ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请从给定的汇编指令推导出C源代码:

```asm
<aml_array_int_4d_free>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,0x18(%rsp)
  mov    %edx,0x14(%rsp)
  test   %esi,%esi
  jle    L31
  mov    %ecx,%r14d
  mov    %rdi,%r12
  lea    -0x1(%rsi),%eax
  lea    0x8(%rdi,%rax,8),%rax
  mov    %rax,0x8(%rsp)
  lea    -0x1(%rdx),%eax
  lea    0x8(,%rax,8),%r15
  lea    -0x1(%rcx),%eax
  lea    0x8(,%rax,8),%r13
  jmp    L76
L9:
  mov    (%r12),%rax
  mov    (%rax,%rbp,1),%rax
  mov    (%rax,%rbx,1),%rdi
  call   <free@plt>
  add    $0x8,%rbx
  cmp    %rbx,%r13
  jne    L9
L80:
  mov    (%r12),%rax
  mov    (%rax,%rbp,1),%rdi
  call   <free@plt>
  add    $0x8,%rbp
  cmp    %r15,%rbp
  je     L77
L36:
  mov    $0x0,%ebx
  test   %r14d,%r14d
  jg     L9
  jmp    L80
L77:
  mov    (%r12),%rdi
  call   <free@plt>
  add    $0x8,%r12
  cmp    0x8(%rsp),%r12
  je     L31
L76:
  mov    $0x0,%ebp
  cmpl   $0x0,0x14(%rsp)
  jg     L36
  jmp    L77
L31:
  mov    0x18(%rsp),%rdi
  call   <free@plt>
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void aml_array_int_4d_free(int ****arr, int s1, int s2, int s3) {
  int i, j, k;
  for (i = 0; i < s1; i++) {
    for (j = 0; j < s2; j++) {
      for (k = 0; k < s3; k++)
        free(arr[i][j][k]);
      free(arr[i][j]);
    }
    free(arr[i]);
  }
  free(arr);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<aml_array_int_4d_free>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,0x18(%rsp)
  mov    %edx,0x14(%rsp)
  test   %esi,%esi
  jle    L34
  mov    %esi,%eax
  mov    %ecx,%r12d
  mov    %rdi,%r15
  sub    $0x1,%eax
  lea    0x8(%rdi,%rax,8),%rax
  mov    %rax,0x8(%rsp)
  lea    -0x1(%rdx),%eax
  lea    0x8(,%rax,8),%r13
  lea    -0x1(%rcx),%eax
  lea    0x8(,%rax,8),%rbp
  xchg   %ax,%ax
L2:
  mov    0x14(%rsp),%eax
  xor    %ebx,%ebx
  test   %eax,%eax
  jle    L6
  nopw   0x0(%rax,%rax,1)
L59:
  xor    %r14d,%r14d
  test   %r12d,%r12d
  jle    L20
  nopl   0x0(%rax,%rax,1)
L35:
  mov    (%r15),%rdx
  mov    (%rdx,%rbx,1),%rdx
  mov    (%rdx,%r14,1),%rdi
  add    $0x8,%r14
  call   <free@plt>
  cmp    %rbp,%r14
  jne    L35
L20:
  mov    (%r15),%rax
  mov    (%rax,%rbx,1),%rdi
  add    $0x8,%rbx
  call   <free@plt>
  cmp    %rbx,%r13
  jne    L59
L6:
  mov    (%r15),%rdi
  add    $0x8,%r15
  call   <free@plt>
  cmp    %r15,0x8(%rsp)
  jne    L2
L34:
  mov    0x18(%rsp),%rdi
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  jmp    1050 <free@plt>
```

