ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  lea    -0xf4000(%rsp),%r11
L11:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L11
  sub    $0x268,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xf4258(%rsp)
  xor    %eax,%eax
  movl   $0x0,0xc(%rsp)
  lea    0xc(%rsp),%rsi
  lea    D20(%rip),%rdi
  call   <scanf@plt>
  lea    D5(%rip),%rbx
  jmp    L58
L8:
  addl   $0x1,0xc(%rsp)
L58:
  lea    0x10(%rsp),%rbp
  mov    0xc(%rsp),%edx
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <sprintf@plt>
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %eax,%ecx
  shr    $0x1f,%ecx
  add    %eax,%ecx
  sar    %ecx
  cmp    $0x1,%eax
  jle    L6
  cltq
  lea    0x10(%rsp,%rax,1),%rdx
  mov    $0x0,%eax
L18:
  movzbl -0x1(%rdx),%esi
  cmp    %sil,0x10(%rsp,%rax,1)
  jne    L8
  add    $0x1,%rax
  sub    $0x1,%rdx
  cmp    %eax,%ecx
  jg     L18
L6:
  lea    0x10(%rsp),%rsi
  lea    D0(%rip),%rdi
  call   <printf@plt>
  mov    0xf4258(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  mov    $0x0,%eax
  add    $0xf4268,%rsp
  pop    %rbx
  pop    %rbp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D20
type_guessed: string
name: retrieve_data
arguments: tag: D5
type_guessed: string
name: retrieve_data
arguments: tag: D0
type_guessed: string

tool: D20: %d

tool: D5: %X

tool: D0: %s

assistant: ```c
int main() {
  int num = 0;
  char str[1000000];
  int index = 0;
  int sec_index = 0;
  int length = 0;
  int ok = 1;
  scanf("%d", &num);
  for (;;) {
    ok = 1;
    sprintf(str, "%X", num);
    length = strlen(str);
    for (index = 0, sec_index = length - 1; index < length / 2; index++, sec_index--) {
      if (str[index] != str[sec_index]) {
        ok = 0;
        break;
      }
    }
    if (ok == 1) {
      break;
    } else {
      num++;
    }
  }
  printf("%s", str);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  lea    -0xf4000(%rsp),%r11
L38:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L38
  sub    $0x268,%rsp
  lea    D18(%rip),%rdi
  lea    D26(%rip),%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0xf4258(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rsi
  movl   $0x0,0xc(%rsp)
  lea    0x10(%rsp),%rbp
  call   <scanf@plt>
  mov    0xc(%rsp),%edx
  nopw   0x0(%rax,%rax,1)
L51:
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <sprintf@plt>
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %eax,%ecx
  shr    $0x1f,%ecx
  add    %eax,%ecx
  sar    %ecx
  cmp    $0x1,%eax
  jle    L62
  cltq
  xor    %edx,%edx
  add    %rbp,%rax
  jmp    L43
  nopw   0x0(%rax,%rax,1)
L22:
  add    $0x1,%rdx
  sub    $0x1,%rax
  cmp    %edx,%ecx
  jle    L62
L43:
  movzbl -0x1(%rax),%esi
  cmp    %sil,0x0(%rbp,%rdx,1)
  je     L22
  mov    0xc(%rsp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,0xc(%rsp)
  jmp    L51
  nopl   0x0(%rax)
L62:
  mov    %rbp,%rsi
  lea    D24(%rip),%rdi
  call   <printf@plt>
  mov    0xf4258(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L41
  add    $0xf4268,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L41:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D18
datatype: string
name: unpack_data
arguments: label: D26
datatype: string
name: unpack_data
arguments: label: D24
datatype: string

