ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将以下的汇编代码转换成C语言代码：

```asm
<Rys_findRoots>:
  endbr64
  sub    $0x8,%rsp
  cmp    $0x5,%edi
  ja     L65
  mov    %edi,%edi
  lea    D13(%rip),%rdx
  movslq (%rdx,%rdi,4),%rax
  add    %rdx,%rax
  notrack jmp *%rax
  lea    D7(%rip),%rdi
  call   <fatalError@plt>
L55:
  add    $0x8,%rsp
  ret
  call   <root1@plt>
  jmp    L55
  call   <root2@plt>
  jmp    L55
  call   <root3@plt>
  jmp    L55
  call   <root4@plt>
  jmp    L55
  call   <root5@plt>
  jmp    L55
L65:
  call   <rootN@plt>
  jmp    L55
```

tool_calls: name: disassemble_data
arguments: label: D7
type_guessed: string

tool: D7: Roots: cannot have rys.roots for zero order

assistant: ```c
void Rys_findRoots(int n, double X) {
  switch (n) {
  case 0:
    fatalError("Roots: cannot have rys.roots for zero order");
    break;
  case 1:
    root1(X);
    break;
  case 2:
    root2(X);
    break;
  case 3:
    root3(X);
    break;
  case 4:
    root4(X);
    break;
  case 5:
    root5(X);
    break;
  default:
    rootN(n, X);
    break;
  }
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<Rys_findRoots>:
  endbr64
  cmp    $0x5,%edi
  ja     L3
  lea    D11(%rip),%rdx
  mov    %edi,%edi
  movslq (%rdx,%rdi,4),%rax
  add    %rdx,%rax
  notrack jmp *%rax
  nopl   0x0(%rax)
  jmp    1110 <root4@plt>
  nopl   (%rax)
  jmp    10b0 <root5@plt>
  nopl   (%rax)
  lea    D9(%rip),%rdi
  jmp    1100 <fatalError@plt>
  nopl   0x0(%rax)
  jmp    10e0 <root1@plt>
  nopl   (%rax)
  jmp    10c0 <root2@plt>
  nopl   (%rax)
  jmp    10d0 <root3@plt>
L3:
  jmp    10f0 <rootN@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D9
datatype: string

tool: D9: Roots: cannot have rys.roots for zero order

assistant: ```c
void Rys_findRoots(int n, double X) {
  switch (n) {
  case 0:
    fatalError("Roots: cannot have rys.roots for zero order");
    break;
  case 1:
    root1(X);
    break;
  case 2:
    root2(X);
    break;
  case 3:
    root3(X);
    break;
  case 4:
    root4(X);
    break;
  case 5:
    root5(X);
    break;
  default:
    rootN(n, X);
    break;
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<Rys_findRoots>:
  endbr64
  cmp    $0x5,%edi
  ja     L5
  lea    D14(%rip),%rdx
  mov    %edi,%edi
  movslq (%rdx,%rdi,4),%rax
  add    %rdx,%rax
  notrack jmp *%rax
  nopl   0x0(%rax)
  jmp    1110 <root4@plt>
  nopl   (%rax)
  jmp    10b0 <root5@plt>
  nopl   (%rax)
  lea    D1(%rip),%rdi
  jmp    1100 <fatalError@plt>
  nopl   0x0(%rax)
  jmp    10e0 <root1@plt>
  nopl   (%rax)
  jmp    10c0 <root2@plt>
  nopl   (%rax)
  jmp    10d0 <root3@plt>
L5:
  jmp    10f0 <rootN@plt>
```

tool_calls: name: read_data
arguments: tag: D1
type: string

tool: D1: Roots: cannot have rys.roots for zero order

assistant: ```c
void Rys_findRoots(int n, double X) {
  switch (n) {
  case 0:
    fatalError("Roots: cannot have rys.roots for zero order");
    break;
  case 1:
    root1(X);
    break;
  case 2:
    root2(X);
    break;
  case 3:
    root3(X);
    break;
  case 4:
    root4(X);
    break;
  case 5:
    root5(X);
    break;
  default:
    rootN(n, X);
    break;
  }
}
```
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
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0xc(%rbp),%rax
  mov    %rax,%rsi
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  jmp    L29
L23:
  lea    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x10(%rbp),%eax
  sub    $0x1,%eax
  shl    $0x2,%eax
  cltq
  add    $0x1,%rax
  mov    %rax,%rsi
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L29:
  mov    -0xc(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0xc(%rbp)
  test   %eax,%eax
  jne    L23
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L21
  call   <__stack_chk_fail@plt>
L21:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: label: D8
data_type: string
name: retrieve_data
arguments: label: D12
data_type: string

