ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0xc(%rbp),%rsi
  lea    -0x10(%rbp),%rcx
  lea    -0x14(%rbp),%rdx
  lea    -0x18(%rbp),%rax
  mov    %rsi,%r8
  mov    %rax,%rsi
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x18(%rbp),%edx
  mov    -0x14(%rbp),%eax
  imul   %edx,%eax
  mov    -0x10(%rbp),%ecx
  mov    -0xc(%rbp),%edx
  imul   %ecx,%edx
  sub    %edx,%eax
  mov    %eax,%esi
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L9
  call   <__stack_chk_fail@plt>
L9:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: tag: D0
type_infered: string
name: unpack_data
arguments: tag: D18
type_infered: string

tool: D0: %d %d %d %d

tool: D18: DIFERENCA = %d


assistant: ```c
int main() {
  int a, b, c, d;
  scanf("%d %d %d %d", &a, &b, &c, &d);
  printf("DIFERENCA = %d\n", (a * b - c * d));
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rcx
  lea    0xc(%rsp),%rdx
  lea    0x8(%rsp),%rsi
  lea    0x14(%rsp),%r8
  lea    D2(%rip),%rdi
  call   <scanf@plt>
  mov    0x8(%rsp),%esi
  imul   0xc(%rsp),%esi
  mov    0x10(%rsp),%eax
  imul   0x14(%rsp),%eax
  sub    %eax,%esi
  lea    D6(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  mov    $0x0,%eax
  add    $0x28,%rsp
  ret
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D2
type: string
name: parse_data
arguments: tag: D6
type: string

tool: D2: %d %d %d %d

tool: D6: DIFERENCA = %d


assistant: ```c
int main() {
  int a, b, c, d;
  scanf("%d %d %d %d", &a, &b, &c, &d);
  printf("DIFERENCA = %d\n", (a * b - c * d));
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  lea    D9(%rip),%rdi
  lea    0x8(%rsp),%rsi
  lea    0x10(%rsp),%rcx
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rdx
  lea    0x14(%rsp),%r8
  call   <scanf@plt>
  mov    0x8(%rsp),%esi
  mov    0x10(%rsp),%eax
  lea    D18(%rip),%rdi
  imul   0xc(%rsp),%esi
  imul   0x14(%rsp),%eax
  sub    %eax,%esi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  xor    %eax,%eax
  add    $0x28,%rsp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D9
type_infered: string
name: disassemble_data
arguments: data_tag: D18
type_infered: string

tool: D9: %d %d %d %d

tool: D18: DIFERENCA = %d


assistant: ```c
int main() {
  int a, b, c, d;
  scanf("%d %d %d %d", &a, &b, &c, &d);
  printf("DIFERENCA = %d\n", (a * b - c * d));
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  lea    D17(%rip),%rdi
  lea    0x8(%rsp),%rsi
  lea    0x10(%rsp),%rcx
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rdx
  lea    0x14(%rsp),%r8
  call   <scanf@plt>
  mov    0x8(%rsp),%esi
  mov    0x10(%rsp),%eax
  lea    D12(%rip),%rdi
  imul   0xc(%rsp),%esi
  imul   0x14(%rsp),%eax
  sub    %eax,%esi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  xor    %eax,%eax
  add    $0x28,%rsp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D17
type_infered: string
name: unpack_data
arguments: data_tag: D12
type_infered: string

