ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<stream_dflt_stdio_fp>:
  endbr64
  test   %rdi,%rdi
  je     L1
  cmp    $0x1,%rdi
  je     L39
  mov    $0x0,%eax
  cmp    $0x2,%rdi
  je     L37
L4:
  ret
L1:
  mov    D5(%rip),%rax
  mov    (%rax),%rax
  ret
L39:
  mov    D7(%rip),%rax
  mov    (%rax),%rax
  ret
L37:
  mov    D0(%rip),%rax
  mov    (%rax),%rax
  jmp    L4
```

assistant: ```c
FILE *stream_dflt_stdio_fp(void *fileno) {
  if (fileno == ((void *)0))
    return stdin;
  else if (fileno == ((void *)1))
    return stdout;
  else if (fileno == ((void *)2))
    return stderr;
  return NULL;
}
```
user: What is the c source code of the assembly code below?

```asm
<stream_dflt_stdio_fp>:
  endbr64
  test   %rdi,%rdi
  je     L4
  cmp    $0x1,%rdi
  je     L8
  xor    %eax,%eax
  cmp    $0x2,%rdi
  je     L18
  ret
  nopl   0x0(%rax,%rax,1)
L18:
  mov    D12(%rip),%rax
  mov    (%rax),%rax
  ret
  nopl   0x0(%rax,%rax,1)
L4:
  mov    D0(%rip),%rax
  mov    (%rax),%rax
  ret
  nopl   0x0(%rax,%rax,1)
L8:
  mov    D7(%rip),%rax
  mov    (%rax),%rax
  ret
```

assistant: ```c
FILE *stream_dflt_stdio_fp(void *fileno) {
  if (fileno == ((void *)0))
    return stdin;
  else if (fileno == ((void *)1))
    return stdout;
  else if (fileno == ((void *)2))
    return stderr;
  return NULL;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<stream_dflt_stdio_fp>:
  endbr64
  test   %rdi,%rdi
  je     L1
  cmp    $0x1,%rdi
  je     L25
  xor    %eax,%eax
  cmp    $0x2,%rdi
  je     L16
  ret
  nopl   0x0(%rax,%rax,1)
L16:
  mov    D25(%rip),%rax
  mov    (%rax),%rax
  ret
  nopl   0x0(%rax,%rax,1)
L1:
  mov    D17(%rip),%rax
  mov    (%rax),%rax
  ret
  nopl   0x0(%rax,%rax,1)
L25:
  mov    D18(%rip),%rax
  mov    (%rax),%rax
  ret
```

assistant: ```c
FILE *stream_dflt_stdio_fp(void *fileno) {
  if (fileno == ((void *)0))
    return stdin;
  else if (fileno == ((void *)1))
    return stdout;
  else if (fileno == ((void *)2))
    return stderr;
  return NULL;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  movq   $0x0,(%rsp)
  movq   $0x0,0x8(%rsp)
  movq   $0x0,0x10(%rsp)
  movq   $0x0,0x18(%rsp)
  cmp    $0x2,%edi
  jne    L93
  mov    0x8(%rsi),%rsi
  lea    0x18(%rsp),%rdi
  call   <initialize_sequential_reads@plt>
  movq   $0x0,0x20(%rsp)
  lea    D28(%rip),%r13
  lea    D30(%rip),%r12
  lea    D22(%rip),%r14
L50:
  lea    0x10(%rsp),%rbp
  jmp    L18
L93:
  mov    (%rsi),%rsi
  lea    D33(%rip),%rdi
  call   <printf@plt>
  mov    $0xffffffff,%ebx
  jmp    L45
L77:
  call   <free@plt>
  movq   $0x0,0x8(%rsp)
L102:
  mov    0x10(%rsp),%rdi
  test   %rdi,%rdi
  jne    L79
L18:
  lea    0x8(%rsp),%rcx
  mov    %rsp,%rdx
  lea    0x20(%rsp),%rsi
  mov    %rbp,%r8
  mov    0x18(%rsp),%rdi
  call   <read_data_sequential@plt>
  mov    %eax,%ebx
  test   %eax,%eax
  jne    L29
  mov    (%rsp),%rdx
  mov    0x20(%rsp),%rsi
