ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What would the C source look like for the assembly code provided here:

```asm
<convtile>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x68,%rsp
  mov    %rsi,0x48(%rsp)
  movslq %edx,%rsi
  mov    %esi,%eax
  mov    %rdi,0x40(%rsp)
  cltd
  mov    %ecx,0xc(%rsp)
  idiv   %ecx
  mov    %r8d,0x14(%rsp)
  mov    %eax,0x2c(%rsp)
  test   %eax,%eax
  jle    L61
  lea    0x0(,%rcx,8),%edx
  movslq %ecx,%r11
  movl   $0x0,0x58(%rsp)
  lea    0x0(,%rsi,8),%r13
  mov    %edx,0x10(%rsp)
  mov    %ecx,%edx
  lea    0x0(,%r11,8),%r15
  imul   %ecx,%edx
  movl   $0x0,0x54(%rsp)
  mov    %r15,%r12
  mov    %r13,%r15
  imul   %esi,%ecx
  movl   $0x0,0x50(%rsp)
  imul   %edx,%eax
  mov    %edx,0x5c(%rsp)
  mov    %ecx,0x34(%rsp)
  mov    %eax,0x30(%rsp)
  nopl   0x0(%rax)
L46:
  movslq 0x54(%rsp),%rax
  movl   $0x0,0x28(%rsp)
  mov    %r12,%r13
  movl   $0x0,0x20(%rsp)
  mov    %rax,0x38(%rsp)
  mov    0x58(%rsp),%eax
  mov    %eax,0x24(%rsp)
  nopl   (%rax)
L74:
  mov    0xc(%rsp),%eax
  test   %eax,%eax
  jle    L53
  movslq 0x24(%rsp),%rbp
  mov    0x48(%rsp),%rax
  xor    %r12d,%r12d
  mov    0x40(%rsp),%rcx
  movslq 0x28(%rsp),%rbx
  add    0x38(%rsp),%rbx
  shl    $0x3,%rbp
  shl    $0x3,%rbx
  lea    (%rcx,%rbp,1),%r14
  add    %rax,%rbp
  lea    (%rax,%rbx,1),%r9
  add    %rcx,%rbx
  mov    %r14,%r8
  mov    %r13,%r14
  mov    %r15,%r13
  mov    %rbp,%r15
  mov    %rbx,%rbp
  mov    %r9,%rbx
  jmp    L7
  nopw   0x0(%rax,%rax,1)
L26:
  mov    0x10(%rsp),%edx
  mov    %r8,%rsi
  mov    %rbx,%rdi
  mov    %r8,0x18(%rsp)
  call   <memcpy@plt>
  mov    0x18(%rsp),%r8
L5:
  add    $0x1,%r12d
  add    %r13,%rbx
  add    %r14,%r8
  add    %r14,%r15
  add    %r13,%rbp
  cmp    %r12d,0xc(%rsp)
  je     L72
L7:
  mov    0x14(%rsp),%edx
  test   %edx,%edx
  jne    L26
  mov    0x10(%rsp),%edx
  mov    %rbp,%rsi
  mov    %r15,%rdi
  mov    %r8,0x18(%rsp)
  call   <memcpy@plt>
  mov    0x18(%rsp),%r8
  jmp    L5
  nopl   0x0(%rax,%rax,1)
L72:
  mov    %r13,%r15
  mov    %r14,%r13
L53:
  addl   $0x1,0x20(%rsp)
  mov    0x30(%rsp),%edi
  mov    0x34(%rsp),%edx
  add    %edi,0x24(%rsp)
  add    %edx,0x28(%rsp)
  mov    0x20(%rsp),%eax
  cmp    %eax,0x2c(%rsp)
  jne    L74
  addl   $0x1,0x50(%rsp)
  mov    0xc(%rsp),%ecx
  mov    %r13,%r12
  mov    0x5c(%rsp),%edx
  add    %ecx,0x54(%rsp)
  add    %edx,0x58(%rsp)
  mov    0x50(%rsp),%eax
  cmp    %eax,0x2c(%rsp)
  jne    L46
L61:
  add    $0x68,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void convtile(double *A, double *B, int n, int bs, int tofrom) {
