ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<is_palindrome>:
  endbr64
  push   %rbx
  mov    %edi,%edx
  lea    D0(%rip),%rsi
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x1(%rsp),%rbx
  mov    %rbx,%rdi
  call   <sprintf@plt>
  movzbl 0x1(%rsp),%ecx
  test   %cl,%cl
  je     L21
  mov    %rbx,%rdx
  nopl   0x0(%rax,%rax,1)
L5:
  mov    %rdx,%rax
  lea    0x1(%rdx),%rdx
  cmpb   $0x0,0x1(%rax)
  jne    L5
  cmp    %rbx,%rax
  jbe    L21
  mov    %rbx,%rdx
  jmp    L4
  nopw   0x0(%rax,%rax,1)
L38:
  add    $0x1,%rdx
  sub    $0x1,%rax
  cmp    %rax,%rdx
  jae    L21
  movzbl (%rdx),%ecx
L4:
  cmp    %cl,(%rax)
  je     L38
  xor    %eax,%eax
  jmp    L33
  nopl   0x0(%rax,%rax,1)
L21:
  mov    $0x1,%eax
L33:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L68
  add    $0x10,%rsp
  pop    %rbx
  ret
L68:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D0
data_type: string

tool: D0: %d

assistant: ```c
int is_palindrome(int value) {
  char *first, *last, string[7];
  sprintf(string, "%d", value);
  first = string;
  last = string;
  while (*last != '\0') {
    last++;
  }
  last--;
  while (last > first) {
    if (*first != *last)
      return 0;
    first++;
    last--;
  }
  return 1;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<is_palindrome>:
  endbr64
  push   %rbx
  mov    %edi,%edx
  lea    D9(%rip),%rsi
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x1(%rsp),%rbx
  mov    %rbx,%rdi
  call   <sprintf@plt>
  cmpb   $0x0,0x1(%rsp)
  je     L4
  mov    %rbx,%rdx
  nopl   0x0(%rax)
L68:
  mov    %rdx,%rax
  lea    0x1(%rdx),%rdx
  cmpb   $0x0,0x1(%rax)
  jne    L68
  cmp    %rbx,%rax
  jbe    L4
  mov    %rbx,%rdx
  jmp    L44
  nopw   0x0(%rax,%rax,1)
L21:
  add    $0x1,%rdx
  sub    $0x1,%rax
  cmp    %rax,%rdx
  jae    L4
L44:
  movzbl (%rax),%ecx
  cmp    %cl,(%rdx)
  je     L21
  xor    %eax,%eax
  jmp    L9
  nopl   0x0(%rax,%rax,1)
L4:
  mov    $0x1,%eax
L9:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L57
  add    $0x10,%rsp
  pop    %rbx
  ret
L57:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D9
type_guessed: string

tool: D9: %d

assistant: ```c
int is_palindrome(int value) {
  char *first, *last, string[7];
  sprintf(string, "%d", value);
  first = string;
  last = string;
  while (*last != '\0') {
    last++;
  }
  last--;
  while (last > first) {
    if (*first != *last)
      return 0;
    first++;
    last--;
  }
  return 1;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x5,-0x14(%rbp)
  lea    -0x14(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,%esi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L6
  call   <__stack_chk_fail@plt>
L6:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: tag: D9
data_type: string

tool: D9: %d

assistant: ```c
int main() {
  int y = 5;
  int *ptrY = &y;
  printf("%d", *ptrY);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x5,%esi
  lea    D6(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

