ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<io_int_record_name>:
  endbr64
  test   %rdi,%rdi
  je     L62
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  sub    $0x8,%rsp
  cmp    $0x4,%rdx
  ja     L69
L30:
  mov    D8(%rip),%rax
  mov    (%rax),%rax
L1:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   (%rax)
L69:
  lea    0x5(%rsi),%r12
  mov    %rdi,%rbx
  mov    %rcx,%r13
  mov    (%rdi,%r12,8),%rdi
  call   <ntohl@plt>
  lea    0x6(%rax),%r14
  cmp    %rbp,%r14
  ja     L30
  lea    0x1(%r12,%rax,1),%rax
  mov    (%rbx,%rax,8),%rdi
  lea    0x0(,%rax,8),%r15
  call   <ntohl@plt>
  add    %rax,%r14
  mov    %rax,%r12
  cmp    %rbp,%r14
  jae    L30
  test   %rax,%rax
  je     L30
  mov    0x0(%r13),%rdi
  mov    $0x84,%edx
  mov    $0x20,%esi
  lea    -0x8(%rbx,%r15,1),%rbp
  xor    %ebx,%ebx
  call   <memset@plt>
  nopl   0x0(%rax)
L67:
  mov    0x10(%rbp,%rbx,8),%rdi
  call   <ntohl@plt>
  mov    0x0(%r13),%rdx
  mov    %al,(%rdx,%rbx,1)
  add    $0x1,%rbx
  cmp    %rbx,%r12
  jne    L67
  mov    D19(%rip),%rax
  mov    (%rax),%rax
  jmp    L1
  nopl   0x0(%rax)
L62:
  mov    D8(%rip),%rax
  mov    (%rax),%rax
  ret
```

assistant: ```c
int32_t io_int_record_name(int32_t *fmap, int32_t pos, int32_t size, char *(name[132])) {
  int32_t doff = 5;
  int32_t dlen = 0;
  int32_t nlen = 0;
  int32_t i = 0;
  if (fmap == NULL) {
    return (EXIT_FAILURE);
  }
  if (size >= doff) {
    dlen = ntohl(fmap[pos + doff]);
  } else {
    return (EXIT_FAILURE);
  }
  if (size >= doff + dlen + 1) {
    nlen = ntohl(fmap[pos + doff + dlen + 1]);
  } else {
    return (EXIT_FAILURE);
  }
  if (size <= (doff + dlen + 1 + nlen)) {
    return (EXIT_FAILURE);
  }
  if (nlen <= 0) {
    return (EXIT_FAILURE);
  }
  memset(*name, 32, 132);
  for (i = 0; i < nlen; ++i) {
    (*name)[i] = (char)ntohl(fmap[pos + doff + dlen + 2 + i]);
  }
  return (EXIT_SUCCESS);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<io_int_record_name>:
  endbr64
  test   %rdi,%rdi
  je     L29
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  sub    $0x8,%rsp
  cmp    $0x4,%rdx
  ja     L28
L4:
  mov    D27(%rip),%rax
  mov    (%rax),%rax
L60:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   (%rax)
L28:
  lea    0x5(%rsi),%r12
  mov    %rdi,%rbx
  mov    %rcx,%r13
  mov    (%rdi,%r12,8),%rdi
  call   <ntohl@plt>
  lea    0x6(%rax),%r14
  cmp    %rbp,%r14
  ja     L4
  lea    0x1(%r12,%rax,1),%rax
  mov    (%rbx,%rax,8),%rdi
  lea    0x0(,%rax,8),%r15
  call   <ntohl@plt>
  add    %rax,%r14
  mov    %rax,%r12
  cmp    %rbp,%r14
  jae    L4
  test   %rax,%rax
  je     L4
  mov    0x0(%r13),%rdi
  mov    $0x84,%edx
  mov    $0x20,%esi
  lea    -0x8(%rbx,%r15,1),%rbp
  xor    %ebx,%ebx
  call   <memset@plt>
  nopl   0x0(%rax)
L5:
  mov    0x10(%rbp,%rbx,8),%rdi
  call   <ntohl@plt>
  mov    0x0(%r13),%rdx
  mov    %al,(%rdx,%rbx,1)
  add    $0x1,%rbx
  cmp    %rbx,%r12
  jne    L5
  mov    D12(%rip),%rax
  mov    (%rax),%rax
  jmp    L60
  nopl   0x0(%rax)
L29:
  mov    D27(%rip),%rax
  mov    (%rax),%rax
  ret
```

assistant: ```c
int32_t io_int_record_name(int32_t *fmap, int32_t pos, int32_t size, char *(name[132])) {
  int32_t doff = 5;
  int32_t dlen = 0;
  int32_t nlen = 0;
  int32_t i = 0;
  if (fmap == NULL) {
    return (EXIT_FAILURE);
  }
  if (size >= doff) {
    dlen = ntohl(fmap[pos + doff]);
  } else {
    return (EXIT_FAILURE);
  }
  if (size >= doff + dlen + 1) {
    nlen = ntohl(fmap[pos + doff + dlen + 1]);
  } else {
    return (EXIT_FAILURE);
  }
