ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0xa8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x98(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rdi
  mov    $0xf,%ecx
  rep stos %rax,%es:(%rdi)
  movl   $0x0,(%rdi)
  movl   $0x1,0x10(%rsp)
  lea    0x10(%rsp),%rax
  lea    0x88(%rsp),%rcx
L87:
  mov    (%rax),%ebx
  lea    (%rbx,%rbx,1),%edx
  mov    %edx,0x4(%rax)
  add    $0x4,%rax
  cmp    %rcx,%rax
  jne    L87
  mov    %rsp,%rsi
  lea    D9(%rip),%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    (%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,(%rsp)
  test   %eax,%eax
  je     L88
  lea    0xc(%rsp),%r13
  lea    0x8(%rsp),%r12
  lea    D13(%rip),%rbp
  lea    D21(%rip),%rbx
  jmp    L110
L39:
  add    %ecx,%edx
  add    %edx,%eax
  cmp    %eax,%ecx
  jge    L56
  movslq %ecx,%rdx
  lea    0xc(%rsp,%rdx,4),%rdx
  mov    $0x0,%esi
L45:
  xor    (%rdx),%esi
  sub    $0x4,%rdx
  sub    $0x1,%eax
  cmp    %eax,%ecx
  jl     L45
L41:
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    (%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,(%rsp)
  test   %eax,%eax
  je     L88
L110:
  lea    0x4(%rsp),%rsi
  mov    %r13,%rcx
  mov    %r12,%rdx
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    0x8(%rsp),%edi
  call   <ones@plt>
  mov    %eax,0x8(%rsp)
  mov    0xc(%rsp),%edi
  call   <ones@plt>
  mov    %eax,0xc(%rsp)
  mov    0x8(%rsp),%edx
  lea    (%rax,%rdx,1),%edi
  mov    0x4(%rsp),%ecx
  cmp    %ecx,%edi
  jle    L39
  lea    (%rcx,%rcx,1),%eax
  mov    $0x0,%esi
  cmp    %eax,%edi
  jge    L41
  movslq %ecx,%rcx
  lea    0xc(%rsp,%rcx,4),%rdx
  mov    $0x0,%esi
L62:
  xor    (%rdx),%esi
  sub    $0x4,%rdx
  sub    $0x1,%eax
  cmp    %eax,%edi
  jl     L62
  jmp    L41
L56:
  mov    $0x0,%esi
  jmp    L41
L88:
  mov    0x98(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L68
  mov    $0x0,%eax
  add    $0xa8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L68:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D9
datatype: string
name: unpack_data
arguments: label: D13
datatype: string
name: unpack_data
arguments: label: D21
datatype: string

