ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  push   %r13
  mov    $0x6,%esi
  lea    D26(%rip),%rdi
  push   %r12
  push   %rbp
  lea    D57(%rip),%rbp
  push   %rbx
  sub    $0x28,%rsp
  movdqa D2(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    D18(%rip),%rax
  mov    %rsp,%rbx
  lea    0x18(%rsp),%r12
  movaps %xmm0,(%rsp)
  mov    %rax,0x10(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x6,%esi
  mov    %rbx,%rdi
  call   <max_area@plt>
  lea    D3(%rip),%rdi
  mov    %eax,%r13d
  xor    %eax,%eax
  call   <printf@plt>
  cs nopw 0x0(%rax,%rax,1)
L14:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %rbx,%r12
  jne    L14
  xor    %eax,%eax
  mov    %r13d,%esi
  lea    D46(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D26
type_guessed: string
name: read_data
arguments: data_label: D57
type_guessed: string
name: read_data
arguments: data_label: D2
type_guessed: byte[16]
name: read_data
arguments: data_label: D18
type_guessed: byte[8]
name: read_data
arguments: data_label: D3
type_guessed: string
name: read_data
arguments: data_label: D46
type_guessed: string

tool: D26: 
 size is %d 

tool: D57:  %d 

tool: {"D2": [0x07,0x00,0x00,0x00,0x0a,0x00,0x00,0x00,0x03,0x00,0x00,0x00,0x06,0x00,0x00,0x00]}

tool: {"D18": [0x04,0x00,0x00,0x00,0x05,0x00,0x00,0x00]}

tool: D3: 
Max Area of 

tool: D46:  is %d.


assistant: ```c
void main() {
  int A[] = {7, 10, 3, 6, 4, 5};
  int i, maxarea = 0;
  int size = sizeof(A) / 4;
  printf("\n size is %d ", size);
  maxarea = max_area(A, size);
  printf("\nMax Area of ");
  for (i = 0; i < size; i++) {
    printf(" %d ", A[i]);
  }
  printf(" is %d.\n", maxarea);
  return;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<getNext>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  movl   $0xffffffff,(%rax)
  movl   $0xffffffff,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L35
L58:
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x8(%rbp)
L28:
  cmpl   $0xffffffff,-0x8(%rbp)
  je     L37
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%edx
  mov    -0x4(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  movzbl (%rax),%eax
  cmp    %al,%dl
  jne    L58
L37:
  addl   $0x1,-0x4(%rbp)
  addl   $0x1,-0x8(%rbp)
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%edx
  mov    -0x4(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  movzbl (%rax),%eax
  cmp    %al,%dl
  jne    L10
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    -0x4(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x20(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  jmp    L35
L10:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rax,%rdx
  mov    -0x8(%rbp),%eax
  mov    %eax,(%rdx)
L35:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L28
  jmp    L78
L39:
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x8(%rbp)
L78:
  cmpl   $0xffffffff,-0x8(%rbp)
  je     L33
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%edx
  mov    -0x4(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  movzbl (%rax),%eax
  cmp    %al,%dl
  jne    L39
L33:
  addl   $0x1,-0x4(%rbp)
  addl   $0x1,-0x8(%rbp)
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rax,%rdx
  mov    -0x8(%rbp),%eax
  mov    %eax,(%rdx)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void getNext(char *pattern, int next[]) {
  next[0] = -1;
  int k = -1, j = 0;
  while (pattern[j]) {
    while (k != -1 && pattern[k] != pattern[j])
      k = next[k];
    j++;
    k++;
    if (pattern[k] == pattern[j])
      next[j] = next[k];
    else
      next[j] = k;
  }
  while (k != -1 && pattern[k] != pattern[j])
    k = next[k];
  j++;
  k++;
  next[j] = k;
}
```
user: What is the c source code of the assembly code below?

