ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  lea    D67(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0xc(%rsp),%rcx
  lea    0x8(%rsp),%rdx
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  lea    0x10(%rsp),%r8
  lea    D45(%rip),%rdi
  call   <scanf@plt>
  lea    D30(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  lea    0x14(%rsp),%rsi
  lea    D81(%rip),%rdi
  call   <scanf@plt>
  movss  0x14(%rsp),%xmm0
  comiss 0x4(%rsp),%xmm0
  jae    L11
  comiss 0x8(%rsp),%xmm0
  jae    L82
  comiss 0xc(%rsp),%xmm0
  jae    L25
  comiss 0x10(%rsp),%xmm0
  jb     L86
  lea    D84(%rip),%rdi
  call   <printf@plt>
  jmp    L59
L82:
  lea    D35(%rip),%rdi
  call   <printf@plt>
L59:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L37
  xor    %eax,%eax
  add    $0x28,%rsp
  ret
L11:
  lea    D16(%rip),%rdi
  call   <printf@plt>
  jmp    L59
L86:
  lea    D85(%rip),%rdi
  call   <printf@plt>
  jmp    L59
L25:
  lea    D39(%rip),%rdi
  call   <printf@plt>
  jmp    L59
L37:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D67
data_type: string
name: retrieve_data
arguments: data_tag: D45
data_type: string
name: retrieve_data
arguments: data_tag: D30
data_type: string
name: retrieve_data
arguments: data_tag: D81
data_type: string
name: retrieve_data
arguments: data_tag: D84
data_type: string
name: retrieve_data
arguments: data_tag: D35
data_type: string
name: retrieve_data
arguments: data_tag: D16
data_type: string
name: retrieve_data
arguments: data_tag: D85
data_type: string
name: retrieve_data
arguments: data_tag: D39
data_type: string

tool: D67: Enter thresholds for A, B, C, D
in that order, decreasing percentages >

tool: D45: %f%f%f%f

tool: D30: Thank you. Now enter student score (percent) >

tool: D81: %f

tool: D84: Student has an D grade


tool: D35: Student has an B grade


tool: D16: Student has an A grade


tool: D85: Student has failed the course


tool: D39: Student has an C grade


assistant: ```c
int main() {
  float a, b, c, d, e;
  printf("Enter thresholds for A, B, C, D\nin that order, decreasing percentages >");
  scanf("%f%f%f%f", &a, &b, &c, &d);
  printf("Thank you. Now enter student score (percent) >");
  scanf("%f", &e);
  if (e >= a)
    printf("Student has an A grade\n");
  else if (e >= b)
    printf("Student has an B grade\n");
  else if (e >= c)
    printf("Student has an C grade\n");
  else if (e >= d)
    printf("Student has an D grade\n");
  else
    printf("Student has failed the course\n");
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  lea    D26(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0xc(%rsp),%rcx
  lea    0x8(%rsp),%rdx
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  lea    0x10(%rsp),%r8
  lea    D5(%rip),%rdi
  call   <scanf@plt>
  lea    D23(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  lea    0x14(%rsp),%rsi
  lea    D24(%rip),%rdi
  call   <scanf@plt>
  movss  0x14(%rsp),%xmm0
  comiss 0x4(%rsp),%xmm0
  jae    L5
  comiss 0x8(%rsp),%xmm0
  jae    L28
  comiss 0xc(%rsp),%xmm0
  jae    L19
  comiss 0x10(%rsp),%xmm0
  jb     L61
  lea    D66(%rip),%rdi
  call   <printf@plt>
  jmp    L70
L28:
  lea    D54(%rip),%rdi
  call   <printf@plt>
L70:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L77
  xor    %eax,%eax
  add    $0x28,%rsp
  ret
L5:
  lea    D37(%rip),%rdi
  call   <printf@plt>
  jmp    L70
L61:
  lea    D50(%rip),%rdi
  call   <printf@plt>
  jmp    L70
L19:
  lea    D46(%rip),%rdi
  call   <printf@plt>
  jmp    L70
L77:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D26
type_infered: string
name: disassemble_data
arguments: data_label: D5
type_infered: string
name: disassemble_data
arguments: data_label: D23
type_infered: string
name: disassemble_data
arguments: data_label: D24
type_infered: string
name: disassemble_data
arguments: data_label: D66
type_infered: string
name: disassemble_data
arguments: data_label: D54
type_infered: string
name: disassemble_data
arguments: data_label: D37
type_infered: string
name: disassemble_data
arguments: data_label: D50
type_infered: string
name: disassemble_data
arguments: data_label: D46
type_infered: string

