:
  mov    %edx,%ecx
  xor    %eax,%eax
  lea    D38(%rip),%rdx
  call   <snprintf@plt>
  mov    %r12,%rax
  pop    %r12
  ret
  cs nopw 0x0(%rax,%rax,1)
L36:
  test   %edx,%edx
  lea    0xfffff(%rdx),%eax
  mov    %edx,%ecx
  cmovs  %eax,%ecx
  lea    D15(%rip),%rdx
  xor    %eax,%eax
  sar    $0x14,%ecx
  call   <snprintf@plt>
  mov    %r12,%rax
  pop    %r12
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D24
type_infered: string
name: disassemble_data
arguments: data_tag: D16
type_infered: string
name: disassemble_data
arguments: data_tag: D35
type_infered: string
name: disassemble_data
arguments: data_tag: D38
type_infered: string
name: disassemble_data
arguments: data_tag: D15
type_infered: string

tool: D24: %uK

tool: D16: %uG

tool: D35: 0

tool: D38: %u

tool: D15: %uM

assistant: ```c
unsigned char *num_to_size_str(unsigned char *buf, size_t buf_size, int num) {
  if (!num)
    snprintf(buf, buf_size, "0");
  else if (!(num % (1024 * 1024 * 1024)))
    snprintf(buf, buf_size, "%uG", num / (1024 * 1024 * 1024));
  else if (!(num % (1024 * 1024)))
    snprintf(buf, buf_size, "%uM", num / (1024 * 1024));
  else if (!(num % (1024)))
    snprintf(buf, buf_size, "%uK", num / (1024));
  else
    snprintf(buf, buf_size, "%u", num);
  return buf;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<num_to_size_str>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  test   %edx,%edx
  je     L24
  test   $0x3fffffff,%edx
  je     L23
  test   $0xfffff,%edx
  je     L6
  test   $0x3ff,%edx
  jne    L3
  test   %edx,%edx
  lea    0x3ff(%rdx),%eax
  mov    %edx,%ecx
  cmovs  %eax,%ecx
  lea    D25(%rip),%rdx
  xor    %eax,%eax
  sar    $0xa,%ecx
  call   <snprintf@plt>
  mov    %r12,%rax
  pop    %r12
  ret
  nopl   (%rax)
L23:
  test   %edx,%edx
  lea    0x3fffffff(%rdx),%eax
  mov    %edx,%ecx
  cmovs  %eax,%ecx
  lea    D16(%rip),%rdx
  xor    %eax,%eax
  sar    $0x1e,%ecx
  call   <snprintf@plt>
  mov    %r12,%rax
  pop    %r12
  ret
  nopl   0x0(%rax)
L24:
  lea    D0(%rip),%rdx
  xor    %eax,%eax
  call   <snprintf@plt>
  mov    %r12,%rax
  pop    %r12
  ret
  nopl   0x0(%rax)
L3:
  mov    %edx,%ecx
  xor    %eax,%eax
  lea    D23(%rip),%rdx
  call   <snprintf@plt>
  mov    %r12,%rax
  pop    %r12
  ret
  cs nopw 0x0(%rax,%rax,1)
L6:
  test   %edx,%edx
  lea    0xfffff(%rdx),%eax
  mov    %edx,%ecx
  cmovs  %eax,%ecx
  lea    D4(%rip),%rdx
  xor    %eax,%eax
  sar    $0x14,%ecx
  call   <snprintf@plt>
  mov    %r12,%rax
  pop    %r12
  ret
```

tool_calls: name: read_data
arguments: label: D25
type: string
name: read_data
arguments: label: D16
type: string
name: read_data
arguments: label: D0
type: string
name: read_data
arguments: label: D23
type: string
name: read_data
arguments: label: D4
type: string

tool: D25: %uK

tool: D16: %uG

tool: D0: 0

tool: D23: %u

tool: D4: %uM

assistant: ```c
unsigned char *num_to_size_str(unsigned char *buf, size_t buf_size, int num) {
  if (!num)
    snprintf(buf, buf_size, "0");
  else if (!(num % (1024 * 1024 * 1024)))
    snprintf(buf, buf_size, "%uG", num / (1024 * 1024 * 1024));
  else if (!(num % (1024 * 1024)))
    snprintf(buf, buf_size, "%uM", num / (1024 * 1024));
  else if (!(num % (1024)))
    snprintf(buf, buf_size, "%uK", num / (1024));
  else
    snprintf(buf, buf_size, "%u", num);
  return buf;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ieee_p_1242562249_sub_4501204497521157593_1035706684>:
  endbr64
  mov    0xc(%rdx),%eax
  cmp    0xc(%r8),%eax
  je     L20
  mov    $0x0,%eax
  ret
L8:
  mov    $0x1,%eax
  ret
L5:
  mov    $0x0,%eax
  ret
L20:
  test   %eax,%eax
  je     L8
  lea    -0x1(%rax),%edi
  mov    $0x0,%eax
  jmp    L32
L23:
  mov    %rdx,%rax
L32:
  movzbl (%rcx,%rax,1),%edx
  cmp    %dl,(%rsi,%rax,1)
  jne    L5
  lea    0x1(%rax),%rdx
  cmp    %rdi,%rax
  jne    L23
  mov    $0x1,%eax
  ret
```

