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
<palindrome>:
  endbr64
  push   %r12
  push   %rbp
  movslq %esi,%rbp
  push   %rbx
  mov    %edi,%ebx
  lea    0x0(,%rbp,4),%edi
  mov    %rbp,%r12
  call   <malloc@plt>
  mov    %rax,%rdi
  test   %ebx,%ebx
  jle    L20
  mov    %rax,%rcx
  mov    $0xcccccccd,%r8d
  nopl   0x0(%rax)
L21:
  mov    %ebx,%eax
  mov    %ebx,%esi
  add    $0x4,%rcx
  imul   %r8,%rax
  shr    $0x23,%rax
  lea    (%rax,%rax,4),%edx
  add    %edx,%edx
  sub    %edx,%esi
  mov    %ebx,%edx
  mov    %eax,%ebx
  mov    %esi,-0x4(%rcx)
  cmp    $0x9,%edx
  jg     L21
L20:
  mov    %r12d,%esi
  shr    $0x1f,%esi
  add    %r12d,%esi
  sar    %esi
  cmp    $0x1,%r12d
  jle    L55
  lea    (%rdi,%rbp,2),%rcx
  mov    %rdi,%rax
  xor    %edx,%edx
  lea    -0x4(%rcx,%rcx,1),%r8
  jmp    L58
  nopl   0x0(%rax)
L28:
  add    $0x1,%edx
  add    $0x4,%rax
  cmp    %esi,%edx
  jge    L55
L58:
  mov    %r8,%rcx
  sub    %rax,%rcx
  mov    (%rcx),%ebx
  cmp    %ebx,(%rax)
  je     L28
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
  cs nopw 0x0(%rax,%rax,1)
L55:
  call   <free@plt>
  pop    %rbx
  mov    $0x1,%eax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int palindrome(int number, int digits) {
  int i, j;
  int *temp;
  temp = (int *)malloc(digits * sizeof(int));
  i = 0;
  while (number > 0) {
    temp[i] = number % 10;
    number = number / 10;
    i++;
  }
  j = digits;
  for (i = 0; i < digits / 2; i++) {
    if (temp[i] != temp[j - 1]) {
      return 0;
    }
    j--;
  }
  free(temp);
  return 1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<palindrome>:
  endbr64
  push   %r12
  push   %rbp
  movslq %esi,%rbp
  push   %rbx
  mov    %edi,%ebx
  lea    0x0(,%rbp,4),%edi
  mov    %rbp,%r12
  call   <malloc@plt>
  mov    %rax,%rdi
  test   %ebx,%ebx
  jle    L21
  mov    %rax,%rcx
  mov    $0xcccccccd,%r8d
  nopl   0x0(%rax)
L48:
  mov    %ebx,%eax
  mov    %ebx,%esi
  add    $0x4,%rcx
  imul   %r8,%rax
  shr    $0x23,%rax
  lea    (%rax,%rax,4),%edx
  add    %edx,%edx
  sub    %edx,%esi
  mov    %ebx,%edx
  mov    %eax,%ebx
  mov    %esi,-0x4(%rcx)
  cmp    $0x9,%edx
  jg     L48
L21:
  mov    %r12d,%esi
  shr    $0x1f,%esi
  add    %r12d,%esi
  sar    %esi
  cmp    $0x1,%r12d
  jle    L12
  lea    (%rdi,%rbp,2),%rcx
  mov    %rdi,%rax
  xor    %edx,%edx
  lea    -0x4(%rcx,%rcx,1),%r8
  jmp    L25
  nopl   0x0(%rax)
L27:
  add    $0x1,%edx
  add    $0x4,%rax
  cmp    %esi,%edx
  jge    L12
L25:
  mov    %r8,%rcx
  sub    %rax,%rcx
  mov    (%rcx),%ebx
  cmp    %ebx,(%rax)
  je     L27
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
  cs nopw 0x0(%rax,%rax,1)
L12:
  call   <free@plt>
  pop    %rbx
  mov    $0x1,%eax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int palindrome(int number, int digits) {
  int i, j;
  int *temp;
  temp = (int *)malloc(digits * sizeof(int));
  i = 0;
  while (number > 0) {
    temp[i] = number % 10;
    number = number / 10;
    i++;
  }
  j = digits;
  for (i = 0; i < digits / 2; i++) {
    if (temp[i] != temp[j - 1]) {
      return 0;
    }
    j--;
  }
  free(temp);
  return 1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<strcmp_0>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  lea    D19(%rip),%rax
  mov    %rax,-0x10(%rbp)
  lea    D19(%rip),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <clang_analyzer_eval@plt>
