ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movl   $0x2,-0x4(%rbp)
  mov    -0x4(%rbp),%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  lea    0x14(%rax),%edx
  mov    -0x4(%rbp),%eax
  sar    $0x2,%eax
  add    %edx,%eax
  sub    $0x7b,%eax
  mov    %eax,-0xc(%rbp)
  cmpl   $0x9,-0xc(%rbp)
  jg     L79
  jmp    L6
L11:
  mov    -0x4(%rbp),%eax
  shl    $0x2,%eax
  add    %eax,-0xc(%rbp)
L6:
  cmpl   $0x13,-0xc(%rbp)
  jle    L11
L79:
  cmpl   $0x14,-0xc(%rbp)
  jle    L20
L61:
  mov    -0x4(%rbp),%eax
  add    %eax,-0xc(%rbp)
  cmpl   $0x63,-0xc(%rbp)
  jle    L61
  jmp    L53
L20:
  movl   $0x0,-0x8(%rbp)
  jmp    L54
L38:
  addl   $0x1,-0x8(%rbp)
L54:
  cmpl   $0x9,-0x8(%rbp)
  jle    L38
L53:
  mov    -0xc(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int main() {
  int d, i, c;
  c = 2;
  d = 20 + c * 10 + (c >> 2) - 123;
  if (d < 10) {
    while (d < 20) {
      d = d + c * 4;
    }
  }
  if (d > 20) {
    do {
      d += c;
    } while (d < 100);
  } else {
    for (i = 0; i < 10; i++) {
    }
  }
  return d;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  mov    $0x28,%eax
L9:
  sub    $0x1,%eax
  jne    L9
  mov    $0x65,%eax
  ret
```

assistant: ```c
int main() {
  int d, i, c;
  c = 2;
  d = 20 + c * 10 + (c >> 2) - 123;
  if (d < 10) {
    while (d < 20) {
      d = d + c * 4;
    }
  }
  if (d > 20) {
    do {
      d += c;
    } while (d < 100);
  } else {
    for (i = 0; i < 10; i++) {
    }
  }
  return d;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  mov    $0x65,%eax
  ret
```

assistant: ```c
int main() {
  int d, i, c;
  c = 2;
  d = 20 + c * 10 + (c >> 2) - 123;
  if (d < 10) {
    while (d < 20) {
      d = d + c * 4;
    }
  }
  if (d > 20) {
    do {
      d += c;
    } while (d < 100);
  } else {
    for (i = 0; i < 10; i++) {
    }
  }
  return d;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  mov    $0x65,%eax
  ret
```

assistant: ```c
int main() {
  int d, i, c;
  c = 2;
  d = 20 + c * 10 + (c >> 2) - 123;
  if (d < 10) {
    while (d < 20) {
      d = d + c * 4;
    }
  }
  if (d > 20) {
    do {
      d += c;
    } while (d < 100);
  } else {
    for (i = 0; i < 10; i++) {
    }
  }
  return d;
}
```
user: Translate the below assembly code into C programming language:

```asm
<sizefix>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movsd  %xmm0,-0x8(%rbp)
  mov    %edi,-0xc(%rbp)
  cmpl   $0x2,-0xc(%rbp)
  jne    L12
  movsd  -0x8(%rbp),%xmm1
  movsd  D6(%rip),%xmm0
  mulsd  %xmm1,%xmm0
  jmp    L2
L12:
  movsd  -0x8(%rbp),%xmm0
L2:
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  pop    %rbp
  ret
```

tool_calls: name: unpack_data
arguments: label: D6
type: byte[8]

