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   %r13
  xor    %eax,%eax
  push   %r12
  push   %rbp
  call   <getppid@plt>
  mov    %eax,%r12d
  xor    %eax,%eax
  call   <getpid@plt>
  mov    %r12d,%edx
  lea    D30(%rip),%r12
  mov    %eax,%esi
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  call   <fork@plt>
  test   %rax,%rax
  js     L16
  je     L28
  xor    %edi,%edi
  lea    D55(%rip),%rbp
  call   <wait@plt>
L34:
  xor    %eax,%eax
  call   <getppid@plt>
  mov    %eax,%r12d
  xor    %eax,%eax
  call   <getpid@plt>
  mov    %r12d,%edx
  mov    %rbp,%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
L21:
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L28:
  xor    %eax,%eax
  lea    D55(%rip),%rbp
  call   <getppid@plt>
  mov    %eax,%r13d
  xor    %eax,%eax
  call   <getpid@plt>
  mov    %r13d,%edx
  mov    %r12,%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %edx,%edx
  lea    D23(%rip),%rsi
  lea    D13(%rip),%rdi
  call   <execl@plt>
  xor    %eax,%eax
  call   <getppid@plt>
  mov    %eax,%r12d
  xor    %eax,%eax
  call   <getpid@plt>
  mov    %r12d,%edx
  mov    %rbp,%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L34
L16:
  lea    D31(%rip),%rdi
  call   <puts@plt>
  mov    $0x1,%eax
  jmp    L21
```

tool_calls: name: parse_data
arguments: label: D30
data_type: string
name: parse_data
arguments: label: D55
data_type: string
name: parse_data
arguments: label: D23
data_type: string
name: parse_data
arguments: label: D13
data_type: string
name: parse_data
arguments: label: D31
data_type: string

tool: D30: Process %d, child of process %d started...


tool: D55: Process %d, child of process %d stopped...


tool: D23: top

tool: D13: /usr/bin/top

tool: D31: Failed fork()

assistant: ```c
int main(int argc, char *argv[]) {
  printf("Process %d, child of process %d started...\n", getpid(), getppid());
  pid_t pid = fork();
  if (pid < 0) {
    printf("Failed fork()\n");
    return 1;
  } else if (pid == 0) {
    printf("Process %d, child of process %d started...\n", getpid(), getppid());
    execl("/usr/bin/top", "top", NULL);
    printf("Process %d, child of process %d stopped...\n", getpid(), getppid());
  } else {
    wait(NULL);
  }
  printf("Process %d, child of process %d stopped...\n", getpid(), getppid());
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<feed_tty_init_cb>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rdx
  mov    -0x20(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <feed_tty_init@plt>
  mov    %al,-0x9(%rbp)
  movzbl -0x9(%rbp),%eax
  leave
  ret
```

assistant: ```c
char feed_tty_init_cb(void *const p_object, struct feed_client *const p_client, void const *const p_descriptor) {
  char b_result;
  struct feed_tty *const p_tty = (struct feed_tty *)(p_object);
  (void)(p_descriptor);
  b_result = feed_tty_init(p_client, p_tty);
  return b_result;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<feed_tty_init_cb>:
  endbr64
  sub    $0x8,%rsp
  mov    %rdi,%rax
  mov    %rsi,%rdi
  mov    %rax,%rsi
