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

```asm
<reverse>:
  endbr64
  sub    $0x3f8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x3e8(%rsp)
  xor    %eax,%eax
  mov    %esi,%edx
  sub    $0x1,%edx
  js     L29
  mov    %esi,%r8d
  movslq %edx,%rdx
  add    %rdi,%rdx
  mov    %rsp,%rax
  lea    -0x1(%rsi),%ecx
  lea    0x1(%rsp,%rcx,1),%rsi
L15:
  movzbl (%rdx),%ecx
  mov    %cl,(%rax)
  sub    $0x1,%rdx
  add    $0x1,%rax
  cmp    %rsi,%rax
  jne    L15
L37:
  movslq %r8d,%rax
  movb   $0x0,(%rsp,%rax,1)
  movzbl (%rsp),%edx
  test   %dl,%dl
  je     L36
  mov    $0x1,%eax
  lea    -0x1(%rsp),%rsi
L39:
  mov    %dl,-0x1(%rdi,%rax,1)
  mov    %eax,%ecx
  add    $0x1,%rax
  movzbl (%rsi,%rax,1),%edx
  test   %dl,%dl
  jne    L39
L61:
  movslq %ecx,%rcx
  movb   $0x0,(%rdi,%rcx,1)
  mov    0x3e8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L64
  mov    %r8d,%eax
  add    $0x3f8,%rsp
  ret
L29:
  mov    $0x0,%r8d
  jmp    L37
L36:
  mov    $0x0,%ecx
  jmp    L61
L64:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int reverse(char s[], int len) {
  int i, j;
  char copy[1000];
  for (i = len - 1, j = 0; i >= 0; --i, j++)
    copy[j] = s[i];
  copy[j] = '\0';
  for (i = 0; copy[i] != '\0'; i++) {
    s[i] = copy[i];
  }
  s[i] = '\0';
  return j;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<reverse>:
  endbr64
  sub    $0x3f8,%rsp
  mov    %esi,%ecx
  mov    %fs:0x28,%rax
  mov    %rax,0x3e8(%rsp)
  xor    %eax,%eax
  sub    $0x1,%ecx
  js     L45
  movslq %ecx,%rdx
  mov    %ecx,%ecx
  mov    %esi,%r8d
  mov    %rsp,%rax
  add    %rdi,%rdx
  lea    0x1(%rsp,%rcx,1),%rsi
  nopl   0x0(%rax)
L59:
  movzbl (%rdx),%ecx
  add    $0x1,%rax
  sub    $0x1,%rdx
  mov    %cl,-0x1(%rax)
  cmp    %rsi,%rax
  jne    L59
L27:
  movslq %r8d,%rax
  movb   $0x0,(%rsp,%rax,1)
  movzbl (%rsp),%edx
  test   %dl,%dl
  je     L46
  mov    $0x1,%eax
  lea    -0x1(%rsp),%rsi
  nopl   0x0(%rax)
L57:
  mov    %dl,-0x1(%rdi,%rax,1)
  movslq %eax,%rcx
  add    $0x1,%rax
  movzbl (%rsi,%rax,1),%edx
  test   %dl,%dl
  jne    L57
  add    %rcx,%rdi
L46:
  movb   $0x0,(%rdi)
  mov    0x3e8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L30
  mov    %r8d,%eax
  add    $0x3f8,%rsp
  ret
  nopw   0x0(%rax,%rax,1)
L45:
  xor    %r8d,%r8d
  jmp    L27
L30:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int reverse(char s[], int len) {
  int i, j;
  char copy[1000];
  for (i = len - 1, j = 0; i >= 0; --i, j++)
    copy[j] = s[i];
  copy[j] = '\0';
  for (i = 0; copy[i] != '\0'; i++) {
    s[i] = copy[i];
  }
  s[i] = '\0';
  return j;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<reverse>:
  endbr64
  sub    $0x3f8,%rsp
  mov    %esi,%ecx
  mov    %fs:0x28,%rax
  mov    %rax,0x3e8(%rsp)
  xor    %eax,%eax
  sub    $0x1,%ecx
  js     L15
  movslq %ecx,%rdx
  mov    %ecx,%ecx
  mov    %esi,%r8d
  mov    %rsp,%rax
  add    %rdi,%rdx
  lea    0x1(%rsp,%rcx,1),%rsi
  nopl   0x0(%rax)
L59:
  movzbl (%rdx),%ecx
  add    $0x1,%rax
  sub    $0x1,%rdx
  mov    %cl,-0x1(%rax)
  cmp    %rsi,%rax
  jne    L59
L39:
  movslq %r8d,%rax
  movb   $0x0,(%rsp,%rax,1)
  movzbl (%rsp),%edx
  test   %dl,%dl
  je     L40
  mov    $0x1,%eax
  lea    -0x1(%rsp),%rsi
  nopl   0x0(%rax)
L48:
  mov    %dl,-0x1(%rdi,%rax,1)
  movslq %eax,%rcx
  add    $0x1,%rax
  movzbl (%rsi,%rax,1),%edx
  test   %dl,%dl
  jne    L48
  add    %rcx,%rdi
L40:
  movb   $0x0,(%rdi)
  mov    0x3e8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L28
  mov    %r8d,%eax
  add    $0x3f8,%rsp
  ret
  nopw   0x0(%rax,%rax,1)
L15:
  xor    %r8d,%r8d
  jmp    L39
L28:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int reverse(char s[], int len) {
  int i, j;
  char copy[1000];
  for (i = len - 1, j = 0; i >= 0; --i, j++)
    copy[j] = s[i];
  copy[j] = '\0';
  for (i = 0; copy[i] != '\0'; i++) {
    s[i] = copy[i];
  }
  s[i] = '\0';
  return j;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<get2bytes>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  shl    $0x8,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x18(%rbp),%rax
  add    $0x4,%rax
  mov    (%rax),%eax
  or     %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
u_int get2bytes(u_char *buf) {
  u_int ret;
  ret = buf[0] << 8;
  ret |= buf[1];
  return ret;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<get2bytes>:
  endbr64
  mov    (%rdi),%eax
  shl    $0x8,%eax
  or     0x4(%rdi),%eax
  ret
```

