ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<toolame_read_stat>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x4a0,%rsp
  mov    %rdi,-0x498(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x410(%rbp),%rax
  mov    $0x400,%edx
  mov    %rax,%rsi
  mov    $0x0,%edi
  call   <read@plt>
  mov    %eax,-0x484(%rbp)
  cmpl   $0x10,-0x484(%rbp)
  jle    L13
  lea    -0x480(%rbp),%rax
  lea    -0x410(%rbp),%rdx
  lea    0x1(%rdx),%rcx
  mov    %rax,%rdx
  lea    D9(%rip),%rax
  mov    %rax,%rsi
  mov    %rcx,%rdi
  call   <sscanf@plt>
  lea    -0x480(%rbp),%rax
  mov    %rax,%rdi
  call   <atoi@plt>
  mov    -0x498(%rbp),%rdx
  mov    %eax,(%rdx)
  mov    $0x0,%eax
  jmp    L9
L13:
  mov    $0xffffffff,%eax
L9:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L15
  call   <__stack_chk_fail@plt>
L15:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D9
type_infered: string

tool: D9: %s]

assistant: ```c
int toolame_read_stat(int *current) {
  char temp[1024];
  char string[100];
  char *string1;
  int bytes_read;
  bytes_read = read(0, (void *)temp, sizeof(temp));
  if (bytes_read > 16) {
    sscanf(&temp[1], "%s]", string);
    *current = atoi(string);
    return (0);
  }
  return (-1);
}
```
user: What is the c source code of the assembly code below?

```asm
<toolame_read_stat>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x488,%rsp
  mov    %rdi,%rbp
  mov    %fs:0x28,%rax
  mov    %rax,0x478(%rsp)
  xor    %eax,%eax
  lea    0x70(%rsp),%rsi
  mov    $0x400,%edx
  mov    $0x0,%edi
  call   <read@plt>
  cmp    $0x10,%eax
  jle    L1
  mov    %rsp,%rbx
  lea    0x71(%rsp),%rdi
  mov    %rbx,%rdx
  lea    D7(%rip),%rsi
  call   <sscanf@plt>
  mov    %rbx,%rdi
  call   <atoi@plt>
  mov    %eax,0x0(%rbp)
  mov    $0x0,%eax
L18:
  mov    0x478(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L8
  add    $0x488,%rsp
  pop    %rbx
  pop    %rbp
  ret
L1:
  mov    $0xffffffff,%eax
  jmp    L18
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D7
datatype: string

tool: D7: %s]

assistant: ```c
int toolame_read_stat(int *current) {
  char temp[1024];
  char string[100];
  char *string1;
  int bytes_read;
  bytes_read = read(0, (void *)temp, sizeof(temp));
  if (bytes_read > 16) {
    sscanf(&temp[1], "%s]", string);
    *current = atoi(string);
    return (0);
  }
  return (-1);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<toolame_read_stat>:
  endbr64
  push   %rbp
  mov    $0x400,%edx
  push   %rbx
  mov    %rdi,%rbx
  xor    %edi,%edi
  sub    $0x488,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x478(%rsp)
  xor    %eax,%eax
  lea    0x70(%rsp),%rsi
  call   <read@plt>
  cmp    $0x10,%eax
  jle    L13
  mov    %rsp,%rbp
  lea    0x71(%rsp),%rdi
  lea    D1(%rip),%rsi
  mov    %rbp,%rdx
  call   <sscanf@plt>
  mov    %rbp,%rdi
  call   <atoi@plt>
  mov    %eax,(%rbx)
  xor    %eax,%eax
L10:
  mov    0x478(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L4
  add    $0x488,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax)
L13:
  mov    $0xffffffff,%eax
  jmp    L10
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D1
type_infered: string

tool: D1: %s]