  nop
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D19
data_type: string

tool: D19: 123

assistant: ```c
void strcmp_0() {
  char *x = "123";
  char *y = "123";
  clang_analyzer_eval(strcmp(x, y) == 0);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<strcmp_0>:
  endbr64
  sub    $0x8,%rsp
  lea    D1(%rip),%rdi
  mov    %rdi,%rsi
  call   <strcmp@plt>
  test   %rax,%rax
  sete   %dil
  movzbl %dil,%edi
  call   <clang_analyzer_eval@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D1
type_guessed: string

tool: D1: 123

assistant: ```c
void strcmp_0() {
  char *x = "123";
  char *y = "123";
  clang_analyzer_eval(strcmp(x, y) == 0);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<strcmp_0>:
  endbr64
  lea    D3(%rip),%rdi
  sub    $0x8,%rsp
  mov    %rdi,%rsi
  call   <strcmp@plt>
  xor    %edi,%edi
  test   %rax,%rax
  sete   %dil
  add    $0x8,%rsp
  jmp    1070 <clang_analyzer_eval@plt>
```

tool_calls: name: read_data
arguments: data_tag: D3
type_guessed: string

tool: D3: 123

assistant: ```c
void strcmp_0() {
  char *x = "123";
  char *y = "123";
  clang_analyzer_eval(strcmp(x, y) == 0);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<strcmp_0>:
  endbr64
  lea    D1(%rip),%rdi
  sub    $0x8,%rsp
  mov    %rdi,%rsi
  call   <strcmp@plt>
  xor    %edi,%edi
  test   %rax,%rax
  sete   %dil
  add    $0x8,%rsp
  jmp    1070 <clang_analyzer_eval@plt>
```

tool_calls: name: disassemble_data
arguments: label: D1
type: string

tool: D1: 123

assistant: ```c
void strcmp_0() {
  char *x = "123";
  char *y = "123";
  clang_analyzer_eval(strcmp(x, y) == 0);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<get_current_memory>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  sub    $0x3000,%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned int get_current_memory(void *stack_limit) {
  void *stack = (void *)stack_limit - 12288;
  return *(uint32_t *)stack;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<get_current_memory>:
  endbr64
  mov    -0x3000(%rdi),%eax
  ret
```

assistant: ```c
unsigned int get_current_memory(void *stack_limit) {
  void *stack = (void *)stack_limit - 12288;
  return *(uint32_t *)stack;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<get_current_memory>:
  endbr64
  mov    -0x3000(%rdi),%eax
  ret
```

assistant: ```c
unsigned int get_current_memory(void *stack_limit) {
  void *stack = (void *)stack_limit - 12288;
  return *(uint32_t *)stack;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<get_current_memory>:
  endbr64
  mov    -0x3000(%rdi),%eax
  ret
```

assistant: ```c
unsigned int get_current_memory(void *stack_limit) {
  void *stack = (void *)stack_limit - 12288;
  return *(uint32_t *)stack;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<createServerSocket>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x60,%rsp
  mov    %rdi,-0x58(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x30(%rbp),%rax
  mov    $0x28,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  mov    D31(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,-0x20(%rbp)
  mov    D22(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,-0x10(%rbp)
  mov    D32(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,-0x24(%rbp)
  mov    D39(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,-0x28(%rbp)