assistant: ```c
u_int get2bytes(u_char *buf) {
  u_int ret;
  ret = buf[0] << 8;
  ret |= buf[1];
  return ret;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<get2bytes>:
  endbr64
  mov    (%rdi),%eax
  shl    $0x8,%eax
  or     0x4(%rdi),%eax
  ret
```

assistant: ```c
u_int get2bytes(u_char *buf) {
  u_int ret;
  ret = buf[0] << 8;
  ret |= buf[1];
  return ret;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<get2bytes>:
  endbr64
  mov    (%rdi),%eax
  shl    $0x8,%eax
  or     0x4(%rdi),%eax
  ret
```

assistant: ```c
u_int get2bytes(u_char *buf) {
  u_int ret;
  ret = buf[0] << 8;
  ret |= buf[1];
  return ret;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<s_wbd>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  movsd  %xmm0,-0x10(%rbp)
  lea    -0x10(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <s_wb64@plt>
  nop
  leave
  ret
```

assistant: ```c
void s_wbd(struct serializer *s, double d) {
  s_wb64(s, *(uint64_t *)&d);
  return;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<s_wbd>:
  endbr64
  sub    $0x8,%rsp
  movq   %xmm0,%rsi
  call   <s_wb64@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void s_wbd(struct serializer *s, double d) {
  s_wb64(s, *(uint64_t *)&d);
  return;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<s_wbd>:
  endbr64
  movq   %xmm0,%rsi
  jmp    1050 <s_wb64@plt>
```

assistant: ```c
void s_wbd(struct serializer *s, double d) {
  s_wb64(s, *(uint64_t *)&d);
  return;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<s_wbd>:
  endbr64
  movq   %xmm0,%rsi
  jmp    1050 <s_wb64@plt>
```

assistant: ```c
void s_wbd(struct serializer *s, double d) {
  s_wb64(s, *(uint64_t *)&d);
  return;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <f@plt>
  cmp    $0x10,%eax
  je     L1
  mov    $0x0,%eax
  call   <abort@plt>
L1:
  mov    $0x0,%edi
  call   <exit@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main() {
  if (f() != 16)
    abort();
  exit(0);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <f@plt>
  cmp    $0x10,%eax
  jne    L17
L10:
  mov    $0x0,%edi
  call   <exit@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
L17:
  mov    $0x0,%eax
  call   <abort@plt>
  jmp    L10
```

