ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D64(%rip),%rdi
  call   <printf@plt>
  lea    D59(%rip),%rdi
  call   <printf@plt>
  lea    D60(%rip),%rdi
  call   <printf@plt>
  mov    $0x3,%edi
  call   <omp_set_num_threads@plt>
  lea    D29(%rip),%rdi
  call   <printf@plt>
  lea    D40(%rip),%rdi
  call   <printf@plt>
  lea    D26(%rip),%rdi
  call   <printf@plt>
  lea    D25(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: label: D64
type_infered: string
name: unpack_data
arguments: label: D59
type_infered: string
name: unpack_data
arguments: label: D60
type_infered: string
name: unpack_data
arguments: label: D29
type_infered: string
name: unpack_data
arguments: label: D40
type_infered: string
name: unpack_data
arguments: label: D26
type_infered: string
name: unpack_data
arguments: label: D25
type_infered: string

tool: D64: 
Beginning


tool: D59: 
Part I

tool: D60: 

Between I and II...


tool: D29: 
Part II...

tool: D40: 

Between II and III...


tool: D26: 
Part III...

tool: D25: 

End



assistant: ```c
int main(int argc, char **argv) {
  printf("\nBeginning\n");
#pragma omp parallel
  printf("\nPart I");
  printf("\n\nBetween I and II...\n");
  omp_set_num_threads(3);
#pragma omp parallel
  printf("\nPart II...");
  printf("\n\nBetween II and III...\n");
#pragma omp parallel num_threads(5)
  printf("\nPart III...");
  printf("\n\nEnd\n\n");
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D0(%rip),%rdi
  call   <printf@plt>
  lea    D58(%rip),%rdi
  call   <printf@plt>
  lea    D39(%rip),%rdi
  call   <printf@plt>
  mov    $0x3,%edi
  call   <omp_set_num_threads@plt>
  lea    D22(%rip),%rdi
  call   <printf@plt>
  lea    D15(%rip),%rdi
  call   <printf@plt>
  lea    D13(%rip),%rdi
  call   <printf@plt>
  lea    D2(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D0
type_infered: string
name: parse_data
arguments: data_tag: D58
type_infered: string
name: parse_data
arguments: data_tag: D39
type_infered: string
name: parse_data
arguments: data_tag: D22
type_infered: string
name: parse_data
arguments: data_tag: D15
type_infered: string
name: parse_data
arguments: data_tag: D13
type_infered: string
name: parse_data
arguments: data_tag: D2
type_infered: string

tool: D0: 
Beginning


tool: D58: 
Part I

tool: D39: 

Between I and II...


tool: D22: 
Part II...

tool: D15: 

Between II and III...


tool: D13: 
Part III...

tool: D2: 

End



assistant: ```c
int main(int argc, char **argv) {
  printf("\nBeginning\n");
#pragma omp parallel
  printf("\nPart I");
  printf("\n\nBetween I and II...\n");
  omp_set_num_threads(3);
#pragma omp parallel
  printf("\nPart II...");
  printf("\n\nBetween II and III...\n");
#pragma omp parallel num_threads(5)
  printf("\nPart III...");
  printf("\n\nEnd\n\n");
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<remove_duplicates_2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  cmpl   $0x2,-0x1c(%rbp)
  jg     L54
  mov    -0x1c(%rbp),%eax
  jmp    L52
L54:
  movl   $0x1,-0x8(%rbp)
  movl   $0x2,-0x4(%rbp)
  jmp    L49
L78:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%eax
  cmp    %eax,%edx
  je     L31
  addl   $0x1,-0x8(%rbp)
  mov    -0x8(%rbp),%eax
  cmp    -0x4(%rbp),%eax
  je     L0
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x8(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  jmp    L0
L31:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x8(%rbp),%eax
  cltq
  shl    $0x2,%rax
  lea    -0x4(%rax),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%eax
  cmp    %eax,%edx
  je     L0
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  addl   $0x1,-0x8(%rbp)
  mov    -0x8(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rax),%eax
  mov    %eax,(%rdx)
L0:
  addl   $0x1,-0x4(%rbp)
L49:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L78
  mov    -0x8(%rbp),%eax
  add    $0x1,%eax
L52:
  pop    %rbp
  ret
```

assistant: ```c
int remove_duplicates_2(int *nums, int nums_size) {
  if (nums_size < 3) {
    return nums_size;
  }
  int i = 1, j;
  for (j = 2; j < nums_size; ++j) {
    if (nums[j] != nums[i]) {
      if (++i != j) {
        nums[i] = nums[j];
      }
    } else {
      if (nums[j] != nums[i - 1]) {
        nums[++i] = nums[j];
      }
    }
  }
  return i + 1;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<remove_duplicates_2>:
  endbr64
  mov    %esi,%eax
  cmp    $0x2,%esi
  jle    L14
  mov    %esi,%esi
  mov    $0x2,%edx
  mov    $0x1,%eax
  jmp    L45
L29:
  cmp    -0x4(%rdi,%r9,1),%r8d
  je     L53
  add    $0x1,%eax
  mov    %r8d,0x4(%rdi,%r9,1)
L53:
  add    $0x1,%rdx
  cmp    %rdx,%rsi
  je     L55
L45:
  mov    (%rdi,%rdx,4),%r8d
  movslq %eax,%rcx
  lea    0x0(,%rcx,4),%r9
  cmp    (%rdi,%rcx,4),%r8d
  je     L29
  add    $0x1,%eax
  cmp    %edx,%eax
  je     L53
  mov    %r8d,0x4(%rdi,%r9,1)
  jmp    L53
L55:
  add    $0x1,%eax
L14:
  ret
```