  lea    -0x40(%rbp),%rcx
  lea    -0x30(%rbp),%rdx
  mov    -0x58(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x0,%edi
  call   <getaddrinfo@plt>
  mov    %eax,-0x44(%rbp)
  cmpl   $0x0,-0x44(%rbp)
  jns    L44
  mov    -0x44(%rbp),%eax
  mov    %eax,%edi
  call   <gai_strerror@plt>
  mov    %eax,%edi
  call   <sysError@plt>
L44:
  movl   $0xffffffff,-0x48(%rbp)
  mov    -0x40(%rbp),%rax
  mov    %rax,-0x38(%rbp)
  jmp    L47
L12:
  mov    -0x38(%rbp),%rax
  mov    0x8(%rax),%edx
  mov    -0x38(%rbp),%rax
  mov    0xc(%rax),%ecx
  mov    -0x38(%rbp),%rax
  mov    0x10(%rax),%eax
  mov    %ecx,%esi
  mov    %eax,%edi
  call   <socket@plt>
  mov    %eax,-0x48(%rbp)
  cmpl   $0x0,-0x48(%rbp)
  js     L56
  mov    -0x38(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x38(%rbp),%rax
  mov    0x4(%rax),%ecx
  mov    -0x48(%rbp),%eax
  mov    %ecx,%esi
  mov    %eax,%edi
  call   <bind@plt>
  test   %rax,%rax
  jne    L30
  mov    -0x48(%rbp),%eax
  mov    $0xa,%esi
  mov    %eax,%edi
  call   <listen@plt>
  test   %rax,%rax
  jne    L30
  mov    -0x48(%rbp),%eax
  mov    %eax,%edi
  call   <getSockName@plt>
  jmp    L24
L30:
  mov    -0x48(%rbp),%eax
  mov    %eax,%edi
  call   <close@plt>
  movl   $0xffffffff,-0x48(%rbp)
  jmp    L63
L56:
  nop
L63:
  mov    -0x38(%rbp),%rax
  mov    0x18(%rax),%rax
  mov    %rax,-0x38(%rbp)
L47:
  cmpq   $0x0,-0x38(%rbp)
  jne    L12
L24:
  mov    -0x40(%rbp),%rax
  mov    %rax,%rdi
  call   <freeaddrinfo@plt>
  mov    -0x48(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L54
  call   <__stack_chk_fail@plt>
L54:
  leave
  ret
```

assistant: ```c
int createServerSocket(const char *service) {
  struct addrinfo addrCriteria;
  memset(&addrCriteria, 0, sizeof(addrCriteria));
  addrCriteria.ai_family = AF_UNSPEC;
  addrCriteria.ai_flags = AI_PASSIVE;
  addrCriteria.ai_socktype = SOCK_STREAM;
  addrCriteria.ai_protocol = IPPROTO_TCP;
  struct addrinfo *servAddr;
  int ret = getaddrinfo(NULL, service, &addrCriteria, &servAddr);
  if (ret < 0)
    sysError(gai_strerror(ret));
  int servSock = -1;
  for (struct addrinfo *addr = servAddr; addr != NULL; addr = addr->ai_next) {
    servSock = socket(addr->ai_family, addr->ai_socktype, addr->ai_protocol);
    if (servSock < 0)
      continue;
    if (bind(servSock, addr->ai_addr, addr->ai_addrlen) == 0 && listen(servSock, 10) == 0) {
      getSockName(servSock);
      break;
    }
    close(servSock);
    servSock = -1;
  }
  freeaddrinfo(servAddr);
  return servSock;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<createServerSocket>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x40,%rsp
  mov    %rdi,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rbp
  mov    $0x28,%edx
  mov    $0x0,%esi
  mov    %rbp,%rdi
  call   <memset@plt>
  mov    D22(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,0x20(%rsp)
  mov    D26(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,0x30(%rsp)
  mov    D29(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,0x1c(%rsp)
  mov    D6(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,0x18(%rsp)
  lea    0x8(%rsp),%rcx
  mov    %rbp,%rdx
  mov    %rbx,%rsi
  mov    $0x0,%edi
  call   <getaddrinfo@plt>
  test   %eax,%eax
  js     L45
L68:
  mov    0x8(%rsp),%rbx
  test   %rbx,%rbx
  je     L57
  mov    $0xffffffff,%r12d
  jmp    L85