tool: D8: %d

tool: D12: %lld


assistant: ```c
int main() {
  int i, j, c, len, num, temp, t;
  long long int res;
  scanf("%d", &t);
  while (t--) {
    scanf("%d", &num);
    printf("%lld\n", (long long int)(4 * (num - 1)) + 1);
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  lea    D11(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,0x4(%rsp)
  test   %eax,%eax
  je     L26
  mov    %rsp,%r12
  lea    D11(%rip),%rbp
  lea    D14(%rip),%rbx
L7:
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    (%rsp),%eax
  lea    -0x4(,%rax,4),%esi
  movslq %esi,%rsi
  add    $0x1,%rsi
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    0x4(%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,0x4(%rsp)
  test   %eax,%eax
  jne    L7
L26:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L21
  mov    $0x0,%eax
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L21:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D11
data_type: string
name: retrieve_data
arguments: data_label: D14
data_type: string

tool: D11: %d

tool: D14: %lld


assistant: ```c
int main() {
  int i, j, c, len, num, temp, t;
  long long int res;
  scanf("%d", &t);
  while (t--) {
    scanf("%d", &num);
    printf("%lld\n", (long long int)(4 * (num - 1)) + 1);
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  lea    D12(%rip),%rbx
  mov    %rbx,%rdi
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,0x4(%rsp)
  test   %eax,%eax
  je     L16
  mov    %rsp,%r12
  lea    D9(%rip),%rbp
  nopl   0x0(%rax)
L26:
  mov    %r12,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    (%rsp),%eax
  mov    %rbp,%rdi
  lea    -0x4(,%rax,4),%esi
  movslq %esi,%rsi
  add    $0x1,%rsi
  call   <printf@plt>
  mov    0x4(%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,0x4(%rsp)
  test   %eax,%eax
  jne    L26
L16:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L25
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L25:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D12
datatype: string
name: read_data
arguments: tag: D9
datatype: string

tool: D12: %d

tool: D9: %lld


assistant: ```c
int main() {
  int i, j, c, len, num, temp, t;
  long long int res;
  scanf("%d", &t);
  while (t--) {
    scanf("%d", &num);
    printf("%lld\n", (long long int)(4 * (num - 1)) + 1);
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  lea    D5(%rip),%rbx
  mov    %rbx,%rdi
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,0x4(%rsp)
  test   %eax,%eax
  je     L26
  mov    %rsp,%r12
  lea    D0(%rip),%rbp
  nopl   0x0(%rax)
L5:
  mov    %r12,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    (%rsp),%eax
  mov    %rbp,%rdi
  lea    -0x4(,%rax,4),%esi
  movslq %esi,%rsi
  add    $0x1,%rsi
  call   <printf@plt>
  mov    0x4(%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,0x4(%rsp)
  test   %eax,%eax
  jne    L5
L26:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L18
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L18:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D5
type_guessed: string
name: parse_data
arguments: data_tag: D0
type_guessed: string

