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

```asm
<section_by_type>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0x0,-0x20(%rbp)
  movq   $0x0,-0x18(%rbp)
  mov    -0x38(%rbp),%rax
  movq   $0x0,(%rax)
  lea    -0x20(%rbp),%rdx
  mov    -0x24(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <read_header@plt>
  test   %rax,%rax
  jne    L26
  mov    -0x20(%rbp),%rcx
  lea    -0x18(%rbp),%rdx
  mov    -0x24(%rbp),%eax
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <read_section_table@plt>
  test   %rax,%rax
  je     L71
L26:
  mov    D13(%rip),%rax
  mov    (%rax),%eax
  jmp    L9
L71:
  movq   $0x0,-0x10(%rbp)
  jmp    L70
L44:
  mov    -0x18(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  shl    $0x4,%rax
  add    %rdx,%rax
  mov    0x8(%rax),%rax
  cmp    %rax,-0x30(%rbp)
  jne    L22
  mov    $0x10,%edi
  call   <malloc@plt>
  mov    %rax,%rdx
  mov    -0x38(%rbp),%rax
  mov    %rdx,(%rax)
  mov    -0x38(%rbp),%rax
  mov    (%rax),%rax
  test   %rax,%rax
  jne    L72
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    D13(%rip),%rax
  mov    (%rax),%eax
  jmp    L9
L72:
  mov    -0x18(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  shl    $0x4,%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0x38(%rbp),%rax
  mov    (%rax),%rax
  mov    $0x10,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  jmp    L8
L22:
  addq   $0x1,-0x10(%rbp)
L70:
  mov    -0x20(%rbp),%rax
  mov    (%rax),%rax
  cmp    %rax,-0x10(%rbp)
  jb     L44
L8:
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    $0x0,%eax
L9:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L60
  call   <__stack_chk_fail@plt>
L60:
  leave
  ret
```

assistant: ```c
int section_by_type(int d, size_t const section_type, Elf64_Shdr **section) {
  Elf64_Ehdr *header = NULL;
  Elf64_Shdr *sections = NULL;
  size_t i;
  *section = NULL;
  if (read_header(d, &header) || read_section_table(d, header, &sections))
    return errno;
  for (i = 0; i < header->e_shnum; ++i)
    if (section_type == sections[i].sh_type) {
      *section = (Elf64_Shdr *)malloc(sizeof(Elf64_Shdr));
      if (NULL == *section) {
        free(header);
        free(sections);
        return errno;
      }
      memcpy(*section, sections + i, sizeof(Elf64_Shdr));
      break;
    }
  free(header);
  free(sections);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<section_by_type>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %edi,%ebx
  mov    %rsi,%rbp
  mov    %rdx,%r12
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movq   $0x0,0x8(%rsp)
  movq   $0x0,0x10(%rsp)
  movq   $0x0,(%rdx)
  lea    0x8(%rsp),%rsi
  call   <read_header@plt>
  test   %rax,%rax
  je     L12
L50:
  mov    D3(%rip),%rax
  mov    (%rax),%eax
L49:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L21
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L12:
  lea    0x10(%rsp),%rdx
  mov    0x8(%rsp),%rsi
  mov    %ebx,%edi
  call   <read_section_table@plt>
  test   %rax,%rax
  jne    L50
  mov    0x8(%rsp),%rax
  mov    (%rax),%rdx
  test   %rdx,%rdx
  je     L13
  mov    0x10(%rsp),%rcx
  mov    $0x0,%ebx
  mov    $0x0,%eax
L83:
  cmp    %rbp,0x8(%rcx,%rbx,1)
  je     L67
  add    $0x1,%rax
  add    $0x10,%rbx
  cmp    %rdx,%rax
  jne    L83
L13:
  mov    0x8(%rsp),%rdi
  call   <free@plt>
  mov    0x10(%rsp),%rdi
  call   <free@plt>
  mov    $0x0,%eax
  jmp    L49
L67:
  mov    $0x10,%edi
  call   <malloc@plt>
  mov    %rax,%rdi
  mov    %rax,(%r12)
  test   %rax,%rax
  je     L77
  mov    %rbx,%rsi
  add    0x10(%rsp),%rsi
  mov    $0x10,%edx
  call   <memcpy@plt>
  jmp    L13
L77:
  mov    0x8(%rsp),%rdi
  call   <free@plt>
  mov    0x10(%rsp),%rdi
