ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<read>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x418,%rsp
  mov    %rdi,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x408(%rsp)
  xor    %eax,%eax
  lea    0x1(%rsi),%edi
  mov    $0x800,%esi
  call   <calloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L112
  lea    D19(%rip),%rsi
  mov    %rbx,%rdi
  call   <fopen@plt>
  mov    %rax,%r14
  test   %rax,%rax
  je     L13
  mov    $0x400,%edi
  call   <malloc@plt>
  mov    %rax,%r13
  mov    $0x0,%ebx
  lea    D10(%rip),%rbp
  test   %rax,%rax
  je     L9
L110:
  mov    %r14,%rcx
  mov    $0x400,%edx
  mov    $0x1,%esi
  mov    %r13,%rdi
  call   <fread@plt>
  test   %rax,%rax
  je     L20
  mov    %rsp,%r15
  mov    %r13,%rsi
  mov    %r15,%rdi
  call   <strcpy@plt>
  mov    %rbp,%rsi
  mov    %r15,%rdi
  call   <strtok@plt>
  test   %rax,%rax
  je     L110
L54:
  mov    %rax,(%r12,%rbx,8)
  add    $0x1,%rbx
  mov    %rbp,%rsi
  mov    $0x0,%edi
  call   <strtok@plt>
  test   %rax,%rax
  jne    L54
  jmp    L110
L112:
  lea    D20(%rip),%rdi
  call   <printf@plt>
  jmp    L40
L9:
  lea    D14(%rip),%rdi
  call   <printf@plt>
  mov    %r13,%r12
  jmp    L40
L20:
  cmpq   $0x0,-0x8(%r12,%rbx,8)
  je     L71
  movq   $0x0,(%r12,%rbx,8)
L71:
  mov    %r13,%rdi
  call   <free@plt>
L40:
  mov    0x408(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L27
  mov    %r12,%rax
  add    $0x418,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L13:
  mov    %rax,%r12
  jmp    L40
L27:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D19
type_guessed: string
name: read_data
arguments: data_label: D10
type_guessed: string
name: read_data
arguments: data_label: D20
type_guessed: string
name: read_data
arguments: data_label: D14
type_guessed: string

tool: D19: r

tool: D10: 


tool: D20: There was a malloc error with ret_lines

tool: D14: There was a malloc error

assistant: ```c
char **read(const char *fname, const int lines) {
  char **ret_lines = calloc(lines + 1, sizeof(char *) * 256);
  if (ret_lines == NULL) {
    printf("There was a malloc error with ret_lines");
    return NULL;
  }
  FILE *file;
  file = fopen(fname, "r");
  if (file) {
    char *buff = malloc(1024);
    size_t nread;
    if (buff == NULL) {
      printf("There was a malloc error");
      return NULL;
    }
    size_t idx = 0;
    while ((nread = fread(buff, 1, 1024, file)) > 0) {
      char tmp[1024];
      strcpy(tmp, buff);
      char *token = strtok(tmp, "\n");
      while (token) {
        ret_lines[idx] = token;
        idx++;
        token = strtok(NULL, "\n");
      }
    }
    if (ret_lines[idx - 1] != NULL)
      ret_lines[idx] = NULL;
    free(buff);
  } else
    return NULL;
  return ret_lines;
}
```
user: What is the c source code of the assembly code below?

```asm
<read>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  lea    0x1(%rsi),%edi
  mov    $0x800,%esi
  push   %rbx
  sub    $0x418,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x408(%rsp)
  xor    %eax,%eax
  call   <calloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L37
  lea    D35(%rip),%rsi
  mov    %rbp,%rdi
  call   <fopen@plt>
  mov    %rax,%r13
  test   %rax,%rax
  je     L51
  mov    $0x400,%edi
  xor    %r14d,%r14d
  lea    D39(%rip),%rbx
  call   <malloc@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  je     L36
  nop
L22:
  mov    %r13,%rcx
  mov    $0x400,%edx
  mov    $0x1,%esi
  mov    %rbp,%rdi
  call   <fread@plt>
  test   %rax,%rax
  je     L12
  mov    %rsp,%r15
  mov    %rbp,%rsi
  mov    %r15,%rdi
  call   <strcpy@plt>
  mov    %rbx,%rsi
  mov    %r15,%rdi
  call   <strtok@plt>
  test   %rax,%rax
  je     L22
  nopl   0x0(%rax,%rax,1)
L119:
  mov    %rax,(%r12,%r14,8)
  xor    %edi,%edi
  mov    %rbx,%rsi
  add    $0x1,%r14
  call   <strtok@plt>
  test   %rax,%rax
  jne    L119
  jmp    L22
  nopl   0x0(%rax)
L12:
  cmpq   $0x0,-0x8(%r12,%r14,8)
  je     L52
  movq   $0x0,(%r12,%r14,8)
L52:
  mov    %rbp,%rdi
  call   <free@plt>
L90:
  mov    0x408(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L63
  add    $0x418,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L51:
  xor    %r12d,%r12d
  jmp    L90
L37:
  lea    D1(%rip),%rdi
  call   <printf@plt>
  jmp    L90
L36:
  lea    D0(%rip),%rdi
  xor    %r12d,%r12d
  call   <printf@plt>
  jmp    L90
L63:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D35
type: string
name: unpack_data
arguments: label: D39
type: string
name: unpack_data
arguments: label: D1
type: string
name: unpack_data
arguments: label: D0
type: string