assistant: ```c
int remove_duplicates_2(int *nums, int nums_size) {
  if (nums_size < 3) {
    return nums_size;
  }
  int i = 1, j;
  for (j = 2; j < nums_size; ++j) {
    if (nums[j] != nums[i]) {
      if (++i != j) {
        nums[i] = nums[j];
      }
    } else {
      if (nums[j] != nums[i - 1]) {
        nums[++i] = nums[j];
      }
    }
  }
  return i + 1;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<remove_duplicates_2>:
  endbr64
  mov    %esi,%eax
  cmp    $0x2,%esi
  jle    L7
  movslq %esi,%rsi
  mov    $0x2,%edx
  mov    $0x1,%r9d
  mov    $0x2,%eax
  jmp    L61
L70:
  cmp    %edx,%eax
  je     L31
L49:
  mov    %r8d,0x4(%rdi,%r10,1)
  mov    %eax,%r9d
  add    $0x1,%eax
L8:
  add    $0x1,%rdx
  cmp    %rdx,%rsi
  je     L7
L61:
  movslq %r9d,%rcx
  mov    (%rdi,%rdx,4),%r8d
  lea    0x0(,%rcx,4),%r10
  cmp    (%rdi,%rcx,4),%r8d
  jne    L70
  cmp    -0x4(%rdi,%r10,1),%r8d
  jne    L49
  add    $0x1,%rdx
  cmp    %rdx,%rsi
  jne    L61
L7:
  ret
  xchg   %ax,%ax
L31:
  mov    %eax,%r9d
  add    $0x1,%eax
  jmp    L8
```

assistant: ```c
int remove_duplicates_2(int *nums, int nums_size) {
  if (nums_size < 3) {
    return nums_size;
  }
  int i = 1, j;
  for (j = 2; j < nums_size; ++j) {
    if (nums[j] != nums[i]) {
      if (++i != j) {
        nums[i] = nums[j];
      }
    } else {
      if (nums[j] != nums[i - 1]) {
        nums[++i] = nums[j];
      }
    }
  }
  return i + 1;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<remove_duplicates_2>:
  endbr64
  mov    %esi,%eax
  cmp    $0x2,%esi
  jle    L13
  movslq %esi,%rsi
  mov    $0x2,%edx
  mov    $0x1,%r9d
  mov    $0x2,%eax
  jmp    L1
L42:
  cmp    %edx,%eax
  je     L62
L72:
  mov    %r8d,0x4(%rdi,%r10,1)
  mov    %eax,%r9d
  add    $0x1,%eax
L57:
  add    $0x1,%rdx
  cmp    %rdx,%rsi
  je     L13
L1:
  movslq %r9d,%rcx
  mov    (%rdi,%rdx,4),%r8d
  lea    0x0(,%rcx,4),%r10
  cmp    (%rdi,%rcx,4),%r8d
  jne    L42
  cmp    -0x4(%rdi,%r10,1),%r8d
  jne    L72
  add    $0x1,%rdx
  cmp    %rdx,%rsi
  jne    L1
L13:
  ret
  xchg   %ax,%ax
L62:
  mov    %eax,%r9d
  add    $0x1,%eax
  jmp    L57
```

