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

```asm
<ell_vformat>:
  endbr64
  push   %r14
  mov    %esi,%r14d
  push   %r13
  mov    %rdi,%r13
  push   %r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  mov    %ebp,%edi
  call   <va_copy@plt>
  xor    %edi,%edi
  xor    %esi,%esi
  mov    %ebp,%ecx
  mov    %r13,%rdx
  call   <vsnprintf@plt>
  mov    %ebp,%edi
  mov    %eax,%ebx
  call   <va_end@plt>
  test   %ebx,%ebx
  js     L25
  lea    0x1(%rbx),%ebp
  mov    %ebp,%edi
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L25
  mov    %r14d,%ecx
  mov    %r13,%rdx
  mov    %ebp,%esi
  mov    %rax,%rdi
  call   <vsnprintf@plt>
  test   %eax,%eax
  jns    L5
  mov    %r12,%rdi
  call   <free@plt>
  nopl   0x0(%rax)
L25:
  xor    %r12d,%r12d
L5:
  pop    %rbx
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
char *ell_vformat(const char *format, va_list ap) {
  va_list aq;
  va_copy(aq, ap);
  int size = vsnprintf(NULL, 0, format, aq);
  va_end(aq);
  if (size < 0)
    return NULL;
  char *buf = malloc(size + 1);
  if (buf && vsnprintf(buf, size + 1, format, ap) < 0) {
    free(buf);
    return NULL;
  }
  return buf;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<ell_vformat>:
  endbr64
  push   %r14
  mov    %esi,%r14d
  push   %r13
  mov    %rdi,%r13
  push   %r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  mov    %ebp,%edi
  call   <va_copy@plt>
  xor    %edi,%edi
  xor    %esi,%esi
  mov    %ebp,%ecx
  mov    %r13,%rdx
  call   <vsnprintf@plt>
  mov    %ebp,%edi
  mov    %eax,%ebx
  call   <va_end@plt>
  test   %ebx,%ebx
  js     L5
  lea    0x1(%rbx),%ebp
  mov    %ebp,%edi
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L5
  mov    %r14d,%ecx
  mov    %r13,%rdx
  mov    %ebp,%esi
  mov    %rax,%rdi
  call   <vsnprintf@plt>
  test   %eax,%eax
  jns    L18
  mov    %r12,%rdi
  call   <free@plt>
  nopl   0x0(%rax)
L5:
  xor    %r12d,%r12d
L18:
  pop    %rbx
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
char *ell_vformat(const char *format, va_list ap) {
  va_list aq;
  va_copy(aq, ap);
  int size = vsnprintf(NULL, 0, format, aq);
  va_end(aq);
  if (size < 0)
    return NULL;
  char *buf = malloc(size + 1);
  if (buf && vsnprintf(buf, size + 1, format, ap) < 0) {
    free(buf);
    return NULL;
  }
  return buf;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<print_list>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <system@plt>
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  movl   $0x0,-0x4(%rbp)
  movl   $0x1,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0xc(%rbp)
  mov    %eax,%edi
  call   <li@plt>
  movl   $0x0,-0x8(%rbp)
  jmp    L11
L21:
  mov    -0x4(%rbp),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <putchar@plt>
  cmpl   $0xa,-0x4(%rbp)
  jne    L14
  mov    -0xc(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0xc(%rbp)
  mov    %eax,%edi
  call   <li@plt>
L14:
  addl   $0x1,-0x8(%rbp)
L11:
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jne    L21
  mov    $0xa,%edi
  call   <putchar@plt>
  nop
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_label: D1
type_infered: string

tool: D1: clear

assistant: ```c
void print_list(const char *items, char *heading) {
  system("clear");
  puts(heading);
  int curchar = 0, item_num = 1;
  li(item_num++);
  for (int i = 0; (curchar = items[i]) != '\0'; i++) {
    putchar(curchar);
    if (curchar == '\n')
      li(item_num++);
  }
  putchar('\n');
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<print_list>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    %rsi,%rbx
  lea    D4(%rip),%rdi
  call   <system@plt>
  mov    %rbx,%rdi
