ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Converte this assembly snippet into its equivalent C code:

```asm
<mak_links>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    $0x1c0,%esi
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <mkdir@plt>
  test   %eax,%eax
  jne    L45
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  nop
L28:
  addq   $0x1,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L28
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  movb   $0x2f,(%rax)
  mov    -0x8(%rbp),%rax
  add    $0x1,%rax
  movb   $0x0,(%rax)
  movl   $0x30,-0xc(%rbp)
L10:
  mov    -0xc(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0xc(%rbp)
  mov    %eax,%edx
  mov    -0x8(%rbp),%rax
  mov    %dl,(%rax)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <symlink@plt>
  cmpl   $0x57,-0xc(%rbp)
  je     L30
  cmpl   $0x3a,-0xc(%rbp)
  jne    L10
  movl   $0x40,-0xc(%rbp)
  jmp    L10
L30:
  nop
  mov    -0x8(%rbp),%rax
  sub    $0x1,%rax
  movb   $0x0,(%rax)
  jmp    L22
L45:
  nop
L22:
  leave
  ret
```

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

tool: D7: .

assistant: ```c
void mak_links(fpath) char *fpath;
{
  char *fname;
  int ch;
  if (mkdir(fpath, 0700))
    return;
  fname = fpath;
  while (*++fname)
    ;
  *fname++ = '/';
  fname[1] = '\0';
  ch = '0';
  for (;;) {
    *fname = ch++;
    symlink(".", fpath);
    if (ch == 'W')
      break;
    if (ch == '9' + 1)
      ch = '@';
  }
  fname[-1] = '\0';
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<mak_links>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r12
  mov    $0x1c0,%esi
  mov    $0x0,%eax
  call   <mkdir@plt>
  test   %eax,%eax
  jne    L13
  mov    %r12,%rbx
L6:
  add    $0x1,%rbx
  cmpb   $0x0,(%rbx)
  jne    L6
  movb   $0x2f,(%rbx)
  movb   $0x0,0x2(%rbx)
  movb   $0x30,0x1(%rbx)
  mov    %r12,%rsi
  lea    D2(%rip),%rdi
  mov    $0x0,%eax
  call   <symlink@plt>
  mov    $0x31,%eax
  lea    D2(%rip),%r13
  jmp    L14
L32:
  cmp    $0x3a,%ebp
  mov    $0x40,%eax
  cmovne %ebp,%eax
L14:
  lea    0x1(%rax),%ebp
  mov    %al,0x1(%rbx)
  mov    %r12,%rsi
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <symlink@plt>
  cmp    $0x57,%ebp
  jne    L32
  movb   $0x0,(%rbx)
L13:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: read_data
arguments: data_label: D2
type: string

tool: D2: .

assistant: ```c
void mak_links(fpath) char *fpath;
{
  char *fname;
  int ch;
  if (mkdir(fpath, 0700))
    return;
  fname = fpath;
  while (*++fname)
    ;
  *fname++ = '/';
  fname[1] = '\0';
  ch = '0';
  for (;;) {
    *fname = ch++;
    symlink(".", fpath);
    if (ch == 'W')
      break;
    if (ch == '9' + 1)
      ch = '@';
  }
  fname[-1] = '\0';
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<mak_links>:
  endbr64
  push   %r14
  xor    %eax,%eax
  mov    $0x1c0,%esi
  push   %r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  call   <mkdir@plt>
  test   %eax,%eax
  jne    L25
  mov    %r12,%rbx
  nopw   0x0(%rax,%rax,1)
L39:
  add    $0x1,%rbx
  cmpb   $0x0,(%rbx)
  jne    L39
  mov    $0x302f,%eax
  movb   $0x0,0x2(%rbx)
  mov    %r12,%rsi
  mov    $0x40,%r14d
  mov    %ax,(%rbx)
  lea    D7(%rip),%r13
  xor    %eax,%eax
  mov    %r13,%rdi
  call   <symlink@plt>
  mov    $0x31,%eax
  jmp    L26
  xchg   %ax,%ax
L15:
  cmp    $0x3a,%ebp
  mov    %ebp,%eax
  cmove  %r14d,%eax
L26:
  mov    %al,0x1(%rbx)
  lea    0x1(%rax),%ebp
  mov    %r12,%rsi
  xor    %eax,%eax
  mov    %r13,%rdi
  call   <symlink@plt>
  cmp    $0x57,%ebp
  jne    L15
  movb   $0x0,(%rbx)
L25:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