L45:
  mov    %eax,%edi
  call   <gai_strerror@plt>
  mov    %eax,%edi
  call   <sysError@plt>
  jmp    L68
L26:
  mov    %ebp,%edi
  call   <close@plt>
  mov    %r12d,%ebp
L14:
  mov    0x18(%rbx),%rbx
  test   %rbx,%rbx
  je     L7
L85:
  mov    0x8(%rbx),%edx
  mov    0xc(%rbx),%esi
  mov    0x10(%rbx),%edi
  call   <socket@plt>
  mov    %eax,%ebp
  test   %eax,%eax
  js     L14
  mov    0x4(%rbx),%esi
  mov    (%rbx),%edx
  mov    %eax,%edi
  call   <bind@plt>
  test   %rax,%rax
  jne    L26
  mov    $0xa,%esi
  mov    %ebp,%edi
  call   <listen@plt>
  test   %rax,%rax
  jne    L26
  mov    %ebp,%edi
  call   <getSockName@plt>
L7:
  mov    0x8(%rsp),%rdi
  call   <freeaddrinfo@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L10
  mov    %ebp,%eax
  add    $0x40,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L57:
  mov    $0xffffffff,%ebp
  jmp    L7
L10:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int createServerSocket(const char *service) {
  struct addrinfo addrCriteria;
  memset(&addrCriteria, 0, sizeof(addrCriteria));
  addrCriteria.ai_family = AF_UNSPEC;
  addrCriteria.ai_flags = AI_PASSIVE;
  addrCriteria.ai_socktype = SOCK_STREAM;
  addrCriteria.ai_protocol = IPPROTO_TCP;
  struct addrinfo *servAddr;
  int ret = getaddrinfo(NULL, service, &addrCriteria, &servAddr);
  if (ret < 0)
    sysError(gai_strerror(ret));
  int servSock = -1;
  for (struct addrinfo *addr = servAddr; addr != NULL; addr = addr->ai_next) {
    servSock = socket(addr->ai_family, addr->ai_socktype, addr->ai_protocol);
    if (servSock < 0)
      continue;
    if (bind(servSock, addr->ai_addr, addr->ai_addrlen) == 0 && listen(servSock, 10) == 0) {
      getSockName(servSock);
      break;
    }
    close(servSock);
    servSock = -1;
  }
  freeaddrinfo(servAddr);
  return servSock;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<createServerSocket>:
  endbr64
  push   %r12
  xor    %esi,%esi
  mov    $0x28,%edx
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x48,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%r12
  mov    %r12,%rdi
  call   <memset@plt>
  xor    %edi,%edi
  lea    0x8(%rsp),%rcx
  mov    %r12,%rdx
  mov    D21(%rip),%rax
  mov    %rbp,%rsi
  mov    (%rax),%eax
  mov    %eax,0x20(%rsp)
  mov    D16(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,0x30(%rsp)
  mov    D11(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,0x1c(%rsp)
  mov    D38(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,0x18(%rsp)
  call   <getaddrinfo@plt>
  test   %eax,%eax
  js     L82
L36:
  mov    0x8(%rsp),%rbp
  test   %rbp,%rbp
  je     L58
  xchg   %ax,%ax
L89:
  mov    0x8(%rbp),%edx
  mov    0xc(%rbp),%esi
  mov    0x10(%rbp),%edi
  call   <socket@plt>
  mov    %eax,%r12d
  test   %eax,%eax
  js     L77
  mov    0x4(%rbp),%esi
  mov    0x0(%rbp),%edx
  mov    %eax,%edi
  call   <bind@plt>
  test   %rax,%rax
  jne    L39
  mov    $0xa,%esi
  mov    %r12d,%edi
  call   <listen@plt>
  test   %rax,%rax
  je     L27
L39:
  mov    %r12d,%edi
  mov    $0xffffffff,%r12d
  call   <close@plt>
L77:
  mov    0x18(%rbp),%rbp
  test   %rbp,%rbp
  jne    L89
  mov    0x8(%rsp),%rbp
L81:
  mov    %rbp,%rdi
  call   <freeaddrinfo@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L28
  add    $0x48,%rsp
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L82:
  mov    %eax,%edi