  call   <puts@plt>
  mov    $0x1,%edi
  call   <li@plt>
  movsbl 0x0(%rbp),%ebx
  test   %ebx,%ebx
  je     L1
  add    $0x1,%rbp
  mov    $0x2,%r12d
  jmp    L49
L39:
  movsbl 0x0(%rbp),%ebx
  add    $0x1,%rbp
  test   %ebx,%ebx
  je     L1
L49:
  mov    %ebx,%edi
  call   <putchar@plt>
  cmp    $0xa,%ebx
  jne    L39
  lea    0x1(%r12),%ebx
  mov    %r12d,%edi
  call   <li@plt>
  mov    %ebx,%r12d
  jmp    L39
L1:
  mov    $0xa,%edi
  call   <putchar@plt>
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

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

tool: D4: clear

assistant: ```c
void print_list(const char *items, char *heading) {
  system("clear");
  puts(heading);
  int curchar = 0, item_num = 1;
  li(item_num++);
  for (int i = 0; (curchar = items[i]) != '\0'; i++) {
    putchar(curchar);
    if (curchar == '\n')
      li(item_num++);
  }
  putchar('\n');
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<print_list>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  lea    D9(%rip),%rdi
  push   %rbx
  call   <system@plt>
  mov    %r12,%rdi
  call   <puts@plt>
  mov    $0x1,%edi
  call   <li@plt>
  movsbl 0x0(%rbp),%ebx
  test   %ebx,%ebx
  je     L26
  add    $0x1,%rbp
  mov    $0x2,%r12d
  jmp    L3
L12:
  movsbl 0x0(%rbp),%ebx
  add    $0x1,%rbp
  test   %ebx,%ebx
  je     L26
L3:
  mov    %ebx,%edi
  call   <putchar@plt>
  cmp    $0xa,%ebx
  jne    L12
  lea    0x1(%r12),%ebx
  mov    %r12d,%edi
  add    $0x1,%rbp
  call   <li@plt>
  mov    %ebx,%r12d
  movsbl -0x1(%rbp),%ebx
  test   %ebx,%ebx
  jne    L3
L26:
  pop    %rbx
  mov    $0xa,%edi
  pop    %rbp
  pop    %r12
  jmp    1080 <putchar@plt>
```

tool_calls: name: retrieve_data
arguments: label: D9
type: string

tool: D9: clear

assistant: ```c
void print_list(const char *items, char *heading) {
  system("clear");
  puts(heading);
  int curchar = 0, item_num = 1;
  li(item_num++);
  for (int i = 0; (curchar = items[i]) != '\0'; i++) {
    putchar(curchar);
    if (curchar == '\n')
      li(item_num++);
  }
  putchar('\n');
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<print_list>:
  endbr64
  push   %r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  lea    D1(%rip),%rdi
  push   %rbx
  sub    $0x8,%rsp
  call   <system@plt>
  mov    %r12,%rdi
  call   <puts@plt>
  mov    $0x1,%edi
  call   <li@plt>
  movsbl 0x0(%rbp),%ebx
  test   %ebx,%ebx
  je     L47
  add    $0x1,%rbp
  mov    $0x2,%r12d
  nopl   0x0(%rax)
L19:
  mov    %ebx,%edi
  call   <putchar@plt>
  cmp    $0xa,%ebx
  je     L27
  movsbl 0x0(%rbp),%ebx
  add    $0x1,%rbp
  test   %ebx,%ebx
  jne    L19
L47:
  add    $0x8,%rsp
  mov    $0xa,%edi
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    1080 <putchar@plt>
  nopl   0x0(%rax)
L27:
  mov    %r12d,%edi
  lea    0x1(%r12),%r13d
  add    $0x1,%rbp
  call   <li@plt>
  movsbl -0x1(%rbp),%ebx
  test   %ebx,%ebx
  je     L47
  mov    %r13d,%r12d
  jmp    L19
```

tool_calls: name: unpack_data
arguments: data_tag: D1
type_guessed: string

tool: D1: clear

assistant: ```c
void print_list(const char *items, char *heading) {
  system("clear");
  puts(heading);
  int curchar = 0, item_num = 1;
  li(item_num++);
  for (int i = 0; (curchar = items[i]) != '\0'; i++) {
    putchar(curchar);
    if (curchar == '\n')
      li(item_num++);
  }
  putchar('\n');
}
```
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    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x2,-0x30(%rbp)
  movl   $0x6,-0x2c(%rbp)
  movl   $0x4,-0x28(%rbp)
  movl   $0x8,-0x24(%rbp)
  movl   $0xa,-0x20(%rbp)
  movl   $0xc,-0x1c(%rbp)
  movl   $0x59,-0x18(%rbp)
  movl   $0x44,-0x14(%rbp)
