ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请从给定的汇编指令推导出C源代码:

```asm
<fa_fast_atan2>:
  endbr64
  movaps %xmm0,%xmm2
  pxor   %xmm0,%xmm0
  sub    $0x8,%rsp
  comiss %xmm0,%xmm2
  movaps %xmm2,%xmm3
  ja     L44
  xorps  D41(%rip),%xmm3
L44:
  comiss %xmm0,%xmm1
  jbe    L40
  comiss %xmm1,%xmm3
  jb     L9
L10:
  divss  %xmm2,%xmm1
  comiss %xmm0,%xmm2
  movaps %xmm1,%xmm0
  jbe    L29
  call   <fast_atan@plt>
  movapd %xmm0,%xmm1
  movsd  D26(%rip),%xmm0
  add    $0x8,%rsp
  subsd  %xmm1,%xmm0
  cvtsd2ss %xmm0,%xmm0
  ret
  nopw   0x0(%rax,%rax,1)
L9:
  divss  %xmm1,%xmm2
  movaps %xmm2,%xmm0
  call   <fast_atan@plt>
  add    $0x8,%rsp
  cvtsd2ss %xmm0,%xmm0
  ret
  nopl   (%rax)
L29:
  call   <fast_atan@plt>
  movapd %xmm0,%xmm1
  movsd  D26(%rip),%xmm0
  subsd  %xmm1,%xmm0
  subsd  D58(%rip),%xmm0
  add    $0x8,%rsp
  cvtsd2ss %xmm0,%xmm0
  ret
  cs nopw 0x0(%rax,%rax,1)
L40:
  movaps %xmm1,%xmm4
  xorps  D41(%rip),%xmm4
  comiss %xmm4,%xmm3
  jae    L10
  divss  %xmm1,%xmm2
  movaps %xmm2,%xmm0
  call   <fast_atan@plt>
  subsd  D58(%rip),%xmm0
  add    $0x8,%rsp
  cvtsd2ss %xmm0,%xmm0
  ret
```

tool_calls: name: read_data
arguments: tag: D41
type_guessed: byte[16]
name: read_data
arguments: tag: D26
type_guessed: f64
name: read_data
arguments: tag: D58
type_guessed: f64

tool: {"D41": [0x00,0x00,0x00,0x80,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00]}

tool: D26: 1.5707963267948966

tool: D58: 3.141592653589793

assistant: ```c
float fa_fast_atan2(float y, float x) {
  float sita;
  float abs_y, abs_x;
  abs_y = ((y) > 0 ? (y) : (-y));
  abs_x = ((x) > 0 ? (x) : (-x));
  if (abs_y >= abs_x) {
    if (y > 0)
      sita = 1.57079632679489661923 - fast_atan(x / y);
    else
      sita = 1.57079632679489661923 - fast_atan(x / y) - 3.14159265358979323846;
  } else {
    if (x > 0)
      sita = fast_atan(y / x);
    else
      sita = fast_atan(y / x) - 3.14159265358979323846;
  }
  return sita;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<wrap_get_iface_nr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x4(%rbp)
  mov    $0x0,%eax
  call   <get_iface_nr@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jle    L16
  mov    -0x4(%rbp),%eax
  add    $0x2,%eax
  jmp    L12
L16:
  mov    $0x0,%eax
L12:
  leave
  ret
```

assistant: ```c
int wrap_get_iface_nr(struct activity *a) {
  int n = 0;
  if ((n = get_iface_nr()) > 0)
    return n + 2;
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<wrap_get_iface_nr>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <get_iface_nr@plt>
  lea    0x2(%rax),%edx
  test   %eax,%eax
  mov    $0x0,%eax
  cmovg  %edx,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int wrap_get_iface_nr(struct activity *a) {
  int n = 0;
  if ((n = get_iface_nr()) > 0)
    return n + 2;
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<wrap_get_iface_nr>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <get_iface_nr@plt>
  test   %eax,%eax
  lea    0x2(%rax),%edx
  mov    $0x0,%eax
  cmovg  %edx,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int wrap_get_iface_nr(struct activity *a) {
  int n = 0;
  if ((n = get_iface_nr()) > 0)
    return n + 2;
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<wrap_get_iface_nr>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <get_iface_nr@plt>
  test   %eax,%eax
  lea    0x2(%rax),%edx
  mov    $0x0,%eax
  cmovg  %edx,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int wrap_get_iface_nr(struct activity *a) {
  int n = 0;
  if ((n = get_iface_nr()) > 0)
    return n + 2;
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<eembed_str_to_long>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0xa,-0x4(%rbp)
  mov    -0x4(%rbp),%edx
  mov    -0x20(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <eembed_str_to_i64@plt>
  leave
  ret
```