assistant: ```c
unsigned char ieee_p_1242562249_sub_4501204497521157593_1035706684(char *t1, char *t2, char *t3, char *t4, char *t5) {
  char t7[40];
  unsigned char t0;
  char *t8;
  unsigned char t9;
  char *t10;
  char *t11;
  unsigned char t12;
  char *t13;
  char *t14;
  unsigned int t15;
  char *t16;
  unsigned int t17;
  unsigned char t18;
  unsigned int t19;
  char *t20;
  char *t21;
LAB0:
  t8 = (t7 + 4U);
  t9 = (t2 != 0);
  if (t9 == 1)
    goto LAB3;
LAB2:
  t10 = (t7 + 12U);
  *((char **)t10) = t3;
  t11 = (t7 + 20U);
  t12 = (t4 != 0);
  if (t12 == 1)
    goto LAB5;
LAB4:
  t13 = (t7 + 28U);
  *((char **)t13) = t5;
  t14 = (t3 + 12U);
  t15 = *((unsigned int *)t14);
  t15 = (t15 * 1U);
  t16 = (t5 + 12U);
  t17 = *((unsigned int *)t16);
  t17 = (t17 * 1U);
  t18 = 1;
  if (t15 == t17)
    goto LAB6;
LAB7:
  t18 = 0;
LAB8:
  t0 = t18;
LAB1:
  return t0;
LAB3:
  *((char **)t8) = t2;
  goto LAB2;
LAB5:
  *((char **)t11) = t4;
  goto LAB4;
LAB6:
  t19 = 0;
LAB9:
  if (t19 < t15)
    goto LAB10;
  else
    goto LAB8;
LAB10:
  t20 = (t2 + t19);
  t21 = (t4 + t19);
  if (*((unsigned char *)t20) != *((unsigned char *)t21))
    goto LAB7;
LAB11:
  t19 = (t19 + 1);
  goto LAB9;
LAB12:;
}
```
user: Translate the below assembly code into C programming language:

```asm
<ieee_p_1242562249_sub_4501204497521157593_1035706684>:
  endbr64
  mov    0xc(%rdx),%eax
  cmp    0xc(%r8),%eax
  je     L32
L25:
  xor    %eax,%eax
  ret
L32:
  test   %eax,%eax
  je     L36
  mov    %eax,%edx
  xor    %eax,%eax
  nopl   0x0(%rax,%rax,1)
L20:
  movzbl (%rcx,%rax,1),%edi
  cmp    %dil,(%rsi,%rax,1)
  jne    L25
  add    $0x1,%rax
  cmp    %rdx,%rax
  jne    L20
L36:
  mov    $0x1,%eax
  ret
```

assistant: ```c
unsigned char ieee_p_1242562249_sub_4501204497521157593_1035706684(char *t1, char *t2, char *t3, char *t4, char *t5) {
  char t7[40];
  unsigned char t0;
  char *t8;
  unsigned char t9;
  char *t10;
  char *t11;
  unsigned char t12;
  char *t13;
  char *t14;
  unsigned int t15;
  char *t16;
  unsigned int t17;
  unsigned char t18;
  unsigned int t19;
  char *t20;
  char *t21;
LAB0:
  t8 = (t7 + 4U);
  t9 = (t2 != 0);
  if (t9 == 1)
    goto LAB3;
LAB2:
  t10 = (t7 + 12U);
  *((char **)t10) = t3;
  t11 = (t7 + 20U);
  t12 = (t4 != 0);
  if (t12 == 1)
    goto LAB5;
LAB4:
  t13 = (t7 + 28U);
  *((char **)t13) = t5;
  t14 = (t3 + 12U);
  t15 = *((unsigned int *)t14);
  t15 = (t15 * 1U);
  t16 = (t5 + 12U);
  t17 = *((unsigned int *)t16);
  t17 = (t17 * 1U);
  t18 = 1;
  if (t15 == t17)
    goto LAB6;
LAB7:
  t18 = 0;
LAB8:
  t0 = t18;
LAB1:
  return t0;
LAB3:
  *((char **)t8) = t2;
  goto LAB2;
LAB5:
  *((char **)t11) = t4;
  goto LAB4;
LAB6:
  t19 = 0;
LAB9:
  if (t19 < t15)
    goto LAB10;
  else
    goto LAB8;
LAB10:
  t20 = (t2 + t19);
  t21 = (t4 + t19);
  if (*((unsigned char *)t20) != *((unsigned char *)t21))
    goto LAB7;
LAB11:
  t19 = (t19 + 1);
  goto LAB9;
LAB12:;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<ieee_p_1242562249_sub_4501204497521157593_1035706684>:
  endbr64
  mov    0xc(%rdx),%eax
  cmp    0xc(%r8),%eax
  je     L16
L36:
  xor    %eax,%eax
  ret
L16:
  test   %eax,%eax
  je     L21
  mov    %eax,%edx
  xor    %eax,%eax
  nopl   0x0(%rax,%rax,1)
L25:
  movzbl (%rcx,%rax,1),%edi
  cmp    %dil,(%rsi,%rax,1)
  jne    L36
  add    $0x1,%rax
  cmp    %rdx,%rax
  jne    L25
L21:
  mov    $0x1,%eax
  ret
```

assistant: ```c
unsigned char ieee_p_1242562249_sub_4501204497521157593_1035706684(char *t1, char *t2, char *t3, char *t4, char *t5) {
  char t7[40];
  unsigned char t0;
  char *t8;
  unsigned char t9;
  char *t10;
  char *t11;
  unsigned char t12;
  char *t13;
  char *t14;
  unsigned int t15;
  char *t16;
  unsigned int t17;
  unsigned char t18;
  unsigned int t19;
  char *t20;
  char *t21;
LAB0:
  t8 = (t7 + 4U);
  t9 = (t2 != 0);
  if (t9 == 1)
    goto LAB3;
LAB2:
  t10 = (t7 + 12U);
  *((char **)t10) = t3;
  t11 = (t7 + 20U);
  t12 = (t4 != 0);
  if (t12 == 1)
    goto LAB5;
LAB4:
  t13 = (t7 + 28U);
  *((char **)t13) = t5;
  t14 = (t3 + 12U);
  t15 = *((unsigned int *)t14);
  t15 = (t15 * 1U);
  t16 = (t5 + 12U);
  t17 = *((unsigned int *)t16);
  t17 = (t17 * 1U);
  t18 = 1;
  if (t15 == t17)
    goto LAB6;
LAB7:
  t18 = 0;
LAB8:
  t0 = t18;
LAB1:
  return t0;
LAB3:
  *((char **)t8) = t2;
  goto LAB2;
LAB5:
  *((char **)t11) = t4;
  goto LAB4;
LAB6:
  t19 = 0;
LAB9:
  if (t19 < t15)
    goto LAB10;
  else
    goto LAB8;
LAB10:
  t20 = (t2 + t19);
  t21 = (t4 + t19);
  if (*((unsigned char *)t20) != *((unsigned char *)t21))
    goto LAB7;
LAB11:
  t19 = (t19 + 1);
  goto LAB9;
LAB12:;
}
```
user: Translate the below assembly code into C programming language:

```asm
<ft_recursive_power>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  jle    L30
  cmpl   $0x0,-0x18(%rbp)
  jg     L12
L30:
  mov    $0x0,%eax
  jmp    L39
L12:
  cmpl   $0x1,-0x18(%rbp)
  jle    L72
  cmpl   $0x0,-0x14(%rbp)
  jle    L72
  mov    -0x18(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    -0x14(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <ft_recursive_power>
  mov    -0x14(%rbp),%edx
  imul   %edx,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  jmp    L39
L72:
  cmpl   $0x1,-0x18(%rbp)
  jne    L65
  cmpl   $0x0,-0x14(%rbp)
  je     L65
  mov    -0x14(%rbp),%eax
  jmp    L39
L65:
  mov    $0x1,%eax
L39:
  leave
  ret
```

assistant: ```c
int ft_recursive_power(int nb, int power) {
  if (nb <= 0 || power <= 0) {
    return (0);
  } else if (power > 1 && nb > 0) {
    int result;
    result = nb * ft_recursive_power(nb, power - 1);
    return (result);
  } else if (power == 1 && nb != 0)
    return (nb);
  else
    return (1);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<ft_recursive_power>:
  endbr64
  push   %rbx
  test   %edi,%edi
  jle    L47
  mov    %edi,%ebx
  test   %esi,%esi
  jle    L47
  cmp    $0x1,%esi
  jg     L3
L28:
  mov    %ebx,%eax
  pop    %rbx
  ret
L3:
  sub    $0x1,%esi
  call   <ft_recursive_power>
  imul   %eax,%ebx
  jmp    L28
L47:
  mov    $0x0,%ebx
  jmp    L28
```

assistant: ```c
int ft_recursive_power(int nb, int power) {
  if (nb <= 0 || power <= 0) {
    return (0);
  } else if (power > 1 && nb > 0) {
    int result;
    result = nb * ft_recursive_power(nb, power - 1);
    return (result);
  } else if (power == 1 && nb != 0)
    return (nb);
  else
    return (1);
}
```
user: Translate the below assembly code into C programming language:

```asm
<ft_recursive_power>:
  endbr64
  test   %esi,%esi
  jle    L39
  test   %edi,%edi
  jle    L39
  cmp    $0x1,%esi
  je     L2
  mov    $0x1,%eax
  cs nopw 0x0(%rax,%rax,1)
L0:
  sub    $0x1,%esi
  imul   %edi,%eax
  cmp    $0x1,%esi
  jne    L0
  imul   %edi,%eax
  ret
L39:
  xor    %eax,%eax
  ret
L2:
  mov    %edi,%eax
  ret
```

assistant: ```c
int ft_recursive_power(int nb, int power) {
  if (nb <= 0 || power <= 0) {
    return (0);
  } else if (power > 1 && nb > 0) {
    int result;
    result = nb * ft_recursive_power(nb, power - 1);
    return (result);
  } else if (power == 1 && nb != 0)
    return (nb);
  else
    return (1);
}
```
user: Translate the below assembly code into C programming language:

```asm
<ft_recursive_power>:
  endbr64
  test   %esi,%esi
  jle    L19
  test   %edi,%edi
  jle    L19
  cmp    $0x1,%esi
  je     L27
  mov    $0x1,%eax
  cs nopw 0x0(%rax,%rax,1)
L30:
  sub    $0x1,%esi
  imul   %edi,%eax
  cmp    $0x1,%esi
  jne    L30
  imul   %edi,%eax
  ret
L19:
  xor    %eax,%eax
  ret
L27:
  mov    %edi,%eax
  ret
```

assistant: ```c
int ft_recursive_power(int nb, int power) {
  if (nb <= 0 || power <= 0) {
    return (0);
  } else if (power > 1 && nb > 0) {
    int result;
    result = nb * ft_recursive_power(nb, power - 1);
    return (result);
  } else if (power == 1 && nb != 0)
    return (nb);
  else
    return (1);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x3,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0xc(%rbp)
  mov    %eax,-0x8(%rbp)
  addl   $0x1,-0x8(%rbp)
  mov    -0x8(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%ecx
  mov    -0x8(%rbp),%edx
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D7
type_guessed: string