tool: {"D6": [0x00,0x00,0x00,0x00,0x00,0x00,0x20,0x40]}

assistant: ```c
double sizefix(double value, int type) {
  if (type == 2)
    return value * 8;
  return value;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<sizefix>:
  endbr64
  cmp    $0x2,%edi
  je     L0
L10:
  ret
L0:
  mulsd  D5(%rip),%xmm0
  jmp    L10
```

tool_calls: name: retrieve_data
arguments: data_label: D5
type: byte[8]

tool: {"D5": [0x00,0x00,0x00,0x00,0x00,0x00,0x20,0x40]}

assistant: ```c
double sizefix(double value, int type) {
  if (type == 2)
    return value * 8;
  return value;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<sizefix>:
  endbr64
  cmp    $0x2,%edi
  je     L9
  ret
  nopw   0x0(%rax,%rax,1)
L9:
  mulsd  D5(%rip),%xmm0
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D5
data_type: byte[8]

tool: {"D5": [0x00,0x00,0x00,0x00,0x00,0x00,0x20,0x40]}

assistant: ```c
double sizefix(double value, int type) {
  if (type == 2)
    return value * 8;
  return value;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<sizefix>:
  endbr64
  cmp    $0x2,%edi
  je     L2
  ret
  nopw   0x0(%rax,%rax,1)
L2:
  mulsd  D5(%rip),%xmm0
  ret
```

tool_calls: name: parse_data
arguments: tag: D5
type_guessed: byte[8]

tool: {"D5": [0x00,0x00,0x00,0x00,0x00,0x00,0x20,0x40]}

assistant: ```c
double sizefix(double value, int type) {
  if (type == 2)
    return value * 8;
  return value;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<jrand48_r>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    -0x10(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <__drand48_iterate@plt>
  test   %eax,%eax
  jns    L8
  mov    $0xffffffff,%eax
  jmp    L5
L8:
  mov    -0x8(%rbp),%rax
  add    $0x4,%rax
  movzwl (%rax),%eax
  movzwl %ax,%eax
  shl    $0x10,%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%rax
  add    $0x2,%rax
  movzwl (%rax),%eax
  movzwl %ax,%eax
  or     %edx,%eax
  cltq
  mov    %eax,%edx
  mov    -0x18(%rbp),%rax
  mov    %rdx,(%rax)
  mov    $0x0,%eax
L5:
  leave
  ret
```

assistant: ```c
int jrand48_r(xsubi, buffer, result)
unsigned short int xsubi[3];
struct drand48_data *buffer;
long int *result;
{
  if (__drand48_iterate(xsubi, buffer) < 0)
    return -1;
  *result = ((xsubi[2] << 16) | xsubi[1]) & 0xffffffffl;
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<jrand48_r>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rdx,%rbp
  mov    $0x0,%eax
  call   <__drand48_iterate@plt>
  test   %eax,%eax
  js     L2
  movzwl 0x4(%rbx),%eax
  shl    $0x10,%eax
  movzwl 0x2(%rbx),%edx
  or     %edx,%eax
  mov    %rax,0x0(%rbp)
  mov    $0x0,%eax
L8:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L2:
  mov    $0xffffffff,%eax
  jmp    L8
```

assistant: ```c
int jrand48_r(xsubi, buffer, result)
unsigned short int xsubi[3];
struct drand48_data *buffer;
long int *result;
{
  if (__drand48_iterate(xsubi, buffer) < 0)
    return -1;
  *result = ((xsubi[2] << 16) | xsubi[1]) & 0xffffffffl;
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<jrand48_r>:
  endbr64
  push   %rbp
  xor    %eax,%eax
  mov    %rdx,%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  call   <__drand48_iterate@plt>
  test   %eax,%eax
  js     L18
  mov    0x2(%rbx),%eax
  mov    %rax,0x0(%rbp)
  xor    %eax,%eax
L5:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L18:
  mov    $0xffffffff,%eax
  jmp    L5
```

assistant: ```c
int jrand48_r(xsubi, buffer, result)
unsigned short int xsubi[3];
struct drand48_data *buffer;
long int *result;
{
  if (__drand48_iterate(xsubi, buffer) < 0)
    return -1;
  *result = ((xsubi[2] << 16) | xsubi[1]) & 0xffffffffl;
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<jrand48_r>:
  endbr64
  push   %rbp
  xor    %eax,%eax
  mov    %rdx,%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  call   <__drand48_iterate@plt>
  test   %eax,%eax
  js     L9
  mov    0x2(%rbx),%eax
  mov    %rax,0x0(%rbp)
  xor    %eax,%eax
L3:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L9:
  mov    $0xffffffff,%eax
  jmp    L3
```