assistant: ```c
int remove_duplicates_2(int *nums, int nums_size) {
  if (nums_size < 3) {
    return nums_size;
  }
  int i = 1, j;
  for (j = 2; j < nums_size; ++j) {
    if (nums[j] != nums[i]) {
      if (++i != j) {
        nums[i] = nums[j];
      }
    } else {
      if (nums[j] != nums[i - 1]) {
        nums[++i] = nums[j];
      }
    }
  }
  return i + 1;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x1,-0xc(%rbp)
  mov    D19(%rip),%rax
  mov    (%rax),%eax
  lea    -0xc(%rbp),%rsi
  mov    $0x0,%r9d
  mov    $0xffffffff,%r8d
  mov    $0xffffffff,%ecx
  mov    $0x0,%edx
  mov    %eax,%edi
  call   <syscall@plt>
  mov    D17(%rip),%rax
  mov    (%rax),%eax
  cmp    $0xe,%eax
  jne    L37
  mov    D20(%rip),%rax
  mov    (%rax),%eax
  cmp    $0x2000,%eax
  je     L5
L37:
  mov    $0xffffffff,%eax
  jmp    L8
L5:
  mov    $0x0,%eax
L8:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L12
  call   <__stack_chk_fail@plt>
L12:
  leave
  ret
```

assistant: ```c
int main(void) {
  struct perf_event_attr attr = {
      .context_switch = 1,
  };
  syscall(SYS_perf_event_open, &attr, 0, -1, -1, 0);
  if (PERF_RECORD_SWITCH != 14 || PERF_RECORD_MISC_SWITCH_OUT != (1 << 13))
    return -1;
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0x1,0x4(%rsp)
  lea    0x4(%rsp),%rsi
  mov    $0x0,%r9d
  mov    $0xffffffff,%r8d
  mov    $0xffffffff,%ecx
  mov    $0x0,%edx
  mov    D28(%rip),%rax
  mov    (%rax),%edi
  call   <syscall@plt>
  mov    D0(%rip),%rax
  cmpl   $0xe,(%rax)
  jne    L8
  mov    D22(%rip),%rax
  cmpl   $0x2000,(%rax)
  setne  %al
  movzbl %al,%eax
  neg    %eax
L10:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L23
  add    $0x18,%rsp
  ret
L8:
  mov    $0xffffffff,%eax
  jmp    L10
L23:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(void) {
  struct perf_event_attr attr = {
      .context_switch = 1,
  };
  syscall(SYS_perf_event_open, &attr, 0, -1, -1, 0);
  if (PERF_RECORD_SWITCH != 14 || PERF_RECORD_MISC_SWITCH_OUT != (1 << 13))
    return -1;
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  xor    %r9d,%r9d
  xor    %edx,%edx
  mov    $0xffffffff,%r8d
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    D13(%rip),%rax
  lea    0x4(%rsp),%rsi
  mov    $0xffffffff,%ecx
  movl   $0x1,0x4(%rsp)
  mov    (%rax),%edi
  call   <syscall@plt>
  mov    D24(%rip),%rax
  cmpl   $0xe,(%rax)
  jne    L20
  mov    D0(%rip),%rax
  cmpl   $0x2000,(%rax)
  setne  %al
  movzbl %al,%eax
  neg    %eax
L13:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L2
  add    $0x18,%rsp
  ret
L20:
  or     $0xffffffff,%eax
  jmp    L13
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(void) {
  struct perf_event_attr attr = {
      .context_switch = 1,
  };
  syscall(SYS_perf_event_open, &attr, 0, -1, -1, 0);
  if (PERF_RECORD_SWITCH != 14 || PERF_RECORD_MISC_SWITCH_OUT != (1 << 13))
    return -1;
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  xor    %r9d,%r9d
  xor    %edx,%edx
  mov    $0xffffffff,%r8d
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    D29(%rip),%rax
  lea    0x4(%rsp),%rsi
  mov    $0xffffffff,%ecx
  movl   $0x1,0x4(%rsp)
  mov    (%rax),%edi
  call   <syscall@plt>
  mov    D8(%rip),%rax
  cmpl   $0xe,(%rax)
  jne    L9
  mov    D6(%rip),%rax
  cmpl   $0x2000,(%rax)
  setne  %al
  movzbl %al,%eax
  neg    %eax
L4:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L25
  add    $0x18,%rsp
  ret
L9:
  or     $0xffffffff,%eax
  jmp    L4
L25:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(void) {
  struct perf_event_attr attr = {
      .context_switch = 1,
  };
  syscall(SYS_perf_event_open, &attr, 0, -1, -1, 0);
  if (PERF_RECORD_SWITCH != 14 || PERF_RECORD_MISC_SWITCH_OUT != (1 << 13))
    return -1;
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<function_with_leak2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0xc,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  nop
  leave
  ret
```

