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

```asm
<true_or_false>:
  endbr64
  test   %edi,%edi
  lea    D6(%rip),%rax
  lea    D11(%rip),%rdx
  cmove  %rdx,%rax
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D6
type_infered: string
name: disassemble_data
arguments: data_label: D11
type_infered: string

tool: D6: True

tool: D11: False

assistant: ```c
char *true_or_false(int flag) {
  if (flag) {
    return ("True");
  } else {
    return ("False");
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

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

tool_calls: name: parse_data
arguments: tag: D1
type_guessed: string
name: parse_data
arguments: tag: D3
type_guessed: string

tool: D1: False

tool: D3: True

assistant: ```c
char *true_or_false(int flag) {
  if (flag) {
    return ("True");
  } else {
    return ("False");
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<true_or_false>:
  endbr64
  test   %edi,%edi
  lea    D12(%rip),%rdx
  lea    D19(%rip),%rax
  cmove  %rdx,%rax
  ret
```

tool_calls: name: read_data
arguments: data_label: D12
type: string
name: read_data
arguments: data_label: D19
type: string

tool: D12: False

tool: D19: True

assistant: ```c
char *true_or_false(int flag) {
  if (flag) {
    return ("True");
  } else {
    return ("False");
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<display_file>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    -0x100000(%rsp),%r11
L12:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L12
  sub    $0x30,%rsp
  mov    %edi,-0x100024(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x100010(%rbp),%rcx
  mov    -0x100024(%rbp),%eax
  mov    $0x100000,%edx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <read@plt>
  mov    %eax,-0x100014(%rbp)
  movl   $0x0,-0x100018(%rbp)
  jmp    L4
L32:
  mov    -0x100018(%rbp),%eax
  cltq
  movzbl -0x100010(%rbp,%rax,1),%eax
  movsbl %al,%eax
  mov    %eax,%esi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x100018(%rbp)
L4:
  mov    -0x100018(%rbp),%eax
  cmp    -0x100014(%rbp),%eax
  jl     L32
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L11
  call   <__stack_chk_fail@plt>
L11:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: label: D5
datatype: string

tool: D5: %c

assistant: ```c
void display_file(int fd) {
  char buf[(1024 * 1024)];
  int i, n = read(fd, buf, (1024 * 1024));
  for (i = 0; i < n; i++)
    printf("%c", buf[i]);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<display_file>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0x100000(%rsp),%r11
L21:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L21
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x100008(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  mov    $0x100000,%edx
  call   <read@plt>
  test   %eax,%eax
  jle    L29
  mov    %rsp,%rbx
  lea    -0x1(%rax),%eax
  lea    0x1(%rsp,%rax,1),%r12
  lea    D6(%rip),%rbp
L38:
  movsbl (%rbx),%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  add    $0x1,%rbx
  cmp    %r12,%rbx
  jne    L38
L29:
  mov    0x100008(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L19
  add    $0x100010,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L19:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D6
datatype: string

tool: D6: %c

assistant: ```c
void display_file(int fd) {
  char buf[(1024 * 1024)];
  int i, n = read(fd, buf, (1024 * 1024));
  for (i = 0; i < n; i++)
    printf("%c", buf[i]);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<display_file>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0x100000(%rsp),%r11
L32:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L32
  sub    $0x10,%rsp
  mov    $0x100000,%edx
  mov    %fs:0x28,%rax
  mov    %rax,0x100008(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  mov    %rbx,%rsi
  call   <read@plt>
  test   %eax,%eax
  jle    L19
  sub    $0x1,%eax
  lea    D1(%rip),%rbp
  lea    0x1(%rsp,%rax,1),%r12
  nopl   0x0(%rax,%rax,1)
L20:
  movsbl (%rbx),%esi
  mov    %rbp,%rdi
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    %r12,%rbx
  jne    L20
L19:
  mov    0x100008(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L29
  add    $0x100010,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L29:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D1
type_guessed: string