tool: D9: %d

tool: D13: %d%d%d

tool: D21: %d


assistant: ```c
int main() {
  int t, n, a, b, p[31] = {1}, i, x;
  for (i = 1; i < 31; ++i)
    p[i] = p[i - 1] << 1;
  scanf("%d", &t);
  while (t--) {
    scanf("%d%d%d", &n, &a, &b);
    a = ones(a);
    b = ones(b);
    x = 0;
    if (a + b <= n)
      for (i = n; i + a + b > n; --i)
        x ^= p[i - 1];
    else
      for (i = n; i + n > a + b; --i)
        x ^= p[i - 1];
    printf("%d\n", x);
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %r15
  mov    $0xf,%ecx
  mov    $0x1,%edx
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0xa8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x98(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rbx
  mov    %rbx,%rdi
  rep stos %rax,%es:(%rdi)
  movl   $0x1,0x10(%rsp)
  lea    0x14(%rsp),%rax
  lea    0x88(%rsp),%rcx
  movl   $0x0,(%rdi)
  jmp    L23
  nopw   0x0(%rax,%rax,1)
L69:
  add    $0x4,%rax
L23:
  add    %edx,%edx
  mov    %edx,(%rax)
  cmp    %rcx,%rax
  jne    L69
  mov    %rsp,%rsi
  lea    D7(%rip),%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    (%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,(%rsp)
  test   %eax,%eax
  je     L33
  lea    0xc(%rsp),%r15
  lea    0x8(%rsp),%r14
  lea    0x4(%rsp),%r13
  lea    D26(%rip),%r12
  lea    D5(%rip),%rbp
  nopl   0x0(%rax)
L6:
  mov    %r15,%rcx
  mov    %r14,%rdx
  mov    %r13,%rsi
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    0x8(%rsp),%edi
  call   <ones@plt>
  mov    0xc(%rsp),%edi
  mov    %eax,0x8(%rsp)
  call   <ones@plt>
  mov    0x8(%rsp),%edx
  movslq 0x4(%rsp),%rcx
  mov    %eax,0xc(%rsp)
  lea    (%rax,%rdx,1),%edi
  cmp    %ecx,%edi
  jle    L66
  lea    (%rcx,%rcx,1),%eax
  cmp    %eax,%edi
  jge    L113
  lea    -0x4(%rbx,%rcx,4),%rdx
  xor    %esi,%esi
  nop
L83:
  sub    $0x1,%eax
  xor    (%rdx),%esi
  sub    $0x4,%rdx
  cmp    %eax,%edi
  jl     L83
L119:
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    (%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,(%rsp)
  test   %eax,%eax
  jne    L6
L33:
  mov    0x98(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L99
  add    $0xa8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L66:
  add    %ecx,%edx
  add    %edx,%eax
  cmp    %eax,%ecx
  jge    L113
  movslq %ecx,%rdx
  xor    %esi,%esi
  lea    -0x4(%rbx,%rdx,4),%rdx
  nopw   0x0(%rax,%rax,1)
L56:
  sub    $0x1,%eax
  xor    (%rdx),%esi
  sub    $0x4,%rdx
  cmp    %eax,%ecx
  jl     L56
  jmp    L119
  nopw   0x0(%rax,%rax,1)
L113:
  xor    %esi,%esi
  jmp    L119
L99:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D7
type_guessed: string
name: extract_symbolic_data
arguments: data_label: D26
type_guessed: string
name: extract_symbolic_data
arguments: data_label: D5
type_guessed: string

tool: D7: %d

tool: D26: %d%d%d

tool: D5: %d


assistant: ```c
int main() {
  int t, n, a, b, p[31] = {1}, i, x;
  for (i = 1; i < 31; ++i)
    p[i] = p[i - 1] << 1;
  scanf("%d", &t);
  while (t--) {
    scanf("%d%d%d", &n, &a, &b);
    a = ones(a);
    b = ones(b);
    x = 0;
    if (a + b <= n)
      for (i = n; i + a + b > n; --i)
        x ^= p[i - 1];
    else
      for (i = n; i + n > a + b; --i)
        x ^= p[i - 1];
    printf("%d\n", x);
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<f6>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x40(%rbp),%rax
  mov    %rax,-0x50(%rbp)
  mov    -0x50(%rbp),%rax
  mov    %rax,-0x48(%rbp)
  mov    -0x48(%rbp),%rax
  movl   $0x64,(%rax)
  mov    -0x48(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,%edi
  call   <putchar@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L3
  call   <__stack_chk_fail@plt>
L3:
  leave
  ret
```

assistant: ```c
void f6(void) {
  int t[4][3];
  typedef int arr[3];
  typedef arr(*brr)[4];
  int(*a)[4][3];
  brr b = &t;
  a = b;
  (*a)[0][0] = 'd';
  putchar((*a)[0][0]);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<f6>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x64,%edi
  call   <putchar@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void f6(void) {
  int t[4][3];
  typedef int arr[3];
  typedef arr(*brr)[4];
  int(*a)[4][3];
  brr b = &t;
  a = b;
  (*a)[0][0] = 'd';
  putchar((*a)[0][0]);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<f6>:
  endbr64
  mov    $0x64,%edi
  jmp    1050 <putchar@plt>
```

