ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D19(%rip),%rsi
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%rdi
  call   <strcpy@plt>
  mov    %rbp,%rdi
  lea    D13(%rip),%rsi
  call   <strtok@plt>
  mov    %rax,%rdi
  call   <puts@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D19
type: string
name: extract_symbolic_data
arguments: data_tag: D13
type: string

tool: D19: dsdsdsdso

tool: D13: .

assistant: ```c
int main() {
  char s[22];
  strcpy(s, "dsdsdsdso");
  char *str = strtok(s, ".");
  puts(str);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<robust_read>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x18(%rbp)
  mov    -0x18(%rbp),%eax
  mov    %eax,-0xc(%rbp)
L33:
  mov    -0x18(%rbp),%eax
  sub    -0xc(%rbp),%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0xc(%rbp),%edx
  mov    -0x14(%rbp),%eax
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <read@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jle    L25
  mov    -0xc(%rbp),%eax
  sub    -0x4(%rbp),%eax
  jmp    L3
L25:
  mov    -0xc(%rbp),%eax
L3:
  mov    %eax,-0xc(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jns    L43
  mov    D1(%rip),%rax
  mov    (%rax),%eax
  neg    %eax
  cmp    %eax,-0x4(%rbp)
  je     L43
  mov    -0x4(%rbp),%eax
  jmp    L19
L43:
  cmpl   $0x0,-0x4(%rbp)
  jne    L33
  mov    -0x18(%rbp),%eax
  sub    -0xc(%rbp),%eax
L19:
  leave
  ret
```

assistant: ```c
int robust_read(int fd, char *buf, int buf_size) {
  int bytes_remain, bytes_read;
  int buf_offset;
  bytes_remain = buf_size;
  while (true) {
    buf_offset = buf_size - bytes_remain;
    bytes_read = read(fd, buf + buf_offset, bytes_remain);
    bytes_remain = (bytes_read > 0) ? bytes_remain - bytes_read : bytes_remain;
    if (bytes_read < 0 && bytes_read != -EINTR) {
      return bytes_read;
    } else if (bytes_read == 0) {
      return buf_size - bytes_remain;
    }
  }
  assert(false);
  return -EINVAL;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<robust_read>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%r14d
  mov    %rsi,%r13
  mov    %edx,%r12d
  mov    %edx,%ebx
  jmp    L19
L27:
  sub    %eax,%ebx
L19:
  mov    %r12d,%ebp
  sub    %ebx,%ebp
  movslq %ebp,%rsi
  add    %r13,%rsi
  mov    %ebx,%edx
  mov    %r14d,%edi
  call   <read@plt>
  test   %eax,%eax
  jg     L27
  jns    L1
  mov    D1(%rip),%rdx
  mov    %eax,%ecx
  add    (%rdx),%ecx
  je     L19
  mov    %eax,%ebp
L1:
  mov    %ebp,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int robust_read(int fd, char *buf, int buf_size) {
  int bytes_remain, bytes_read;
  int buf_offset;
  bytes_remain = buf_size;
  while (true) {
    buf_offset = buf_size - bytes_remain;
    bytes_read = read(fd, buf + buf_offset, bytes_remain);
    bytes_remain = (bytes_read > 0) ? bytes_remain - bytes_read : bytes_remain;
    if (bytes_read < 0 && bytes_read != -EINTR) {
      return bytes_read;
    } else if (bytes_read == 0) {
      return buf_size - bytes_remain;
    }
  }
  assert(false);
  return -EINVAL;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<robust_read>:
  endbr64
  push   %r14
  mov    %edx,%r14d
  push   %r13
  mov    %edi,%r13d
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %edx,%ebp
  push   %rbx
  nopw   0x0(%rax,%rax,1)
L12:
  mov    %ebp,%ebx
  mov    %r14d,%edx
  mov    %r13d,%edi
  sub    %r14d,%ebx
  movslq %ebx,%rsi
  add    %r12,%rsi
  call   <read@plt>
  test   %eax,%eax
  jle    L28
  sub    %eax,%r14d
  jmp    L12
  nop
L28:
  je     L42
  mov    D5(%rip),%rdx
  mov    %eax,%ecx
  add    (%rdx),%ecx
  je     L12
L1:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L42:
  mov    %ebx,%eax
  jmp    L1
```

