ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What would the C source look like for the assembly code provided here:

```asm
<RSA2048_load_pubkey_E>:
  endbr64
  sub    $0x8,%rsp
  mov    %rsi,%rdx
  mov    $0x4,%ecx
  lea    D6(%rip),%rsi
  call   <RSA2048_load@plt>
  test   %eax,%eax
  setne  %al
  add    $0x8,%rsp
  movzbl %al,%eax
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D6
datatype: string

tool: D6: BF51000004

assistant: ```c
int RSA2048_load_pubkey_E(int fd, unsigned char *RSA2048_E) {
  int ret;
  ret = RSA2048_load(fd, "BF51000004", RSA2048_E, 4);
  if (ret != 0)
    return 1;
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<RSA2048_load_pubkey_E>:
  endbr64
  sub    $0x8,%rsp
  mov    %rsi,%rdx
  mov    $0x4,%ecx
  lea    D0(%rip),%rsi
  call   <RSA2048_load@plt>
  test   %eax,%eax
  setne  %al
  add    $0x8,%rsp
  movzbl %al,%eax
  ret
```

tool_calls: name: read_data
arguments: data_tag: D0
type_guessed: string

tool: D0: BF51000004

assistant: ```c
int RSA2048_load_pubkey_E(int fd, unsigned char *RSA2048_E) {
  int ret;
  ret = RSA2048_load(fd, "BF51000004", RSA2048_E, 4);
  if (ret != 0)
    return 1;
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<file_error>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <perror@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  nop
  leave
  ret
```

assistant: ```c
void file_error(const char *call_name) {
  perror(call_name);
  exit(1);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<file_error>:
  endbr64
  sub    $0x8,%rsp
  call   <perror@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void file_error(const char *call_name) {
  perror(call_name);
  exit(1);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<file_error>:
  endbr64
  sub    $0x8,%rsp
  call   <perror@plt>
  mov    $0x1,%edi
  add    $0x8,%rsp
  jmp    1070 <exit@plt>
```

assistant: ```c
void file_error(const char *call_name) {
  perror(call_name);
  exit(1);
}
```
user: What is the c source code of the assembly code below?

