ylfeng/ReF-Decompile-dataset huggingface dataset:
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
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x14(%rbp)
  lea    D35(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  jmp    L58
L28:
  mov    -0x18(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x18(%rbp),%edx
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%ecx
  mov    %ecx,%eax
  shl    $0x2,%eax
  add    %ecx,%eax
  add    %eax,%eax
  sub    %eax,%edx
  mov    %edx,-0xc(%rbp)
  jmp    L21
L61:
  mov    -0x10(%rbp),%ecx
  movslq %ecx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %ecx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  sub    %eax,%ecx
  mov    %ecx,%edx
  cmp    %edx,-0xc(%rbp)
  jne    L30
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D16(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x1,-0x14(%rbp)
L30:
  mov    -0x10(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x10(%rbp)
L21:
  cmpl   $0x0,-0x10(%rbp)
  jg     L61
  mov    -0x18(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x18(%rbp)
L58:
  mov    -0x18(%rbp),%eax
  test   %eax,%eax
  jg     L28
  cmpl   $0x0,-0x14(%rbp)
  jne    L24
  lea    D31(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L24:
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L10
  call   <__stack_chk_fail@plt>
L10:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: tag: D35
datatype: string
name: unpack_data
arguments: tag: D18
datatype: string
name: unpack_data
arguments: tag: D16
datatype: string
name: unpack_data
arguments: tag: D31
datatype: string

tool: D35: 
 input number:

tool: D18: %d

tool: D16: 
 %d

tool: D31: 
 no equal digits

assistant: ```c
void main(void) {
  int i = 0, n, j, a;
  printf("\n input number:");
  scanf("%d", &n);
  while (n > 0) {
    a = n / 10;
    j = n % 10;
    while (a > 0) {
      if (j == a % 10) {
        printf("\n %d", j);
        i = 1;
      }
      a = a / 10;
    }
    n = n / 10;
  }
  if (i == 0)
    printf("\n no equal digits");
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D11(%rip),%rdi
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D12(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  test   %eax,%eax
  jle    L33
  mov    $0x0,%ecx
  lea    D3(%rip),%r12
  jmp    L17
L108:
  mov    %ebp,%esi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x1,%ecx
L12:
  movslq %ebx,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  mov    %ebx,%edx
  sar    $0x1f,%edx
  sub    %edx,%eax
  cmp    $0x9,%ebx
  jle    L68
  mov    %eax,%ebx
L52:
  movslq %ebx,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  mov    %ebx,%edx
  sar    $0x1f,%edx
  sub    %edx,%eax
  lea    (%rax,%rax,4),%eax
  add    %eax,%eax
  mov    %ebx,%edx
  sub    %eax,%edx
  cmp    %ebp,%edx
  jne    L12
  jmp    L108
L68:
  mov    0x4(%rsp),%edx
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  mov    %edx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,0x4(%rsp)
  cmp    $0x9,%edx
  jle    L89
L17:
  movslq %eax,%rbx
  imul   $0x66666667,%rbx,%rbx
  sar    $0x22,%rbx
  cltd
  sub    %edx,%ebx
  lea    (%rbx,%rbx,4),%edx
  add    %edx,%edx
  mov    %eax,%ebp
  sub    %edx,%ebp
  cmp    $0x9,%eax
  jg     L52
  jmp    L68
L89:
  test   %ecx,%ecx
  je     L33
L34:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L84
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L33:
  lea    D17(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L34
L84:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D11
data_type: string
name: retrieve_data
arguments: data_tag: D12
data_type: string
name: retrieve_data
arguments: data_tag: D3
data_type: string
name: retrieve_data
arguments: data_tag: D17
data_type: string

tool: D11: 
 input number:

tool: D12: %d

tool: D3: 
 %d

tool: D17: 
 no equal digits

assistant: ```c
void main(void) {
  int i = 0, n, j, a;
  printf("\n input number:");
  scanf("%d", &n);
  while (n > 0) {
    a = n / 10;
    j = n % 10;
    while (a > 0) {
      if (j == a % 10) {
        printf("\n %d", j);
        i = 1;
      }
      a = a / 10;
    }
    n = n / 10;
  }
  if (i == 0)
    printf("\n no equal digits");
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  push   %r13
  lea    D10(%rip),%rdi
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D35(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%edx
  test   %edx,%edx
  jle    L71
  mov    %edx,%r13d
  xor    %ecx,%ecx
  mov    $0xcccccccd,%r12d
  lea    D18(%rip),%rbx
  nopl   0x0(%rax)
L89:
  movslq %r13d,%rax
  mov    %r13d,%esi
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%esi
  sar    $0x22,%rax
  sub    %esi,%eax
  lea    (%rax,%rax,4),%esi
  add    %esi,%esi
  sub    %esi,%r13d
  test   %eax,%eax
  jne    L27
  jmp    L20
  nopl   0x0(%rax)
L54:
  imul   %r12,%rbp
  shr    $0x23,%rbp
  mov    %ebp,%eax
  test   %ebp,%ebp
  je     L44
L27:
  mov    %eax,%ebp
  mov    %rbp,%rdx
  imul   %r12,%rdx
  shr    $0x23,%rdx
  lea    (%rdx,%rdx,4),%edx
  add    %edx,%edx
  sub    %edx,%eax
  cmp    %r13d,%eax
  jne    L54
  imul   %r12,%rbp
  mov    %r13d,%esi
  mov    %rbx,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x1,%ecx
  shr    $0x23,%rbp
  mov    %ebp,%eax
  test   %ebp,%ebp
  jne    L27
L44:
  mov    0x4(%rsp),%edx
L20:
  movslq %edx,%r13
  mov    %r13,%rax
  imul   $0x66666667,%r13,%r13
  cltd
  sar    $0x22,%r13
  sub    %edx,%r13d
  mov    %r13d,0x4(%rsp)
  mov    %r13d,%edx
  cmp    $0x9,%eax
  jg     L89
  test   %ecx,%ecx
  je     L71
L33:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L53
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L71:
  lea    D27(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L33
L53:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D10
data_type: string
name: retrieve_data
arguments: data_tag: D35
data_type: string
name: retrieve_data
arguments: data_tag: D18
data_type: string
name: retrieve_data
arguments: data_tag: D27
data_type: string

tool: D10: 
 input number:

tool: D35: %d

tool: D18: 
 %d

tool: D27: 
 no equal digits

assistant: ```c
void main(void) {
  int i = 0, n, j, a;
  printf("\n input number:");
  scanf("%d", &n);
  while (n > 0) {
    a = n / 10;
    j = n % 10;
    while (a > 0) {
      if (j == a % 10) {
        printf("\n %d", j);
        i = 1;
      }
      a = a / 10;
    }
    n = n / 10;
  }
  if (i == 0)
    printf("\n no equal digits");
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %r13
  lea    D21(%rip),%rdi
  push   %r12
  mov    $0xcccccccd,%r12d
  push   %rbp
  push   %rbx
  lea    D32(%rip),%rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D11(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%edx
  xor    %ecx,%ecx
  test   %edx,%edx
  jle    L37
  cs nopw 0x0(%rax,%rax,1)
L5:
  movslq %edx,%rax
  mov    %edx,%esi
  mov    %edx,%r13d
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%esi
  sar    $0x22,%rax
  sub    %esi,%eax
  lea    (%rax,%rax,4),%esi
  add    %esi,%esi
  sub    %esi,%r13d
  test   %eax,%eax
  jne    L1
  jmp    L38
  nopl   0x0(%rax)
L92:
  imul   %r12,%rbp
  shr    $0x23,%rbp
  mov    %ebp,%eax
  test   %ebp,%ebp
  je     L53
L1:
  mov    %eax,%ebp
  mov    %rbp,%rdx
  imul   %r12,%rdx
  shr    $0x23,%rdx
  lea    (%rdx,%rdx,4),%edx
  add    %edx,%edx
  sub    %edx,%eax
  cmp    %r13d,%eax
  jne    L92
  imul   %r12,%rbp
  mov    %r13d,%esi
  mov    %rbx,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x1,%ecx
  shr    $0x23,%rbp
  mov    %ebp,%eax
  test   %ebp,%ebp
  jne    L1
L53:
  mov    0x4(%rsp),%eax
L72:
  movslq %eax,%rdx
  mov    %eax,%esi
  imul   $0x66666667,%rdx,%rdx
  sar    $0x1f,%esi
  sar    $0x22,%rdx
  sub    %esi,%edx
  mov    %edx,0x4(%rsp)
  cmp    $0x9,%eax
  jg     L5
  test   %ecx,%ecx
  je     L37
L73:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L97
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L38:
  mov    %edx,%eax
  jmp    L72
L37:
  lea    D2(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L73
L97:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D21
type: string
name: parse_data
arguments: data_tag: D32
type: string
name: parse_data
arguments: data_tag: D11
type: string
name: parse_data
arguments: data_tag: D2
type: string