tool: D7: a = %d b = %d c = %d


assistant: ```c
int main(int argc, char **argv) {
  int a = 3;
  int b = a++;
  int c = ++b;
  printf("a = %d b = %d c = %d\n", a, b, c);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x4,%ecx
  mov    $0x4,%edx
  mov    $0x4,%esi
  lea    D0(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: tag: D0
type_infered: string

tool: D0: a = %d b = %d c = %d


assistant: ```c
int main(int argc, char **argv) {
  int a = 3;
  int b = a++;
  int c = ++b;
  printf("a = %d b = %d c = %d\n", a, b, c);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x4,%ecx
  mov    $0x4,%edx
  mov    $0x4,%esi
  lea    D2(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: label: D2
data_type: string

tool: D2: a = %d b = %d c = %d


assistant: ```c
int main(int argc, char **argv) {
  int a = 3;
  int b = a++;
  int c = ++b;
  printf("a = %d b = %d c = %d\n", a, b, c);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x4,%ecx
  mov    $0x4,%edx
  mov    $0x4,%esi
  lea    D0(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

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

tool: D0: a = %d b = %d c = %d


assistant: ```c
int main(int argc, char **argv) {
  int a = 3;
  int b = a++;
  int c = ++b;
  printf("a = %d b = %d c = %d\n", a, b, c);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<find_maximum_row>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %esi,-0x3c(%rbp)
  mov    %edx,-0x40(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  mov    %rsp,%rax
  mov    %rax,%rbx
  mov    -0x3c(%rbp),%eax
  movslq %eax,%rdx
  sub    $0x1,%rdx
  mov    %rdx,-0x28(%rbp)
  movslq %eax,%rdx
  mov    %rdx,%r10
  mov    $0x0,%r11d
  movslq %eax,%rdx
  mov    %rdx,%r8
  mov    $0x0,%r9d
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    $0x10,%eax
  sub    $0x1,%rax
  add    %rdx,%rax
  mov    $0x10,%esi
  mov    $0x0,%edx
  div    %rsi
  imul   $0x10,%rax,%rax
  mov    %rax,%rcx
  and    $0xfffffffffffff000,%rcx
  mov    %rsp,%rdx
  sub    %rcx,%rdx
L20:
  cmp    %rdx,%rsp
  je     L24
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L20
L24:
  mov    %rax,%rdx
  and    $0xfff,%edx
  sub    %rdx,%rsp
  mov    %rax,%rdx
  and    $0xfff,%edx
  test   %rdx,%rdx
  je     L23
  and    $0xfff,%eax
  sub    $0x8,%rax
  add    %rsp,%rax
  orq    $0x0,(%rax)
L23:
  mov    %rsp,%rax
  add    $0x3,%rax
  shr    $0x2,%rax
  shl    $0x2,%rax
  mov    %rax,-0x20(%rbp)
  movl   $0x0,-0x2c(%rbp)
  jmp    L51
L26:
  mov    -0x40(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x2c(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rdx,%rax
  mov    (%rax),%ecx
  mov    -0x20(%rbp),%rax
  mov    -0x2c(%rbp),%edx
  movslq %edx,%rdx
  mov    %ecx,(%rax,%rdx,4)
  addl   $0x1,-0x2c(%rbp)
L51:
  mov    -0x2c(%rbp),%eax
  cmp    -0x3c(%rbp),%eax
  jl     L26
  mov    -0x3c(%rbp),%edx
  mov    -0x20(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <find_max_subsequence@plt>
  mov    %rbx,%rsp
  mov    -0x18(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L17
  call   <__stack_chk_fail@plt>
L17:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int find_maximum_row(int **matrix, int size, int row) {
  int values[size];
  for (int i = 0; i < size; i++)
    values[i] = matrix[row][i];
  return find_max_subsequence(values, size);
}
```