  int i, j, ii, jj, i2, j2, nb = n / bs;
  for (i = 0; i < nb; i++) {
    for (j = 0; j < nb; j++) {
      for (jj = 0; jj < bs; jj++) {
        if (tofrom)
          memcpy(B + i * bs + (j * bs + jj) * n, A + (i + j * nb) * bs * bs + jj * bs, bs * sizeof(double));
        else
          memcpy(B + (i + j * nb) * bs * bs + jj * bs, A + i * bs + (j * bs + jj) * n, bs * sizeof(double));
      }
    }
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<convtile>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x58,%rsp
  mov    %rsi,0x30(%rsp)
  movslq %edx,%rsi
  mov    %esi,%eax
  mov    %rdi,0x28(%rsp)
  cltd
  mov    %ecx,0x8(%rsp)
  idiv   %ecx
  mov    %r8d,0x3c(%rsp)
  mov    %eax,0x20(%rsp)
  test   %eax,%eax
  jle    L22
  mov    %ecx,%edi
  mov    %eax,%ecx
  movl   $0x0,0x44(%rsp)
  lea    0x0(,%rsi,8),%rbx
  movslq %edi,%rax
  mov    %edi,%edx
  lea    0x0(,%rdi,8),%ebp
  movl   $0x0,0x48(%rsp)
  imul   %edi,%edi
  movl   $0x0,0x40(%rsp)
  lea    0x0(,%rax,8),%r15
  imul   %esi,%edx
  imul   %edi,%ecx
  mov    %edi,0x4c(%rsp)
  mov    %edx,0x24(%rsp)
  mov    %ecx,0x38(%rsp)
  cs nopw 0x0(%rax,%rax,1)
L25:
  movslq 0x48(%rsp),%rax
  movl   $0x0,0x10(%rsp)
  movl   $0x0,0xc(%rsp)
  mov    %rax,0x18(%rsp)
  mov    0x44(%rsp),%eax
  mov    %eax,0x14(%rsp)
  nopw   0x0(%rax,%rax,1)
L0:
  mov    0x8(%rsp),%eax
  test   %eax,%eax
  jle    L37
  mov    0x3c(%rsp),%ecx
  movslq 0x14(%rsp),%rdx
  test   %ecx,%ecx
  jne    L6
  mov    0x30(%rsp),%rax
  mov    0x28(%rsp),%rcx
  xor    %r12d,%r12d
  lea    (%rax,%rdx,8),%r14
  movslq 0x10(%rsp),%rax
  add    0x18(%rsp),%rax
  lea    (%rcx,%rax,8),%r13
L43:
  mov    %r13,%rsi
  mov    %r14,%rdi
  mov    %ebp,%edx
  add    $0x1,%r12d
  call   <memcpy@plt>
  add    %r15,%r14
  add    %rbx,%r13
  cmp    %r12d,0x8(%rsp)
  jne    L43
L37:
  addl   $0x1,0xc(%rsp)
  mov    0x38(%rsp),%edi
  mov    0x24(%rsp),%esi
  add    %edi,0x14(%rsp)
  add    %esi,0x10(%rsp)
  mov    0xc(%rsp),%eax
  cmp    %eax,0x20(%rsp)
  jne    L0
  addl   $0x1,0x40(%rsp)
  mov    0x8(%rsp),%edi
  mov    0x4c(%rsp),%ecx
  add    %edi,0x48(%rsp)
  add    %ecx,0x44(%rsp)
  mov    0x40(%rsp),%eax
  cmp    %eax,0x20(%rsp)
  jne    L25
L22:
  add    $0x58,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nop
L6:
  mov    0x30(%rsp),%rcx
  movslq 0x10(%rsp),%rax
  xor    %r14d,%r14d
  add    0x18(%rsp),%rax
  lea    (%rcx,%rax,8),%r12
  mov    0x28(%rsp),%rax
  lea    (%rax,%rdx,8),%r13
  nop
L78:
  mov    %r13,%rsi
  mov    %r12,%rdi
  mov    %ebp,%edx
  add    $0x1,%r14d
  call   <memcpy@plt>
  add    %rbx,%r12
  add    %r15,%r13
  cmp    %r14d,0x8(%rsp)
  jne    L78
  jmp    L37
```

assistant: ```c
void convtile(double *A, double *B, int n, int bs, int tofrom) {
  int i, j, ii, jj, i2, j2, nb = n / bs;
  for (i = 0; i < nb; i++) {
    for (j = 0; j < nb; j++) {
      for (jj = 0; jj < bs; jj++) {
        if (tofrom)
          memcpy(B + i * bs + (j * bs + jj) * n, A + (i + j * nb) * bs * bs + jj * bs, bs * sizeof(double));
        else
          memcpy(B + (i + j * nb) * bs * bs + jj * bs, A + i * bs + (j * bs + jj) * n, bs * sizeof(double));
      }
    }
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<long_float_lt_xxxx>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jns    L11
  mov    -0x8(%rbp),%rdx
  movabs $0x8000000000000000,%rax
  sub    %rdx,%rax
  mov    %rax,-0x8(%rbp)
L11:
  cmpq   $0x0,-0x10(%rbp)
  jns    L2
  mov    -0x10(%rbp),%rdx
  movabs $0x8000000000000000,%rax
  sub    %rdx,%rax
  mov    %rax,-0x10(%rbp)
L2:
  mov    -0x8(%rbp),%rax
  cmp    -0x10(%rbp),%rax
  setl   %al
  movzbl %al,%eax
  pop    %rbp
  ret
```

assistant: ```c
int long_float_lt_xxxx(long a, long b) {
  if (a < 0) {
    a = 0x8000000000000000lu - a;
  }
  if (b < 0) {
    b = 0x8000000000000000lu - b;
  }
  return a < b;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<long_float_lt_xxxx>:
  endbr64
  test   %rdi,%rdi
  js     L15
L8:
  test   %rsi,%rsi
  js     L30
L33:
  cmp    %rsi,%rdi
  setl   %al
  movzbl %al,%eax
  ret
L15:
  movabs $0x8000000000000000,%rax
  sub    %rdi,%rax
  mov    %rax,%rdi
  jmp    L8
L30:
  movabs $0x8000000000000000,%rax
  sub    %rsi,%rax
  mov    %rax,%rsi
  jmp    L33
```

assistant: ```c
int long_float_lt_xxxx(long a, long b) {
  if (a < 0) {
    a = 0x8000000000000000lu - a;
  }
  if (b < 0) {
    b = 0x8000000000000000lu - b;
  }
  return a < b;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<long_float_lt_xxxx>:
  endbr64
  test   %rdi,%rdi
  jns    L7
  movabs $0x8000000000000000,%rax
  sub    %rdi,%rax
  mov    %rax,%rdi
L7:
  test   %rsi,%rsi
  jns    L3
  movabs $0x8000000000000000,%rax
  sub    %rsi,%rax
  mov    %rax,%rsi
L3:
  xor    %eax,%eax
  cmp    %rsi,%rdi
  setl   %al
  ret
```

assistant: ```c
int long_float_lt_xxxx(long a, long b) {
  if (a < 0) {
    a = 0x8000000000000000lu - a;
  }
  if (b < 0) {
    b = 0x8000000000000000lu - b;
  }
  return a < b;
}
```
user: What is the c source code of the assembly code below?