  call   <gai_strerror@plt>
  mov    %eax,%edi
  call   <sysError@plt>
  jmp    L36
  nopl   0x0(%rax,%rax,1)
L27:
  mov    %r12d,%edi
  call   <getSockName@plt>
  mov    0x8(%rsp),%rbp
  jmp    L81
L58:
  mov    $0xffffffff,%r12d
  jmp    L81
L28:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int createServerSocket(const char *service) {
  struct addrinfo addrCriteria;
  memset(&addrCriteria, 0, sizeof(addrCriteria));
  addrCriteria.ai_family = AF_UNSPEC;
  addrCriteria.ai_flags = AI_PASSIVE;
  addrCriteria.ai_socktype = SOCK_STREAM;
  addrCriteria.ai_protocol = IPPROTO_TCP;
  struct addrinfo *servAddr;
  int ret = getaddrinfo(NULL, service, &addrCriteria, &servAddr);
  if (ret < 0)
    sysError(gai_strerror(ret));
  int servSock = -1;
  for (struct addrinfo *addr = servAddr; addr != NULL; addr = addr->ai_next) {
    servSock = socket(addr->ai_family, addr->ai_socktype, addr->ai_protocol);
    if (servSock < 0)
      continue;
    if (bind(servSock, addr->ai_addr, addr->ai_addrlen) == 0 && listen(servSock, 10) == 0) {
      getSockName(servSock);
      break;
    }
    close(servSock);
    servSock = -1;
  }
  freeaddrinfo(servAddr);
  return servSock;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<createServerSocket>:
  endbr64
  push   %r12
  xor    %esi,%esi
  mov    $0x28,%edx
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x48,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%r12
  mov    %r12,%rdi
  call   <memset@plt>
  xor    %edi,%edi
  lea    0x8(%rsp),%rcx
  mov    %rbp,%rsi
  mov    D20(%rip),%rax
  mov    D33(%rip),%rdx
  mov    (%rax),%eax
  movd   (%rdx),%xmm0
  mov    %r12,%rdx
  mov    %eax,0x20(%rsp)
  mov    D34(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,0x30(%rsp)
  mov    D28(%rip),%rax
  mov    (%rax),%eax
  movd   %eax,%xmm1
  punpckldq %xmm1,%xmm0
  movq   %xmm0,0x18(%rsp)
  call   <getaddrinfo@plt>
  test   %eax,%eax
  js     L63
L6:
  mov    0x8(%rsp),%rbp
  test   %rbp,%rbp
  je     L30
  cs nopw 0x0(%rax,%rax,1)
L93:
  mov    0x8(%rbp),%edx
  mov    0xc(%rbp),%esi
  mov    0x10(%rbp),%edi
  call   <socket@plt>
  mov    %eax,%r12d
  test   %eax,%eax
  js     L48
  mov    0x4(%rbp),%esi
  mov    0x0(%rbp),%edx
  mov    %eax,%edi
  call   <bind@plt>
  test   %rax,%rax
  jne    L42
  mov    $0xa,%esi
  mov    %r12d,%edi
  call   <listen@plt>
  test   %rax,%rax
  je     L11
L42:
  mov    %r12d,%edi
  mov    $0xffffffff,%r12d
  call   <close@plt>
L48:
  mov    0x18(%rbp),%rbp
  test   %rbp,%rbp
  jne    L93
  mov    0x8(%rsp),%rbp
L94:
  mov    %rbp,%rdi
  call   <freeaddrinfo@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L45
  add    $0x48,%rsp
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L63:
  mov    %eax,%edi
  call   <gai_strerror@plt>
  mov    %eax,%edi
  call   <sysError@plt>
  jmp    L6
  nopl   0x0(%rax,%rax,1)
L11:
  mov    %r12d,%edi
  call   <getSockName@plt>
  mov    0x8(%rsp),%rbp
  jmp    L94
L30:
  mov    $0xffffffff,%r12d
  jmp    L94
L45:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int createServerSocket(const char *service) {
  struct addrinfo addrCriteria;
  memset(&addrCriteria, 0, sizeof(addrCriteria));
  addrCriteria.ai_family = AF_UNSPEC;
  addrCriteria.ai_flags = AI_PASSIVE;
  addrCriteria.ai_socktype = SOCK_STREAM;
  addrCriteria.ai_protocol = IPPROTO_TCP;
  struct addrinfo *servAddr;