tool: D35: r

tool: D39: 


tool: D1: There was a malloc error with ret_lines

tool: D0: There was a malloc error

assistant: ```c
char **read(const char *fname, const int lines) {
  char **ret_lines = calloc(lines + 1, sizeof(char *) * 256);
  if (ret_lines == NULL) {
    printf("There was a malloc error with ret_lines");
    return NULL;
  }
  FILE *file;
  file = fopen(fname, "r");
  if (file) {
    char *buff = malloc(1024);
    size_t nread;
    if (buff == NULL) {
      printf("There was a malloc error");
      return NULL;
    }
    size_t idx = 0;
    while ((nread = fread(buff, 1, 1024, file)) > 0) {
      char tmp[1024];
      strcpy(tmp, buff);
      char *token = strtok(tmp, "\n");
      while (token) {
        ret_lines[idx] = token;
        idx++;
        token = strtok(NULL, "\n");
      }
    }
    if (ret_lines[idx - 1] != NULL)
      ret_lines[idx] = NULL;
    free(buff);
  } else
    return NULL;
  return ret_lines;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<read>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  lea    0x1(%rsi),%edi
  mov    $0x800,%esi
  push   %rbx
  sub    $0x418,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x408(%rsp)
  xor    %eax,%eax
  call   <calloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L76
  lea    D27(%rip),%rsi
  mov    %rbp,%rdi
  call   <fopen@plt>
  mov    %rax,%r13
  test   %rax,%rax
  je     L22
  mov    $0x400,%edi
  xor    %r14d,%r14d
  lea    D35(%rip),%rbx
  call   <malloc@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  je     L44
  nop
L106:
  mov    %r13,%rcx
  mov    $0x400,%edx
  mov    $0x1,%esi
  mov    %rbp,%rdi
  call   <fread@plt>
  test   %rax,%rax
  je     L105
  mov    %rsp,%r15
  mov    %rbp,%rsi
  mov    %r15,%rdi
  call   <strcpy@plt>
  mov    %rbx,%rsi
  mov    %r15,%rdi
  call   <strtok@plt>
  test   %rax,%rax
  je     L106
  nopl   0x0(%rax,%rax,1)
L81:
  mov    %rax,(%r12,%r14,8)
  xor    %edi,%edi
  mov    %rbx,%rsi
  add    $0x1,%r14
  call   <strtok@plt>
  test   %rax,%rax
  jne    L81
  jmp    L106
  nopl   0x0(%rax)
L105:
  cmpq   $0x0,-0x8(%r12,%r14,8)
  je     L51
  movq   $0x0,(%r12,%r14,8)
L51:
  mov    %rbp,%rdi
  call   <free@plt>
L118:
  mov    0x408(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L15
  add    $0x418,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L22:
  xor    %r12d,%r12d
  jmp    L118
L76:
  lea    D24(%rip),%rdi
  call   <printf@plt>
  jmp    L118
L44:
  lea    D29(%rip),%rdi
  xor    %r12d,%r12d
  call   <printf@plt>
  jmp    L118
L15:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D27
type: string
name: extract_symbolic_data
arguments: tag: D35
type: string
name: extract_symbolic_data
arguments: tag: D24
type: string
name: extract_symbolic_data
arguments: tag: D29
type: string

tool: D27: r

tool: D35: 


