ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  push   %rbp
  lea    D19(%rip),%rsi
  lea    D21(%rip),%rdi
  mov    %rsp,%rbp
  push   %r14
  lea    D16(%rip),%r14
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x28(%rbp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    -0x2c(%rbp),%rsi
  mov    %r14,%rdi
  call   <scanf@plt>
  movslq -0x2c(%rbp),%rax
  mov    %rsp,%rcx
  lea    0xf(,%rax,4),%rax
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rcx
  and    $0xfffffffffffffff0,%rdx
  cmp    %rcx,%rsp
  je     L24
L59:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rcx,%rsp
  jne    L59
L24:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  jne    L36
L38:
  lea    D54(%rip),%rdi
  mov    %rsp,%r13
  call   <puts@plt>
  mov    -0x2c(%rbp),%esi
  test   %esi,%esi
  jle    L50
  mov    %r13,%r12
  xor    %ebx,%ebx
  xchg   %ax,%ax
L12:
  mov    %r12,%rsi
  mov    %r14,%rdi
  add    $0x1,%ebx
  add    $0x4,%r12
  call   <scanf@plt>
  mov    -0x2c(%rbp),%esi
  cmp    %ebx,%esi
  jg     L12
L50:
  mov    %r13,%rdi
  call   <quicksort@plt>
  lea    D11(%rip),%rdi
  call   <puts@plt>
  mov    -0x2c(%rbp),%eax
  test   %eax,%eax
  jle    L2
  xor    %ebx,%ebx
  lea    D1(%rip),%r12
  nopl   (%rax)
L34:
  mov    0x0(%r13,%rbx,4),%esi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    %ebx,-0x2c(%rbp)
  jg     L34
L2:
  mov    -0x28(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L20
  lea    -0x20(%rbp),%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %rbp
  ret
L36:
  orq    $0x0,-0x8(%rsp,%rdx,1)
  jmp    L38
L20:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D19
datatype: string
name: parse_data
arguments: data_tag: D21
datatype: string
name: parse_data
arguments: data_tag: D16
datatype: string
name: parse_data
arguments: data_tag: D54
datatype: string
name: parse_data
arguments: data_tag: D11
datatype: string
name: parse_data
arguments: data_tag: D1
datatype: string

tool: D19: Enter the length of your array:

tool: D21: %s

tool: D16: %d

tool: D54: Enter the numbers in your array:

tool: D11: After quicksort:

tool: D1: %d


assistant: ```c
int main(int argc, char const *argv[]) {
  int size;
  printf("%s", "Enter the length of your array:");
  scanf("%d", &size);
  int i;
  int array[size];
  printf("%s\n", "Enter the numbers in your array:");
  for (i = 0; i < size; i++) {
    scanf("%d", &array[i]);
  }
  quicksort(array, size);
  printf("%s\n", "After quicksort:");
  for (i = 0; i < size; i++) {
    printf("%d\n", array[i]);
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D1(%rip),%rsi
  lea    D23(%rip),%rdi
  mov    %rsp,%rbp
  push   %r14
  lea    D39(%rip),%r14
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x28(%rbp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    -0x2c(%rbp),%rsi
  mov    %r14,%rdi
  call   <scanf@plt>
  movslq -0x2c(%rbp),%rax
  mov    %rsp,%rcx
  lea    0xf(,%rax,4),%rax
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rcx
  and    $0xfffffffffffffff0,%rdx
  cmp    %rcx,%rsp
  je     L14
L28:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rcx,%rsp
  jne    L28
L14:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  jne    L75
L13:
  lea    D26(%rip),%rdi
  mov    %rsp,%r13
  call   <puts@plt>
  mov    -0x2c(%rbp),%esi
  test   %esi,%esi
  jle    L39
  mov    %r13,%r12
  xor    %ebx,%ebx
  xchg   %ax,%ax
L63:
  mov    %r12,%rsi
  mov    %r14,%rdi
  add    $0x1,%ebx
  add    $0x4,%r12
  call   <scanf@plt>
  mov    -0x2c(%rbp),%esi
  cmp    %ebx,%esi
  jg     L63
L39:
  mov    %r13,%rdi
  call   <quicksort@plt>
  lea    D49(%rip),%rdi
  call   <puts@plt>
  mov    -0x2c(%rbp),%eax
  test   %eax,%eax
  jle    L23
  xor    %ebx,%ebx
  lea    D4(%rip),%r12
  nopl   (%rax)
L22:
  mov    0x0(%r13,%rbx,4),%esi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    %ebx,-0x2c(%rbp)
  jg     L22
L23:
  mov    -0x28(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L55
  lea    -0x20(%rbp),%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %rbp
  ret
L75:
  orq    $0x0,-0x8(%rsp,%rdx,1)
  jmp    L13
L55:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D1
type_infered: string
name: unpack_data
arguments: tag: D23
type_infered: string
name: unpack_data
arguments: tag: D39
type_infered: string
name: unpack_data
arguments: tag: D26
type_infered: string
name: unpack_data
arguments: tag: D49
type_infered: string
name: unpack_data
arguments: tag: D4
type_infered: string