  call   <feed_tty_init@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
char feed_tty_init_cb(void *const p_object, struct feed_client *const p_client, void const *const p_descriptor) {
  char b_result;
  struct feed_tty *const p_tty = (struct feed_tty *)(p_object);
  (void)(p_descriptor);
  b_result = feed_tty_init(p_client, p_tty);
  return b_result;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<feed_tty_init_cb>:
  endbr64
  mov    %rdi,%r8
  mov    %rsi,%rdi
  mov    %r8,%rsi
  jmp    1050 <feed_tty_init@plt>
```

assistant: ```c
char feed_tty_init_cb(void *const p_object, struct feed_client *const p_client, void const *const p_descriptor) {
  char b_result;
  struct feed_tty *const p_tty = (struct feed_tty *)(p_object);
  (void)(p_descriptor);
  b_result = feed_tty_init(p_client, p_tty);
  return b_result;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<feed_tty_init_cb>:
  endbr64
  mov    %rdi,%r8
  mov    %rsi,%rdi
  mov    %r8,%rsi
  jmp    1050 <feed_tty_init@plt>
```

assistant: ```c
char feed_tty_init_cb(void *const p_object, struct feed_client *const p_client, void const *const p_descriptor) {
  char b_result;
  struct feed_tty *const p_tty = (struct feed_tty *)(p_object);
  (void)(p_descriptor);
  b_result = feed_tty_init(p_client, p_tty);
  return b_result;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<next_permutaion>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    -0x1c(%rbp),%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x8(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x10(%rbp)
  jmp    L36
L63:
  subl   $0x1,-0x10(%rbp)
L36:
  cmpl   $0x0,-0x10(%rbp)
  jle    L21
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x10(%rbp),%eax
  cltq
  shl    $0x2,%rax
  lea    -0x4(%rax),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%eax
  cmp    %eax,%edx
  jl     L63
L21:
  cmpl   $0x0,-0x10(%rbp)
  jne    L39
  mov    $0x0,%eax
  jmp    L1
L39:
  mov    -0x8(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0xc(%rbp)
  jmp    L58
L76:
  subl   $0x1,-0xc(%rbp)
L58:
  mov    -0xc(%rbp),%eax
  cmp    -0x10(%rbp),%eax
  jle    L28
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x10(%rbp),%eax
  cltq
  shl    $0x2,%rax
  lea    -0x4(%rax),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%eax
  cmp    %eax,%edx
  jl     L76
L28:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x10(%rbp),%eax
  cltq
  shl    $0x2,%rax
  lea    -0x4(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0xc(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  mov    -0x10(%rbp),%eax
  cltq
  shl    $0x2,%rax
  lea    -0x4(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  mov    %eax,(%rdx)
  mov    -0x8(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0xc(%rbp)
  jmp    L71
L26:
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x10(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  mov    %eax,(%rdx)
  subl   $0x1,-0xc(%rbp)
  addl   $0x1,-0x10(%rbp)
L71:
  mov    -0x10(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jl     L26
  mov    $0x1,%eax
L1:
  pop    %rbp
  ret