```asm
<getNext>:
  endbr64
  movl   $0xffffffff,(%rsi)
  mov    $0x1,%r8d
  mov    $0xffffffff,%eax
  mov    $0x0,%r9d
  cmpb   $0x0,(%rdi)
  jne    L99
  jmp    L70
L84:
  add    $0x1,%r8
L99:
  mov    %r8d,%r9d
  add    $0x1,%eax
  movslq %eax,%rdx
  movzbl (%rdi,%r8,1),%ecx
  cmp    %cl,(%rdi,%rdx,1)
  je     L64
  mov    %eax,(%rsi,%r8,4)
L56:
  movzbl (%rdi,%r8,1),%ecx
  test   %cl,%cl
  je     L18
  cmp    $0xffffffff,%eax
  je     L84
L51:
  movslq %eax,%rdx
  cmp    %cl,(%rdi,%rdx,1)
  je     L84
  mov    (%rsi,%rdx,4),%eax
  cmp    $0xffffffff,%eax
  jne    L51
  jmp    L84
L64:
  mov    (%rsi,%rdx,4),%edx
  mov    %edx,(%rsi,%r8,4)
  jmp    L56
L18:
  cmp    $0xffffffff,%eax
  je     L70
L117:
  movslq %eax,%rdx
  cmpb   $0x0,(%rdi,%rdx,1)
  je     L70
  mov    (%rsi,%rdx,4),%eax
  cmp    $0xffffffff,%eax
  jne    L117
L70:
  lea    0x1(%r9),%edx
  movslq %edx,%rdx
  add    $0x1,%eax
  mov    %eax,(%rsi,%rdx,4)
  ret
```

assistant: ```c
void getNext(char *pattern, int next[]) {
  next[0] = -1;
  int k = -1, j = 0;
  while (pattern[j]) {
    while (k != -1 && pattern[k] != pattern[j])
      k = next[k];
    j++;
    k++;
    if (pattern[k] == pattern[j])
      next[j] = next[k];
    else
      next[j] = k;
  }
  while (k != -1 && pattern[k] != pattern[j])
    k = next[k];
  j++;
  k++;
  next[j] = k;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<getNext>:
  endbr64
  movl   $0xffffffff,(%rsi)
  movzbl (%rdi),%r9d
  mov    %rdi,%rcx
  xor    %edx,%edx
  mov    $0x1,%r8d
  xor    %eax,%eax
  mov    $0x4,%edi
  test   %r9b,%r9b
  je     L8
  nopl   0x0(%rax)
L90:
  mov    %r8d,%r10d
  mov    %edx,%eax
  mov    %edx,%edi
  cmp    %r9b,(%rcx,%r8,1)
  jne    L86
  movslq %edx,%rdi
  mov    (%rsi,%rdi,4),%edi
L86:
  mov    %edi,(%rsi,%r8,4)
  movzbl (%rcx,%r8,1),%edi
  test   %dil,%dil
  je     L40
  cmp    $0xffffffff,%edx
  jne    L126
  jmp    L9
  nopl   0x0(%rax,%rax,1)
L50:
  mov    (%rsi,%rdx,4),%eax
  cmp    $0xffffffff,%eax
  je     L9
L126:
  movslq %eax,%rdx
  cmp    %dil,(%rcx,%rdx,1)
  jne    L50
  lea    0x1(%rax),%edx
  movslq %edx,%rax
  add    %rcx,%rax
L62:
  movzbl (%rax),%r9d
  add    $0x1,%r8
  jmp    L90
  nopl   0x0(%rax)
L9:
  mov    %rcx,%rax
  xor    %edx,%edx
  jmp    L62
  nop
L40:
  lea    0x1(%r10),%edi
  movslq %edi,%rdi
  shl    $0x2,%rdi
  cmp    $0xffffffff,%edx
  jne    L24
  jmp    L58
  nopw   0x0(%rax,%rax,1)
L30:
  mov    (%rsi,%rdx,4),%eax
  cmp    $0xffffffff,%eax
  je     L58
L24:
  movslq %eax,%rdx
  cmpb   $0x0,(%rcx,%rdx,1)
  jne    L30
  add    $0x1,%eax
L8:
  mov    %eax,(%rsi,%rdi,1)
  ret
L58:
  xor    %eax,%eax
  mov    %eax,(%rsi,%rdi,1)
  ret
```