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

tool: D6: %d

assistant: ```c
int main() {
  int y = 5;
  int *ptrY = &y;
  printf("%d", *ptrY);
}
```
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    $0x5,%esi
  lea    D0(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D0
type: string

tool: D0: %d

assistant: ```c
int main() {
  int y = 5;
  int *ptrY = &y;
  printf("%d", *ptrY);
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x5,%esi
  lea    D5(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D5
data_type: string

tool: D5: %d

assistant: ```c
int main() {
  int y = 5;
  int *ptrY = &y;
  printf("%d", *ptrY);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<av_int2dbl>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jns    L2
  pxor   %xmm0,%xmm0
  divsd  %xmm0,%xmm0
  movq   %xmm0,%rax
  jmp    L7
L2:
  mov    $0x34,%edx
  mov    -0x4(%rbp),%eax
  mov    %edx,%ecx
  sar    %cl,%eax
  and    $0x7ff,%eax
  sub    $0x433,%eax
  mov    -0x4(%rbp),%edx
  movslq %edx,%rdx
  movabs $0xfffffffffffff,%rcx
  and    %rdx,%rcx
  movabs $0x10000000000000,%rdx
  add    %rcx,%rdx
  mov    %edx,%esi
  mov    $0x3f,%ecx
  mov    -0x4(%rbp),%edx
  sar    %cl,%edx
  or     $0x1,%edx
  imul   %esi,%edx
  mov    %eax,%esi
  mov    %edx,%edi
  call   <ldexp@plt>
  movq   %xmm0,%rax
L7:
  movq   %rax,%xmm0
  leave
  ret
```

assistant: ```c
double av_int2dbl(int64_t v) {
  if (v + v > 0xFFEULL << 52)
    return 0.0 / 0.0;
  return ldexp(((v & ((1LL << 52) - 1)) + (1LL << 52)) * (v >> 63 | 1), (v >> 52 & 0x7FF) - 1075);
}
```
user: Translate the below assembly code into C programming language:

```asm
<av_int2dbl>:
  endbr64
  test   %edi,%edi
  js     L8
  sub    $0x8,%rsp
  mov    %edi,%esi
  mov    $0x34,%ecx
  sar    %cl,%esi
  and    $0x7ff,%esi
  sub    $0x433,%esi
  mov    %edi,%eax
  mov    $0x3f,%ecx
  sar    %cl,%eax
  or     $0x1,%eax
  imul   %eax,%edi
  call   <ldexp@plt>
  add    $0x8,%rsp
  ret
L8:
  pxor   %xmm0,%xmm0
  divsd  %xmm0,%xmm0
  ret
```

assistant: ```c
double av_int2dbl(int64_t v) {
  if (v + v > 0xFFEULL << 52)
    return 0.0 / 0.0;
  return ldexp(((v & ((1LL << 52) - 1)) + (1LL << 52)) * (v >> 63 | 1), (v >> 52 & 0x7FF) - 1075);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<av_int2dbl>:
  endbr64
  test   %edi,%edi
  js     L0
  mov    %edi,%esi
  mov    $0x34,%ecx
  mov    %edi,%eax
  sar    %cl,%esi
  mov    $0x3f,%ecx
  sar    %cl,%eax
  and    $0x7ff,%esi
  or     $0x1,%eax
  sub    $0x433,%esi
  imul   %eax,%edi
  jmp    1050 <ldexp@plt>
  nopl   0x0(%rax)
L0:
  pxor   %xmm0,%xmm0
  divsd  %xmm0,%xmm0
  ret
```

assistant: ```c
double av_int2dbl(int64_t v) {
  if (v + v > 0xFFEULL << 52)
    return 0.0 / 0.0;
  return ldexp(((v & ((1LL << 52) - 1)) + (1LL << 52)) * (v >> 63 | 1), (v >> 52 & 0x7FF) - 1075);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<av_int2dbl>:
  endbr64
  test   %edi,%edi
  js     L0
  mov    %edi,%esi
  mov    $0x34,%ecx
  mov    %edi,%eax
  sar    %cl,%esi
  mov    $0x3f,%ecx
  sar    %cl,%eax
  and    $0x7ff,%esi
  or     $0x1,%eax
  sub    $0x433,%esi
  imul   %eax,%edi
  jmp    1050 <ldexp@plt>
  nopl   0x0(%rax)
L0:
  pxor   %xmm0,%xmm0
  divsd  %xmm0,%xmm0
  ret
```

assistant: ```c
double av_int2dbl(int64_t v) {
  if (v + v > 0xFFEULL << 52)
    return 0.0 / 0.0;
  return ldexp(((v & ((1LL << 52) - 1)) + (1LL << 52)) * (v >> 63 | 1), (v >> 52 & 0x7FF) - 1075);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<init_robot>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <init@plt>
  cmp    $0xffffffff,%eax
  jne    L1
  mov    $0x1,%edi
  call   <exit@plt>
L1:
  mov    $0x1,%edi
  call   <led_on@plt>
  nop
  pop    %rbp
  ret
```