tool: D5: %d

tool: D0: %lld


assistant: ```c
int main() {
  int i, j, c, len, num, temp, t;
  long long int res;
  scanf("%d", &t);
  while (t--) {
    scanf("%d", &num);
    printf("%lld\n", (long long int)(4 * (num - 1)) + 1);
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<fgets_dynamic>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    -0x38(%rbp),%rax
  mov    %rax,%rdi
  call   <ftell@plt>
  mov    %rax,-0x18(%rbp)
  mov    -0x30(%rbp),%rax
  mov    (%rax),%rcx
  mov    -0x28(%rbp),%rax
  mov    (%rax),%rax
  mov    -0x38(%rbp),%rdx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <fgets@plt>
  test   %eax,%eax
  jne    L64
  mov    $0x0,%eax
  jmp    L66
L64:
  mov    -0x38(%rbp),%rax
  mov    %rax,%rdi
  call   <ftell@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  sub    -0x18(%rbp),%rax
  mov    %rax,-0x20(%rbp)
  jmp    L63
L74:
  mov    D8(%rip),%rax
  mov    (%rax),%edx
  mov    -0x18(%rbp),%rcx
  mov    -0x38(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <fseek@plt>
  mov    -0x30(%rbp),%rax
  mov    (%rax),%rax
  lea    (%rax,%rax,1),%rdx
  mov    -0x30(%rbp),%rax
  mov    %rdx,(%rax)
  mov    -0x30(%rbp),%rax
  mov    (%rax),%rdx
  mov    -0x28(%rbp),%rax
  mov    (%rax),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <realloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L49
  mov    -0x28(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    $0x0,%eax
  jmp    L66
L49:
  mov    -0x28(%rbp),%rax
  mov    -0x8(%rbp),%rdx
  mov    %rdx,(%rax)
  mov    -0x30(%rbp),%rax
  mov    (%rax),%rcx
  mov    -0x28(%rbp),%rax
  mov    (%rax),%rax
  mov    -0x38(%rbp),%rdx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <fgets@plt>
  test   %eax,%eax
  je     L19
  mov    -0x38(%rbp),%rax
  mov    %rax,%rdi
  call   <ftell@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  sub    -0x18(%rbp),%rax
  mov    %rax,-0x20(%rbp)
L63:
  mov    -0x30(%rbp),%rax
  mov    (%rax),%rax
  sub    $0x1,%rax
  cmp    %rax,-0x20(%rbp)
  jne    L12
  mov    -0x28(%rbp),%rax
  mov    (%rax),%rdx
  mov    -0x30(%rbp),%rax
  mov    (%rax),%rax
  sub    $0x1,%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0xa,%al
  jne    L74
  jmp    L12
L19:
  nop
L12:
  mov    -0x28(%rbp),%rax
  mov    (%rax),%rax
L66:
  leave
  ret
```

assistant: ```c
char *fgets_dynamic(char **buf, size_t *buf_len, FILE *fp) {
  long pos_from = ftell(fp);
  if (!fgets(*buf, *buf_len, fp)) {
    return NULL;
  }
  long pos_to = ftell(fp);
  unsigned long chars_read = pos_to - pos_from;
  while (chars_read == *buf_len - 1 && (*buf)[*buf_len - 1] != '\n') {
    fseek(fp, pos_from, SEEK_SET);
    *buf_len = *buf_len * 2;
    char *new_buf = realloc(*buf, *buf_len);
    if (!new_buf) {
      free(*buf);
      return NULL;
    }
    *buf = new_buf;
    if (!fgets(*buf, *buf_len, fp)) {
      break;
    }
    pos_to = ftell(fp);
    chars_read = pos_to - pos_from;
  }
  return *buf;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<fgets_dynamic>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r13
  mov    %rsi,%rbp
  mov    %rdx,%r12
  mov    %rdx,%rdi
  call   <ftell@plt>
  mov    %rax,%r14
  mov    %r12,%rdx
  mov    0x0(%rbp),%rsi
  mov    0x0(%r13),%rdi
  call   <fgets@plt>
  mov    $0x0,%ebx
  test   %eax,%eax
  je     L65
  mov    %r12,%rdi
  call   <ftell@plt>
  mov    0x0(%rbp),%rcx
  lea    -0x1(%rcx),%rdx
  sub    %r14,%rax
  cmp    %rdx,%rax
  jne    L10
L66:
  mov    0x0(%r13),%rax
  cmpb   $0xa,(%rax,%rdx,1)
  je     L10
  mov    D7(%rip),%rax
  mov    (%rax),%edx
  mov    %r14,%rsi
  mov    %r12,%rdi
  call   <fseek@plt>
  mov    0x0(%rbp),%rax
  lea    (%rax,%rax,1),%rsi
  mov    %rsi,0x0(%rbp)
  mov    0x0(%r13),%rdi
  call   <realloc@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L48
  mov    %rax,0x0(%r13)
  mov    %r12,%rdx
  mov    0x0(%rbp),%rsi
  mov    %rax,%rdi
  call   <fgets@plt>
  test   %eax,%eax
  je     L10
  mov    %r12,%rdi
  call   <ftell@plt>
  mov    0x0(%rbp),%rcx
  lea    -0x1(%rcx),%rdx
  sub    %r14,%rax
  cmp    %rax,%rdx
  je     L66
L10:
  mov    0x0(%r13),%rbx
L65:
  mov    %rbx,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L48:
  mov    0x0(%r13),%rdi
  call   <free@plt>
  jmp    L65
```