assistant: ```c
void f6(void) {
  int t[4][3];
  typedef int arr[3];
  typedef arr(*brr)[4];
  int(*a)[4][3];
  brr b = &t;
  a = b;
  (*a)[0][0] = 'd';
  putchar((*a)[0][0]);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<f6>:
  endbr64
  mov    $0x64,%edi
  jmp    1050 <putchar@plt>
```

assistant: ```c
void f6(void) {
  int t[4][3];
  typedef int arr[3];
  typedef arr(*brr)[4];
  int(*a)[4][3];
  brr b = &t;
  a = b;
  (*a)[0][0] = 'd';
  putchar((*a)[0][0]);
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rsi,%rbp
  cmp    $0x2,%edi
  jle    L34
  mov    $0x0,%r15d
  cmp    $0x3,%edi
  jle    L49
  mov    0x18(%rsi),%rax
  cmpb   $0x6e,(%rax)
  sete   %r15b
  movzbl %r15b,%r15d
L49:
  mov    0x8(%rbp),%rdi
  call   <strlen@plt>
  lea    0x1(%rax),%ebx
  mov    0x10(%rbp),%rdi
  call   <strlen@plt>
  mov    %eax,0x8(%rsp)
  test   %r15d,%r15d
  sete   %r12b
  movzbl %r12b,%r12d
  mov    0x10(%rbp),%rdx
  mov    0x8(%rbp),%rdi
  mov    %r12d,%r9d
  mov    $0xa,%r8d
  mov    %eax,%ecx
  mov    %ebx,%esi
  call   <dummy_crypt_seed@plt>
  mov    %rax,%r14
  mov    %r12d,%edx
  mov    $0xa,%esi
  mov    %ebx,%edi
  call   <dummy_crypt_size_seed@plt>
  mov    %eax,%ebx
  mov    %eax,0xc(%rsp)
  lea    D31(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %ebx,%eax
  test   %ebx,%ebx
  jle    L27
  mov    %r14,%rbx
  lea    -0x1(%rax),%eax
  lea    0x1(%r14,%rax,1),%r13
  lea    D39(%rip),%r12
L46:
  movsbl (%rbx),%esi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x1,%rbx
  cmp    %r13,%rbx
  jne    L46
L27:
  mov    $0xa,%edi
  call   <putchar@plt>
  test   %r15d,%r15d
  mov    0x10(%rbp),%rdx
  mov    $0xa,%r8d
  cmove  %r15d,%r8d
  mov    0x8(%rsp),%ecx
  mov    0xc(%rsp),%esi
  mov    %r14,%rdi
  call   <dummy_decrypt_seed@plt>
  mov    %rax,%rbx
  mov    %rax,%rsi
  lea    D38(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %r14,%rdi
  call   <free@plt>
  mov    %rbx,%rdi
  call   <free@plt>
  mov    $0x0,%eax
L5:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L34:
  mov    (%rsi),%rsi
  lea    D21(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x1,%eax
  jmp    L5
```

tool_calls: name: read_data
arguments: data_tag: D31
data_type: string
name: read_data
arguments: data_tag: D39
data_type: string
name: read_data
arguments: data_tag: D38
data_type: string
name: read_data
arguments: data_tag: D21
data_type: string

tool: D31: Crypted  : 

tool: D39: %.2X 

tool: D38: Decrypted: %s


tool: D21: Usage: %s msg key [n]