  if (size <= (doff + dlen + 1 + nlen)) {
    return (EXIT_FAILURE);
  }
  if (nlen <= 0) {
    return (EXIT_FAILURE);
  }
  memset(*name, 32, 132);
  for (i = 0; i < nlen; ++i) {
    (*name)[i] = (char)ntohl(fmap[pos + doff + dlen + 2 + i]);
  }
  return (EXIT_SUCCESS);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<get_int>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L15
L42:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x2f,%al
  jle    L30
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x39,%al
  jle    L4
L30:
  mov    -0x1c(%rbp),%eax
  jmp    L44
L4:
  mov    -0x8(%rbp),%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  mov    %eax,%ecx
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  add    %ecx,%eax
  sub    $0x30,%eax
  mov    %eax,-0x8(%rbp)
  addl   $0x1,-0x4(%rbp)
L15:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L42
  mov    -0x8(%rbp),%eax
L44:
  pop    %rbp
  ret
```

assistant: ```c
int get_int(char string[], int error) {
  int result = 0, i = 0;
  while (string[i] != '\0') {
    if (string[i] < '0' || string[i] > '9') {
      return error;
    } else {
      result = 10 * result + string[i] - '0';
      ++i;
    }
  }
  return result;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<get_int>:
  endbr64
  movzbl (%rdi),%eax
  test   %al,%al
  je     L46
  add    $0x1,%rdi
  mov    $0x0,%edx
L23:
  lea    -0x30(%rax),%ecx
  cmp    $0x9,%cl
  ja     L4
  lea    (%rdx,%rdx,4),%edx
  movsbl %al,%eax
  lea    -0x30(%rax,%rdx,2),%edx
  add    $0x1,%rdi
  movzbl -0x1(%rdi),%eax
  test   %al,%al
  jne    L23
  jmp    L42
L46:
  mov    $0x0,%edx
  jmp    L42
L4:
  mov    %esi,%edx
L42:
  mov    %edx,%eax
  ret
```

assistant: ```c
int get_int(char string[], int error) {
  int result = 0, i = 0;
  while (string[i] != '\0') {
    if (string[i] < '0' || string[i] > '9') {
      return error;
    } else {
      result = 10 * result + string[i] - '0';
      ++i;
    }
  }
  return result;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<get_int>:
  endbr64
  movsbl (%rdi),%eax
  test   %al,%al
  je     L13
  add    $0x1,%rdi
  xor    %r8d,%r8d
  jmp    L4
  nopl   0x0(%rax)
L29:
  lea    (%r8,%r8,4),%edx
  add    $0x1,%rdi
  lea    -0x30(%rax,%rdx,2),%r8d
  movsbl -0x1(%rdi),%eax
  test   %al,%al
  je     L3
L4:
  lea    -0x30(%rax),%edx
  cmp    $0x9,%dl
  jbe    L29
  mov    %esi,%r8d
L3:
  mov    %r8d,%eax
  ret
L13:
  xor    %r8d,%r8d
  jmp    L3
```

assistant: ```c
int get_int(char string[], int error) {
  int result = 0, i = 0;
  while (string[i] != '\0') {
    if (string[i] < '0' || string[i] > '9') {
      return error;
    } else {
      result = 10 * result + string[i] - '0';
      ++i;
    }
  }
  return result;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<get_int>:
  endbr64
  movsbl (%rdi),%eax
  test   %al,%al
  je     L6
  add    $0x1,%rdi
  xor    %r8d,%r8d
  jmp    L48
  nopl   0x0(%rax)
L40:
  lea    (%r8,%r8,4),%edx
  add    $0x1,%rdi
  lea    -0x30(%rax,%rdx,2),%r8d
  movsbl -0x1(%rdi),%eax
  test   %al,%al
  je     L28
L48:
  lea    -0x30(%rax),%edx
  cmp    $0x9,%dl
  jbe    L40
  mov    %esi,%r8d
L28:
  mov    %r8d,%eax
  ret
L6:
  xor    %r8d,%r8d
  jmp    L28
```

assistant: ```c
int get_int(char string[], int error) {
  int result = 0, i = 0;
  while (string[i] != '\0') {
    if (string[i] < '0' || string[i] > '9') {
      return error;
    } else {
      result = 10 * result + string[i] - '0';
      ++i;
    }
  }
  return result;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<freeWindow>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L2