tool: D17: %d %d %d %d

tool: D12: DIFERENCA = %d


assistant: ```c
int main() {
  int a, b, c, d;
  scanf("%d %d %d %d", &a, &b, &c, &d);
  printf("DIFERENCA = %d\n", (a * b - c * d));
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<from_hex>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <toupper@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0x2f,-0x4(%rbp)
  jg     L48
  mov    $0x0,%eax
  jmp    L9
L48:
  cmpl   $0x46,-0x4(%rbp)
  jle    L28
  mov    $0xf,%eax
  jmp    L9
L28:
  subl   $0x30,-0x4(%rbp)
  cmpl   $0x9,-0x4(%rbp)
  jle    L7
  subl   $0x7,-0x4(%rbp)
L7:
  mov    -0x4(%rbp),%eax
L9:
  leave
  ret
```

assistant: ```c
int from_hex(int letter) {
  letter = toupper(letter);
  if (letter < '0')
    return 0;
  if (letter > 'F')
    return 15;
  letter -= '0';
  if (letter > 9)
    letter -= 7;
  return letter;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<from_hex>:
  endbr64
  sub    $0x8,%rsp
  call   <toupper@plt>
  mov    $0x0,%edx
  cmp    $0x2f,%eax
  jle    L1
  mov    $0xf,%edx
  cmp    $0x46,%eax
  jg     L1
  lea    -0x30(%rax),%edx
  sub    $0x37,%eax
  cmp    $0x9,%edx
  cmovg  %eax,%edx
L1:
  mov    %edx,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int from_hex(int letter) {
  letter = toupper(letter);
  if (letter < '0')
    return 0;
  if (letter > 'F')
    return 15;
  letter -= '0';
  if (letter > 9)
    letter -= 7;
  return letter;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<from_hex>:
  endbr64
  sub    $0x8,%rsp
  call   <toupper@plt>
  xor    %r8d,%r8d
  cmp    $0x2f,%eax
  jle    L10
  mov    $0xf,%r8d
  cmp    $0x46,%eax
  jg     L10
  lea    -0x30(%rax),%r8d
  sub    $0x37,%eax
  cmp    $0x9,%r8d
  cmovg  %eax,%r8d
L10:
  mov    %r8d,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int from_hex(int letter) {
  letter = toupper(letter);
  if (letter < '0')
    return 0;
  if (letter > 'F')
    return 15;
  letter -= '0';
  if (letter > 9)
    letter -= 7;
  return letter;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<from_hex>:
  endbr64
  sub    $0x8,%rsp
  call   <toupper@plt>
  xor    %r8d,%r8d
  cmp    $0x2f,%eax
  jle    L19
  mov    $0xf,%r8d
  cmp    $0x46,%eax
  jg     L19
  lea    -0x30(%rax),%r8d
  sub    $0x37,%eax
  cmp    $0x9,%r8d
  cmovg  %eax,%r8d
L19:
  mov    %r8d,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int from_hex(int letter) {
  letter = toupper(letter);
  if (letter < '0')
    return 0;
  if (letter > 'F')
    return 15;
  letter -= '0';
  if (letter > 9)
    letter -= 7;
  return letter;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<kitoa>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x90,%rsp
  mov    %edi,%ebx
  mov    %rsi,%r12
  mov    %rdx,%rbp
  mov    %ecx,%r14d
  mov    %fs:0x28,%rax
  mov    %rax,0x88(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  mov    $0x80,%edx
  mov    $0x0,%esi
  call   <memset@plt>
  mov    %rbp,%rdx
  mov    $0x0,%esi
  mov    %r12,%rdi
  call   <memset@plt>
  cmp    $0x1,%rbp
  jbe    L101
  test   %ebx,%ebx
  je     L1
  jns    L98
  cmp    $0x10,%r14d
  je     L98
  neg    %ebx
  mov    $0x1,%r13d
L29:
  mov    $0x1,%ecx
  lea    -0x1(%rsp),%rsi
L64:
  mov    %ebx,%eax
  cltd
  idiv   %r14d
  lea    0x30(%rdx),%edi
  lea    0x37(%rdx),%eax
  cmp    $0x9,%edx
  cmovle %edi,%eax
  mov    %al,(%rsi,%rcx,1)
  mov    %ebx,%eax
  cltd
  idiv   %r14d
  mov    %eax,%ebx
  cmp    $0x7f,%ecx
  setle  %dl
  test   %eax,%eax
  setg   %al
  add    $0x1,%rcx
  test   %al,%dl
  jne    L64
  test   %r13d,%r13d
  je     L103
  movb   $0x2d,(%r12)
L103:
  mov    %rsp,%rdi
  call   <strlen@plt>
  movslq %r13d,%r13
  test   %eax,%eax