assistant: ```c
void getNext(char *pattern, int next[]) {
  next[0] = -1;
  int k = -1, j = 0;
  while (pattern[j]) {
    while (k != -1 && pattern[k] != pattern[j])
      k = next[k];
    j++;
    k++;
    if (pattern[k] == pattern[j])
      next[j] = next[k];
    else
      next[j] = k;
  }
  while (k != -1 && pattern[k] != pattern[j])
    k = next[k];
  j++;
  k++;
  next[j] = k;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<getNext>:
  endbr64
  movl   $0xffffffff,(%rsi)
  movzbl (%rdi),%r9d
  mov    %rdi,%rcx
  xor    %edx,%edx
  mov    $0x1,%r8d
  xor    %eax,%eax
  mov    $0x4,%edi
  test   %r9b,%r9b
  je     L23
  nopl   0x0(%rax)
L90:
  mov    %r8d,%r10d
  mov    %edx,%eax
  mov    %edx,%edi
  cmp    %r9b,(%rcx,%r8,1)
  jne    L113
  movslq %edx,%rdi
  mov    (%rsi,%rdi,4),%edi
L113:
  mov    %edi,(%rsi,%r8,4)
  movzbl (%rcx,%r8,1),%edi
  test   %dil,%dil
  je     L80
  cmp    $0xffffffff,%edx
  jne    L127
  jmp    L106
  nopl   0x0(%rax,%rax,1)
L78:
  mov    (%rsi,%rdx,4),%eax
  cmp    $0xffffffff,%eax
  je     L106
L127:
  movslq %eax,%rdx
  cmp    %dil,(%rcx,%rdx,1)
  jne    L78
  lea    0x1(%rax),%edx
  movslq %edx,%rax
  add    %rcx,%rax
L112:
  movzbl (%rax),%r9d
  add    $0x1,%r8
  jmp    L90
  nopl   0x0(%rax)
L106:
  mov    %rcx,%rax
  xor    %edx,%edx
  jmp    L112
  nop
L80:
  lea    0x1(%r10),%edi
  movslq %edi,%rdi
  shl    $0x2,%rdi
  cmp    $0xffffffff,%edx
  jne    L46
  jmp    L5
  nopw   0x0(%rax,%rax,1)
L69:
  mov    (%rsi,%rdx,4),%eax
  cmp    $0xffffffff,%eax
  je     L5
L46:
  movslq %eax,%rdx
  cmpb   $0x0,(%rcx,%rdx,1)
  jne    L69
  add    $0x1,%eax
L23:
  mov    %eax,(%rsi,%rdi,1)
  ret
L5:
  xor    %eax,%eax
  mov    %eax,(%rsi,%rdi,1)
  ret
```

assistant: ```c
void getNext(char *pattern, int next[]) {
  next[0] = -1;
  int k = -1, j = 0;
  while (pattern[j]) {
    while (k != -1 && pattern[k] != pattern[j])
      k = next[k];
    j++;
    k++;
    if (pattern[k] == pattern[j])
      next[j] = next[k];
    else
      next[j] = k;
  }
  while (k != -1 && pattern[k] != pattern[j])
    k = next[k];
  j++;
  k++;
  next[j] = k;
}
```
user: What is the c source code of the assembly code below:

```asm
<bmp180_main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    $0x0,%eax
  call   <bmp180_init@plt>
  mov    $0x0,%eax
  call   <bmp180_get_t@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x8(%rbp)
  mov    $0x0,%eax
  call   <bmp180_get_p@plt>
  mov    %eax,%edx
  movsd  -0x8(%rbp),%xmm0
  movsd  D6(%rip),%xmm1
  divsd  %xmm1,%xmm0
  movq   %xmm0,%rax
  mov    %edx,%esi
  movq   %rax,%xmm0
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D6
data_type: byte[8]
name: unpack_data
arguments: data_tag: D8
data_type: string

tool: {"D6": [0x00,0x00,0x00,0x00,0x00,0x00,0x24,0x40]}

tool: D8: get temp = %.1f, pressure = %d


assistant: ```c
int bmp180_main(int argc, const char *argv[]) {
  bmp180_init();
  double t = bmp180_get_t();
  printf("get temp = %.1f, pressure = %d\n", t / 10, bmp180_get_p());
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<bmp180_main>:
  endbr64
  push   %rbx
  mov    $0x0,%eax
  call   <bmp180_init@plt>
  mov    $0x0,%eax
  call   <bmp180_get_t@plt>
  movq   %xmm0,%rbx
  mov    $0x0,%eax
  call   <bmp180_get_p@plt>
  mov    %eax,%esi
  movq   %rbx,%xmm0
  divsd  D19(%rip),%xmm0
  lea    D6(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  pop    %rbx
  ret
```