tool: D1: Enter the length of your array:

tool: D23: %s

tool: D39: %d

tool: D26: Enter the numbers in your array:

tool: D49: After quicksort:

tool: D4: %d


assistant: ```c
int main(int argc, char const *argv[]) {
  int size;
  printf("%s", "Enter the length of your array:");
  scanf("%d", &size);
  int i;
  int array[size];
  printf("%s\n", "Enter the numbers in your array:");
  for (i = 0; i < size; i++) {
    scanf("%d", &array[i]);
  }
  quicksort(array, size);
  printf("%s\n", "After quicksort:");
  for (i = 0; i < size; i++) {
    printf("%d\n", array[i]);
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<HTS_calloc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  imul   -0x20(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x10(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  cmpq   $0x0,-0x8(%rbp)
  jne    L8
  lea    D5(%rip),%rax
  mov    %rax,%rsi
  mov    $0x1,%edi
  call   <HTS_error@plt>
L8:
  mov    -0x8(%rbp),%rax
  leave
  ret
```

tool_calls: name: read_data
arguments: tag: D5
data_type: string

tool: D5: HTS_calloc: Cannot allocate memory.


assistant: ```c
void *HTS_calloc(const size_t num, const size_t size) {
  size_t n = num * size;
  void *mem = (void *)malloc(n);
  memset(mem, 0, n);
  if (mem == NULL)
    HTS_error(1, "HTS_calloc: Cannot allocate memory.\n");
  return mem;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<HTS_calloc>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  imul   %rsi,%rdi
  mov    %rdi,%rbp
  call   <malloc@plt>
  mov    %rax,%rbx
  mov    %rbp,%rdx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  test   %rbx,%rbx
  je     L7
L15:
  mov    %rbx,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L7:
  lea    D3(%rip),%rsi
  mov    $0x1,%edi
  call   <HTS_error@plt>
  jmp    L15
```

tool_calls: name: extract_symbolic_data
arguments: label: D3
data_type: string

tool: D3: HTS_calloc: Cannot allocate memory.


assistant: ```c
void *HTS_calloc(const size_t num, const size_t size) {
  size_t n = num * size;
  void *mem = (void *)malloc(n);
  memset(mem, 0, n);
  if (mem == NULL)
    HTS_error(1, "HTS_calloc: Cannot allocate memory.\n");
  return mem;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<HTS_calloc>:
  endbr64
  push   %r13
  imul   %rsi,%rdi
  push   %r12
  mov    %rdi,%r13
  sub    $0x8,%rsp
  call   <malloc@plt>
  xor    %esi,%esi
  mov    %r13,%rdx
  mov    %rax,%r12
  mov    %rax,%rdi
  call   <memset@plt>
  test   %r12,%r12
  je     L9
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L9:
  lea    D2(%rip),%rsi
  mov    $0x1,%edi
  call   <HTS_error@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: unpack_data
arguments: tag: D2
type_infered: string

tool: D2: HTS_calloc: Cannot allocate memory.


