ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D16(%rip),%r13
  lea    0x4(%rsp),%r12
  lea    D22(%rip),%rbp
  lea    D19(%rip),%rbx
L39:
  mov    %r13,%rdi
  call   <puts@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    0x4(%rsp),%esi
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x4(%rsp),%edi
  test   %edi,%edi
  jle    L20
L14:
  mov    %edi,%eax
  shr    $0x1f,%eax
  add    %eax,%edi
  and    $0x1,%edi
  sub    %eax,%edi
  call   <putchar@plt>
  mov    0x4(%rsp),%edx
  mov    %edx,%eax
  shr    $0x1f,%eax
  add    %edx,%eax
  sar    %eax
  mov    %eax,%edi
  mov    %eax,0x4(%rsp)
  cmp    $0x1,%edx
  jg     L14
L20:
  test   %edi,%edi
  jne    L39
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D16
data_type: string
name: unpack_data
arguments: tag: D22
data_type: string
name: unpack_data
arguments: tag: D19
data_type: string

tool: D16: Dwse neo arithmo

tool: D22: %d

tool: D19: H antistrofi dyadiki anaparastasi tou %d einai: 

assistant: ```c
int main() {
  int n, d;
  const a = 2;
  do {
    printf("Dwse neo arithmo\n");
    scanf("%d", &n);
    d = getchar();
    printf("H antistrofi dyadiki anaparastasi tou %d einai: ", n);
    for (n = n; n > 0; n = n / a) {
      d = n % a;
      n > 0 ? putchar(d) : printf("Lathos epilogi\n");
    }
  } while (n != 0);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r13
  lea    D0(%rip),%r13
  push   %r12
  push   %rbp
  lea    D28(%rip),%rbp
  push   %rbx
  lea    D17(%rip),%rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%r12
  nopl   0x0(%rax,%rax,1)
L24:
  mov    %r13,%rdi
  call   <puts@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  xor    %eax,%eax
  call   <getchar@plt>
  mov    0x4(%rsp),%esi
  mov    %rbx,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x4(%rsp),%edi
  test   %edi,%edi
  jle    L0
L22:
  and    $0x1,%edi
  call   <putchar@plt>
  mov    0x4(%rsp),%eax
  mov    %eax,%edi
  shr    $0x1f,%edi
  add    %eax,%edi
  sar    %edi
  mov    %edi,0x4(%rsp)
  cmp    $0x1,%eax
  jg     L22
L0:
  test   %edi,%edi
  jne    L24
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L20
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L20:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D0
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D28
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D17
type_infered: string

tool: D0: Dwse neo arithmo

tool: D28: %d

tool: D17: H antistrofi dyadiki anaparastasi tou %d einai: 

assistant: ```c
int main() {
  int n, d;
  const a = 2;
  do {
    printf("Dwse neo arithmo\n");
    scanf("%d", &n);
    d = getchar();
    printf("H antistrofi dyadiki anaparastasi tou %d einai: ", n);
    for (n = n; n > 0; n = n / a) {
      d = n % a;
      n > 0 ? putchar(d) : printf("Lathos epilogi\n");
    }
  } while (n != 0);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r13
  lea    D24(%rip),%r13
  push   %r12
  push   %rbp
  lea    D25(%rip),%rbp
  push   %rbx
  lea    D27(%rip),%rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%r12
  nopl   0x0(%rax,%rax,1)
L0:
  mov    %r13,%rdi
  call   <puts@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  xor    %eax,%eax
  call   <getchar@plt>
  mov    0x4(%rsp),%esi
  mov    %rbx,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x4(%rsp),%edi
  test   %edi,%edi
  jle    L12
L32:
  and    $0x1,%edi
  call   <putchar@plt>
  mov    0x4(%rsp),%eax
  mov    %eax,%edi
  shr    $0x1f,%edi
  add    %eax,%edi
  sar    %edi
  mov    %edi,0x4(%rsp)
  cmp    $0x1,%eax
  jg     L32
L12:
  test   %edi,%edi
  jne    L0
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L24
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L24:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D24
type_guessed: string
name: parse_data
arguments: tag: D25
type_guessed: string
name: parse_data
arguments: tag: D27
type_guessed: string