```asm
<file_error>:
  endbr64
  sub    $0x8,%rsp
  call   <perror@plt>
  mov    $0x1,%edi
  add    $0x8,%rsp
  jmp    1070 <exit@plt>
```

assistant: ```c
void file_error(const char *call_name) {
  perror(call_name);
  exit(1);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<foo>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <trace@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D0
data_type: string

tool: D0: libfoo

assistant: ```c
int foo(void) {
  trace("libfoo");
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<foo>:
  endbr64
  sub    $0x8,%rsp
  lea    D1(%rip),%rdi
  call   <trace@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: label: D1
type_infered: string

tool: D1: libfoo

assistant: ```c
int foo(void) {
  trace("libfoo");
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<foo>:
  endbr64
  sub    $0x8,%rsp
  lea    D7(%rip),%rdi
  call   <trace@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D7
type: string

tool: D7: libfoo

assistant: ```c
int foo(void) {
  trace("libfoo");
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<foo>:
  endbr64
  sub    $0x8,%rsp
  lea    D8(%rip),%rdi
  call   <trace@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: data_label: D8
type: string

tool: D8: libfoo

assistant: ```c
int foo(void) {
  trace("libfoo");
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0xc(%rbp)
  lea    D24(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x10(%rbp),%rdx
  lea    -0x14(%rbp),%rax
  mov    %rax,%rsi
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x14(%rbp),%edx
  mov    -0x10(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L6
  call   <__stack_chk_fail@plt>
L6:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D24
type_guessed: string
name: retrieve_data
arguments: data_tag: D3
type_guessed: string
name: retrieve_data
arguments: data_tag: D4
type_guessed: string

tool: D24: Entre com dois numeros: 

tool: D3: %d %d

tool: D4: Resultado: %d

assistant: ```c
int main() {
  int x, y, soma = 0;
  printf("Entre com dois numeros: ");
  scanf("%d %d", &x, &y);
  soma = x + y;
  printf("Resultado: %d", soma);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D20(%rip),%rdi
  call   <printf@plt>
  lea    0x4(%rsp),%rdx
  mov    %rsp,%rsi
  lea    D0(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%esi
  add    (%rsp),%esi
  lea    D25(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  mov    $0x0,%eax
  add    $0x18,%rsp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D20
type: string
name: parse_data
arguments: label: D0
type: string
name: parse_data
arguments: label: D25
type: string

tool: D20: Entre com dois numeros: 

tool: D0: %d %d

tool: D25: Resultado: %d

assistant: ```c
int main() {
  int x, y, soma = 0;
  printf("Entre com dois numeros: ");
  scanf("%d %d", &x, &y);
  soma = x + y;
  printf("Resultado: %d", soma);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D20(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rsp,%rsi
  lea    0x4(%rsp),%rdx
  lea    D24(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%esi
  xor    %eax,%eax
  add    (%rsp),%esi
  lea    D25(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D20
type: string
name: read_data
arguments: data_label: D24
type: string
name: read_data
arguments: data_label: D25
type: string

tool: D20: Entre com dois numeros: 

tool: D24: %d %d

tool: D25: Resultado: %d

assistant: ```c
int main() {
  int x, y, soma = 0;
  printf("Entre com dois numeros: ");
  scanf("%d %d", &x, &y);
  soma = x + y;
  printf("Resultado: %d", soma);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D27(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rsp,%rsi
  lea    0x4(%rsp),%rdx
  lea    D18(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%esi
  xor    %eax,%eax
  add    (%rsp),%esi
  lea    D4(%rip),%rdi
  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: parse_data
arguments: data_label: D27
type_infered: string
name: parse_data
arguments: data_label: D18
type_infered: string
name: parse_data
arguments: data_label: D4
type_infered: string

tool: D27: Entre com dois numeros: 

tool: D18: %d %d

tool: D4: Resultado: %d

assistant: ```c
int main() {
  int x, y, soma = 0;
  printf("Entre com dois numeros: ");
  scanf("%d %d", &x, &y);
  soma = x + y;
  printf("Resultado: %d", soma);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<chtostr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,%eax
  mov    %rsi,-0x10(%rbp)
  mov    %al,-0x4(%rbp)
  mov    -0x10(%rbp),%rax
  movzbl -0x4(%rbp),%edx
  mov    %dl,(%rax)
  mov    -0x10(%rbp),%rax
  add    $0x1,%rax
  movb   $0x0,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void chtostr(char ch, char *str) {
  str[0] = ch;
  str[1] = '\0';
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<chtostr>:
  endbr64
  mov    %dil,(%rsi)
  movb   $0x0,0x1(%rsi)
  ret
```

assistant: ```c
void chtostr(char ch, char *str) {
  str[0] = ch;
  str[1] = '\0';
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<chtostr>:
  endbr64
  mov    %dil,(%rsi)
  movb   $0x0,0x1(%rsi)
  ret
```

assistant: ```c
void chtostr(char ch, char *str) {
  str[0] = ch;
  str[1] = '\0';
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<chtostr>:
  endbr64
  mov    %dil,(%rsi)
  movb   $0x0,0x1(%rsi)
  ret
```

assistant: ```c
void chtostr(char ch, char *str) {
  str[0] = ch;
  str[1] = '\0';
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<file_base>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %rax,-0x8(%rbp)
  jmp    L19
L10:
  subq   $0x1,-0x8(%rbp)
L19:
  cmpq   $0x0,-0x8(%rbp)
  je     L12
  mov    -0x8(%rbp),%rax
  lea    -0x1(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x2f,%al
  jne    L10
L12:
  mov    -0x18(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  leave
  ret
```

assistant: ```c
const char *file_base(const char *filename) {
  size_t i = strlen(filename);
  while (i && filename[i - 1] != '/')
    --i;
  return &filename[i];
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<file_base>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  jmp    L6
L1:
  mov    %rdx,%rax
L6:
  test   %rax,%rax
  je     L4
  lea    -0x1(%rax),%rdx
  cmpb   $0x2f,-0x1(%rbx,%rax,1)
  jne    L1
L4:
  add    %rbx,%rax
  pop    %rbx
  ret
```