```asm
<long_float_lt_xxxx>:
  endbr64
  test   %rdi,%rdi
  jns    L14
  movabs $0x8000000000000000,%rax
  sub    %rdi,%rax
  mov    %rax,%rdi
L14:
  test   %rsi,%rsi
  jns    L3
  movabs $0x8000000000000000,%rax
  sub    %rsi,%rax
  mov    %rax,%rsi
L3:
  xor    %eax,%eax
  cmp    %rsi,%rdi
  setl   %al
  ret
```

assistant: ```c
int long_float_lt_xxxx(long a, long b) {
  if (a < 0) {
    a = 0x8000000000000000lu - a;
  }
  if (b < 0) {
    b = 0x8000000000000000lu - b;
  }
  return a < b;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<wait4children>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  nop
L16:
  mov    D2(%rip),%rax
  mov    (%rax),%edx
  lea    -0xc(%rbp),%rax
  mov    %rax,%rsi
  mov    $0xffffffff,%edi
  call   <waitpid@plt>
  test   %rax,%rax
  jg     L16
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L10
  call   <__stack_chk_fail@plt>
L10:
  leave
  ret
```

assistant: ```c
void wait4children(int signo) {
  int status;
  while (waitpid(-1, &status, WNOHANG) > 0)
    ;
}
```
user: What is the c source code of the assembly code below:

```asm
<wait4children>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    D6(%rip),%rbx
L19:
  lea    0x4(%rsp),%rsi
  mov    (%rbx),%edx
  mov    $0xffffffff,%edi
  call   <waitpid@plt>
  test   %rax,%rax
  jg     L19
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x10,%rsp
  pop    %rbx
  ret
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void wait4children(int signo) {
  int status;
  while (waitpid(-1, &status, WNOHANG) > 0)
    ;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<wait4children>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    D1(%rip),%rbp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rbx
  cs nopw 0x0(%rax,%rax,1)
L0:
  mov    0x0(%rbp),%edx
  mov    %rbx,%rsi
  mov    $0xffffffff,%edi
  call   <waitpid@plt>
  test   %rax,%rax
  jg     L0
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L16
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L16:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void wait4children(int signo) {
  int status;
  while (waitpid(-1, &status, WNOHANG) > 0)
    ;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<wait4children>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    D2(%rip),%rbp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rbx
  cs nopw 0x0(%rax,%rax,1)
L14:
  mov    0x0(%rbp),%edx
  mov    %rbx,%rsi
  mov    $0xffffffff,%edi
  call   <waitpid@plt>
  test   %rax,%rax
  jg     L14
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L4:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void wait4children(int signo) {
  int status;
  while (waitpid(-1, &status, WNOHANG) > 0)
    ;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<p3n>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rsi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D7
type_infered: string

tool: D7: 
***%d


assistant: ```c
int p3n(long n) {
  printf("\n***%d\n", n);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<p3n>:
  endbr64
  sub    $0x8,%rsp
  mov    %rdi,%rsi
  lea    D0(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: data_tag: D0
data_type: string

tool: D0: 
***%d


assistant: ```c
int p3n(long n) {
  printf("\n***%d\n", n);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<p3n>:
  endbr64
  sub    $0x8,%rsp
  mov    %rdi,%rsi
  lea    D2(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

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

tool: D2: 
***%d


assistant: ```c
int p3n(long n) {
  printf("\n***%d\n", n);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<p3n>:
  endbr64
  sub    $0x8,%rsp
  mov    %rdi,%rsi
  lea    D2(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

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

tool: D2: 
***%d


assistant: ```c
int p3n(long n) {
  printf("\n***%d\n", n);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<memory_patient_next_id>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    D27(%rip),%rax
  add    $0x1,%rax
  mov    %rax,D19(%rip)
  mov    D19(%rip),%rax
  pop    %rbp
  ret
```

assistant: ```c
unsigned long long memory_patient_next_id() {
  ;
  static unsigned long long id = 0;
  id++;
  return id;
}
```
user: Translate the below assembly code into C programming language:

```asm
<memory_patient_next_id>:
  endbr64
  mov    D19(%rip),%rax
  add    $0x1,%rax
  mov    %rax,D0(%rip)
  ret
```

assistant: ```c
unsigned long long memory_patient_next_id() {
  ;
  static unsigned long long id = 0;
  id++;
  return id;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<memory_patient_next_id>:
  endbr64
  mov    D18(%rip),%rax
  add    $0x1,%rax
  mov    %rax,D16(%rip)
  ret
```

assistant: ```c
unsigned long long memory_patient_next_id() {
  ;
  static unsigned long long id = 0;
  id++;
  return id;
}
```
user: Translate the below assembly code into C programming language:

```asm
<memory_patient_next_id>:
  endbr64
  mov    D7(%rip),%rax
  add    $0x1,%rax
  mov    %rax,D15(%rip)
  ret
```

assistant: ```c
unsigned long long memory_patient_next_id() {
  ;
  static unsigned long long id = 0;
  id++;
  return id;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <bufoverflow@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main() {
  bufoverflow();
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <bufoverflow@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  bufoverflow();
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <bufoverflow@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  bufoverflow();
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <bufoverflow@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  bufoverflow();
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<getFileModTime>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0xc(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <stat@plt>
  test   %eax,%eax
  jne    L2
  mov    -0xc(%rbp),%eax
  jmp    L16
L2:
  mov    $0x0,%eax
L16:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L23
  call   <__stack_chk_fail@plt>
L23:
  leave
  ret
```

assistant: ```c
time_t getFileModTime(const char *filename) {
  struct stat st;
  if (!stat(filename, &st))
    return st.st_mtime;
  else
    return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<getFileModTime>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <stat@plt>
  test   %eax,%eax
  mov    $0x0,%eax
  cmove  0x4(%rsp),%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L2
  add    $0x18,%rsp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
time_t getFileModTime(const char *filename) {
  struct stat st;
  if (!stat(filename, &st))
    return st.st_mtime;
  else
    return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<getFileModTime>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <stat@plt>
  test   %eax,%eax
  mov    $0x0,%eax
  cmove  0x4(%rsp),%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L0
  add    $0x18,%rsp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
time_t getFileModTime(const char *filename) {
  struct stat st;
  if (!stat(filename, &st))
    return st.st_mtime;
  else
    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
<getFileModTime>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <stat@plt>
  test   %eax,%eax
  mov    $0x0,%eax
  cmove  0x4(%rsp),%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L3
  add    $0x18,%rsp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
time_t getFileModTime(const char *filename) {
  struct stat st;
  if (!stat(filename, &st))
    return st.st_mtime;
  else
    return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<hash_string>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,-0x28(%rbp)
  movl   $0x0,-0x14(%rbp)
  jmp    L4
L14:
  mov    -0x14(%rbp),%eax
  rol    $0x5,%eax
  mov    %eax,%ebx
  mov    -0x28(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x28(%rbp)
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <toupper@plt>
  xor    %ebx,%eax
  mov    %eax,-0x14(%rbp)
L4:
  mov    -0x28(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L14
  mov    -0x14(%rbp),%edx
  mov    %edx,%ecx
  mov    $0xbacf914d,%eax
  imul   %rcx,%rax
  shr    $0x20,%rax
  mov    %rax,%rcx
  mov    %edx,%eax
  sub    %ecx,%eax
  shr    %eax
  add    %ecx,%eax
  shr    $0x5,%eax
  mov    %eax,%ecx
  mov    %ecx,%eax
  shl    $0x3,%eax
  add    %ecx,%eax
  shl    $0x2,%eax
  add    %ecx,%eax
  sub    %eax,%edx
  mov    %edx,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
unsigned int hash_string(const char *key) {
  unsigned int hash = 0;
  while (*key) {
    hash = ((hash << 5) ^ (hash >> 27)) ^ toupper(*key++);
  }
  hash = hash % 37;
  return hash;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<hash_string>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  movzbl (%rdi),%edi
  test   %dil,%dil
  je     L16
  mov    $0x0,%edx
L12:
  mov    %edx,%ebx
  rol    $0x5,%ebx
  add    $0x1,%rbp
  movsbl %dil,%edi
  call   <toupper@plt>
  mov    %ebx,%edx
  xor    %eax,%edx
  movzbl 0x0(%rbp),%edi
  test   %dil,%dil
  jne    L12
L29:
  mov    %edx,%ecx
  mov    $0xbacf914d,%eax
  imul   %rax,%rcx
  shr    $0x20,%rcx
  mov    %edx,%eax
  sub    %ecx,%eax
  shr    %eax