assistant: ```c
void aml_array_int_4d_free(int ****arr, int s1, int s2, int s3) {
  int i, j, k;
  for (i = 0; i < s1; i++) {
    for (j = 0; j < s2; j++) {
      for (k = 0; k < s3; k++)
        free(arr[i][j][k]);
      free(arr[i][j]);
    }
    free(arr[i]);
  }
  free(arr);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<aml_array_int_4d_free>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,0x18(%rsp)
  mov    %edx,0x14(%rsp)
  test   %esi,%esi
  jle    L10
  mov    %esi,%eax
  mov    %ecx,%r12d
  mov    %rdi,%r15
  sub    $0x1,%eax
  lea    0x8(%rdi,%rax,8),%rax
  mov    %rax,0x8(%rsp)
  lea    -0x1(%rdx),%eax
  lea    0x8(,%rax,8),%r13
  lea    -0x1(%rcx),%eax
  lea    0x8(,%rax,8),%rbp
  xchg   %ax,%ax
L58:
  mov    0x14(%rsp),%eax
  xor    %ebx,%ebx
  test   %eax,%eax
  jle    L30
  nopw   0x0(%rax,%rax,1)
L40:
  xor    %r14d,%r14d
  test   %r12d,%r12d
  jle    L22
  nopl   0x0(%rax,%rax,1)
L29:
  mov    (%r15),%rdx
  mov    (%rdx,%rbx,1),%rdx
  mov    (%rdx,%r14,1),%rdi
  add    $0x8,%r14
  call   <free@plt>
  cmp    %rbp,%r14
  jne    L29
L22:
  mov    (%r15),%rax
  mov    (%rax,%rbx,1),%rdi
  add    $0x8,%rbx
  call   <free@plt>
  cmp    %rbx,%r13
  jne    L40
L30:
  mov    (%r15),%rdi
  add    $0x8,%r15
  call   <free@plt>
  cmp    0x8(%rsp),%r15
  jne    L58
L10:
  mov    0x18(%rsp),%rdi
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  jmp    1050 <free@plt>
```

assistant: ```c
void aml_array_int_4d_free(int ****arr, int s1, int s2, int s3) {
  int i, j, k;
  for (i = 0; i < s1; i++) {
    for (j = 0; j < s2; j++) {
      for (k = 0; k < s3; k++)
        free(arr[i][j][k]);
      free(arr[i][j]);
    }
    free(arr[i]);
  }
  free(arr);
}
```
user: Translate the below assembly code into C programming language:

```asm
<arch_decomp_puts>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  jmp    L23
L18:
  cmpb   $0xa,-0x1(%rbp)
  jne    L28
  mov    $0xd,%edi
  call   <putc@plt>
L28:
  movsbl -0x1(%rbp),%eax
  mov    %eax,%edi
  call   <putc@plt>
L23:
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  movzbl (%rax),%eax
  mov    %al,-0x1(%rbp)
  cmpb   $0x0,-0x1(%rbp)
  jne    L18
  nop
  nop
  leave
  ret
```

assistant: ```c
void arch_decomp_puts(const char *ptr) {
  char c;
  while ((c = *ptr++) != '\0') {
    if (c == '\n')
      putc('\r');
    putc(c);
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<arch_decomp_puts>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  lea    0x1(%rdi),%rbp
  movzbl (%rdi),%ebx
  test   %bl,%bl
  jne    L29
L8:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L30:
  movsbl %bl,%edi
  call   <putc@plt>
  add    $0x1,%rbp
  movzbl -0x1(%rbp),%ebx
  test   %bl,%bl
  je     L8
L29:
  cmp    $0xa,%bl
  jne    L30
  mov    $0xd,%edi
  call   <putc@plt>
  jmp    L30
```

assistant: ```c
void arch_decomp_puts(const char *ptr) {
  char c;
  while ((c = *ptr++) != '\0') {
    if (c == '\n')
      putc('\r');
    putc(c);
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<arch_decomp_puts>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  movzbl (%rdi),%ebx
  test   %bl,%bl
  je     L26
  lea    0x1(%rdi),%rbp
  jmp    L4
  nopw   0x0(%rax,%rax,1)
L16:
  movsbl %bl,%edi
  add    $0x1,%rbp
  call   <putc@plt>
  movzbl -0x1(%rbp),%ebx
  test   %bl,%bl
  je     L26
L4:
  cmp    $0xa,%bl
  jne    L16
  mov    $0xd,%edi
  call   <putc@plt>
  jmp    L16
  nopl   (%rax)
L26:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void arch_decomp_puts(const char *ptr) {
  char c;
  while ((c = *ptr++) != '\0') {
    if (c == '\n')
      putc('\r');
    putc(c);
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<arch_decomp_puts>:
  endbr64
  push   %rbx
  lea    0x1(%rdi),%rbx
  movsbl (%rdi),%edi
  test   %dil,%dil
  jne    L6
  jmp    L5
  nopl   0x0(%rax,%rax,1)
L39:
  call   <putc@plt>
  movsbl (%rbx),%edi
  add    $0x1,%rbx
  test   %dil,%dil
  je     L5
L6:
  cmp    $0xa,%dil
  jne    L39
  mov    $0xd,%edi
  call   <putc@plt>
  mov    $0xa,%edi
  jmp    L39
L5:
  pop    %rbx
  ret
```