tool: D18: %d

tool: D26: %X

tool: D24: %s

assistant: ```c
int main() {
  int num = 0;
  char str[1000000];
  int index = 0;
  int sec_index = 0;
  int length = 0;
  int ok = 1;
  scanf("%d", &num);
  for (;;) {
    ok = 1;
    sprintf(str, "%X", num);
    length = strlen(str);
    for (index = 0, sec_index = length - 1; index < length / 2; index++, sec_index--) {
      if (str[index] != str[sec_index]) {
        ok = 0;
        break;
      }
    }
    if (ok == 1) {
      break;
    } else {
      num++;
    }
  }
  printf("%s", str);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  lea    -0xf4000(%rsp),%r11
L49:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L49
  sub    $0x268,%rsp
  lea    D10(%rip),%rdi
  lea    D13(%rip),%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0xf4258(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rsi
  movl   $0x0,0xc(%rsp)
  lea    0x10(%rsp),%rbp
  call   <scanf@plt>
  mov    0xc(%rsp),%edx
  nopw   0x0(%rax,%rax,1)
L64:
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <sprintf@plt>
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %eax,%ecx
  shr    $0x1f,%ecx
  add    %eax,%ecx
  sar    %ecx
  cmp    $0x1,%eax
  jle    L61
  cltq
  xor    %edx,%edx
  add    %rbp,%rax
  jmp    L50
  nopw   0x0(%rax,%rax,1)
L1:
  add    $0x1,%rdx
  sub    $0x1,%rax
  cmp    %edx,%ecx
  jle    L61
L50:
  movzbl -0x1(%rax),%esi
  cmp    %sil,0x0(%rbp,%rdx,1)
  je     L1
  mov    0xc(%rsp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,0xc(%rsp)
  jmp    L64
  nopl   0x0(%rax)
L61:
  mov    %rbp,%rsi
  lea    D0(%rip),%rdi
  call   <printf@plt>
  mov    0xf4258(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0xf4268,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D10
type_infered: string
name: disassemble_data
arguments: data_label: D13
type_infered: string
name: disassemble_data
arguments: data_label: D0
type_infered: string