assistant: ```c
int toolame_read_stat(int *current) {
  char temp[1024];
  char string[100];
  char *string1;
  int bytes_read;
  bytes_read = read(0, (void *)temp, sizeof(temp));
  if (bytes_read > 16) {
    sscanf(&temp[1], "%s]", string);
    *current = atoi(string);
    return (0);
  }
  return (-1);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<toolame_read_stat>:
  endbr64
  push   %rbp
  mov    $0x400,%edx
  push   %rbx
  mov    %rdi,%rbx
  xor    %edi,%edi
  sub    $0x488,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x478(%rsp)
  xor    %eax,%eax
  lea    0x70(%rsp),%rsi
  call   <read@plt>
  cmp    $0x10,%eax
  jle    L0
  mov    %rsp,%rbp
  lea    0x71(%rsp),%rdi
  lea    D1(%rip),%rsi
  mov    %rbp,%rdx
  call   <sscanf@plt>
  mov    %rbp,%rdi
  call   <atoi@plt>
  mov    %eax,(%rbx)
  xor    %eax,%eax
L10:
  mov    0x478(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L8
  add    $0x488,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax)
L0:
  mov    $0xffffffff,%eax
  jmp    L10
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D1
type: string

tool: D1: %s]

assistant: ```c
int toolame_read_stat(int *current) {
  char temp[1024];
  char string[100];
  char *string1;
  int bytes_read;
  bytes_read = read(0, (void *)temp, sizeof(temp));
  if (bytes_read > 16) {
    sscanf(&temp[1], "%s]", string);
    *current = atoi(string);
    return (0);
  }
  return (-1);
}
```
user: What is the c source code of the assembly code below:

```asm
<jo_calcBits>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,%edx
  neg    %edx
  cmovns %edx,%eax
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  jns    L37
  mov    -0x14(%rbp),%eax
  sub    $0x1,%eax
  jmp    L2
L37:
  mov    -0x14(%rbp),%eax
L2:
  mov    %eax,-0x14(%rbp)
  mov    -0x20(%rbp),%rax
  add    $0x2,%rax
  movw   $0x1,(%rax)
  jmp    L0
L39:
  mov    -0x20(%rbp),%rax
  add    $0x2,%rax
  movzwl (%rax),%edx
  add    $0x1,%edx
  mov    %dx,(%rax)
L0:
  sarl   -0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jne    L39
  mov    -0x20(%rbp),%rax
  add    $0x2,%rax
  movzwl (%rax),%eax
  movzwl %ax,%eax
  mov    $0x1,%edx
  mov    %eax,%ecx
  shl    %cl,%edx
  mov    %edx,%eax
  sub    $0x1,%eax
  mov    %eax,%edx
  mov    -0x14(%rbp),%eax
  and    %edx,%eax
  mov    %eax,%edx
  mov    -0x20(%rbp),%rax
  mov    %dx,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void jo_calcBits(int val, unsigned short bits[2]) {
  int tmp1 = val < 0 ? -val : val;
  val = val < 0 ? val - 1 : val;
  bits[1] = 1;
  while (tmp1 >>= 1) {
    ++bits[1];
  }
  bits[0] = val & ((1 << bits[1]) - 1);
}
```
user: Translate the below assembly code into C programming language:

```asm
<jo_calcBits>:
  endbr64
  mov    %edi,%eax
  neg    %eax
  cmovs  %edi,%eax
  cmp    $0x80000000,%edi
  adc    $0xffffffff,%edi
  movw   $0x1,0x2(%rsi)
  sar    %eax
  je     L8
  mov    $0x2,%edx
L5:
  mov    %edx,%ecx
  add    $0x1,%edx
  sar    %eax
  jne    L5
  mov    %cx,0x2(%rsi)
L8:
  movzwl 0x2(%rsi),%ecx
  mov    $0x1,%eax
  shl    %cl,%eax
  sub    $0x1,%eax
  and    %edi,%eax
  mov    %ax,(%rsi)
  ret
```

assistant: ```c
void jo_calcBits(int val, unsigned short bits[2]) {
  int tmp1 = val < 0 ? -val : val;
  val = val < 0 ? val - 1 : val;
  bits[1] = 1;
  while (tmp1 >>= 1) {
    ++bits[1];
  }
  bits[0] = val & ((1 << bits[1]) - 1);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<jo_calcBits>:
  endbr64
  mov    %edi,%eax
  mov    $0x1,%edx
  neg    %eax
  mov    %dx,0x2(%rsi)
  cmovs  %edi,%eax
  cmp    $0x80000000,%edi
  adc    $0xffffffff,%edi
  sar    %eax
  je     L12
  mov    $0x2,%edx
  cs nopw 0x0(%rax,%rax,1)
L5:
  mov    %edx,%ecx
  add    $0x1,%edx
  sar    %eax
  jne    L5
  mov    $0x1,%eax
  mov    %cx,0x2(%rsi)
  shl    %cl,%eax
  sub    $0x1,%eax
  and    %edi,%eax
  mov    %ax,(%rsi)
  ret
  nopl   (%rax)
L12:
  mov    $0x1,%eax
  and    %edi,%eax
  mov    %ax,(%rsi)
  ret
```

assistant: ```c
void jo_calcBits(int val, unsigned short bits[2]) {
  int tmp1 = val < 0 ? -val : val;
  val = val < 0 ? val - 1 : val;
  bits[1] = 1;
  while (tmp1 >>= 1) {
    ++bits[1];
  }
  bits[0] = val & ((1 << bits[1]) - 1);
}
```
user: Translate the below assembly code into C programming language:

```asm
<jo_calcBits>:
  endbr64
  mov    %edi,%eax
  mov    $0x1,%edx
  neg    %eax
  mov    %dx,0x2(%rsi)
  cmovs  %edi,%eax
  cmp    $0x80000000,%edi
  adc    $0xffffffff,%edi
  sar    %eax
  je     L16
  mov    $0x2,%edx
  cs nopw 0x0(%rax,%rax,1)
L7:
  mov    %edx,%ecx
  add    $0x1,%edx
  sar    %eax
  jne    L7
  mov    $0x1,%eax
  mov    %cx,0x2(%rsi)
  shl    %cl,%eax
  sub    $0x1,%eax
  and    %edi,%eax
  mov    %ax,(%rsi)
  ret
  nopl   (%rax)
L16:
  mov    $0x1,%eax
  and    %edi,%eax
  mov    %ax,(%rsi)
  ret
```

assistant: ```c
void jo_calcBits(int val, unsigned short bits[2]) {
  int tmp1 = val < 0 ? -val : val;
  val = val < 0 ? val - 1 : val;
  bits[1] = 1;
  while (tmp1 >>= 1) {
    ++bits[1];
  }
  bits[0] = val & ((1 << bits[1]) - 1);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<permutations>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x48,%rsp
  mov    %rdi,%r12
  mov    %rdi,0x28(%rsp)
  mov    %rsi,%rbp
  mov    %rsi,(%rsp)
  mov    %edx,%r13d
  mov    %ecx,%ebx
  mov    %ecx,0x34(%rsp)
  mov    %r8d,%r14d
  mov    %r8d,0xc(%rsp)
  mov    %edx,%edi
  call   <factorial@plt>
  mov    %eax,%r15d
  mov    %eax,%r8d
  mov    %r14d,%ecx
  mov    %ebx,%edx
  mov    %r13d,%esi
  lea    D8(%rip),%rdi
  call   <printf@plt>
  mov    %r13d,%esi
  mov    %rbp,%rdi
  call   <print_array@plt>
  test   %r15d,%r15d
  jle    L51
  movslq %ebx,%rax
  shl    $0x2,%rax
  mov    %rax,0x10(%rsp)
  mov    %r15d,%eax
  cltd
  idiv   %r13d
  mov    %eax,0x20(%rsp)
  lea    0x0(,%r13,4),%eax
  mov    %eax,0x24(%rsp)
  mov    %r15d,%eax
  mov    %rax,0x18(%rsp)
  movslq %r14d,%rax
  lea    (%r12,%rax,8),%r14
  mov    $0x0,%ebp
  mov    $0x0,%ebx
  lea    -0x1(%r13),%eax
  mov    %eax,0x30(%rsp)
  mov    %r13d,0x8(%rsp)
  jmp    L44
L58:
  add    $0x1,%rbx
  add    0x8(%rsp),%ebp
  cmp    0x18(%rsp),%rbx
  je     L51
L44:
  mov    %ebp,%eax
  cltd
  idiv   %r15d
  mov    %eax,%r12d
  mov    0xc(%rsp),%eax
  lea    (%rax,%rbx,1),%r13d
  movslq %r12d,%rax
  mov    (%rsp),%rcx
  mov    (%rcx,%rax,4),%edx
  mov    (%r14,%rbx,8),%rax
  mov    0x10(%rsp),%rsi
  mov    %edx,(%rax,%rsi,1)
  mov    %ebx,%eax
  cltd
  idivl  0x20(%rsp)
  test   %edx,%edx
  jne    L58
  cmpl   $0x1,0x8(%rsp)
  jle    L58
  mov    0x24(%rsp),%edi
  call   <malloc@plt>
  mov    0x24(%rsp),%edx
  mov    (%rsp),%rsi
  mov    %rax,0x38(%rsp)
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    0x30(%rsp),%edx
  mov    %r12d,%esi
  mov    0x38(%rsp),%r12
  mov    %r12,%rdi
  call   <exchange@plt>
  mov    0x34(%rsp),%eax
  lea    0x1(%rax),%ecx
  mov    %r13d,%r8d
  mov    0x30(%rsp),%edx
  mov    %r12,%rsi
  mov    0x28(%rsp),%rdi
  call   <permutations>
  mov    %r12,%rdi
  call   <free@plt>
  jmp    L58
L51:
  add    $0x48,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: read_data
arguments: data_label: D8
type: string

tool: D8: nums_size: %d, col: %d, row: %d, n_perms: %d


assistant: ```c
void permutations(int **perms, int *nums, int nums_size, int col, int row) {