  add    %ecx,%eax
  shr    $0x5,%eax
  lea    (%rax,%rax,8),%ecx
  lea    (%rax,%rcx,4),%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L16:
  mov    $0x0,%edx
  jmp    L29
```

assistant: ```c
unsigned int hash_string(const char *key) {
  unsigned int hash = 0;
  while (*key) {
    hash = ((hash << 5) ^ (hash >> 27)) ^ toupper(*key++);
  }
  hash = hash % 37;
  return hash;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<hash_string>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  movsbl (%rdi),%edi
  test   %dil,%dil
  je     L6
  xor    %edx,%edx
  nopw   0x0(%rax,%rax,1)
L0:
  mov    %edx,%ebx
  add    $0x1,%rbp
  call   <toupper@plt>
  rol    $0x5,%ebx
  movsbl 0x0(%rbp),%edi
  mov    %ebx,%edx
  xor    %eax,%edx
  test   %dil,%dil
  jne    L0
  mov    $0xbacf914d,%eax
  mov    %edx,%ecx
  add    $0x8,%rsp
  imul   %rax,%rcx
  mov    %edx,%eax
  pop    %rbx
  pop    %rbp
  shr    $0x20,%rcx
  sub    %ecx,%eax
  shr    %eax
  add    %ecx,%eax
  shr    $0x5,%eax
  lea    (%rax,%rax,8),%ecx
  lea    (%rax,%rcx,4),%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  ret
  cs nopw 0x0(%rax,%rax,1)
L6:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
unsigned int hash_string(const char *key) {
  unsigned int hash = 0;
  while (*key) {
    hash = ((hash << 5) ^ (hash >> 27)) ^ toupper(*key++);
  }
  hash = hash % 37;
  return hash;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<hash_string>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  movsbl (%rdi),%edi
  test   %dil,%dil
  je     L7
  xor    %edx,%edx
  nopw   0x0(%rax,%rax,1)
L12:
  mov    %edx,%ebx
  add    $0x1,%rbp
  call   <toupper@plt>
  rol    $0x5,%ebx
  movsbl 0x0(%rbp),%edi
  mov    %ebx,%edx
  xor    %eax,%edx
  test   %dil,%dil
  jne    L12
  mov    $0xbacf914d,%eax
  mov    %edx,%ecx
  add    $0x8,%rsp
  imul   %rax,%rcx
  mov    %edx,%eax
  pop    %rbx
  pop    %rbp
  shr    $0x20,%rcx
  sub    %ecx,%eax
  shr    %eax
  add    %ecx,%eax
  shr    $0x5,%eax
  lea    (%rax,%rax,8),%ecx
  lea    (%rax,%rcx,4),%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  ret
  cs nopw 0x0(%rax,%rax,1)
L7:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
unsigned int hash_string(const char *key) {
  unsigned int hash = 0;
  while (*key) {
    hash = ((hash << 5) ^ (hash >> 27)) ^ toupper(*key++);
  }
  hash = hash % 37;
  return hash;
}
```
user: What is the c source code of the assembly code below:

```asm
<setDbgLevel>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  cmpl   $0x2,-0x4(%rbp)
  jne    L14
  mov    -0x10(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  mov    $0x2,%edx
  lea    D34(%rip),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strncmp@plt>
  test   %rax,%rax
  jne    L14
  mov    -0x10(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,%rsi
  lea    D26(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x0,%eax
  jmp    L61
L14:
  cmpl   $0x2,-0x4(%rbp)
  je     L5
  lea    D36(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x0,%eax
  jmp    L61
L5:
  mov    -0x10(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <isaHexNum@plt>
  test   %eax,%eax
  jne    L33
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0xffffffff,%eax
  jmp    L61
L33:
  mov    -0x10(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  mov    $0x10,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <strtoul@plt>
  mov    %rax,%rsi
  lea    D41(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
L61:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: label: D34
datatype: string
name: disassemble_data
arguments: label: D26
datatype: string
name: disassemble_data
arguments: label: D19
datatype: string
name: disassemble_data
arguments: label: D36
datatype: string
name: disassemble_data
arguments: label: D1
datatype: string
name: disassemble_data
arguments: label: D41
datatype: string

tool: D34: -h

tool: D26: Usage:  %-10s 

tool: D19: <hex debug mask>


tool: D36: Parm Error: single 'debugLevel' parameter required

tool: D1: Parm Error: Invalid debugLevel

tool: D41: New debug mask is:0x%lx


assistant: ```c
int setDbgLevel(int argc, char **argv) {
  {
    if ((argc == 2) && (strncmp(argv[1], "-h", 2) == 0)) {
      printf("Usage:  %-10s ", argv[0]);
      printf("%s\n", "<hex debug mask>\n");
      return (0);
    }
  };
  if (argc != 2) {
    printf("Parm Error: single 'debugLevel' parameter required\n");
    return (0);
  }
  if (!(isaHexNum(argv[1]))) {
    printf("Parm Error: Invalid debugLevel\n");
    return (-1);
  }
  printf("New debug mask is:0x%lx\n", strtoul((argv[1]), NULL, 16));
  return (0);
}
```