tool: D10: %d

tool: D13: %X

tool: D0: %s

assistant: ```c
int main() {
  int num = 0;
  char str[1000000];
  int index = 0;
  int sec_index = 0;
  int length = 0;
  int ok = 1;
  scanf("%d", &num);
  for (;;) {
    ok = 1;
    sprintf(str, "%X", num);
    length = strlen(str);
    for (index = 0, sec_index = length - 1; index < length / 2; index++, sec_index--) {
      if (str[index] != str[sec_index]) {
        ok = 0;
        break;
      }
    }
    if (ok == 1) {
      break;
    } else {
      num++;
    }
  }
  printf("%s", str);
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<ReplaceFirstSubstring>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    %rcx,-0x40(%rbp)
  cmpq   $0x0,-0x28(%rbp)
  je     L51
  cmpq   $0x0,-0x30(%rbp)
  je     L51
  cmpq   $0x0,-0x38(%rbp)
  je     L51
  cmpq   $0x0,-0x40(%rbp)
  jne    L15
L51:
  mov    D2(%rip),%rax
  mov    (%rax),%edx
  mov    D17(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xfffffffe,%eax
  jmp    L55
L15:
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x14(%rbp)
  mov    -0x30(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x10(%rbp)
  mov    -0x38(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0xc(%rbp)
  movl   $0x0,-0x8(%rbp)
  mov    -0x8(%rbp),%edx
  mov    -0x30(%rbp),%rcx
  mov    -0x28(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <FindSubstring@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  js     L58
  mov    -0x4(%rbp),%edx
  mov    -0x28(%rbp),%rcx
  mov    -0x40(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strncpy@plt>
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x40(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0xc(%rbp),%edx
  mov    -0x38(%rbp),%rax
  mov    %rax,%rsi
  mov    %rcx,%rdi
  call   <strncpy@plt>
  mov    -0x14(%rbp),%eax
  sub    -0x4(%rbp),%eax
  sub    -0x10(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    -0x4(%rbp),%ecx
  mov    -0x10(%rbp),%eax
  add    %ecx,%eax
  movslq %eax,%rcx
  mov    -0x28(%rbp),%rax
  add    %rax,%rcx
  mov    -0x4(%rbp),%esi
  mov    -0xc(%rbp),%eax
  add    %esi,%eax
  movslq %eax,%rsi
  mov    -0x40(%rbp),%rax
  add    %rsi,%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strncpy@plt>
  mov    $0x1,%eax
  jmp    L55
L58:
  mov    -0x14(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    -0x28(%rbp),%rcx
  mov    -0x40(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strncpy@plt>
  mov    $0x0,%eax
L55:
  leave
  ret
```

assistant: ```c
int ReplaceFirstSubstring(const char *source, const char *target, const char *replac, char *OutResult) {
  if (!source || !target || !replac || !OutResult) {
    errno = EINVAL;
    return -2;
  }
  int sourceLen = strlen(source);
  int targetLen = strlen(target);
  int replacLen = strlen(replac);
  int start = 0;
  int i = FindSubstring(source, target, start);
  if (i >= 0) {
    strncpy(OutResult, source, i);
    strncpy(&OutResult[i], replac, replacLen);
    strncpy(&OutResult[i + replacLen], &source[i + targetLen], sourceLen - i - targetLen + 1);
    return 1;
  }
  strncpy(OutResult, source, sourceLen + 1);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ReplaceFirstSubstring>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  test   %rdi,%rdi
  je     L18
  mov    %rdi,%rbp
  mov    %rsi,%r13
  mov    %rdx,%r15
  mov    %rcx,%r14
  test   %rsi,%rsi
  je     L18
  test   %rdx,%rdx
  je     L18
  test   %rcx,%rcx
  je     L18
  call   <strlen@plt>
  mov    %eax,%ebx
  mov    %r13,%rdi
  call   <strlen@plt>
  mov    %eax,0xc(%rsp)
  mov    %r15,%rdi
  call   <strlen@plt>
  mov    %eax,%r12d
  mov    $0x0,%edx
  mov    %r13,%rsi
  mov    %rbp,%rdi
  call   <FindSubstring@plt>
  mov    %eax,%r13d
  test   %eax,%eax
  jns    L22
  lea    0x1(%rbx),%edx
  mov    %rbp,%rsi
  mov    %r14,%rdi
  call   <strncpy@plt>
  mov    $0x0,%eax
L6:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L18:
  mov    D13(%rip),%rax
  mov    (%rax),%edx
  mov    D5(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xfffffffe,%eax
  jmp    L6
L22:
  mov    %eax,%edx
  mov    %rbp,%rsi
  mov    %r14,%rdi
  call   <strncpy@plt>
  movslq %r13d,%rdi
  add    %r14,%rdi
  mov    %r12d,%edx
  mov    %r15,%rsi
  call   <strncpy@plt>
  sub    %r13d,%ebx
  mov    0xc(%rsp),%eax
  sub    %eax,%ebx
  lea    0x1(%rbx),%edx
  add    %r13d,%eax
  movslq %eax,%rsi
  add    %rbp,%rsi
  lea    (%r12,%r13,1),%edi
  movslq %edi,%rdi
  add    %r14,%rdi
  call   <strncpy@plt>
  mov    $0x1,%eax
  jmp    L6
```