assistant: ```c
void function_with_leak2() {
  char *x = (char *)malloc(12);
  int m = 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<function_with_leak2>:
  endbr64
  sub    $0x8,%rsp
  mov    $0xc,%edi
  call   <malloc@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void function_with_leak2() {
  char *x = (char *)malloc(12);
  int m = 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<function_with_leak2>:
  endbr64
  mov    $0xc,%edi
  jmp    1050 <malloc@plt>
```

assistant: ```c
void function_with_leak2() {
  char *x = (char *)malloc(12);
  int m = 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<function_with_leak2>:
  endbr64
  mov    $0xc,%edi
  jmp    1050 <malloc@plt>
```

assistant: ```c
void function_with_leak2() {
  char *x = (char *)malloc(12);
  int m = 0;
}
```
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    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x0,-0x4(%rbp)
  cmpl   $0x1,-0x14(%rbp)
  jne    L43
  lea    D11(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  jmp    L22
L43:
  cmpl   $0x1,-0x14(%rbp)
  jle    L36
  cmpl   $0x3,-0x14(%rbp)
  jg     L36
  lea    D32(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  movl   $0x0,-0x8(%rbp)
  jmp    L59
L56:
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    %rax,%rsi
  lea    D30(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x8(%rbp)
L59:
  mov    -0x8(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L56
  mov    $0xa,%edi
  call   <putchar@plt>
  jmp    L22
L36:
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
L22:
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: read_data
arguments: tag: D11
type_guessed: string
name: read_data
arguments: tag: D32
type_guessed: string
name: read_data
arguments: tag: D30
type_guessed: string
name: read_data
arguments: tag: D19
type_guessed: string

tool: D11: You only have one argument. What ya doing bud???

tool: D32: Here are your arguments:

tool: D30: %s 

tool: D19: You have too many arguments. What ya doing bud???

assistant: ```c
int main(int argc, char *argv[]) {
  int i = 0;
  if (argc == 1) {
    printf("You only have one argument. What ya doing bud???\n");
  } else if (argc > 1 && argc < 4) {
    printf("Here are your arguments:\n");
    for (int i = 0; i < argc; i++)
      printf("%s ", argv[i]);
    printf("\n");
  } else {
    printf("You have too many arguments. What ya doing bud???\n");
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  cmp    $0x1,%edi
  je     L8
  mov    %edi,%ebp
  mov    %rsi,%r12
  lea    -0x2(%rdi),%eax
  cmp    $0x1,%eax
  jbe    L41
  lea    D37(%rip),%rdi
  call   <puts@plt>
L48:
  mov    $0x0,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L8:
  lea    D34(%rip),%rdi
  call   <puts@plt>
  jmp    L48
L41:
  lea    D29(%rip),%rdi
  call   <puts@plt>
  mov    %r12,%rbx
  lea    -0x1(%rbp),%eax
  lea    0x8(%r12,%rax,8),%r12
  lea    D27(%rip),%rbp
L1:
  mov    (%rbx),%rsi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x8,%rbx
  cmp    %r12,%rbx
  jne    L1
  mov    $0xa,%edi
  call   <putchar@plt>
  jmp    L48
```

tool_calls: name: retrieve_data
arguments: data_label: D37
type: string
name: retrieve_data
arguments: data_label: D34
type: string
name: retrieve_data
arguments: data_label: D29
type: string
name: retrieve_data
arguments: data_label: D27
type: string

tool: D37: You have too many arguments. What ya doing bud???

tool: D34: You only have one argument. What ya doing bud???