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rsi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x10(%rsp),%rsi
  test   %rsi,%rsi
  je     L8
  mov    %r14,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L8:
  mov    (%rsp),%rdi
  test   %rdi,%rdi
  je     L76
  call   <free@plt>
  movq   $0x0,(%rsp)
L76:
  mov    0x8(%rsp),%rdi
  test   %rdi,%rdi
  jne    L77
  jmp    L102
L79:
  call   <free@plt>
  movq   $0x0,0x10(%rsp)
  jmp    L50
L29:
  mov    0x18(%rsp),%rdi
  call   <close_sequential_reads@plt>
L45:
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L73
  mov    %ebx,%eax
  add    $0x30,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L73:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D28
data_type: string
name: parse_data
arguments: data_tag: D30
data_type: string
name: parse_data
arguments: data_tag: D22
data_type: string
name: parse_data
arguments: data_tag: D33
data_type: string

tool: D28: >%s
%s


tool: D30: %s


tool: D22: extras:%s


tool: D33: Usage %s fbin_file



assistant: ```c
int main(int argc, char *argv[]) {
  char *fasta = NULL;
  char *qual = NULL;
  char *extras = NULL;
  int size = 5000;
  int res = 0;
  struct file_data *filed = NULL;
  if (argc != 2) {
    printf("Usage %s fbin_file\n\n", argv[0]);
    return -1;
  }
  initialize_sequential_reads(&filed, argv[1]);
  char *sname = NULL;
  while ((res = read_data_sequential(filed, &sname, &fasta, &qual, &extras)) == 0) {
    if (res == 0) {
      printf(">%s\n%s\n", sname, fasta);
      printf("%s\n", qual);
      if (extras != NULL)
        printf("extras:%s\n", extras);
    }
    if (fasta != NULL) {
      free(fasta);
      fasta = NULL;
    }
    if (qual != NULL) {
      free(qual);
      qual = NULL;
    }
    if (extras != NULL) {
      free(extras);
      extras = NULL;
    }
  }
  close_sequential_reads(filed);
  return res;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x48,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  movq   $0x0,0x10(%rsp)
  movq   $0x0,0x18(%rsp)
  movq   $0x0,0x20(%rsp)
  movq   $0x0,0x28(%rsp)
  cmp    $0x2,%edi
  jne    L51
  mov    0x8(%rsi),%rsi
  lea    0x28(%rsp),%rdi
  lea    0x20(%rsp),%r13
  lea    0x18(%rsp),%r12
  lea    0x10(%rsp),%rbp
  call   <initialize_sequential_reads@plt>
  lea    0x30(%rsp),%rbx
  movq   $0x0,0x30(%rsp)
  lea    D1(%rip),%r15
  lea    D11(%rip),%r14
  jmp    L67
  nopw   0x0(%rax,%rax,1)
L6:
  mov    0x10(%rsp),%rdx
  mov    0x30(%rsp),%rsi
  mov    %r15,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x18(%rsp),%rsi
  mov    %r14,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x20(%rsp),%rsi
  test   %rsi,%rsi
  je     L31
  lea    D18(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
L31:
  mov    0x10(%rsp),%rdi
  test   %rdi,%rdi
  je     L46
  call   <free@plt>
  movq   $0x0,0x10(%rsp)
L46:
  mov    0x18(%rsp),%rdi
  test   %rdi,%rdi
  je     L86
  call   <free@plt>
  movq   $0x0,0x18(%rsp)
L86:
  mov    0x20(%rsp),%rdi
  test   %rdi,%rdi
  jne    L96
L67:
  mov    0x28(%rsp),%rdi
  mov    %r13,%r8
  mov    %r12,%rcx
  mov    %rbp,%rdx
  mov    %rbx,%rsi
  call   <read_data_sequential@plt>
  test   %eax,%eax
  je     L6
  mov    0x28(%rsp),%rdi
  mov    %eax,0xc(%rsp)
  call   <close_sequential_reads@plt>
  mov    0xc(%rsp),%r8d
L49:
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L48
  add    $0x48,%rsp
  mov    %r8d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax,%rax,1)
L96:
  call   <free@plt>
  movq   $0x0,0x20(%rsp)
  jmp    L67
L51:
  mov    (%rsi),%rsi
  lea    D3(%rip),%rdi
  call   <printf@plt>
  or     $0xffffffff,%r8d
  jmp    L49
L48:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D1
data_type: string
name: disassemble_data
arguments: data_label: D11
data_type: string
name: disassemble_data
arguments: data_label: D18
data_type: string
name: disassemble_data
arguments: data_label: D3
data_type: string

tool: D1: >%s
%s


tool: D11: %s


tool: D18: extras:%s


tool: D3: Usage %s fbin_file



assistant: ```c
int main(int argc, char *argv[]) {
  char *fasta = NULL;
  char *qual = NULL;
  char *extras = NULL;
  int size = 5000;
  int res = 0;
  struct file_data *filed = NULL;
  if (argc != 2) {
    printf("Usage %s fbin_file\n\n", argv[0]);
    return -1;
  }
  initialize_sequential_reads(&filed, argv[1]);
  char *sname = NULL;
  while ((res = read_data_sequential(filed, &sname, &fasta, &qual, &extras)) == 0) {
    if (res == 0) {
      printf(">%s\n%s\n", sname, fasta);
      printf("%s\n", qual);
      if (extras != NULL)
        printf("extras:%s\n", extras);
    }
    if (fasta != NULL) {
      free(fasta);
      fasta = NULL;
    }
    if (qual != NULL) {
      free(qual);
      qual = NULL;
    }
    if (extras != NULL) {
      free(extras);
      extras = NULL;
    }
  }
  close_sequential_reads(filed);
  return res;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<threadqueue_free>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <threadqueue_exit@plt>
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  nop
  leave
  ret
```

assistant: ```c
void threadqueue_free(struct ThreadQueue *tq) {
  threadqueue_exit(tq);
  free(tq);
}
```
user: What is the c source code of the assembly code below?

```asm
<threadqueue_free>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <threadqueue_exit@plt>
  mov    %rbx,%rdi
  call   <free@plt>
  pop    %rbx
  ret
```

assistant: ```c
void threadqueue_free(struct ThreadQueue *tq) {
  threadqueue_exit(tq);
  free(tq);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<threadqueue_free>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  call   <threadqueue_exit@plt>
  mov    %rbp,%rdi
  pop    %rbp
  jmp    1060 <free@plt>
```

assistant: ```c
void threadqueue_free(struct ThreadQueue *tq) {
  threadqueue_exit(tq);
  free(tq);
}
```
user: What is the c source code of the assembly code below?