tool: D24: Dwse neo arithmo

tool: D25: %d

tool: D27: H antistrofi dyadiki anaparastasi tou %d einai: 

assistant: ```c
int main() {
  int n, d;
  const a = 2;
  do {
    printf("Dwse neo arithmo\n");
    scanf("%d", &n);
    d = getchar();
    printf("H antistrofi dyadiki anaparastasi tou %d einai: ", n);
    for (n = n; n > 0; n = n / a) {
      d = n % a;
      n > 0 ? putchar(d) : printf("Lathos epilogi\n");
    }
  } while (n != 0);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x64,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D6
type_guessed: string

tool: D6:  You are %d miles away. 


assistant: ```c
int main(int argc, char *argv[]) {
  int distance = 100;
  printf(" You are %d miles away. \n", distance);
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x64,%esi
  lea    D6(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: label: D6
datatype: string

tool: D6:  You are %d miles away. 


assistant: ```c
int main(int argc, char *argv[]) {
  int distance = 100;
  printf(" You are %d miles away. \n", distance);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x64,%esi
  lea    D6(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D6
data_type: string

tool: D6:  You are %d miles away. 


assistant: ```c
int main(int argc, char *argv[]) {
  int distance = 100;
  printf(" You are %d miles away. \n", distance);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x64,%esi
  lea    D9(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: data_label: D9
type: string

tool: D9:  You are %d miles away. 


assistant: ```c
int main(int argc, char *argv[]) {
  int distance = 100;
  printf(" You are %d miles away. \n", distance);
  return 0;
}
```
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    $0xb0,%rsp
  mov    %edi,-0xa4(%rbp)
  mov    %rsi,-0xb0(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x90(%rbp),%rax
  mov    $0x80,%esi
  mov    %rax,%rdi
  call   <getcwd@plt>
  mov    %rax,-0x98(%rbp)
  cmpq   $0x0,-0x98(%rbp)
  jne    L16
  mov    $0x1,%eax
  jmp    L23
L16:
  mov    -0x98(%rbp),%rax
  mov    %rax,%rsi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
L23:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L10
  call   <__stack_chk_fail@plt>
L10:
  leave
  ret
```

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

tool: D7: %s

assistant: ```c
int main(int argc, char **argv) {
  char tmp[128];
  char *p;
  int ret;
  p = getcwd(tmp, 128);
  if (!p) {
    return 1;
  }
  printf("%s", p);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x98,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x88(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  mov    $0x80,%esi
  call   <getcwd@plt>
  mov    %rax,%rsi
  mov    $0x1,%eax
  test   %rsi,%rsi
  je     L8
  lea    D1(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
L8:
  mov    0x88(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L12
  add    $0x98,%rsp
  ret
L12:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D1
data_type: string

tool: D1: %s

assistant: ```c
int main(int argc, char **argv) {
  char tmp[128];
  char *p;
  int ret;
  p = getcwd(tmp, 128);
  if (!p) {
    return 1;
  }
  printf("%s", p);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  sub    $0x98,%rsp
  mov    $0x80,%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x88(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  call   <getcwd@plt>
  mov    %rax,%rsi
  mov    $0x1,%eax
  test   %rsi,%rsi
  je     L11
  lea    D2(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
L11:
  mov    0x88(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L12
  add    $0x98,%rsp
  ret
L12:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D2
data_type: string

tool: D2: %s

assistant: ```c
int main(int argc, char **argv) {
  char tmp[128];
  char *p;
  int ret;
  p = getcwd(tmp, 128);
  if (!p) {
    return 1;
  }
  printf("%s", p);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  sub    $0x98,%rsp
  mov    $0x80,%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x88(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  call   <getcwd@plt>
  mov    %rax,%rsi
  mov    $0x1,%eax
  test   %rsi,%rsi
  je     L11
  lea    D8(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
L11:
  mov    0x88(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L5
  add    $0x98,%rsp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D8
type: string