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

tool: D7: .

assistant: ```c
void mak_links(fpath) char *fpath;
{
  char *fname;
  int ch;
  if (mkdir(fpath, 0700))
    return;
  fname = fpath;
  while (*++fname)
    ;
  *fname++ = '/';
  fname[1] = '\0';
  ch = '0';
  for (;;) {
    *fname = ch++;
    symlink(".", fpath);
    if (ch == 'W')
      break;
    if (ch == '9' + 1)
      ch = '@';
  }
  fname[-1] = '\0';
}
```
user: Translate the below assembly code into C programming language:

```asm
<mak_links>:
  endbr64
  push   %r14
  xor    %eax,%eax
  mov    $0x1c0,%esi
  push   %r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  call   <mkdir@plt>
  test   %eax,%eax
  jne    L31
  mov    %r12,%rbx
  nopw   0x0(%rax,%rax,1)
L16:
  add    $0x1,%rbx
  cmpb   $0x0,(%rbx)
  jne    L16
  mov    $0x302f,%eax
  movb   $0x0,0x2(%rbx)
  mov    %r12,%rsi
  mov    $0x40,%r14d
  mov    %ax,(%rbx)
  lea    D2(%rip),%r13
  xor    %eax,%eax
  mov    %r13,%rdi
  call   <symlink@plt>
  mov    $0x31,%eax
  jmp    L36
  xchg   %ax,%ax
L30:
  cmp    $0x3a,%ebp
  mov    %ebp,%eax
  cmove  %r14d,%eax
L36:
  mov    %al,0x1(%rbx)
  lea    0x1(%rax),%ebp
  mov    %r12,%rsi
  xor    %eax,%eax
  mov    %r13,%rdi
  call   <symlink@plt>
  cmp    $0x57,%ebp
  jne    L30
  movb   $0x0,(%rbx)
L31:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

tool_calls: name: parse_data
arguments: tag: D2
type: string

tool: D2: .

assistant: ```c
void mak_links(fpath) char *fpath;
{
  char *fname;
  int ch;
  if (mkdir(fpath, 0700))
    return;
  fname = fpath;
  while (*++fname)
    ;
  *fname++ = '/';
  fname[1] = '\0';
  ch = '0';
  for (;;) {
    *fname = ch++;
    symlink(".", fpath);
    if (ch == 'W')
      break;
    if (ch == '9' + 1)
      ch = '@';
  }
  fname[-1] = '\0';
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<rdpmc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x10000,%eax
  mov    %eax,%ecx
  rdpmc
  mov    %eax,-0x8(%rbp)
  mov    %edx,-0x4(%rbp)
  mov    -0x8(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
uint64_t rdpmc(void) {
  union {
    uint64_t tsc_64;
    struct {
      uint32_t lo_32;
      uint32_t hi_32;
    } t32;
  } tsc;
  asm volatile("rdpmc" : "=a"(tsc.t32.lo_32), "=d"(tsc.t32.hi_32) : "c"(0x10000));
  return tsc.tsc_64;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<rdpmc>:
  endbr64
  mov    $0x10000,%ecx
  rdpmc
  shl    $0x20,%rdx
  mov    %eax,%eax
  or     %rdx,%rax
  ret
```

assistant: ```c
uint64_t rdpmc(void) {
  union {
    uint64_t tsc_64;
    struct {
      uint32_t lo_32;
      uint32_t hi_32;
    } t32;
  } tsc;
  asm volatile("rdpmc" : "=a"(tsc.t32.lo_32), "=d"(tsc.t32.hi_32) : "c"(0x10000));
  return tsc.tsc_64;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<rdpmc>:
  endbr64
  mov    $0x10000,%ecx
  rdpmc
  shl    $0x20,%rdx
  mov    %eax,%eax
  or     %rdx,%rax
  ret
```