assistant: ```c
int main() {
  if (f() != 16)
    abort();
  exit(0);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <f@plt>
  cmp    $0x10,%eax
  je     L5
  xor    %eax,%eax
  call   <abort@plt>
L5:
  xor    %edi,%edi
  call   <exit@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  if (f() != 16)
    abort();
  exit(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    $0x8,%rsp
  xor    %eax,%eax
  call   <f@plt>
  cmp    $0x10,%eax
  je     L2
  xor    %eax,%eax
  call   <abort@plt>
L2:
  xor    %edi,%edi
  call   <exit@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  if (f() != 16)
    abort();
  exit(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
<strdup_wchar_to_char>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <wcslen@plt>
  add    $0x1,%rax
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L11
  mov    $0x0,%eax
  jmp    L15
L11:
  mov    -0x18(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy_wchar_to_char@plt>
  mov    -0x8(%rbp),%rax
L15:
  leave
  ret
```

assistant: ```c
char *strdup_wchar_to_char(const wchar_t *src) {
  char *dst = malloc((wcslen(src) + 1));
  if (!dst)
    return NULL;
  strcpy_wchar_to_char(dst, src);
  return dst;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<strdup_wchar_to_char>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  call   <wcslen@plt>
  lea    0x1(%rax),%rdi
  call   <malloc@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L5
  mov    %rbp,%rsi
  mov    %rax,%rdi
  call   <strcpy_wchar_to_char@plt>
L5:
  mov    %rbx,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
char *strdup_wchar_to_char(const wchar_t *src) {
  char *dst = malloc((wcslen(src) + 1));
  if (!dst)
    return NULL;
  strcpy_wchar_to_char(dst, src);
  return dst;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<strdup_wchar_to_char>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <wcslen@plt>
  lea    0x1(%rax),%rdi
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L9
  mov    %rbp,%rsi
  mov    %rax,%rdi
  call   <strcpy_wchar_to_char@plt>
L9:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *strdup_wchar_to_char(const wchar_t *src) {
  char *dst = malloc((wcslen(src) + 1));
  if (!dst)
    return NULL;
  strcpy_wchar_to_char(dst, src);
  return dst;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<strdup_wchar_to_char>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <wcslen@plt>
  lea    0x1(%rax),%rdi
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L8
  mov    %rbp,%rsi
  mov    %rax,%rdi
  call   <strcpy_wchar_to_char@plt>
L8:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *strdup_wchar_to_char(const wchar_t *src) {
  char *dst = malloc((wcslen(src) + 1));
  if (!dst)
    return NULL;
  strcpy_wchar_to_char(dst, src);
  return dst;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<uart_send_byte>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,%eax
  mov    %al,-0x4(%rbp)
  nop
L5:
  mov    $0x50000010,%eax
  mov    (%rax),%rax
  and    $0x4,%eax
  test   %rax,%rax
  je     L5
  mov    $0x50000020,%edx
  movsbq -0x4(%rbp),%rax
  mov    %rax,(%rdx)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void uart_send_byte(char ch) {
  while (!((*(volatile unsigned long *)(0x50000000 + 0x10)) & 4))
    ;
  (*(volatile unsigned long *)(0x50000000 + 0x20)) = ch;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<uart_send_byte>:
  endbr64
L1:
  mov    0x50000010,%rax
  test   $0x4,%al
  je     L1
  movsbq %dil,%rdi
  mov    %rdi,0x50000020
  ret
```