```

assistant: ```c
int next_permutaion(int *list, int lsize) {
  int size = lsize;
  int i, j;
  int temp;
  for (i = size - 1; (i > 0) && (list[i] < list[i - 1]); --i)
    ;
  if (0 == i)
    return 0;
  for (j = size - 1; (j > i) && (list[j] < list[i - 1]); --j)
    ;
  temp = list[j];
  list[j] = list[i - 1];
  list[i - 1] = temp;
  for (j = size - 1; i < j; --j, ++i) {
    temp = list[i];
    list[i] = list[j];
    list[j] = temp;
  }
  return 1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<next_permutaion>:
  endbr64
  mov    %rdi,%rcx
  sub    $0x1,%esi
  test   %esi,%esi
  jle    L63
  movslq %esi,%rax
  lea    (%rdi,%rax,4),%rax
  mov    %esi,%edx
L73:
  mov    -0x4(%rax),%edi
  cmp    %edi,(%rax)
  jge    L83
  sub    $0x4,%rax
  sub    $0x1,%edx
  jne    L73
  jmp    L35
L83:
  test   %edx,%edx
  je     L35
  cmp    %edx,%esi
  jle    L107
  movslq %edx,%rdi
  lea    0x0(,%rdi,4),%r10
  mov    -0x4(%rcx,%r10,1),%r9d
  movslq %esi,%rsi
  mov    %rsi,%rax
L110:
  mov    %eax,%r8d
  cmp    %r9d,(%rcx,%rax,4)
  jge    L80
  lea    -0x1(%rax),%r8d
  sub    $0x1,%rax
  cmp    %eax,%edx
  jl     L110
L80:
  movslq %r8d,%r8
  lea    (%rcx,%r8,4),%rdx
  mov    (%rdx),%r8d
  lea    -0x4(%rcx,%r10,1),%rax
  mov    (%rax),%r9d
  mov    %r9d,(%rdx)
  mov    %r8d,(%rax)
L23:
  mov    (%rcx,%rdi,4),%eax
  mov    (%rcx,%rsi,4),%edx
  mov    %edx,(%rcx,%rdi,4)
  mov    %eax,(%rcx,%rsi,4)
  add    $0x1,%rdi
  sub    $0x1,%rsi
  cmp    %esi,%edi
  jl     L23
  mov    $0x1,%edx
L35:
  mov    %edx,%eax
  ret
L16:
  mov    %esi,%edx
  jmp    L35
L63:
  mov    %esi,%edx
  je     L16
L107:
  movslq %esi,%rsi
  lea    (%rcx,%rsi,4),%rsi
  mov    (%rsi),%edi
  movslq %edx,%rdx
  lea    -0x4(%rcx,%rdx,4),%rax
  mov    (%rax),%edx
  mov    %edx,(%rsi)
  mov    %edi,(%rax)
  mov    $0x1,%edx
  jmp    L35
```

assistant: ```c
int next_permutaion(int *list, int lsize) {
  int size = lsize;
  int i, j;
  int temp;
  for (i = size - 1; (i > 0) && (list[i] < list[i - 1]); --i)
    ;
  if (0 == i)
    return 0;
  for (j = size - 1; (j > i) && (list[j] < list[i - 1]); --j)
    ;
  temp = list[j];
  list[j] = list[i - 1];
  list[i - 1] = temp;
  for (j = size - 1; i < j; --j, ++i) {
    temp = list[i];
    list[i] = list[j];
    list[j] = temp;
  }
  return 1;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<next_permutaion>:
  endbr64
  sub    $0x1,%esi
  mov    %rdi,%r8
  test   %esi,%esi
  jle    L38
  movslq %esi,%rdi
  mov    %esi,%eax
  lea    -0x4(%r8,%rdi,4),%rdx
  jmp    L90
  xchg   %ax,%ax
L3:
  sub    $0x4,%rdx
  sub    $0x1,%eax
  je     L4
L90:
  mov    (%rdx),%r9d
  mov    %rdx,%r10
  cmp    %r9d,0x4(%rdx)
  jl     L3
  lea    (%r8,%rdi,4),%rcx
  mov    (%rcx),%r11d
  cmp    %eax,%esi
  jg     L58
  jmp    L49
  nopw   0x0(%rax,%rax,1)
L40:
  sub    $0x1,%esi
  sub    $0x4,%rcx
  cmp    %eax,%esi
  je     L99
L58:
  mov    (%rcx),%r10d
  cmp    %r10d,%r9d
  jg     L40
  cltq
L80:
  mov    %r9d,(%rcx)
  mov    %r10d,(%rdx)
  nopl   0x0(%rax,%rax,1)
L34:
  mov    (%r8,%rax,4),%edx
  mov    (%r8,%rdi,4),%ecx