tool: D1: %c

assistant: ```c
void display_file(int fd) {
  char buf[(1024 * 1024)];
  int i, n = read(fd, buf, (1024 * 1024));
  for (i = 0; i < n; i++)
    printf("%c", buf[i]);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<display_file>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0x100000(%rsp),%r11
L2:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L2
  sub    $0x10,%rsp
  mov    $0x100000,%edx
  mov    %fs:0x28,%rax
  mov    %rax,0x100008(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  mov    %rbx,%rsi
  call   <read@plt>
  test   %eax,%eax
  jle    L7
  sub    $0x1,%eax
  lea    D6(%rip),%rbp
  lea    0x1(%rsp,%rax,1),%r12
  nopl   0x0(%rax,%rax,1)
L13:
  movsbl (%rbx),%esi
  mov    %rbp,%rdi
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    %r12,%rbx
  jne    L13
L7:
  mov    0x100008(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L39
  add    $0x100010,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L39:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D6
data_type: string

tool: D6: %c

assistant: ```c
void display_file(int fd) {
  char buf[(1024 * 1024)];
  int i, n = read(fd, buf, (1024 * 1024));
  for (i = 0; i < n; i++)
    printf("%c", buf[i]);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<get_letter_idx>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,%eax
  mov    %al,-0x4(%rbp)
  cmpb   $0x60,-0x4(%rbp)
  jle    L34
  cmpb   $0x7a,-0x4(%rbp)
  jg     L34
  movsbl -0x4(%rbp),%eax
  sub    $0x61,%eax
  jmp    L13
L34:
  cmpb   $0x2f,-0x4(%rbp)
  jle    L47
  cmpb   $0x39,-0x4(%rbp)
  jg     L47
  movsbl -0x4(%rbp),%eax
  sub    $0x30,%eax
  jmp    L13
L47:
  mov    $0x0,%eax
L13:
  pop    %rbp
  ret
```

assistant: ```c
int get_letter_idx(char letter) {
  if (letter >= 'a' && letter <= 'z')
    return letter - 'a';
  if (letter >= '0' && letter <= '9')
    return letter - '0';
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<get_letter_idx>:
  endbr64
  lea    -0x61(%rdi),%eax
  cmp    $0x19,%al
  jbe    L2
  lea    -0x30(%rdi),%edx
  movsbl %dil,%eax
  sub    $0x30,%eax
  cmp    $0xa,%dl
  mov    $0x0,%edx
  cmovae %edx,%eax
  ret
L2:
  movsbl %dil,%edi
  lea    -0x61(%rdi),%eax
  ret
```

assistant: ```c
int get_letter_idx(char letter) {
  if (letter >= 'a' && letter <= 'z')
    return letter - 'a';
  if (letter >= '0' && letter <= '9')
    return letter - '0';
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<get_letter_idx>:
  endbr64
  lea    -0x61(%rdi),%eax
  cmp    $0x19,%al
  jbe    L1
  sub    $0x30,%edi
  xor    %edx,%edx
  movsbl %dil,%eax
  cmp    $0xa,%dil
  cmovae %edx,%eax
  ret
  nopl   0x0(%rax)
L1:
  movsbl %al,%eax
  ret
```

assistant: ```c
int get_letter_idx(char letter) {
  if (letter >= 'a' && letter <= 'z')
    return letter - 'a';
  if (letter >= '0' && letter <= '9')
    return letter - '0';
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<get_letter_idx>:
  endbr64
  lea    -0x61(%rdi),%eax
  cmp    $0x19,%al
  jbe    L1
  sub    $0x30,%edi
  xor    %edx,%edx
  movsbl %dil,%eax
  cmp    $0xa,%dil
  cmovae %edx,%eax
  ret
  nopl   0x0(%rax)
L1:
  movsbl %al,%eax
  ret
```