assistant: ```c
void *HTS_calloc(const size_t num, const size_t size) {
  size_t n = num * size;
  void *mem = (void *)malloc(n);
  memset(mem, 0, n);
  if (mem == NULL)
    HTS_error(1, "HTS_calloc: Cannot allocate memory.\n");
  return mem;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<HTS_calloc>:
  endbr64
  push   %r13
  imul   %rsi,%rdi
  push   %r12
  mov    %rdi,%r13
  sub    $0x8,%rsp
  call   <malloc@plt>
  xor    %esi,%esi
  mov    %r13,%rdx
  mov    %rax,%r12
  mov    %rax,%rdi
  call   <memset@plt>
  test   %r12,%r12
  je     L2
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L2:
  lea    D8(%rip),%rsi
  mov    $0x1,%edi
  call   <HTS_error@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D8
data_type: string

tool: D8: HTS_calloc: Cannot allocate memory.


assistant: ```c
void *HTS_calloc(const size_t num, const size_t size) {
  size_t n = num * size;
  void *mem = (void *)malloc(n);
  memset(mem, 0, n);
  if (mem == NULL)
    HTS_error(1, "HTS_calloc: Cannot allocate memory.\n");
  return mem;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0x0,(%rsp)
  mov    D19(%rip),%r12
  lea    0x4(%rsp),%rbx
  mov    (%r12),%ecx
  mov    $0x1,%edx
  mov    $0x4,%esi
  mov    %rbx,%rdi
  call   <fread@plt>
  mov    D15(%rip),%rbp
  mov    %rsp,%r13
  mov    0x0(%rbp),%ecx
  mov    $0x1,%edx
  mov    $0x4,%esi
  mov    %r13,%rdi
  call   <fwrite@plt>
  mov    0x0(%rbp),%ecx
  mov    $0x1,%edx
  mov    $0x4,%esi
  mov    %rbx,%rdi
  call   <fwrite@plt>
  mov    (%r12),%ecx
  mov    $0x1,%edx
  mov    $0x4,%esi
  mov    %rbx,%rdi
  call   <fread@plt>
  mov    0x0(%rbp),%ecx
  mov    $0x1,%edx
  mov    $0x4,%esi
  mov    %r13,%rdi
  call   <fwrite@plt>
  mov    0x0(%rbp),%ecx
  mov    $0x1,%edx
  mov    $0x4,%esi
  mov    %rbx,%rdi
  call   <fwrite@plt>
  mov    (%r12),%ecx
  mov    $0x1,%edx
  mov    $0x4,%esi
  mov    %rbx,%rdi
  call   <fread@plt>
  mov    0x0(%rbp),%ecx
  mov    $0x1,%edx
  mov    $0x4,%esi
  mov    %rbx,%rdi
  call   <fwrite@plt>
  mov    (%r12),%ecx
  mov    $0x1,%edx
  mov    $0x4,%esi
  mov    %rbx,%rdi
  call   <fread@plt>
  mov    0x0(%rbp),%ecx
  mov    $0x1,%edx
  mov    $0x4,%esi
  mov    %rbx,%rdi
  call   <fwrite@plt>
  movl   $0x1,0x4(%rsp)
  mov    0x0(%rbp),%ecx
  mov    $0x1,%edx
  mov    $0x4,%esi
  mov    %rbx,%rdi
  call   <fwrite@plt>
  mov    0x0(%rbp),%ecx
  mov    $0x1,%edx
  mov    $0x4,%esi
  mov    %rbx,%rdi
  call   <fwrite@plt>
  mov    D19(%rip),%rbp
L25:
  mov    0x0(%rbp),%edi
  call   <feof@plt>
  test   %eax,%eax
  jne    L27
  mov    D19(%rip),%rbx
  lea    0x4(%rsp),%rdi
  mov    (%rbx),%ecx
  mov    $0x1,%edx
  mov    $0x4,%esi
  call   <fread@plt>
  mov    (%rbx),%edi
  call   <feof@plt>
  test   %eax,%eax
  jne    L25
  lea    0x4(%rsp),%rdi
  mov    D15(%rip),%rax
  mov    (%rax),%ecx
  mov    $0x1,%edx
  mov    $0x4,%esi
  call   <fwrite@plt>
  jmp    L25
L27:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L33
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L33:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(int argc, char *argv[]) {
  uint8_t zero = 0;
  uint8_t val;
  fread(&val, sizeof(val), 1, stdin);
  fwrite(&zero, sizeof(zero), 1, stdout);
  fwrite(&val, sizeof(val), 1, stdout);
  fread(&val, sizeof(val), 1, stdin);
  fwrite(&zero, sizeof(zero), 1, stdout);
  fwrite(&val, sizeof(val), 1, stdout);
  fread(&val, sizeof(val), 1, stdin);
  fwrite(&val, sizeof(val), 1, stdout);
  fread(&val, sizeof(val), 1, stdin);
  fwrite(&val, sizeof(val), 1, stdout);
  val = 1;
  fwrite(&val, sizeof(val), 1, stdout);
  fwrite(&val, sizeof(val), 1, stdout);
  while (!feof(stdin)) {
    fread(&val, sizeof(val), 1, stdin);
    if (!feof(stdin)) {
      fwrite(&val, sizeof(val), 1, stdout);
    }
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  push   %r13
  mov    $0x1,%edx
  mov    $0x4,%esi