tool: D24: There was a malloc error with ret_lines

tool: D29: There was a malloc error

assistant: ```c
char **read(const char *fname, const int lines) {
  char **ret_lines = calloc(lines + 1, sizeof(char *) * 256);
  if (ret_lines == NULL) {
    printf("There was a malloc error with ret_lines");
    return NULL;
  }
  FILE *file;
  file = fopen(fname, "r");
  if (file) {
    char *buff = malloc(1024);
    size_t nread;
    if (buff == NULL) {
      printf("There was a malloc error");
      return NULL;
    }
    size_t idx = 0;
    while ((nread = fread(buff, 1, 1024, file)) > 0) {
      char tmp[1024];
      strcpy(tmp, buff);
      char *token = strtok(tmp, "\n");
      while (token) {
        ret_lines[idx] = token;
        idx++;
        token = strtok(NULL, "\n");
      }
    }
    if (ret_lines[idx - 1] != NULL)
      ret_lines[idx] = NULL;
    free(buff);
  } else
    return NULL;
  return ret_lines;
}
```
user: What is the c source code of the assembly code below?

```asm
<wrap>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  jmp    L32
L28:
  mov    -0x8(%rbp),%eax
  add    %eax,-0x4(%rbp)
L32:
  cmpl   $0x0,-0x4(%rbp)
  js     L28
  jmp    L30
L16:
  mov    -0x8(%rbp),%eax
  sub    %eax,-0x4(%rbp)
L30:
  mov    -0x4(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jge    L16
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
__attribute__((__const__)) int wrap(int x, int max) {
  while (x < 0)
    x += max;
  while (x >= max)
    x -= max;
  return x;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<wrap>:
  endbr64
  mov    %edi,%eax
  test   %edi,%edi
  jns    L38
L16:
  add    %esi,%eax
  js     L16
L38:
  cmp    %eax,%esi
  jg     L9
L5:
  sub    %esi,%eax
  cmp    %eax,%esi
  jle    L5
L9:
  ret
```

assistant: ```c
__attribute__((__const__)) int wrap(int x, int max) {
  while (x < 0)
    x += max;
  while (x >= max)
    x -= max;
  return x;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<wrap>:
  endbr64
  mov    %edi,%eax
  test   %edi,%edi
  jns    L13
  nopw   0x0(%rax,%rax,1)
L38:
  add    %esi,%eax
  js     L38
  cmp    %eax,%esi
  jg     L28
  nopl   0x0(%rax,%rax,1)
L19:
  sub    %esi,%eax
L13:
  cmp    %eax,%esi
  jle    L19
L28:
  ret
```

assistant: ```c
__attribute__((__const__)) int wrap(int x, int max) {
  while (x < 0)
    x += max;
  while (x >= max)
    x -= max;
  return x;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<wrap>:
  endbr64
  mov    %edi,%eax
  test   %edi,%edi
  jns    L4
  nopw   0x0(%rax,%rax,1)
L19:
  add    %esi,%eax
  js     L19
  cmp    %eax,%esi
  jg     L27
  nopl   0x0(%rax,%rax,1)
L30:
  sub    %esi,%eax
L4:
  cmp    %eax,%esi
  jle    L30
L27:
  ret
```

assistant: ```c
__attribute__((__const__)) int wrap(int x, int max) {
  while (x < 0)
    x += max;
  while (x >= max)
    x -= max;
  return x;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<sortStringArray>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %esi,-0x3c(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    0x10(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  lea    -0x30(%rbp),%rax
  lea    -0x30(%rbp),%rdx
  movl   $0xfa1e0ff3,(%rax)
  movw   $0xbb49,0x4(%rax)
  lea    -0xf1(%rip),%rcx        # 1139 <compareStrings.0>
  mov    %rcx,0x6(%rax)
  movw   $0xba49,0xe(%rax)
  mov    %rdx,0x10(%rax)
  movl   $0x90e3ff49,0x18(%rax)
  lea    -0x30(%rbp),%rax
  mov    %rax,%rdx
  mov    -0x3c(%rbp),%esi
  mov    -0x38(%rbp),%rax
  mov    %rdx,%rcx
  mov    $0x8,%edx
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <qsort@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L8
  call   <__stack_chk_fail@plt>
L8:
  leave
  ret
```