  mov    %ecx,(%r8,%rax,4)
  add    $0x1,%rax
  mov    %edx,(%r8,%rdi,4)
  sub    $0x1,%rdi
  cmp    %edi,%eax
  jl     L34
L118:
  mov    $0x1,%eax
L4:
  ret
  nopw   0x0(%rax,%rax,1)
L99:
  movslq %esi,%rax
  lea    (%r8,%rax,4),%rcx
  mov    (%rcx),%r10d
  jmp    L80
L38:
  je     L27
  movslq %esi,%rsi
  shl    $0x2,%rsi
  lea    -0x4(%rdi,%rsi,1),%r10
  lea    (%rdi,%rsi,1),%rcx
  mov    (%r10),%r9d
  mov    (%rcx),%r11d
L49:
  mov    %r9d,(%rcx)
  mov    %r11d,(%r10)
  jmp    L118
L27:
  xor    %eax,%eax
  ret
```

assistant: ```c
int next_permutaion(int *list, int lsize) {
  int size = lsize;
  int i, j;
  int temp;
  for (i = size - 1; (i > 0) && (list[i] < list[i - 1]); --i)
    ;
  if (0 == i)
    return 0;
  for (j = size - 1; (j > i) && (list[j] < list[i - 1]); --j)
    ;
  temp = list[j];
  list[j] = list[i - 1];
  list[i - 1] = temp;
  for (j = size - 1; i < j; --j, ++i) {
    temp = list[i];
    list[i] = list[j];
    list[j] = temp;
  }
  return 1;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<puthex>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    $0x30,%edi
  call   <putc@plt>
  mov    $0x78,%edi
  call   <putc@plt>
  movl   $0x0,-0x4(%rbp)
  jmp    L57
L12:
  mov    $0x7,%eax
  sub    -0x4(%rbp),%eax
  shl    $0x2,%eax
  mov    -0x18(%rbp),%rdx
  mov    %eax,%ecx
  shr    %cl,%rdx
  mov    %rdx,%rax
  and    $0xf,%eax
  mov    %al,-0x5(%rbp)
  cmpb   $0x9,-0x5(%rbp)
  ja     L58
  addb   $0x30,-0x5(%rbp)
  jmp    L11
L58:
  cmpb   $0x9,-0x5(%rbp)
  jbe    L11
  cmpb   $0xf,-0x5(%rbp)
  ja     L11
  addb   $0x37,-0x5(%rbp)
L11:
  movzbl -0x5(%rbp),%eax
  mov    %eax,%edi
  call   <putc@plt>
  addl   $0x1,-0x4(%rbp)
L57:
  cmpl   $0x7,-0x4(%rbp)
  jle    L12
  nop
  nop
  leave
  ret
```

assistant: ```c
void puthex(unsigned long val) {
  unsigned char c;
  int i;
  putc('0');
  putc('x');
  for (i = 0; i < 8; i++) {
    c = (val >> ((7 - i) * 4)) & 0xF;
    if ((c >= 0) && (c <= 9)) {
      c = '0' + c;
    } else if ((c >= 0xA) && (c <= 0xF)) {
      c = 'A' + (c - 0xA);
    }
    putc(c);
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<puthex>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    $0x30,%edi
  call   <putc@plt>
  mov    $0x78,%edi
  call   <putc@plt>
  mov    $0x1c,%ebx
L3:
  mov    %rbp,%rax
  mov    %ebx,%ecx
  shr    %cl,%rax
  and    $0xf,%eax
  lea    0x30(%rax),%edx
  lea    0x37(%rax),%edi
  cmp    $0x9,%al
  cmovbe %edx,%edi
  movzbl %dil,%edi
  call   <putc@plt>
  sub    $0x4,%ebx
  cmp    $0xfffffffc,%ebx
  jne    L3
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void puthex(unsigned long val) {
  unsigned char c;
  int i;
  putc('0');
  putc('x');
  for (i = 0; i < 8; i++) {
    c = (val >> ((7 - i) * 4)) & 0xF;
    if ((c >= 0) && (c <= 9)) {
      c = '0' + c;
    } else if ((c >= 0xA) && (c <= 0xF)) {
      c = 'A' + (c - 0xA);
    }
    putc(c);
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<puthex>:
  endbr64
  push   %rbp
  mov    $0x1c,%ebp
  push   %rbx
  mov    %rdi,%rbx
  mov    $0x30,%edi
  sub    $0x8,%rsp
  call   <putc@plt>