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    D0(%rip),%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rbp
  movl   $0x0,(%rsp)
  mov    %rsp,%r13
  mov    (%rbx),%ecx
  mov    %rbp,%rdi
  call   <fread@plt>
  mov    D18(%rip),%r12
  mov    $0x1,%edx
  mov    %r13,%rdi
  mov    $0x4,%esi
  mov    (%r12),%ecx
  call   <fwrite@plt>
  mov    (%r12),%ecx
  mov    $0x1,%edx
  mov    %rbp,%rdi
  mov    $0x4,%esi
  call   <fwrite@plt>
  mov    (%rbx),%ecx
  mov    $0x1,%edx
  mov    %rbp,%rdi
  mov    $0x4,%esi
  call   <fread@plt>
  mov    (%r12),%ecx
  mov    $0x1,%edx
  mov    %r13,%rdi
  mov    $0x4,%esi
  call   <fwrite@plt>
  mov    (%r12),%ecx
  mov    $0x1,%edx
  mov    %rbp,%rdi
  mov    $0x4,%esi
  call   <fwrite@plt>
  mov    (%rbx),%ecx
  mov    $0x1,%edx
  mov    %rbp,%rdi
  mov    $0x4,%esi
  call   <fread@plt>
  mov    (%r12),%ecx
  mov    $0x1,%edx
  mov    %rbp,%rdi
  mov    $0x4,%esi
  call   <fwrite@plt>
  mov    (%rbx),%ecx
  mov    $0x1,%edx
  mov    %rbp,%rdi
  mov    $0x4,%esi
  call   <fread@plt>
  mov    (%r12),%ecx
  mov    $0x1,%edx
  mov    %rbp,%rdi
  mov    $0x4,%esi
  call   <fwrite@plt>
  mov    (%r12),%ecx
  mov    $0x1,%edx
  mov    %rbp,%rdi
  mov    $0x4,%esi
  movl   $0x1,0x4(%rsp)
  call   <fwrite@plt>
L19:
  mov    (%r12),%ecx
  mov    $0x1,%edx
  mov    $0x4,%esi
  mov    %rbp,%rdi
  call   <fwrite@plt>
  jmp    L27
  nopl   (%rax)
L26:
  mov    (%rbx),%ecx
  mov    %rbp,%rdi
  mov    $0x1,%edx
  mov    $0x4,%esi
  call   <fread@plt>
  mov    (%rbx),%edi
  call   <feof@plt>
  test   %eax,%eax
  je     L19
L27:
  mov    (%rbx),%edi
  call   <feof@plt>
  test   %eax,%eax
  je     L26
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L36
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L36:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(int argc, char *argv[]) {
  uint8_t zero = 0;
  uint8_t val;
  fread(&val, sizeof(val), 1, stdin);
  fwrite(&zero, sizeof(zero), 1, stdout);
  fwrite(&val, sizeof(val), 1, stdout);
  fread(&val, sizeof(val), 1, stdin);
  fwrite(&zero, sizeof(zero), 1, stdout);
  fwrite(&val, sizeof(val), 1, stdout);
  fread(&val, sizeof(val), 1, stdin);
  fwrite(&val, sizeof(val), 1, stdout);
  fread(&val, sizeof(val), 1, stdin);
  fwrite(&val, sizeof(val), 1, stdout);
  val = 1;
  fwrite(&val, sizeof(val), 1, stdout);
  fwrite(&val, sizeof(val), 1, stdout);
  while (!feof(stdin)) {
    fread(&val, sizeof(val), 1, stdin);
    if (!feof(stdin)) {
      fwrite(&val, sizeof(val), 1, stdout);
    }
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  push   %r13
  mov    $0x1,%edx
  mov    $0x4,%esi
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    D1(%rip),%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rbp
  movl   $0x0,(%rsp)
  mov    %rsp,%r13
  mov    (%rbx),%ecx
  mov    %rbp,%rdi
  call   <fread@plt>
  mov    D19(%rip),%r12
  mov    $0x1,%edx
  mov    %r13,%rdi
  mov    $0x4,%esi
  mov    (%r12),%ecx
  call   <fwrite@plt>
  mov    (%r12),%ecx
  mov    $0x1,%edx
  mov    %rbp,%rdi
  mov    $0x4,%esi
  call   <fwrite@plt>
  mov    (%rbx),%ecx
  mov    $0x1,%edx
  mov    %rbp,%rdi
  mov    $0x4,%esi
  call   <fread@plt>
  mov    (%r12),%ecx
  mov    $0x1,%edx
  mov    %r13,%rdi
  mov    $0x4,%esi
  call   <fwrite@plt>
  mov    (%r12),%ecx
  mov    $0x1,%edx
  mov    %rbp,%rdi
  mov    $0x4,%esi
  call   <fwrite@plt>
  mov    (%rbx),%ecx
  mov    $0x1,%edx
  mov    %rbp,%rdi
  mov    $0x4,%esi
  call   <fread@plt>
  mov    (%r12),%ecx
  mov    $0x1,%edx
  mov    %rbp,%rdi
  mov    $0x4,%esi
  call   <fwrite@plt>
  mov    (%rbx),%ecx
  mov    $0x1,%edx
  mov    %rbp,%rdi
  mov    $0x4,%esi
  call   <fread@plt>
  mov    (%r12),%ecx
  mov    $0x1,%edx
  mov    %rbp,%rdi
  mov    $0x4,%esi
  call   <fwrite@plt>
  mov    (%r12),%ecx
  mov    $0x1,%edx
  mov    %rbp,%rdi
  mov    $0x4,%esi
  movl   $0x1,0x4(%rsp)
  call   <fwrite@plt>
L19:
  mov    (%r12),%ecx
  mov    $0x1,%edx
  mov    $0x4,%esi
  mov    %rbp,%rdi
  call   <fwrite@plt>
  jmp    L6
  nopl   (%rax)
L29:
  mov    (%rbx),%ecx
  mov    %rbp,%rdi
  mov    $0x1,%edx
  mov    $0x4,%esi
  call   <fread@plt>
  mov    (%rbx),%edi
  call   <feof@plt>
  test   %eax,%eax
  je     L19
L6:
  mov    (%rbx),%edi
  call   <feof@plt>
  test   %eax,%eax
  je     L29
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L27
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L27:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(int argc, char *argv[]) {
  uint8_t zero = 0;
  uint8_t val;
  fread(&val, sizeof(val), 1, stdin);
  fwrite(&zero, sizeof(zero), 1, stdout);
  fwrite(&val, sizeof(val), 1, stdout);
  fread(&val, sizeof(val), 1, stdin);
  fwrite(&zero, sizeof(zero), 1, stdout);
  fwrite(&val, sizeof(val), 1, stdout);
  fread(&val, sizeof(val), 1, stdin);
  fwrite(&val, sizeof(val), 1, stdout);
  fread(&val, sizeof(val), 1, stdin);
  fwrite(&val, sizeof(val), 1, stdout);
  val = 1;
  fwrite(&val, sizeof(val), 1, stdout);
  fwrite(&val, sizeof(val), 1, stdout);
  while (!feof(stdin)) {
    fread(&val, sizeof(val), 1, stdin);
    if (!feof(stdin)) {
      fwrite(&val, sizeof(val), 1, stdout);
    }
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<lonelyinteger>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,-0x4(%rbp)
  movl   $0x1,-0x8(%rbp)
  jmp    L14
L2:
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  xor    %eax,-0x4(%rbp)
  addl   $0x1,-0x8(%rbp)
L14:
  mov    -0x8(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L2
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int lonelyinteger(int a_size, int *a) {
  int i, val;
  val = a[0];
  for (i = 1; i < a_size; i++) {
    val = val ^ a[i];
  }
  return val;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<lonelyinteger>:
  endbr64
  mov    (%rsi),%edx
  cmp    $0x1,%edi
  jle    L2
  lea    0x4(%rsi),%rax
  lea    -0x2(%rdi),%ecx
  lea    0x8(%rsi,%rcx,4),%rcx
L16:
  xor    (%rax),%edx
  add    $0x4,%rax
  cmp    %rcx,%rax
  jne    L16
L2:
  mov    %edx,%eax
  ret
```

assistant: ```c
int lonelyinteger(int a_size, int *a) {
  int i, val;
  val = a[0];
  for (i = 1; i < a_size; i++) {
    val = val ^ a[i];
  }
  return val;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<lonelyinteger>:
  endbr64
  mov    (%rsi),%r8d
  cmp    $0x1,%edi
  jle    L12
  lea    -0x2(%rdi),%edx
  lea    0x4(%rsi),%rax
  lea    0x8(%rsi,%rdx,4),%rdx
  nopl   0x0(%rax,%rax,1)
L13:
  xor    (%rax),%r8d
  add    $0x4,%rax
  cmp    %rdx,%rax
  jne    L13
L12:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int lonelyinteger(int a_size, int *a) {
  int i, val;
  val = a[0];
  for (i = 1; i < a_size; i++) {
    val = val ^ a[i];
  }
  return val;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<lonelyinteger>:
  endbr64
  mov    (%rsi),%r8d
  mov    %rsi,%rdx
  cmp    $0x1,%edi
  jle    L41
  lea    -0x2(%rdi),%eax
  lea    -0x1(%rdi),%esi
  cmp    $0x2,%eax
  jbe    L51
  mov    %esi,%ecx
  lea    0x4(%rdx),%rax
  pxor   %xmm0,%xmm0
  shr    $0x2,%ecx
  sub    $0x1,%ecx
  shl    $0x4,%rcx
  lea    0x14(%rdx,%rcx,1),%rcx
  nopl   0x0(%rax,%rax,1)
L36:
  movdqu (%rax),%xmm2
  add    $0x10,%rax
  pxor   %xmm2,%xmm0
  cmp    %rcx,%rax
  jne    L36
  movdqa %xmm0,%xmm1
  mov    %esi,%ecx
  psrldq $0x8,%xmm1
  and    $0xfffffffc,%ecx
  pxor   %xmm1,%xmm0
  movdqa %xmm0,%xmm1
  psrldq $0x4,%xmm1
  pxor   %xmm1,%xmm0
  movd   %xmm0,%eax
  xor    %eax,%r8d
  lea    0x1(%rcx),%eax
  cmp    %ecx,%esi
  je     L41
L11:
  movslq %eax,%rcx
  lea    0x0(,%rcx,4),%rsi
  xor    (%rdx,%rcx,4),%r8d
  lea    0x1(%rax),%ecx
  cmp    %ecx,%edi
  jle    L41
  add    $0x2,%eax
  xor    0x4(%rdx,%rsi,1),%r8d
  cmp    %eax,%edi
  jle    L41
  xor    0x8(%rdx,%rsi,1),%r8d
L41:
  mov    %r8d,%eax
  ret
L51:
  mov    $0x1,%eax
  jmp    L11
```