assistant: ```c
int64_t eembed_str_to_long(const char *str, char **endptr) {
  int base10 = 10;
  return eembed_str_to_i64(str, endptr, base10);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<eembed_str_to_long>:
  endbr64
  sub    $0x8,%rsp
  mov    $0xa,%edx
  call   <eembed_str_to_i64@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
int64_t eembed_str_to_long(const char *str, char **endptr) {
  int base10 = 10;
  return eembed_str_to_i64(str, endptr, base10);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<eembed_str_to_long>:
  endbr64
  mov    $0xa,%edx
  jmp    1050 <eembed_str_to_i64@plt>
```

assistant: ```c
int64_t eembed_str_to_long(const char *str, char **endptr) {
  int base10 = 10;
  return eembed_str_to_i64(str, endptr, base10);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<eembed_str_to_long>:
  endbr64
  mov    $0xa,%edx
  jmp    1050 <eembed_str_to_i64@plt>
```

assistant: ```c
int64_t eembed_str_to_long(const char *str, char **endptr) {
  int base10 = 10;
  return eembed_str_to_i64(str, endptr, base10);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<hash1>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movq   $0x1505,-0x8(%rbp)
  jmp    L6
L10:
  mov    -0x8(%rbp),%rax
  shl    $0x5,%rax
  mov    %rax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rax,%rdx
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movsbq %al,%rax
  add    %rdx,%rax
  mov    %rax,-0x8(%rbp)
  addq   $0x1,-0x18(%rbp)
L6:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L10
  mov    -0x8(%rbp),%rax
  cqto
  idivq  -0x20(%rbp)
  mov    %rdx,%rax
  pop    %rbp
  ret
```

assistant: ```c
long hash1(const char *s, long sz) {
  long h;
  for (h = 5381; *s != '\0'; s++) {
    h = ((h << 5) + h) + (long)*s;
  }
  return h % sz;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<hash1>:
  endbr64
  movzbl (%rdi),%edx
  test   %dl,%dl
  je     L0
  mov    $0x1505,%eax
L25:
  mov    %rax,%rcx
  shl    $0x5,%rcx
  add    %rcx,%rax
  movsbq %dl,%rdx
  add    %rdx,%rax
  add    $0x1,%rdi
  movzbl (%rdi),%edx
  test   %dl,%dl
  jne    L25
L4:
  cqto
  idiv   %rsi
  mov    %rdx,%rax
  ret
L0:
  mov    $0x1505,%eax
  jmp    L4
```

assistant: ```c
long hash1(const char *s, long sz) {
  long h;
  for (h = 5381; *s != '\0'; s++) {
    h = ((h << 5) + h) + (long)*s;
  }
  return h % sz;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<hash1>:
  endbr64
  movsbq (%rdi),%rdx
  mov    $0x1505,%eax
  test   %dl,%dl
  je     L1
  nopl   0x0(%rax)
L2:
  mov    %rax,%rcx
  add    $0x1,%rdi
  shl    $0x5,%rcx
  add    %rcx,%rax
  add    %rdx,%rax
  movsbq (%rdi),%rdx
  test   %dl,%dl
  jne    L2
L1:
  cqto
  idiv   %rsi
  mov    %rdx,%rax
  ret
```

assistant: ```c
long hash1(const char *s, long sz) {
  long h;
  for (h = 5381; *s != '\0'; s++) {
    h = ((h << 5) + h) + (long)*s;
  }
  return h % sz;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<hash1>:
  endbr64
  movsbq (%rdi),%rdx
  mov    $0x1505,%eax
  test   %dl,%dl
  je     L2
  nopl   0x0(%rax)
L9:
  mov    %rax,%rcx
  add    $0x1,%rdi
  shl    $0x5,%rcx
  add    %rcx,%rax
  add    %rdx,%rax
  movsbq (%rdi),%rdx
  test   %dl,%dl
  jne    L9
L2:
  cqto
  idiv   %rsi
  mov    %rdx,%rax
  ret
```