assistant: ```c
int jrand48_r(xsubi, buffer, result)
unsigned short int xsubi[3];
struct drand48_data *buffer;
long int *result;
{
  if (__drand48_iterate(xsubi, buffer) < 0)
    return -1;
  *result = ((xsubi[2] << 16) | xsubi[1]) & 0xffffffffl;
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<valid>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  movl   $0x0,-0xc(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L23
L20:
  movl   $0x0,-0x4(%rbp)
  jmp    L2
L22:
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%edx
  mov    -0x4(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x20(%rbp),%rax
  add    %rcx,%rax
  movzbl (%rax),%eax
  cmp    %al,%dl
  jne    L11
  addl   $0x1,-0xc(%rbp)
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movb   $0x9,(%rax)
L11:
  addl   $0x1,-0x4(%rbp)
L2:
  mov    -0x4(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jl     L22
  addl   $0x1,-0x8(%rbp)
L23:
  mov    -0x8(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jl     L20
  mov    -0xc(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int valid(char *s1, char *s2, int n) {
  int flag = 0;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if (*(s1 + i) == *(s2 + j)) {
        flag++;
        *(s2 + j) = 9;
      }
    }
  }
  return flag;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<valid>:
  endbr64
  test   %edx,%edx
  jle    L69
  mov    %rdi,%rcx
  lea    -0x1(%rdx),%eax
  lea    0x1(%rdi,%rax,1),%r8
  mov    $0x0,%edi
  lea    0x1(%rsi,%rax,1),%rdx
  jmp    L35
L21:
  add    $0x1,%rax
  cmp    %rax,%rdx
  je     L4
L11:
  movzbl (%rax),%r9d
  cmp    %r9b,(%rcx)
  jne    L21
  add    $0x1,%edi
  movb   $0x9,(%rax)
  jmp    L21
L4:
  add    $0x1,%rcx
  cmp    %r8,%rcx
  je     L13
L35:
  mov    %rsi,%rax
  jmp    L11
L69:
  mov    $0x0,%edi
L13:
  mov    %edi,%eax
  ret
```

assistant: ```c
int valid(char *s1, char *s2, int n) {
  int flag = 0;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if (*(s1 + i) == *(s2 + j)) {
        flag++;
        *(s2 + j) = 9;
      }
    }
  }
  return flag;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<valid>:
  endbr64
  test   %edx,%edx
  jle    L22
  lea    -0x1(%rdx),%eax
  xor    %r8d,%r8d
  lea    0x1(%rdi,%rax,1),%rcx
  lea    0x1(%rsi,%rax,1),%rdx
  nopl   0x0(%rax,%rax,1)
L15:
  mov    %rsi,%rax
  nopl   0x0(%rax,%rax,1)
L11:
  movzbl (%rax),%r9d
  cmp    %r9b,(%rdi)
  jne    L0
  movb   $0x9,(%rax)
  add    $0x1,%r8d
L0:
  add    $0x1,%rax
  cmp    %rax,%rdx
  jne    L11
  add    $0x1,%rdi
  cmp    %rcx,%rdi
  jne    L15
  mov    %r8d,%eax
  ret
L22:
  xor    %r8d,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
int valid(char *s1, char *s2, int n) {
  int flag = 0;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if (*(s1 + i) == *(s2 + j)) {
        flag++;
        *(s2 + j) = 9;
      }
    }
  }
  return flag;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<valid>:
  endbr64
  test   %edx,%edx
  jle    L6
  lea    -0x1(%rdx),%eax
  xor    %r8d,%r8d
  lea    0x1(%rdi,%rax,1),%rcx
  lea    0x1(%rsi,%rax,1),%rdx
  nopl   0x0(%rax,%rax,1)
L28:
  mov    %rsi,%rax
  nopl   0x0(%rax,%rax,1)
L24:
  movzbl (%rax),%r9d
  cmp    %r9b,(%rdi)
  jne    L21
  movb   $0x9,(%rax)
  add    $0x1,%r8d
L21:
  add    $0x1,%rax
  cmp    %rax,%rdx
  jne    L24
  add    $0x1,%rdi
  cmp    %rcx,%rdi
  jne    L28
  mov    %r8d,%eax
  ret
L6:
  xor    %r8d,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
int valid(char *s1, char *s2, int n) {
  int flag = 0;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if (*(s1 + i) == *(s2 + j)) {
        flag++;
        *(s2 + j) = 9;
      }
    }
  }
  return flag;
}
```
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   <cipher_correct_test@plt>
  mov    $0x0,%eax
  call   <cipher_speed_test@plt>
  nop
  pop    %rbp
  ret
```

assistant: ```c
void main(void) {
  cipher_correct_test();
  cipher_speed_test();
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <cipher_correct_test@plt>
  mov    $0x0,%eax
  call   <cipher_speed_test@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void main(void) {
  cipher_correct_test();
  cipher_speed_test();
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <cipher_correct_test@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1070 <cipher_speed_test@plt>
```