assistant: ```c
void init_robot() {
  if (init() == -1) {
    exit(1);
  }
  led_on(1);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<init_robot>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <init@plt>
  cmp    $0xffffffff,%eax
  je     L14
L15:
  mov    $0x1,%edi
  call   <led_on@plt>
  add    $0x8,%rsp
  ret
L14:
  mov    $0x1,%edi
  call   <exit@plt>
  jmp    L15
```

assistant: ```c
void init_robot() {
  if (init() == -1) {
    exit(1);
  }
  led_on(1);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<init_robot>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <init@plt>
  cmp    $0xffffffff,%eax
  je     L5
  mov    $0x1,%edi
  add    $0x8,%rsp
  jmp    1070 <led_on@plt>
  nopw   0x0(%rax,%rax,1)
L5:
  mov    $0x1,%edi
  call   <exit@plt>
  mov    $0x1,%edi
  add    $0x8,%rsp
  jmp    1070 <led_on@plt>
```

assistant: ```c
void init_robot() {
  if (init() == -1) {
    exit(1);
  }
  led_on(1);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<init_robot>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <init@plt>
  cmp    $0xffffffff,%eax
  je     L5
  mov    $0x1,%edi
  add    $0x8,%rsp
  jmp    1070 <led_on@plt>
  nopw   0x0(%rax,%rax,1)
L5:
  mov    $0x1,%edi
  call   <exit@plt>
  mov    $0x1,%edi
  add    $0x8,%rsp
  jmp    1070 <led_on@plt>
```

assistant: ```c
void init_robot() {
  if (init() == -1) {
    exit(1);
  }
  led_on(1);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<generateMovesFile>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %edi,-0x24(%rbp)
  lea    D26(%rip),%rax
  mov    %rax,%rsi
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  call   <fopen@plt>
  mov    %rax,-0x18(%rbp)
  movl   $0x0,-0x1c(%rbp)
  jmp    L7
L12:
  mov    $0x0,%eax
  call   <rand@plt>
  cltd
  shr    $0x1e,%edx
  add    %edx,%eax
  and    $0x3,%eax
  sub    %edx,%eax
  mov    %eax,%ebx
  mov    $0x0,%eax
  call   <rand@plt>
  mov    %eax,%edx
  movslq %edx,%rax
  imul   $0xffffffff92492493,%rax,%rax
  shr    $0x20,%rax
  add    %edx,%eax
  sar    $0x2,%eax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  mov    %eax,%ecx
  shl    $0x3,%ecx
  sub    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    -0x18(%rbp),%rdi
  mov    %ebx,%ecx
  mov    %eax,%edx
  lea    D28(%rip),%rax
  mov    %rax,%rsi
  call   <fprintf@plt>
  addl   $0x1,-0x1c(%rbp)
L7:
  mov    -0x1c(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jl     L12
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <fclose@plt>
  nop
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D26
type_guessed: string
name: disassemble_data
arguments: data_tag: D18
type_guessed: string
name: disassemble_data
arguments: data_tag: D28
type_guessed: string

tool: D26: w

tool: D18: moves

tool: D28: %d,%d


assistant: ```c
void generateMovesFile(int size) {
  FILE *f = fopen("moves", "w");
  int i;
  for (i = 0; i < size; i++)
    fprintf(f, "%d,%d\n", rand() % 7, rand() % 4);
  fclose(f);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<generateMovesFile>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%r12d
  lea    D16(%rip),%rsi
  lea    D4(%rip),%rdi
  call   <fopen@plt>
  mov    %rax,%rbp
  test   %r12d,%r12d
  jle    L15
  mov    $0x0,%ebx
  lea    D6(%rip),%r13
L7:
  mov    $0x0,%eax
  call   <rand@plt>
  mov    %eax,%r14d
  mov    $0x0,%eax
  call   <rand@plt>
  mov    %eax,%edx
  mov    %r14d,%eax
  sar    $0x1f,%eax
  shr    $0x1e,%eax
  lea    (%r14,%rax,1),%ecx
  and    $0x3,%ecx
  sub    %eax,%ecx
  movslq %edx,%rax
  imul   $0xffffffff92492493,%rax,%rax
  shr    $0x20,%rax
  add    %edx,%eax
  sar    $0x2,%eax
  mov    %edx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  lea    0x0(,%rax,8),%esi
  sub    %eax,%esi
  sub    %esi,%edx
  mov    %r13,%rsi
  mov    %rbp,%rdi
  call   <fprintf@plt>
  add    $0x1,%ebx
  cmp    %ebx,%r12d
  jne    L7
L15:
  mov    %rbp,%rdi
  call   <fclose@plt>
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

tool_calls: name: parse_data
arguments: data_label: D16
type_guessed: string
name: parse_data
arguments: data_label: D4
type_guessed: string
name: parse_data
arguments: data_label: D6
type_guessed: string

tool: D16: w

tool: D4: moves

tool: D6: %d,%d


assistant: ```c
void generateMovesFile(int size) {
  FILE *f = fopen("moves", "w");
  int i;
  for (i = 0; i < size; i++)
    fprintf(f, "%d,%d\n", rand() % 7, rand() % 4);
  fclose(f);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<generateMovesFile>:
  endbr64
  push   %r14
  lea    D12(%rip),%rsi
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%ebx
  lea    D17(%rip),%rdi
  call   <fopen@plt>
  mov    %rax,%rbp
  test   %ebx,%ebx
  jle    L19
  xor    %r13d,%r13d
  lea    D8(%rip),%r12
  nopw   0x0(%rax,%rax,1)
L7:
  xor    %eax,%eax
  add    $0x1,%r13d
  call   <rand@plt>
  mov    %eax,%r14d
  xor    %eax,%eax
  call   <rand@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  movslq %eax,%rcx
  mov    %r14d,%eax
  mov    %rcx,%rdx
  sar    $0x1f,%eax
  imul   $0xffffffff92492493,%rcx,%rcx
  shr    $0x1e,%eax
  add    %eax,%r14d
  and    $0x3,%r14d
  shr    $0x20,%rcx
  sub    %eax,%r14d
  add    %edx,%ecx
  mov    %edx,%eax
  sar    $0x1f,%eax
  sar    $0x2,%ecx
  sub    %eax,%ecx
  lea    0x0(,%rcx,8),%eax
  sub    %ecx,%eax
  mov    %r14d,%ecx
  sub    %eax,%edx
  call   <fprintf@plt>
  cmp    %r13d,%ebx
  jne    L7
L19:
  pop    %rbx
  mov    %rbp,%rdi
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  jmp    1080 <fclose@plt>
```