assistant: ```c
void arch_decomp_puts(const char *ptr) {
  char c;
  while ((c = *ptr++) != '\0') {
    if (c == '\n')
      putc('\r');
    putc(c);
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x60,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    -0x60(%rbp),%rax
  mov    %rax,%rdi
  call   <gets@plt>
  lea    -0x40(%rbp),%rax
  mov    %rax,%rdi
  call   <gets@plt>
  lea    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <gets@plt>
  lea    -0x40(%rbp),%rdx
  lea    -0x60(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jle    L15
  lea    -0x40(%rbp),%rdx
  lea    -0x60(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <swap@plt>
L15:
  lea    -0x20(%rbp),%rdx
  lea    -0x60(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jle    L34
  lea    -0x20(%rbp),%rdx
  lea    -0x60(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <swap@plt>
L34:
  lea    -0x20(%rbp),%rdx
  lea    -0x40(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jle    L28
  lea    -0x20(%rbp),%rdx
  lea    -0x40(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <swap@plt>
L28:
  lea    D11(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    -0x20(%rbp),%rcx
  lea    -0x40(%rbp),%rdx
  lea    -0x60(%rbp),%rax
  mov    %rax,%rsi
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L8
  call   <__stack_chk_fail@plt>
L8:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: tag: D19
datatype: string
name: unpack_data
arguments: tag: D11
datatype: string
name: unpack_data
arguments: tag: D0
datatype: string

tool: D19: input three line:

tool: D11: New order is:

tool: D0: %s
%s
%s


assistant: ```c
int main() {
  void swap(char *, char *);
  char str1[20], str2[20], str3[20];
  printf("input three line:\n");
  gets(str1);
  gets(str2);
  gets(str3);
  if (strcmp(str1, str2) > 0)
    swap(str1, str2);
  if (strcmp(str1, str3) > 0)
    swap(str1, str3);
  if (strcmp(str2, str3) > 0)
    swap(str2, str3);
  printf("New order is:\n");
  printf("%s\n%s\n%s\n", str1, str2, str3);
  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
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x68,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x58(%rsp)
  xor    %eax,%eax
  lea    D5(%rip),%rdi
  call   <puts@plt>
  mov    %rsp,%rbx
  mov    %rbx,%rdi
  call   <gets@plt>
  lea    0x20(%rsp),%rbp
  mov    %rbp,%rdi
  call   <gets@plt>
  lea    0x40(%rsp),%rdi
  call   <gets@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jg     L43
L26:
  lea    0x40(%rsp),%rsi
  mov    %rsp,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jg     L31
L48:
  lea    0x40(%rsp),%rsi
  lea    0x20(%rsp),%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jg     L33
L56:
  lea    D8(%rip),%rdi
  call   <puts@plt>
  lea    0x40(%rsp),%rcx
  lea    0x20(%rsp),%rdx
  mov    %rsp,%rsi
  lea    D20(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x58(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  mov    $0x0,%eax
  add    $0x68,%rsp
  pop    %rbx
  pop    %rbp
  ret
L43:
  mov    %rbp,%rsi
  mov    %rsp,%rdi
  call   <swap@plt>
  jmp    L26
L31:
  lea    0x40(%rsp),%rsi
  mov    %rsp,%rdi
  call   <swap@plt>
  jmp    L48
L33:
  lea    0x40(%rsp),%rsi
  lea    0x20(%rsp),%rdi
  call   <swap@plt>
  jmp    L56
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D5
type_guessed: string
name: disassemble_data
arguments: data_tag: D8
type_guessed: string
name: disassemble_data
arguments: data_tag: D20
type_guessed: string