assistant: ```c
void main(void) {
  cipher_correct_test();
  cipher_speed_test();
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <cipher_correct_test@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1070 <cipher_speed_test@plt>
```

assistant: ```c
void main(void) {
  cipher_correct_test();
  cipher_speed_test();
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<is_prime>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x8(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cltq
  lea    -0x1(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0xa,%al
  jne    L0
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  sub    $0x1,%eax
  jmp    L87
L0:
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
L87:
  mov    %eax,-0x4(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L144
L169:
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x2f,%al
  jle    L89
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x39,%al
  jle    L56
L89:
  mov    $0xffffffff,%eax
  jmp    L57
L56:
  addl   $0x1,-0xc(%rbp)
L144:
  mov    -0xc(%rbp),%eax
  cmp    -0x4(%rbp),%eax
  jl     L169
  movl   $0x0,-0xc(%rbp)
  jmp    L90
L35:
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  sub    $0x30,%eax
  add    %eax,-0x8(%rbp)
  mov    -0x4(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0xc(%rbp)
  je     L147
  mov    -0x8(%rbp),%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  mov    %eax,-0x8(%rbp)
L147:
  addl   $0x1,-0xc(%rbp)
L90:
  mov    -0xc(%rbp),%eax
  cmp    -0x4(%rbp),%eax
  jl     L35
  cmpl   $0x0,-0x8(%rbp)
  je     L127
  cmpl   $0x1,-0x8(%rbp)
  jne    L97
L127:
  mov    $0x0,%eax
  jmp    L57
L97:
  movl   $0x2,-0xc(%rbp)
  jmp    L123
L157:
  mov    -0x8(%rbp),%eax
  cltd
  idivl  -0xc(%rbp)
  mov    %edx,%eax
  test   %eax,%eax
  jne    L13
  mov    -0xc(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  je     L13
  mov    $0x0,%eax
  jmp    L57
L13:
  addl   $0x1,-0xc(%rbp)
L123:
  mov    -0xc(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jl     L157
  mov    $0x1,%eax
L57:
  leave
  ret
```