```asm
<threadqueue_free>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  call   <threadqueue_exit@plt>
  mov    %rbp,%rdi
  pop    %rbp
  jmp    1060 <free@plt>
```

assistant: ```c
void threadqueue_free(struct ThreadQueue *tq) {
  threadqueue_exit(tq);
  free(tq);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<output>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L19
L10:
  mov    -0x18(%rbp),%rdx
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x20(%rbp),%eax
  cltq
  add    %rax,-0x18(%rbp)
  addl   $0x1,-0x4(%rbp)
L19:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L10
  nop
  nop
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D5
data_type: string

tool: D5: %d. %s


assistant: ```c
void output(char *pbase, int n, int s) {
  int i;
  for (i = 0; i < n; i++) {
    printf("%d. %s\n", i, pbase);
    pbase += s;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<output>:
  endbr64
  test   %esi,%esi
  jle    L3
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    %esi,%r12d
  mov    %edx,%r13d
  mov    $0x0,%ebx
  lea    D0(%rip),%r14
L9:
  mov    %rbp,%rdx
  mov    %ebx,%esi
  mov    %r14,%rdi
  call   <printf@plt>
  movslq %r13d,%rax
  add    %rax,%rbp
  add    $0x1,%ebx
  cmp    %ebx,%r12d
  jne    L9
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L3:
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D0
type_guessed: string

tool: D0: %d. %s


assistant: ```c
void output(char *pbase, int n, int s) {
  int i;
  for (i = 0; i < n; i++) {
    printf("%d. %s\n", i, pbase);
    pbase += s;
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<output>:
  endbr64
  test   %esi,%esi
  jle    L10
  push   %r14
  lea    D3(%rip),%r14
  push   %r13
  movslq %edx,%r13
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  nopw   0x0(%rax,%rax,1)
L19:
  mov    %rbp,%rdx
  mov    %ebx,%esi
  mov    %r14,%rdi
  add    $0x1,%ebx
  call   <printf@plt>
  add    %r13,%rbp
  cmp    %ebx,%r12d
  jne    L19
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
L10:
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D3
datatype: string

tool: D3: %d. %s


assistant: ```c
void output(char *pbase, int n, int s) {
  int i;
  for (i = 0; i < n; i++) {
    printf("%d. %s\n", i, pbase);
    pbase += s;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<output>:
  endbr64
  test   %esi,%esi
  jle    L19
  push   %r14
  lea    D2(%rip),%r14
  push   %r13
  movslq %edx,%r13
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  nopw   0x0(%rax,%rax,1)
L16:
  mov    %rbp,%rdx
  mov    %ebx,%esi
  mov    %r14,%rdi
  add    $0x1,%ebx
  call   <printf@plt>
  add    %r13,%rbp
  cmp    %ebx,%r12d
  jne    L16
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
L19:
  ret
```

tool_calls: name: read_data
arguments: label: D2
data_type: string

tool: D2: %d. %s


assistant: ```c
void output(char *pbase, int n, int s) {
  int i;
  for (i = 0; i < n; i++) {
    printf("%d. %s\n", i, pbase);
    pbase += s;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<xwrite>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    -0x30(%rbp),%rax
  mov    %rax,-0x18(%rbp)
  movq   $0x0,-0x10(%rbp)
  jmp    L16
L65:
  mov    -0x38(%rbp),%rax
  sub    -0x10(%rbp),%rax
  mov    %rax,%rdx
  mov    -0x18(%rbp),%rcx
  mov    -0x24(%rbp),%eax
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <write@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jns    L71
  mov    D27(%rip),%rax
  mov    (%rax),%rdx
  mov    D35(%rip),%rax
  mov    (%rax),%rax
  cmp    %rax,%rdx
  je     L16
  mov    D27(%rip),%rax
  mov    (%rax),%rdx
  mov    D36(%rip),%rax
  mov    (%rax),%rax
  cmp    %rax,%rdx
  jne    L71
  jmp    L16
L71:
  cmpq   $0x0,-0x8(%rbp)
  jns    L56
  mov    $0xffffffff,%eax
  jmp    L77
L56:
  mov    -0x8(%rbp),%rax
  add    %rax,-0x10(%rbp)
  mov    -0x8(%rbp),%rax
  add    %rax,-0x18(%rbp)
L16:
  mov    -0x10(%rbp),%rax
  cmp    -0x38(%rbp),%rax
  jb     L65
  mov    -0x10(%rbp),%rax
L77:
  leave
  ret
```

assistant: ```c
int xwrite(int sd, const void *buf, size_t len) {
  char *p = (char *)buf;
  size_t nsent = 0;
  ssize_t rv;
  while (nsent < len) {
    rv = write(sd, p, len - nsent);
    if (0 > rv && (errno == EINTR || errno == EAGAIN))
      continue;
    if (0 > rv)
      return -1;
    nsent += rv;
    p += rv;
  }
  return nsent;
}
```
user: What is the c source code of the assembly code below?