assistant: ```c
void uart_send_byte(char ch) {
  while (!((*(volatile unsigned long *)(0x50000000 + 0x10)) & 4))
    ;
  (*(volatile unsigned long *)(0x50000000 + 0x20)) = ch;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<uart_send_byte>:
  endbr64
  nopl   0x0(%rax)
L7:
  mov    0x50000010,%rax
  test   $0x4,%al
  je     L7
  movsbq %dil,%rdi
  mov    %rdi,0x50000020
  ret
```

assistant: ```c
void uart_send_byte(char ch) {
  while (!((*(volatile unsigned long *)(0x50000000 + 0x10)) & 4))
    ;
  (*(volatile unsigned long *)(0x50000000 + 0x20)) = ch;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<uart_send_byte>:
  endbr64
  nopl   0x0(%rax)
L0:
  mov    0x50000010,%rax
  test   $0x4,%al
  je     L0
  movsbq %dil,%rdi
  mov    %rdi,0x50000020
  ret
```

assistant: ```c
void uart_send_byte(char ch) {
  while (!((*(volatile unsigned long *)(0x50000000 + 0x10)) & 4))
    ;
  (*(volatile unsigned long *)(0x50000000 + 0x20)) = ch;
}
```
user: What is the c source code of the assembly code below?

```asm
<count_matches>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,%eax
  mov    %al,-0x1c(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
L14:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    %al,-0x1c(%rbp)
  jne    L10
  addl   $0x1,-0xc(%rbp)
L10:
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  movzbl (%rax),%eax
  test   %al,%al
  jne    L14
  mov    -0xc(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int count_matches(const char *s1, char c) {
  const char *ptr = s1;
  int n = 0;
  do {
    if (*ptr == c)
      n++;
  } while (*(ptr++));
  return n;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<count_matches>:
  endbr64
  mov    $0x0,%edx
L2:
  movzbl (%rdi),%eax
  cmp    %sil,%al
  sete   %cl
  movzbl %cl,%ecx
  add    %ecx,%edx
  add    $0x1,%rdi
  test   %al,%al
  jne    L2
  mov    %edx,%eax
  ret
```

assistant: ```c
int count_matches(const char *s1, char c) {
  const char *ptr = s1;
  int n = 0;
  do {
    if (*ptr == c)
      n++;
  } while (*(ptr++));
  return n;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<count_matches>:
  endbr64
  xor    %r8d,%r8d
  nopw   0x0(%rax,%rax,1)
L1:
  movzbl (%rdi),%eax
  xor    %edx,%edx
  cmp    %sil,%al
  sete   %dl
  add    $0x1,%rdi
  add    %edx,%r8d
  test   %al,%al
  jne    L1
  mov    %r8d,%eax
  ret
```

assistant: ```c
int count_matches(const char *s1, char c) {
  const char *ptr = s1;
  int n = 0;
  do {
    if (*ptr == c)
      n++;
  } while (*(ptr++));
  return n;
}
```
user: What is the c source code of the assembly code below?

```asm
<count_matches>:
  endbr64
  xor    %r8d,%r8d
  nopw   0x0(%rax,%rax,1)
L1:
  movzbl (%rdi),%eax
  xor    %edx,%edx
  cmp    %al,%sil
  sete   %dl
  add    $0x1,%rdi
  add    %edx,%r8d
  test   %al,%al
  jne    L1
  mov    %r8d,%eax
  ret
```

assistant: ```c
int count_matches(const char *s1, char c) {
  const char *ptr = s1;
  int n = 0;
  do {
    if (*ptr == c)
      n++;
  } while (*(ptr++));
  return n;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0xa0000000,-0xc(%rbp)
  movl   $0x0,-0x14(%rbp)
  movl   $0x0,-0x10(%rbp)
  mov    -0xc(%rbp),%eax
  mov    %eax,-0x14(%rbp)
  mov    %eax,-0x10(%rbp)
  mov    -0x14(%rbp),%eax
  cmp    %eax,-0xc(%rbp)
  jne    L8
  mov    -0x10(%rbp),%eax
  cmp    %eax,-0xc(%rbp)
  jne    L8
  mov    $0x1,%eax
  jmp    L18
L8:
  mov    $0x0,%eax
L18:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L28
  call   <__stack_chk_fail@plt>
L28:
  leave
  ret
```