  movl   $0x2d,-0x10(%rbp)
  movl   $0x25,-0xc(%rbp)
  lea    D37(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  movq   $0x0,-0x38(%rbp)
  jmp    L1
L12:
  mov    -0x38(%rbp),%rax
  mov    -0x30(%rbp,%rax,4),%eax
  mov    %eax,%esi
  lea    D48(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addq   $0x1,-0x38(%rbp)
L1:
  cmpq   $0x9,-0x38(%rbp)
  jbe    L12
  lea    -0x30(%rbp),%rax
  mov    $0xa,%esi
  mov    %rax,%rdi
  call   <bubbleSort@plt>
  lea    D14(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  movq   $0x0,-0x38(%rbp)
  jmp    L34
L27:
  mov    -0x38(%rbp),%rax
  mov    -0x30(%rbp,%rax,4),%eax
  mov    %eax,%esi
  lea    D48(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addq   $0x1,-0x38(%rbp)
L34:
  cmpq   $0x9,-0x38(%rbp)
  jbe    L27
  lea    D41(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L44
  call   <__stack_chk_fail@plt>
L44:
  leave
  ret
```

tool_calls: name: read_data
arguments: label: D37
datatype: string
name: read_data
arguments: label: D48
datatype: string
name: read_data
arguments: label: D14
datatype: string
name: read_data
arguments: label: D41
datatype: string

tool: D37: Dati nell'ordine originario

tool: D48: %4d

tool: D14: 
Dati ordinati in modo crescente

tool: 

assistant: ```c
int main(void) {
  int a[10] = {2, 6, 4, 8, 10, 12, 89, 68, 45, 37};
  size_t i;
  puts("Dati nell'ordine originario");
  for (i = 0; i < 10; ++i) {
    printf("%4d", a[i]);
  }
  bubbleSort(a, 10);
  puts("\nDati ordinati in modo crescente");
  for (i = 0; i < 10; ++i) {
    printf("%4d", a[i]);
  }
  puts("");
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  movl   $0x2,(%rsp)
  movl   $0x6,0x4(%rsp)
  movl   $0x4,0x8(%rsp)
  movl   $0x8,0xc(%rsp)
  movl   $0xa,0x10(%rsp)
  movl   $0xc,0x14(%rsp)
  movl   $0x59,0x18(%rsp)
  movl   $0x44,0x1c(%rsp)
  movl   $0x2d,0x20(%rsp)
  movl   $0x25,0x24(%rsp)
  lea    D26(%rip),%rdi
  call   <puts@plt>
  mov    %rsp,%rbx
  lea    0x28(%rsp),%r12
  mov    %rbx,%rbp
  lea    D13(%rip),%r13
L7:
  mov    0x0(%rbp),%esi
  mov    %r13,%rdi
  call   <printf@plt>
  add    $0x4,%rbp
  cmp    %r12,%rbp
  jne    L7
  mov    %rsp,%rdi
  mov    $0xa,%esi
  call   <bubbleSort@plt>
  lea    D19(%rip),%rdi
  call   <puts@plt>
  lea    D13(%rip),%rbp
L28:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  add    $0x4,%rbx
  cmp    %r12,%rbx
  jne    L28
  lea    D8(%rip),%rdi
  call   <puts@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L24
  mov    $0x0,%eax
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L24:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D26
datatype: string
name: read_data
arguments: label: D13
datatype: string
name: read_data
arguments: label: D19
datatype: string
name: read_data
arguments: label: D8
datatype: string

tool: D26: Dati nell'ordine originario

tool: D13: %4d

tool: D19: 
Dati ordinati in modo crescente

tool: 

assistant: ```c
int main(void) {
  int a[10] = {2, 6, 4, 8, 10, 12, 89, 68, 45, 37};
  size_t i;
  puts("Dati nell'ordine originario");
  for (i = 0; i < 10; ++i) {
    printf("%4d", a[i]);
  }
  bubbleSort(a, 10);
  puts("\nDati ordinati in modo crescente");
  for (i = 0; i < 10; ++i) {
    printf("%4d", a[i]);
  }
  puts("");
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r14
  lea    D18(%rip),%rdi
  push   %r13
  lea    D34(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  movabs $0x600000002,%rax
  mov    %rsp,%r14
  lea    0x28(%rsp),%r12
  mov    %r14,%rbx
  mov    %r14,%rbp
  mov    %rax,(%rsp)
  movabs $0x800000004,%rax
  mov    %rax,0x8(%rsp)
  movabs $0xc0000000a,%rax
  mov    %rax,0x10(%rsp)
  movabs $0x4400000059,%rax