  int n_perms = factorial(nums_size);
  printf("nums_size: %d, col: %d, row: %d, n_perms: %d\n", nums_size, col, row, n_perms);
  print_array(nums, nums_size);
  for (int i = 0; i < n_perms; i++) {
    int current_index = i * nums_size / n_perms;
    perms[i + row][col] = nums[current_index];
    if (i % (n_perms / nums_size) == 0 && nums_size - 1 > 0) {
      int *nums_cpy = malloc(nums_size * sizeof(int));
      memcpy(nums_cpy, nums, nums_size * sizeof(int));
      exchange(nums_cpy, current_index, nums_size - 1);
      permutations(perms, nums_cpy, nums_size - 1, col + 1, i + row);
      free(nums_cpy);
    }
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<permutations>:
  endbr64
  push   %r15
  mov    %rsi,%r15
  push   %r14
  mov    %r8d,%r14d
  push   %r13
  mov    %edx,%r13d
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %ecx,%ebx
  sub    $0x48,%rsp
  mov    %rdi,0x18(%rsp)
  mov    %edx,%edi
  mov    %ecx,0x24(%rsp)
  call   <factorial@plt>
  mov    %r13d,%esi
  mov    %r14d,%ecx
  mov    %ebx,%edx
  lea    D7(%rip),%rdi
  mov    %eax,%r8d
  mov    %eax,%r12d
  call   <printf@plt>
  mov    %r13d,%esi
  mov    %r15,%rdi
  call   <print_array@plt>
  test   %r12d,%r12d
  jle    L22
  movslq %ebx,%rax
  lea    0x0(,%rax,4),%r10
  mov    %r12d,%eax
  cltd
  idiv   %r13d
  cmp    $0x1,%r13d
  setg   0x13(%rsp)
  xor    %ebx,%ebx
  mov    %eax,%r9d
  lea    0x0(,%r13,4),%eax
  mov    %eax,0x14(%rsp)
  movslq %r12d,%rax
  mov    %rax,(%rsp)
  movslq %r14d,%rax
  lea    0x0(%rbp,%rax,8),%r11
  lea    -0x1(%r13),%eax
  mov    %r9d,0xc(%rsp)
  xor    %ebp,%ebp
  mov    %eax,0x20(%rsp)
  mov    %r14d,%eax
  mov    %r13d,%r9d
  mov    %r10,%r14
  mov    %eax,%r10d
L18:
  mov    %ebp,%eax
  lea    (%r10,%rbx,1),%r8d
  cltd
  idiv   %r12d
  cltq
  mov    (%r15,%rax,4),%edx
  mov    %rax,%rcx
  mov    (%r11,%rbx,8),%rax
  mov    %edx,(%rax,%r14,1)
  mov    %ebx,%eax
  cltd
  idivl  0xc(%rsp)
  test   %edx,%edx
  jne    L6
  cmpb   $0x0,0x13(%rsp)
  je     L6
  mov    0x14(%rsp),%edi
  mov    %r10d,0x3c(%rsp)
  mov    %r9d,0x38(%rsp)
  mov    %r11,0x30(%rsp)
  mov    %ecx,0x2c(%rsp)
  mov    %r8d,0x28(%rsp)
  call   <malloc@plt>
  mov    0x14(%rsp),%edx
  mov    %r15,%rsi
  mov    %rax,%rdi
  mov    %rax,%r13
  call   <memcpy@plt>
  mov    0x20(%rsp),%edx
  mov    0x2c(%rsp),%esi
  mov    %r13,%rdi
  call   <exchange@plt>
  mov    0x24(%rsp),%eax
  mov    0x20(%rsp),%edx
  mov    %r13,%rsi
  mov    0x28(%rsp),%r8d
  mov    0x18(%rsp),%rdi
  lea    0x1(%rax),%ecx
  call   <permutations>
  mov    %r13,%rdi
  call   <free@plt>
  mov    0x3c(%rsp),%r10d
  mov    0x38(%rsp),%r9d
  mov    0x30(%rsp),%r11
L6:
  add    $0x1,%rbx
  add    %r9d,%ebp
  cmp    %rbx,(%rsp)
  jne    L18
L22:
  add    $0x48,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D7
data_type: string

tool: D7: nums_size: %d, col: %d, row: %d, n_perms: %d


assistant: ```c
void permutations(int **perms, int *nums, int nums_size, int col, int row) {
  int n_perms = factorial(nums_size);