assistant: ```c
char *fgets_dynamic(char **buf, size_t *buf_len, FILE *fp) {
  long pos_from = ftell(fp);
  if (!fgets(*buf, *buf_len, fp)) {
    return NULL;
  }
  long pos_to = ftell(fp);
  unsigned long chars_read = pos_to - pos_from;
  while (chars_read == *buf_len - 1 && (*buf)[*buf_len - 1] != '\n') {
    fseek(fp, pos_from, SEEK_SET);
    *buf_len = *buf_len * 2;
    char *new_buf = realloc(*buf, *buf_len);
    if (!new_buf) {
      free(*buf);
      return NULL;
    }
    *buf = new_buf;
    if (!fgets(*buf, *buf_len, fp)) {
      break;
    }
    pos_to = ftell(fp);
    chars_read = pos_to - pos_from;
  }
  return *buf;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<fgets_dynamic>:
  endbr64
  push   %r14
  push   %r13
  mov    %rdi,%r13
  mov    %rdx,%rdi
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  mov    %rsi,%rbx
  call   <ftell@plt>
  mov    (%rbx),%rsi
  mov    0x0(%r13),%rdi
  mov    %r12,%rdx
  mov    %rax,%r14
  call   <fgets@plt>
  test   %eax,%eax
  je     L44
  mov    %r12,%rdi
  call   <ftell@plt>
  mov    (%rbx),%rcx
  sub    %r14,%rax
  lea    -0x1(%rcx),%rdx
  cmp    %rdx,%rax
  je     L21
  jmp    L42
  nopw   0x0(%rax,%rax,1)
L7:
  mov    D5(%rip),%rax
  mov    %r14,%rsi
  mov    %r12,%rdi
  mov    (%rax),%edx
  call   <fseek@plt>
  mov    (%rbx),%rax
  mov    0x0(%r13),%rdi
  lea    (%rax,%rax,1),%rsi
  mov    %rsi,(%rbx)
  call   <realloc@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  je     L34
  mov    %rax,0x0(%r13)
  mov    (%rbx),%rsi
  mov    %r12,%rdx
  mov    %rax,%rdi
  call   <fgets@plt>
  test   %eax,%eax
  je     L42
  mov    %r12,%rdi
  call   <ftell@plt>
  mov    (%rbx),%rcx
  sub    %r14,%rax
  lea    -0x1(%rcx),%rdx
  cmp    %rax,%rdx
  jne    L42
L21:
  mov    0x0(%r13),%rbp
  cmpb   $0xa,0x0(%rbp,%rdx,1)
  jne    L7
L44:
  pop    %rbx
  mov    %rbp,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax,%rax,1)
L42:
  mov    0x0(%r13),%rbp
  pop    %rbx
  mov    %rbp,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L34:
  mov    0x0(%r13),%rdi
  call   <free@plt>
  pop    %rbx
  mov    %rbp,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
char *fgets_dynamic(char **buf, size_t *buf_len, FILE *fp) {
  long pos_from = ftell(fp);
  if (!fgets(*buf, *buf_len, fp)) {
    return NULL;
  }
  long pos_to = ftell(fp);
  unsigned long chars_read = pos_to - pos_from;
  while (chars_read == *buf_len - 1 && (*buf)[*buf_len - 1] != '\n') {
    fseek(fp, pos_from, SEEK_SET);
    *buf_len = *buf_len * 2;
    char *new_buf = realloc(*buf, *buf_len);
    if (!new_buf) {
      free(*buf);
      return NULL;
    }
    *buf = new_buf;
    if (!fgets(*buf, *buf_len, fp)) {
      break;
    }
    pos_to = ftell(fp);
    chars_read = pos_to - pos_from;
  }
  return *buf;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<fgets_dynamic>:
  endbr64
  push   %r14
  push   %r13
  mov    %rdi,%r13
  mov    %rdx,%rdi
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  mov    %rsi,%rbx
  call   <ftell@plt>
  mov    (%rbx),%rsi
  mov    0x0(%r13),%rdi
  mov    %r12,%rdx
  mov    %rax,%r14
  call   <fgets@plt>
  test   %eax,%eax
  je     L42
  mov    %r12,%rdi