assistant: ```c
int robust_read(int fd, char *buf, int buf_size) {
  int bytes_remain, bytes_read;
  int buf_offset;
  bytes_remain = buf_size;
  while (true) {
    buf_offset = buf_size - bytes_remain;
    bytes_read = read(fd, buf + buf_offset, bytes_remain);
    bytes_remain = (bytes_read > 0) ? bytes_remain - bytes_read : bytes_remain;
    if (bytes_read < 0 && bytes_read != -EINTR) {
      return bytes_read;
    } else if (bytes_read == 0) {
      return buf_size - bytes_remain;
    }
  }
  assert(false);
  return -EINVAL;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<robust_read>:
  endbr64
  push   %r14
  mov    %edx,%r14d
  push   %r13
  mov    %edi,%r13d
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %edx,%ebp
  push   %rbx
  nopw   0x0(%rax,%rax,1)
L39:
  mov    %ebp,%ebx
  mov    %r14d,%edx
  mov    %r13d,%edi
  sub    %r14d,%ebx
  movslq %ebx,%rsi
  add    %r12,%rsi
  call   <read@plt>
  test   %eax,%eax
  jle    L10
  sub    %eax,%r14d
  jmp    L39
  nop
L10:
  je     L3
  mov    D9(%rip),%rdx
  mov    %eax,%ecx
  add    (%rdx),%ecx
  je     L39
L49:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L3:
  mov    %ebx,%eax
  jmp    L49
```

assistant: ```c
int robust_read(int fd, char *buf, int buf_size) {
  int bytes_remain, bytes_read;
  int buf_offset;
  bytes_remain = buf_size;
  while (true) {
    buf_offset = buf_size - bytes_remain;
    bytes_read = read(fd, buf + buf_offset, bytes_remain);
    bytes_remain = (bytes_read > 0) ? bytes_remain - bytes_read : bytes_remain;
    if (bytes_read < 0 && bytes_read != -EINTR) {
      return bytes_read;
    } else if (bytes_read == 0) {
      return buf_size - bytes_remain;
    }
  }
  assert(false);
  return -EINVAL;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<printf>:
  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     L0
  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)
L0:
  cmpq   $0x0,-0xc8(%rbp)
  setne  %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  mov    -0xc8(%rbp),%rdx
  mov    -0xb8(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <va_start@plt>
  mov    -0xb8(%rbp),%edx
  mov    -0xc8(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x1,%edi
  call   <vfdprintf@plt>
  mov    %eax,-0xb4(%rbp)
  mov    -0xb8(%rbp),%eax
  mov    %eax,%edi
  call   <va_end@plt>
  mov    -0xb4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int printf(const char *format, ...) {
  int ret;
  va_list args;
  assert(format != NULL);
  va_start(args, format);
  ret = vfdprintf(1, format, args);
  va_end(args);
  return ret;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<printf>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    $0x1,%edi
  call   <assert@plt>
  mov    %rbx,%rsi
  mov    $0x0,%ebp
  mov    %ebp,%edi
  call   <va_start@plt>
  mov    %ebp,%edx
  mov    %rbx,%rsi
  mov    $0x1,%edi
  call   <vfdprintf@plt>
  mov    %eax,%ebx
  mov    %ebp,%edi
  call   <va_end@plt>
  mov    %ebx,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int printf(const char *format, ...) {
  int ret;
  va_list args;
  assert(format != NULL);
  va_start(args, format);
  ret = vfdprintf(1, format, args);
  va_end(args);
  return ret;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<printf>:
  endbr64
  push   %r13
  xor    %r13d,%r13d
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  mov    $0x1,%edi
  call   <assert@plt>
  mov    %rbp,%rsi
  mov    %r13d,%edi
  call   <va_start@plt>
  mov    %r13d,%edx
  mov    %rbp,%rsi
  mov    $0x1,%edi
  call   <vfdprintf@plt>
  mov    %r13d,%edi
  mov    %eax,%r12d
  call   <va_end@plt>
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int printf(const char *format, ...) {
  int ret;
  va_list args;
  assert(format != NULL);
  va_start(args, format);
  ret = vfdprintf(1, format, args);
  va_end(args);
  return ret;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<printf>:
  endbr64
  push   %r13
  xor    %r13d,%r13d
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  mov    $0x1,%edi
  call   <assert@plt>
  mov    %rbp,%rsi
  mov    %r13d,%edi
  call   <va_start@plt>
  mov    %r13d,%edx
  mov    %rbp,%rsi
  mov    $0x1,%edi
  call   <vfdprintf@plt>
  mov    %r13d,%edi
  mov    %eax,%r12d
  call   <va_end@plt>
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int printf(const char *format, ...) {
  int ret;
  va_list args;
  assert(format != NULL);
  va_start(args, format);
  ret = vfdprintf(1, format, args);
  va_end(args);
  return ret;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D29(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    -0x10(%rbp),%rdx
  lea    -0x14(%rbp),%rax
  mov    %rax,%rsi
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x14(%rbp),%edx
  mov    %edx,%eax
  add    %eax,%eax
  add    %edx,%eax
  shl    $0x2,%eax
  mov    %eax,%edx
  mov    -0x10(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x10(%rbp),%eax
  imul   $0x1e,%eax,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D28(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L1
  call   <__stack_chk_fail@plt>
L1:
  leave
  ret
```