tool_calls: name: read_data
arguments: data_label: D19
type_guessed: byte[8]
name: read_data
arguments: data_label: D6
type_guessed: string

tool: {"D19": [0x00,0x00,0x00,0x00,0x00,0x00,0x24,0x40]}

tool: D6: get temp = %.1f, pressure = %d


assistant: ```c
int bmp180_main(int argc, const char *argv[]) {
  bmp180_init();
  double t = bmp180_get_t();
  printf("get temp = %.1f, pressure = %d\n", t / 10, bmp180_get_p());
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<bmp180_main>:
  endbr64
  sub    $0x18,%rsp
  xor    %eax,%eax
  call   <bmp180_init@plt>
  xor    %eax,%eax
  call   <bmp180_get_t@plt>
  xor    %eax,%eax
  movsd  %xmm0,0x8(%rsp)
  call   <bmp180_get_p@plt>
  lea    D7(%rip),%rdi
  movsd  0x8(%rsp),%xmm0
  divsd  D0(%rip),%xmm0
  mov    %eax,%esi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: tag: D7
type_guessed: string
name: unpack_data
arguments: tag: D0
type_guessed: byte[8]

tool: D7: get temp = %.1f, pressure = %d


tool: {"D0": [0x00,0x00,0x00,0x00,0x00,0x00,0x24,0x40]}

assistant: ```c
int bmp180_main(int argc, const char *argv[]) {
  bmp180_init();
  double t = bmp180_get_t();
  printf("get temp = %.1f, pressure = %d\n", t / 10, bmp180_get_p());
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<bmp180_main>:
  endbr64
  sub    $0x18,%rsp
  xor    %eax,%eax
  call   <bmp180_init@plt>
  xor    %eax,%eax
  call   <bmp180_get_t@plt>
  xor    %eax,%eax
  movsd  %xmm0,0x8(%rsp)
  call   <bmp180_get_p@plt>
  lea    D14(%rip),%rdi
  movsd  0x8(%rsp),%xmm0
  divsd  D17(%rip),%xmm0
  mov    %eax,%esi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
```

tool_calls: name: parse_data
arguments: data_label: D14
data_type: string
name: parse_data
arguments: data_label: D17
data_type: byte[8]

tool: D14: get temp = %.1f, pressure = %d


tool: {"D17": [0x00,0x00,0x00,0x00,0x00,0x00,0x24,0x40]}

assistant: ```c
int bmp180_main(int argc, const char *argv[]) {
  bmp180_init();
  double t = bmp180_get_t();
  printf("get temp = %.1f, pressure = %d\n", t / 10, bmp180_get_p());
  return 0;
}
```
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
  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
  mov    $0x0,%eax
  call   <megaInit@plt>
  cmpl   $0x2,-0x24(%rbp)
  je     L48
  mov    $0x1,%esi
  lea    D26(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0xffffffff,%edi
  call   <exit@plt>
L48:
  movl   $0x0,-0x1c(%rbp)
  jmp    L66
L59:
  mov    -0x1c(%rbp),%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    $0xa,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <strtoul@plt>
  mov    %eax,-0x14(%rbp)
  mov    -0x1c(%rbp),%eax
  cltq
  mov    -0x14(%rbp),%edx
  mov    %edx,-0x10(%rbp,%rax,4)
  addl   $0x1,-0x1c(%rbp)
L66:
  cmpl   $0x0,-0x1c(%rbp)
  jle    L59
  lea    -0xc(%rbp),%rdx
  lea    -0x10(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <RandomFunc@plt>
  mov    -0xc(%rbp),%eax
  cmp    $0x57fffdff,%eax
  jne    L7
  lea    D15(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
L7:
  movl   $0x0,-0x18(%rbp)
  jmp    L46
L25:
  mov    -0x18(%rbp),%eax
  cltq
  mov    -0xc(%rbp,%rax,4),%eax
  mov    %eax,%esi
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x18(%rbp)
L46:
  cmpl   $0x0,-0x18(%rbp)
  jle    L25
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L58
  call   <__stack_chk_fail@plt>
L58:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D26
type: string
name: extract_symbolic_data
arguments: tag: D15
type: string
name: extract_symbolic_data
arguments: tag: D2
type: string

tool: D26: Call this program with %i arguments


tool: D15: You win!