  call   <free@plt>
  mov    D3(%rip),%rax
  mov    (%rax),%eax
  jmp    L49
L21:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int section_by_type(int d, size_t const section_type, Elf64_Shdr **section) {
  Elf64_Ehdr *header = NULL;
  Elf64_Shdr *sections = NULL;
  size_t i;
  *section = NULL;
  if (read_header(d, &header) || read_section_table(d, header, &sections))
    return errno;
  for (i = 0; i < header->e_shnum; ++i)
    if (section_type == sections[i].sh_type) {
      *section = (Elf64_Shdr *)malloc(sizeof(Elf64_Shdr));
      if (NULL == *section) {
        free(header);
        free(sections);
        return errno;
      }
      memcpy(*section, sections + i, sizeof(Elf64_Shdr));
      break;
    }
  free(header);
  free(sections);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<section_by_type>:
  endbr64
  push   %r13
  mov    %rdx,%r13
  push   %r12
  mov    %edi,%r12d
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movq   $0x0,(%rdx)
  lea    0x8(%rsp),%rsi
  movq   $0x0,0x8(%rsp)
  movq   $0x0,0x10(%rsp)
  call   <read_header@plt>
  test   %rax,%rax
  je     L63
L6:
  mov    D7(%rip),%rax
  mov    (%rax),%eax
L28:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L12
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L63:
  mov    0x8(%rsp),%rsi
  lea    0x10(%rsp),%rdx
  mov    %r12d,%edi
  call   <read_section_table@plt>
  test   %rax,%rax
  jne    L6
  mov    0x8(%rsp),%rdi
  mov    (%rdi),%rdx
  test   %rdx,%rdx
  je     L26
  mov    0x10(%rsp),%rcx
  xor    %ebx,%ebx
  xor    %eax,%eax
  jmp    L97
  nop
L84:
  add    $0x1,%rax
  add    $0x10,%rbx
  cmp    %rdx,%rax
  je     L26
L97:
  cmp    %rbp,0x8(%rcx,%rbx,1)
  jne    L84
  mov    $0x10,%edi
  call   <malloc@plt>
  mov    %rax,0x0(%r13)
  mov    %rax,%rdi
  test   %rax,%rax
  je     L55
  add    0x10(%rsp),%rbx
  mov    $0x10,%edx
  mov    %rbx,%rsi
  call   <memcpy@plt>
  mov    0x8(%rsp),%rdi
L26:
  call   <free@plt>
  mov    0x10(%rsp),%rdi
  call   <free@plt>
  xor    %eax,%eax
  jmp    L28
L55:
  mov    0x8(%rsp),%rdi
  call   <free@plt>
  mov    0x10(%rsp),%rdi
  call   <free@plt>
  jmp    L6
L12:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int section_by_type(int d, size_t const section_type, Elf64_Shdr **section) {
  Elf64_Ehdr *header = NULL;
  Elf64_Shdr *sections = NULL;
  size_t i;
  *section = NULL;
  if (read_header(d, &header) || read_section_table(d, header, &sections))
    return errno;
  for (i = 0; i < header->e_shnum; ++i)
    if (section_type == sections[i].sh_type) {
      *section = (Elf64_Shdr *)malloc(sizeof(Elf64_Shdr));
      if (NULL == *section) {
        free(header);
        free(sections);
        return errno;
      }
      memcpy(*section, sections + i, sizeof(Elf64_Shdr));
      break;
    }
  free(header);
  free(sections);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<section_by_type>:
  endbr64
  push   %r13
  mov    %rdx,%r13
  push   %r12
  mov    %edi,%r12d
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movq   $0x0,(%rdx)
  lea    0x8(%rsp),%rsi
  movq   $0x0,0x8(%rsp)
  movq   $0x0,0x10(%rsp)
  call   <read_header@plt>
  test   %rax,%rax
  je     L83
L52:
  mov    D4(%rip),%rax
  mov    (%rax),%eax
L25:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L93
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L83:
  mov    0x8(%rsp),%rsi
  lea    0x10(%rsp),%rdx
  mov    %r12d,%edi
  call   <read_section_table@plt>
  test   %rax,%rax
  jne    L52
  mov    0x8(%rsp),%rdi
  mov    (%rdi),%rdx
  test   %rdx,%rdx
  je     L23
  mov    0x10(%rsp),%rcx
  xor    %ebx,%ebx
  xor    %eax,%eax
  jmp    L71
  nop
L86:
  add    $0x1,%rax
  add    $0x10,%rbx