  jle    L89
  sub    $0x2,%eax
  cltq
  lea    0x1(%rsp),%rsi
L77:
  movzbl (%rsi,%rax,1),%edx
  mov    %dl,(%r12,%r13,1)
  add    $0x1,%r13
  mov    %eax,%edx
  not    %edx
  shr    $0x1f,%edx
  cmp    %rbp,%r13
  setb   %cl
  sub    $0x1,%rax
  test   %dl,%cl
  jne    L77
L89:
  movb   $0x0,-0x1(%r12,%rbp,1)
L101:
  mov    0x88(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L130
  add    $0x90,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L1:
  movb   $0x30,(%r12)
  movb   $0x0,0x1(%r12)
  jmp    L101
L98:
  cmp    $0x10,%r14d
  jne    L62
  mov    $0x0,%r13d
  test   %ebx,%ebx
  jg     L29
  jmp    L103
L62:
  mov    $0x0,%r13d
  test   %ebx,%ebx
  jg     L29
  jmp    L103
L130:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void kitoa(int val_signed, char *dst_c, size_t sz, int radix) {
  char dst[128];
  memset(dst, 0, 128);
  memset(dst_c, 0, sz);
  if (sz <= 1)
    return;
  if (val_signed == 0) {
    dst_c[0] = '0';
    dst_c[1] = 0;
    return;
  }
  int neg = 0;
  if (val_signed < 0 && radix != 16) {
    neg = 1;
    val_signed = -val_signed;
  }
  if (radix == 16)
    neg = 0;
  int val = val_signed;
  int x;
  for (x = 0; x < 128 && val > 0; x++) {
    int mod = val % radix;
    char character = mod + '0';
    if (mod >= 10)
      character = (mod - 10) + 'A';
    dst[x] = character;
    val /= radix;
  }
  x = 0;
  if (neg) {
    dst_c[0] = '-';
    x = 1;
  }
  int pos = strlen(dst) - 1;
  for (; x < sz && pos >= 0; x++, pos--)
    dst_c[x] = dst[pos];
  dst_c[sz - 1] = 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<kitoa>:
  endbr64
  push   %r15
  push   %r14
  mov    %ecx,%r14d
  push   %r13
  mov    %rsi,%r13
  xor    %esi,%esi
  push   %r12
  mov    %rdx,%r12
  mov    $0x80,%edx
  push   %rbp
  push   %rbx
  mov    %edi,%ebx
  sub    $0x98,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x88(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r15
  mov    %r15,%rdi
  call   <memset@plt>
  xor    %esi,%esi
  mov    %r12,%rdx
  mov    %r13,%rdi
  call   <memset@plt>
  cmp    $0x1,%r12
  jbe    L108
  test   %ebx,%ebx
  je     L67
  jns    L70
  cmp    $0x10,%r14d
  je     L70
  neg    %ebx
  mov    $0x1,%ebp
L26:
  mov    $0x1,%ecx
  lea    -0x1(%rsp),%rdi
  nopl   (%rax)
L7:
  mov    %ebx,%eax
  cltd
  idiv   %r14d
  cmp    $0x9,%edx
  mov    %eax,%ebx
  lea    0x37(%rdx),%esi
  lea    0x30(%rdx),%eax
  cmovle %eax,%esi
  cmp    $0x7f,%ecx
  setle  %dl
  test   %ebx,%ebx
  setg   %al
  mov    %sil,(%rdi,%rcx,1)
  add    $0x1,%rcx
  test   %al,%dl
  jne    L7
  test   %ebp,%ebp
  je     L118
  movb   $0x2d,0x0(%r13)
  mov    $0x1,%ebx
L115:
  mov    %r15,%rdi
  call   <strlen@plt>
  test   %eax,%eax
  jle    L25
  sub    $0x2,%eax
  lea    0x1(%rbp),%edx
  lea    0x1(%rsp),%rdi
  cltq
  movslq %edx,%rdx
  nopl   0x0(%rax)
L82:
  movzbl (%rdi,%rax,1),%ecx
  mov    %cl,0x0(%r13,%rbx,1)
  mov    %eax,%ecx
  mov    %rdx,%rbx
  not    %ecx
  shr    $0x1f,%ecx
  cmp    %rdx,%r12
  seta   %sil
  sub    $0x1,%rax
  add    $0x1,%rdx
  test   %cl,%sil
  jne    L82
L25:
  movb   $0x0,-0x1(%r13,%r12,1)
L108:
  mov    0x88(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L98
  add    $0x98,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L67:
  mov    $0x30,%eax
  mov    %ax,0x0(%r13)
  jmp    L108
  nopl   0x0(%rax)
L70:
  xor    %ebp,%ebp
  test   %ebx,%ebx
  jg     L26
  nopw   0x0(%rax,%rax,1)
L118:
  xor    %ebx,%ebx
  xor    %ebp,%ebp
  jmp    L115
L98:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void kitoa(int val_signed, char *dst_c, size_t sz, int radix) {
  char dst[128];
  memset(dst, 0, 128);
  memset(dst_c, 0, sz);
  if (sz <= 1)