tool: D2: %u


assistant: ```c
int main(int argc, char *argv[]) {
  unsigned int input[1];
  unsigned int output[1];
  int randomFuns_i5;
  unsigned int randomFuns_value6;
  int randomFuns_main_i7;
  {
    megaInit();
    if (argc != 2) {
      printf("Call this program with %i arguments\n", 1);
      exit(-1);
    } else {
    }
    randomFuns_i5 = 0;
    while (randomFuns_i5 < 1) {
      randomFuns_value6 = (unsigned int)strtoul(argv[randomFuns_i5 + 1], 0, 10);
      input[randomFuns_i5] = randomFuns_value6;
      randomFuns_i5++;
    }
    RandomFunc(input, output);
    if (output[0] == 1476394495U) {
      printf("You win!\n");
    } else {
    }
    randomFuns_main_i7 = 0;
    while (randomFuns_main_i7 < 1) {
      printf("%u\n", output[randomFuns_main_i7]);
      randomFuns_main_i7++;
    }
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %edi,%ebp
  mov    %rsi,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <megaInit@plt>
  cmp    $0x2,%ebp
  jne    L12
L39:
  mov    0x8(%rbx),%rdi
  mov    $0xa,%edx
  mov    $0x0,%esi
  call   <strtoul@plt>
  mov    %eax,(%rsp)
  lea    0x4(%rsp),%rsi
  mov    %rsp,%rdi
  call   <RandomFunc@plt>
  cmpl   $0x57fffdff,0x4(%rsp)
  je     L3
L18:
  mov    0x4(%rsp),%esi
  lea    D11(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L43
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L12:
  mov    $0x1,%esi
  lea    D28(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0xffffffff,%edi
  call   <exit@plt>
  jmp    L39
L3:
  lea    D15(%rip),%rdi
  call   <puts@plt>
  jmp    L18
L43:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D11
type_infered: string
name: retrieve_data
arguments: tag: D28
type_infered: string
name: retrieve_data
arguments: tag: D15
type_infered: string

tool: D11: %u


tool: D28: Call this program with %i arguments


tool: D15: You win!

assistant: ```c
int main(int argc, char *argv[]) {
  unsigned int input[1];
  unsigned int output[1];
  int randomFuns_i5;
  unsigned int randomFuns_value6;
  int randomFuns_main_i7;
  {
    megaInit();
    if (argc != 2) {
      printf("Call this program with %i arguments\n", 1);
      exit(-1);
    } else {
    }
    randomFuns_i5 = 0;
    while (randomFuns_i5 < 1) {
      randomFuns_value6 = (unsigned int)strtoul(argv[randomFuns_i5 + 1], 0, 10);
      input[randomFuns_i5] = randomFuns_value6;
      randomFuns_i5++;
    }
    RandomFunc(input, output);
    if (output[0] == 1476394495U) {
      printf("You win!\n");
    } else {
    }
    randomFuns_main_i7 = 0;
    while (randomFuns_main_i7 < 1) {
      printf("%u\n", output[randomFuns_main_i7]);
      randomFuns_main_i7++;
    }
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <megaInit@plt>
  cmp    $0x2,%ebp
  je     L8
  lea    D21(%rip),%rdi
  mov    $0x1,%esi
  xor    %eax,%eax
  call   <printf@plt>
  or     $0xffffffff,%edi
  call   <exit@plt>
L8:
  mov    0x8(%rbx),%rdi
  mov    $0xa,%edx
  xor    %esi,%esi
  call   <strtoul@plt>
  lea    0x4(%rsp),%rsi
  mov    %rsp,%rdi
  mov    %eax,(%rsp)
  call   <RandomFunc@plt>
  mov    0x4(%rsp),%esi
  cmp    $0x57fffdff,%esi
  je     L14
L19:
  xor    %eax,%eax
  lea    D4(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L18
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L14:
  lea    D15(%rip),%rdi
  call   <puts@plt>
  mov    0x4(%rsp),%esi
  jmp    L19
L18:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D21
type: string
name: retrieve_data
arguments: data_tag: D4
type: string
name: retrieve_data
arguments: data_tag: D15
type: string

tool: D21: Call this program with %i arguments


tool: D4: %u


tool: D15: You win!