assistant: ```c
int get_letter_idx(char letter) {
  if (letter >= 'a' && letter <= 'z')
    return letter - 'a';
  if (letter >= '0' && letter <= '9')
    return letter - '0';
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<isSubstr>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  test   %rsi,%rsi
  je     L125
  mov    %rdi,%rbp
  mov    %rsi,%rbx
  test   %rdi,%rdi
  je     L125
  mov    $0x1,%eax
  cmpb   $0x0,(%rdi)
  je     L83
L81:
  mov    %eax,%r12d
  add    $0x1,%rax
  cmpb   $0x0,-0x1(%rbp,%rax,1)
  jne    L81
  cmpb   $0x0,(%rbx)
  je     L78
L103:
  mov    $0x1,%eax
L150:
  mov    %eax,%r13d
  add    $0x1,%rax
  cmpb   $0x0,-0x1(%rbx,%rax,1)
  jne    L150
L179:
  mov    %r13d,%edx
  mov    %r12d,%esi
  lea    D38(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  cmp    %r13d,%r12d
  jl     L110
L12:
  movzbl 0x0(%rbp),%eax
  test   %al,%al
  je     L101
  movzbl (%rbx),%esi
  mov    $0x0,%edx
  jmp    L52
L125:
  lea    D15(%rip),%rdi
  call   <puts@plt>
  jmp    L127
L78:
  mov    $0x0,%r13d
  jmp    L179
L110:
  lea    D15(%rip),%rdi
  call   <puts@plt>
  jmp    L127
L158:
  add    $0x1,%rdx
  movzbl 0x0(%rbp,%rdx,1),%eax
  test   %al,%al
  je     L101
L52:
  mov    %edx,%edi
  cmp    %al,%sil
  jne    L158
  mov    $0x0,%eax
  lea    0x0(%rbp,%rdx,1),%rcx
L113:
  movzbl (%rcx,%rax,1),%r8d
  cmp    %r8b,(%rbx,%rax,1)
  jne    L158
  add    $0x1,%rax
  cmpb   $0x0,(%rbx,%rax,1)
  jne    L113
  mov    %edi,%esi
  lea    D28(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L127:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L101:
  lea    D21(%rip),%rdi
  call   <puts@plt>
  jmp    L127
L83:
  mov    $0x0,%r12d
  cmpb   $0x0,(%rsi)
  jne    L103
  mov    $0x0,%edx
  mov    $0x0,%esi
  lea    D38(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L12
```

tool_calls: name: unpack_data
arguments: tag: D38
datatype: string
name: unpack_data
arguments: tag: D15
datatype: string
name: unpack_data
arguments: tag: D28
datatype: string
name: unpack_data
arguments: tag: D21
datatype: string

tool: D38: %d, %d


tool: D15: Error!

tool: D28: Yes is Substring from %d


tool: D21: Not a substring

assistant: ```c
void isSubstr(char *str, char *target) {
  if ((target == NULL) || (str == NULL)) {
    printf("Error!\n");
    return;
  }
  int strLen = 0, targetLen = 0;
  while (str[strLen] != '\0')
    strLen++;
  while (target[targetLen] != '\0')
    targetLen++;
  printf("%d, %d\n", strLen, targetLen);
  if (targetLen > strLen) {
    printf("Error!\n");
    return;
  }
  int i = 0, j = 0;
  while (str[i] != '\0') {
    j = 0;
    if (str[i] == target[j]) {
      int newi = i;
      while (1) {
        if (target[j] != str[newi])
          break;
        newi++;
        j++;
        if (target[j] == '\0') {
          printf("Yes is Substring from %d\n", i);
          return;
        }
      }
    }
    i++;
  }
  printf("Not a substring\n");
  return;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<isSubstr>:
  endbr64
  test   %rsi,%rsi
  je     L103
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  test   %rdi,%rdi
  je     L7
  xor    %r12d,%r12d
  cmpb   $0x0,(%rdi)
  mov    %rsi,%rbp
  mov    $0x1,%eax
  je     L139