assistant: ```c
uint64_t rdpmc(void) {
  union {
    uint64_t tsc_64;
    struct {
      uint32_t lo_32;
      uint32_t hi_32;
    } t32;
  } tsc;
  asm volatile("rdpmc" : "=a"(tsc.t32.lo_32), "=d"(tsc.t32.hi_32) : "c"(0x10000));
  return tsc.tsc_64;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<rdpmc>:
  endbr64
  mov    $0x10000,%ecx
  rdpmc
  shl    $0x20,%rdx
  mov    %eax,%eax
  or     %rdx,%rax
  ret
```

assistant: ```c
uint64_t rdpmc(void) {
  union {
    uint64_t tsc_64;
    struct {
      uint32_t lo_32;
      uint32_t hi_32;
    } t32;
  } tsc;
  asm volatile("rdpmc" : "=a"(tsc.t32.lo_32), "=d"(tsc.t32.hi_32) : "c"(0x10000));
  return tsc.tsc_64;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %edi,-0x34(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x64,-0x28(%rbp)
  movss  D36(%rip),%xmm0
  movss  %xmm0,-0x24(%rbp)
  movsd  D46(%rip),%xmm0
  movsd  %xmm0,-0x20(%rbp)
  movb   $0x4b,-0x29(%rbp)
  movl   $0x72656d4f,-0x13(%rbp)
  movb   $0x0,-0xf(%rbp)
  movl   $0x697a6157,-0xe(%rbp)
  movw   $0x72,-0xa(%rbp)
  mov    -0x28(%rbp),%eax
  mov    %eax,%esi
  lea    D74(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  pxor   %xmm1,%xmm1
  cvtss2sd -0x24(%rbp),%xmm1
  movq   %xmm1,%rax
  movq   %rax,%xmm0
  lea    D47(%rip),%rax
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    -0x20(%rbp),%rax
  movq   %rax,%xmm0
  lea    D61(%rip),%rax
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  movsbl -0x29(%rbp),%eax
  mov    %eax,%esi
  lea    D86(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x13(%rbp),%rax
  mov    %rax,%rsi
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0xe(%rbp),%rax
  mov    %rax,%rsi
  lea    D32(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movsbl -0x29(%rbp),%edx
  lea    -0xe(%rbp),%rcx
  lea    -0x13(%rbp),%rax
  mov    %rax,%rsi
  lea    D23(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L6
  call   <__stack_chk_fail@plt>
L6:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D36
datatype: byte[4]
name: unpack_data
arguments: data_label: D46
datatype: f64
name: unpack_data
arguments: data_label: D74
datatype: string
name: unpack_data
arguments: data_label: D47
datatype: string
name: unpack_data
arguments: data_label: D61
datatype: string
name: unpack_data
arguments: data_label: D86
datatype: string
name: unpack_data
arguments: data_label: D10
datatype: string
name: unpack_data
arguments: data_label: D32
datatype: string
name: unpack_data
arguments: data_label: D23
datatype: string

tool: {"D36": [0x7b,0x14,0x16,0x40]}

tool: D46: 56789.4532

tool: D74: You are %d miles away.


tool: D47: You have %f levels of power.


tool: D61: You have %f awesome super powers.


tool: D86: I have an initial %c.


tool: D10: I have a first name %s.


tool: D32: I have a last name %s.


tool: D23: 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 = 'K';
  char first_name[] = "Omer";
  char last_name[] = "Wazir";
  printf("You are %d miles 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("I have a first name %s.\n", first_name);
  printf("I have a last name %s.\n", last_name);
  printf("My whole name is %s %c %s.\n", first_name, initial, last_name);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movl   $0x72656d4f,0xd(%rsp)
  movb   $0x0,0x11(%rsp)
  movl   $0x697a6157,0x12(%rsp)
  movw   $0x72,0x16(%rsp)
  mov    $0x64,%esi
  lea    D22(%rip),%rdi
  call   <printf@plt>
  movsd  D29(%rip),%xmm0