tool: D26: Enter thresholds for A, B, C, D
in that order, decreasing percentages >

tool: D5: %f%f%f%f

tool: D23: Thank you. Now enter student score (percent) >

tool: D24: %f

tool: D66: Student has an D grade


tool: D54: Student has an B grade


tool: D37: Student has an A grade


tool: D50: Student has failed the course


tool: D46: Student has an C grade


assistant: ```c
int main() {
  float a, b, c, d, e;
  printf("Enter thresholds for A, B, C, D\nin that order, decreasing percentages >");
  scanf("%f%f%f%f", &a, &b, &c, &d);
  printf("Thank you. Now enter student score (percent) >");
  scanf("%f", &e);
  if (e >= a)
    printf("Student has an A grade\n");
  else if (e >= b)
    printf("Student has an B grade\n");
  else if (e >= c)
    printf("Student has an C grade\n");
  else if (e >= d)
    printf("Student has an D grade\n");
  else
    printf("Student has failed the course\n");
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<str_sub>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    -0x20(%rbp),%eax
  sub    -0x1c(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  mov    %eax,%edi
  call   <new_str@plt>
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0x10(%rbp)
  jmp    L3
L9:
  mov    -0x1c(%rbp),%edx
  mov    -0x10(%rbp),%eax
  add    %edx,%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x10(%rbp),%edx
  movslq %edx,%rcx
  mov    -0x8(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  addl   $0x1,-0x10(%rbp)
L3:
  mov    -0x10(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jl     L9
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *str_sub(const char *str, int l, int r) {
  int len = r - l + 1;
  char *sub = new_str(len);
  for (int i = 0; i < len; i++)
    sub[i] = str[l + i];
  sub[len] = '\0';
  return sub;
}
```
user: Translate the below assembly code into C programming language:

```asm
<str_sub>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r12
  mov    %esi,%r13d
  mov    %edx,%ebx
  sub    %esi,%ebx
  lea    0x1(%rbx),%ebp
  mov    %ebp,%edi
  call   <new_str@plt>
  test   %ebp,%ebp
  jle    L15
  mov    %ebx,%edx
  mov    $0x0,%ecx
  movslq %r13d,%r13
  add    %r13,%r12
L16:
  movzbl (%r12,%rcx,1),%esi
  mov    %sil,(%rax,%rcx,1)
  mov    %rcx,%rsi
  add    $0x1,%rcx
  cmp    %rdx,%rsi
  jne    L16
L15:
  movslq %ebp,%rbp
  movb   $0x0,(%rax,%rbp,1)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *str_sub(const char *str, int l, int r) {
  int len = r - l + 1;
  char *sub = new_str(len);
  for (int i = 0; i < len; i++)
    sub[i] = str[l + i];
  sub[len] = '\0';
  return sub;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<str_sub>:
  endbr64
  push   %r13
  push   %r12
  movslq %esi,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %edx,%ebx
  sub    %r12d,%ebx
  lea    0x1(%rbx),%r13d
  sub    $0x8,%rsp
  mov    %r13d,%edi
  call   <new_str@plt>
  test   %r13d,%r13d
  jle    L19
  mov    %ebx,%edx
  xor    %ecx,%ecx
  add    %r12,%rbp
  nopl   0x0(%rax)
L6:
  movzbl 0x0(%rbp,%rcx,1),%esi
  mov    %sil,(%rax,%rcx,1)
  mov    %rcx,%rsi
  add    $0x1,%rcx
  cmp    %rsi,%rdx
  jne    L6
L19:
  movslq %r13d,%r13
  movb   $0x0,(%rax,%r13,1)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *str_sub(const char *str, int l, int r) {
  int len = r - l + 1;
  char *sub = new_str(len);
  for (int i = 0; i < len; i++)
    sub[i] = str[l + i];
  sub[len] = '\0';
  return sub;
}
```
user: Translate the below assembly code into C programming language:

```asm
<str_sub>:
  endbr64
  push   %r13
  mov    %rdi,%r13
  push   %r12
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %edx,%ebx
  sub    %esi,%ebx
  lea    0x1(%rbx),%r12d
  sub    $0x8,%rsp
  mov    %r12d,%edi
  call   <new_str@plt>
  mov    %rax,%r8
  test   %r12d,%r12d
  jle    L89
  movslq %ebp,%rdx
  lea    0x1(%r13,%rdx,1),%rcx
  sub    %rcx,%rax
  cmp    $0xe,%rax
  jbe    L34
  cmp    $0x6,%ebx
  jbe    L34
  cmp    $0xe,%ebx
  jbe    L11
  mov    %r12d,%ecx
  add    %r13,%rdx
  xor    %eax,%eax
  shr    $0x4,%ecx
  shl    $0x4,%rcx
  nopl   (%rax)
L137:
  movdqu (%rdx,%rax,1),%xmm0
  movups %xmm0,(%r8,%rax,1)
  add    $0x10,%rax
  cmp    %rcx,%rax
  jne    L137
  mov    %r12d,%ecx
  and    $0xfffffff0,%ecx
  mov    %ecx,%eax
  cmp    %ecx,%r12d
  je     L89
  mov    %ebx,%edi
  sub    %ecx,%edi
  lea    0x1(%rdi),%esi
  cmp    $0x6,%edi
  jbe    L1
L8:
  mov    (%rdx,%rcx,1),%rdx
  mov    %rdx,(%r8,%rcx,1)
  mov    %esi,%edx
  and    $0xfffffff8,%edx
  add    %edx,%eax
  cmp    %edx,%esi
  je     L89
L1:
  lea    0x0(%rbp,%rax,1),%edx
  movslq %edx,%rdx
  movzbl 0x0(%r13,%rdx,1),%ecx
  movslq %eax,%rdx
  mov    %cl,(%r8,%rdx,1)
  lea    0x1(%rax),%edx
  cmp    %eax,%ebx
  jle    L89
  lea    0x0(%rbp,%rdx,1),%ecx
  movslq %ecx,%rcx
  movzbl 0x0(%r13,%rcx,1),%esi
  movslq %edx,%rcx
  mov    %sil,(%r8,%rcx,1)
  lea    0x2(%rax),%ecx
  cmp    %edx,%ebx
  jle    L89
  lea    0x0(%rbp,%rcx,1),%edx
  movslq %edx,%rdx
  movzbl 0x0(%r13,%rdx,1),%esi
  movslq %ecx,%rdx
  mov    %sil,(%r8,%rdx,1)
  lea    0x3(%rax),%edx
  cmp    %ecx,%ebx
  jle    L89
  lea    0x0(%rbp,%rdx,1),%ecx
  movslq %ecx,%rcx
  movzbl 0x0(%r13,%rcx,1),%esi
  movslq %edx,%rcx
  mov    %sil,(%r8,%rcx,1)
  lea    0x4(%rax),%ecx
  cmp    %edx,%ebx
  jle    L89
  lea    0x0(%rbp,%rcx,1),%edx
  movslq %edx,%rdx
  movzbl 0x0(%r13,%rdx,1),%esi
  movslq %ecx,%rdx
  mov    %sil,(%r8,%rdx,1)
  lea    0x5(%rax),%edx
  cmp    %ecx,%ebx
  jle    L89
  lea    0x0(%rbp,%rdx,1),%ecx
  add    $0x6,%eax
  movslq %ecx,%rcx
  movzbl 0x0(%r13,%rcx,1),%esi
  movslq %edx,%rcx
  mov    %sil,(%r8,%rcx,1)
  cmp    %edx,%ebx
  jle    L89
  add    %eax,%ebp
  cltq
  movslq %ebp,%rbp
  movzbl 0x0(%r13,%rbp,1),%edx
  mov    %dl,(%r8,%rax,1)
L89:
  movslq %r12d,%r12
  mov    %r8,%rax
  movb   $0x0,(%r8,%r12,1)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L34:
  xor    %eax,%eax
  add    %r13,%rdx
  nopl   (%rax)
L5:
  movzbl (%rdx,%rax,1),%ecx
  mov    %cl,(%r8,%rax,1)
  mov    %rax,%rcx
  add    $0x1,%rax
  cmp    %rbx,%rcx
  jne    L5
  jmp    L89
L11:
  mov    %r12d,%esi
  xor    %ecx,%ecx
  xor    %eax,%eax
  add    %r13,%rdx
  jmp    L8
```

assistant: ```c
char *str_sub(const char *str, int l, int r) {
  int len = r - l + 1;
  char *sub = new_str(len);
  for (int i = 0; i < len; i++)
    sub[i] = str[l + i];
  sub[len] = '\0';
  return sub;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<loggerGetFileName>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    -0x18(%rbp),%rax
  mov    $0x2f,%esi
  mov    %rax,%rdi
  call   <strrchr@plt>
  mov    %rax,-0x10(%rbp)
  cmpq   $0x0,-0x10(%rbp)
  je     L50
  mov    -0x10(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L29
L50:
  mov    $0x0,%eax
  jmp    L6
L29:
  addq   $0x1,-0x10(%rbp)
  cmpl   $0x0,-0x1c(%rbp)
  je     L25
  mov    -0x10(%rbp),%rax
  mov    $0x2e,%esi
  mov    %rax,%rdi
  call   <strrchr@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L25
  mov    -0x8(%rbp),%rax
  sub    -0x10(%rbp),%rax
  mov    %eax,%edx
  mov    -0x10(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <loggerStrNDup@plt>
  jmp    L6
L25:
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <loggerStrDup@plt>
L6:
  leave
  ret
```