  nopl   0x0(%rax,%rax,1)
L86:
  mov    %eax,%r12d
  add    $0x1,%rax
  cmpb   $0x0,-0x1(%rbx,%rax,1)
  jne    L86
L139:
  cmpb   $0x0,0x0(%rbp)
  je     L83
  mov    $0x1,%eax
  nopl   (%rax)
L100:
  mov    %eax,%r13d
  add    $0x1,%rax
  cmpb   $0x0,-0x1(%rbp,%rax,1)
  jne    L100
  xor    %eax,%eax
  mov    %r13d,%edx
  lea    D15(%rip),%rdi
  mov    %r12d,%esi
  call   <printf@plt>
  cmp    %r12d,%r13d
  jg     L7
L130:
  movzbl (%rbx),%eax
  test   %al,%al
  je     L108
  movzbl 0x0(%rbp),%edi
  xor    %ecx,%ecx
  jmp    L80
  xchg   %ax,%ax
L48:
  add    $0x1,%rcx
  movzbl (%rbx,%rcx,1),%eax
  test   %al,%al
  je     L108
L80:
  mov    %ecx,%r8d
  cmp    %al,%dil
  jne    L48
  mov    %rbp,%rsi
  mov    %rcx,%rax
  mov    %edi,%edx
  sub    %rcx,%rsi
  nop
L77:
  cmp    %dl,(%rbx,%rax,1)
  jne    L48
  add    $0x1,%rax
  movzbl (%rsi,%rax,1),%edx
  test   %dl,%dl
  jne    L77
  add    $0x8,%rsp
  mov    %r8d,%esi
  lea    D44(%rip),%rdi
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    1070 <printf@plt>
  nopl   0x0(%rax)
L7:
  add    $0x8,%rsp
  lea    D45(%rip),%rdi
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    1060 <puts@plt>
  cs nopw 0x0(%rax,%rax,1)
L108:
  add    $0x8,%rsp
  lea    D51(%rip),%rdi
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    1060 <puts@plt>
  cs nopw 0x0(%rax,%rax,1)
L83:
  xor    %edx,%edx
  mov    %r12d,%esi
  lea    D15(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L130
  nopl   0x0(%rax,%rax,1)
L103:
  lea    D45(%rip),%rdi
  jmp    1060 <puts@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D15
datatype: string
name: retrieve_data
arguments: data_tag: D44
datatype: string
name: retrieve_data
arguments: data_tag: D45
datatype: string
name: retrieve_data
arguments: data_tag: D51
datatype: string

tool: D15: %d, %d


tool: D44: Yes is Substring from %d


tool: D45: Error!

tool: D51: Not a substring

assistant: ```c
void isSubstr(char *str, char *target) {
  if ((target == NULL) || (str == NULL)) {
    printf("Error!\n");
    return;
  }
  int strLen = 0, targetLen = 0;
  while (str[strLen] != '\0')
    strLen++;
  while (target[targetLen] != '\0')
    targetLen++;
  printf("%d, %d\n", strLen, targetLen);
  if (targetLen > strLen) {
    printf("Error!\n");
    return;
  }
  int i = 0, j = 0;
  while (str[i] != '\0') {
    j = 0;
    if (str[i] == target[j]) {
      int newi = i;
      while (1) {
        if (target[j] != str[newi])
          break;
        newi++;
        j++;
        if (target[j] == '\0') {
          printf("Yes is Substring from %d\n", i);
          return;
        }
      }
    }
    i++;
  }
  printf("Not a substring\n");
  return;
}
```
user: What is the c source code of the assembly code below?

```asm
<isSubstr>:
  endbr64
  test   %rsi,%rsi
  je     L118
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  test   %rdi,%rdi
  je     L16
  xor    %r12d,%r12d
  cmpb   $0x0,(%rdi)
  mov    %rsi,%rbx
  mov    $0x1,%eax
  je     L66
  nopl   0x0(%rax,%rax,1)
L124:
  mov    %eax,%r12d
  add    $0x1,%rax
  cmpb   $0x0,-0x1(%rbp,%rax,1)
  jne    L124
L66:
  cmpb   $0x0,(%rbx)
  je     L3
  mov    $0x1,%eax
  nopl   0x0(%rax)
L79:
  mov    %eax,%r13d
  add    $0x1,%rax
  cmpb   $0x0,-0x1(%rbx,%rax,1)