  cmp    %rdx,%rax
  je     L23
L71:
  cmp    %rbp,0x8(%rcx,%rbx,1)
  jne    L86
  mov    $0x10,%edi
  call   <malloc@plt>
  mov    %rax,0x0(%r13)
  mov    %rax,%rdi
  test   %rax,%rax
  je     L6
  add    0x10(%rsp),%rbx
  mov    $0x10,%edx
  mov    %rbx,%rsi
  call   <memcpy@plt>
  mov    0x8(%rsp),%rdi
L23:
  call   <free@plt>
  mov    0x10(%rsp),%rdi
  call   <free@plt>
  xor    %eax,%eax
  jmp    L25
L6:
  mov    0x8(%rsp),%rdi
  call   <free@plt>
  mov    0x10(%rsp),%rdi
  call   <free@plt>
  jmp    L52
L93:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int section_by_type(int d, size_t const section_type, Elf64_Shdr **section) {
  Elf64_Ehdr *header = NULL;
  Elf64_Shdr *sections = NULL;
  size_t i;
  *section = NULL;
  if (read_header(d, &header) || read_section_table(d, header, &sections))
    return errno;
  for (i = 0; i < header->e_shnum; ++i)
    if (section_type == sections[i].sh_type) {
      *section = (Elf64_Shdr *)malloc(sizeof(Elf64_Shdr));
      if (NULL == *section) {
        free(header);
        free(sections);
        return errno;
      }
      memcpy(*section, sections + i, sizeof(Elf64_Shdr));
      break;
    }
  free(header);
  free(sections);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<join>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r12
  push   %rbx
  mov    %rdi,%rbx
  mov    %rsi,%r12
  jmp    L32
L5:
  movzbl (%rbx),%eax
  movsbl %al,%eax
  mov    %r12,%rsi
  mov    %eax,%edi
  mov    $0x0,%eax
  call   <any@plt>
  test   %eax,%eax
  je     L25
  mov    $0x1,%eax
  jmp    L6
L25:
  add    $0x1,%rbx
L32:
  movzbl (%rbx),%eax
  test   %al,%al
  jne    L5
  mov    $0x0,%eax
L6:
  pop    %rbx
  pop    %r12
  pop    %rbp
  ret
```

assistant: ```c
join(s, t) register char *s, *t;
{
  for (; *s; s++)
    if (any(*s, t))
      return (1);
  return (0);
}
```
user: What is the c source code of the assembly code below?

```asm
<join>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  movzbl (%rdi),%edi
  test   %dil,%dil
  je     L26
L16:
  movsbl %dil,%edi
  mov    %rbp,%rsi
  mov    $0x0,%eax
  call   <any@plt>
  test   %eax,%eax
  jne    L32
  add    $0x1,%rbx
  movzbl (%rbx),%edi
  test   %dil,%dil
  jne    L16
  jmp    L21
L26:
  mov    $0x0,%eax
  jmp    L21
L32:
  mov    $0x1,%eax
L21:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
join(s, t) register char *s, *t;
{
  for (; *s; s++)
    if (any(*s, t))
      return (1);
  return (0);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<join>:
  endbr64
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  movsbl (%rdi),%edi
  test   %dil,%dil
  je     L37
  mov    %rsi,%rbp
  jmp    L36
  nopw   0x0(%rax,%rax,1)
L18:
  movsbl 0x1(%rbx),%edi
  add    $0x1,%rbx
  test   %dil,%dil
  je     L37
L36:
  xor    %eax,%eax
  mov    %rbp,%rsi
  call   <any@plt>
  test   %eax,%eax
  je     L18
  add    $0x8,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  ret
  nopw   0x0(%rax,%rax,1)
L37:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
join(s, t) register char *s, *t;
{
  for (; *s; s++)
    if (any(*s, t))
      return (1);
  return (0);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<join>:
  endbr64
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  movsbl (%rdi),%edi
  test   %dil,%dil
  je     L39
  mov    %rsi,%rbp
  jmp    L12
  nopw   0x0(%rax,%rax,1)
L14:
  movsbl 0x1(%rbx),%edi
  add    $0x1,%rbx
  test   %dil,%dil
  je     L39
L12:
  xor    %eax,%eax
  mov    %rbp,%rsi
  call   <any@plt>
  test   %eax,%eax
  je     L14
  add    $0x8,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  ret
  nopw   0x0(%rax,%rax,1)
L39:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