  int ret = getaddrinfo(NULL, service, &addrCriteria, &servAddr);
  if (ret < 0)
    sysError(gai_strerror(ret));
  int servSock = -1;
  for (struct addrinfo *addr = servAddr; addr != NULL; addr = addr->ai_next) {
    servSock = socket(addr->ai_family, addr->ai_socktype, addr->ai_protocol);
    if (servSock < 0)
      continue;
    if (bind(servSock, addr->ai_addr, addr->ai_addrlen) == 0 && listen(servSock, 10) == 0) {
      getSockName(servSock);
      break;
    }
    close(servSock);
    servSock = -1;
  }
  freeaddrinfo(servAddr);
  return servSock;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<isPrime>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %edi,-0x24(%rbp)
  cmpl   $0x1,-0x24(%rbp)
  jg     L6
  mov    $0x0,%eax
  jmp    L69
L6:
  movl   $0x2,-0x14(%rbp)
  jmp    L21
L57:
  mov    -0x24(%rbp),%eax
  cltd
  idivl  -0x14(%rbp)
  mov    %edx,%eax
  test   %eax,%eax
  je     L37
  addl   $0x1,-0x14(%rbp)
L21:
  mov    -0x24(%rbp),%eax
  mov    %eax,%edi
  call   <sqrt@plt>
  cmp    %eax,-0x14(%rbp)
  jle    L57
  jmp    L42
L37:
  nop
L42:
  mov    -0x14(%rbp),%eax
  lea    -0x1(%rax),%ebx
  mov    -0x24(%rbp),%eax
  mov    %eax,%edi
  call   <sqrt@plt>
  cmp    %eax,%ebx
  jne    L41
  mov    $0x1,%eax
  jmp    L69
L41:
  mov    $0x0,%eax
L69:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int isPrime(int x) {
  if (x < 2)
    return 0;
  int i;
  for (i = 2; i <= (int)sqrt(x); i++)
    if (x % i == 0)
      break;
  if (i - 1 == (int)sqrt(x))
    return 1;
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<isPrime>:
  endbr64
  mov    $0x0,%eax
  cmp    $0x1,%edi
  jle    L17
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebp
  mov    $0x2,%ebx
L4:
  mov    %ebp,%edi
  call   <sqrt@plt>
  cmp    %eax,%ebx
  jg     L23
  mov    %ebp,%eax
  cltd
  idiv   %ebx
  test   %edx,%edx
  je     L23
  add    $0x1,%ebx
  jmp    L4
L23:
  mov    %ebp,%edi
  call   <sqrt@plt>
  sub    $0x1,%ebx
  cmp    %eax,%ebx
  sete   %al
  movzbl %al,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L17:
  ret
```

assistant: ```c
int isPrime(int x) {
  if (x < 2)
    return 0;
  int i;
  for (i = 2; i <= (int)sqrt(x); i++)
    if (x % i == 0)
      break;
  if (i - 1 == (int)sqrt(x))
    return 1;
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<isPrime>:
  endbr64
  cmp    $0x1,%edi
  jle    L16
  push   %r12
  mov    %edi,%r12d
  push   %rbx
  mov    $0x2,%ebx
  sub    $0x8,%rsp
  jmp    L27
  nopw   0x0(%rax,%rax,1)
L29:
  mov    %r12d,%eax
  cltd
  idiv   %ebx
  test   %edx,%edx
  je     L14
  add    $0x1,%ebx
L27:
  mov    %r12d,%edi
  call   <sqrt@plt>
  cmp    %eax,%ebx
  jle    L29
L14:
  mov    %r12d,%edi
  sub    $0x1,%ebx
  call   <sqrt@plt>
  cmp    %eax,%ebx
  sete   %al
  add    $0x8,%rsp
  movzbl %al,%eax
  pop    %rbx
  pop    %r12
  ret
  nopl   0x0(%rax)
L16:
  xor    %eax,%eax
  ret
```

assistant: ```c
int isPrime(int x) {
  if (x < 2)
    return 0;
  int i;
  for (i = 2; i <= (int)sqrt(x); i++)
    if (x % i == 0)
      break;
  if (i - 1 == (int)sqrt(x))
    return 1;
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<isPrime>:
  endbr64
  cmp    $0x1,%edi
  jle    L21
  push   %r12
  mov    %edi,%r12d
  push   %rbx
  mov    $0x2,%ebx
  sub    $0x8,%rsp
  jmp    L20
  nopw   0x0(%rax,%rax,1)
L10:
  mov    %r12d,%eax
  cltd
  idiv   %ebx