tool: D8: %s

assistant: ```c
int main(int argc, char **argv) {
  char tmp[128];
  char *p;
  int ret;
  p = getcwd(tmp, 128);
  if (!p) {
    return 1;
  }
  printf("%s", p);
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<set_ints>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  jne    L8
  mov    $0x1,%eax
  jmp    L33
L8:
  movl   $0x0,-0x4(%rbp)
  jmp    L6
L28:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  mov    %eax,(%rdx)
  addl   $0x1,-0x4(%rbp)
L6:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L28
  mov    $0x0,%eax
L33:
  pop    %rbp
  ret
```

assistant: ```c
int set_ints(int *ints, int nr) {
  int i;
  if (ints == NULL) {
    return 1;
  }
  for (i = 0; i < nr; i++) {
    ints[i] = i;
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<set_ints>:
  endbr64
  test   %rdi,%rdi
  je     L28
  test   %esi,%esi
  jle    L3
  mov    %esi,%esi
  mov    $0x0,%eax
L16:
  mov    %eax,(%rdi,%rax,4)
  add    $0x1,%rax
  cmp    %rsi,%rax
  jne    L16
  mov    $0x0,%eax
  ret
L28:
  mov    $0x1,%eax
  ret
L3:
  mov    $0x0,%eax
  ret
```

assistant: ```c
int set_ints(int *ints, int nr) {
  int i;
  if (ints == NULL) {
    return 1;
  }
  for (i = 0; i < nr; i++) {
    ints[i] = i;
  }
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<set_ints>:
  endbr64
  test   %rdi,%rdi
  je     L2
  movslq %esi,%rdx
  xor    %eax,%eax
  test   %esi,%esi
  jle    L24
  nopw   0x0(%rax,%rax,1)
L26:
  mov    %eax,(%rdi,%rax,4)
  add    $0x1,%rax
  cmp    %rdx,%rax
  jne    L26
L24:
  xor    %eax,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L2:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int set_ints(int *ints, int nr) {
  int i;
  if (ints == NULL) {
    return 1;
  }
  for (i = 0; i < nr; i++) {
    ints[i] = i;
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<set_ints>:
  endbr64
  test   %rdi,%rdi
  je     L5
  test   %esi,%esi
  jle    L2
  lea    -0x1(%rsi),%eax
  cmp    $0x2,%eax
  jbe    L52
  mov    %esi,%edx
  movdqa D5(%rip),%xmm0
  movdqa D12(%rip),%xmm2
  mov    %rdi,%rax
  shr    $0x2,%edx
  sub    $0x1,%edx
  shl    $0x4,%rdx
  lea    0x10(%rdi,%rdx,1),%rdx
  nopl   (%rax)
L66:
  movdqa %xmm0,%xmm1
  add    $0x10,%rax
  paddd  %xmm2,%xmm0
  movups %xmm1,-0x10(%rax)
  cmp    %rdx,%rax
  jne    L66
  mov    %esi,%eax
  and    $0xfffffffc,%eax
  test   $0x3,%sil
  je     L2
L29:
  movslq %eax,%rdx
  mov    %eax,(%rdi,%rdx,4)
  lea    0x0(,%rdx,4),%rcx
  lea    0x1(%rax),%edx
  cmp    %esi,%edx
  jge    L2
  add    $0x2,%eax
  mov    %edx,0x4(%rdi,%rcx,1)
  cmp    %eax,%esi
  jle    L2
  mov    %eax,0x8(%rdi,%rcx,1)
L2:
  xor    %eax,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L5:
  mov    $0x1,%eax
  ret
L52:
  xor    %eax,%eax
  jmp    L29
```

tool_calls: name: disassemble_data
arguments: data_label: D5
type: byte[16]
name: disassemble_data
arguments: data_label: D12
type: byte[16]

tool: {"D5": [0x00,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x02,0x00,0x00,0x00,0x03,0x00,0x00,0x00]}

tool: {"D12": [0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00]}

assistant: ```c
int set_ints(int *ints, int nr) {
  int i;
  if (ints == NULL) {
    return 1;
  }
  for (i = 0; i < nr; i++) {
    ints[i] = i;
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<addDigits>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  add    $0x9,%eax
  cmp    $0x12,%eax
  ja     L23
  mov    -0x14(%rbp),%eax
  jmp    L17
L23:
  movl   $0x0,-0x4(%rbp)
  jmp    L30
L4:
  mov    -0x14(%rbp),%ecx
  movslq %ecx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %ecx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  sub    %eax,%ecx
  mov    %ecx,%edx
  add    %edx,-0x4(%rbp)
  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)
L30:
  cmpl   $0x0,-0x14(%rbp)
  jne    L4
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <addDigits>
L17:
  leave
  ret
```