  mov    %rax,0x18(%rsp)
  movabs $0x250000002d,%rax
  mov    %rax,0x20(%rsp)
  call   <puts@plt>
  nopl   0x0(%rax)
L15:
  mov    0x0(%rbp),%esi
  mov    %r13,%rdi
  add    $0x4,%rbp
  call   <printf@plt>
  cmp    %r12,%rbp
  jne    L15
  mov    %r14,%rdi
  mov    $0xa,%esi
  call   <bubbleSort@plt>
  lea    D39(%rip),%rdi
  call   <puts@plt>
  nopl   (%rax)
L16:
  mov    (%rbx),%esi
  mov    %r13,%rdi
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %r12,%rbx
  jne    L16
  lea    D6(%rip),%rdi
  call   <puts@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L13
  add    $0x30,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L13:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D18
datatype: string
name: unpack_data
arguments: data_label: D34
datatype: string
name: unpack_data
arguments: data_label: D39
datatype: string
name: unpack_data
arguments: data_label: D6
datatype: string

tool: D18: Dati nell'ordine originario

tool: D34: %4d

tool: D39: 
Dati ordinati in modo crescente

tool: 

assistant: ```c
int main(void) {
  int a[10] = {2, 6, 4, 8, 10, 12, 89, 68, 45, 37};
  size_t i;
  puts("Dati nell'ordine originario");
  for (i = 0; i < 10; ++i) {
    printf("%4d", a[i]);
  }
  bubbleSort(a, 10);
  puts("\nDati ordinati in modo crescente");
  for (i = 0; i < 10; ++i) {
    printf("%4d", a[i]);
  }
  puts("");
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r14
  lea    D16(%rip),%rdi
  push   %r13
  lea    D18(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x30,%rsp
  movdqa D69(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    D12(%rip),%rax
  mov    %rsp,%r14
  lea    0x28(%rsp),%r12
  movaps %xmm0,(%rsp)
  movdqa D54(%rip),%xmm0
  mov    %r14,%rbx
  mov    %r14,%rbp
  mov    %rax,0x20(%rsp)
  movaps %xmm0,0x10(%rsp)
  call   <puts@plt>
  cs nopw 0x0(%rax,%rax,1)
L0:
  mov    0x0(%rbp),%esi
  mov    %r13,%rdi
  add    $0x4,%rbp
  call   <printf@plt>
  cmp    %r12,%rbp
  jne    L0
  mov    %r14,%rdi
  mov    $0xa,%esi
  call   <bubbleSort@plt>
  lea    D40(%rip),%rdi
  call   <puts@plt>
  nopl   (%rax)
L11:
  mov    (%rbx),%esi
  mov    %r13,%rdi
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %r12,%rbx
  jne    L11
  lea    D0(%rip),%rdi
  call   <puts@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L26
  add    $0x30,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L26:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D16
type_guessed: string
name: parse_data
arguments: label: D18
type_guessed: string
name: parse_data
arguments: label: D69
type_guessed: byte[16]
name: parse_data
arguments: label: D12
type_guessed: byte[8]
name: parse_data
arguments: label: D54
type_guessed: byte[16]
name: parse_data
arguments: label: D40
type_guessed: string
name: parse_data
arguments: label: D0
type_guessed: string

tool: D16: Dati nell'ordine originario

tool: D18: %4d

tool: {"D69": [0x02,0x00,0x00,0x00,0x06,0x00,0x00,0x00,0x04,0x00,0x00,0x00,0x08,0x00,0x00,0x00]}

tool: {"D12": [0x2d,0x00,0x00,0x00,0x25,0x00,0x00,0x00]}

tool: {"D54": [0x0a,0x00,0x00,0x00,0x0c,0x00,0x00,0x00,0x59,0x00,0x00,0x00,0x44,0x00,0x00,0x00]}

tool: D40: 
Dati ordinati in modo crescente

tool: 

assistant: ```c
int main(void) {
  int a[10] = {2, 6, 4, 8, 10, 12, 89, 68, 45, 37};
  size_t i;
  puts("Dati nell'ordine originario");
  for (i = 0; i < 10; ++i) {
    printf("%4d", a[i]);
  }
  bubbleSort(a, 10);
  puts("\nDati ordinati in modo crescente");
  for (i = 0; i < 10; ++i) {
    printf("%4d", a[i]);
  }
  puts("");
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<recvn>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x18(%rbp)