  lea    D68(%rip),%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  movsd  D49(%rip),%xmm0
  lea    D5(%rip),%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    $0x4b,%esi
  lea    D42(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0xd(%rsp),%rbx
  mov    %rbx,%rsi
  lea    D32(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0x12(%rsp),%rbp
  mov    %rbp,%rsi
  lea    D26(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %rbp,%rcx
  mov    $0x4b,%edx
  mov    %rbx,%rsi
  lea    D6(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  mov    $0x0,%eax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D22
datatype: string
name: unpack_data
arguments: data_label: D29
datatype: byte[8]
name: unpack_data
arguments: data_label: D68
datatype: string
name: unpack_data
arguments: data_label: D49
datatype: f64
name: unpack_data
arguments: data_label: D5
datatype: string
name: unpack_data
arguments: data_label: D42
datatype: string
name: unpack_data
arguments: data_label: D32
datatype: string
name: unpack_data
arguments: data_label: D26
datatype: string
name: unpack_data
arguments: data_label: D6
datatype: string

tool: D22: You are %d miles away.


tool: {"D29": [0x00,0x00,0x00,0x60,0x8f,0xc2,0x02,0x40]}

tool: D68: You have %f levels of power.


tool: D49: 56789.4532

tool: D5: You have %f awesome super powers.


tool: D42: I have an initial %c.


tool: D32: I have a first name %s.


tool: D26: I have a last name %s.


tool: D6: 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 = 'K';
  char first_name[] = "Omer";
  char last_name[] = "Wazir";
  printf("You are %d miles 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("I have a first name %s.\n", first_name);
  printf("I have a last name %s.\n", last_name);
  printf("My whole name is %s %c %s.\n", first_name, initial, last_name);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  push   %r12
  mov    $0x72,%edx
  mov    $0x64,%esi
  lea    D31(%rip),%rdi
  push   %rbp
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %dx,0x16(%rsp)
  lea    0xd(%rsp),%rbp
  lea    0x12(%rsp),%r12
  movl   $0x72656d4f,0xd(%rsp)
  movb   $0x0,0x11(%rsp)
  movl   $0x697a6157,0x12(%rsp)
  call   <printf@plt>
  movsd  D16(%rip),%xmm0
  mov    $0x1,%eax
  lea    D17(%rip),%rdi
  call   <printf@plt>
  movsd  D22(%rip),%xmm0
  mov    $0x1,%eax
  lea    D64(%rip),%rdi
  call   <printf@plt>
  mov    $0x4b,%esi
  lea    D0(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rsi
  lea    D79(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r12,%rsi
  lea    D63(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  mov    %r12,%rcx
  mov    $0x4b,%edx
  mov    %rbp,%rsi
  lea    D75(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D31
type: string
name: disassemble_data
arguments: tag: D16
type: byte[8]
name: disassemble_data
arguments: tag: D17
type: string
name: disassemble_data
arguments: tag: D22
type: f64
name: disassemble_data
arguments: tag: D64
type: string
name: disassemble_data
arguments: tag: D0
type: string
name: disassemble_data
arguments: tag: D79
type: string
name: disassemble_data
arguments: tag: D63
type: string
name: disassemble_data
arguments: tag: D75
type: string

tool: D31: You are %d miles away.


tool: {"D16": [0x00,0x00,0x00,0x60,0x8f,0xc2,0x02,0x40]}

tool: D17: You have %f levels of power.


tool: D22: 56789.4532

tool: D64: You have %f awesome super powers.


tool: D0: I have an initial %c.