assistant: ```c
int addDigits(int num) {
  if (num / 10 == 0) {
    return num;
  }
  int sum = 0;
  while (num != 0) {
    sum += num % 10;
    num /= 10;
  }
  return addDigits(sum);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<addDigits>:
  endbr64
  lea    0x9(%rdi),%edx
  mov    %edi,%eax
  cmp    $0x12,%edx
  jbe    L4
  sub    $0x8,%rsp
  mov    $0x0,%ecx
L19:
  movslq %edi,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  mov    %edi,%edx
  sar    $0x1f,%edx
  sub    %edx,%eax
  lea    (%rax,%rax,4),%edx
  add    %edx,%edx
  sub    %edx,%edi
  add    %edi,%ecx
  mov    %eax,%edi
  test   %eax,%eax
  jne    L19
  mov    %ecx,%edi
  call   <addDigits>
  add    $0x8,%rsp
  ret
L4:
  ret
```

assistant: ```c
int addDigits(int num) {
  if (num / 10 == 0) {
    return num;
  }
  int sum = 0;
  while (num != 0) {
    sum += num % 10;
    num /= 10;
  }
  return addDigits(sum);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<addDigits>:
  endbr64
  lea    0x9(%rdi),%eax
  mov    %edi,%r8d
  cmp    $0x12,%eax
  jbe    L2
  nop
L10:
  mov    %r8d,%edx
  xor    %r8d,%r8d
  cs nopw 0x0(%rax,%rax,1)
L1:
  movslq %edx,%rax
  mov    %edx,%ecx
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%ecx
  sar    $0x22,%rax
  sub    %ecx,%eax
  lea    (%rax,%rax,4),%ecx
  add    %ecx,%ecx
  sub    %ecx,%edx
  add    %edx,%r8d
  mov    %eax,%edx
  test   %eax,%eax
  jne    L1
  lea    0x9(%r8),%eax
  cmp    $0x12,%eax
  ja     L10
L2:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int addDigits(int num) {
  if (num / 10 == 0) {
    return num;
  }
  int sum = 0;
  while (num != 0) {
    sum += num % 10;
    num /= 10;
  }
  return addDigits(sum);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<addDigits>:
  endbr64
  lea    0x9(%rdi),%eax
  mov    %edi,%r8d
  cmp    $0x12,%eax
  jbe    L1
  nop
L5:
  mov    %r8d,%edx
  xor    %r8d,%r8d
  cs nopw 0x0(%rax,%rax,1)
L23:
  movslq %edx,%rax
  mov    %edx,%ecx
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%ecx
  sar    $0x22,%rax
  sub    %ecx,%eax
  lea    (%rax,%rax,4),%ecx
  add    %ecx,%ecx
  sub    %ecx,%edx
  add    %edx,%r8d
  mov    %eax,%edx
  test   %eax,%eax
  jne    L23
  lea    0x9(%r8),%eax
  cmp    $0x12,%eax
  ja     L5
L1:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int addDigits(int num) {
  if (num / 10 == 0) {
    return num;
  }
  int sum = 0;
  while (num != 0) {
    sum += num % 10;
    num /= 10;
  }
  return addDigits(sum);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<replaced>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x48,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    %rdx,-0x48(%rbp)
  mov    -0x38(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cltq
  mov    %rax,-0x28(%rbp)
  mov    -0x40(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cltq
  mov    %rax,-0x20(%rbp)
  mov    -0x48(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,%ebx
  mov    -0x38(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  sub    %eax,%ebx
  mov    -0x40(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  add    %ebx,%eax
  add    $0x1,%eax
  mov    %eax,%edi
  call   <xmalloc@plt>
  mov    %rax,-0x18(%rbp)
  mov    -0x48(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,%ebx
  mov    -0x38(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  sub    %eax,%ebx
  mov    -0x40(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  add    %ebx,%eax
  lea    0x1(%rax),%edx
  mov    -0x18(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  movl   $0x0,-0x2c(%rbp)
  jmp    L1
L6:
  addl   $0x1,-0x2c(%rbp)
L1:
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cmp    %eax,-0x2c(%rbp)
  jl     L6
  nop
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
char *replaced(const char *s, const char *d, const char *src) {
  char *dst;
  unsigned long slen = strlen(s);
  unsigned long dlen = strlen(d);
  dst = xmalloc(strlen(src) - strlen(s) + strlen(d) + 1);