tool_calls: name: parse_data
arguments: label: D29
datatype: string
name: parse_data
arguments: label: D19
datatype: string
name: parse_data
arguments: label: D28
datatype: string

tool: D29: Quantos dias tem em...?
Digite, primeiro, um numero qualquer de anos
Depois, digite um numero qualquer de dias
Apos isso irei te dizer quanto isso equivale em dias!

tool: D19: %d%d

tool: D28: Voce me deu um total de %d dias


assistant: ```c
int main() {
  int ano, dia, mes;
  printf("Quantos dias tem em...?\nDigite, primeiro, um numero qualquer de anos\nDepois, digite um numero qualquer de dias\nApos isso irei te dizer quanto isso equivale em dias!\n");
  scanf("%d"
        "%d",
        &ano, &mes);
  mes = mes + ano * 12;
  dia = mes * 30;
  printf("Voce me deu um total de %d dias\n", dia);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D0(%rip),%rdi
  call   <puts@plt>
  lea    0x4(%rsp),%rdx
  mov    %rsp,%rsi
  lea    D28(%rip),%rdi
  call   <scanf@plt>
  mov    (%rsp),%eax
  lea    (%rax,%rax,2),%edx
  mov    0x4(%rsp),%eax
  lea    (%rax,%rdx,4),%esi
  mov    %esi,0x4(%rsp)
  imul   $0x1e,%esi,%esi
  lea    D6(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  mov    $0x0,%eax
  add    $0x18,%rsp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D0
datatype: string
name: read_data
arguments: tag: D28
datatype: string
name: read_data
arguments: tag: D6
datatype: string

tool: D0: Quantos dias tem em...?
Digite, primeiro, um numero qualquer de anos
Depois, digite um numero qualquer de dias
Apos isso irei te dizer quanto isso equivale em dias!

tool: D28: %d%d

tool: D6: Voce me deu um total de %d dias


assistant: ```c
int main() {
  int ano, dia, mes;
  printf("Quantos dias tem em...?\nDigite, primeiro, um numero qualquer de anos\nDepois, digite um numero qualquer de dias\nApos isso irei te dizer quanto isso equivale em dias!\n");
  scanf("%d"
        "%d",
        &ano, &mes);
  mes = mes + ano * 12;
  dia = mes * 30;
  printf("Voce me deu um total de %d dias\n", dia);
  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
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D15(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  lea    0x4(%rsp),%rdx
  mov    %rsp,%rsi
  lea    D6(%rip),%rdi
  call   <scanf@plt>
  mov    (%rsp),%eax
  lea    D0(%rip),%rdi
  lea    (%rax,%rax,2),%edx
  mov    0x4(%rsp),%eax
  lea    (%rax,%rdx,4),%esi
  xor    %eax,%eax
  mov    %esi,0x4(%rsp)
  imul   $0x1e,%esi,%esi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D15
type_infered: string
name: read_data
arguments: tag: D6
type_infered: string
name: read_data
arguments: tag: D0
type_infered: string

tool: D15: Quantos dias tem em...?
Digite, primeiro, um numero qualquer de anos
Depois, digite um numero qualquer de dias
Apos isso irei te dizer quanto isso equivale em dias!

tool: D6: %d%d

tool: D0: Voce me deu um total de %d dias


assistant: ```c
int main() {
  int ano, dia, mes;
  printf("Quantos dias tem em...?\nDigite, primeiro, um numero qualquer de anos\nDepois, digite um numero qualquer de dias\nApos isso irei te dizer quanto isso equivale em dias!\n");
  scanf("%d"
        "%d",
        &ano, &mes);
  mes = mes + ano * 12;
  dia = mes * 30;
  printf("Voce me deu um total de %d dias\n", dia);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D19(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  lea    0x4(%rsp),%rdx
  mov    %rsp,%rsi
  lea    D16(%rip),%rdi
  call   <scanf@plt>
  mov    (%rsp),%eax
  lea    D18(%rip),%rdi
  lea    (%rax,%rax,2),%edx
  mov    0x4(%rsp),%eax
  lea    (%rax,%rdx,4),%esi
  xor    %eax,%eax
  mov    %esi,0x4(%rsp)
  imul   $0x1e,%esi,%esi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D19
type_infered: string
name: unpack_data
arguments: data_tag: D16
type_infered: string
name: unpack_data
arguments: data_tag: D18
type_infered: string

tool: D19: Quantos dias tem em...?
Digite, primeiro, um numero qualquer de anos
Depois, digite um numero qualquer de dias
Apos isso irei te dizer quanto isso equivale em dias!