assistant: ```c
int ReplaceFirstSubstring(const char *source, const char *target, const char *replac, char *OutResult) {
  if (!source || !target || !replac || !OutResult) {
    errno = EINVAL;
    return -2;
  }
  int sourceLen = strlen(source);
  int targetLen = strlen(target);
  int replacLen = strlen(replac);
  int start = 0;
  int i = FindSubstring(source, target, start);
  if (i >= 0) {
    strncpy(OutResult, source, i);
    strncpy(&OutResult[i], replac, replacLen);
    strncpy(&OutResult[i + replacLen], &source[i + targetLen], sourceLen - i - targetLen + 1);
    return 1;
  }
  strncpy(OutResult, source, sourceLen + 1);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<ReplaceFirstSubstring>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  test   %rdi,%rdi
  je     L20
  mov    %rsi,%r14
  test   %rsi,%rsi
  je     L20
  mov    %rdx,%r15
  test   %rdx,%rdx
  je     L20
  mov    %rcx,%r13
  test   %rcx,%rcx
  je     L20
  mov    %rdi,%rbp
  call   <strlen@plt>
  mov    %r14,%rdi
  mov    %eax,%ebx
  call   <strlen@plt>
  mov    %r15,%rdi
  mov    %eax,0xc(%rsp)
  call   <strlen@plt>
  mov    %r14,%rsi
  xor    %edx,%edx
  mov    %rbp,%rdi
  mov    %eax,%r12d
  call   <FindSubstring@plt>
  mov    %eax,%r14d
  test   %eax,%eax
  jns    L15
  lea    0x1(%rbx),%edx
  mov    %rbp,%rsi
  mov    %r13,%rdi
  call   <strncpy@plt>
  xor    %eax,%eax
L2:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax,%rax,1)
L15:
  mov    %rbp,%rsi
  mov    %r13,%rdi
  mov    %eax,%edx
  sub    %r14d,%ebx
  call   <strncpy@plt>
  movslq %r14d,%rdi
  mov    %r12d,%edx
  mov    %r15,%rsi
  add    %r13,%rdi
  call   <strncpy@plt>
  mov    0xc(%rsp),%eax
  lea    (%r12,%r14,1),%edi
  movslq %edi,%rdi
  sub    %eax,%ebx
  add    %r14d,%eax
  add    %r13,%rdi
  movslq %eax,%rsi
  lea    0x1(%rbx),%edx
  add    %rbp,%rsi
  call   <strncpy@plt>
  add    $0x18,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  xchg   %ax,%ax
L20:
  mov    D4(%rip),%rax
  mov    (%rax),%edx
  mov    D19(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xfffffffe,%eax
  jmp    L2
```

assistant: ```c
int ReplaceFirstSubstring(const char *source, const char *target, const char *replac, char *OutResult) {
  if (!source || !target || !replac || !OutResult) {
    errno = EINVAL;
    return -2;
  }
  int sourceLen = strlen(source);
  int targetLen = strlen(target);
  int replacLen = strlen(replac);
  int start = 0;
  int i = FindSubstring(source, target, start);
  if (i >= 0) {
    strncpy(OutResult, source, i);
    strncpy(&OutResult[i], replac, replacLen);
    strncpy(&OutResult[i + replacLen], &source[i + targetLen], sourceLen - i - targetLen + 1);
    return 1;
  }
  strncpy(OutResult, source, sourceLen + 1);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<ReplaceFirstSubstring>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  test   %rdi,%rdi
  je     L43
  mov    %rsi,%r14
  test   %rsi,%rsi
  je     L43
  mov    %rdx,%r15
  test   %rdx,%rdx
  je     L43
  mov    %rcx,%r13
  test   %rcx,%rcx
  je     L43
  mov    %rdi,%rbp
  call   <strlen@plt>
  mov    %r14,%rdi
  mov    %eax,%ebx
  call   <strlen@plt>
  mov    %r15,%rdi
  mov    %eax,0xc(%rsp)
  call   <strlen@plt>
  mov    %r14,%rsi
  xor    %edx,%edx
  mov    %rbp,%rdi
  mov    %eax,%r12d
  call   <FindSubstring@plt>
  mov    %eax,%r14d
  test   %eax,%eax
  jns    L28
  lea    0x1(%rbx),%edx
  mov    %rbp,%rsi
  mov    %r13,%rdi
  call   <strncpy@plt>
  xor    %eax,%eax
L3:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax,%rax,1)
L28:
  mov    %rbp,%rsi
  mov    %r13,%rdi
  mov    %eax,%edx
  sub    %r14d,%ebx
  call   <strncpy@plt>
  movslq %r14d,%rdi
  mov    %r12d,%edx
  mov    %r15,%rsi
  add    %r13,%rdi
  call   <strncpy@plt>
  mov    0xc(%rsp),%eax
  lea    (%r12,%r14,1),%edi
  movslq %edi,%rdi
  sub    %eax,%ebx
  add    %r14d,%eax
  add    %r13,%rdi
  movslq %eax,%rsi
  lea    0x1(%rbx),%edx
  add    %rbp,%rsi
  call   <strncpy@plt>
  add    $0x18,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  xchg   %ax,%ax
L43:
  mov    D11(%rip),%rax
  mov    (%rax),%edx
  mov    D4(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xfffffffe,%eax
  jmp    L3
```