assistant: ```c
char *loggerGetFileName(const char *absPath_, int withoutExt_) {
  const char *fileName = strrchr(absPath_, '/');
  if (!fileName || fileName[1] == 0) {
    return NULL;
  }
  ++fileName;
  if (withoutExt_) {
    const char *extpos = strrchr(fileName, '.');
    if (extpos) {
      return loggerStrNDup(fileName, extpos - fileName);
    }
  }
  return loggerStrDup(fileName);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<loggerGetFileName>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %esi,%ebp
  mov    $0x2f,%esi
  call   <strrchr@plt>
  test   %rax,%rax
  je     L34
  cmpb   $0x0,0x1(%rax)
  je     L11
  lea    0x1(%rax),%rbx
  test   %ebp,%ebp
  je     L17
  mov    $0x2e,%esi
  mov    %rbx,%rdi
  call   <strrchr@plt>
  test   %rax,%rax
  jne    L6
L17:
  mov    %rbx,%rdi
  call   <loggerStrDup@plt>
L34:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L6:
  mov    %eax,%esi
  sub    %ebx,%esi
  mov    %rbx,%rdi
  call   <loggerStrNDup@plt>
  jmp    L34
L11:
  mov    $0x0,%eax
  jmp    L34
```

assistant: ```c
char *loggerGetFileName(const char *absPath_, int withoutExt_) {
  const char *fileName = strrchr(absPath_, '/');
  if (!fileName || fileName[1] == 0) {
    return NULL;
  }
  ++fileName;
  if (withoutExt_) {
    const char *extpos = strrchr(fileName, '.');
    if (extpos) {
      return loggerStrNDup(fileName, extpos - fileName);
    }
  }
  return loggerStrDup(fileName);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<loggerGetFileName>:
  endbr64
  push   %rbp
  push   %rbx
  mov    %esi,%ebx
  mov    $0x2f,%esi
  sub    $0x8,%rsp
  call   <strrchr@plt>
  test   %rax,%rax
  je     L4
  cmpb   $0x0,0x1(%rax)
  je     L4
  lea    0x1(%rax),%rbp
  test   %ebx,%ebx
  je     L31
  mov    $0x2e,%esi
  mov    %rbp,%rdi
  call   <strrchr@plt>
  test   %rax,%rax
  jne    L8
L31:
  add    $0x8,%rsp
  mov    %rbp,%rdi
  pop    %rbx
  pop    %rbp
  jmp    1070 <loggerStrDup@plt>
  nopl   0x0(%rax)
L4:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax)
L8:
  add    $0x8,%rsp
  mov    %eax,%esi
  mov    %rbp,%rdi
  sub    %ebp,%esi
  pop    %rbx
  pop    %rbp
  jmp    1090 <loggerStrNDup@plt>
```