tool_calls: name: read_data
arguments: data_label: D12
type: string
name: read_data
arguments: data_label: D17
type: string
name: read_data
arguments: data_label: D8
type: string

tool: D12: w

tool: D17: moves

tool: D8: %d,%d


assistant: ```c
void generateMovesFile(int size) {
  FILE *f = fopen("moves", "w");
  int i;
  for (i = 0; i < size; i++)
    fprintf(f, "%d,%d\n", rand() % 7, rand() % 4);
  fclose(f);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<generateMovesFile>:
  endbr64
  push   %r14
  lea    D24(%rip),%rsi
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%ebx
  lea    D2(%rip),%rdi
  call   <fopen@plt>
  mov    %rax,%rbp
  test   %ebx,%ebx
  jle    L2
  xor    %r13d,%r13d
  lea    D29(%rip),%r12
  nopw   0x0(%rax,%rax,1)
L17:
  xor    %eax,%eax
  add    $0x1,%r13d
  call   <rand@plt>
  mov    %eax,%r14d
  xor    %eax,%eax
  call   <rand@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  movslq %eax,%rcx
  mov    %r14d,%eax
  mov    %rcx,%rdx
  sar    $0x1f,%eax
  imul   $0xffffffff92492493,%rcx,%rcx
  shr    $0x1e,%eax
  add    %eax,%r14d
  and    $0x3,%r14d
  shr    $0x20,%rcx
  sub    %eax,%r14d
  add    %edx,%ecx
  mov    %edx,%eax
  sar    $0x1f,%eax
  sar    $0x2,%ecx
  sub    %eax,%ecx
  lea    0x0(,%rcx,8),%eax
  sub    %ecx,%eax
  mov    %r14d,%ecx
  sub    %eax,%edx
  call   <fprintf@plt>
  cmp    %r13d,%ebx
  jne    L17
L2:
  pop    %rbx
  mov    %rbp,%rdi
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  jmp    1080 <fclose@plt>
```

tool_calls: name: retrieve_data
arguments: label: D24
type: string
name: retrieve_data
arguments: label: D2
type: string
name: retrieve_data
arguments: label: D29
type: string

tool: D24: w

tool: D2: moves

tool: D29: %d,%d


assistant: ```c
void generateMovesFile(int size) {
  FILE *f = fopen("moves", "w");
  int i;
  for (i = 0; i < size; i++)
    fprintf(f, "%d,%d\n", rand() % 7, rand() % 4);
  fclose(f);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<KheBuffAddStr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %rcx,-0x30(%rbp)
  mov    -0x30(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L5
L14:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%ecx
  mov    -0x28(%rbp),%rdx
  mov    -0x20(%rbp),%rsi
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <KheBuffAddChar@plt>
  addq   $0x1,-0x8(%rbp)
L5:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L14
  nop
  nop
  leave
  ret
```