  jne    L79
  xor    %eax,%eax
  mov    %r13d,%edx
  lea    D25(%rip),%rdi
  mov    %r12d,%esi
  call   <printf@plt>
  cmp    %r12d,%r13d
  jg     L16
L0:
  movzbl 0x0(%rbp),%eax
  test   %al,%al
  je     L132
  movzbl (%rbx),%esi
  xor    %edx,%edx
  jmp    L22
  xchg   %ax,%ax
L102:
  add    $0x1,%rdx
  movzbl 0x0(%rbp,%rdx,1),%eax
  test   %al,%al
  je     L132
L22:
  mov    %edx,%r8d
  cmp    %al,%sil
  jne    L102
  xor    %eax,%eax
  lea    0x0(%rbp,%rdx,1),%rcx
  nopl   0x0(%rax)
L119:
  movzbl (%rcx,%rax,1),%edi
  cmp    %dil,(%rbx,%rax,1)
  jne    L102
  add    $0x1,%rax
  cmpb   $0x0,(%rbx,%rax,1)
  jne    L119
  add    $0x8,%rsp
  mov    %r8d,%esi
  lea    D12(%rip),%rdi
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    1070 <printf@plt>
  nop
L16:
  add    $0x8,%rsp
  lea    D33(%rip),%rdi
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    1060 <puts@plt>
  cs nopw 0x0(%rax,%rax,1)
L132:
  add    $0x8,%rsp
  lea    D40(%rip),%rdi
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    1060 <puts@plt>
  cs nopw 0x0(%rax,%rax,1)
L3:
  xor    %edx,%edx
  mov    %r12d,%esi
  lea    D25(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L0
  nopl   0x0(%rax,%rax,1)
L118:
  lea    D33(%rip),%rdi
  jmp    1060 <puts@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D25
datatype: string
name: unpack_data
arguments: data_tag: D12
datatype: string
name: unpack_data
arguments: data_tag: D33
datatype: string
name: unpack_data
arguments: data_tag: D40
datatype: string

tool: D25: %d, %d


tool: D12: Yes is Substring from %d


tool: D33: Error!

tool: D40: Not a substring

assistant: ```c
void isSubstr(char *str, char *target) {
  if ((target == NULL) || (str == NULL)) {
    printf("Error!\n");
    return;
  }
  int strLen = 0, targetLen = 0;
  while (str[strLen] != '\0')
    strLen++;
  while (target[targetLen] != '\0')
    targetLen++;
  printf("%d, %d\n", strLen, targetLen);
  if (targetLen > strLen) {
    printf("Error!\n");
    return;
  }
  int i = 0, j = 0;
  while (str[i] != '\0') {
    j = 0;
    if (str[i] == target[j]) {
      int newi = i;
      while (1) {
        if (target[j] != str[newi])
          break;
        newi++;
        j++;
        if (target[j] == '\0') {
          printf("Yes is Substring from %d\n", i);
          return;
        }
      }
    }
    i++;
  }
  printf("Not a substring\n");
  return;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<L3_find_bitrate_index>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  cmpl   $0x1,-0x18(%rbp)
  jne    L74
  movl   $0x0,-0x4(%rbp)
  jmp    L13
L93:
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rcx
  lea    D15(%rip),%rax
  mov    (%rcx,%rax,1),%rax
  cmp    %rax,%rdx
  jne    L63
  mov    -0x4(%rbp),%eax
  jmp    L94
L63:
  addl   $0x1,-0x4(%rbp)
L13:
  cmpl   $0xe,-0x4(%rbp)
  jle    L93
  jmp    L29
L74:
  movl   $0x0,-0x4(%rbp)
  jmp    L81
L25:
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rcx
  lea    D13(%rip),%rax
  mov    (%rcx,%rax,1),%rax
  cmp    %rax,%rdx
  jne    L27
  mov    -0x4(%rbp),%eax
  jmp    L94
L27:
  addl   $0x1,-0x4(%rbp)
L81:
  cmpl   $0xe,-0x4(%rbp)
  jle    L25
L29:
  mov    $0xffffffff,%eax
L94:
  pop    %rbp
  ret
```