assistant: ```c
void sortStringArray(char **arrString, int len) {
  int compareStrings(const void *a, const void *b) {
    const char **A = (const char **)a;
    const char **B = (const char **)b;
    int i;
    for (i = 0; *(*A + i) == *(*B + i); i++) {
      if (*(*A + i) == '\0') {
        return 0;
      }
    }
    return 2 * (*(*A + i) > *(*B + i)) - 1;
  }
  qsort(arrString, len, sizeof(char *), compareStrings);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<sortStringArray>:
  endbr64
  sub    $0x8,%rsp
  lea    -0x53(%rip),%rcx        # 1119 <compareStrings.0>
  mov    $0x8,%edx
  mov    $0x0,%eax
  call   <qsort@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void sortStringArray(char **arrString, int len) {
  int compareStrings(const void *a, const void *b) {
    const char **A = (const char **)a;
    const char **B = (const char **)b;
    int i;
    for (i = 0; *(*A + i) == *(*B + i); i++) {
      if (*(*A + i) == '\0') {
        return 0;
      }
    }
    return 2 * (*(*A + i) > *(*B + i)) - 1;
  }
  qsort(arrString, len, sizeof(char *), compareStrings);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<sortStringArray>:
  endbr64
  lea    -0x5b(%rip),%rcx        # 1120 <compareStrings.0>
  mov    $0x8,%edx
  xor    %eax,%eax
  jmp    1050 <qsort@plt>
```

assistant: ```c
void sortStringArray(char **arrString, int len) {
  int compareStrings(const void *a, const void *b) {
    const char **A = (const char **)a;
    const char **B = (const char **)b;
    int i;
    for (i = 0; *(*A + i) == *(*B + i); i++) {
      if (*(*A + i) == '\0') {
        return 0;
      }
    }
    return 2 * (*(*A + i) > *(*B + i)) - 1;
  }
  qsort(arrString, len, sizeof(char *), compareStrings);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<sortStringArray>:
  endbr64
  lea    -0x5b(%rip),%rcx        # 1120 <compareStrings.0>
  mov    $0x8,%edx
  xor    %eax,%eax
  jmp    1050 <qsort@plt>
```

assistant: ```c
void sortStringArray(char **arrString, int len) {
  int compareStrings(const void *a, const void *b) {
    const char **A = (const char **)a;
    const char **B = (const char **)b;
    int i;
    for (i = 0; *(*A + i) == *(*B + i); i++) {
      if (*(*A + i) == '\0') {
        return 0;
      }
    }
    return 2 * (*(*A + i) > *(*B + i)) - 1;
  }
  qsort(arrString, len, sizeof(char *), compareStrings);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <atoi@plt>
  mov    %eax,-0x8(%rbp)
  mov    -0x8(%rbp),%eax
  mov    %eax,%edi
  call   <siguienteNumeroPrimo@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%edx
  mov    -0x8(%rbp),%eax
  mov    %eax,%esi
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: read_data
arguments: label: D3
data_type: string

tool: D3: El siguiente numero primo despues de %d es: %d


assistant: ```c
int main(int argc, char const *argv[]) {
  int primo;
  int anterior;
  anterior = atoi(argv[1]);
  primo = siguienteNumeroPrimo(anterior);
  printf("El siguiente numero primo despues de %d es: %d\n", anterior, primo);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbx
  mov    0x8(%rsi),%rdi
  call   <atoi@plt>
  mov    %eax,%ebx
  mov    %eax,%edi
  call   <siguienteNumeroPrimo@plt>
  mov    %eax,%edx
  mov    %ebx,%esi
  lea    D4(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  pop    %rbx
  ret
```

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

tool: D4: El siguiente numero primo despues de %d es: %d


assistant: ```c
int main(int argc, char const *argv[]) {
  int primo;
  int anterior;
  anterior = atoi(argv[1]);
  primo = siguienteNumeroPrimo(anterior);
  printf("El siguiente numero primo despues de %d es: %d\n", anterior, primo);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  push   %rbp
  mov    0x8(%rsi),%rdi
  call   <atoi@plt>
  mov    %eax,%edi
  mov    %eax,%ebp
  call   <siguienteNumeroPrimo@plt>
  mov    %ebp,%esi
  lea    D2(%rip),%rdi
  mov    %eax,%edx
  call   <printf@plt>
  xor    %eax,%eax
  pop    %rbp
  ret
```