  mov    $0x78,%edi
  call   <putc@plt>
  cs nopw 0x0(%rax,%rax,1)
L0:
  mov    %ebp,%ecx
  mov    %rbx,%rax
  shr    %cl,%rax
  and    $0xf,%eax
  cmp    $0x9,%al
  lea    0x30(%rax),%edx
  lea    0x37(%rax),%edi
  cmovbe %edx,%edi
  sub    $0x4,%ebp
  movzbl %dil,%edi
  call   <putc@plt>
  cmp    $0xfffffffc,%ebp
  jne    L0
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void puthex(unsigned long val) {
  unsigned char c;
  int i;
  putc('0');
  putc('x');
  for (i = 0; i < 8; i++) {
    c = (val >> ((7 - i) * 4)) & 0xF;
    if ((c >= 0) && (c <= 9)) {
      c = '0' + c;
    } else if ((c >= 0xA) && (c <= 0xF)) {
      c = 'A' + (c - 0xA);
    }
    putc(c);
  }
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<puthex>:
  endbr64
  push   %rbp
  mov    $0x1c,%ebp
  push   %rbx
  mov    %rdi,%rbx
  mov    $0x30,%edi
  sub    $0x8,%rsp
  call   <putc@plt>
  mov    $0x78,%edi
  call   <putc@plt>
  cs nopw 0x0(%rax,%rax,1)
L1:
  mov    %ebp,%ecx
  mov    %rbx,%rax
  shr    %cl,%rax
  and    $0xf,%eax
  cmp    $0x9,%al
  lea    0x30(%rax),%edx
  lea    0x37(%rax),%edi
  cmovbe %edx,%edi
  sub    $0x4,%ebp
  movzbl %dil,%edi
  call   <putc@plt>
  cmp    $0xfffffffc,%ebp
  jne    L1
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void puthex(unsigned long val) {
  unsigned char c;
  int i;
  putc('0');
  putc('x');
  for (i = 0; i < 8; i++) {
    c = (val >> ((7 - i) * 4)) & 0xF;
    if ((c >= 0) && (c <= 9)) {
      c = '0' + c;
    } else if ((c >= 0xA) && (c <= 0xF)) {
      c = 'A' + (c - 0xA);
    }
    putc(c);
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<base10len>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L11
L18:
  mov    -0x14(%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,-0x14(%rbp)
  addl   $0x1,-0x4(%rbp)
L11:
  cmpl   $0x0,-0x14(%rbp)
  jne    L18
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int base10len(int n) {
  int len = 0;
  while (n) {
    n /= 10;
    ++len;
  }
  return len;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<base10len>:
  endbr64
  test   %edi,%edi
  je     L2
  mov    $0x0,%edx
L22:
  movslq %edi,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  sar    $0x1f,%edi
  sub    %edi,%eax
  mov    %eax,%edi
  add    $0x1,%edx
  test   %eax,%eax
  jne    L22
L25:
  mov    %edx,%eax
  ret
L2:
  mov    %edi,%edx
  jmp    L25
```

assistant: ```c
int base10len(int n) {
  int len = 0;
  while (n) {
    n /= 10;
    ++len;
  }
  return len;
}
```
user: Translate the below assembly code into C programming language:

```asm
<base10len>:
  endbr64
  xor    %r8d,%r8d
  test   %edi,%edi
  je     L17
  nopl   0x0(%rax,%rax,1)
L18:
  movslq %edi,%rax
  mov    %edi,%edx
  add    $0x1,%r8d
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%edx
  sar    $0x22,%rax
  sub    %edx,%eax
  mov    %eax,%edi
  jne    L18
L17:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int base10len(int n) {
  int len = 0;
  while (n) {
    n /= 10;
    ++len;
  }
  return len;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<base10len>:
  endbr64
  xor    %r8d,%r8d