assistant: ```c
int main(int argc, char **argv) {
  if (argc <= 2) {
    printf("Usage: %s msg key [n]\n", argv[0]);
    return 1;
  }
  uint32_t no_seed = 0;
  if (argc >= 4) {
    if (argv[3][0] == 'n') {
      no_seed = 1;
    }
  }
  uint32_t seed = 10;
  uint32_t srclen = strlen(argv[1]) + 1;
  uint32_t key_len = strlen(argv[2]);
  char *m = dummy_crypt_seed(argv[1], srclen, argv[2], key_len, seed, no_seed ? 0 : 1);
  uint32_t s = dummy_crypt_size_seed(srclen, seed, no_seed ? 0 : 1);
  printf("Crypted  : ");
  for (uint32_t i = 0; i < s; ++i) {
    printf("%.2X ", (uint8_t)m[i]);
  }
  printf("\n");
  char *p = dummy_decrypt_seed(m, s, argv[2], key_len, no_seed ? seed : 0);
  printf("Decrypted: %s\n", p);
  free(m);
  free(p);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x18,%rsp
  cmp    $0x2,%edi
  jle    L1
  movl   $0x0,0x8(%rsp)
  cmp    $0x3,%edi
  je     L45
  mov    0x18(%rsi),%rax
  cmpb   $0x6e,(%rax)
  sete   %al
  movzbl %al,%eax
  mov    %eax,0x8(%rsp)
L45:
  mov    0x8(%rbx),%rdi
  call   <strlen@plt>
  mov    0x10(%rbx),%rdi
  lea    0x1(%rax),%ebp
  call   <strlen@plt>
  mov    0x8(%rsp),%r12d
  mov    0x10(%rbx),%rdx
  mov    %ebp,%esi
  mov    0x8(%rbx),%rdi
  mov    $0xa,%r8d
  mov    %eax,%ecx
  mov    %eax,0xc(%rsp)
  xor    $0x1,%r12d
  mov    %r12d,%r9d
  call   <dummy_crypt_seed@plt>
  mov    %ebp,%edi
  mov    %r12d,%edx
  mov    $0xa,%esi
  mov    %rax,%r13
  call   <dummy_crypt_size_seed@plt>
  lea    D24(%rip),%rdi
  mov    %eax,%r14d
  xor    %eax,%eax
  call   <printf@plt>
  test   %r14d,%r14d
  jle    L37
  lea    -0x1(%r14),%eax
  mov    %r13,%r15
  lea    D16(%rip),%rbp
  lea    0x1(%r13,%rax,1),%r12
  nopl   0x0(%rax,%rax,1)
L30:
  movsbl (%r15),%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x1,%r15
  call   <printf@plt>
  cmp    %r12,%r15
  jne    L30
L37:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x8(%rsp),%eax
  mov    0x10(%rbx),%rdx
  mov    %r14d,%esi
  mov    $0xa,%r8d
  mov    0xc(%rsp),%ecx
  mov    %r13,%rdi
  test   %eax,%eax
  cmove  %eax,%r8d
  call   <dummy_decrypt_seed@plt>
  lea    D26(%rip),%rdi
  mov    %rax,%rsi
  mov    %rax,%rbp
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r13,%rdi
  call   <free@plt>
  mov    %rbp,%rdi
  call   <free@plt>
  xor    %eax,%eax
L15:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L1:
  mov    (%rsi),%rsi
  lea    D28(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x1,%eax
  jmp    L15
```

tool_calls: name: retrieve_data
arguments: data_tag: D24
datatype: string
name: retrieve_data
arguments: data_tag: D16
datatype: string
name: retrieve_data
arguments: data_tag: D26
datatype: string
name: retrieve_data
arguments: data_tag: D28
datatype: string

tool: D24: Crypted  : 

tool: D16: %.2X 

tool: D26: Decrypted: %s


tool: D28: Usage: %s msg key [n]


assistant: ```c
int main(int argc, char **argv) {
  if (argc <= 2) {
    printf("Usage: %s msg key [n]\n", argv[0]);
    return 1;
  }
  uint32_t no_seed = 0;
  if (argc >= 4) {
    if (argv[3][0] == 'n') {
      no_seed = 1;
    }
  }
  uint32_t seed = 10;
  uint32_t srclen = strlen(argv[1]) + 1;
  uint32_t key_len = strlen(argv[2]);
  char *m = dummy_crypt_seed(argv[1], srclen, argv[2], key_len, seed, no_seed ? 0 : 1);
  uint32_t s = dummy_crypt_size_seed(srclen, seed, no_seed ? 0 : 1);
  printf("Crypted  : ");
  for (uint32_t i = 0; i < s; ++i) {
    printf("%.2X ", (uint8_t)m[i]);