  printf("nums_size: %d, col: %d, row: %d, n_perms: %d\n", nums_size, col, row, n_perms);
  print_array(nums, nums_size);
  for (int i = 0; i < n_perms; i++) {
    int current_index = i * nums_size / n_perms;
    perms[i + row][col] = nums[current_index];
    if (i % (n_perms / nums_size) == 0 && nums_size - 1 > 0) {
      int *nums_cpy = malloc(nums_size * sizeof(int));
      memcpy(nums_cpy, nums, nums_size * sizeof(int));
      exchange(nums_cpy, current_index, nums_size - 1);
      permutations(perms, nums_cpy, nums_size - 1, col + 1, i + row);
      free(nums_cpy);
    }
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<permutations>:
  endbr64
  push   %r15
  mov    %rsi,%r15
  push   %r14
  mov    %r8d,%r14d
  push   %r13
  mov    %edx,%r13d
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %ecx,%ebx
  sub    $0x48,%rsp
  mov    %rdi,0x18(%rsp)
  mov    %edx,%edi
  mov    %ecx,0x24(%rsp)
  call   <factorial@plt>
  mov    %r13d,%esi
  mov    %r14d,%ecx
  mov    %ebx,%edx
  lea    D3(%rip),%rdi
  mov    %eax,%r8d
  mov    %eax,%r12d
  call   <printf@plt>
  mov    %r13d,%esi
  mov    %r15,%rdi
  call   <print_array@plt>
  test   %r12d,%r12d
  jle    L7
  movslq %ebx,%rax
  lea    0x0(,%rax,4),%r10
  mov    %r12d,%eax
  cltd
  idiv   %r13d
  cmp    $0x1,%r13d
  setg   0x13(%rsp)
  xor    %ebx,%ebx
  mov    %eax,%r9d
  lea    0x0(,%r13,4),%eax
  mov    %eax,0x14(%rsp)
  movslq %r12d,%rax
  mov    %rax,(%rsp)
  movslq %r14d,%rax
  lea    0x0(%rbp,%rax,8),%r11
  lea    -0x1(%r13),%eax
  mov    %r9d,0xc(%rsp)
  xor    %ebp,%ebp
  mov    %eax,0x20(%rsp)
  mov    %r14d,%eax
  mov    %r13d,%r9d
  mov    %r10,%r14
  mov    %eax,%r10d
L27:
  mov    %ebp,%eax
  lea    (%r10,%rbx,1),%r8d
  cltd
  idiv   %r12d
  cltq
  mov    (%r15,%rax,4),%edx
  mov    %rax,%rcx
  mov    (%r11,%rbx,8),%rax
  mov    %edx,(%rax,%r14,1)
  mov    %ebx,%eax
  cltd
  idivl  0xc(%rsp)
  test   %edx,%edx
  jne    L36
  cmpb   $0x0,0x13(%rsp)
  je     L36
  mov    0x14(%rsp),%edi
  mov    %r10d,0x3c(%rsp)
  mov    %r9d,0x38(%rsp)
  mov    %r11,0x30(%rsp)
  mov    %ecx,0x2c(%rsp)
  mov    %r8d,0x28(%rsp)
  call   <malloc@plt>
  mov    0x14(%rsp),%edx
  mov    %r15,%rsi
  mov    %rax,%rdi
  mov    %rax,%r13
  call   <memcpy@plt>
  mov    0x20(%rsp),%edx
  mov    0x2c(%rsp),%esi
  mov    %r13,%rdi
  call   <exchange@plt>
  mov    0x24(%rsp),%eax
  mov    0x20(%rsp),%edx
  mov    %r13,%rsi
  mov    0x28(%rsp),%r8d
  mov    0x18(%rsp),%rdi
  lea    0x1(%rax),%ecx
  call   <permutations>
  mov    %r13,%rdi
  call   <free@plt>
  mov    0x3c(%rsp),%r10d
  mov    0x38(%rsp),%r9d
  mov    0x30(%rsp),%r11
L36:
  add    $0x1,%rbx
  add    %r9d,%ebp
  cmp    %rbx,(%rsp)
  jne    L27
L7:
  add    $0x48,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D3
datatype: string

tool: D3: nums_size: %d, col: %d, row: %d, n_perms: %d


assistant: ```c
void permutations(int **perms, int *nums, int nums_size, int col, int row) {
  int n_perms = factorial(nums_size);
  printf("nums_size: %d, col: %d, row: %d, n_perms: %d\n", nums_size, col, row, n_perms);