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
L2:
  nop
  leave
  ret
```

assistant: ```c
void freeWindow(struct window *thisWindow) {
  if (thisWindow) {
    free(thisWindow);
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<freeWindow>:
  endbr64
  test   %rdi,%rdi
  je     L2
  sub    $0x8,%rsp
  call   <free@plt>
  add    $0x8,%rsp
  ret
L2:
  ret
```

assistant: ```c
void freeWindow(struct window *thisWindow) {
  if (thisWindow) {
    free(thisWindow);
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<freeWindow>:
  endbr64
  test   %rdi,%rdi
  je     L8
  jmp    1050 <free@plt>
  xchg   %ax,%ax
L8:
  ret
```

assistant: ```c
void freeWindow(struct window *thisWindow) {
  if (thisWindow) {
    free(thisWindow);
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<freeWindow>:
  endbr64
  test   %rdi,%rdi
  je     L4
  jmp    1050 <free@plt>
  xchg   %ax,%ax
L4:
  ret
```

assistant: ```c
void freeWindow(struct window *thisWindow) {
  if (thisWindow) {
    free(thisWindow);
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<wrapsum>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  not    %eax
  and    $0xffff,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <htons@plt>
  leave
  ret
```

assistant: ```c
u_int32_t wrapsum(u_int32_t sum) {
  sum = ~sum & 0xFFFF;
  return htons(sum);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<wrapsum>:
  endbr64
  sub    $0x8,%rsp
  not    %edi
  movzwl %di,%edi
  call   <htons@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
u_int32_t wrapsum(u_int32_t sum) {
  sum = ~sum & 0xFFFF;
  return htons(sum);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<wrapsum>:
  endbr64
  not    %edi
  movzwl %di,%edi
  jmp    1050 <htons@plt>
```

assistant: ```c
u_int32_t wrapsum(u_int32_t sum) {
  sum = ~sum & 0xFFFF;
  return htons(sum);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<wrapsum>:
  endbr64
  not    %edi
  movzwl %di,%edi
  jmp    1050 <htons@plt>
```

assistant: ```c
u_int32_t wrapsum(u_int32_t sum) {
  sum = ~sum & 0xFFFF;
  return htons(sum);
}
```
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    $0x50,%rsp
  mov    %edi,-0x44(%rbp)
  mov    %rsi,-0x50(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x28(%rbp),%rcx
  lea    -0x30(%rbp),%rdx
  lea    -0x38(%rbp),%rax
  mov    %rax,%rsi
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x30(%rbp),%rax
  mov    -0x38(%rbp),%rdi
  cqto
  idiv   %rdi
  mov    %rax,-0x20(%rbp)
  mov    -0x30(%rbp),%rax
  test   %rax,%rax
  jns    L19
  mov    -0x30(%rbp),%rax
  mov    -0x38(%rbp),%rcx
  cqto
  idiv   %rcx
  mov    %rdx,%rax
  test   %rax,%rax
  je     L19
  subq   $0x1,-0x20(%rbp)
L19:
  mov    -0x28(%rbp),%rax
  mov    -0x38(%rbp),%rsi
  cqto
  idiv   %rsi
  mov    %rax,-0x18(%rbp)
  mov    -0x28(%rbp),%rax
  test   %rax,%rax
  jns    L10
  mov    -0x28(%rbp),%rax
  mov    -0x38(%rbp),%rcx
  cqto
  idiv   %rcx
  mov    %rdx,%rax
  test   %rax,%rax
  je     L10
  subq   $0x1,-0x18(%rbp)
L10:
  mov    -0x18(%rbp),%rax
  sub    -0x20(%rbp),%rax
  mov    %rax,%rsi
  mov    -0x30(%rbp),%rax
  mov    -0x38(%rbp),%rcx
  cqto
  idiv   %rcx
  mov    %rdx,%rax
  test   %rax,%rax
  sete   %al
  movzbl %al,%eax
  add    %rsi,%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L12
  call   <__stack_chk_fail@plt>
L12:
  leave
  ret
```

tool_calls: name: parse_data
arguments: label: D19
type_guessed: string
name: parse_data
arguments: label: D10
type_guessed: string