assistant: ```c
char *loggerGetFileName(const char *absPath_, int withoutExt_) {
  const char *fileName = strrchr(absPath_, '/');
  if (!fileName || fileName[1] == 0) {
    return NULL;
  }
  ++fileName;
  if (withoutExt_) {
    const char *extpos = strrchr(fileName, '.');
    if (extpos) {
      return loggerStrNDup(fileName, extpos - fileName);
    }
  }
  return loggerStrDup(fileName);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<loggerGetFileName>:
  endbr64
  push   %rbp
  push   %rbx
  mov    %esi,%ebx
  mov    $0x2f,%esi
  sub    $0x8,%rsp
  call   <strrchr@plt>
  test   %rax,%rax
  je     L7
  cmpb   $0x0,0x1(%rax)
  je     L7
  lea    0x1(%rax),%rbp
  test   %ebx,%ebx
  je     L37
  mov    $0x2e,%esi
  mov    %rbp,%rdi
  call   <strrchr@plt>
  test   %rax,%rax
  jne    L11
L37:
  add    $0x8,%rsp
  mov    %rbp,%rdi
  pop    %rbx
  pop    %rbp
  jmp    1070 <loggerStrDup@plt>
  nopl   0x0(%rax)
L7:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax)
L11:
  add    $0x8,%rsp
  mov    %eax,%esi
  mov    %rbp,%rdi
  sub    %ebp,%esi
  pop    %rbx
  pop    %rbp
  jmp    1090 <loggerStrNDup@plt>
```

assistant: ```c
char *loggerGetFileName(const char *absPath_, int withoutExt_) {
  const char *fileName = strrchr(absPath_, '/');
  if (!fileName || fileName[1] == 0) {
    return NULL;
  }
  ++fileName;
  if (withoutExt_) {
    const char *extpos = strrchr(fileName, '.');
    if (extpos) {
      return loggerStrNDup(fileName, extpos - fileName);
    }
  }
  return loggerStrDup(fileName);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<atou64>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    D16(%rip),%rax
  movq   $0x0,(%rax)
  lea    -0x10(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  mov    $0x0,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strtoull@plt>
  mov    -0x20(%rbp),%rdx
  mov    %eax,(%rdx)
  mov    D16(%rip),%rax
  mov    (%rax),%rdx
  mov    D13(%rip),%rax
  mov    (%rax),%rax
  cmp    %rax,%rdx
  jne    L78
  mov    -0x20(%rbp),%rax
  mov    (%rax),%edx
  mov    D25(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,%edx
  je     L31
L78:
  mov    D16(%rip),%rax
  mov    (%rax),%rax
  test   %rax,%rax
  je     L50
  mov    -0x20(%rbp),%rax
  mov    (%rax),%eax
  test   %eax,%eax
  je     L31
L50:
  mov    -0x10(%rbp),%rax
  cmp    %rax,-0x18(%rbp)
  je     L31
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L22
L31:
  mov    $0xffffffff,%eax
  jmp    L6
L22:
  mov    $0x1,%eax
L6:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L36
  call   <__stack_chk_fail@plt>
L36:
  leave
  ret
```