    return;
  if (val_signed == 0) {
    dst_c[0] = '0';
    dst_c[1] = 0;
    return;
  }
  int neg = 0;
  if (val_signed < 0 && radix != 16) {
    neg = 1;
    val_signed = -val_signed;
  }
  if (radix == 16)
    neg = 0;
  int val = val_signed;
  int x;
  for (x = 0; x < 128 && val > 0; x++) {
    int mod = val % radix;
    char character = mod + '0';
    if (mod >= 10)
      character = (mod - 10) + 'A';
    dst[x] = character;
    val /= radix;
  }
  x = 0;
  if (neg) {
    dst_c[0] = '-';
    x = 1;
  }
  int pos = strlen(dst) - 1;
  for (; x < sz && pos >= 0; x++, pos--)
    dst_c[x] = dst[pos];
  dst_c[sz - 1] = 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<kitoa>:
  endbr64
  push   %r15
  push   %r14
  mov    %ecx,%r14d
  push   %r13
  mov    %rsi,%r13
  xor    %esi,%esi
  push   %r12
  mov    %rdx,%r12
  mov    $0x80,%edx
  push   %rbp
  push   %rbx
  mov    %edi,%ebx
  sub    $0x98,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x88(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r15
  mov    %r15,%rdi
  call   <memset@plt>
  xor    %esi,%esi
  mov    %r12,%rdx
  mov    %r13,%rdi
  call   <memset@plt>
  cmp    $0x1,%r12
  jbe    L80
  test   %ebx,%ebx
  je     L54
  jns    L30
  cmp    $0x10,%r14d
  je     L30
  neg    %ebx
  mov    $0x1,%ebp
L45:
  mov    $0x1,%ecx
  lea    -0x1(%rsp),%rdi
  nopl   (%rax)
L42:
  mov    %ebx,%eax
  cltd
  idiv   %r14d
  cmp    $0xa,%edx
  mov    %eax,%ebx
  lea    0x30(%rdx),%esi
  lea    0x37(%rdx),%eax
  cmovge %eax,%esi
  test   %ebx,%ebx
  setg   %dl
  cmp    $0x7f,%ecx
  setle  %al
  mov    %sil,(%rdi,%rcx,1)
  add    $0x1,%rcx
  test   %al,%dl
  jne    L42
  test   %ebp,%ebp
  je     L101
  movb   $0x2d,0x0(%r13)
  mov    $0x1,%ebx
L113:
  mov    %r15,%rdi
  call   <strlen@plt>
  test   %eax,%eax
  jle    L21
  sub    $0x2,%eax
  lea    0x1(%rbp),%edx
  lea    0x1(%rsp),%rdi
  cltq
  movslq %edx,%rdx
  nopl   0x0(%rax)
L108:
  movzbl (%rdi,%rax,1),%ecx
  mov    %cl,0x0(%r13,%rbx,1)
  mov    %eax,%ecx
  mov    %rdx,%rbx
  not    %ecx
  shr    $0x1f,%ecx
  cmp    %rdx,%r12
  seta   %sil
  sub    $0x1,%rax
  add    $0x1,%rdx
  test   %cl,%sil
  jne    L108
L21:
  movb   $0x0,-0x1(%r13,%r12,1)
L80:
  mov    0x88(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L19
  add    $0x98,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L54:
  mov    $0x30,%eax
  mov    %ax,0x0(%r13)
  jmp    L80
  nopl   0x0(%rax)
L30:
  xor    %ebp,%ebp
  test   %ebx,%ebx
  jg     L45
  nopw   0x0(%rax,%rax,1)
L101:
  xor    %ebx,%ebx
  xor    %ebp,%ebp
  jmp    L113
L19:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void kitoa(int val_signed, char *dst_c, size_t sz, int radix) {
  char dst[128];
  memset(dst, 0, 128);
  memset(dst_c, 0, sz);
  if (sz <= 1)
    return;
  if (val_signed == 0) {
    dst_c[0] = '0';
    dst_c[1] = 0;
    return;
  }
  int neg = 0;
  if (val_signed < 0 && radix != 16) {
    neg = 1;
    val_signed = -val_signed;
  }
  if (radix == 16)
    neg = 0;
  int val = val_signed;
  int x;
  for (x = 0; x < 128 && val > 0; x++) {
    int mod = val % radix;
    char character = mod + '0';
    if (mod >= 10)
      character = (mod - 10) + 'A';
    dst[x] = character;
    val /= radix;
  }
  x = 0;
  if (neg) {
    dst_c[0] = '-';
    x = 1;
  }
  int pos = strlen(dst) - 1;
  for (; x < sz && pos >= 0; x++, pos--)
    dst_c[x] = dst[pos];
  dst_c[sz - 1] = 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<readshort>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movzbl %al,%edx
  mov    -0x8(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  shl    $0x8,%eax
  or     %edx,%eax
  pop    %rbp
  ret
```