  call   <ftell@plt>
  mov    (%rbx),%rcx
  sub    %r14,%rax
  lea    -0x1(%rcx),%rdx
  cmp    %rdx,%rax
  je     L22
  jmp    L49
  nopw   0x0(%rax,%rax,1)
L4:
  mov    D2(%rip),%rax
  mov    %r14,%rsi
  mov    %r12,%rdi
  mov    (%rax),%edx
  call   <fseek@plt>
  mov    (%rbx),%rax
  mov    0x0(%r13),%rdi
  lea    (%rax,%rax,1),%rsi
  mov    %rsi,(%rbx)
  call   <realloc@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  je     L11
  mov    %rax,0x0(%r13)
  mov    (%rbx),%rsi
  mov    %r12,%rdx
  mov    %rax,%rdi
  call   <fgets@plt>
  test   %eax,%eax
  je     L49
  mov    %r12,%rdi
  call   <ftell@plt>
  mov    (%rbx),%rcx
  sub    %r14,%rax
  lea    -0x1(%rcx),%rdx
  cmp    %rax,%rdx
  jne    L49
L22:
  mov    0x0(%r13),%rbp
  cmpb   $0xa,0x0(%rbp,%rdx,1)
  jne    L4
L42:
  pop    %rbx
  mov    %rbp,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax,%rax,1)
L49:
  mov    0x0(%r13),%rbp
  pop    %rbx
  mov    %rbp,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L11:
  mov    0x0(%r13),%rdi
  call   <free@plt>
  pop    %rbx
  mov    %rbp,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
char *fgets_dynamic(char **buf, size_t *buf_len, FILE *fp) {
  long pos_from = ftell(fp);
  if (!fgets(*buf, *buf_len, fp)) {
    return NULL;
  }
  long pos_to = ftell(fp);
  unsigned long chars_read = pos_to - pos_from;
  while (chars_read == *buf_len - 1 && (*buf)[*buf_len - 1] != '\n') {
    fseek(fp, pos_from, SEEK_SET);
    *buf_len = *buf_len * 2;
    char *new_buf = realloc(*buf, *buf_len);
    if (!new_buf) {
      free(*buf);
      return NULL;
    }
    *buf = new_buf;
    if (!fgets(*buf, *buf_len, fp)) {
      break;
    }
    pos_to = ftell(fp);
    chars_read = pos_to - pos_from;
  }
  return *buf;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<permute>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %ecx,-0x2c(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L7
L3:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cltq
  lea    -0x1(%rax),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    -0x4(%rbp),%edx
  movslq %edx,%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  addl   $0x1,-0x4(%rbp)
L7:
  mov    -0x4(%rbp),%eax
  cmp    -0x2c(%rbp),%eax
  jl     L3
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void permute(char *out, char *in, int *p, int n) {
  int i;
  for (i = 0; i < n; i++)
    out[i] = in[p[i] - 1];
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<permute>:
  endbr64
  test   %ecx,%ecx
  jle    L3
  mov    %ecx,%ecx
  mov    $0x0,%eax
L12:
  movslq (%rdx,%rax,4),%r8
  movzbl -0x1(%rsi,%r8,1),%r8d
  mov    %r8b,(%rdi,%rax,1)
  add    $0x1,%rax
  cmp    %rcx,%rax
  jne    L12
L3:
  ret
```

assistant: ```c
void permute(char *out, char *in, int *p, int n) {
  int i;
  for (i = 0; i < n; i++)
    out[i] = in[p[i] - 1];
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<permute>:
  endbr64
  test   %ecx,%ecx
  jle    L10
  movslq %ecx,%rcx
  xor    %eax,%eax
  nopl   (%rax)
L12:
  movslq (%rdx,%rax,4),%r8
  movzbl -0x1(%rsi,%r8,1),%r8d
  mov    %r8b,(%rdi,%rax,1)
  add    $0x1,%rax
  cmp    %rax,%rcx
  jne    L12
L10:
  ret
```