assistant: ```c
void KheBuffAddStr(char **buff, int *curr_len, int *max_len, char *str) {
  char *p;
  for (p = str; *p != '\0'; p++)
    KheBuffAddChar(buff, curr_len, max_len, *p);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<KheBuffAddStr>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r13
  mov    %rsi,%r12
  mov    %rdx,%rbp
  mov    %rcx,%rbx
  movzbl (%rcx),%ecx
  test   %cl,%cl
  je     L19
L3:
  movsbl %cl,%ecx
  mov    %rbp,%rdx
  mov    %r12,%rsi
  mov    %r13,%rdi
  call   <KheBuffAddChar@plt>
  add    $0x1,%rbx
  movzbl (%rbx),%ecx
  test   %cl,%cl
  jne    L3
L19:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
void KheBuffAddStr(char **buff, int *curr_len, int *max_len, char *str) {
  char *p;
  for (p = str; *p != '\0'; p++)
    KheBuffAddChar(buff, curr_len, max_len, *p);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<KheBuffAddStr>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rcx,%rbx
  sub    $0x8,%rsp
  movsbl (%rcx),%ecx
  test   %cl,%cl
  je     L18
  mov    %rdi,%r13
  mov    %rsi,%r12
  mov    %rdx,%rbp
  nopl   0x0(%rax)
L5:
  mov    %rbp,%rdx
  mov    %r12,%rsi
  mov    %r13,%rdi
  add    $0x1,%rbx
  call   <KheBuffAddChar@plt>
  movsbl (%rbx),%ecx
  test   %cl,%cl
  jne    L5
L18:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
void KheBuffAddStr(char **buff, int *curr_len, int *max_len, char *str) {
  char *p;
  for (p = str; *p != '\0'; p++)
    KheBuffAddChar(buff, curr_len, max_len, *p);
}
```
user: What is the c source code of the assembly code below?