assistant: ```c
long hash1(const char *s, long sz) {
  long h;
  for (h = 5381; *s != '\0'; s++) {
    h = ((h << 5) + h) + (long)*s;
  }
  return h % sz;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<setPixelRGBCol>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  mov    %ecx,-0x28(%rbp)
  mov    %r8d,-0x2c(%rbp)
  cmpl   $0x0,-0x24(%rbp)
  je     L38
  mov    -0x20(%rbp),%rax
  mov    0x10(%rax),%eax
  mov    $0x1,%edx
  mov    %eax,%ecx
  shl    %cl,%edx
  mov    %edx,%eax
  cmp    %eax,-0x24(%rbp)
  jb     L3
  mov    -0x20(%rbp),%rax
  mov    0x10(%rax),%eax
  mov    $0x1,%edx
  mov    %eax,%ecx
  shl    %cl,%edx
  mov    %edx,%eax
  sub    $0x1,%eax
  jmp    L99
L3:
  mov    -0x24(%rbp),%eax
L99:
  mov    -0x20(%rbp),%rdx
  mov    0x14(%rdx),%edx
  mov    %edx,%ecx
  shl    %cl,%eax
  mov    %eax,%esi
  jmp    L72
L38:
  mov    $0x0,%esi
L72:
  cmpl   $0x0,-0x28(%rbp)
  je     L42
  mov    -0x20(%rbp),%rax
  mov    0x8(%rax),%eax
  mov    $0x1,%edx
  mov    %eax,%ecx
  shl    %cl,%edx
  mov    %edx,%eax
  cmp    %eax,-0x28(%rbp)
  jb     L74
  mov    -0x20(%rbp),%rax
  mov    0x8(%rax),%eax
  mov    $0x1,%edx
  mov    %eax,%ecx
  shl    %cl,%edx
  mov    %edx,%eax
  sub    $0x1,%eax
  jmp    L101
L74:
  mov    -0x28(%rbp),%eax
L101:
  mov    -0x20(%rbp),%rdx
  mov    0xc(%rdx),%edx
  mov    %edx,%ecx
  shl    %cl,%eax
  jmp    L108
L42:
  mov    $0x0,%eax
L108:
  or     %eax,%esi
  cmpl   $0x0,-0x2c(%rbp)
  je     L54
  mov    -0x20(%rbp),%rax
  mov    (%rax),%eax
  mov    $0x1,%edx
  mov    %eax,%ecx
  shl    %cl,%edx
  mov    %edx,%eax
  cmp    %eax,-0x2c(%rbp)
  jb     L8
  mov    -0x20(%rbp),%rax
  mov    (%rax),%eax
  mov    $0x1,%edx
  mov    %eax,%ecx
  shl    %cl,%edx
  mov    %edx,%eax
  sub    $0x1,%eax
  jmp    L21
L8:
  mov    -0x2c(%rbp),%eax
L21:
  mov    -0x20(%rbp),%rdx
  mov    0x4(%rdx),%edx
  mov    %edx,%ecx
  shl    %cl,%eax
  jmp    L56
L54:
  mov    $0x0,%eax
L56:
  or     %esi,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%edx
  mov    -0x20(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <setPixelRGBRaw@plt>
  leave
  ret
```

assistant: ```c
void *setPixelRGBCol(void *buf_v, struct fb_var_screeninfo const *info, uint8_t r, uint8_t g, uint8_t b) {
  uint32_t val = ((r) == 0 ? 0 : (((r) >= (1u << info->red.length) ? ((1u << info->red.length) - 1u) : (r)) << (info->red.offset))) | ((g) == 0 ? 0 : (((g) >= (1u << info->green.length) ? ((1u << info->green.length) - 1u) : (g)) << (info->green.offset))) | ((b) == 0 ? 0 : (((b) >= (1u << info->blue.length) ? ((1u << info->blue.length) - 1u) : (b)) << (info->blue.offset)));