  }
  printf("\n");
  char *p = dummy_decrypt_seed(m, s, argv[2], key_len, no_seed ? seed : 0);
  printf("Decrypted: %s\n", p);
  free(m);
  free(p);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x18,%rsp
  cmp    $0x2,%edi
  jle    L37
  movl   $0x0,0x8(%rsp)
  cmp    $0x3,%edi
  je     L48
  mov    0x18(%rsi),%rax
  cmpb   $0x6e,(%rax)
  sete   %al
  movzbl %al,%eax
  mov    %eax,0x8(%rsp)
L48:
  mov    0x8(%rbx),%rdi
  call   <strlen@plt>
  mov    0x10(%rbx),%rdi
  lea    0x1(%rax),%ebp
  call   <strlen@plt>
  mov    0x8(%rsp),%r12d
  mov    0x10(%rbx),%rdx
  mov    %ebp,%esi
  mov    0x8(%rbx),%rdi
  mov    $0xa,%r8d
  mov    %eax,%ecx
  mov    %eax,0xc(%rsp)
  xor    $0x1,%r12d
  mov    %r12d,%r9d
  call   <dummy_crypt_seed@plt>
  mov    %ebp,%edi
  mov    %r12d,%edx
  mov    $0xa,%esi
  mov    %rax,%r13
  call   <dummy_crypt_size_seed@plt>
  lea    D10(%rip),%rdi
  mov    %eax,%r14d
  xor    %eax,%eax
  call   <printf@plt>
  test   %r14d,%r14d
  jle    L36
  lea    -0x1(%r14),%eax
  mov    %r13,%r15
  lea    D1(%rip),%rbp
  lea    0x1(%r13,%rax,1),%r12
  nopl   0x0(%rax,%rax,1)
L2:
  movsbl (%r15),%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x1,%r15
  call   <printf@plt>
  cmp    %r12,%r15
  jne    L2
L36:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x8(%rsp),%eax
  mov    0x10(%rbx),%rdx
  mov    %r14d,%esi
  mov    $0xa,%r8d
  mov    0xc(%rsp),%ecx
  mov    %r13,%rdi
  test   %eax,%eax
  cmove  %eax,%r8d
  call   <dummy_decrypt_seed@plt>
  lea    D22(%rip),%rdi
  mov    %rax,%rsi
  mov    %rax,%rbp
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r13,%rdi
  call   <free@plt>
  mov    %rbp,%rdi
  call   <free@plt>
  xor    %eax,%eax
L0:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L37:
  mov    (%rsi),%rsi
  lea    D17(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x1,%eax
  jmp    L0
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D10
data_type: string
name: extract_symbolic_data
arguments: data_tag: D1
data_type: string
name: extract_symbolic_data
arguments: data_tag: D22
data_type: string
name: extract_symbolic_data
arguments: data_tag: D17
data_type: string

tool: D10: Crypted  : 

tool: D1: %.2X 

tool: D22: Decrypted: %s


tool: D17: Usage: %s msg key [n]


assistant: ```c
int main(int argc, char **argv) {
  if (argc <= 2) {
    printf("Usage: %s msg key [n]\n", argv[0]);
    return 1;
  }
  uint32_t no_seed = 0;
  if (argc >= 4) {
    if (argv[3][0] == 'n') {
      no_seed = 1;
    }
  }
  uint32_t seed = 10;
  uint32_t srclen = strlen(argv[1]) + 1;
  uint32_t key_len = strlen(argv[2]);
  char *m = dummy_crypt_seed(argv[1], srclen, argv[2], key_len, seed, no_seed ? 0 : 1);
  uint32_t s = dummy_crypt_size_seed(srclen, seed, no_seed ? 0 : 1);
  printf("Crypted  : ");
  for (uint32_t i = 0; i < s; ++i) {
    printf("%.2X ", (uint8_t)m[i]);
  }
  printf("\n");
  char *p = dummy_decrypt_seed(m, s, argv[2], key_len, no_seed ? seed : 0);
  printf("Decrypted: %s\n", p);
  free(m);
  free(p);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<place_length>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    -0x20(%rbp),%rax
  lea    -0x8(%rax),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <padd@plt>
  mov    %rax,%rbx
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <ident64@plt>
  mov    %eax,(%rbx)
  nop
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