tool: D79: I have a first name %s.


tool: D63: I have a last name %s.


tool: D75: 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 = 'K';
  char first_name[] = "Omer";
  char last_name[] = "Wazir";
  printf("You are %d miles 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("I have a first name %s.\n", first_name);
  printf("I have a last name %s.\n", last_name);
  printf("My whole name is %s %c %s.\n", first_name, initial, last_name);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  push   %r12
  mov    $0x72,%edx
  mov    $0x64,%esi
  lea    D39(%rip),%rdi
  push   %rbp
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %dx,0x16(%rsp)
  lea    0xd(%rsp),%rbp
  lea    0x12(%rsp),%r12
  movl   $0x72656d4f,0xd(%rsp)
  movb   $0x0,0x11(%rsp)
  movl   $0x697a6157,0x12(%rsp)
  call   <printf@plt>
  movsd  D61(%rip),%xmm0
  mov    $0x1,%eax
  lea    D53(%rip),%rdi
  call   <printf@plt>
  movsd  D8(%rip),%xmm0
  mov    $0x1,%eax
  lea    D23(%rip),%rdi
  call   <printf@plt>
  mov    $0x4b,%esi
  lea    D42(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rsi
  lea    D60(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r12,%rsi
  lea    D40(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  mov    %r12,%rcx
  mov    $0x4b,%edx
  mov    %rbp,%rsi
  lea    D57(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D39
type: string
name: extract_symbolic_data
arguments: tag: D61
type: byte[8]
name: extract_symbolic_data
arguments: tag: D53
type: string
name: extract_symbolic_data
arguments: tag: D8
type: f64
name: extract_symbolic_data
arguments: tag: D23
type: string
name: extract_symbolic_data
arguments: tag: D42
type: string
name: extract_symbolic_data
arguments: tag: D60
type: string
name: extract_symbolic_data
arguments: tag: D40
type: string
name: extract_symbolic_data
arguments: tag: D57
type: string

tool: D39: You are %d miles away.


tool: {"D61": [0x00,0x00,0x00,0x60,0x8f,0xc2,0x02,0x40]}

tool: D53: You have %f levels of power.


tool: D8: 56789.4532

tool: D23: You have %f awesome super powers.


tool: D42: I have an initial %c.


tool: D60: I have a first name %s.


tool: D40: I have a last name %s.


tool: D57: 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 = 'K';
  char first_name[] = "Omer";
  char last_name[] = "Wazir";
  printf("You are %d miles 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("I have a first name %s.\n", first_name);
  printf("I have a last name %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
<gleIntLoad_Version_3_3>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x1,-0x4(%rbp)
  mov    $0x0,%eax
  call   <gleIntClear@plt>
  mov    $0x0,%eax
  call   <gleIntLoadCore_1_1_Version_3_1@plt>
  test   %eax,%eax
  jne    L86
  movl   $0x0,-0x4(%rbp)
L86:
  mov    $0x0,%eax
  call   <gleIntLoadCore_1_2_Version_3_1@plt>
  test   %eax,%eax
  jne    L109
  movl   $0x0,-0x4(%rbp)
L109:
  mov    $0x0,%eax
  call   <gleIntLoadCore_1_3_Version_3_1@plt>
  test   %eax,%eax
  jne    L62
  movl   $0x0,-0x4(%rbp)
L62:
  mov    $0x0,%eax
  call   <gleIntLoadCore_1_4_Version_3_1@plt>
  test   %eax,%eax
  jne    L55
  movl   $0x0,-0x4(%rbp)
L55:
  mov    $0x0,%eax
  call   <gleIntLoadCore_1_5_Version_3_1@plt>
  test   %eax,%eax
  jne    L99
  movl   $0x0,-0x4(%rbp)
L99:
  mov    $0x0,%eax
  call   <gleIntLoadCore_2_0_Version_3_1@plt>
  test   %eax,%eax
  jne    L72
  movl   $0x0,-0x4(%rbp)
L72:
  mov    $0x0,%eax
  call   <gleIntLoadCore_2_1_Version_3_1@plt>