  test   %edi,%edi
  je     L13
  nopl   0x0(%rax,%rax,1)
L6:
  movslq %edi,%rax
  mov    %edi,%edx
  add    $0x1,%r8d
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%edx
  sar    $0x22,%rax
  sub    %edx,%eax
  mov    %eax,%edi
  jne    L6
L13:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int base10len(int n) {
  int len = 0;
  while (n) {
    n /= 10;
    ++len;
  }
  return len;
}
```
user: What is the c source code of the assembly code below:

```asm
<get_ram_size>:
  endbr64
  push   %rbx
  sub    $0x110,%rsp
  mov    %rsi,%r10
  mov    %fs:0x28,%rax
  mov    %rax,0x108(%rsp)
  xor    %eax,%eax
  mov    %rsi,%r8
  shr    $0x3,%r8
  mov    %rsi,%rax
  shr    $0x4,%rax
  je     L10
  mov    $0x1,%edx
L104:
  lea    (%rdi,%rax,8),%rsi
  mov    %edx,%ecx
  mov    (%rsi),%r9
  mov    %r9,-0x8(%rsp,%rdx,8)
  mov    %rax,%r9
  not    %r9
  mov    %r9,(%rsi)
  sar    %rax
  add    $0x1,%rdx
  test   %rax,%rax
  jg     L104
L29:
  mov    (%rdi),%rax
  movslq %ecx,%rdx
  mov    %rax,(%rsp,%rdx,8)
  movq   $0x0,(%rdi)
  mov    (%rdi),%rdx
  test   %rdx,%rdx
  jne    L32
  mov    %r10,%rax
  cmp    $0xf,%r10
  jbe    L82
  sub    $0x1,%ecx
  movslq %ecx,%rcx
  mov    $0x1,%edx
  mov    $0x1,%eax
L45:
  shl    $0x3,%rax
  lea    (%rdi,%rax,1),%rsi
  mov    (%rsi),%r9
  mov    (%rsp,%rcx,8),%r11
  mov    %r11,(%rsi)
  mov    %rdx,%rsi
  not    %rsi
  cmp    %r9,%rsi
  jne    L103
  add    %rdx,%rdx
  mov    %rdx,%rax
  sub    $0x1,%rcx
  cmp    %rdx,%r8
  ja     L45
  mov    %r10,%rax
L82:
  mov    0x108(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L112
  add    $0x110,%rsp
  pop    %rbx
  ret
L10:
  mov    $0x0,%ecx
  jmp    L29
L32:
  mov    %rax,(%rdi)
  cmp    $0xf,%r10
  jbe    L2
  movslq %ecx,%rcx
  lea    (%rsp,%rcx,8),%rdx
  mov    $0x1,%eax
  mov    $0x1,%ecx
L44:
  lea    (%rdi,%rcx,8),%rcx
  mov    -0x8(%rdx),%rsi
  mov    %rsi,(%rcx)
  add    %rax,%rax
  mov    %rax,%rcx
  sub    $0x8,%rdx
  cmp    %rax,%r8
  ja     L44
  mov    $0x0,%eax
  jmp    L82
L103:
  add    %rdx,%rdx
  mov    %rdx,%rsi
  cmp    %rdx,%r8
  jbe    L82
  movslq %ecx,%rbx
  lea    (%rsp,%rbx,8),%rcx
L68:
  lea    (%rdi,%rsi,8),%rsi
  mov    -0x8(%rcx),%r9
  mov    %r9,(%rsi)
  add    %rdx,%rdx
  mov    %rdx,%rsi
  sub    $0x8,%rcx
  cmp    %rdx,%r8
  ja     L68
  jmp    L82
L2:
  mov    $0x0,%eax
  jmp    L82
L112:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
long get_ram_size(volatile long *base, long maxsize) {
  volatile long *addr;
  long save[32];
  long cnt;
  long val;
  long size;
  int i = 0;
  for (cnt = (maxsize / sizeof(long)) >> 1; cnt > 0; cnt >>= 1) {
    addr = base + cnt;
    save[i++] = *addr;
    *addr = ~cnt;
  }
  addr = base;
  save[i] = *addr;
  *addr = 0;
  if ((val = *addr) != 0) {
    *addr = save[i];
    for (cnt = 1; cnt < maxsize / sizeof(long); cnt <<= 1) {
      addr = base + cnt;
      *addr = save[--i];
    }
    return (0);
  }
  for (cnt = 1; cnt < maxsize / sizeof(long); cnt <<= 1) {
    addr = base + cnt;
    val = *addr;
    *addr = save[--i];
    if (val != ~cnt) {