assistant: ```c
int L3_find_bitrate_index(int bitr, int mpegtype) {
  static long mpeg1[15] = {0, 32, 40, 48, 56, 64, 80, 96, 112, 128, 160, 192, 224, 256, 320};
  static long mpeg2[15] = {0, 8, 16, 24, 32, 40, 48, 56, 64, 80, 96, 112, 128, 144, 160};
  int i;
  if (mpegtype == 1) {
    for (i = 0; i < 15; i++)
      if (bitr == mpeg1[i])
        return i;
  } else {
    for (i = 0; i < 15; i++)
      if (bitr == mpeg2[i])
        return i;
  }
  return -1;
}
```
user: What is the c source code of the assembly code below?

```asm
<L3_find_bitrate_index>:
  endbr64
  mov    %esi,%edx
  mov    $0x0,%eax
  movslq %edi,%rsi
  lea    D4(%rip),%rcx
  cmp    $0x1,%edx
  je     L45
L36:
  mov    %eax,%edx
  cmp    (%rcx,%rax,8),%rsi
  je     L17
  add    $0x1,%rax
  cmp    $0xf,%rax
  jne    L36
  mov    $0xffffffff,%edx
L17:
  mov    %edx,%eax
  ret
L45:
  mov    %rsi,%rcx
  lea    D1(%rip),%rsi
L1:
  mov    %eax,%edx
  cmp    (%rsi,%rax,8),%rcx
  je     L17
  add    $0x1,%rax
  cmp    $0xf,%rax
  jne    L1
  mov    $0xffffffff,%edx
  jmp    L17
```

tool_calls: name: unpack_data
arguments: label: D4
type: string
name: unpack_data
arguments: label: D1
type: string

tool: 

tool: 

assistant: ```c
int L3_find_bitrate_index(int bitr, int mpegtype) {
  static long mpeg1[15] = {0, 32, 40, 48, 56, 64, 80, 96, 112, 128, 160, 192, 224, 256, 320};
  static long mpeg2[15] = {0, 8, 16, 24, 32, 40, 48, 56, 64, 80, 96, 112, 128, 144, 160};
  int i;
  if (mpegtype == 1) {
    for (i = 0; i < 15; i++)
      if (bitr == mpeg1[i])
        return i;
  } else {
    for (i = 0; i < 15; i++)
      if (bitr == mpeg2[i])
        return i;
  }
  return -1;
}
```
user: What is the c source code of the assembly code below:

```asm
<L3_find_bitrate_index>:
  endbr64
  movslq %edi,%rdi
  xor    %eax,%eax
  lea    D7(%rip),%rcx
  xor    %edx,%edx
  cmp    $0x1,%esi
  jne    L13
  jmp    L59
  nopl   0x0(%rax)
L65:
  mov    (%rcx,%rax,8),%rdx
L13:
  mov    %eax,%r8d
  cmp    %rdi,%rdx
  je     L20
  add    $0x1,%rax
  cmp    $0xf,%rax
  jne    L65
L79:
  mov    $0xffffffff,%r8d
L20:
  mov    %r8d,%eax
  ret
L59:
  lea    D17(%rip),%rcx
  jmp    L11
  nopl   0x0(%rax)
L3:
  mov    (%rcx,%rax,8),%rdx
L11:
  mov    %eax,%r8d
  cmp    %rdx,%rdi
  je     L20
  add    $0x1,%rax
  cmp    $0xf,%rax
  jne    L3
  jmp    L79
```