assistant: ```c
int is_prime(char *input) {
  int i, length, number = 0;
  length = input[strlen(input) - 1] == '\n' ? strlen(input) - 1 : strlen(input);
  for (i = 0; i < length; i++) {
    if (input[i] < '0' || input[i] > '9') {
      return -1;
    }
  }
  for (i = 0; i < length; i++) {
    number += input[i] - '0';
    if (i != length - 1) {
      number *= 10;
    }
  }
  if (number == 0 || number == 1) {
    return 0;
  }
  for (i = 2; i < number; i++) {
    if (number % i == 0 && i != number) {
      return 0;
    }
  }
  return 1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<is_prime>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  cltq
  cmpb   $0xa,-0x1(%rbx,%rax,1)
  je     L118
  mov    %rbx,%rdi
  call   <strlen@plt>
L158:
  test   %eax,%eax
  jle    L188
  mov    %rbx,%rcx
  mov    %eax,%edi
  lea    -0x1(%rax),%edx
  lea    0x1(%rbx,%rdx,1),%rsi
L77:
  movzbl (%rcx),%edx
  sub    $0x30,%edx
  cmp    $0x9,%dl
  ja     L133
  add    $0x1,%rcx
  cmp    %rsi,%rcx
  jne    L77
  mov    %edi,%edi
  mov    $0x0,%edx
  mov    $0x0,%ecx
  sub    $0x1,%eax
  jmp    L101
L118:
  mov    %rbx,%rdi
  call   <strlen@plt>
  sub    $0x1,%eax
  jmp    L158
L9:
  add    $0x1,%rdx
  cmp    %rdi,%rdx
  je     L159
L101:
  movsbl (%rbx,%rdx,1),%esi
  lea    -0x30(%rcx,%rsi,1),%ecx
  cmp    %edx,%eax
  je     L9
  lea    (%rcx,%rcx,4),%ecx
  add    %ecx,%ecx
  jmp    L9
L159:
  mov    %ecx,%eax
  cmp    $0x1,%ecx
  jbe    L10
  cmp    $0x2,%ecx
  jle    L165
  test   $0x1,%al
  je     L160
  mov    $0x2,%esi
L60:
  add    $0x1,%esi
  cmp    %esi,%ecx
  je     L16
  mov    %ecx,%eax
  cltd
  idiv   %esi
  mov    %edx,%eax
  test   %edx,%edx
  jne    L60
  jmp    L124
L16:
  mov    $0x1,%eax
  jmp    L124
L188:
  mov    $0x0,%eax
  jmp    L124
L133:
  mov    $0xffffffff,%eax
L124:
  pop    %rbx
  ret
L10:
  mov    $0x0,%eax
  jmp    L124
L165:
  mov    $0x1,%eax
  jmp    L124
L160:
  mov    $0x0,%eax
  jmp    L124
```

assistant: ```c
int is_prime(char *input) {
  int i, length, number = 0;
  length = input[strlen(input) - 1] == '\n' ? strlen(input) - 1 : strlen(input);
  for (i = 0; i < length; i++) {
    if (input[i] < '0' || input[i] > '9') {
      return -1;
    }
  }
  for (i = 0; i < length; i++) {
    number += input[i] - '0';
    if (i != length - 1) {
      number *= 10;
    }
  }
  if (number == 0 || number == 1) {
    return 0;
  }
  for (i = 2; i < number; i++) {
    if (number % i == 0 && i != number) {
      return 0;
    }
  }
  return 1;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<is_prime>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  mov    %rbx,%rdi
  cltq
  cmpb   $0xa,-0x1(%rbx,%rax,1)
  je     L6
  call   <strlen@plt>
L115:
  test   %eax,%eax
  jle    L26
  lea    -0x1(%rax),%edx
  mov    %rbx,%rcx
  mov    %rdx,%rdi
  lea    0x1(%rbx,%rdx,1),%rsi
  jmp    L116
  nopw   0x0(%rax,%rax,1)
L57:
  add    $0x1,%rcx
  cmp    %rsi,%rcx
  je     L18
L116:
  movzbl (%rcx),%edx
  sub    $0x30,%edx
  cmp    $0x9,%dl
  jbe    L57
  mov    $0xffffffff,%edx
L37:
  mov    %edx,%eax
  pop    %rbx
  ret
  nopl   (%rax)
L18:
  mov    %eax,%edx
  xor    %ecx,%ecx
  xor    %eax,%eax
  cs nopw 0x0(%rax,%rax,1)
L34:
  movsbl (%rbx,%rax,1),%esi
  lea    -0x30(%rcx,%rsi,1),%ecx
  cmp    %eax,%edi
  je     L60
  lea    (%rcx,%rcx,4),%ecx
  add    %ecx,%ecx
L60:
  add    $0x1,%rax
  cmp    %rax,%rdx
  jne    L34
  xor    %edx,%edx
  cmp    $0x1,%ecx
  jbe    L37
  cmp    $0x2,%ecx
  jle    L142
  test   $0x1,%cl
  je     L37
  mov    $0x2,%esi
  jmp    L64
  nopw   0x0(%rax,%rax,1)
L49:
  mov    %ecx,%eax
  cltd
  idiv   %esi
  test   %edx,%edx
  je     L37
L64:
  add    $0x1,%esi
  cmp    %esi,%ecx
  jne    L49
L142:
  mov    $0x1,%edx
  pop    %rbx
  mov    %edx,%eax
  ret
  nopl   0x0(%rax)
L6:
  call   <strlen@plt>
  sub    $0x1,%eax
  jmp    L115
L26:
  xor    %edx,%edx
  jmp    L37
```