assistant: ```c
int ReplaceFirstSubstring(const char *source, const char *target, const char *replac, char *OutResult) {
  if (!source || !target || !replac || !OutResult) {
    errno = EINVAL;
    return -2;
  }
  int sourceLen = strlen(source);
  int targetLen = strlen(target);
  int replacLen = strlen(replac);
  int start = 0;
  int i = FindSubstring(source, target, start);
  if (i >= 0) {
    strncpy(OutResult, source, i);
    strncpy(&OutResult[i], replac, replacLen);
    strncpy(&OutResult[i + replacLen], &source[i + targetLen], sourceLen - i - targetLen + 1);
    return 1;
  }
  strncpy(OutResult, source, sourceLen + 1);
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<obterUptime>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x150,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <lerArquivo@plt>
  mov    %eax,%edx
  lea    -0x140(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <strcpy@plt>
  lea    -0x140(%rbp),%rax
  lea    D34(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strtok@plt>
  lea    -0x140(%rbp),%rax
  lea    D9(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strtok@plt>
  lea    -0x140(%rbp),%rax
  mov    %rax,%rdi
  call   <atoi@plt>
  mov    %eax,-0x150(%rbp)
  mov    -0x150(%rbp),%eax
  movslq %eax,%rdx
  imul   $0xffffffff91a2b3c5,%rdx,%rdx
  shr    $0x20,%rdx
  add    %eax,%edx
  sar    $0xb,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x14c(%rbp)
  mov    -0x14c(%rbp),%eax
  imul   $0xfffff1f0,%eax,%edx
  mov    -0x150(%rbp),%eax
  add    %edx,%eax
  movslq %eax,%rdx
  imul   $0xffffffff88888889,%rdx,%rdx
  shr    $0x20,%rdx
  add    %eax,%edx
  sar    $0x5,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x148(%rbp)
  mov    -0x14c(%rbp),%eax
  imul   $0xfffff1f0,%eax,%edx
  mov    -0x150(%rbp),%eax
  lea    (%rdx,%rax,1),%ecx
  mov    -0x148(%rbp),%edx
  mov    %edx,%eax
  shl    $0x4,%edx
  sub    %edx,%eax
  shl    $0x2,%eax
  add    %ecx,%eax
  mov    %eax,-0x144(%rbp)
  mov    -0x144(%rbp),%esi
  mov    -0x148(%rbp),%ecx
  mov    -0x14c(%rbp),%edx
  lea    -0xd0(%rbp),%rax
  mov    %esi,%r8d
  lea    D13(%rip),%rsi
  mov    %rax,%rdi
  call   <sprintf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L4
  call   <__stack_chk_fail@plt>
L4:
  leave
  ret
```

tool_calls: name: read_data
arguments: data_tag: D7
data_type: string
name: read_data
arguments: data_tag: D34
data_type: string
name: read_data
arguments: data_tag: D9
data_type: string
name: read_data
arguments: data_tag: D13
data_type: string

tool: D7: /proc/uptime

tool: D34:  

tool: D9: .

tool: D13: %dh : %dm :%ds

assistant: ```c
char *obterUptime() {
  char uptime[100];
  char tempo[200];
  int tempoSegundos;
  int horas;
  int minutos;
  int segundos;
  strcpy(uptime, lerArquivo("/proc/uptime"));
  strtok(uptime, " ");
  strtok(uptime, ".");
  tempoSegundos = atoi(uptime);
  horas = (tempoSegundos / 3600);
  minutos = (tempoSegundos - (3600 * horas)) / 60;
  segundos = (tempoSegundos - (3600 * horas) - (minutos * 60));
  sprintf(tempo, "%dh : %dm :%ds", horas, minutos, segundos);
  return tempo;
}
```
user: What is the c source code of the assembly code below:

```asm
<obterUptime>:
  endbr64
  push   %rbx
  sub    $0x140,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x138(%rsp)
  xor    %eax,%eax
  lea    D17(%rip),%rdi
  call   <lerArquivo@plt>
  mov    %eax,%esi
  mov    %rsp,%rbx
  mov    %rbx,%rdi
  call   <strcpy@plt>
  lea    D27(%rip),%rsi
  mov    %rbx,%rdi
  call   <strtok@plt>
  lea    D15(%rip),%rsi
  mov    %rbx,%rdi