tool_calls: name: extract_symbolic_data
arguments: tag: D7
type: string
name: extract_symbolic_data
arguments: tag: D17
type: string

tool: 

tool: 

assistant: ```c
int L3_find_bitrate_index(int bitr, int mpegtype) {
  static long mpeg1[15] = {0, 32, 40, 48, 56, 64, 80, 96, 112, 128, 160, 192, 224, 256, 320};
  static long mpeg2[15] = {0, 8, 16, 24, 32, 40, 48, 56, 64, 80, 96, 112, 128, 144, 160};
  int i;
  if (mpegtype == 1) {
    for (i = 0; i < 15; i++)
      if (bitr == mpeg1[i])
        return i;
  } else {
    for (i = 0; i < 15; i++)
      if (bitr == mpeg2[i])
        return i;
  }
  return -1;
}
```
user: What is the c source code of the assembly code below?

```asm
<L3_find_bitrate_index>:
  endbr64
  movslq %edi,%rdi
  cmp    $0x1,%esi
  je     L183
  test   %rdi,%rdi
  je     L283
  cmp    $0x8,%rdi
  je     L116
  cmp    $0x10,%rdi
  je     L193
  cmp    $0x18,%rdi
  je     L19
  cmp    $0x20,%rdi
  je     L113
  cmp    $0x28,%rdi
  je     L15
  cmp    $0x30,%rdi
  je     L236
  cmp    $0x38,%rdi
  je     L138
  cmp    $0x40,%rdi
  je     L276
  cmp    $0x50,%rdi
  je     L200
  cmp    $0x60,%rdi
  je     L157
  cmp    $0x70,%rdi
  je     L8
  cmp    $0x80,%rdi
  je     L281
  cmp    $0x90,%rdi
  je     L96
  cmp    $0xa0,%rdi
  mov    $0xe,%edx
  mov    $0xffffffff,%eax
  cmove  %edx,%eax
  ret
  cs nopw 0x0(%rax,%rax,1)
L183:
  test   %rdi,%rdi
  je     L283
  cmp    $0x20,%rdi
  je     L116
  cmp    $0x28,%rdi
  je     L193
  cmp    $0x30,%rdi
  je     L19
  cmp    $0x38,%rdi
  je     L113
  cmp    $0x40,%rdi
  je     L15
  cmp    $0x50,%rdi
  je     L236
  cmp    $0x60,%rdi
  je     L138
  cmp    $0x70,%rdi
  je     L276
  cmp    $0x80,%rdi
  je     L200
  cmp    $0xa0,%rdi
  je     L157
  cmp    $0xc0,%rdi
  je     L8
  cmp    $0xe0,%rdi
  je     L281
  cmp    $0x100,%rdi
  je     L96
  cmp    $0x140,%rdi
  mov    $0xffffffff,%edx
  mov    $0xe,%eax
  cmovne %edx,%eax
  ret
L283:
  xor    %eax,%eax
  ret
L116:
  mov    $0x1,%eax
  ret
L113:
  mov    $0x4,%eax
  ret
L200:
  mov    $0x9,%eax
  ret
L193:
  mov    $0x2,%eax
  ret
L19:
  mov    $0x3,%eax
  ret
L15:
  mov    $0x5,%eax
  ret
L236:
  mov    $0x6,%eax
  ret
L138:
  mov    $0x7,%eax
  ret
L276:
  mov    $0x8,%eax
  ret
L157:
  mov    $0xa,%eax
  ret
L8:
  mov    $0xb,%eax
  ret
L281:
  mov    $0xc,%eax
  ret
L96:
  mov    $0xd,%eax
  ret
```

assistant: ```c
int L3_find_bitrate_index(int bitr, int mpegtype) {
  static long mpeg1[15] = {0, 32, 40, 48, 56, 64, 80, 96, 112, 128, 160, 192, 224, 256, 320};
  static long mpeg2[15] = {0, 8, 16, 24, 32, 40, 48, 56, 64, 80, 96, 112, 128, 144, 160};
  int i;
  if (mpegtype == 1) {
    for (i = 0; i < 15; i++)
      if (bitr == mpeg1[i])
        return i;
  } else {
    for (i = 0; i < 15; i++)
      if (bitr == mpeg2[i])
        return i;
  }
  return -1;
}
```
user: 这段汇编代码用C语言表示是什么样的:

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

assistant: ```c
main() {
  if (!L'\400' != 0)
    ;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  mov    $0x0,%eax
  ret
```

assistant: ```c
main() {
  if (!L'\400' != 0)
    ;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
main() {
  if (!L'\400' != 0)
    ;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
main() {
  if (!L'\400' != 0)
    ;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<f>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  je     L1
  mov    -0x4(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,%edi
  call   <f>
  imul   -0x4(%rbp),%eax
  jmp    L13
L1:
  mov    $0x1,%eax
L13:
  leave
  ret
```

assistant: ```c
int f(int n) {
  if (n)
    return n * f(n - 1);
  return 1;
}
```