tool: D19: %I64d %I64d %I64d

tool: D10: %I64d


assistant: ```c
int main(int args, char **argv) {
  long long k, a, b;
  scanf("%I64d %I64d %I64d", &k, &a, &b);
  long long total;
  long long specialStepA = a / k;
  if (a < 0 && a % k != 0)
    specialStepA--;
  long long specialStepB = b / k;
  if (b < 0 && b % k != 0)
    specialStepB--;
  total = (specialStepB - specialStepA + (a % k == 0));
  printf("%I64d\n", total);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rcx
  lea    0x8(%rsp),%rdx
  mov    %rsp,%rsi
  lea    D14(%rip),%rdi
  call   <scanf@plt>
  mov    0x8(%rsp),%rsi
  mov    (%rsp),%rcx
  mov    %rsi,%rax
  cqto
  idiv   %rcx
  mov    %rax,%r9
  test   %rsi,%rsi
  js     L2
L62:
  mov    0x10(%rsp),%rdi
  mov    %rdi,%rax
  cqto
  idiv   %rcx
  mov    %rax,%r10
  mov    %rax,%r8
  test   %rdi,%rdi
  js     L5
L23:
  mov    %rsi,%rax
  cqto
  idiv   %rcx
  test   %rdx,%rdx
  sete   %sil
  movzbl %sil,%esi
  sub    %r9,%r8
  add    %r8,%rsi
  lea    D3(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L18
  mov    $0x0,%eax
  add    $0x28,%rsp
  ret
L2:
  mov    %rsi,%rax
  cqto
  idiv   %rcx
  test   %rdx,%rdx
  je     L62
  lea    -0x1(%r9),%r9
  jmp    L62
L5:
  mov    %rdi,%rax
  cqto
  idiv   %rcx
  test   %rdx,%rdx
  je     L23
  lea    -0x1(%r10),%r8
  jmp    L23
L18:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D14
datatype: string
name: extract_symbolic_data
arguments: tag: D3
datatype: string

tool: D14: %I64d %I64d %I64d

tool: D3: %I64d


assistant: ```c
int main(int args, char **argv) {
  long long k, a, b;
  scanf("%I64d %I64d %I64d", &k, &a, &b);
  long long total;
  long long specialStepA = a / k;
  if (a < 0 && a % k != 0)
    specialStepA--;
  long long specialStepB = b / k;
  if (b < 0 && b % k != 0)
    specialStepB--;
  total = (specialStepB - specialStepA + (a % k == 0));
  printf("%I64d\n", total);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  lea    D7(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rcx
  lea    0x8(%rsp),%rdx
  mov    %rsp,%rsi
  call   <scanf@plt>
  mov    0x8(%rsp),%rsi
  mov    (%rsp),%rcx
  mov    %rsi,%rax
  cqto
  idiv   %rcx
  mov    %rdx,%rdi
  mov    %rax,%r8
  test   %rsi,%rsi
  jns    L15
  cmp    $0x1,%rdx
  adc    $0xffffffffffffffff,%r8
L15:
  mov    0x10(%rsp),%rsi
  mov    %rsi,%rax
  cqto
  idiv   %rcx
  test   %rsi,%rsi
  jns    L6
  cmp    $0x1,%rdx
  adc    $0xffffffffffffffff,%rax
L6:
  sub    %r8,%rax
  cmp    $0x1,%rdi
  lea    D9(%rip),%rdi
  adc    $0x0,%rax
  mov    %rax,%rsi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  xor    %eax,%eax
  add    $0x28,%rsp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D7
type_infered: string
name: read_data
arguments: tag: D9
type_infered: string

tool: D7: %I64d %I64d %I64d

tool: D9: %I64d


assistant: ```c
int main(int args, char **argv) {
  long long k, a, b;
  scanf("%I64d %I64d %I64d", &k, &a, &b);
  long long total;
  long long specialStepA = a / k;
  if (a < 0 && a % k != 0)
    specialStepA--;
  long long specialStepB = b / k;
  if (b < 0 && b % k != 0)
    specialStepB--;
  total = (specialStepB - specialStepA + (a % k == 0));
  printf("%I64d\n", total);
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  lea    D16(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rcx
  lea    0x8(%rsp),%rdx
  mov    %rsp,%rsi
  call   <scanf@plt>
  mov    0x8(%rsp),%rsi
  mov    (%rsp),%rcx
  mov    %rsi,%rax
  cqto
  idiv   %rcx
  mov    %rdx,%rdi
  mov    %rax,%r8
  test   %rsi,%rsi
  jns    L4
  cmp    $0x1,%rdx