assistant: ```c
const char *file_base(const char *filename) {
  size_t i = strlen(filename);
  while (i && filename[i - 1] != '/')
    --i;
  return &filename[i];
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<file_base>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  call   <strlen@plt>
  jmp    L9
L21:
  cmpb   $0x2f,-0x1(%r12,%rax,1)
  lea    -0x1(%rax),%rdx
  je     L11
  mov    %rdx,%rax
L9:
  test   %rax,%rax
  jne    L21
  mov    %r12,%rax
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L11:
  add    %rax,%r12
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
const char *file_base(const char *filename) {
  size_t i = strlen(filename);
  while (i && filename[i - 1] != '/')
    --i;
  return &filename[i];
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<file_base>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  call   <strlen@plt>
  jmp    L3
L26:
  cmpb   $0x2f,-0x1(%r12,%rax,1)
  lea    -0x1(%rax),%rdx
  je     L15
  mov    %rdx,%rax
L3:
  test   %rax,%rax
  jne    L26
  mov    %r12,%rax
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L15:
  add    %rax,%r12
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
const char *file_base(const char *filename) {
  size_t i = strlen(filename);
  while (i && filename[i - 1] != '/')
    --i;
  return &filename[i];
}
```
user: Translate the below assembly code into C programming language:

```asm
<benchmark>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    $0x0,%eax
  call   <tic@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x18(%rbp)
  movl   $0x0,-0x24(%rbp)
  jmp    L35
L5:
  mov    $0x0,%eax
  call   <rand_double@plt>
  movsd  D31(%rip),%xmm1
  mulsd  %xmm1,%xmm0
  movsd  D12(%rip),%xmm1
  subsd  %xmm1,%xmm0
  movsd  %xmm0,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  movq   %rax,%xmm0
  call   <stdlib_base_inv@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x20(%rbp)
  movsd  -0x20(%rbp),%xmm0
  ucomisd -0x20(%rbp),%xmm0
  jnp    L24
  lea    D29(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L36
L24:
  addl   $0x1,-0x24(%rbp)
L35:
  cmpl   $0xf423f,-0x24(%rbp)
  jle    L5
L36:
  mov    $0x0,%eax
  call   <tic@plt>
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  subsd  -0x18(%rbp),%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movsd  -0x20(%rbp),%xmm0
  ucomisd -0x20(%rbp),%xmm0
  jnp    L39
  lea    D29(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L39:
  movsd  -0x8(%rbp),%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D31
datatype: f64
name: retrieve_data
arguments: tag: D12
datatype: f64
name: retrieve_data
arguments: tag: D29
datatype: string