  test   %edx,%edx
  je     L19
  add    $0x1,%ebx
L20:
  mov    %r12d,%edi
  call   <sqrt@plt>
  cmp    %eax,%ebx
  jle    L10
L19:
  mov    %r12d,%edi
  sub    $0x1,%ebx
  call   <sqrt@plt>
  cmp    %eax,%ebx
  sete   %al
  add    $0x8,%rsp
  movzbl %al,%eax
  pop    %rbx
  pop    %r12
  ret
  nopl   0x0(%rax)
L21:
  xor    %eax,%eax
  ret
```

assistant: ```c
int isPrime(int x) {
  if (x < 2)
    return 0;
  int i;
  for (i = 2; i <= (int)sqrt(x); i++)
    if (x % i == 0)
      break;
  if (i - 1 == (int)sqrt(x))
    return 1;
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main1>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x440,%rsp
  mov    %edi,-0x424(%rbp)
  mov    %rsi,-0x430(%rbp)
  mov    %rdx,-0x438(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x414(%rbp)
  jmp    L13
L5:
  mov    -0x414(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x430(%rbp),%rax
  add    %rdx,%rax
  movss  (%rax),%xmm1
  mov    -0x414(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x438(%rbp),%rax
  add    %rdx,%rax
  movss  (%rax),%xmm0
  mulss  %xmm1,%xmm0
  mov    -0x414(%rbp),%eax
  cltq
  movss  %xmm0,-0x410(%rbp,%rax,4)
  addl   $0x1,-0x414(%rbp)
L13:
  mov    -0x414(%rbp),%eax
  cmp    -0x424(%rbp),%eax
  jl     L5
  mov    -0x438(%rbp),%rdx
  mov    -0x430(%rbp),%rcx
  lea    -0x410(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <bar@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L26
  call   <__stack_chk_fail@plt>
L26:
  leave
  ret
```

assistant: ```c
__attribute__((no)) int main1(int n, float *pb, float *pc) {
  float pa[256] __attribute__((__aligned__(16)));
  int i;
  for (i = 0; i < n; i++) {
    pa[i] = pb[i] * pc[i];
  }
  bar(pa, pb, pc);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main1>:
  endbr64
  sub    $0x418,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x408(%rsp)
  xor    %eax,%eax
  test   %edi,%edi
  jle    L5
  mov    %edi,%edi
L19:
  movss  (%rsi,%rax,4),%xmm0
  mulss  (%rdx,%rax,4),%xmm0
  movss  %xmm0,(%rsp,%rax,4)
  add    $0x1,%rax
  cmp    %rdi,%rax
  jne    L19
L5:
  mov    %rsp,%rdi
  call   <bar@plt>
  mov    0x408(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  mov    $0x0,%eax
  add    $0x418,%rsp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
__attribute__((no)) int main1(int n, float *pb, float *pc) {
  float pa[256] __attribute__((__aligned__(16)));
  int i;
  for (i = 0; i < n; i++) {
    pa[i] = pb[i] * pc[i];
  }
  bar(pa, pb, pc);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main1>:
  endbr64
  sub    $0x418,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x408(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r8
  test   %edi,%edi
  jle    L29
  movslq %edi,%rdi
  xor    %eax,%eax
  mov    %rsp,%r8
  nopl   (%rax)
L11:
  movss  (%rsi,%rax,4),%xmm0
  mulss  (%rdx,%rax,4),%xmm0
  movss  %xmm0,(%r8,%rax,4)
  add    $0x1,%rax
  cmp    %rax,%rdi
  jne    L11
L29:
  mov    %r8,%rdi
  call   <bar@plt>
  mov    0x408(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L12
  xor    %eax,%eax
  add    $0x418,%rsp
  ret
L12:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
__attribute__((no)) int main1(int n, float *pb, float *pc) {
  float pa[256] __attribute__((__aligned__(16)));
  int i;
  for (i = 0; i < n; i++) {
    pa[i] = pb[i] * pc[i];
  }
  bar(pa, pb, pc);
  return 0;
}
```