assistant: ```c
int atou64(const char *str, uint64_t *val) {
  char *endptr;
  errno = 0;
  *val = strtoull(str, &endptr, 0);
  if ((errno == ERANGE && *val == ULLONG_MAX) || (errno != 0 && *val == 0) || endptr == str || *endptr != '\0')
    return -1;
  return 1;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<atou64>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    D2(%rip),%r12
  movq   $0x0,(%r12)
  mov    %rsp,%rsi
  mov    $0x0,%edx
  call   <strtoull@plt>
  mov    %eax,0x0(%rbp)
  mov    (%r12),%rcx
  mov    D7(%rip),%rdx
  cmp    (%rdx),%rcx
  je     L62
L19:
  test   %eax,%eax
  jne    L72
  mov    $0xffffffff,%edx
  test   %rcx,%rcx
  jne    L38
L72:
  mov    (%rsp),%rax
  cmp    %rbx,%rax
  je     L46
  cmpb   $0x1,(%rax)
  sbb    %edx,%edx
  and    $0x2,%edx
  sub    $0x1,%edx
L38:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L14
  mov    %edx,%eax
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L62:
  mov    $0xffffffff,%edx
  mov    D3(%rip),%rsi
  cmp    (%rsi),%eax
  jne    L19
  jmp    L38
L46:
  mov    $0xffffffff,%edx
  jmp    L38
L14:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int atou64(const char *str, uint64_t *val) {
  char *endptr;
  errno = 0;
  *val = strtoull(str, &endptr, 0);
  if ((errno == ERANGE && *val == ULLONG_MAX) || (errno != 0 && *val == 0) || endptr == str || *endptr != '\0')
    return -1;
  return 1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<atou64>:
  endbr64
  push   %r12
  xor    %edx,%edx
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x10,%rsp
  mov    D4(%rip),%r12
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  movq   $0x0,(%r12)
  call   <strtoull@plt>
  mov    (%r12),%rcx
  mov    %eax,0x0(%rbp)
  mov    %eax,%edx
  mov    D15(%rip),%rax
  cmp    (%rax),%rcx
  je     L6
L39:
  test   %rcx,%rcx
  je     L42
  mov    $0xffffffff,%eax
  test   %edx,%edx
  je     L71
L42:
  mov    (%rsp),%rax
  cmp    %rbx,%rax
  je     L11
  cmpb   $0x1,(%rax)
  sbb    %eax,%eax
  and    $0x2,%eax
  sub    $0x1,%eax
L71:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L18
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L6:
  mov    D17(%rip),%rsi
  mov    $0xffffffff,%eax
  cmp    (%rsi),%edx
  jne    L39
  jmp    L71
  nopw   0x0(%rax,%rax,1)
L11:
  mov    $0xffffffff,%eax
  jmp    L71
L18:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int atou64(const char *str, uint64_t *val) {
  char *endptr;
  errno = 0;
  *val = strtoull(str, &endptr, 0);
  if ((errno == ERANGE && *val == ULLONG_MAX) || (errno != 0 && *val == 0) || endptr == str || *endptr != '\0')
    return -1;
  return 1;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<atou64>:
  endbr64
  push   %r12
  xor    %edx,%edx
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x10,%rsp
  mov    D20(%rip),%r12
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  movq   $0x0,(%r12)
  call   <strtoull@plt>
  mov    (%r12),%rcx
  mov    %eax,0x0(%rbp)
  mov    %eax,%edx
  mov    D5(%rip),%rax
  cmp    (%rax),%rcx
  je     L55
L42:
  test   %rcx,%rcx
  je     L46
  mov    $0xffffffff,%eax
  test   %edx,%edx
  je     L27
L46:
  mov    (%rsp),%rax
  cmp    %rbx,%rax
  je     L66
  cmpb   $0x1,(%rax)
  sbb    %eax,%eax
  and    $0x2,%eax
  sub    $0x1,%eax
L27:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L79
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L55:
  mov    D22(%rip),%rsi
  mov    $0xffffffff,%eax
  cmp    (%rsi),%edx
  jne    L42
  jmp    L27
  nopw   0x0(%rax,%rax,1)
L66:
  mov    $0xffffffff,%eax
  jmp    L27
L79:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int atou64(const char *str, uint64_t *val) {
  char *endptr;
  errno = 0;
  *val = strtoull(str, &endptr, 0);
  if ((errno == ERANGE && *val == ULLONG_MAX) || (errno != 0 && *val == 0) || endptr == str || *endptr != '\0')
    return -1;
  return 1;
}
```
user: What is the c source code of the assembly code below:

```asm
<strerrno>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jne    L8
  lea    D3(%rip),%rax
  jmp    L17
L8:
  lea    D14(%rip),%rax
L17:
  pop    %rbp
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D3
datatype: string
name: unpack_data
arguments: data_tag: D14
datatype: string

tool: D3: OK

tool: D14: E??

assistant: ```c
const char *strerrno(int err) {
  switch (err) {
  case 0:
    return "OK";
  default:
    return "E??";
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<strerrno>:
  endbr64
  test   %edi,%edi
  lea    D1(%rip),%rax
  lea    D3(%rip),%rdx
  cmovne %rdx,%rax
  ret
```

tool_calls: name: unpack_data
arguments: label: D1
type: string
name: unpack_data
arguments: label: D3
type: string

tool: D1: OK

tool: D3: E??

assistant: ```c
const char *strerrno(int err) {
  switch (err) {
  case 0:
    return "OK";
  default:
    return "E??";
  }
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<strerrno>:
  endbr64
  test   %edi,%edi
  lea    D3(%rip),%rdx
  lea    D12(%rip),%rax
  cmovne %rdx,%rax
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D3
data_type: string
name: disassemble_data
arguments: data_label: D12
data_type: string

tool: D3: E??

tool: D12: OK

assistant: ```c
const char *strerrno(int err) {
  switch (err) {
  case 0:
    return "OK";
  default:
    return "E??";
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<strerrno>:
  endbr64
  test   %edi,%edi
  lea    D1(%rip),%rdx
  lea    D16(%rip),%rax
  cmovne %rdx,%rax
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D1
type_infered: string
name: disassemble_data
arguments: tag: D16
type_infered: string

tool: D1: E??

tool: D16: OK

assistant: ```c
const char *strerrno(int err) {
  switch (err) {
  case 0:
    return "OK";
  default:
    return "E??";
  }
}
```
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   <init_field@plt>
  mov    $0x1,%edi
  call   <generate_mines@plt>
  mov    $0x0,%eax
  call   <print_field@plt>
  mov    $0x0,%eax
  call   <game_loop@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
main(void) {
  init_field();
  generate_mines(1);
  print_field();
  game_loop();
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <init_field@plt>
  mov    $0x1,%edi
  call   <generate_mines@plt>
  mov    $0x0,%eax
  call   <print_field@plt>
  mov    $0x0,%eax
  call   <game_loop@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
main(void) {
  init_field();
  generate_mines(1);
  print_field();
  game_loop();
  return 0;
}
```