  return setPixelRGBRaw(buf_v, info, val);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<setPixelRGBCol>:
  endbr64
  sub    $0x8,%rsp
  mov    %ecx,%eax
  test   %edx,%edx
  je     L3
  mov    0x10(%rsi),%ecx
  mov    $0x1,%r9d
  shl    %cl,%r9d
  mov    %r9d,%ecx
  lea    -0x1(%r9),%r9d
  cmp    %edx,%ecx
  cmovbe %r9d,%edx
  mov    0x14(%rsi),%ecx
  shl    %cl,%edx
L3:
  test   %eax,%eax
  je     L18
  mov    0x8(%rsi),%ecx
  mov    $0x1,%r9d
  shl    %cl,%r9d
  mov    %r9d,%ecx
  lea    -0x1(%r9),%r9d
  cmp    %eax,%ecx
  cmovbe %r9d,%eax
  mov    0xc(%rsi),%ecx
  shl    %cl,%eax
L18:
  or     %eax,%edx
  test   %r8d,%r8d
  je     L19
  mov    (%rsi),%ecx
  mov    $0x1,%eax
  shl    %cl,%eax
  lea    -0x1(%rax),%ecx
  cmp    %r8d,%eax
  cmovbe %ecx,%r8d
  mov    0x4(%rsi),%ecx
  shl    %cl,%r8d
L19:
  or     %r8d,%edx
  call   <setPixelRGBRaw@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void *setPixelRGBCol(void *buf_v, struct fb_var_screeninfo const *info, uint8_t r, uint8_t g, uint8_t b) {
  uint32_t val = ((r) == 0 ? 0 : (((r) >= (1u << info->red.length) ? ((1u << info->red.length) - 1u) : (r)) << (info->red.offset))) | ((g) == 0 ? 0 : (((g) >= (1u << info->green.length) ? ((1u << info->green.length) - 1u) : (g)) << (info->green.offset))) | ((b) == 0 ? 0 : (((b) >= (1u << info->blue.length) ? ((1u << info->blue.length) - 1u) : (b)) << (info->blue.offset)));
  return setPixelRGBRaw(buf_v, info, val);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<setPixelRGBCol>:
  endbr64
  mov    %ecx,%eax
  test   %edx,%edx
  je     L11
  mov    0x10(%rsi),%ecx
  mov    $0x1,%r9d
  shl    %cl,%r9d
  mov    %r9d,%ecx
  lea    -0x1(%r9),%r9d
  cmp    %edx,%ecx
  mov    0x14(%rsi),%ecx
  cmovbe %r9d,%edx
  shl    %cl,%edx
L11:
  test   %eax,%eax
  je     L0
  mov    0x8(%rsi),%ecx
  mov    $0x1,%r9d
  shl    %cl,%r9d
  mov    %r9d,%ecx
  lea    -0x1(%r9),%r9d
  cmp    %eax,%ecx
  mov    0xc(%rsi),%ecx
  cmovbe %r9d,%eax
  shl    %cl,%eax
  or     %eax,%edx
L0:
  test   %r8d,%r8d
  je     L26
  mov    (%rsi),%ecx
  mov    $0x1,%eax
  shl    %cl,%eax
  cmp    %r8d,%eax
  lea    -0x1(%rax),%ecx
  cmovbe %ecx,%r8d
  mov    0x4(%rsi),%ecx
  shl    %cl,%r8d
  or     %r8d,%edx
L26:
  jmp    1050 <setPixelRGBRaw@plt>
```