```asm
<KheBuffAddStr>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rcx,%rbx
  sub    $0x8,%rsp
  movsbl (%rcx),%ecx
  test   %cl,%cl
  je     L13
  mov    %rdi,%r13
  mov    %rsi,%r12
  mov    %rdx,%rbp
  nopl   0x0(%rax)
L16:
  mov    %rbp,%rdx
  mov    %r12,%rsi
  mov    %r13,%rdi
  add    $0x1,%rbx
  call   <KheBuffAddChar@plt>
  movsbl (%rbx),%ecx
  test   %cl,%cl
  jne    L16
L13:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
void KheBuffAddStr(char **buff, int *curr_len, int *max_len, char *str) {
  char *p;
  for (p = str; *p != '\0'; p++)
    KheBuffAddChar(buff, curr_len, max_len, *p);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<build_max_heap>:
  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)
  mov    -0x1c(%rbp),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  mov    %eax,-0x4(%rbp)
  jmp    L6
L15:
  mov    -0x4(%rbp),%edx
  mov    -0x20(%rbp),%ecx
  mov    -0x18(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <max_heapify@plt>
  subl   $0x1,-0x4(%rbp)
L6:
  cmpl   $0x0,-0x4(%rbp)
  jg     L15
  nop
  nop
  leave
  ret
```

assistant: ```c
void build_max_heap(int *arr, int length, int heap_size) {
  int i = 0;
  for (i = length / 2; i >= 1; i--) {
    max_heapify(arr, heap_size, i);
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<build_max_heap>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    %edx,%ebp
  mov    %esi,%ebx
  shr    $0x1f,%ebx
  add    %esi,%ebx
  sar    %ebx
  cmp    $0x1,%esi
  jle    L5
L11:
  mov    %ebx,%edx
  mov    %ebp,%esi
  mov    %r12,%rdi
  call   <max_heapify@plt>
  sub    $0x1,%ebx
  test   %ebx,%ebx
  jg     L11
L5:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void build_max_heap(int *arr, int length, int heap_size) {
  int i = 0;
  for (i = length / 2; i >= 1; i--) {
    max_heapify(arr, heap_size, i);
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<build_max_heap>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %esi,%ebx
  shr    $0x1f,%ebx
  add    %esi,%ebx
  sar    %ebx
  cmp    $0x1,%esi
  jle    L15
  mov    %rdi,%r12
  mov    %edx,%ebp
  nopl   0x0(%rax,%rax,1)
L11:
  mov    %ebx,%edx
  mov    %ebp,%esi
  mov    %r12,%rdi
  call   <max_heapify@plt>
  sub    $0x1,%ebx
  jne    L11
L15:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void build_max_heap(int *arr, int length, int heap_size) {
  int i = 0;
  for (i = length / 2; i >= 1; i--) {
    max_heapify(arr, heap_size, i);
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<build_max_heap>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %esi,%ebx
  shr    $0x1f,%ebx
  add    %esi,%ebx
  sar    %ebx
  cmp    $0x1,%esi
  jle    L12
  mov    %rdi,%r12
  mov    %edx,%ebp
  nopl   0x0(%rax,%rax,1)
L14:
  mov    %ebx,%edx
  mov    %ebp,%esi
  mov    %r12,%rdi
  call   <max_heapify@plt>
  sub    $0x1,%ebx
  jne    L14
L12:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void build_max_heap(int *arr, int length, int heap_size) {
  int i = 0;
  for (i = length / 2; i >= 1; i--) {
    max_heapify(arr, heap_size, i);
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<add_tracks>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    %ecx,-0x24(%rbp)
  mov    -0x24(%rbp),%eax
  imul   -0x1c(%rbp),%eax
  cltq
  shl    $0x2,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L12
L0:
  mov    -0x20(%rbp),%eax
  cltq
  shl    $0x2,%eax
  mov    %eax,%edx
  mov    -0xc(%rbp),%eax
  imul   -0x20(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x18(%rbp),%rax
  add    %rax,%rcx
  mov    -0xc(%rbp),%eax
  imul   -0x24(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rsi
  mov    -0x8(%rbp),%rax
  add    %rsi,%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0x24(%rbp),%eax
  sub    -0x20(%rbp),%eax
  cltq
  shl    $0x2,%eax
  mov    %eax,%edx
  mov    -0xc(%rbp),%eax
  imul   -0x24(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x20(%rbp),%eax
  cltq
  add    %rcx,%rax
  lea    0x0(,%rax,4),%rcx
  mov    -0x8(%rbp),%rax
  add    %rcx,%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  addl   $0x1,-0xc(%rbp)
L12:
  mov    -0xc(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L0
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
int *add_tracks(int *old_pattern, int pattern_length, int old_num_tracks, int new_num_tracks) {
  int *new_pattern = malloc(new_num_tracks * pattern_length * sizeof(int));
  for (int p = 0; p < pattern_length; p++) {
    memcpy(new_pattern + p * new_num_tracks, old_pattern + p * old_num_tracks, sizeof(int) * old_num_tracks);
    memset(new_pattern + p * new_num_tracks + old_num_tracks, 0, sizeof(int) * (new_num_tracks - old_num_tracks));
  }
  free(old_pattern);
  return new_pattern;
}
```