join(s, t) register char *s, *t;
{
  for (; *s; s++)
    if (any(*s, t))
      return (1);
  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
<__puts>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%edx
  mov    -0x18(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x0,%edi
  call   <write@plt>
  nop
  leave
  ret
```

assistant: ```c
void __puts(const char *buffer) {
  int len = strlen(buffer);
  write(0, buffer, len);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<__puts>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  mov    %eax,%edx
  mov    %rbx,%rsi
  mov    $0x0,%edi
  call   <write@plt>
  pop    %rbx
  ret
```

assistant: ```c
void __puts(const char *buffer) {
  int len = strlen(buffer);
  write(0, buffer, len);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<__puts>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  call   <strlen@plt>
  mov    %rbp,%rsi
  xor    %edi,%edi
  pop    %rbp
  mov    %eax,%edx
  jmp    1060 <write@plt>
```

assistant: ```c
void __puts(const char *buffer) {
  int len = strlen(buffer);
  write(0, buffer, len);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<__puts>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  call   <strlen@plt>
  mov    %rbp,%rsi
  xor    %edi,%edi
  pop    %rbp
  mov    %eax,%edx
  jmp    1060 <write@plt>
```

assistant: ```c
void __puts(const char *buffer) {
  int len = strlen(buffer);
  write(0, buffer, len);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<max3>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  mov    %edx,-0xc(%rbp)
  mov    -0x4(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jle    L12
  mov    -0x4(%rbp),%eax
  mov    %eax,-0x8(%rbp)
L12:
  mov    -0x8(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jle    L17
  mov    -0x8(%rbp),%eax
  mov    %eax,-0xc(%rbp)
L17:
  mov    -0xc(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int max3(int a, int b, int c) {
  if (a > b)
    b = a;
  if (b > c)
    c = b;
  return c;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<max3>:
  endbr64
  cmp    %edx,%esi
  cmovl  %edx,%esi
  cmp    %edi,%esi
  mov    %edi,%eax
  cmovge %esi,%eax
  ret
```

assistant: ```c
int max3(int a, int b, int c) {
  if (a > b)
    b = a;
  if (b > c)
    c = b;
  return c;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<max3>:
  endbr64
  cmp    %edx,%esi
  mov    %edi,%eax
  cmovl  %edx,%esi
  cmp    %edi,%esi
  cmovge %esi,%eax
  ret
```

assistant: ```c
int max3(int a, int b, int c) {
  if (a > b)
    b = a;
  if (b > c)
    c = b;
  return c;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<max3>:
  endbr64
  cmp    %edx,%esi
  mov    %edi,%eax
  cmovl  %edx,%esi
  cmp    %edi,%esi
  cmovge %esi,%eax
  ret
```

assistant: ```c
int max3(int a, int b, int c) {
  if (a > b)
    b = a;
  if (b > c)
    c = b;
  return c;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  jmp    L0
L1:
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <Buffer_with_interleave@plt>
L0:
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %eax,-0x4(%rbp)
  mov    D2(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,-0x4(%rbp)
  jne    L1
  mov    $0x0,%eax
  call   <PrintInterleavedBuf@plt>
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main(void) {
  int c;
  while ((c = getchar()) != EOF) {
    Buffer_with_interleave(c);
  }
  PrintInterleavedBuf();
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  jmp    L12
L2:
  call   <Buffer_with_interleave@plt>
L12:
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %eax,%edi
  mov    D9(%rip),%rax
  cmp    %edi,(%rax)
  jne    L2
  mov    $0x0,%eax
  call   <PrintInterleavedBuf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  int c;