assistant: ```c
void *setPixelRGBCol(void *buf_v, struct fb_var_screeninfo const *info, uint8_t r, uint8_t g, uint8_t b) {
  uint32_t val = ((r) == 0 ? 0 : (((r) >= (1u << info->red.length) ? ((1u << info->red.length) - 1u) : (r)) << (info->red.offset))) | ((g) == 0 ? 0 : (((g) >= (1u << info->green.length) ? ((1u << info->green.length) - 1u) : (g)) << (info->green.offset))) | ((b) == 0 ? 0 : (((b) >= (1u << info->blue.length) ? ((1u << info->blue.length) - 1u) : (b)) << (info->blue.offset)));
  return setPixelRGBRaw(buf_v, info, val);
}
```
user: Translate the below assembly code into C programming language:

```asm
<setPixelRGBCol>:
  endbr64
  mov    %ecx,%eax
  test   %edx,%edx
  je     L23
  mov    0x10(%rsi),%ecx
  mov    $0x1,%r9d
  shl    %cl,%r9d
  mov    %r9d,%ecx
  lea    -0x1(%r9),%r9d
  cmp    %edx,%ecx
  mov    0x14(%rsi),%ecx
  cmovbe %r9d,%edx
  shl    %cl,%edx
L23:
  test   %eax,%eax
  je     L1
  mov    0x8(%rsi),%ecx
  mov    $0x1,%r9d
  shl    %cl,%r9d
  mov    %r9d,%ecx
  lea    -0x1(%r9),%r9d
  cmp    %eax,%ecx
  mov    0xc(%rsi),%ecx
  cmovbe %r9d,%eax
  shl    %cl,%eax
  or     %eax,%edx
L1:
  test   %r8d,%r8d
  je     L6
  mov    (%rsi),%ecx
  mov    $0x1,%eax
  shl    %cl,%eax
  cmp    %r8d,%eax
  lea    -0x1(%rax),%ecx
  cmovbe %ecx,%r8d
  mov    0x4(%rsi),%ecx
  shl    %cl,%r8d
  or     %r8d,%edx
L6:
  jmp    1050 <setPixelRGBRaw@plt>
```

assistant: ```c
void *setPixelRGBCol(void *buf_v, struct fb_var_screeninfo const *info, uint8_t r, uint8_t g, uint8_t b) {
  uint32_t val = ((r) == 0 ? 0 : (((r) >= (1u << info->red.length) ? ((1u << info->red.length) - 1u) : (r)) << (info->red.offset))) | ((g) == 0 ? 0 : (((g) >= (1u << info->green.length) ? ((1u << info->green.length) - 1u) : (g)) << (info->green.offset))) | ((b) == 0 ? 0 : (((b) >= (1u << info->blue.length) ? ((1u << info->blue.length) - 1u) : (b)) << (info->blue.offset)));
  return setPixelRGBRaw(buf_v, info, val);
}
```
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    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D46(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D49(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    -0x18(%rbp),%eax
  shl    $0x2,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x10(%rbp)
  cmpq   $0x0,-0x10(%rbp)
  jne    L58
  mov    $0xffffffff,%eax
  jmp    L55
L58:
  lea    D34(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x0,-0x14(%rbp)
  jmp    L21
L4:
  mov    -0x14(%rbp),%eax
  lea    0x0(,%rax,4),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,%rsi
  lea    D49(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  addl   $0x1,-0x14(%rbp)
L21:
  mov    -0x18(%rbp),%eax
  cmp    %eax,-0x14(%rbp)
  jb     L4
  mov    -0x18(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    -0x10(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <quickSort@plt>
  movl   $0x0,-0x14(%rbp)
  jmp    L9
L12:
  mov    -0x14(%rbp),%eax
  lea    0x0(,%rax,4),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,%esi
  lea    D39(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x14(%rbp)
L9:
  mov    -0x18(%rbp),%eax
  cmp    %eax,-0x14(%rbp)
  jb     L12
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    $0x0,%eax
L55:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L23
  call   <__stack_chk_fail@plt>
L23:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: tag: D46
data_type: string
name: unpack_data
arguments: tag: D49
data_type: string
name: unpack_data
arguments: tag: D34
data_type: string
name: unpack_data
arguments: tag: D39
data_type: string