tool: D5: input three line:

tool: D8: New order is:

tool: D20: %s
%s
%s


assistant: ```c
int main() {
  void swap(char *, char *);
  char str1[20], str2[20], str3[20];
  printf("input three line:\n");
  gets(str1);
  gets(str2);
  gets(str3);
  if (strcmp(str1, str2) > 0)
    swap(str1, str2);
  if (strcmp(str1, str3) > 0)
    swap(str1, str3);
  if (strcmp(str2, str3) > 0)
    swap(str2, str3);
  printf("New order is:\n");
  printf("%s\n%s\n%s\n", str1, str2, str3);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r13
  lea    D10(%rip),%rdi
  push   %r12
  push   %rbp
  sub    $0x60,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x58(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  lea    0x20(%rsp),%r12
  lea    0x40(%rsp),%r13
  call   <puts@plt>
  mov    %rbp,%rdi
  call   <gets@plt>
  mov    %r12,%rdi
  call   <gets@plt>
  mov    %r13,%rdi
  call   <gets@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jle    L0
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <swap@plt>
L0:
  mov    %r13,%rsi
  mov    %rbp,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jle    L18
  mov    %r13,%rsi
  mov    %rbp,%rdi
  call   <swap@plt>
L18:
  mov    %r13,%rsi
  mov    %r12,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jle    L4
  mov    %r13,%rsi
  mov    %r12,%rdi
  call   <swap@plt>
L4:
  lea    D22(%rip),%rdi
  call   <puts@plt>
  xor    %eax,%eax
  mov    %r13,%rcx
  mov    %r12,%rdx
  mov    %rbp,%rsi
  lea    D9(%rip),%rdi
  call   <printf@plt>
  mov    0x58(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x60,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D10
type_guessed: string
name: parse_data
arguments: tag: D22
type_guessed: string
name: parse_data
arguments: tag: D9
type_guessed: string

tool: D10: input three line:

tool: D22: New order is:

tool: D9: %s
%s
%s


assistant: ```c
int main() {
  void swap(char *, char *);
  char str1[20], str2[20], str3[20];
  printf("input three line:\n");
  gets(str1);
  gets(str2);
  gets(str3);
  if (strcmp(str1, str2) > 0)
    swap(str1, str2);
  if (strcmp(str1, str3) > 0)
    swap(str1, str3);
  if (strcmp(str2, str3) > 0)
    swap(str2, str3);
  printf("New order is:\n");
  printf("%s\n%s\n%s\n", str1, str2, str3);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r13
  lea    D2(%rip),%rdi
  push   %r12
  push   %rbp
  sub    $0x60,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x58(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  lea    0x20(%rsp),%r12
  lea    0x40(%rsp),%r13
  call   <puts@plt>
  mov    %rbp,%rdi
  call   <gets@plt>
  mov    %r12,%rdi
  call   <gets@plt>
  mov    %r13,%rdi
  call   <gets@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jle    L9
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <swap@plt>
L9:
  mov    %r13,%rsi
  mov    %rbp,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jle    L11
  mov    %r13,%rsi
  mov    %rbp,%rdi
  call   <swap@plt>
L11:
  mov    %r13,%rsi
  mov    %r12,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jle    L1
  mov    %r13,%rsi
  mov    %r12,%rdi
  call   <swap@plt>
L1:
  lea    D22(%rip),%rdi
  call   <puts@plt>
  xor    %eax,%eax
  mov    %r13,%rcx
  mov    %r12,%rdx
  mov    %rbp,%rsi
  lea    D18(%rip),%rdi
  call   <printf@plt>
  mov    0x58(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L20
  add    $0x60,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L20:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D2
type_guessed: string
name: parse_data
arguments: label: D22
type_guessed: string
name: parse_data
arguments: label: D18
type_guessed: string