void place_length(uint8_t *buffer, size_t bufsize, size_t length) {
  if (sizeof(size_t) == 4) {
    *(uint32_t *)padd(buffer, bufsize - 4) = ident32((uint32_t)length);
  } else {
    *(uint64_t *)padd(buffer, bufsize - 8) = ident64(length);
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<place_length>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdx,%rbp
  sub    $0x8,%rsi
  call   <padd@plt>
  mov    %rax,%rbx
  mov    %rbp,%rdi
  call   <ident64@plt>
  mov    %eax,(%rbx)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void place_length(uint8_t *buffer, size_t bufsize, size_t length) {
  if (sizeof(size_t) == 4) {
    *(uint32_t *)padd(buffer, bufsize - 4) = ident32((uint32_t)length);
  } else {
    *(uint64_t *)padd(buffer, bufsize - 8) = ident64(length);
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<place_length>:
  endbr64
  push   %rbp
  sub    $0x8,%rsi
  mov    %rdx,%rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <padd@plt>
  mov    %rbp,%rdi
  mov    %rax,%rbx
  call   <ident64@plt>
  mov    %eax,(%rbx)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void place_length(uint8_t *buffer, size_t bufsize, size_t length) {
  if (sizeof(size_t) == 4) {
    *(uint32_t *)padd(buffer, bufsize - 4) = ident32((uint32_t)length);
  } else {
    *(uint64_t *)padd(buffer, bufsize - 8) = ident64(length);
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<place_length>:
  endbr64
  push   %rbp
  sub    $0x8,%rsi
  mov    %rdx,%rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <padd@plt>
  mov    %rbp,%rdi
  mov    %rax,%rbx
  call   <ident64@plt>
  mov    %eax,(%rbx)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void place_length(uint8_t *buffer, size_t bufsize, size_t length) {
  if (sizeof(size_t) == 4) {
    *(uint32_t *)padd(buffer, bufsize - 4) = ident32((uint32_t)length);
  } else {
    *(uint64_t *)padd(buffer, bufsize - 8) = ident64(length);
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<sum_to_zero>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    %rcx,-0x40(%rbp)
  mov    %r8,-0x48(%rbp)
  mov    -0x30(%rbp),%rdx
  mov    -0x28(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <quicksort@plt>
  movq   $0x0,-0x18(%rbp)
  movq   $0x0,-0x18(%rbp)
  jmp    L36
L68:
  mov    -0x18(%rbp),%rax
  add    $0x1,%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x28(%rbp),%rax
  sub    $0x1,%rax
  mov    %rax,-0x8(%rbp)
  jmp    L25
L57:
  mov    -0x18(%rbp),%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x10(%rbp),%rax
  lea    0x0(,%rax,4),%rcx
  mov    -0x30(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%eax
  add    %eax,%edx
  mov    -0x8(%rbp),%rax
  lea    0x0(,%rax,4),%rcx
  mov    -0x30(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%eax
  add    %edx,%eax
  mov    %eax,-0x1c(%rbp)
  cmpl   $0x0,-0x1c(%rbp)
  jne    L12
  mov    -0x38(%rbp),%rax
  mov    -0x18(%rbp),%rdx
  mov    %rdx,(%rax)
  mov    -0x40(%rbp),%rax
  mov    -0x10(%rbp),%rdx
  mov    %rdx,(%rax)
  mov    -0x48(%rbp),%rax
  mov    -0x8(%rbp),%rdx
  mov    %rdx,(%rax)
  mov    $0x1,%eax
  jmp    L73
L12:
  cmpl   $0x0,-0x1c(%rbp)
  jns    L5
  addq   $0x1,-0x10(%rbp)
  jmp    L25
L5:
  subq   $0x1,-0x8(%rbp)
L25:
  mov    -0x10(%rbp),%rax
  cmp    -0x8(%rbp),%rax
  jb     L57
  addq   $0x1,-0x18(%rbp)
L36:
  mov    -0x18(%rbp),%rax
  cmp    -0x28(%rbp),%rax
  jb     L68
  mov    $0x0,%eax
L73:
  leave
  ret
```