tool: D16: %d%d

tool: D18: Voce me deu um total de %d dias


assistant: ```c
int main() {
  int ano, dia, mes;
  printf("Quantos dias tem em...?\nDigite, primeiro, um numero qualquer de anos\nDepois, digite um numero qualquer de dias\nApos isso irei te dizer quanto isso equivale em dias!\n");
  scanf("%d"
        "%d",
        &ano, &mes);
  mes = mes + ano * 12;
  dia = mes * 30;
  printf("Voce me deu um total de %d dias\n", dia);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D17(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D16(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: data_label: D17
type_infered: string
name: parse_data
arguments: data_label: D16
type_infered: string

tool: D17: Hola Mundo


tool: D16: Nananana Batman

assistant: ```c
void main() {
  printf("Hola Mundo\n");
  printf("Nananana Batman");
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D9(%rip),%rdi
  call   <printf@plt>
  lea    D7(%rip),%rdi
  call   <printf@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: label: D9
type: string
name: retrieve_data
arguments: label: D7
type: string

tool: D9: Hola Mundo


tool: D7: Nananana Batman

assistant: ```c
void main() {
  printf("Hola Mundo\n");
  printf("Nananana Batman");
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D15(%rip),%rdi
  call   <printf@plt>
  lea    D13(%rip),%rdi
  add    $0x8,%rsp
  jmp    1050 <printf@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D15
type_infered: string
name: retrieve_data
arguments: data_label: D13
type_infered: string

tool: D15: Hola Mundo


tool: D13: Nananana Batman

assistant: ```c
void main() {
  printf("Hola Mundo\n");
  printf("Nananana Batman");
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D3(%rip),%rdi
  call   <printf@plt>
  lea    D0(%rip),%rdi
  add    $0x8,%rsp
  jmp    1050 <printf@plt>
```

tool_calls: name: read_data
arguments: label: D3
data_type: string
name: read_data
arguments: label: D0
data_type: string

tool: D3: Hola Mundo


tool: D0: Nananana Batman

assistant: ```c
void main() {
  printf("Hola Mundo\n");
  printf("Nananana Batman");
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<get_allocation_size>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  sub    $0x10,%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rax
  pop    %rbp
  ret
```

assistant: ```c
size_t get_allocation_size(const void *mem) {
  char *buf = (char *)mem - (sizeof(size_t) * 2);
  return ((size_t *)buf)[0];
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<get_allocation_size>:
  endbr64
  mov    -0x10(%rdi),%rax
  ret
```

assistant: ```c
size_t get_allocation_size(const void *mem) {
  char *buf = (char *)mem - (sizeof(size_t) * 2);
  return ((size_t *)buf)[0];
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<get_allocation_size>:
  endbr64
  mov    -0x10(%rdi),%rax
  ret
```

assistant: ```c
size_t get_allocation_size(const void *mem) {
  char *buf = (char *)mem - (sizeof(size_t) * 2);
  return ((size_t *)buf)[0];
}
```
user: What is the c source code of the assembly code below:

```asm
<get_allocation_size>:
  endbr64
  mov    -0x10(%rdi),%rax
  ret
```

assistant: ```c
size_t get_allocation_size(const void *mem) {
  char *buf = (char *)mem - (sizeof(size_t) * 2);
  return ((size_t *)buf)[0];
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<inw>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,%eax
  mov    %ax,-0x14(%rbp)
  movzwl -0x14(%rbp),%eax
  mov    %eax,%edx
  in     (%dx),%ax
  mov    %ax,-0x2(%rbp)
  movzwl -0x2(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned int inw(unsigned short _port) {
  unsigned short rv;
  __asm__ __volatile__("inw %1,%0" : "=a"(rv) : "dN"(_port));
  return rv;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<inw>:
  endbr64
  mov    %edi,%edx
  in     (%dx),%ax
  movzwl %ax,%eax
  ret
```