  movl   $0x0,-0x4(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L5
L29:
  mov    -0x18(%rbp),%eax
  sub    -0x8(%rbp),%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x20(%rbp),%rax
  lea    (%rcx,%rax,1),%rsi
  mov    -0x14(%rbp),%eax
  mov    $0x0,%ecx
  mov    %eax,%edi
  call   <recv@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jg     L4
  mov    $0xffffffff,%eax
  jmp    L8
L4:
  mov    -0x4(%rbp),%eax
  add    %eax,-0x8(%rbp)
L5:
  mov    -0x8(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jl     L29
  mov    $0x0,%eax
L8:
  leave
  ret
```

assistant: ```c
int recvn(int fd, char *strBuf, int size) {
  int iRet = 0, iRead = 0;
  while (iRead < size) {
    iRet = recv(fd, strBuf + iRead, size - iRead, 0);
    if (iRet <= 0) {
      return -1;
    }
    iRead += iRet;
  }
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<recvn>:
  endbr64
  test   %edx,%edx
  jle    L27
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%r12d
  mov    %rsi,%r13
  mov    %edx,%ebp
  mov    $0x0,%ebx
L29:
  mov    %ebp,%edx
  sub    %ebx,%edx
  movslq %ebx,%rsi
  add    %r13,%rsi
  mov    $0x0,%ecx
  mov    %r12d,%edi
  call   <recv@plt>
  test   %eax,%eax
  jle    L2
  add    %eax,%ebx
  cmp    %ebx,%ebp
  jg     L29
  mov    $0x0,%eax
  jmp    L3
L27:
  mov    $0x0,%eax
  ret
L2:
  mov    $0xffffffff,%eax
L3:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int recvn(int fd, char *strBuf, int size) {
  int iRet = 0, iRead = 0;
  while (iRead < size) {
    iRet = recv(fd, strBuf + iRead, size - iRead, 0);
    if (iRet <= 0) {
      return -1;
    }
    iRead += iRet;
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<recvn>:
  endbr64
  test   %edx,%edx
  jle    L2
  push   %r13
  mov    %rsi,%r13
  push   %r12
  mov    %edi,%r12d
  push   %rbp
  mov    %edx,%ebp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  jmp    L29
  xchg   %ax,%ax
L10:
  add    %eax,%ebx
  cmp    %ebx,%ebp
  jle    L12
L29:
  mov    %ebp,%edx
  movslq %ebx,%rsi
  xor    %ecx,%ecx
  mov    %r12d,%edi
  sub    %ebx,%edx
  add    %r13,%rsi
  call   <recv@plt>
  test   %eax,%eax
  jg     L10
  add    $0x8,%rsp
  mov    $0xffffffff,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  xchg   %ax,%ax
L12:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L2:
  xor    %eax,%eax
  ret
```

assistant: ```c
int recvn(int fd, char *strBuf, int size) {
  int iRet = 0, iRead = 0;
  while (iRead < size) {
    iRet = recv(fd, strBuf + iRead, size - iRead, 0);
    if (iRet <= 0) {
      return -1;
    }
    iRead += iRet;
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<recvn>:
  endbr64
  test   %edx,%edx
  jle    L24
  push   %r13
  mov    %rsi,%r13
  push   %r12
  mov    %edi,%r12d
  push   %rbp
  mov    %edx,%ebp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  jmp    L3
  xchg   %ax,%ax
L10:
  add    %eax,%ebx
  cmp    %ebx,%ebp
  jle    L35
L3:
  mov    %ebp,%edx
  movslq %ebx,%rsi
  xor    %ecx,%ecx
  mov    %r12d,%edi
  sub    %ebx,%edx
  add    %r13,%rsi
  call   <recv@plt>
  test   %eax,%eax
  jg     L10
  add    $0x8,%rsp
  mov    $0xffffffff,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  xchg   %ax,%ax
L35:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L24:
  xor    %eax,%eax
  ret
```

assistant: ```c
int recvn(int fd, char *strBuf, int size) {
  int iRet = 0, iRead = 0;
  while (iRead < size) {
    iRet = recv(fd, strBuf + iRead, size - iRead, 0);
    if (iRet <= 0) {
      return -1;
    }
    iRead += iRet;
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<raw_voltage>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  add    $0x30,%rax
  mov    (%rax),%rax
  mov    %eax,%edx