  while ((c = getchar()) != EOF) {
    Buffer_with_interleave(c);
  }
  PrintInterleavedBuf();
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbx
  mov    D4(%rip),%rbx
  jmp    L1
  xchg   %ax,%ax
L8:
  call   <Buffer_with_interleave@plt>
L1:
  xor    %eax,%eax
  call   <getchar@plt>
  mov    %eax,%edi
  cmp    %eax,(%rbx)
  jne    L8
  xor    %eax,%eax
  call   <PrintInterleavedBuf@plt>
  xor    %eax,%eax
  pop    %rbx
  ret
```

assistant: ```c
int main(void) {
  int c;
  while ((c = getchar()) != EOF) {
    Buffer_with_interleave(c);
  }
  PrintInterleavedBuf();
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbx
  mov    D8(%rip),%rbx
  jmp    L12
  xchg   %ax,%ax
L3:
  call   <Buffer_with_interleave@plt>
L12:
  xor    %eax,%eax
  call   <getchar@plt>
  mov    %eax,%edi
  cmp    %eax,(%rbx)
  jne    L3
  xor    %eax,%eax
  call   <PrintInterleavedBuf@plt>
  xor    %eax,%eax
  pop    %rbx
  ret
```

assistant: ```c
int main(void) {
  int c;
  while ((c = getchar()) != EOF) {
    Buffer_with_interleave(c);
  }
  PrintInterleavedBuf();
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<xp_asprintf_start>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  add    $0x11,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L21
  mov    $0x0,%eax
  jmp    L79
L21:
  mov    -0x8(%rbp),%rax
  lea    0x10(%rax),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rax,%rsi
  mov    %rdx,%rdi
  call   <strcpy@plt>
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    -0x8(%rbp),%rdx
  add    $0x8,%rdx
  cltq
  mov    %rax,(%rdx)
  mov    -0x8(%rbp),%rax
  add    $0x10,%rax
  mov    %rax,-0x10(%rbp)
  jmp    L69
L9:
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x25,%al
  jne    L28
  mov    -0x10(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  cmp    $0x25,%al
  jne    L17
  addq   $0x1,-0x10(%rbp)
L28:
  addq   $0x1,-0x10(%rbp)
L69:
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L9
  jmp    L80
L17:
  nop
L80:
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L83
  mov    -0x8(%rbp),%rax
  movq   $0x0,(%rax)
  jmp    L36
L83:
  mov    -0x10(%rbp),%rax
  sub    -0x8(%rbp),%rax
  mov    %rax,%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,(%rax)
L36:
  mov    -0x8(%rbp),%rax
L79:
  leave
  ret
```

assistant: ```c
char *xp_asprintf_start(const char *format) {
  char *ret;
  char *p;
  ret = (char *)malloc(strlen(format) + 1 + ((sizeof(size_t) * 2)));
  if (ret == NULL)
    return (NULL);
  strcpy(ret + sizeof(size_t) * 2, format);
  *(size_t *)(ret + sizeof(size_t)) = strlen(format);
  for (p = ret + sizeof(size_t) * 2; *p; p++) {
    if (*p == '%') {
      if (*(p + 1) == '%')
        p++;
      else
        break;
    }
  }
  if (!*p)
    *(size_t *)ret = 0;
  else
    *(size_t *)ret = p - ret;
  return (ret);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<xp_asprintf_start>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r13
  call   <strlen@plt>
  lea    0x11(%rax),%edi
  call   <malloc@plt>
  mov    $0x0,%r12d
  test   %rax,%rax
  je     L32
  mov    %rax,%rbp
  mov    %rax,%r12
  lea    0x10(%rax),%rbx
  mov    %r13,%rsi
  mov    %rbx,%rdi
  call   <strcpy@plt>
  mov    %r13,%rdi
  call   <strlen@plt>
  cltq
  mov    %rax,0x8(%rbp)
  movzbl 0x10(%rbp),%edx
  test   %dl,%dl
  jne    L43
L1:
  movq   $0x0,0x0(%rbp)
L32:
  mov    %r12,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L56:
  lea    0x1(%rax),%rbx
  movzbl 0x1(%rax),%edx
  test   %dl,%dl
  je     L1
L43:
  mov    %rbx,%rax
  cmp    $0x25,%dl
  jne    L56
  cmpb   $0x25,0x1(%rbx)
  jne    L46
  lea    0x1(%rbx),%rax
  jmp    L56
L46:
  sub    %rbp,%rbx
  mov    %rbx,0x0(%rbp)