tool: D31: 1000

tool: D12: 500

tool: D29: should not return NaN


assistant: ```c
double benchmark() {
  double elapsed;
  double x;
  double y;
  double t;
  int i;
  t = tic();
  for (i = 0; i < 1000000; i++) {
    x = (1000.0 * rand_double()) - 500.0;
    y = stdlib_base_inv(x);
    if (y != y) {
      printf("should not return NaN\n");
      break;
    }
  }
  elapsed = tic() - t;
  if (y != y) {
    printf("should not return NaN\n");
  }
  return elapsed;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<benchmark>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    $0x0,%eax
  call   <tic@plt>
  movq   %xmm0,%r12
  mov    $0xf4240,%ebx
L8:
  mov    $0x0,%eax
  call   <rand_double@plt>
  mulsd  D13(%rip),%xmm0
  subsd  D12(%rip),%xmm0
  call   <stdlib_base_inv@plt>
  ucomisd %xmm0,%xmm0
  jp     L20
  sub    $0x1,%ebx
  jne    L8
  mov    $0x0,%eax
  call   <tic@plt>
  movq   %r12,%xmm2
  subsd  %xmm2,%xmm0
  movq   %xmm0,%rbx
  jmp    L21
L20:
  lea    D24(%rip),%rbp
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  call   <tic@plt>
  movq   %r12,%xmm1
  subsd  %xmm1,%xmm0
  movq   %xmm0,%rbx
  mov    %rbp,%rdi
  call   <printf@plt>
L21:
  movq   %rbx,%xmm0
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: unpack_data
arguments: label: D13
type_guessed: f64
name: unpack_data
arguments: label: D12
type_guessed: f64
name: unpack_data
arguments: label: D24
type_guessed: string

tool: D13: 1000

tool: D12: 500

tool: D24: should not return NaN


assistant: ```c
double benchmark() {
  double elapsed;
  double x;
  double y;
  double t;
  int i;
  t = tic();
  for (i = 0; i < 1000000; i++) {
    x = (1000.0 * rand_double()) - 500.0;
    y = stdlib_base_inv(x);
    if (y != y) {
      printf("should not return NaN\n");
      break;
    }
  }
  elapsed = tic() - t;
  if (y != y) {
    printf("should not return NaN\n");
  }
  return elapsed;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<benchmark>:
  endbr64
  push   %rbp
  xor    %eax,%eax
  push   %rbx
  mov    $0xf4240,%ebx
  sub    $0x18,%rsp
  call   <tic@plt>
  movsd  %xmm0,0x8(%rsp)
  jmp    L4
  xchg   %ax,%ax
L1:
  sub    $0x1,%ebx
  je     L18
L4:
  xor    %eax,%eax
  call   <rand_double@plt>
  mulsd  D0(%rip),%xmm0
  subsd  D19(%rip),%xmm0
  call   <stdlib_base_inv@plt>
  ucomisd %xmm0,%xmm0
  jnp    L1
  lea    D12(%rip),%rbp
  mov    %rbp,%rdi
  call   <printf@plt>
  xor    %eax,%eax
  call   <tic@plt>
  subsd  0x8(%rsp),%xmm0
  mov    %rbp,%rdi
  movsd  %xmm0,0x8(%rsp)
  call   <printf@plt>
  movsd  0x8(%rsp),%xmm0
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
  xchg   %ax,%ax
L18:
  xor    %eax,%eax
  call   <tic@plt>
  subsd  0x8(%rsp),%xmm0
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D0
type_infered: f64
name: retrieve_data
arguments: tag: D19
type_infered: f64
name: retrieve_data
arguments: tag: D12
type_infered: string

tool: D0: 1000

tool: D19: 500

tool: D12: should not return NaN


assistant: ```c
double benchmark() {
  double elapsed;
  double x;
  double y;
  double t;
  int i;
  t = tic();
  for (i = 0; i < 1000000; i++) {
    x = (1000.0 * rand_double()) - 500.0;
    y = stdlib_base_inv(x);
    if (y != y) {
      printf("should not return NaN\n");
      break;
    }
  }
  elapsed = tic() - t;
  if (y != y) {
    printf("should not return NaN\n");
  }
  return elapsed;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<benchmark>:
  endbr64
  push   %rbp
  xor    %eax,%eax
  push   %rbx
  mov    $0xf4240,%ebx
  sub    $0x18,%rsp
  call   <tic@plt>
  movsd  %xmm0,0x8(%rsp)
  jmp    L20
  xchg   %ax,%ax
L21:
  sub    $0x1,%ebx
  je     L27
L20:
  xor    %eax,%eax
  call   <rand_double@plt>
  mulsd  D5(%rip),%xmm0
  subsd  D21(%rip),%xmm0
  call   <stdlib_base_inv@plt>
  ucomisd %xmm0,%xmm0
  jnp    L21
  lea    D16(%rip),%rbp
  mov    %rbp,%rdi
  call   <printf@plt>
  xor    %eax,%eax
  call   <tic@plt>
  subsd  0x8(%rsp),%xmm0
  mov    %rbp,%rdi
  movsd  %xmm0,0x8(%rsp)
  call   <printf@plt>
  movsd  0x8(%rsp),%xmm0
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
  xchg   %ax,%ax
L27:
  xor    %eax,%eax
  call   <tic@plt>
  subsd  0x8(%rsp),%xmm0
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D5
datatype: f64
name: extract_symbolic_data
arguments: data_label: D21
datatype: f64
name: extract_symbolic_data
arguments: data_label: D16
datatype: string