assistant: ```c
int main(int argc, char *argv[]) {
  unsigned int input[1];
  unsigned int output[1];
  int randomFuns_i5;
  unsigned int randomFuns_value6;
  int randomFuns_main_i7;
  {
    megaInit();
    if (argc != 2) {
      printf("Call this program with %i arguments\n", 1);
      exit(-1);
    } else {
    }
    randomFuns_i5 = 0;
    while (randomFuns_i5 < 1) {
      randomFuns_value6 = (unsigned int)strtoul(argv[randomFuns_i5 + 1], 0, 10);
      input[randomFuns_i5] = randomFuns_value6;
      randomFuns_i5++;
    }
    RandomFunc(input, output);
    if (output[0] == 1476394495U) {
      printf("You win!\n");
    } else {
    }
    randomFuns_main_i7 = 0;
    while (randomFuns_main_i7 < 1) {
      printf("%u\n", output[randomFuns_main_i7]);
      randomFuns_main_i7++;
    }
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <megaInit@plt>
  cmp    $0x2,%ebp
  je     L16
  lea    D14(%rip),%rdi
  mov    $0x1,%esi
  xor    %eax,%eax
  call   <printf@plt>
  or     $0xffffffff,%edi
  call   <exit@plt>
L16:
  mov    0x8(%rbx),%rdi
  mov    $0xa,%edx
  xor    %esi,%esi
  call   <strtoul@plt>
  lea    0x4(%rsp),%rsi
  mov    %rsp,%rdi
  mov    %eax,(%rsp)
  call   <RandomFunc@plt>
  mov    0x4(%rsp),%esi
  cmp    $0x57fffdff,%esi
  je     L1
L13:
  xor    %eax,%eax
  lea    D5(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L37
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L1:
  lea    D8(%rip),%rdi
  call   <puts@plt>
  mov    0x4(%rsp),%esi
  jmp    L13
L37:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D14
type_infered: string
name: extract_symbolic_data
arguments: tag: D5
type_infered: string
name: extract_symbolic_data
arguments: tag: D8
type_infered: string

tool: D14: Call this program with %i arguments


tool: D5: %u


tool: D8: You win!

assistant: ```c
int main(int argc, char *argv[]) {
  unsigned int input[1];
  unsigned int output[1];
  int randomFuns_i5;
  unsigned int randomFuns_value6;
  int randomFuns_main_i7;
  {
    megaInit();
    if (argc != 2) {
      printf("Call this program with %i arguments\n", 1);
      exit(-1);
    } else {
    }
    randomFuns_i5 = 0;
    while (randomFuns_i5 < 1) {
      randomFuns_value6 = (unsigned int)strtoul(argv[randomFuns_i5 + 1], 0, 10);
      input[randomFuns_i5] = randomFuns_value6;
      randomFuns_i5++;
    }
    RandomFunc(input, output);
    if (output[0] == 1476394495U) {
      printf("You win!\n");
    } else {
    }
    randomFuns_main_i7 = 0;
    while (randomFuns_main_i7 < 1) {
      printf("%u\n", output[randomFuns_main_i7]);
      randomFuns_main_i7++;
    }
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<evil_pivot_root>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    D14(%rip),%rax
  mov    (%rax),%edx
  mov    D1(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xffffffff,%eax
  pop    %rbp
  ret
```

assistant: ```c
int evil_pivot_root(const char *new_root, const char *old_put) {
  errno = EPERM;
  return -1;
}
```
user: What is the c source code of the assembly code below:

```asm
<evil_pivot_root>:
  endbr64
  mov    D16(%rip),%rax
  mov    (%rax),%edx
  mov    D10(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int evil_pivot_root(const char *new_root, const char *old_put) {
  errno = EPERM;
  return -1;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<evil_pivot_root>:
  endbr64
  mov    D12(%rip),%rax
  mov    (%rax),%edx
  mov    D0(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int evil_pivot_root(const char *new_root, const char *old_put) {
  errno = EPERM;
  return -1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<evil_pivot_root>:
  endbr64
  mov    D17(%rip),%rax
  mov    (%rax),%edx
  mov    D6(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int evil_pivot_root(const char *new_root, const char *old_put) {
  errno = EPERM;
  return -1;
}
```