  jmp    L32
```

assistant: ```c
char *xp_asprintf_start(const char *format) {
  char *ret;
  char *p;
  ret = (char *)malloc(strlen(format) + 1 + ((sizeof(size_t) * 2)));
  if (ret == NULL)
    return (NULL);
  strcpy(ret + sizeof(size_t) * 2, format);
  *(size_t *)(ret + sizeof(size_t)) = strlen(format);
  for (p = ret + sizeof(size_t) * 2; *p; p++) {
    if (*p == '%') {
      if (*(p + 1) == '%')
        p++;
      else
        break;
    }
  }
  if (!*p)
    *(size_t *)ret = 0;
  else
    *(size_t *)ret = p - ret;
  return (ret);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<xp_asprintf_start>:
  endbr64
  push   %r13
  xor    %r13d,%r13d
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  lea    0x11(%rax),%edi
  call   <malloc@plt>
  test   %rax,%rax
  je     L8
  lea    0x10(%rax),%rbx
  mov    %r12,%rsi
  mov    %rax,%rbp
  mov    %rax,%r13
  mov    %rbx,%rdi
  call   <strcpy@plt>
  mov    %r12,%rdi
  call   <strlen@plt>
  movzbl 0x10(%rbp),%edx
  cltq
  mov    %rax,0x8(%rbp)
  test   %dl,%dl
  jne    L45
  jmp    L65
  nopl   0x0(%rax,%rax,1)
L58:
  movzbl 0x1(%rax),%edx
  lea    0x1(%rax),%rbx
  test   %dl,%dl
  je     L65
L45:
  mov    %rbx,%rax
  cmp    $0x25,%dl
  jne    L58
  cmpb   $0x25,0x1(%rbx)
  jne    L66
  movzbl 0x2(%rax),%edx
  add    $0x1,%rax
  lea    0x1(%rax),%rbx
  test   %dl,%dl
  jne    L45
L65:
  movq   $0x0,0x0(%rbp)
L8:
  add    $0x8,%rsp
  mov    %r13,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax,%rax,1)
L66:
  sub    %rbp,%rbx
  mov    %r13,%rax
  mov    %rbx,0x0(%rbp)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *xp_asprintf_start(const char *format) {
  char *ret;
  char *p;
  ret = (char *)malloc(strlen(format) + 1 + ((sizeof(size_t) * 2)));
  if (ret == NULL)
    return (NULL);
  strcpy(ret + sizeof(size_t) * 2, format);
  *(size_t *)(ret + sizeof(size_t)) = strlen(format);
  for (p = ret + sizeof(size_t) * 2; *p; p++) {
    if (*p == '%') {
      if (*(p + 1) == '%')
        p++;
      else
        break;
    }
  }
  if (!*p)
    *(size_t *)ret = 0;
  else
    *(size_t *)ret = p - ret;
  return (ret);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<xp_asprintf_start>:
  endbr64
  push   %r13
  xor    %r13d,%r13d
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  lea    0x11(%rax),%edi
  call   <malloc@plt>
  test   %rax,%rax
  je     L31
  lea    0x10(%rax),%rbx
  mov    %r12,%rsi
  mov    %rax,%rbp
  mov    %rax,%r13
  mov    %rbx,%rdi
  call   <strcpy@plt>
  mov    %r12,%rdi
  call   <strlen@plt>
  movzbl 0x10(%rbp),%edx
  cltq
  mov    %rax,0x8(%rbp)
  test   %dl,%dl
  jne    L69
  jmp    L27
  nopl   0x0(%rax,%rax,1)
L35:
  movzbl 0x1(%rax),%edx
  lea    0x1(%rax),%rbx
  test   %dl,%dl
  je     L27
L69:
  mov    %rbx,%rax
  cmp    $0x25,%dl
  jne    L35
  cmpb   $0x25,0x1(%rbx)
  jne    L66
  movzbl 0x2(%rax),%edx
  add    $0x1,%rax
  lea    0x1(%rax),%rbx
  test   %dl,%dl
  jne    L69
L27:
  movq   $0x0,0x0(%rbp)
L31:
  add    $0x8,%rsp
  mov    %r13,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax,%rax,1)
L66:
  sub    %rbp,%rbx
  mov    %r13,%rax
  mov    %rbx,0x0(%rbp)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *xp_asprintf_start(const char *format) {
  char *ret;
  char *p;
  ret = (char *)malloc(strlen(format) + 1 + ((sizeof(size_t) * 2)));
  if (ret == NULL)
    return (NULL);
  strcpy(ret + sizeof(size_t) * 2, format);
  *(size_t *)(ret + sizeof(size_t)) = strlen(format);
  for (p = ret + sizeof(size_t) * 2; *p; p++) {
    if (*p == '%') {
      if (*(p + 1) == '%')
        p++;
      else
        break;
    }
  }
  if (!*p)
    *(size_t *)ret = 0;
  else
    *(size_t *)ret = p - ret;
  return (ret);
}
```