assistant: ```c
short readshort(void *value) {
  unsigned char *bytes = (unsigned char *)value;
  return bytes[0] | (bytes[1] << 8);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<readshort>:
  endbr64
  movzbl 0x1(%rdi),%edx
  shl    $0x8,%edx
  movzbl (%rdi),%eax
  or     %edx,%eax
  ret
```

assistant: ```c
short readshort(void *value) {
  unsigned char *bytes = (unsigned char *)value;
  return bytes[0] | (bytes[1] << 8);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<readshort>:
  endbr64
  movzwl (%rdi),%eax
  ret
```

assistant: ```c
short readshort(void *value) {
  unsigned char *bytes = (unsigned char *)value;
  return bytes[0] | (bytes[1] << 8);
}
```
user: Translate the below assembly code into C programming language:

```asm
<readshort>:
  endbr64
  movzwl (%rdi),%eax
  ret
```

assistant: ```c
short readshort(void *value) {
  unsigned char *bytes = (unsigned char *)value;
  return bytes[0] | (bytes[1] << 8);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<gethostname>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0xc(%rbp),%rax
  mov    %rax,%rdi
  call   <uname@plt>
  cmp    $0xffffffff,%eax
  jne    L23
  mov    $0xffffffff,%eax
  jmp    L21
L23:
  mov    -0xc(%rbp),%ecx
  mov    -0x20(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <strncpy@plt>
  mov    $0x0,%eax
L21:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L0
  call   <__stack_chk_fail@plt>
L0:
  leave
  ret
```