```asm
<xwrite>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdx,%r12
  test   %rdx,%rdx
  je     L37
  mov    %edi,%r13d
  mov    %rsi,%rbp
  mov    $0x0,%ebx
  jmp    L14
L11:
  add    %rax,%rbx
  add    %rax,%rbp
L66:
  cmp    %r12,%rbx
  jae    L54
L14:
  mov    %r12,%rdx
  sub    %rbx,%rdx
  mov    %rbp,%rsi
  mov    %r13d,%edi
  call   <write@plt>
  test   %rax,%rax
  jns    L11
  mov    D1(%rip),%rax
  mov    (%rax),%rax
  mov    D19(%rip),%rdx
  cmp    (%rdx),%rax
  je     L66
  mov    D14(%rip),%rdx
  cmp    (%rdx),%rax
  je     L66
  mov    $0xffffffff,%eax
L47:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L37:
  mov    %rdx,%rbx
L54:
  mov    %ebx,%eax
  jmp    L47
```

assistant: ```c
int xwrite(int sd, const void *buf, size_t len) {
  char *p = (char *)buf;
  size_t nsent = 0;
  ssize_t rv;
  while (nsent < len) {
    rv = write(sd, p, len - nsent);
    if (0 > rv && (errno == EINTR || errno == EAGAIN))
      continue;
    if (0 > rv)
      return -1;
    nsent += rv;
    p += rv;
  }
  return nsent;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<xwrite>:
  endbr64
  test   %rdx,%rdx
  je     L10
  push   %r13
  mov    %rdx,%r13
  push   %r12
  mov    %edi,%r12d
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  jmp    L42
L31:
  add    %rax,%rbx
  add    %rax,%rbp
L53:
  cmp    %r13,%rbx
  jae    L38
L42:
  mov    %r13,%rdx
  mov    %rbp,%rsi
  mov    %r12d,%edi
  sub    %rbx,%rdx
  call   <write@plt>
  test   %rax,%rax
  jns    L31
  mov    D6(%rip),%rax
  mov    D28(%rip),%rdx
  mov    (%rax),%rax
  cmp    (%rdx),%rax
  je     L53
  mov    D18(%rip),%rdx
  cmp    (%rdx),%rax
  je     L53
  add    $0x8,%rsp
  mov    $0xffffffff,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax,%rax,1)
L38:
  add    $0x8,%rsp
  mov    %ebx,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L10:
  xor    %eax,%eax
  ret
```

assistant: ```c
int xwrite(int sd, const void *buf, size_t len) {
  char *p = (char *)buf;
  size_t nsent = 0;
  ssize_t rv;
  while (nsent < len) {
    rv = write(sd, p, len - nsent);
    if (0 > rv && (errno == EINTR || errno == EAGAIN))
      continue;
    if (0 > rv)
      return -1;
    nsent += rv;
    p += rv;
  }
  return nsent;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<xwrite>:
  endbr64
  test   %rdx,%rdx
  je     L10
  push   %r13
  mov    %rdx,%r13
  push   %r12
  mov    %edi,%r12d
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  jmp    L25
L1:
  add    %rax,%rbx
  add    %rax,%rbp
L47:
  cmp    %r13,%rbx
  jae    L13
L25:
  mov    %r13,%rdx
  mov    %rbp,%rsi
  mov    %r12d,%edi
  sub    %rbx,%rdx
  call   <write@plt>
  test   %rax,%rax
  jns    L1
  mov    D4(%rip),%rax
  mov    D5(%rip),%rdx
  mov    (%rax),%rax
  cmp    (%rdx),%rax
  je     L47
  mov    D29(%rip),%rdx
  cmp    (%rdx),%rax
  je     L47
  add    $0x8,%rsp
  mov    $0xffffffff,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax,%rax,1)
L13:
  add    $0x8,%rsp
  mov    %ebx,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L10:
  xor    %eax,%eax