tool: D29: Here are your arguments:

tool: D27: %s 

assistant: ```c
int main(int argc, char *argv[]) {
  int i = 0;
  if (argc == 1) {
    printf("You only have one argument. What ya doing bud???\n");
  } else if (argc > 1 && argc < 4) {
    printf("Here are your arguments:\n");
    for (int i = 0; i < argc; i++)
      printf("%s ", argv[i]);
    printf("\n");
  } else {
    printf("You have too many arguments. What ya doing bud???\n");
  }
  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
  push   %rbp
  push   %rbx
  cmp    $0x1,%edi
  je     L42
  lea    -0x2(%rdi),%eax
  mov    %edi,%ebp
  cmp    $0x1,%eax
  jbe    L47
  lea    D12(%rip),%rdi
  call   <puts@plt>
L23:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
L47:
  lea    D5(%rip),%rdi
  mov    %rsi,%rbx
  call   <puts@plt>
  lea    -0x1(%rbp),%eax
  lea    D27(%rip),%rbp
  lea    0x8(%rbx,%rax,8),%r12
L24:
  mov    (%rbx),%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x8,%rbx
  call   <printf@plt>
  cmp    %r12,%rbx
  jne    L24
  mov    $0xa,%edi
  call   <putchar@plt>
  jmp    L23
L42:
  lea    D30(%rip),%rdi
  call   <puts@plt>
  jmp    L23
```

tool_calls: name: extract_symbolic_data
arguments: tag: D12
type_guessed: string
name: extract_symbolic_data
arguments: tag: D5
type_guessed: string
name: extract_symbolic_data
arguments: tag: D27
type_guessed: string
name: extract_symbolic_data
arguments: tag: D30
type_guessed: string

tool: D12: You have too many arguments. What ya doing bud???

tool: D5: Here are your arguments:

tool: D27: %s 

tool: D30: You only have one argument. What ya doing bud???

assistant: ```c
int main(int argc, char *argv[]) {
  int i = 0;
  if (argc == 1) {
    printf("You only have one argument. What ya doing bud???\n");
  } else if (argc > 1 && argc < 4) {
    printf("Here are your arguments:\n");
    for (int i = 0; i < argc; i++)
      printf("%s ", argv[i]);
    printf("\n");
  } else {
    printf("You have too many arguments. What ya doing bud???\n");
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  cmp    $0x1,%edi
  je     L33
  lea    -0x2(%rdi),%eax
  mov    %edi,%ebp
  cmp    $0x1,%eax
  jbe    L38
  lea    D6(%rip),%rdi
  call   <puts@plt>
L28:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
L38:
  lea    D33(%rip),%rdi
  mov    %rsi,%rbx
  call   <puts@plt>
  lea    -0x1(%rbp),%eax
  lea    D17(%rip),%rbp
  lea    0x8(%rbx,%rax,8),%r12
L44:
  mov    (%rbx),%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x8,%rbx
  call   <printf@plt>
  cmp    %r12,%rbx
  jne    L44
  mov    $0xa,%edi
  call   <putchar@plt>
  jmp    L28
L33:
  lea    D25(%rip),%rdi
  call   <puts@plt>
  jmp    L28
```

tool_calls: name: read_data
arguments: data_tag: D6
data_type: string
name: read_data
arguments: data_tag: D33
data_type: string
name: read_data
arguments: data_tag: D17
data_type: string
name: read_data
arguments: data_tag: D25
data_type: string

tool: D6: You have too many arguments. What ya doing bud???

tool: D33: Here are your arguments:

tool: D17: %s 

tool: D25: You only have one argument. What ya doing bud???