assistant: ```c
int lonelyinteger(int a_size, int *a) {
  int i, val;
  val = a[0];
  for (i = 1; i < a_size; i++) {
    val = val ^ a[i];
  }
  return val;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<uart1GetchBlock>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  nop
L4:
  mov    $0xe0010014,%eax
  movzbl (%rax),%eax
  movzbl %al,%eax
  and    $0x1,%eax
  test   %eax,%eax
  je     L4
  mov    $0xe0010000,%eax
  movzbl (%rax),%eax
  movzbl %al,%eax
  pop    %rbp
  ret
```

assistant: ```c
int uart1GetchBlock(void) {
  while (!((*((volatile unsigned char *)0xE0010014)) & (1 << 0)))
    ;
  return (*((volatile unsigned char *)0xE0010000));
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<uart1GetchBlock>:
  endbr64
  mov    $0xe0010014,%edx
L3:
  movzbl (%rdx),%eax
  test   $0x1,%al
  je     L3
  movabs 0xe0010000,%al
  movzbl %al,%eax
  ret
```

assistant: ```c
int uart1GetchBlock(void) {
  while (!((*((volatile unsigned char *)0xE0010014)) & (1 << 0)))
    ;
  return (*((volatile unsigned char *)0xE0010000));
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<uart1GetchBlock>:
  endbr64
  mov    $0xe0010014,%edx
  nopl   0x0(%rax)
L9:
  movzbl (%rdx),%eax
  test   $0x1,%al
  je     L9
  movabs 0xe0010000,%al
  movzbl %al,%eax
  ret
```

assistant: ```c
int uart1GetchBlock(void) {
  while (!((*((volatile unsigned char *)0xE0010014)) & (1 << 0)))
    ;
  return (*((volatile unsigned char *)0xE0010000));
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<uart1GetchBlock>:
  endbr64
  mov    $0xe0010014,%edx
  nopl   0x0(%rax)
L0:
  movzbl (%rdx),%eax
  test   $0x1,%al
  je     L0
  movabs 0xe0010000,%al
  movzbl %al,%eax
  ret
```