assistant: ```c
int main() {
  int arg = 0xA0000000;
  int out1 = 0;
  int out2 = 0;
  __asm__("movl %2, %0; movl %2, %1;" : "=m"(out1), "=m"(out2) : "r"(arg));
  return (arg == out1) && (arg == out2);
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    $0xa0000000,%eax
  mov    %eax,(%rsp)
  mov    %eax,0x4(%rsp)
  mov    $0x0,%eax
  cmpl   $0xa0000000,(%rsp)
  je     L3
L27:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L12
  add    $0x18,%rsp
  ret
L3:
  cmpl   $0xa0000000,0x4(%rsp)
  sete   %al
  movzbl %al,%eax
  jmp    L27
L12:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main() {
  int arg = 0xA0000000;
  int out1 = 0;
  int out2 = 0;
  __asm__("movl %2, %0; movl %2, %1;" : "=m"(out1), "=m"(out2) : "r"(arg));
  return (arg == out1) && (arg == out2);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  mov    $0xa0000000,%eax
  mov    %eax,(%rsp)
  mov    %eax,0x4(%rsp)
  xor    %eax,%eax
  cmpl   $0xa0000000,(%rsp)
  jne    L13
  xor    %eax,%eax
  cmpl   $0xa0000000,0x4(%rsp)
  sete   %al
L13:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L17
  add    $0x18,%rsp
  ret
L17:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main() {
  int arg = 0xA0000000;
  int out1 = 0;
  int out2 = 0;
  __asm__("movl %2, %0; movl %2, %1;" : "=m"(out1), "=m"(out2) : "r"(arg));
  return (arg == out1) && (arg == out2);
}
```
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)
  mov    $0xa0000000,%eax
  mov    %eax,(%rsp)
  mov    %eax,0x4(%rsp)
  xor    %eax,%eax
  cmpl   $0xa0000000,(%rsp)
  jne    L13
  xor    %eax,%eax
  cmpl   $0xa0000000,0x4(%rsp)
  sete   %al
L13:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L16
  add    $0x18,%rsp
  ret
L16:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main() {
  int arg = 0xA0000000;
  int out1 = 0;
  int out2 = 0;
  __asm__("movl %2, %0; movl %2, %1;" : "=m"(out1), "=m"(out2) : "r"(arg));
  return (arg == out1) && (arg == out2);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %rsi,-0x30(%rbp)
  cmpl   $0x2,-0x24(%rbp)
  je     L7
  mov    $0x0,%edi
  call   <exit@plt>
L7:
  mov    -0x30(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  mov    $0xa,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <strtoul@plt>
  mov    %eax,-0x18(%rbp)
  mov    -0x18(%rbp),%eax
  shl    $0x14,%eax
  shr    %eax
  mov    %eax,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
L15:
  mov    -0x14(%rbp),%edx
  mov    -0x10(%rbp),%rcx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0x14(%rbp),%edx
  mov    -0x8(%rbp),%rcx
  mov    -0x10(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    $0x2,%edi
  call   <sleep@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    -0x14(%rbp),%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  jmp    L15
```