assistant: ```c
int main(int argc, char *argv[]) {
  int i = 0;
  if (argc == 1) {
    printf("You only have one argument. What ya doing bud???\n");
  } else if (argc > 1 && argc < 4) {
    printf("Here are your arguments:\n");
    for (int i = 0; i < argc; i++)
      printf("%s ", argv[i]);
    printf("\n");
  } else {
    printf("You have too many arguments. What ya doing bud???\n");
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<__test2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x60,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x1,-0x5c(%rbp)
  movl   $0x3,-0x58(%rbp)
  movl   $0x7,-0x54(%rbp)
  movl   $0x2,-0x50(%rbp)
  movl   $0x4,-0x4c(%rbp)
  movl   $0x5,-0x48(%rbp)
  movl   $0x6,-0x44(%rbp)
  movl   $0x8,-0x40(%rbp)
  lea    -0x50(%rbp),%rcx
  lea    -0x5c(%rbp),%rdx
  lea    -0x30(%rbp),%rax
  sub    $0x8,%rsp
  push   $0x4
  mov    $0x5,%r9d
  mov    %rcx,%r8
  mov    $0x3,%ecx
  mov    $0x8,%esi
  mov    %rax,%rdi
  call   <mmmerge@plt>
  add    $0x10,%rsp
  movl   $0x0,-0x60(%rbp)
  jmp    L19
L20:
  mov    -0x60(%rbp),%eax
  cltq
  mov    -0x30(%rbp,%rax,4),%eax
  mov    %eax,%esi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x60(%rbp)
L19:
  cmpl   $0x7,-0x60(%rbp)
  jle    L20
  mov    $0xa,%edi
  call   <putchar@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L15
  call   <__stack_chk_fail@plt>
L15:
  leave
  ret
```

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

tool: D7: %d 

assistant: ```c
void __test2(void) {
  int v1[3] = {1, 3, 7};
  int v2[5] = {2, 4, 5, 6, 8};
  int r[8];
  int i;
  mmmerge(r, 8, v1, 3, v2, 5, sizeof(int));
  for (i = 0; i < 8; ++i)
    printf("%d ", r[i]);
  printf("\n");
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<__test2>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x60,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x58(%rsp)
  xor    %eax,%eax
  movl   $0x1,0x4(%rsp)
  movl   $0x3,0x8(%rsp)
  movl   $0x7,0xc(%rsp)
  movl   $0x2,0x10(%rsp)
  movl   $0x4,0x14(%rsp)
  movl   $0x5,0x18(%rsp)
  movl   $0x6,0x1c(%rsp)
  movl   $0x8,0x20(%rsp)
  lea    0x4(%rsp),%rdx
  lea    0x30(%rsp),%rbx
  sub    $0x8,%rsp
  push   $0x4
  mov    $0x5,%r9d
  lea    0x20(%rsp),%r8
  mov    $0x3,%ecx
  mov    $0x8,%esi
  mov    %rbx,%rdi
  call   <mmmerge@plt>
  lea    0x60(%rsp),%r12
  add    $0x10,%rsp
  lea    D2(%rip),%rbp
L1:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x4,%rbx
  cmp    %r12,%rbx
  jne    L1
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x58(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L19
  add    $0x60,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L19:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D2
data_type: string

tool: D2: %d 

assistant: ```c
void __test2(void) {
  int v1[3] = {1, 3, 7};
  int v2[5] = {2, 4, 5, 6, 8};
  int r[8];
  int i;
  mmmerge(r, 8, v1, 3, v2, 5, sizeof(int));
  for (i = 0; i < 8; ++i)
    printf("%d ", r[i]);
  printf("\n");
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<__test2>:
  endbr64
  push   %r12
  mov    $0x5,%r9d
  mov    $0x3,%ecx
  mov    $0x8,%esi
  push   %rbp
  lea    D2(%rip),%rbp
  push   %rbx
  sub    $0x60,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x58(%rsp)
  movabs $0x300000001,%rax
  lea    0x4(%rsp),%rdx
  lea    0x30(%rsp),%rbx
  movl   $0x7,0xc(%rsp)
  movl   $0x8,0x20(%rsp)
  sub    $0x8,%rsp
  mov    %rbx,%rdi
  mov    %rax,0xc(%rsp)
  movabs $0x400000002,%rax
  mov    %rax,0x18(%rsp)
  movabs $0x600000005,%rax
  mov    %rax,0x20(%rsp)
  push   $0x4
  lea    0x20(%rsp),%r8
  lea    0x60(%rsp),%r12
  call   <mmmerge@plt>
  pop    %rax
  pop    %rdx
  nopw   0x0(%rax,%rax,1)
L11:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %r12,%rbx
  jne    L11
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x58(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x60,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D2
type_infered: string

tool: D2: %d 

assistant: ```c
void __test2(void) {
  int v1[3] = {1, 3, 7};
  int v2[5] = {2, 4, 5, 6, 8};
  int r[8];
  int i;
  mmmerge(r, 8, v1, 3, v2, 5, sizeof(int));
  for (i = 0; i < 8; ++i)
    printf("%d ", r[i]);
  printf("\n");
}
```