tool: D2: input three line:

tool: D22: New order is:

tool: D18: %s
%s
%s


assistant: ```c
int main() {
  void swap(char *, char *);
  char str1[20], str2[20], str3[20];
  printf("input three line:\n");
  gets(str1);
  gets(str2);
  gets(str3);
  if (strcmp(str1, str2) > 0)
    swap(str1, str2);
  if (strcmp(str1, str3) > 0)
    swap(str1, str3);
  if (strcmp(str2, str3) > 0)
    swap(str2, str3);
  printf("New order is:\n");
  printf("%s\n%s\n%s\n", str1, str2, str3);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<parse_time>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    %rcx,-0x20(%rbp)
  lea    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <a2i@plt>
  mov    -0x10(%rbp),%rdx
  mov    %eax,(%rdx)
  mov    -0x8(%rbp),%rax
  mov    $0x3a,%esi
  mov    %rax,%rdi
  call   <strchr@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  test   %rax,%rax
  jne    L17
  mov    -0x20(%rbp),%rax
  movl   $0x0,(%rax)
  mov    -0x20(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x18(%rbp),%rax
  mov    %edx,(%rax)
  jmp    L33
L17:
  mov    -0x8(%rbp),%rax
  add    $0x1,%rax
  mov    %rax,-0x8(%rbp)
  lea    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <a2i@plt>
  mov    -0x18(%rbp),%rdx
  mov    %eax,(%rdx)
  mov    -0x8(%rbp),%rax
  mov    $0x3a,%esi
  mov    %rax,%rdi
  call   <strchr@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  test   %rax,%rax
  je     L18
  mov    -0x8(%rbp),%rax
  add    $0x1,%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <atoi@plt>
  jmp    L28
L18:
  mov    $0x0,%eax
L28:
  mov    -0x20(%rbp),%rdx
  mov    %eax,(%rdx)
L33:
  nop
  leave
  ret
```

assistant: ```c
void parse_time(char const *str, int *hour, int *min, int *sec) {
  *hour = a2i(&str);
  if ((str = strchr(str, ':')) == NULL)
    *min = *sec = 0;
  else {
    ++str;
    *min = a2i(&str);
    *sec = ((str = strchr(str, ':')) == NULL) ? 0 : atoi(++str);
  }
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<parse_time>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %rdi,0x8(%rsp)
  mov    %rsi,%rbx
  mov    %rdx,%r12
  mov    %rcx,%rbp
  lea    0x8(%rsp),%rdi
  call   <a2i@plt>
  mov    %eax,(%rbx)
  mov    $0x3a,%esi
  mov    0x8(%rsp),%rdi
  call   <strchr@plt>
  test   %rax,%rax
  je     L1
  add    $0x1,%rax
  mov    %rax,0x8(%rsp)
  lea    0x8(%rsp),%rdi
  call   <a2i@plt>
  mov    %eax,(%r12)
  mov    $0x3a,%esi
  mov    0x8(%rsp),%rdi
  call   <strchr@plt>
  mov    $0x0,%edx
  test   %rax,%rax
  je     L12
  lea    0x1(%rax),%rdi
  mov    %rdi,0x8(%rsp)
  call   <atoi@plt>
  mov    %eax,%edx
L12:
  mov    %edx,0x0(%rbp)
L9:
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L1:
  movl   $0x0,0x0(%rbp)
  movl   $0x0,(%r12)
  jmp    L9
```