  call   <strtok@plt>
  mov    %rbx,%rdi
  call   <atoi@plt>
  movslq %eax,%rdx
  imul   $0xffffffff91a2b3c5,%rdx,%rdx
  shr    $0x20,%rdx
  add    %eax,%edx
  sar    $0xb,%edx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  imul   $0xfffff1f0,%edx,%ecx
  add    %ecx,%eax
  movslq %eax,%rcx
  imul   $0xffffffff88888889,%rcx,%rcx
  shr    $0x20,%rcx
  add    %eax,%ecx
  sar    $0x5,%ecx
  mov    %eax,%esi
  sar    $0x1f,%esi
  sub    %esi,%ecx
  mov    %ecx,%edi
  shl    $0x4,%edi
  mov    %ecx,%esi
  sub    %edi,%esi
  lea    0x70(%rsp),%rdi
  lea    (%rax,%rsi,4),%r8d
  lea    D18(%rip),%rsi
  call   <sprintf@plt>
  mov    0x138(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  mov    $0x0,%eax
  add    $0x140,%rsp
  pop    %rbx
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D17
type: string
name: retrieve_data
arguments: label: D27
type: string
name: retrieve_data
arguments: label: D15
type: string
name: retrieve_data
arguments: label: D18
type: string

tool: D17: /proc/uptime

tool: D27:  

tool: D15: .

tool: D18: %dh : %dm :%ds

assistant: ```c
char *obterUptime() {
  char uptime[100];
  char tempo[200];
  int tempoSegundos;
  int horas;
  int minutos;
  int segundos;
  strcpy(uptime, lerArquivo("/proc/uptime"));
  strtok(uptime, " ");
  strtok(uptime, ".");
  tempoSegundos = atoi(uptime);
  horas = (tempoSegundos / 3600);
  minutos = (tempoSegundos - (3600 * horas)) / 60;
  segundos = (tempoSegundos - (3600 * horas) - (minutos * 60));
  sprintf(tempo, "%dh : %dm :%ds", horas, minutos, segundos);
  return tempo;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<obterUptime>:
  endbr64
  push   %rbp
  lea    D24(%rip),%rdi
  sub    $0x140,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x138(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  call   <lerArquivo@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  call   <strcpy@plt>
  lea    D2(%rip),%rsi
  mov    %rbp,%rdi
  call   <strtok@plt>
  lea    D35(%rip),%rsi
  mov    %rbp,%rdi
  call   <strtok@plt>
  mov    %rbp,%rdi
  call   <atoi@plt>
  movslq %eax,%rdx
  mov    %eax,%ecx
  imul   $0xffffffff91a2b3c5,%rdx,%rdx
  sar    $0x1f,%ecx
  shr    $0x20,%rdx
  add    %eax,%edx
  sar    $0xb,%edx
  sub    %ecx,%edx
  imul   $0xfffff1f0,%edx,%ecx
  add    %ecx,%eax
  movslq %eax,%rcx
  mov    %eax,%esi
  imul   $0xffffffff88888889,%rcx,%rcx
  sar    $0x1f,%esi
  shr    $0x20,%rcx
  add    %eax,%ecx
  sar    $0x5,%ecx
  sub    %esi,%ecx
  mov    %ecx,%edi
  mov    %ecx,%esi
  shl    $0x4,%edi
  sub    %edi,%esi
  lea    0x70(%rsp),%rdi
  lea    (%rax,%rsi,4),%r8d
  lea    D30(%rip),%rsi
  call   <sprintf@plt>
  mov    0x138(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x140,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D24
type_infered: string
name: retrieve_data
arguments: label: D2
type_infered: string
name: retrieve_data
arguments: label: D35
type_infered: string
name: retrieve_data
arguments: label: D30
type_infered: string

tool: D24: /proc/uptime

tool: D2:  

tool: D35: .

tool: D30: %dh : %dm :%ds

assistant: ```c
char *obterUptime() {
  char uptime[100];
  char tempo[200];
  int tempoSegundos;
  int horas;
  int minutos;
  int segundos;
  strcpy(uptime, lerArquivo("/proc/uptime"));
  strtok(uptime, " ");
  strtok(uptime, ".");
  tempoSegundos = atoi(uptime);
  horas = (tempoSegundos / 3600);
  minutos = (tempoSegundos - (3600 * horas)) / 60;
  segundos = (tempoSegundos - (3600 * horas) - (minutos * 60));
  sprintf(tempo, "%dh : %dm :%ds", horas, minutos, segundos);
  return tempo;
}
```