  adc    $0xffffffffffffffff,%r8
L4:
  mov    0x10(%rsp),%rsi
  mov    %rsi,%rax
  cqto
  idiv   %rcx
  test   %rsi,%rsi
  jns    L12
  cmp    $0x1,%rdx
  adc    $0xffffffffffffffff,%rax
L12:
  sub    %r8,%rax
  cmp    $0x1,%rdi
  lea    D9(%rip),%rdi
  adc    $0x0,%rax
  mov    %rax,%rsi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L28
  xor    %eax,%eax
  add    $0x28,%rsp
  ret
L28:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D16
datatype: string
name: retrieve_data
arguments: data_label: D9
datatype: string

tool: D16: %I64d %I64d %I64d

tool: D9: %I64d


assistant: ```c
int main(int args, char **argv) {
  long long k, a, b;
  scanf("%I64d %I64d %I64d", &k, &a, &b);
  long long total;
  long long specialStepA = a / k;
  if (a < 0 && a % k != 0)
    specialStepA--;
  long long specialStepB = b / k;
  if (b < 0 && b % k != 0)
    specialStepB--;
  total = (specialStepB - specialStepA + (a % k == 0));
  printf("%I64d\n", total);
  return 0;
}
```
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    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    $0x0,%eax
  call   <sieve@plt>
  jmp    L42
L87:
  mov    -0x18(%rbp),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  mov    %eax,-0xc(%rbp)
  movl   $0x0,-0x14(%rbp)
  mov    -0x18(%rbp),%eax
  test   %eax,%eax
  je     L86
  movl   $0x1,-0x10(%rbp)
  jmp    L46
L50:
  mov    -0x10(%rbp),%eax
  mov    %eax,%edi
  call   <prime@plt>
  cmp    $0x1,%eax
  jne    L28
  mov    -0x18(%rbp),%eax
  sub    -0x10(%rbp),%eax
  mov    %eax,%edi
  call   <prime@plt>
  cmp    $0x1,%eax
  jne    L28
  addl   $0x1,-0x14(%rbp)
L28:
  addl   $0x1,-0x10(%rbp)
L46:
  mov    -0x10(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jle    L50
  mov    -0x14(%rbp),%eax
  mov    %eax,%esi
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L42:
  lea    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  cmp    $0x1,%eax
  je     L87
  jmp    L52
L86:
  nop
L52:
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L11
  call   <__stack_chk_fail@plt>
L11:
  leave
  ret
```

tool_calls: name: read_data
arguments: label: D10
type_guessed: string
name: read_data
arguments: label: D5
type_guessed: string

tool: D10: %d


tool: D5: %d

assistant: ```c
int main() {
  int n, count, p;
  sieve();
  while (scanf("%d", &n) == 1) {
    int half = n / 2;
    count = 0;
    if (n == 0)
      break;
    for (p = 1; p <= half; p++) {
      if (prime(p) == 1 && prime(n - p) == 1)
        count++;
    }
    printf("%d\n", count);
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <sieve@plt>
  lea    0x4(%rsp),%r14
  lea    D0(%rip),%r13
  lea    D18(%rip),%r15
  jmp    L86
L65:
  add    $0x1,%ebx
  cmp    %ebx,%ebp
  jl     L69
L0:
  mov    %ebx,%edi
  call   <prime@plt>
  cmp    $0x1,%eax
  jne    L65
  mov    0x4(%rsp),%edi
  sub    %ebx,%edi
  call   <prime@plt>
  cmp    $0x1,%eax
  sete   %al
  movzbl %al,%eax
  add    %eax,%r12d
  jmp    L65
L43:
  mov    $0x0,%r12d
L69:
  mov    %r12d,%esi
  mov    %r15,%rdi
  call   <printf@plt>
L86:
  mov    %r14,%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
  mov    %eax,%ebx
  cmp    $0x1,%eax
  jne    L9
  mov    0x4(%rsp),%eax
  mov    %eax,%ebp
  shr    $0x1f,%ebp
  add    %eax,%ebp
  sar    %ebp
  test   %eax,%eax
  je     L9
  cmp    $0x1,%eax
  jle    L43
  mov    $0x0,%r12d
  jmp    L0
L9:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L20
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L20:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D0
type: string