tool: D5: 1000

tool: D21: 500

tool: D16: should not return NaN


assistant: ```c
double benchmark() {
  double elapsed;
  double x;
  double y;
  double t;
  int i;
  t = tic();
  for (i = 0; i < 1000000; i++) {
    x = (1000.0 * rand_double()) - 500.0;
    y = stdlib_base_inv(x);
    if (y != y) {
      printf("should not return NaN\n");
      break;
    }
  }
  elapsed = tic() - t;
  if (y != y) {
    printf("should not return NaN\n");
  }
  return elapsed;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<isprime>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movl   $0x1,-0xc(%rbp)
  movq   $0x2,-0x8(%rbp)
  jmp    L27
L20:
  mov    -0x18(%rbp),%rax
  cqto
  idivq  -0x8(%rbp)
  mov    %rdx,%rax
  test   %rax,%rax
  jne    L11
  movl   $0x0,-0xc(%rbp)
L11:
  addq   $0x2,-0x8(%rbp)
L27:
  mov    -0x8(%rbp),%rax
  imul   %rax,%rax
  cmp    %rax,-0x18(%rbp)
  jge    L20
  mov    -0xc(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int isprime(long int n) {
  long int i, j;
  int r = 1;
  for (i = 2; i * i <= n; i += 2)
    if (n % i == 0)
      r = 0;
  return r;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<isprime>:
  endbr64
  cmp    $0x3,%rdi
  jle    L5
  mov    $0x1,%esi
  mov    $0x2,%ecx
  mov    $0x0,%r8d
L1:
  mov    %rdi,%rax
  cqto
  idiv   %rcx
  test   %rdx,%rdx
  cmove  %r8d,%esi
  add    $0x2,%rcx
  mov    %rcx,%rax
  imul   %rcx,%rax
  cmp    %rdi,%rax
  jle    L1
L27:
  mov    %esi,%eax
  ret
L5:
  mov    $0x1,%esi
  jmp    L27
```

assistant: ```c
int isprime(long int n) {
  long int i, j;
  int r = 1;
  for (i = 2; i * i <= n; i += 2)
    if (n % i == 0)
      r = 0;
  return r;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<isprime>:
  endbr64
  mov    $0x1,%r8d
  cmp    $0x3,%rdi
  jle    L5
  mov    $0x2,%ecx
  xor    %esi,%esi
  nopw   0x0(%rax,%rax,1)
L1:
  mov    %rdi,%rax
  cqto
  idiv   %rcx
  test   %rdx,%rdx
  cmove  %esi,%r8d
  add    $0x2,%rcx
  mov    %rcx,%rax
  imul   %rcx,%rax
  cmp    %rdi,%rax
  jle    L1
L5:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int isprime(long int n) {
  long int i, j;
  int r = 1;
  for (i = 2; i * i <= n; i += 2)
    if (n % i == 0)
      r = 0;
  return r;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<isprime>:
  endbr64
  mov    $0x1,%r8d
  cmp    $0x3,%rdi
  jle    L8
  mov    $0x2,%ecx
  xor    %esi,%esi
  nopw   0x0(%rax,%rax,1)
L9:
  mov    %rdi,%rax
  cqto
  idiv   %rcx
  test   %rdx,%rdx
  cmove  %esi,%r8d
  add    $0x2,%rcx
  mov    %rcx,%rax
  imul   %rcx,%rax
  cmp    %rax,%rdi
  jge    L9
L8:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int isprime(long int n) {
  long int i, j;
  int r = 1;
  for (i = 2; i * i <= n; i += 2)
    if (n % i == 0)
      r = 0;
  return r;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<poly1305aes_test_authenticate>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    %rcx,-0x40(%rbp)
  mov    %r8d,-0x44(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x38(%rbp),%rdx
  mov    -0x30(%rbp),%rcx
  lea    -0x20(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <aes_openssl@plt>
  mov    -0x30(%rbp),%rax
  lea    0x10(%rax),%rsi
  mov    -0x44(%rbp),%edi
  mov    -0x40(%rbp),%rcx
  lea    -0x20(%rbp),%rdx
  mov    -0x28(%rbp),%rax
  mov    %edi,%r8d
  mov    %rax,%rdi
  call   <poly1305_gmp@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L0
  call   <__stack_chk_fail@plt>
L0:
  leave
  ret
```