assistant: ```c
int uart1GetchBlock(void) {
  while (!((*((volatile unsigned char *)0xE0010014)) & (1 << 0)))
    ;
  return (*((volatile unsigned char *)0xE0010000));
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<ai_free>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  setne  %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rax
  test   %rax,%rax
  je     L5
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <freeaddrinfo@plt>
L5:
  nop
  leave
  ret
```

assistant: ```c
void ai_free(struct addrinfo **const ai) {
  assert(ai != NULL);
  if (*ai != NULL)
    freeaddrinfo(*ai);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ai_free>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  test   %rdi,%rdi
  setne  %dil
  movzbl %dil,%edi
  call   <assert@plt>
  mov    (%rbx),%rdi
  test   %rdi,%rdi
  je     L8
  call   <freeaddrinfo@plt>
L8:
  pop    %rbx
  ret
```

assistant: ```c
void ai_free(struct addrinfo **const ai) {
  assert(ai != NULL);
  if (*ai != NULL)
    freeaddrinfo(*ai);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<ai_free>:
  endbr64
  push   %rbx
  test   %rdi,%rdi
  mov    %rdi,%rbx
  setne  %dil
  movzbl %dil,%edi
  call   <assert@plt>
  mov    (%rbx),%rdi
  test   %rdi,%rdi
  je     L7
  pop    %rbx
  jmp    1070 <freeaddrinfo@plt>
  cs nopw 0x0(%rax,%rax,1)
L7:
  pop    %rbx
  ret
```

assistant: ```c
void ai_free(struct addrinfo **const ai) {
  assert(ai != NULL);
  if (*ai != NULL)
    freeaddrinfo(*ai);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<ai_free>:
  endbr64
  push   %rbx
  test   %rdi,%rdi
  mov    %rdi,%rbx
  setne  %dil
  movzbl %dil,%edi
  call   <assert@plt>
  mov    (%rbx),%rdi
  test   %rdi,%rdi
  je     L2
  pop    %rbx
  jmp    1070 <freeaddrinfo@plt>
  cs nopw 0x0(%rax,%rax,1)
L2:
  pop    %rbx
  ret
```

assistant: ```c
void ai_free(struct addrinfo **const ai) {
  assert(ai != NULL);
  if (*ai != NULL)
    freeaddrinfo(*ai);
}
```