assistant: ```c
unsigned int inw(unsigned short _port) {
  unsigned short rv;
  __asm__ __volatile__("inw %1,%0" : "=a"(rv) : "dN"(_port));
  return rv;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<inw>:
  endbr64
  mov    %edi,%edx
  in     (%dx),%ax
  movzwl %ax,%eax
  ret
```

assistant: ```c
unsigned int inw(unsigned short _port) {
  unsigned short rv;
  __asm__ __volatile__("inw %1,%0" : "=a"(rv) : "dN"(_port));
  return rv;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<inw>:
  endbr64
  mov    %edi,%edx
  in     (%dx),%ax
  movzwl %ax,%eax
  ret
```

assistant: ```c
unsigned int inw(unsigned short _port) {
  unsigned short rv;
  __asm__ __volatile__("inw %1,%0" : "=a"(rv) : "dN"(_port));
  return rv;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<ft_ultimate_range>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  movl   $0x0,-0xc(%rbp)
  mov    -0x1c(%rbp),%eax
  cmp    -0x20(%rbp),%eax
  jge    L8
  mov    -0x20(%rbp),%eax
  sub    -0x1c(%rbp),%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x10(%rbp),%eax
  add    $0x1,%eax
  cltq
  shl    $0x3,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  jmp    L14
L15:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    -0x1c(%rbp),%ecx
  mov    -0xc(%rbp),%edx
  add    %ecx,%edx
  mov    %edx,(%rax)
  subl   $0x1,-0x10(%rbp)
  addl   $0x1,-0xc(%rbp)
L14:
  cmpl   $0x0,-0x10(%rbp)
  jg     L15
  mov    -0x18(%rbp),%rax
  mov    -0x8(%rbp),%rdx
  mov    %rdx,(%rax)
  mov    -0xc(%rbp),%eax
  jmp    L24
L8:
  mov    -0x18(%rbp),%rax
  movq   $0x0,(%rax)
  mov    $0x0,%eax
L24:
  leave
  ret
```

assistant: ```c
int ft_ultimate_range(int **range, int min, int max) {
  int len;
  int i;
  int *ptr_range;
  i = 0;
  if (min < max) {
    len = max - min;
    ptr_range = malloc(sizeof(int *) * (len + 1));
    while (len > 0) {
      ptr_range[i] = min + i;
      len--;
      i++;
    }
    *range = ptr_range;
  } else {
    *range = NULL;
    return (0);
  }
  return (i);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<ft_ultimate_range>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  cmp    %edx,%esi
  jl     L16
  mov    $0x0,%eax
  mov    $0x0,%ebp
L2:
  mov    %rax,(%r12)
  mov    %ebp,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L16:
  mov    %esi,%ebx
  mov    %edx,%ebp
  sub    %esi,%ebp
  lea    0x8(,%rbp,8),%edi
  call   <malloc@plt>
  mov    %ebp,%edx
  mov    $0x0,%ecx
L4:
  lea    (%rbx,%rcx,1),%esi
  mov    %esi,(%rax,%rcx,4)
  add    $0x1,%rcx
  cmp    %rdx,%rcx
  jne    L4
  jmp    L2
```

assistant: ```c
int ft_ultimate_range(int **range, int min, int max) {
  int len;
  int i;
  int *ptr_range;
  i = 0;
  if (min < max) {
    len = max - min;
    ptr_range = malloc(sizeof(int *) * (len + 1));
    while (len > 0) {
      ptr_range[i] = min + i;
      len--;
      i++;
    }
    *range = ptr_range;
  } else {
    *range = NULL;
    return (0);
  }
  return (i);
}
```
user: Translate the below assembly code into C programming language:

```asm
<ft_ultimate_range>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  cmp    %edx,%esi
  jl     L14
  xor    %eax,%eax
  xor    %r12d,%r12d
  pop    %rbx
  mov    %rax,0x0(%rbp)
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  ret
L14:
  mov    %edx,%r12d
  mov    %esi,%ebx
  sub    %esi,%r12d
  lea    0x8(,%r12,8),%edi
  call   <malloc@plt>
  movslq %r12d,%rdx
  xor    %ecx,%ecx
  nopw   0x0(%rax,%rax,1)
L19:
  lea    (%rbx,%rcx,1),%esi
  mov    %esi,(%rax,%rcx,4)
  add    $0x1,%rcx
  cmp    %rdx,%rcx
  jne    L19
  mov    %rax,0x0(%rbp)
  pop    %rbx
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int ft_ultimate_range(int **range, int min, int max) {
  int len;
  int i;
  int *ptr_range;
  i = 0;
  if (min < max) {
    len = max - min;
    ptr_range = malloc(sizeof(int *) * (len + 1));
    while (len > 0) {
      ptr_range[i] = min + i;
      len--;
      i++;
    }
    *range = ptr_range;
  } else {
    *range = NULL;
    return (0);
  }
  return (i);
}
```