assistant: ```c
void poly1305aes_test_authenticate(unsigned char out[16], const unsigned char kr[32], const unsigned char n[16], const unsigned char m[], unsigned int l) {
  unsigned char aeskn[16];
  aes_openssl(aeskn, (kr + 0), n);
  poly1305_gmp(out, (kr + 16), aeskn, m, l);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<poly1305aes_test_authenticate>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %rdi,%rbp
  mov    %rsi,%rbx
  mov    %rcx,%r12
  mov    %r8d,%r13d
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r14
  mov    %r14,%rdi
  call   <aes_openssl@plt>
  lea    0x10(%rbx),%rsi
  mov    %r13d,%r8d
  mov    %r12,%rcx
  mov    %r14,%rdx
  mov    %rbp,%rdi
  call   <poly1305_gmp@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void poly1305aes_test_authenticate(unsigned char out[16], const unsigned char kr[32], const unsigned char n[16], const unsigned char m[], unsigned int l) {
  unsigned char aeskn[16];
  aes_openssl(aeskn, (kr + 0), n);
  poly1305_gmp(out, (kr + 16), aeskn, m, l);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<poly1305aes_test_authenticate>:
  endbr64
  push   %r14
  push   %r13
  mov    %r8d,%r13d
  push   %r12
  mov    %rcx,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r14
  mov    %r14,%rdi
  call   <aes_openssl@plt>
  lea    0x10(%rbx),%rsi
  mov    %r13d,%r8d
  mov    %r12,%rcx
  mov    %r14,%rdx
  mov    %rbp,%rdi
  call   <poly1305_gmp@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L6:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void poly1305aes_test_authenticate(unsigned char out[16], const unsigned char kr[32], const unsigned char n[16], const unsigned char m[], unsigned int l) {
  unsigned char aeskn[16];
  aes_openssl(aeskn, (kr + 0), n);
  poly1305_gmp(out, (kr + 16), aeskn, m, l);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<poly1305aes_test_authenticate>:
  endbr64
  push   %r14
  push   %r13
  mov    %r8d,%r13d
  push   %r12
  mov    %rcx,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r14
  mov    %r14,%rdi
  call   <aes_openssl@plt>
  lea    0x10(%rbx),%rsi
  mov    %r13d,%r8d
  mov    %r12,%rcx
  mov    %r14,%rdx
  mov    %rbp,%rdi
  call   <poly1305_gmp@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L6:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void poly1305aes_test_authenticate(unsigned char out[16], const unsigned char kr[32], const unsigned char n[16], const unsigned char m[], unsigned int l) {
  unsigned char aeskn[16];
  aes_openssl(aeskn, (kr + 0), n);
  poly1305_gmp(out, (kr + 16), aeskn, m, l);
}
```