      size = cnt * sizeof(long);
      for (cnt <<= 1; cnt < maxsize / sizeof(long); cnt <<= 1) {
        addr = base + cnt;
        *addr = save[--i];
      }
      return (size);
    }
  }
  return (maxsize);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<get_ram_size>:
  endbr64
  push   %rbp
  mov    %rsi,%r11
  mov    %rsi,%rax
  push   %rbx
  shr    $0x3,%r11
  sub    $0x118,%rsp
  mov    %fs:0x28,%rdx
  mov    %rdx,0x108(%rsp)
  xor    %edx,%edx
  mov    %rsi,%rdx
  shr    $0x4,%rdx
  je     L15
  mov    $0x1,%esi
  mov    %rsp,%r10
  nop
L37:
  lea    (%rdi,%rdx,8),%r8
  mov    %esi,%ecx
  mov    (%r8),%r9
  mov    %r9,-0x8(%r10,%rsi,8)
  mov    %rdx,%r9
  add    $0x1,%rsi
  sar    %rdx
  not    %r9
  mov    %r9,(%r8)
  jne    L37
L32:
  mov    (%rdi),%rdx
  movq   $0x0,(%rdi)
  movslq %ecx,%rsi
  mov    (%rdi),%r8
  mov    %rdx,(%rsp,%rsi,8)
  test   %r8,%r8
  jne    L0
  cmp    $0x1,%r11
  jbe    L104
  sub    $0x1,%ecx
  mov    $0x1,%edx
  mov    $0x1,%esi
  mov    %rsp,%r10
  movslq %ecx,%rcx
  jmp    L58
  nopl   0x0(%rax,%rax,1)
L44:
  sub    $0x1,%rcx
  cmp    %rdx,%r11
  jbe    L104
L58:
  mov    (%r10,%rcx,8),%rbx
  lea    0x0(,%rsi,8),%r8
  movslq %ecx,%rbp
  lea    (%rdi,%r8,1),%rsi
  mov    (%rsi),%r9
  mov    %rbx,(%rsi)
  mov    %rdx,%rbx
  add    %rdx,%rdx
  not    %rbx
  mov    %rdx,%rsi
  cmp    %r9,%rbx
  je     L44
  mov    %r8,%rax
  cmp    %rdx,%r11
  jbe    L104
  lea    (%r10,%rbp,8),%rcx
  nop
L47:
  mov    -0x8(%rcx),%r8
  lea    (%rdi,%rdx,8),%rsi
  add    %rdx,%rdx
  sub    $0x8,%rcx
  mov    %r8,(%rsi)
  cmp    %rdx,%r11
  ja     L47
L104:
  mov    0x108(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L50
  add    $0x118,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax)
L0:
  mov    %rdx,(%rdi)
  cmp    $0x1,%r11
  jbe    L2
  lea    (%rsp,%rsi,8),%rdx
  mov    $0x1,%eax
  mov    $0x1,%ecx
  nop
L10:
  mov    -0x8(%rdx),%rsi
  lea    (%rdi,%rcx,8),%rcx
  add    %rax,%rax
  sub    $0x8,%rdx
  mov    %rsi,(%rcx)
  mov    %rax,%rcx
  cmp    %rax,%r11
  ja     L10
L2:
  xor    %eax,%eax
  jmp    L104
  xchg   %ax,%ax
L15:
  xor    %ecx,%ecx
  jmp    L32
L50:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
long get_ram_size(volatile long *base, long maxsize) {
  volatile long *addr;
  long save[32];
  long cnt;
  long val;
  long size;
  int i = 0;
  for (cnt = (maxsize / sizeof(long)) >> 1; cnt > 0; cnt >>= 1) {
    addr = base + cnt;
    save[i++] = *addr;
    *addr = ~cnt;
  }
  addr = base;
  save[i] = *addr;
  *addr = 0;
  if ((val = *addr) != 0) {
    *addr = save[i];
    for (cnt = 1; cnt < maxsize / sizeof(long); cnt <<= 1) {
      addr = base + cnt;
      *addr = save[--i];
    }
    return (0);
  }
  for (cnt = 1; cnt < maxsize / sizeof(long); cnt <<= 1) {
    addr = base + cnt;
    val = *addr;
    *addr = save[--i];
    if (val != ~cnt) {
      size = cnt * sizeof(long);
      for (cnt <<= 1; cnt < maxsize / sizeof(long); cnt <<= 1) {
        addr = base + cnt;
        *addr = save[--i];
      }
      return (size);
    }
  }
  return (maxsize);
}
```