  memset(dst, 0, strlen(src) - strlen(s) + strlen(d) + 1);
  for (int i = 0; i < strlen(dst); ++i) {
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<replaced>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r13
  mov    %rsi,%r12
  mov    %rdx,%r14
  call   <strlen@plt>
  mov    %r12,%rdi
  call   <strlen@plt>
  mov    %r14,%rdi
  call   <strlen@plt>
  mov    %eax,%ebx
  mov    %r13,%rdi
  call   <strlen@plt>
  sub    %eax,%ebx
  mov    %r12,%rdi
  call   <strlen@plt>
  lea    0x1(%rbx,%rax,1),%edi
  call   <xmalloc@plt>
  mov    %rax,%rbp
  mov    %r14,%rdi
  call   <strlen@plt>
  mov    %eax,%ebx
  mov    %r13,%rdi
  call   <strlen@plt>
  sub    %eax,%ebx
  mov    %r12,%rdi
  call   <strlen@plt>
  lea    0x1(%rbx,%rax,1),%edx
  mov    $0x0,%esi
  mov    %rbp,%rdi
  call   <memset@plt>
  mov    $0x0,%ebx
  jmp    L10
L9:
  add    $0x1,%ebx
L10:
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jg     L9
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
char *replaced(const char *s, const char *d, const char *src) {
  char *dst;
  unsigned long slen = strlen(s);
  unsigned long dlen = strlen(d);
  dst = xmalloc(strlen(src) - strlen(s) + strlen(d) + 1);
  memset(dst, 0, strlen(src) - strlen(s) + strlen(d) + 1);
  for (int i = 0; i < strlen(dst); ++i) {
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<replaced>:
  endbr64
  push   %r14
  mov    %rdx,%r14
  push   %r13
  mov    %rdi,%r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  push   %rbx
  call   <strlen@plt>
  mov    %r12,%rdi
  call   <strlen@plt>
  mov    %r14,%rdi
  call   <strlen@plt>
  mov    %r13,%rdi
  mov    %eax,%ebx
  call   <strlen@plt>
  mov    %r12,%rdi
  sub    %eax,%ebx
  call   <strlen@plt>
  lea    0x1(%rbx,%rax,1),%edi
  call   <xmalloc@plt>
  mov    %r14,%rdi
  mov    %rax,%rbp
  call   <strlen@plt>
  mov    %r13,%rdi
  mov    %eax,%ebx
  call   <strlen@plt>
  mov    %r12,%rdi
  sub    %eax,%ebx
  call   <strlen@plt>
  xor    %esi,%esi
  mov    %rbp,%rdi
  lea    0x1(%rbx,%rax,1),%edx
  xor    %ebx,%ebx
  call   <memset@plt>
  jmp    L13
  nopl   0x0(%rax,%rax,1)
L19:
  add    $0x1,%ebx
L13:
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jg     L19
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
char *replaced(const char *s, const char *d, const char *src) {
  char *dst;
  unsigned long slen = strlen(s);
  unsigned long dlen = strlen(d);
  dst = xmalloc(strlen(src) - strlen(s) + strlen(d) + 1);
  memset(dst, 0, strlen(src) - strlen(s) + strlen(d) + 1);
  for (int i = 0; i < strlen(dst); ++i) {
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<replaced>:
  endbr64
  push   %r14
  mov    %rdx,%r14
  push   %r13
  mov    %rdi,%r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  push   %rbx
  call   <strlen@plt>
  mov    %r12,%rdi
  call   <strlen@plt>
  mov    %r14,%rdi
  call   <strlen@plt>
  mov    %r13,%rdi
  mov    %eax,%ebx
  call   <strlen@plt>
  mov    %r12,%rdi
  sub    %eax,%ebx
  call   <strlen@plt>
  lea    0x1(%rbx,%rax,1),%edi
  call   <xmalloc@plt>
  mov    %r14,%rdi
  mov    %rax,%rbp
  call   <strlen@plt>
  mov    %r13,%rdi
  mov    %eax,%ebx
  call   <strlen@plt>
  mov    %r12,%rdi
  sub    %eax,%ebx
  call   <strlen@plt>
  xor    %esi,%esi
  mov    %rbp,%rdi
  lea    0x1(%rbx,%rax,1),%edx
  xor    %ebx,%ebx
  call   <memset@plt>
  jmp    L9
  nopl   0x0(%rax,%rax,1)
L1:
  add    $0x1,%ebx
L9:
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jg     L1
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
char *replaced(const char *s, const char *d, const char *src) {
  char *dst;
  unsigned long slen = strlen(s);
  unsigned long dlen = strlen(d);
  dst = xmalloc(strlen(src) - strlen(s) + strlen(d) + 1);
  memset(dst, 0, strlen(src) - strlen(s) + strlen(d) + 1);
  for (int i = 0; i < strlen(dst); ++i) {
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<__assert_func>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    %rcx,-0x20(%rbp)
L3:
  jmp    L3
```