  print_array(nums, nums_size);
  for (int i = 0; i < n_perms; i++) {
    int current_index = i * nums_size / n_perms;
    perms[i + row][col] = nums[current_index];
    if (i % (n_perms / nums_size) == 0 && nums_size - 1 > 0) {
      int *nums_cpy = malloc(nums_size * sizeof(int));
      memcpy(nums_cpy, nums, nums_size * sizeof(int));
      exchange(nums_cpy, current_index, nums_size - 1);
      permutations(perms, nums_cpy, nums_size - 1, col + 1, i + row);
      free(nums_cpy);
    }
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  jmp    L95
L39:
  mov    -0x18(%rbp),%eax
  mov    -0x20(%rbp),%edx
  sub    %edx,%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x14(%rbp),%eax
  mov    -0x1c(%rbp),%edx
  sub    %edx,%eax
  mov    %eax,-0xc(%rbp)
  cmpl   $0x0,-0xc(%rbp)
  jns    L13
  addl   $0x3c,-0xc(%rbp)
  subl   $0x1,-0x10(%rbp)
L13:
  cmpl   $0x0,-0x10(%rbp)
  jns    L65
  addl   $0x18,-0x10(%rbp)
L65:
  mov    -0x10(%rbp),%eax
  imul   $0x3c,%eax,%edx
  mov    -0xc(%rbp),%eax
  add    %edx,%eax
  mov    %eax,%esi
  lea    D11(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L95:
  lea    -0x14(%rbp),%rsi
  lea    -0x18(%rbp),%rcx
  lea    -0x1c(%rbp),%rdx
  lea    -0x20(%rbp),%rax
  mov    %rsi,%r8
  mov    %rax,%rsi
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x20(%rbp),%eax
  test   %eax,%eax
  jne    L61
  mov    -0x1c(%rbp),%eax
  test   %eax,%eax
  jne    L61
  mov    -0x18(%rbp),%eax
  test   %eax,%eax
  jne    L61
  mov    -0x14(%rbp),%eax
  test   %eax,%eax
  je     L60
L61:
  mov    $0x1,%eax
  jmp    L23
L60:
  mov    $0x0,%eax
L23:
  test   %eax,%eax
  jne    L39
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L25
  call   <__stack_chk_fail@plt>
L25:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D11
type: string
name: extract_symbolic_data
arguments: tag: D12
type: string

tool: D11: %d


tool: D12: %d %d %d %d

assistant: ```c
int main(void) {
  int horaInicial, minutoInicial, horaFinal, minutoFinal, dtHora, dtMinuto;
  while (scanf("%d %d %d %d", &horaInicial, &minutoInicial, &horaFinal, &minutoFinal), horaInicial != 0 || minutoInicial != 0 || horaFinal != 0 || minutoFinal != 0) {
    dtHora = horaFinal - horaInicial;
    dtMinuto = minutoFinal - minutoInicial;
    if (dtMinuto < 0) {
      dtMinuto = 60 + dtMinuto;
      dtHora -= 1;
    }
    if (dtHora < 0) {
      dtHora = 24 + dtHora;
    }
    printf("%d\n", dtHora * 60 + dtMinuto);
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x14(%rsp),%r12
  lea    0x10(%rsp),%rbp
  lea    D10(%rip),%rbx
  lea    D4(%rip),%r13
  jmp    L45
L55:
  sub    %ecx,%eax
  mov    %eax,%esi
  mov    0x14(%rsp),%eax
  sub    %edx,%eax
  js     L20
L14:
  lea    0x18(%rsi),%edx
  test   %esi,%esi
  cmovs  %edx,%esi
L0:
  imul   $0x3c,%esi,%esi
  add    %eax,%esi
  mov    %r13,%rdi
  call   <printf@plt>
L45:
  lea    0xc(%rsp),%rdx
  lea    0x8(%rsp),%rsi
  mov    %r12,%r8
  mov    %rbp,%rcx
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    0x8(%rsp),%ecx