tool: D21: 
 input number:

tool: D32: 
 %d

tool: D11: %d

tool: D2: 
 no equal digits

assistant: ```c
void main(void) {
  int i = 0, n, j, a;
  printf("\n input number:");
  scanf("%d", &n);
  while (n > 0) {
    a = n / 10;
    j = n % 10;
    while (a > 0) {
      if (j == a % 10) {
        printf("\n %d", j);
        i = 1;
      }
      a = a / 10;
    }
    n = n / 10;
  }
  if (i == 0)
    printf("\n no equal digits");
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<cpu_init>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <cortexm_init@plt>
  mov    $0x0,%eax
  call   <clk_init@plt>
  nop
  pop    %rbp
  ret
```

assistant: ```c
void cpu_init(void) {
  cortexm_init();
  clk_init();
}
```
user: Please show me the C implementation that matches the assembly code listed below:

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

assistant: ```c
void cpu_init(void) {
  cortexm_init();
  clk_init();
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<cpu_init>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <cortexm_init@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1070 <clk_init@plt>
```

assistant: ```c
void cpu_init(void) {
  cortexm_init();
  clk_init();
}
```
user: What is the c source code of the assembly code below:

```asm
<cpu_init>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <cortexm_init@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1070 <clk_init@plt>
```

assistant: ```c
void cpu_init(void) {
  cortexm_init();
  clk_init();
}
```
user: What would the C source look like for the assembly code provided here:

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

assistant: ```c
int main() {
  start_tests();
  return 0xABCDDCBA;
}
```
user: What is the c source code of the assembly code below?

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

assistant: ```c
int main() {
  start_tests();
  return 0xABCDDCBA;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

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

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

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

assistant: ```c
int main() {
  start_tests();
  return 0xABCDDCBA;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<PushString>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movq   $0x0,-0x8(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,%edx
  mov    -0x20(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <PushNString@plt>
  leave
  ret
```

assistant: ```c
char *PushString(struct Push *push, const char *append) {
  char *string = NULL;
  return PushNString(push, append, strlen(append));
  return string;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<PushString>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    %rsi,%rbx
  mov    %rsi,%rdi
  call   <strlen@plt>
  mov    %eax,%edx
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <PushNString@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
char *PushString(struct Push *push, const char *append) {
  char *string = NULL;
  return PushNString(push, append, strlen(append));
  return string;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<PushString>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  mov    %rsi,%rdi
  push   %rbp
  mov    %rsi,%rbp
  sub    $0x8,%rsp
  call   <strlen@plt>
  add    $0x8,%rsp
  mov    %rbp,%rsi
  mov    %r12,%rdi
  pop    %rbp
  mov    %eax,%edx
  pop    %r12
  jmp    1070 <PushNString@plt>
```

assistant: ```c
char *PushString(struct Push *push, const char *append) {
  char *string = NULL;
  return PushNString(push, append, strlen(append));
  return string;
}
```
user: What is the c source code of the assembly code below:

```asm
<PushString>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  mov    %rsi,%rdi
  push   %rbp
  mov    %rsi,%rbp
  sub    $0x8,%rsp
  call   <strlen@plt>
  add    $0x8,%rsp
  mov    %rbp,%rsi
  mov    %r12,%rdi
  pop    %rbp
  mov    %eax,%edx
  pop    %r12
  jmp    1070 <PushNString@plt>
```

assistant: ```c
char *PushString(struct Push *push, const char *append) {
  char *string = NULL;
  return PushNString(push, append, strlen(append));
  return string;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<copy_string_array>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    -0x30(%rbp),%rax
  mov    $0x8,%edx
  mov    %rax,%rsi
  mov    $0x0,%edi
  call   <reallocarray@plt>
  mov    %rax,-0x18(%rbp)
  movq   $0x0,-0x20(%rbp)
  jmp    L18
L6:
  mov    -0x20(%rbp),%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    -0x20(%rbp),%rdx
  lea    0x0(,%rdx,8),%rcx
  mov    -0x18(%rbp),%rdx
  lea    (%rcx,%rdx,1),%rbx
  mov    %rax,%rdi
  call   <strdup@plt>
  mov    %rax,(%rbx)
  addq   $0x1,-0x20(%rbp)
L18:
  mov    -0x20(%rbp),%rax
  cmp    -0x30(%rbp),%rax
  jb     L6
  mov    -0x18(%rbp),%rax
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
char **copy_string_array(char **array, size_t len) {
  char **result = reallocarray(NULL, len, sizeof(char *));
  size_t i;
  for (i = 0U; i < len; ++i) {
    result[i] = strdup(array[i]);
  }
  return result;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<copy_string_array>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r13
  mov    %rsi,%rbp
  mov    $0x8,%edx
  mov    $0x0,%edi
  call   <reallocarray@plt>
  mov    %rax,%r12
  test   %rbp,%rbp
  je     L7
  mov    $0x0,%ebx
L15:
  mov    0x0(%r13,%rbx,8),%rdi
  call   <strdup@plt>
  mov    %rax,(%r12,%rbx,8)
  add    $0x1,%rbx
  cmp    %rbx,%rbp
  jne    L15
L7:
  mov    %r12,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char **copy_string_array(char **array, size_t len) {
  char **result = reallocarray(NULL, len, sizeof(char *));
  size_t i;
  for (i = 0U; i < len; ++i) {
    result[i] = strdup(array[i]);
  }
  return result;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<copy_string_array>:
  endbr64
  push   %r13
  mov    $0x8,%edx
  mov    %rdi,%r13
  xor    %edi,%edi
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <reallocarray@plt>
  mov    %rax,%r12
  test   %rbp,%rbp
  je     L3
  xor    %ebx,%ebx
  nopw   0x0(%rax,%rax,1)
L10:
  mov    0x0(%r13,%rbx,8),%rdi
  call   <strdup@plt>
  mov    %rax,(%r12,%rbx,8)
  add    $0x1,%rbx
  cmp    %rbx,%rbp
  jne    L10
L3:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char **copy_string_array(char **array, size_t len) {
  char **result = reallocarray(NULL, len, sizeof(char *));
  size_t i;
  for (i = 0U; i < len; ++i) {
    result[i] = strdup(array[i]);
  }
  return result;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<copy_string_array>:
  endbr64
  push   %r13
  mov    $0x8,%edx
  mov    %rdi,%r13
  xor    %edi,%edi