  test   %eax,%eax
  jne    L5
  movl   $0x0,-0x4(%rbp)
L5:
  mov    $0x0,%eax
  call   <gleIntLoadCore_3_0_Version_3_1@plt>
  test   %eax,%eax
  jne    L101
  movl   $0x0,-0x4(%rbp)
L101:
  mov    $0x0,%eax
  call   <gleIntLoadCore_3_1_Base@plt>
  test   %eax,%eax
  jne    L82
  movl   $0x0,-0x4(%rbp)
L82:
  mov    $0x0,%eax
  call   <gleIntLoadCore_3_2_Base@plt>
  test   %eax,%eax
  jne    L4
  movl   $0x0,-0x4(%rbp)
L4:
  mov    $0x0,%eax
  call   <gleIntLoadCore_3_3_Base@plt>
  test   %eax,%eax
  jne    L41
  movl   $0x0,-0x4(%rbp)
L41:
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int gleIntLoad_Version_3_3() {
  int bIsLoaded = 1;
  gleIntClear();
  if (!gleIntLoadCore_1_1_Version_3_1())
    bIsLoaded = 0;
  if (!gleIntLoadCore_1_2_Version_3_1())
    bIsLoaded = 0;
  if (!gleIntLoadCore_1_3_Version_3_1())
    bIsLoaded = 0;
  if (!gleIntLoadCore_1_4_Version_3_1())
    bIsLoaded = 0;
  if (!gleIntLoadCore_1_5_Version_3_1())
    bIsLoaded = 0;
  if (!gleIntLoadCore_2_0_Version_3_1())
    bIsLoaded = 0;
  if (!gleIntLoadCore_2_1_Version_3_1())
    bIsLoaded = 0;
  if (!gleIntLoadCore_3_0_Version_3_1())
    bIsLoaded = 0;
  if (!gleIntLoadCore_3_1_Base())
    bIsLoaded = 0;
  if (!gleIntLoadCore_3_2_Base())
    bIsLoaded = 0;
  if (!gleIntLoadCore_3_3_Base())
    bIsLoaded = 0;
  return bIsLoaded;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<gleIntLoad_Version_3_3>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <gleIntClear@plt>
  mov    $0x0,%eax
  call   <gleIntLoadCore_1_1_Version_3_1@plt>
  mov    %eax,%ebp
  mov    $0x0,%eax
  call   <gleIntLoadCore_1_2_Version_3_1@plt>
  mov    %eax,%ebx
  test   %eax,%eax
  je     L8
  test   %ebp,%ebp
  setne  %bl
  movzbl %bl,%ebx
L8:
  mov    $0x0,%eax
  call   <gleIntLoadCore_1_3_Version_3_1@plt>
  test   %eax,%eax
  cmove  %eax,%ebx
  mov    $0x0,%eax
  call   <gleIntLoadCore_1_4_Version_3_1@plt>
  test   %eax,%eax
  cmove  %eax,%ebx
  mov    $0x0,%eax
  call   <gleIntLoadCore_1_5_Version_3_1@plt>
  test   %eax,%eax
  cmove  %eax,%ebx
  mov    $0x0,%eax
  call   <gleIntLoadCore_2_0_Version_3_1@plt>
  test   %eax,%eax
  cmove  %eax,%ebx
  mov    $0x0,%eax
  call   <gleIntLoadCore_2_1_Version_3_1@plt>
  test   %eax,%eax
  cmove  %eax,%ebx
  mov    $0x0,%eax
  call   <gleIntLoadCore_3_0_Version_3_1@plt>
  test   %eax,%eax
  cmove  %eax,%ebx
  mov    $0x0,%eax
  call   <gleIntLoadCore_3_1_Base@plt>
  test   %eax,%eax
  cmove  %eax,%ebx
  mov    $0x0,%eax
  call   <gleIntLoadCore_3_2_Base@plt>
  test   %eax,%eax
  cmove  %eax,%ebx
  mov    $0x0,%eax
  call   <gleIntLoadCore_3_3_Base@plt>
  test   %eax,%eax
  cmove  %eax,%ebx
  mov    %ebx,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int gleIntLoad_Version_3_3() {
  int bIsLoaded = 1;
  gleIntClear();
  if (!gleIntLoadCore_1_1_Version_3_1())
    bIsLoaded = 0;
  if (!gleIntLoadCore_1_2_Version_3_1())
    bIsLoaded = 0;
  if (!gleIntLoadCore_1_3_Version_3_1())
    bIsLoaded = 0;
  if (!gleIntLoadCore_1_4_Version_3_1())
    bIsLoaded = 0;
  if (!gleIntLoadCore_1_5_Version_3_1())
    bIsLoaded = 0;
  if (!gleIntLoadCore_2_0_Version_3_1())
    bIsLoaded = 0;
  if (!gleIntLoadCore_2_1_Version_3_1())
    bIsLoaded = 0;
  if (!gleIntLoadCore_3_0_Version_3_1())
    bIsLoaded = 0;
  if (!gleIntLoadCore_3_1_Base())
    bIsLoaded = 0;
  if (!gleIntLoadCore_3_2_Base())
    bIsLoaded = 0;
  if (!gleIntLoadCore_3_3_Base())
    bIsLoaded = 0;
  return bIsLoaded;
}
```