assistant: ```c
int gethostname(char *name, size_t len) {
  struct utsname utn;
  if (-1 == uname(&utn)) {
    return -1;
  }
  strncpy(name, utn.nodename, len);
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<gethostname>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%rbp
  mov    %rsi,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rdi
  call   <uname@plt>
  cmp    $0xffffffff,%eax
  je     L2
  mov    %rbx,%rdx
  mov    0x4(%rsp),%esi
  mov    %rbp,%rdi
  call   <strncpy@plt>
  mov    $0x0,%eax
L2:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L17
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L17:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int gethostname(char *name, size_t len) {
  struct utsname utn;
  if (-1 == uname(&utn)) {
    return -1;
  }
  strncpy(name, utn.nodename, len);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<gethostname>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rdi
  call   <uname@plt>
  cmp    $0xffffffff,%eax
  je     L10
  mov    0x4(%rsp),%esi
  mov    %r12,%rdx
  mov    %rbp,%rdi
  call   <strncpy@plt>
  xor    %eax,%eax
L10:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L2
  add    $0x18,%rsp
  pop    %rbp
  pop    %r12
  ret
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int gethostname(char *name, size_t len) {
  struct utsname utn;
  if (-1 == uname(&utn)) {
    return -1;
  }
  strncpy(name, utn.nodename, len);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<gethostname>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rdi
  call   <uname@plt>
  cmp    $0xffffffff,%eax
  je     L16
  mov    0x4(%rsp),%esi
  mov    %r12,%rdx
  mov    %rbp,%rdi
  call   <strncpy@plt>
  xor    %eax,%eax
L16:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L12
  add    $0x18,%rsp
  pop    %rbp
  pop    %r12
  ret
L12:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int gethostname(char *name, size_t len) {
  struct utsname utn;
  if (-1 == uname(&utn)) {
    return -1;
  }
  strncpy(name, utn.nodename, len);
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<warn>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xd0,%rsp
  mov    %rdi,-0xc8(%rbp)
  mov    %rsi,-0xa8(%rbp)
  mov    %rdx,-0xa0(%rbp)
  mov    %rcx,-0x98(%rbp)
  mov    %r8,-0x90(%rbp)
  mov    %r9,-0x88(%rbp)
  test   %al,%al
  je     L6
  movaps %xmm0,-0x80(%rbp)
  movaps %xmm1,-0x70(%rbp)
  movaps %xmm2,-0x60(%rbp)
  movaps %xmm3,-0x50(%rbp)
  movaps %xmm4,-0x40(%rbp)
  movaps %xmm5,-0x30(%rbp)
  movaps %xmm6,-0x20(%rbp)
  movaps %xmm7,-0x10(%rbp)
L6:
  mov    -0xc8(%rbp),%rdx
  mov    -0xb4(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <va_start@plt>
  mov    -0xb4(%rbp),%edx
  mov    -0xc8(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <vwarn@plt>
  mov    -0xb4(%rbp),%eax
  mov    %eax,%edi
  call   <va_end@plt>
  nop
  leave
  ret
```

assistant: ```c
void warn(const char *format, ...) {
  va_list ap;
  va_start(ap, format);
  vwarn(format, ap);
  va_end(ap);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<warn>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rdi,%rsi
  mov    $0x0,%ebp
  mov    %ebp,%edi
  call   <va_start@plt>
  mov    %ebp,%esi
  mov    %rbx,%rdi
  call   <vwarn@plt>
  mov    %ebp,%edi
  call   <va_end@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void warn(const char *format, ...) {
  va_list ap;
  va_start(ap, format);
  vwarn(format, ap);
  va_end(ap);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<warn>:
  endbr64
  push   %r12
  xor    %r12d,%r12d
  mov    %rdi,%rsi
  push   %rbp
  mov    %rdi,%rbp
  mov    %r12d,%edi
  sub    $0x8,%rsp
  call   <va_start@plt>
  mov    %r12d,%esi
  mov    %rbp,%rdi
  call   <vwarn@plt>
  mov    %r12d,%edi
  call   <va_end@plt>
  add    $0x8,%rsp
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void warn(const char *format, ...) {
  va_list ap;
  va_start(ap, format);
  vwarn(format, ap);
  va_end(ap);
}
```
user: Translate the below assembly code into C programming language:

```asm
<warn>:
  endbr64
  push   %r12
  xor    %r12d,%r12d
  mov    %rdi,%rsi
  push   %rbp
  mov    %rdi,%rbp
  mov    %r12d,%edi
  sub    $0x8,%rsp
  call   <va_start@plt>
  mov    %r12d,%esi
  mov    %rbp,%rdi
  call   <vwarn@plt>
  mov    %r12d,%edi
  call   <va_end@plt>
  add    $0x8,%rsp
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void warn(const char *format, ...) {
  va_list ap;
  va_start(ap, format);
  vwarn(format, ap);
  va_end(ap);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<fill_array>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x48,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    %rdx,-0x48(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  lea    -0x20(%rbp),%rcx
  mov    -0x38(%rbp),%rax
  mov    $0xa,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strtoll@plt>
  mov    -0x40(%rbp),%rdx
  mov    %eax,(%rdx)
  movl   $0x1,-0x24(%rbp)
  jmp    L21
L0:
  mov    -0x20(%rbp),%rax
  mov    -0x24(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x40(%rbp),%rdx
  lea    (%rcx,%rdx,1),%rbx
  lea    -0x20(%rbp),%rcx
  mov    $0xa,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strtoll@plt>
  mov    %eax,(%rbx)
  addl   $0x1,-0x24(%rbp)
L21:
  mov    -0x24(%rbp),%eax
  cltq
  cmp    %rax,-0x48(%rbp)
  ja     L0
  nop
  mov    -0x18(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L20
  call   <__stack_chk_fail@plt>
L20:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
void fill_array(const char src[], uint64_t dest[], size_t size) {
  char *token;
  dest[0] = strtoll(src, &token, 10);
  int i;
  for (i = 1; i < size; i++) {
    dest[i] = strtoll(token, &token, 10);
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<fill_array>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rsi,%rbp
  mov    %rdx,%r12
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  mov    $0xa,%edx
  call   <strtoll@plt>
  mov    %eax,0x0(%rbp)
  cmp    $0x1,%r12
  jbe    L13
  mov    $0x1,%ebx
  mov    %rsp,%r13
L20:
  mov    $0xa,%edx
  mov    %r13,%rsi
  mov    (%rsp),%rdi
  call   <strtoll@plt>
  mov    %eax,0x0(%rbp,%rbx,4)
  add    $0x1,%rbx
  cmp    %r12,%rbx
  jne    L20
L13:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L23
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L23:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void fill_array(const char src[], uint64_t dest[], size_t size) {
  char *token;
  dest[0] = strtoll(src, &token, 10);
  int i;
  for (i = 1; i < size; i++) {
    dest[i] = strtoll(token, &token, 10);
  }
}
```