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <reallocarray@plt>
  mov    %rax,%r12
  test   %rbp,%rbp
  je     L0
  xor    %ebx,%ebx
  nopw   0x0(%rax,%rax,1)
L10:
  mov    0x0(%r13,%rbx,8),%rdi
  call   <strdup@plt>
  mov    %rax,(%r12,%rbx,8)
  add    $0x1,%rbx
  cmp    %rbx,%rbp
  jne    L10
L0:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char **copy_string_array(char **array, size_t len) {
  char **result = reallocarray(NULL, len, sizeof(char *));
  size_t i;
  for (i = 0U; i < len; ++i) {
    result[i] = strdup(array[i]);
  }
  return result;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<estostr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%r12
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  call   <strlen@plt>
  add    $0x1,%eax
  movslq %eax,%rdx
  add    $0xf,%rdx
  mov    %rdx,%rsi
  and    $0xfffffffffffffff0,%rsi
  and    $0xfffffffffffff000,%rdx
  mov    %rsp,%rcx
  sub    %rdx,%rcx
L149:
  cmp    %rcx,%rsp
  je     L15
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L149
L15:
  mov    %rsi,%rdx
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  je     L24
  orq    $0x0,-0x8(%rsp,%rdx,1)
L24:
  mov    %rsp,%r15
  mov    %eax,%edx
  mov    $0x0,%esi
  mov    %rsp,%rdi
  call   <memset@plt>
  mov    $0x0,%ebx
  mov    $0x0,%r14d
  mov    $0x0,%r13d
  jmp    L159
L160:
  cmp    $0x6e,%al
  je     L3
  cmp    $0x6e,%al
  jle    L17
  cmp    $0x72,%al
  je     L1
  cmp    $0x74,%al
  jne    L102
  movslq %r13d,%rax
  movb   $0x9,(%r15,%rax,1)
  jmp    L55
L17:
  cmp    $0x30,%al
  je     L121
L102:
  movslq %r13d,%rdx
  movb   $0x5c,(%r15,%rdx,1)
  add    $0x1,%r13d
  movslq %r13d,%rdx
  mov    %al,(%r15,%rdx,1)
  jmp    L55
L3:
  movslq %r13d,%rax
  movb   $0xa,(%r15,%rax,1)
L55:
  add    $0x1,%r13d
  mov    $0x0,%r14d
L179:
  add    $0x1,%rbx
L159:
  mov    %r12,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jl     L167
  movzbl (%r12,%rbx,1),%eax
  cmp    $0x5c,%al
  je     L171
  test   %r14d,%r14d
  jne    L160
  movslq %r13d,%rdx
  mov    %al,(%r15,%rdx,1)
  add    $0x1,%r13d
  jmp    L179
L1:
  movslq %r13d,%rax
  movb   $0xd,(%r15,%rax,1)
  jmp    L55
L121:
  movslq %r13d,%rax
  movb   $0x0,(%r15,%rax,1)
  jmp    L55
L171:
  mov    $0x1,%r14d
  jmp    L179
L167:
  mov    %r15,%rsi
  mov    %r12,%rdi
  call   <strcpy@plt>
  mov    -0x38(%rbp),%rdx
  sub    %fs:0x28,%rdx
  jne    L146
  lea    -0x28(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
L146:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char *estostr(char *s) {
  int i, j = 0;
  int f = 0;
  char ss[strlen(s) + 1];
  memset(ss, 0, sizeof ss);
  for (i = 0; i < strlen(s) + 1; i++) {
    if (s[i] == '\\') {
      f = 1;
      continue;
    }
    if (f) {
      switch (s[i]) {
      case 'n':
        ss[j] = '\n';
        break;
      case 'r':
        ss[j] = '\r';
        break;
      case 't':
        ss[j] = '\t';
        break;
      case '0':
        ss[j] = '\0';
        break;
      case '\\':