assistant: ```c
void parse_time(char const *str, int *hour, int *min, int *sec) {
  *hour = a2i(&str);
  if ((str = strchr(str, ':')) == NULL)
    *min = *sec = 0;
  else {
    ++str;
    *min = a2i(&str);
    *sec = ((str = strchr(str, ':')) == NULL) ? 0 : atoi(++str);
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<parse_time>:
  endbr64
  push   %r13
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  mov    %rcx,%rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x18,%rsp
  lea    0x8(%rsp),%r13
  mov    %rdi,0x8(%rsp)
  mov    %r13,%rdi
  call   <a2i@plt>
  mov    0x8(%rsp),%rdi
  mov    $0x3a,%esi
  mov    %eax,(%rbx)
  call   <strchr@plt>
  test   %rax,%rax
  je     L11
  add    $0x1,%rax
  mov    %r13,%rdi
  mov    %rax,0x8(%rsp)
  call   <a2i@plt>
  mov    0x8(%rsp),%rdi
  mov    $0x3a,%esi
  mov    %eax,(%r12)
  call   <strchr@plt>
  xor    %edx,%edx
  test   %rax,%rax
  je     L5
  lea    0x1(%rax),%rdi
  mov    %rdi,0x8(%rsp)
  call   <atoi@plt>
  mov    %eax,%edx
L5:
  mov    %edx,0x0(%rbp)
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax,%rax,1)
L11:
  movl   $0x0,0x0(%rbp)
  movl   $0x0,(%r12)
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
void parse_time(char const *str, int *hour, int *min, int *sec) {
  *hour = a2i(&str);
  if ((str = strchr(str, ':')) == NULL)
    *min = *sec = 0;
  else {
    ++str;
    *min = a2i(&str);
    *sec = ((str = strchr(str, ':')) == NULL) ? 0 : atoi(++str);
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<parse_time>:
  endbr64
  push   %r13
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  mov    %rcx,%rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x18,%rsp
  lea    0x8(%rsp),%r13
  mov    %rdi,0x8(%rsp)
  mov    %r13,%rdi
  call   <a2i@plt>
  mov    0x8(%rsp),%rdi
  mov    $0x3a,%esi
  mov    %eax,(%rbx)
  call   <strchr@plt>
  test   %rax,%rax
  je     L16
  add    $0x1,%rax
  mov    %r13,%rdi
  mov    %rax,0x8(%rsp)
  call   <a2i@plt>
  mov    0x8(%rsp),%rdi
  mov    $0x3a,%esi
  mov    %eax,(%r12)
  call   <strchr@plt>
  xor    %edx,%edx
  test   %rax,%rax
  je     L19
  lea    0x1(%rax),%rdi
  mov    %rdi,0x8(%rsp)
  call   <atoi@plt>
  mov    %eax,%edx
L19:
  mov    %edx,0x0(%rbp)
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax,%rax,1)
L16:
  movl   $0x0,0x0(%rbp)
  movl   $0x0,(%r12)
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
void parse_time(char const *str, int *hour, int *min, int *sec) {
  *hour = a2i(&str);
  if ((str = strchr(str, ':')) == NULL)
    *min = *sec = 0;
  else {
    ++str;
    *min = a2i(&str);
    *sec = ((str = strchr(str, ':')) == NULL) ? 0 : atoi(++str);
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x0,%eax
  call   <zero@plt>
  mov    %eax,-0x8(%rbp)
  addl   $0x1,-0x8(%rbp)
  mov    -0x8(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  cmpl   $0x1,-0x8(%rbp)
  je     L132
  mov    $0x1,%eax
  jmp    L126
L132:
  cmpl   $0x1,-0x4(%rbp)
  je     L137
  mov    $0x1,%eax
  jmp    L126
L137:
  mov    $0x0,%eax
  call   <one@plt>
  mov    %eax,-0x8(%rbp)
  subl   $0x1,-0x8(%rbp)
  mov    -0x8(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x8(%rbp)
  je     L130
  mov    $0x1,%eax
  jmp    L126
L130:
  cmpl   $0x0,-0x4(%rbp)
  je     L94
  mov    $0x1,%eax
  jmp    L126
L94:
  mov    $0x0,%eax
  call   <zero@plt>
  mov    %eax,-0x8(%rbp)
  mov    -0x8(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x8(%rbp)
  mov    %eax,-0x4(%rbp)
  cmpl   $0x1,-0x8(%rbp)
  je     L89
  mov    $0x1,%eax
  jmp    L126
L89:
  cmpl   $0x0,-0x4(%rbp)
  je     L65
  mov    $0x1,%eax
  jmp    L126
L65:
  mov    $0x0,%eax
  call   <one@plt>
  mov    %eax,-0x8(%rbp)
  mov    -0x8(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0x8(%rbp)
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x8(%rbp)
  je     L16
  mov    $0x1,%eax
  jmp    L126
L16:
  cmpl   $0x1,-0x4(%rbp)
  je     L59
  mov    $0x1,%eax
  jmp    L126
L59:
  mov    $0x0,%eax
L126:
  leave
  ret
```

assistant: ```c
int main() {
  int x;
  int y;
  x = zero();
  y = ++x;
  if (x != 1)
    return 1;
  if (y != 1)
    return 1;
  x = one();
  y = --x;
  if (x != 0)
    return 1;
  if (y != 0)
    return 1;
  x = zero();
  y = x++;
  if (x != 1)
    return 1;
  if (y != 0)
    return 1;
  x = one();
  y = x--;
  if (x != 0)
    return 1;
  if (y != 1)
    return 1;
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  push   %rbx
  mov    $0x0,%eax
  call   <zero@plt>
  mov    $0x1,%ebx
  test   %eax,%eax
  je     L44
L42:
  mov    %ebx,%eax
  pop    %rbx
  ret
L44:
  call   <one@plt>
  mov    %eax,%ebx
  cmp    $0x1,%eax
  je     L38
  mov    $0x1,%ebx
  jmp    L42
L38:
  mov    $0x0,%eax
  call   <zero@plt>
  test   %eax,%eax
  jne    L42
  call   <one@plt>
  cmp    $0x1,%eax
  setne  %bl
  movzbl %bl,%ebx
  jmp    L42
```