assistant: ```c
int is_prime(char *input) {
  int i, length, number = 0;
  length = input[strlen(input) - 1] == '\n' ? strlen(input) - 1 : strlen(input);
  for (i = 0; i < length; i++) {
    if (input[i] < '0' || input[i] > '9') {
      return -1;
    }
  }
  for (i = 0; i < length; i++) {
    number += input[i] - '0';
    if (i != length - 1) {
      number *= 10;
    }
  }
  if (number == 0 || number == 1) {
    return 0;
  }
  for (i = 2; i < number; i++) {
    if (number % i == 0 && i != number) {
      return 0;
    }
  }
  return 1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<is_prime>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  mov    %rbx,%rdi
  cltq
  cmpb   $0xa,-0x1(%rbx,%rax,1)
  je     L140
  call   <strlen@plt>
L95:
  test   %eax,%eax
  jle    L128
  lea    -0x1(%rax),%edx
  mov    %rbx,%rcx
  mov    %rdx,%rdi
  lea    0x1(%rbx,%rdx,1),%rsi
  jmp    L112
  nopw   0x0(%rax,%rax,1)
L44:
  add    $0x1,%rcx
  cmp    %rsi,%rcx
  je     L33
L112:
  movzbl (%rcx),%edx
  sub    $0x30,%edx
  cmp    $0x9,%dl
  jbe    L44
  mov    $0xffffffff,%edx
L143:
  mov    %edx,%eax
  pop    %rbx
  ret
  nopl   (%rax)
L33:
  mov    %eax,%edx
  xor    %ecx,%ecx
  xor    %eax,%eax
  cs nopw 0x0(%rax,%rax,1)
L116:
  movsbl (%rbx,%rax,1),%esi
  lea    -0x30(%rcx,%rsi,1),%ecx
  cmp    %eax,%edi
  je     L25
  lea    (%rcx,%rcx,4),%ecx
  add    %ecx,%ecx
L25:
  add    $0x1,%rax
  cmp    %rax,%rdx
  jne    L116
  xor    %edx,%edx
  cmp    $0x1,%ecx
  jbe    L143
  cmp    $0x2,%ecx
  jle    L149
  test   $0x1,%cl
  je     L143
  mov    $0x2,%esi
  jmp    L133
  nopw   0x0(%rax,%rax,1)
L31:
  mov    %ecx,%eax
  cltd
  idiv   %esi
  test   %edx,%edx
  je     L143
L133:
  add    $0x1,%esi
  cmp    %ecx,%esi
  jne    L31
L149:
  mov    $0x1,%edx
  pop    %rbx
  mov    %edx,%eax
  ret
  nopl   0x0(%rax)
L140:
  call   <strlen@plt>
  sub    $0x1,%eax
  jmp    L95
L128:
  xor    %edx,%edx
  jmp    L143
```

assistant: ```c
int is_prime(char *input) {
  int i, length, number = 0;
  length = input[strlen(input) - 1] == '\n' ? strlen(input) - 1 : strlen(input);
  for (i = 0; i < length; i++) {
    if (input[i] < '0' || input[i] > '9') {
      return -1;
    }
  }
  for (i = 0; i < length; i++) {
    number += input[i] - '0';
    if (i != length - 1) {
      number *= 10;
    }
  }
  if (number == 0 || number == 1) {
    return 0;
  }
  for (i = 2; i < number; i++) {
    if (number % i == 0 && i != number) {
      return 0;
    }
  }
  return 1;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<reverse>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xfd0,%rsp
  mov    %rdi,-0xfc8(%rbp)
  mov    %esi,-0xfcc(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0xfb4(%rbp)
  jmp    L37
L5:
  mov    -0xfcc(%rbp),%eax
  sub    $0x1,%eax
  sub    -0xfb4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0xfc8(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0xfb4(%rbp),%eax
  cltq
  mov    %edx,-0xfb0(%rbp,%rax,4)
  addl   $0x1,-0xfb4(%rbp)
L37:
  mov    -0xfb4(%rbp),%eax
  cmp    -0xfcc(%rbp),%eax
  jl     L5
  movl   $0x0,-0xfb4(%rbp)
  jmp    L19
L21:
  mov    -0xfb4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0xfc8(%rbp),%rax
  add    %rax,%rdx
  mov    -0xfb4(%rbp),%eax
  cltq
  mov    -0xfb0(%rbp,%rax,4),%eax
  mov    %eax,(%rdx)
  addl   $0x1,-0xfb4(%rbp)
L19:
  mov    -0xfb4(%rbp),%eax
  cmp    -0xfcc(%rbp),%eax
  jl     L21
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L22
  call   <__stack_chk_fail@plt>
L22:
  leave
  ret
```