  mov    -0x18(%rbp),%rax
  add    $0x38,%rax
  mov    (%rax),%rax
  shl    $0x8,%eax
  or     %eax,%edx
  mov    -0x18(%rbp),%rax
  add    $0x40,%rax
  mov    (%rax),%rax
  shl    $0x10,%eax
  or     %edx,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  and    $0x800000,%eax
  test   %eax,%eax
  je     L5
  mov    -0x4(%rbp),%eax
  or     $0xff000000,%eax
  mov    %eax,-0x4(%rbp)
L5:
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int32_t raw_voltage(uint8_t *pkt) {
  uint32_t i;
  i = (uint32_t)pkt[6] | (uint32_t)pkt[7] << 8 | (uint32_t)pkt[8] << 16;
  if (i & 0x00800000)
    i |= 0xff000000;
  return (int32_t)i;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<raw_voltage>:
  endbr64
  mov    0x38(%rdi),%eax
  shl    $0x8,%eax
  mov    0x40(%rdi),%edx
  shl    $0x10,%edx
  or     %edx,%eax
  mov    0x30(%rdi),%rcx
  mov    %eax,%edx
  or     %ecx,%edx
  mov    %edx,%esi
  or     $0xff000000,%esi
  test   $0x800000,%edx
  cmove  %edx,%esi
  mov    %esi,%eax
  ret
```

assistant: ```c
int32_t raw_voltage(uint8_t *pkt) {
  uint32_t i;
  i = (uint32_t)pkt[6] | (uint32_t)pkt[7] << 8 | (uint32_t)pkt[8] << 16;
  if (i & 0x00800000)
    i |= 0xff000000;
  return (int32_t)i;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<raw_voltage>:
  endbr64
  mov    0x38(%rdi),%eax
  mov    0x40(%rdi),%edx
  shl    $0x8,%eax
  shl    $0x10,%edx
  or     %edx,%eax
  or     0x30(%rdi),%eax
  mov    %eax,%edx
  or     $0xff000000,%edx
  test   $0x800000,%eax
  cmovne %edx,%eax
  ret
```

assistant: ```c
int32_t raw_voltage(uint8_t *pkt) {
  uint32_t i;
  i = (uint32_t)pkt[6] | (uint32_t)pkt[7] << 8 | (uint32_t)pkt[8] << 16;
  if (i & 0x00800000)
    i |= 0xff000000;
  return (int32_t)i;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<raw_voltage>:
  endbr64
  mov    0x38(%rdi),%eax
  mov    0x40(%rdi),%edx
  shl    $0x8,%eax
  shl    $0x10,%edx
  or     %edx,%eax
  or     0x30(%rdi),%eax
  mov    %eax,%edx
  or     $0xff000000,%edx
  test   $0x800000,%eax
  cmovne %edx,%eax
  ret
```

assistant: ```c
int32_t raw_voltage(uint8_t *pkt) {
  uint32_t i;
  i = (uint32_t)pkt[6] | (uint32_t)pkt[7] << 8 | (uint32_t)pkt[8] << 16;
  if (i & 0x00800000)
    i |= 0xff000000;
  return (int32_t)i;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<mempcpy3>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x4030201,-0xc(%rbp)
  lea    -0xc(%rbp),%rcx
  add    $0x2,%rcx
  lea    -0xf(%rbp),%rax
  add    $0x1,%rax
  mov    $0x2,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <mempcpy@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L1
  call   <__stack_chk_fail@plt>
L1:
  leave
  ret
```

assistant: ```c
void mempcpy3() {
  char src[] = {1, 2, 3, 4};
  char dst[3];
  mempcpy(dst + 1, src + 2, 2);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<mempcpy3>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0x4030201,0x4(%rsp)
  lea    0x6(%rsp),%rsi
  lea    0x2(%rsp),%rdi
  mov    $0x2,%edx
  call   <mempcpy@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x18,%rsp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void mempcpy3() {
  char src[] = {1, 2, 3, 4};
  char dst[3];
  mempcpy(dst + 1, src + 2, 2);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<mempcpy3>:
  endbr64
  sub    $0x18,%rsp
  mov    $0x2,%edx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x6(%rsp),%rsi
  lea    0x2(%rsp),%rdi
  movl   $0x4030201,0x4(%rsp)
  call   <mempcpy@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x18,%rsp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void mempcpy3() {
  char src[] = {1, 2, 3, 4};
  char dst[3];
  mempcpy(dst + 1, src + 2, 2);
}
```