tool_calls: name: unpack_data
arguments: tag: D2
type_guessed: string

tool: D2: El siguiente numero primo despues de %d es: %d


assistant: ```c
int main(int argc, char const *argv[]) {
  int primo;
  int anterior;
  anterior = atoi(argv[1]);
  primo = siguienteNumeroPrimo(anterior);
  printf("El siguiente numero primo despues de %d es: %d\n", anterior, primo);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  push   %rbp
  mov    0x8(%rsi),%rdi
  call   <atoi@plt>
  mov    %eax,%edi
  mov    %eax,%ebp
  call   <siguienteNumeroPrimo@plt>
  mov    %ebp,%esi
  lea    D1(%rip),%rdi
  mov    %eax,%edx
  call   <printf@plt>
  xor    %eax,%eax
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: tag: D1
datatype: string

tool: D1: El siguiente numero primo despues de %d es: %d


assistant: ```c
int main(int argc, char const *argv[]) {
  int primo;
  int anterior;
  anterior = atoi(argv[1]);
  primo = siguienteNumeroPrimo(anterior);
  printf("El siguiente numero primo despues de %d es: %d\n", anterior, primo);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<rc_treeout1>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  mov    %ecx,-0x28(%rbp)
  subq   $0x4,-0x20(%rbp)
  mov    -0x24(%rbp),%eax
  mov    $0x1,%edx
  mov    %eax,%ecx
  shl    %cl,%edx
  mov    %edx,%eax
  or     %eax,-0x28(%rbp)
  mov    -0x24(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  jmp    L5
L16:
  mov    -0x4(%rbp),%eax
  sub    $0x1,%eax
  mov    -0x28(%rbp),%edx
  mov    %eax,%ecx
  shr    %cl,%edx
  mov    %edx,%eax
  and    $0x1,%eax
  mov    %eax,%edx
  mov    -0x4(%rbp),%eax
  mov    -0x28(%rbp),%esi
  mov    %eax,%ecx
  shr    %cl,%esi
  mov    %esi,%eax
  mov    %eax,%eax
  lea    0x0(,%rax,4),%rcx
  mov    -0x20(%rbp),%rax
  add    %rax,%rcx
  mov    -0x18(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <rc_encode1@plt>
  subl   $0x1,-0x4(%rbp)
L5:
  cmpl   $0x0,-0x4(%rbp)
  jg     L16
  nop
  leave
  ret
```

assistant: ```c
void rc_treeout1(struct RC *rc, unsigned int *prb, int len, unsigned int data) {
  int i;
  prb--;
  data |= 1 << len;
  for (i = len; i > 0; i--)
    rc_encode1(rc, &prb[data >> i], (data >> (i - 1)) & 1);
  return;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<rc_treeout1>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r13
  mov    %rsi,%r12
  mov    %edx,%ebx
  bts    %edx,%ecx
  mov    %ecx,%ebp
  test   %edx,%edx
  jle    L11
L17:
  mov    %ebx,%eax
  sub    $0x1,%ebx
  mov    %ebp,%edx
  mov    %ebx,%ecx
  shr    %cl,%edx
  and    $0x1,%edx
  mov    %ebp,%edi
  mov    %eax,%ecx
  shr    %cl,%edi
  lea    -0x4(%r12,%rdi,4),%rsi
  mov    %r13,%rdi
  call   <rc_encode1@plt>
  test   %ebx,%ebx
  jne    L17
L11:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
void rc_treeout1(struct RC *rc, unsigned int *prb, int len, unsigned int data) {
  int i;
  prb--;
  data |= 1 << len;
  for (i = len; i > 0; i--)
    rc_encode1(rc, &prb[data >> i], (data >> (i - 1)) & 1);
  return;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<rc_treeout1>:
  endbr64
  bts    %edx,%ecx
  test   %edx,%edx
  jle    L4
  push   %r13
  mov    %ecx,%r13d
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %edx,%ebx
  sub    $0x8,%rsp
L14:
  mov    %ebx,%eax
  sub    $0x1,%ebx
  mov    %r13d,%edx
  mov    %r13d,%edi
  mov    %ebx,%ecx
  shr    %cl,%edx
  mov    %eax,%ecx
  shr    %cl,%edi
  and    $0x1,%edx
  lea    -0x4(%rbp,%rdi,4),%rsi
  mov    %r12,%rdi
  call   <rc_encode1@plt>
  test   %ebx,%ebx
  jne    L14
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L4:
  ret
```

assistant: ```c
void rc_treeout1(struct RC *rc, unsigned int *prb, int len, unsigned int data) {
  int i;
  prb--;
  data |= 1 << len;
  for (i = len; i > 0; i--)
    rc_encode1(rc, &prb[data >> i], (data >> (i - 1)) & 1);
  return;
}
```
user: Translate the below assembly code into C programming language:

```asm
<rc_treeout1>:
  endbr64
  bts    %edx,%ecx
  test   %edx,%edx
  jle    L2
  push   %r13
  mov    %ecx,%r13d
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %edx,%ebx
  sub    $0x8,%rsp
L5:
  mov    %ebx,%eax
  sub    $0x1,%ebx
  mov    %r13d,%edx
  mov    %r13d,%edi
  mov    %ebx,%ecx
  shr    %cl,%edx
  mov    %eax,%ecx
  shr    %cl,%edi
  and    $0x1,%edx
  lea    -0x4(%rbp,%rdi,4),%rsi
  mov    %r12,%rdi
  call   <rc_encode1@plt>
  test   %ebx,%ebx
  jne    L5
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L2:
  ret
```

assistant: ```c
void rc_treeout1(struct RC *rc, unsigned int *prb, int len, unsigned int data) {
  int i;
  prb--;
  data |= 1 << len;
  for (i = len; i > 0; i--)
    rc_encode1(rc, &prb[data >> i], (data >> (i - 1)) & 1);
  return;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<newton>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  je     L46
  mov    -0x14(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jne    L61
L46:
  mov    $0x1,%eax
  jmp    L48
L61:
  cmpl   $0x0,-0x14(%rbp)
  jle    L48
  cmpl   $0x0,-0x18(%rbp)
  jle    L48
  mov    -0x14(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jle    L48
  mov    -0x18(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    -0x14(%rbp),%eax
  sub    $0x1,%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <newton>
  mov    %eax,%ebx
  mov    -0x14(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    -0x18(%rbp),%eax
  mov    %eax,%esi
  mov    %edx,%edi
  call   <newton>
  add    %ebx,%eax
  jmp    L48
L48:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
newton(int n, int k) {
  if (n == 0 || n == k) {
    return 1;
  }
  if (n > 0 && k > 0 && n > k) {
    return newton(n - 1, k - 1) + newton(n - 1, k);
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<newton>:
  endbr64
  test   %edi,%edi
  je     L24
  push   %r12
  push   %rbp
  push   %rbx
  mov    %esi,%ebp
  cmp    %esi,%edi
  je     L0
  test   %esi,%esi
  mov    $0x0,%eax
  cmovns %esi,%eax
  cmp    %edi,%eax
  jge    L9
  test   %esi,%esi
  jg     L57
L9:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L57:
  lea    -0x1(%rdi),%ebx
  lea    -0x1(%rsi),%esi
  mov    %ebx,%edi
  call   <newton>
  mov    %eax,%r12d
  mov    %ebp,%esi
  mov    %ebx,%edi
  call   <newton>
  add    %r12d,%eax
  jmp    L9
L0:
  mov    $0x1,%eax
  jmp    L9
L24:
  mov    $0x1,%eax
  ret
```

assistant: ```c
newton(int n, int k) {
  if (n == 0 || n == k) {
    return 1;
  }
  if (n > 0 && k > 0 && n > k) {
    return newton(n - 1, k - 1) + newton(n - 1, k);
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<newton>:
  endbr64
  cmp    %esi,%edi
  je     L12
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%ebx
  sub    $0x8,%rsp
  test   %edi,%edi
  je     L56
  xor    %eax,%eax
  test   %esi,%esi
  mov    %esi,%ebp
  cmovns %esi,%eax
  cmp    %edi,%eax
  jge    L62
  test   %esi,%esi
  jle    L62
  lea    -0x1(%rsi),%r13d
  xor    %r12d,%r12d
L0:
  sub    $0x1,%ebx
  mov    %r13d,%esi
  mov    %ebx,%edi
  call   <newton>
  add    %eax,%r12d
  cmp    %ebp,%ebx
  je     L39
  jg     L0
L62:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L39:
  add    $0x8,%rsp
  lea    0x1(%r12),%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L56:
  mov    $0x1,%eax
  jmp    L62
L12:
  mov    $0x1,%eax
  ret
```

assistant: ```c
newton(int n, int k) {
  if (n == 0 || n == k) {
    return 1;
  }
  if (n > 0 && k > 0 && n > k) {
    return newton(n - 1, k - 1) + newton(n - 1, k);
  }
}
```