assistant: ```c
void __attribute__((noreturn)) __assert_func(const char __attribute__((unused)) * file, int __attribute__((unused)) line, const char __attribute__((unused)) * func, const char __attribute__((unused)) * failedexpr) {
  while (1)
    ;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<__assert_func>:
  endbr64
L7:
  jmp    L7
```

assistant: ```c
void __attribute__((noreturn)) __assert_func(const char __attribute__((unused)) * file, int __attribute__((unused)) line, const char __attribute__((unused)) * func, const char __attribute__((unused)) * failedexpr) {
  while (1)
    ;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<__assert_func>:
  endbr64
  nopl   0x0(%rax)
L4:
  jmp    L4
```

assistant: ```c
void __attribute__((noreturn)) __assert_func(const char __attribute__((unused)) * file, int __attribute__((unused)) line, const char __attribute__((unused)) * func, const char __attribute__((unused)) * failedexpr) {
  while (1)
    ;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<__assert_func>:
  endbr64
  nopl   0x0(%rax)
L4:
  jmp    L4
```

assistant: ```c
void __attribute__((noreturn)) __assert_func(const char __attribute__((unused)) * file, int __attribute__((unused)) line, const char __attribute__((unused)) * func, const char __attribute__((unused)) * failedexpr) {
  while (1)
    ;
}
```