  mov    0xc(%rsp),%edx
  mov    0x10(%rsp),%eax
  mov    %ecx,%esi
  or     %edx,%esi
  or     %eax,%esi
  jne    L55
  mov    0x14(%rsp),%eax
  test   %eax,%eax
  je     L31
  jns    L0
L20:
  add    $0x3c,%eax
  sub    $0x1,%esi
  jmp    L14
L31:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L32
  mov    $0x0,%eax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L32:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D10
datatype: string
name: parse_data
arguments: data_label: D4
datatype: string

tool: D10: %d %d %d %d

tool: D4: %d


assistant: ```c
int main(void) {
  int horaInicial, minutoInicial, horaFinal, minutoFinal, dtHora, dtMinuto;
  while (scanf("%d %d %d %d", &horaInicial, &minutoInicial, &horaFinal, &minutoFinal), horaInicial != 0 || minutoInicial != 0 || horaFinal != 0 || minutoFinal != 0) {
    dtHora = horaFinal - horaInicial;
    dtMinuto = minutoFinal - minutoInicial;
    if (dtMinuto < 0) {
      dtMinuto = 60 + dtMinuto;
      dtHora -= 1;
    }
    if (dtHora < 0) {
      dtHora = 24 + dtHora;
    }
    printf("%d\n", dtHora * 60 + dtMinuto);
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r15
  lea    D8(%rip),%r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    D2(%rip),%rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x14(%rsp),%r14
  lea    0x10(%rsp),%r13
  lea    0xc(%rsp),%r12
  lea    0x8(%rsp),%rbp
  jmp    L55
  cs nopw 0x0(%rax,%rax,1)
L13:
  mov    0x14(%rsp),%eax
  test   %eax,%eax
  je     L53
  js     L69
  nopw   0x0(%rax,%rax,1)
L76:
  imul   $0x3c,%esi,%esi
  mov    %r15,%rdi
  add    %eax,%esi
  call   <printf@plt>
L55:
  mov    %r13,%rcx
  mov    %r12,%rdx
  mov    %rbp,%rsi
  mov    %r14,%r8
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    0x8(%rsp),%ecx
  mov    0xc(%rsp),%edx
  mov    0x10(%rsp),%eax
  mov    %ecx,%esi
  or     %edx,%esi
  or     %eax,%esi
  je     L13
  sub    %ecx,%eax
  mov    %eax,%esi
  mov    0x14(%rsp),%eax
  sub    %edx,%eax
  js     L69
L10:
  lea    0x18(%rsi),%edx
  test   %esi,%esi
  cmovs  %edx,%esi
  jmp    L76
  nopl   0x0(%rax,%rax,1)
L69:
  add    $0x3c,%eax
  sub    $0x1,%esi
  jmp    L10
  nopl   0x0(%rax,%rax,1)
L53:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L63
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L63:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D8
datatype: string
name: disassemble_data
arguments: tag: D2
datatype: string

tool: D8: %d


tool: D2: %d %d %d %d

assistant: ```c
int main(void) {
  int horaInicial, minutoInicial, horaFinal, minutoFinal, dtHora, dtMinuto;
  while (scanf("%d %d %d %d", &horaInicial, &minutoInicial, &horaFinal, &minutoFinal), horaInicial != 0 || minutoInicial != 0 || horaFinal != 0 || minutoFinal != 0) {
    dtHora = horaFinal - horaInicial;
    dtMinuto = minutoFinal - minutoInicial;
    if (dtMinuto < 0) {
      dtMinuto = 60 + dtMinuto;
      dtHora -= 1;
    }
    if (dtHora < 0) {
      dtHora = 24 + dtHora;
    }
    printf("%d\n", dtHora * 60 + dtMinuto);
  }
  return 0;
}
```