assistant: ```c
int main() {
  int x;
  int y;
  x = zero();
  y = ++x;
  if (x != 1)
    return 1;
  if (y != 1)
    return 1;
  x = one();
  y = --x;
  if (x != 0)
    return 1;
  if (y != 0)
    return 1;
  x = zero();
  y = x++;
  if (x != 1)
    return 1;
  if (y != 0)
    return 1;
  x = one();
  y = x--;
  if (x != 0)
    return 1;
  if (y != 1)
    return 1;
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <zero@plt>
  test   %eax,%eax
  je     L26
L36:
  mov    $0x1,%eax
L30:
  add    $0x8,%rsp
  ret
L26:
  xor    %eax,%eax
  call   <one@plt>
  sub    $0x1,%eax
  jne    L36
  xor    %eax,%eax
  call   <zero@plt>
  test   %eax,%eax
  jne    L36
  call   <one@plt>
  sub    $0x1,%eax
  setne  %al
  movzbl %al,%eax
  jmp    L30
```

assistant: ```c
int main() {
  int x;
  int y;
  x = zero();
  y = ++x;
  if (x != 1)
    return 1;
  if (y != 1)
    return 1;
  x = one();
  y = --x;
  if (x != 0)
    return 1;
  if (y != 0)
    return 1;
  x = zero();
  y = x++;
  if (x != 1)
    return 1;
  if (y != 0)
    return 1;
  x = one();
  y = x--;
  if (x != 0)
    return 1;
  if (y != 1)
    return 1;
  return 0;
}
```