assistant: ```c
void permute(char *out, char *in, int *p, int n) {
  int i;
  for (i = 0; i < n; i++)
    out[i] = in[p[i] - 1];
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<permute>:
  endbr64
  test   %ecx,%ecx
  jle    L10
  movslq %ecx,%rcx
  xor    %eax,%eax
  nopl   (%rax)
L3:
  movslq (%rdx,%rax,4),%r8
  movzbl -0x1(%rsi,%r8,1),%r8d
  mov    %r8b,(%rdi,%rax,1)
  add    $0x1,%rax
  cmp    %rax,%rcx
  jne    L3
L10:
  ret
```

assistant: ```c
void permute(char *out, char *in, int *p, int n) {
  int i;
  for (i = 0; i < n; i++)
    out[i] = in[p[i] - 1];
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<leftShift>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    $0x5,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  movw   $0x0,-0xa(%rbp)
  jmp    L17
L11:
  movswl -0xa(%rbp),%edx
  mov    -0x1c(%rbp),%eax
  add    %eax,%edx
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    %eax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  mov    %eax,%ecx
  shl    $0x2,%ecx
  add    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  lea    (%rdx,%rax,1),%rsi
  movzwl -0xa(%rbp),%edx
  movswl %dx,%eax
  imul   $0x6667,%eax,%eax
  shr    $0x10,%eax
  sar    %ax
  mov    %edx,%ecx
  sar    $0xf,%cx
  sub    %ecx,%eax
  mov    %eax,%ecx
  shl    $0x2,%ecx
  add    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  movswq %ax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rax,%rdx
  movzbl (%rsi),%eax
  mov    %al,(%rdx)
  movzwl -0xa(%rbp),%eax
  add    $0x1,%eax
  mov    %ax,-0xa(%rbp)
L17:
  cmpw   $0x4,-0xa(%rbp)
  jle    L11
  movswq -0xa(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *leftShift(char text[], int shiftValue) {
  char *shiftedText = malloc(sizeof(char) * 5);
  short i;
  for (i = 0; i < 5; i++) {
    shiftedText[i % 5] = text[(i + shiftValue) % 5];
  }
  shiftedText[i] = '\0';
  return shiftedText;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<leftShift>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    %esi,%ebx
  mov    $0x5,%edi
  call   <malloc@plt>
  mov    %rax,%rsi
  mov    $0x0,%ecx
L8:
  lea    (%rbx,%rcx,1),%edx
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x21,%rax
  mov    %edx,%edi
  sar    $0x1f,%edi
  sub    %edi,%eax
  lea    (%rax,%rax,4),%eax
  sub    %eax,%edx
  movslq %edx,%rdx
  movzbl 0x0(%rbp,%rdx,1),%edi
  movswl %cx,%eax
  imul   $0x6667,%eax,%eax
  sar    $0x11,%eax
  mov    %ecx,%edx
  sar    $0xf,%dx
  sub    %edx,%eax
  lea    (%rax,%rax,4),%eax
  mov    %ecx,%edx
  sub    %eax,%edx
  movswq %dx,%rax
  mov    %dil,(%rsi,%rax,1)
  add    $0x1,%ecx
  cmp    $0x5,%ecx
  jne    L8
  movb   $0x0,0x5(%rsi)
  mov    %rsi,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
char *leftShift(char text[], int shiftValue) {
  char *shiftedText = malloc(sizeof(char) * 5);
  short i;
  for (i = 0; i < 5; i++) {
    shiftedText[i % 5] = text[(i + shiftValue) % 5];
  }
  shiftedText[i] = '\0';
  return shiftedText;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<leftShift>:
  endbr64
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %rdi,%rbx
  mov    $0x5,%edi
  sub    $0x8,%rsp
  call   <malloc@plt>
  mov    %ebp,%esi
  mov    %rax,%r8
  lea    0x5(%rax),%r9
  mov    %rax,%rcx
  sub    %eax,%esi
L8:
  lea    (%rsi,%rcx,1),%edx
  add    $0x1,%rcx
  movslq %edx,%rax
  mov    %edx,%edi
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%edi
  sar    $0x21,%rax
  sub    %edi,%eax
  lea    (%rax,%rax,4),%eax
  sub    %eax,%edx
  movslq %edx,%rdx
  movzbl (%rbx,%rdx,1),%eax
  mov    %al,-0x1(%rcx)
  cmp    %r9,%rcx
  jne    L8
  movb   $0x0,0x5(%r8)
  add    $0x8,%rsp
  mov    %r8,%rax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
char *leftShift(char text[], int shiftValue) {
  char *shiftedText = malloc(sizeof(char) * 5);
  short i;
  for (i = 0; i < 5; i++) {
    shiftedText[i % 5] = text[(i + shiftValue) % 5];
  }
  shiftedText[i] = '\0';
  return shiftedText;
}
```