name: parse_data
arguments: data_tag: D18
type: string

tool: D0: %d

tool: D18: %d


assistant: ```c
int main() {
  int n, count, p;
  sieve();
  while (scanf("%d", &n) == 1) {
    int half = n / 2;
    count = 0;
    if (n == 0)
      break;
    for (p = 1; p <= half; p++) {
      if (prime(p) == 1 && prime(n - p) == 1)
        count++;
    }
    printf("%d\n", count);
  }
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  lea    D7(%rip),%r14
  push   %r13
  push   %r12
  lea    D19(%rip),%r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%r13
  call   <sieve@plt>
  nopw   0x0(%rax,%rax,1)
L29:
  mov    %r13,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  mov    %eax,%ebx
  cmp    $0x1,%eax
  jne    L30
  mov    0x4(%rsp),%eax
  mov    %eax,%ebp
  shr    $0x1f,%ebp
  add    %eax,%ebp
  sar    %ebp
  test   %eax,%eax
  je     L30
  xor    %r15d,%r15d
  cmp    $0x1,%eax
  jg     L28
  jmp    L5
  nopl   (%rax)
L27:
  add    $0x1,%ebx
  cmp    %ebx,%ebp
  jl     L5
L28:
  mov    %ebx,%edi
  call   <prime@plt>
  cmp    $0x1,%eax
  jne    L27
  mov    0x4(%rsp),%edi
  sub    %ebx,%edi
  call   <prime@plt>
  cmp    $0x1,%eax
  sete   %al
  add    $0x1,%ebx
  movzbl %al,%eax
  add    %eax,%r15d
  cmp    %ebx,%ebp
  jge    L28
  nopl   0x0(%rax)
L5:
  mov    %r15d,%esi
  mov    %r14,%rdi
  call   <printf@plt>
  jmp    L29
  nopl   (%rax)
L30:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L55
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L55:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D7
data_type: string
name: retrieve_data
arguments: tag: D19
data_type: string

tool: D7: %d


tool: D19: %d

assistant: ```c
int main() {
  int n, count, p;
  sieve();
  while (scanf("%d", &n) == 1) {
    int half = n / 2;
    count = 0;
    if (n == 0)
      break;
    for (p = 1; p <= half; p++) {
      if (prime(p) == 1 && prime(n - p) == 1)
        count++;
    }
    printf("%d\n", count);
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  lea    D11(%rip),%r14
  push   %r13
  push   %r12
  lea    D10(%rip),%r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%r13
  call   <sieve@plt>
  nopw   0x0(%rax,%rax,1)
L61:
  mov    %r13,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  mov    %eax,%ebx
  cmp    $0x1,%eax
  jne    L0
  mov    0x4(%rsp),%eax
  mov    %eax,%ebp
  shr    $0x1f,%ebp
  add    %eax,%ebp
  sar    %ebp
  test   %eax,%eax
  je     L0
  xor    %r15d,%r15d
  cmp    $0x1,%eax
  jg     L51
  jmp    L16
  nopl   (%rax)
L80:
  add    $0x1,%ebx
  cmp    %ebx,%ebp
  jl     L16
L51:
  mov    %ebx,%edi
  call   <prime@plt>
  cmp    $0x1,%eax
  jne    L80
  mov    0x4(%rsp),%edi
  sub    %ebx,%edi
  call   <prime@plt>
  cmp    $0x1,%eax
  sete   %al
  add    $0x1,%ebx
  movzbl %al,%eax
  add    %eax,%r15d
  cmp    %ebx,%ebp
  jge    L51
  nopl   0x0(%rax)
L16:
  mov    %r15d,%esi
  mov    %r14,%rdi
  call   <printf@plt>
  jmp    L61
  nopl   (%rax)
L0:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L76
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L76:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D11
type_infered: string
name: extract_symbolic_data
arguments: tag: D10
type_infered: string

tool: D11: %d


tool: D10: %d

assistant: ```c
int main() {
  int n, count, p;
  sieve();
  while (scanf("%d", &n) == 1) {
    int half = n / 2;
    count = 0;
    if (n == 0)
      break;
    for (p = 1; p <= half; p++) {
      if (prime(p) == 1 && prime(n - p) == 1)
        count++;
    }
    printf("%d\n", count);
  }
  return 0;
}
```