assistant: ```c
int sum_to_zero(size_t N, int arr[N], size_t *i, size_t *j, size_t *k) {
  quicksort(N, arr);
  size_t arr_i = 0;
  for (arr_i = 0; arr_i < N; arr_i++) {
    size_t front_cursor = arr_i + 1;
    size_t back_cursor = N - 1;
    while (front_cursor < back_cursor) {
      int sum = arr[arr_i] + arr[front_cursor] + arr[back_cursor];
      if (sum == 0) {
        *i = arr_i;
        *j = front_cursor;
        *k = back_cursor;
        return 1;
      } else if (sum < 0) {
        front_cursor++;
      } else {
        back_cursor--;
      }
    }
  }
  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
<sum_to_zero>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    %rsi,%rbx
  mov    %rdx,%r14
  mov    %rcx,%r13
  mov    %r8,%r12
  call   <quicksort@plt>
  test   %rbp,%rbp
  je     L55
  mov    $0x0,%r8d
L6:
  lea    0x1(%r8),%rdi
  lea    -0x1(%rbp),%rdx
  cmp    %rdx,%rdi
  jae    L12
  mov    -0x4(%rbx,%rdi,4),%esi
  mov    %rdi,%rcx
  jmp    L107
L28:
  mov    %r8,(%r14)
  mov    %rcx,0x0(%r13)
  mov    %rdx,(%r12)
  mov    $0x1,%eax
L72:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L41:
  add    $0x1,%rcx
L31:
  cmp    %rcx,%rdx
  jbe    L12
L107:
  mov    %esi,%eax
  add    (%rbx,%rcx,4),%eax
  add    (%rbx,%rdx,4),%eax
  je     L28
  test   %eax,%eax
  js     L41
  sub    $0x1,%rdx
  jmp    L31
L12:
  cmp    %rdi,%rbp
  je     L5
  mov    %rdi,%r8
  jmp    L6
L5:
  mov    $0x0,%eax
  jmp    L72
L55:
  mov    $0x0,%eax
  jmp    L72
```

assistant: ```c
int sum_to_zero(size_t N, int arr[N], size_t *i, size_t *j, size_t *k) {
  quicksort(N, arr);
  size_t arr_i = 0;
  for (arr_i = 0; arr_i < N; arr_i++) {
    size_t front_cursor = arr_i + 1;
    size_t back_cursor = N - 1;
    while (front_cursor < back_cursor) {
      int sum = arr[arr_i] + arr[front_cursor] + arr[back_cursor];
      if (sum == 0) {
        *i = arr_i;
        *j = front_cursor;
        *k = back_cursor;
        return 1;
      } else if (sum < 0) {
        front_cursor++;
      } else {
        back_cursor--;
      }
    }
  }
  return 0;
}
```