assistant: ```c
int main(int argc, char *argv[]) {
  if (argc != 2) {
    exit(0);
  }
  unsigned int use_memory;
  unsigned int max_memory;
  use_memory = strtoul(argv[1], NULL, 10);
  max_memory = use_memory * 1024 * 1024 / 2;
  int *a;
  a = (int *)malloc(max_memory);
  int *c;
  c = (int *)malloc(max_memory);
  while (1) {
    memcpy(c, a, max_memory);
    memcpy(a, c, max_memory);
    sleep(2);
    free(a);
    free(c);
    a = (int *)malloc(max_memory);
    c = (int *)malloc(max_memory);
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  cmp    $0x2,%edi
  jne    L1
L7:
  mov    0x8(%rbx),%rdi
  mov    $0xa,%edx
  mov    $0x0,%esi
  call   <strtoul@plt>
  shl    $0x14,%eax
  mov    %eax,%r12d
  shr    %r12d
  mov    %r12d,%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  mov    %r12d,%edi
  call   <malloc@plt>
  mov    %rax,%rbx
L13:
  mov    %r12d,%edx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <memcpy@plt>
  mov    %r12d,%edx
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <memcpy@plt>
  mov    $0x2,%edi
  call   <sleep@plt>
  mov    %rbp,%rdi
  call   <free@plt>
  mov    %rbx,%rdi
  call   <free@plt>
  mov    %r12d,%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  mov    %r12d,%edi
  call   <malloc@plt>
  mov    %rax,%rbx
  jmp    L13
L1:
  mov    $0x0,%edi
  call   <exit@plt>
  jmp    L7
```

assistant: ```c
int main(int argc, char *argv[]) {
  if (argc != 2) {
    exit(0);
  }
  unsigned int use_memory;
  unsigned int max_memory;
  use_memory = strtoul(argv[1], NULL, 10);
  max_memory = use_memory * 1024 * 1024 / 2;
  int *a;
  a = (int *)malloc(max_memory);
  int *c;
  c = (int *)malloc(max_memory);
  while (1) {
    memcpy(c, a, max_memory);
    memcpy(a, c, max_memory);
    sleep(2);
    free(a);
    free(c);
    a = (int *)malloc(max_memory);
    c = (int *)malloc(max_memory);
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  cmp    $0x2,%edi
  je     L4
  xor    %edi,%edi
  call   <exit@plt>
L4:
  mov    0x8(%rbx),%rdi
  mov    $0xa,%edx
  xor    %esi,%esi
  call   <strtoul@plt>
  shl    $0x14,%eax
  mov    %eax,%ebx
  shr    %ebx
  xchg   %ax,%ax
L0:
  mov    %ebx,%edi
  call   <malloc@plt>
  mov    %ebx,%edi
  mov    %rax,%r12
  call   <malloc@plt>
  mov    %ebx,%edx
  mov    %r12,%rsi
  mov    %rax,%rbp
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    %ebx,%edx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <memcpy@plt>
  mov    $0x2,%edi
  call   <sleep@plt>
  mov    %r12,%rdi
  call   <free@plt>
  mov    %rbp,%rdi
  call   <free@plt>
  jmp    L0
```

assistant: ```c
int main(int argc, char *argv[]) {
  if (argc != 2) {
    exit(0);
  }
  unsigned int use_memory;
  unsigned int max_memory;
  use_memory = strtoul(argv[1], NULL, 10);
  max_memory = use_memory * 1024 * 1024 / 2;
  int *a;
  a = (int *)malloc(max_memory);
  int *c;
  c = (int *)malloc(max_memory);
  while (1) {
    memcpy(c, a, max_memory);
    memcpy(a, c, max_memory);
    sleep(2);
    free(a);
    free(c);
    a = (int *)malloc(max_memory);
    c = (int *)malloc(max_memory);
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  cmp    $0x2,%edi
  je     L2
  xor    %edi,%edi
  call   <exit@plt>
L2:
  mov    0x8(%rbx),%rdi
  mov    $0xa,%edx
  xor    %esi,%esi
  call   <strtoul@plt>
  shl    $0x14,%eax
  mov    %eax,%ebx
  shr    %ebx
  xchg   %ax,%ax
L5:
  mov    %ebx,%edi
  call   <malloc@plt>
  mov    %ebx,%edi
  mov    %rax,%r12
  call   <malloc@plt>
  mov    %ebx,%edx
  mov    %r12,%rsi
  mov    %rax,%rbp
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    %ebx,%edx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <memcpy@plt>
  mov    $0x2,%edi
  call   <sleep@plt>
  mov    %r12,%rdi
  call   <free@plt>
  mov    %rbp,%rdi
  call   <free@plt>
  jmp    L5
```

assistant: ```c
int main(int argc, char *argv[]) {
  if (argc != 2) {
    exit(0);
  }
  unsigned int use_memory;
  unsigned int max_memory;
  use_memory = strtoul(argv[1], NULL, 10);
  max_memory = use_memory * 1024 * 1024 / 2;
  int *a;
  a = (int *)malloc(max_memory);
  int *c;
  c = (int *)malloc(max_memory);
  while (1) {
    memcpy(c, a, max_memory);
    memcpy(a, c, max_memory);
    sleep(2);
    free(a);
    free(c);
    a = (int *)malloc(max_memory);
    c = (int *)malloc(max_memory);
  }
  return 0;
}
```