tool: D46: number of int:

tool: D49: %d

tool: D34: numbers:

tool: D39: %d 

assistant: ```c
int main(int argc, char *argv[]) {
  unsigned int num;
  unsigned int i;
  int *p;
  printf("number of int:");
  scanf("%d", &num);
  p = (int *)malloc(sizeof(int) * num);
  if (!p)
    return -1;
  printf("numbers:");
  for (i = 0; i < num; i++) {
    scanf("%d", &p[i]);
  }
  quickSort(p, 0, num - 1);
  for (i = 0; i < num; i++) {
    printf("%d ", p[i]);
  }
  printf("\n");
  free(p);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D40(%rip),%rdi
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D33(%rip),%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  lea    0x0(,%rax,4),%edi
  call   <malloc@plt>
  test   %rax,%rax
  je     L65
  mov    %rax,%rbp
  lea    D41(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x4(%rsp),%edx
  test   %edx,%edx
  je     L24
  mov    $0x0,%ebx
  lea    D33(%rip),%r12
L31:
  mov    %ebx,%eax
  lea    0x0(%rbp,%rax,4),%rsi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  add    $0x1,%ebx
  mov    0x4(%rsp),%edx
  cmp    %ebx,%edx
  ja     L31
L24:
  sub    $0x1,%edx
  mov    $0x0,%esi
  mov    %rbp,%rdi
  call   <quickSort@plt>
  cmpl   $0x0,0x4(%rsp)
  je     L20
  mov    $0x0,%ebx
  lea    D7(%rip),%r12
L10:
  mov    %ebx,%eax
  mov    0x0(%rbp,%rax,4),%esi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x1,%ebx
  cmp    %ebx,0x4(%rsp)
  ja     L10
L20:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    %rbp,%rdi
  call   <free@plt>
  mov    $0x0,%eax
L35:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L6
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L65:
  mov    $0xffffffff,%eax
  jmp    L35
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D40
type_guessed: string
name: retrieve_data
arguments: label: D33
type_guessed: string
name: retrieve_data
arguments: label: D41
type_guessed: string
name: retrieve_data
arguments: label: D7
type_guessed: string

tool: D40: number of int:

tool: D33: %d

tool: D41: numbers:

tool: D7: %d 

assistant: ```c
int main(int argc, char *argv[]) {
  unsigned int num;
  unsigned int i;
  int *p;
  printf("number of int:");
  scanf("%d", &num);
  p = (int *)malloc(sizeof(int) * num);
  if (!p)
    return -1;
  printf("numbers:");
  for (i = 0; i < num; i++) {
    scanf("%d", &p[i]);
  }
  quickSort(p, 0, num - 1);
  for (i = 0; i < num; i++) {
    printf("%d ", p[i]);
  }
  printf("\n");
  free(p);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %r12
  lea    D23(%rip),%rdi
  lea    D36(%rip),%r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r12,%rdi
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  lea    0x0(,%rax,4),%edi
  call   <malloc@plt>
  test   %rax,%rax
  je     L34
  mov    %rax,%rbp
  lea    D24(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x4(%rsp),%edx
  test   %edx,%edx
  je     L49
  xor    %ebx,%ebx
  nopw   0x0(%rax,%rax,1)
L28:
  mov    %ebx,%eax
  mov    %r12,%rdi
  add    $0x1,%ebx
  lea    0x0(%rbp,%rax,4),%rsi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    0x4(%rsp),%edx
  cmp    %ebx,%edx
  ja     L28
L49:
  sub    $0x1,%edx
  xor    %esi,%esi
  mov    %rbp,%rdi
  call   <quickSort@plt>
  mov    0x4(%rsp),%eax
  test   %eax,%eax
  je     L23
  xor    %ebx,%ebx
  lea    D1(%rip),%r12
  nopw   0x0(%rax,%rax,1)
L60:
  mov    %ebx,%eax
  mov    %r12,%rdi
  add    $0x1,%ebx
  mov    0x0(%rbp,%rax,4),%esi
  xor    %eax,%eax
  call   <printf@plt>
  cmp    %ebx,0x4(%rsp)
  ja     L60
L23:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    %rbp,%rdi
  call   <free@plt>
  xor    %eax,%eax
L62:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L58
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L34:
  or     $0xffffffff,%eax
  jmp    L62
L58:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D23
datatype: string
name: disassemble_data
arguments: data_tag: D36
datatype: string
name: disassemble_data
arguments: data_tag: D24
datatype: string
name: disassemble_data
arguments: data_tag: D1
datatype: string