assistant: ```c
void reverse(int a[], int len) {
  int b[1000];
  int k;
  for (k = 0; k < len; k++)
    b[k] = a[len - 1 - k];
  for (k = 0; k < len; k++)
    a[k] = b[k];
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<reverse>:
  endbr64
  sub    $0xfb8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xfa8(%rsp)
  xor    %eax,%eax
  test   %esi,%esi
  jle    L23
  movslq %esi,%rax
  lea    -0x4(%rdi,%rax,4),%rdx
  mov    %rsp,%rax
  lea    -0x1(%rsi),%ecx
  lea    0x4(%rsp,%rcx,4),%r8
L11:
  mov    (%rdx),%ecx
  mov    %ecx,(%rax)
  sub    $0x4,%rdx
  add    $0x4,%rax
  cmp    %r8,%rax
  jne    L11
  mov    %esi,%esi
  mov    $0x0,%eax
L24:
  mov    (%rsp,%rax,4),%edx
  mov    %edx,(%rdi,%rax,4)
  add    $0x1,%rax
  cmp    %rax,%rsi
  jne    L24
L23:
  mov    0xfa8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L22
  add    $0xfb8,%rsp
  ret
L22:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void reverse(int a[], int len) {
  int b[1000];
  int k;
  for (k = 0; k < len; k++)
    b[k] = a[len - 1 - k];
  for (k = 0; k < len; k++)
    a[k] = b[k];
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<reverse>:
  endbr64
  sub    $0xfb8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xfa8(%rsp)
  xor    %eax,%eax
  test   %esi,%esi
  jle    L12
  lea    -0x1(%rsi),%edx
  mov    %rsp,%r9
  movslq %edx,%rax
  mov    %edx,%edx
  lea    (%rdi,%rax,4),%rcx
  lea    0x4(%rsp,%rdx,4),%rdx
  mov    %r9,%rax
  nopl   0x0(%rax)
L3:
  mov    (%rcx),%r8d
  add    $0x4,%rax
  sub    $0x4,%rcx
  mov    %r8d,-0x4(%rax)
  cmp    %rdx,%rax
  jne    L3
  mov    %esi,%esi
  lea    0x0(,%rsi,4),%rdx
  mov    %r9,%rsi
  call   <memcpy@plt>
L12:
  mov    0xfa8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L18
  add    $0xfb8,%rsp
  ret
L18:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void reverse(int a[], int len) {
  int b[1000];
  int k;
  for (k = 0; k < len; k++)
    b[k] = a[len - 1 - k];
  for (k = 0; k < len; k++)
    a[k] = b[k];
}
```