  ret
```

assistant: ```c
int xwrite(int sd, const void *buf, size_t len) {
  char *p = (char *)buf;
  size_t nsent = 0;
  ssize_t rv;
  while (nsent < len) {
    rv = write(sd, p, len - nsent);
    if (0 > rv && (errno == EINTR || errno == EAGAIN))
      continue;
    if (0 > rv)
      return -1;
    nsent += rv;
    p += rv;
  }
  return nsent;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<htol>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  movb   $0x0,-0xb(%rbp)
  movb   $0x10,-0xa(%rbp)
  movq   $0x0,-0x8(%rbp)
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x30,%al
  jne    L17
  mov    -0x18(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  cmp    $0x78,%al
  je     L7
  mov    -0x18(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  cmp    $0x58,%al
  jne    L17
L7:
  movb   $0x2,-0xb(%rbp)
  addb   $0x2,-0xa(%rbp)
  jmp    L17
L11:
  movzbl -0xb(%rbp),%edx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <scandigit@plt>
  mov    %al,-0x9(%rbp)
  cmpb   $0x0,-0x9(%rbp)
  jns    L18
  mov    $0x0,%eax
  jmp    L73
L18:
  shlq   $0x4,-0x8(%rbp)
  movsbq -0x9(%rbp),%rax
  add    %rax,-0x8(%rbp)
  addb   $0x1,-0xb(%rbp)
L17:
  movzbl -0xb(%rbp),%edx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jle    L46
  movzbl -0xb(%rbp),%eax
  cmp    -0xa(%rbp),%al
  jbe    L11
L46:
  mov    -0x8(%rbp),%rax
L73:
  leave
  ret
```

assistant: ```c
unsigned long htol(const char *nptr) {
  unsigned char i = 0;
  unsigned char max = sizeof(long) * 2;
  unsigned long val = 0;
  if (nptr[0] == '0' && (nptr[1] == 'x' || nptr[1] == 'X')) {
    i = 2;
    max += 2;
  }
  for (i; nptr[i] > '\0' && i <= max; ++i) {
    signed char d = scandigit(nptr[i]);
    if (d < 0)
      return 0;
    val *= 0x10;
    val += d;
  }
  return val;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<htol>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  movzbl (%rdi),%edi
  mov    $0x10,%r12d
  mov    $0x0,%ebx
  cmp    $0x30,%dil
  je     L7
L45:
  movzbl %bl,%eax
  movzbl 0x0(%rbp,%rax,1),%edi
  mov    $0x0,%r13d
  test   %dil,%dil
  jle    L54
L51:
  mov    $0x0,%r13d
L31:
  movsbl %dil,%edi
  call   <scandigit@plt>
  test   %al,%al
  js     L56
  shl    $0x4,%r13
  movsbq %al,%rax
  add    %rax,%r13
  add    $0x1,%ebx
  movzbl %bl,%eax
  movzbl 0x0(%rbp,%rax,1),%edi
  cmp    %r12b,%bl
  ja     L54
  test   %dil,%dil
  jg     L31
  jmp    L54
L7:
  movzbl 0x1(%rbp),%eax
  and    $0xffffffdf,%eax
  cmp    $0x58,%al
  jne    L51
  mov    $0x12,%r12d
  mov    $0x2,%ebx
  jmp    L45
L56:
  mov    $0x0,%r13d
L54:
  mov    %r13,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
unsigned long htol(const char *nptr) {
  unsigned char i = 0;
  unsigned char max = sizeof(long) * 2;
  unsigned long val = 0;
  if (nptr[0] == '0' && (nptr[1] == 'x' || nptr[1] == 'X')) {
    i = 2;
    max += 2;
  }
  for (i; nptr[i] > '\0' && i <= max; ++i) {
    signed char d = scandigit(nptr[i]);
    if (d < 0)
      return 0;
    val *= 0x10;
    val += d;
  }
  return val;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<htol>:
  endbr64
  push   %r13
  xor    %r13d,%r13d
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    $0x10,%ebx
  sub    $0x8,%rsp
  movsbl (%rdi),%edi
  cmp    $0x30,%dil
  je     L58
L27:
  test   %dil,%dil
  jle    L32
L64:
  lea    0x1(%r13),%eax
  xor    %r12d,%r12d
  movzbl %al,%eax
  add    %rax,%rbp
  jmp    L4
  cs nopw 0x0(%rax,%rax,1)
L17:
  shl    $0x4,%r12
  movsbq %al,%rax
  add    $0x1,%r13d
  movsbl 0x0(%rbp),%edi
  add    %rax,%r12
  add    $0x1,%rbp
  cmp    %bl,%r13b
  ja     L46
  test   %dil,%dil
  jle    L46
L4:
  call   <scandigit@plt>
  test   %al,%al
  jns    L17
L32:
  xor    %r12d,%r12d
L46:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax,%rax,1)
L58:
  movzbl 0x1(%rbp),%eax
  and    $0xffffffdf,%eax
  cmp    $0x58,%al
  jne    L64
  movsbl 0x2(%rbp),%edi
  mov    $0x12,%ebx
  mov    $0x2,%r13d
  jmp    L27
```

assistant: ```c
unsigned long htol(const char *nptr) {
  unsigned char i = 0;
  unsigned char max = sizeof(long) * 2;
  unsigned long val = 0;
  if (nptr[0] == '0' && (nptr[1] == 'x' || nptr[1] == 'X')) {
    i = 2;
    max += 2;
  }
  for (i; nptr[i] > '\0' && i <= max; ++i) {
    signed char d = scandigit(nptr[i]);
    if (d < 0)
      return 0;
    val *= 0x10;
    val += d;
  }
  return val;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<htol>:
  endbr64
  push   %r13
  xor    %r13d,%r13d
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    $0x10,%ebx
  sub    $0x8,%rsp
  movsbl (%rdi),%edi
  cmp    $0x30,%dil
  je     L36
L5:
  test   %dil,%dil
  jle    L65
L9:
  lea    0x1(%r13),%eax
  xor    %r12d,%r12d
  movzbl %al,%eax
  add    %rax,%rbp
  jmp    L61
  cs nopw 0x0(%rax,%rax,1)
L51:
  shl    $0x4,%r12
  movsbq %al,%rax
  add    $0x1,%r13d
  movsbl 0x0(%rbp),%edi
  add    %rax,%r12
  add    $0x1,%rbp
  cmp    %bl,%r13b
  ja     L55
  test   %dil,%dil
  jle    L55
L61:
  call   <scandigit@plt>
  test   %al,%al
  jns    L51
L65:
  xor    %r12d,%r12d
L55:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax,%rax,1)
L36:
  movzbl 0x1(%rbp),%eax
  and    $0xffffffdf,%eax
  cmp    $0x58,%al
  jne    L9
  movsbl 0x2(%rbp),%edi
  mov    $0x12,%ebx
  mov    $0x2,%r13d
  jmp    L5
```

assistant: ```c
unsigned long htol(const char *nptr) {
  unsigned char i = 0;
  unsigned char max = sizeof(long) * 2;
  unsigned long val = 0;
  if (nptr[0] == '0' && (nptr[1] == 'x' || nptr[1] == 'X')) {
    i = 2;
    max += 2;
  }
  for (i; nptr[i] > '\0' && i <= max; ++i) {
    signed char d = scandigit(nptr[i]);
    if (d < 0)
      return 0;
    val *= 0x10;
    val += d;
  }
  return val;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<strlen>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movq   $0x0,-0x8(%rbp)
  jmp    L19
L0:
  addq   $0x1,-0x8(%rbp)
  addq   $0x1,-0x18(%rbp)
L19:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L0
  mov    -0x8(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
size_t strlen(const char *str) {
  size_t answer = 0;
  while (*str) {
    ++answer;
    ++str;
  }
  return answer;
}
```