        ss[j] = '\\';
        break;
      default:
        ss[j] = '\\';
        j++;
        ss[j] = s[i];
        break;
      }
      j++;
      f = 0;
      continue;
    }
    ss[j] = s[i];
    j++;
  }
  return strcpy(s, ss);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<estostr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  mov    %rdi,%r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  call   <strlen@plt>
  mov    %rsp,%rcx
  lea    0x1(%rax),%r8d
  movslq %r8d,%rdx
  add    $0xf,%rdx
  mov    %rdx,%rax
  and    $0xfffffffffffff000,%rdx
  sub    %rdx,%rcx
  and    $0xfffffffffffffff0,%rax
  cmp    %rcx,%rsp
  je     L45
L6:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rcx,%rsp
  jne    L6
L45:
  and    $0xfff,%eax
  sub    %rax,%rsp
  test   %rax,%rax
  jne    L149
L96:
  mov    %r8d,%edx
  xor    %esi,%esi
  mov    %rsp,%rdi
  mov    %rsp,%r13
  call   <memset@plt>
  xor    %ebx,%ebx
  xor    %r14d,%r14d
  xor    %r12d,%r12d
  jmp    L95
L67:
  mov    %al,0x0(%r13,%r12,1)
  movslq %edx,%r12
L69:
  add    $0x1,%rbx
L95:
  mov    %r15,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jl     L164
  movzbl (%r15,%rbx,1),%eax
  cmp    $0x5c,%al
  je     L91
  lea    0x1(%r12),%edx
  test   %r14d,%r14d
  je     L67
  cmp    $0x72,%al
  je     L192
  jg     L146
  cmp    $0x30,%al
  je     L58
  cmp    $0x6e,%al
  jne    L188
  movb   $0xa,0x0(%r13,%r12,1)
L18:
  movslq %edx,%r12
  xor    %r14d,%r14d
  jmp    L69
  nopl   0x0(%rax,%rax,1)
L146:
  cmp    $0x74,%al
  jne    L188
  movb   $0x9,0x0(%r13,%r12,1)
  jmp    L18
  nopl   0x0(%rax)
L91:
  mov    $0x1,%r14d
  jmp    L69
  nopl   0x0(%rax,%rax,1)
L164:
  mov    %r13,%rsi
  mov    %r15,%rdi
  call   <strcpy@plt>
  mov    -0x38(%rbp),%rdx
  sub    %fs:0x28,%rdx
  jne    L3
  lea    -0x28(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
  nopl   0x0(%rax)
L192:
  movb   $0xd,0x0(%r13,%r12,1)
  jmp    L18
  nopl   0x0(%rax,%rax,1)
L188:
  movslq %edx,%rcx
  movb   $0x5c,0x0(%r13,%r12,1)
  add    $0x1,%edx
  mov    %al,0x0(%r13,%rcx,1)
  jmp    L18
  nopl   0x0(%rax,%rax,1)
L58:
  movb   $0x0,0x0(%r13,%r12,1)
  jmp    L18
  nopl   0x0(%rax,%rax,1)
L149:
  orq    $0x0,-0x8(%rsp,%rax,1)
  jmp    L96
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char *estostr(char *s) {
  int i, j = 0;
  int f = 0;
  char ss[strlen(s) + 1];
  memset(ss, 0, sizeof ss);
  for (i = 0; i < strlen(s) + 1; i++) {
    if (s[i] == '\\') {
      f = 1;
      continue;
    }
    if (f) {
      switch (s[i]) {
      case 'n':
        ss[j] = '\n';
        break;
      case 'r':
        ss[j] = '\r';
        break;
      case 't':
        ss[j] = '\t';
        break;
      case '0':
        ss[j] = '\0';
        break;
      case '\\':
        ss[j] = '\\';
        break;
      default:
        ss[j] = '\\';
        j++;
        ss[j] = s[i];
        break;
      }
      j++;
      f = 0;
      continue;
    }
    ss[j] = s[i];
    j++;
  }
  return strcpy(s, ss);
}
```