tool: D23: number of int:

tool: D36: %d

tool: D24: numbers:

tool: D1: %d 

assistant: ```c
int main(int argc, char *argv[]) {
  unsigned int num;
  unsigned int i;
  int *p;
  printf("number of int:");
  scanf("%d", &num);
  p = (int *)malloc(sizeof(int) * num);
  if (!p)
    return -1;
  printf("numbers:");
  for (i = 0; i < num; i++) {
    scanf("%d", &p[i]);
  }
  quickSort(p, 0, num - 1);
  for (i = 0; i < num; i++) {
    printf("%d ", p[i]);
  }
  printf("\n");
  free(p);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r12
  lea    D27(%rip),%rdi
  lea    D39(%rip),%r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r12,%rdi
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  lea    0x0(,%rax,4),%edi
  call   <malloc@plt>
  test   %rax,%rax
  je     L20
  mov    %rax,%rbp
  lea    D4(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x4(%rsp),%edx
  test   %edx,%edx
  je     L26
  xor    %ebx,%ebx
  nopw   0x0(%rax,%rax,1)
L54:
  mov    %ebx,%eax
  mov    %r12,%rdi
  add    $0x1,%ebx
  lea    0x0(%rbp,%rax,4),%rsi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    0x4(%rsp),%edx
  cmp    %ebx,%edx
  ja     L54
L26:
  sub    $0x1,%edx
  xor    %esi,%esi
  mov    %rbp,%rdi
  call   <quickSort@plt>
  mov    0x4(%rsp),%eax
  test   %eax,%eax
  je     L36
  xor    %ebx,%ebx
  lea    D17(%rip),%r12
  nopw   0x0(%rax,%rax,1)
L46:
  mov    %ebx,%eax
  mov    %r12,%rdi
  add    $0x1,%ebx
  mov    0x0(%rbp,%rax,4),%esi
  xor    %eax,%eax
  call   <printf@plt>
  cmp    %ebx,0x4(%rsp)
  ja     L46
L36:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    %rbp,%rdi
  call   <free@plt>
  xor    %eax,%eax
L42:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L4
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L20:
  or     $0xffffffff,%eax
  jmp    L42
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D27
type_guessed: string
name: parse_data
arguments: label: D39
type_guessed: string
name: parse_data
arguments: label: D4
type_guessed: string
name: parse_data
arguments: label: D17
type_guessed: string

tool: D27: number of int:

tool: D39: %d

tool: D4: numbers:

tool: D17: %d 

assistant: ```c
int main(int argc, char *argv[]) {
  unsigned int num;
  unsigned int i;
  int *p;
  printf("number of int:");
  scanf("%d", &num);
  p = (int *)malloc(sizeof(int) * num);
  if (!p)
    return -1;
  printf("numbers:");
  for (i = 0; i < num; i++) {
    scanf("%d", &p[i]);
  }
  quickSort(p, 0, num - 1);
  for (i = 0; i < num; i++) {
    printf("%d ", p[i]);
  }
  printf("\n");
  free(p);
  return 0;
}
```