ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Demonstrate the C code that generates the assembly output below:

```asm
<missingNumber>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x90,%rsp
  mov    %rdi,%rbp
  mov    %esi,%r12d
  mov    %fs:0x28,%rax
  mov    %rax,0x88(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  mov    $0x80,%edx
  mov    $0x0,%esi
  mov    %rbx,%rdi
  call   <memset@plt>
  mov    %rbx,%rax
  mov    $0x0,%ecx
L50:
  mov    %r12d,%edx
  sar    %cl,%edx
  and    $0x1,%edx
  add    %edx,(%rax)
  add    $0x1,%ecx
  add    $0x4,%rax
  cmp    $0x20,%ecx
  jne    L50
  test   %r12d,%r12d
  jle    L15
  mov    %r12d,%r12d
  mov    $0x0,%r8d
L19:
  mov    %r8d,%edi
  mov    0x0(%rbp,%r8,4),%esi
  mov    %rbx,%rax
  mov    $0x0,%ecx
L52:
  mov    %esi,%edx
  sar    %cl,%edx
  and    $0x1,%edx
  add    %edx,(%rax)
  add    $0x1,%ecx
  add    $0x4,%rax
  cmp    $0x20,%ecx
  jne    L52
  mov    %rbx,%rax
  mov    $0x0,%ecx
L3:
  mov    %edi,%edx
  sar    %cl,%edx
  and    $0x1,%edx
  add    %edx,(%rax)
  add    $0x1,%ecx
  add    $0x4,%rax
  cmp    $0x20,%ecx
  jne    L3
  add    $0x1,%r8
  cmp    %r12,%r8
  jne    L19
L15:
  mov    $0x0,%ecx
  mov    $0x0,%esi
L55:
  mov    (%rsp,%rcx,4),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  and    $0x1,%eax
  sub    %edx,%eax
  shl    %cl,%eax
  add    %eax,%esi
  add    $0x1,%rcx
  cmp    $0x20,%rcx
  jne    L55
  mov    0x88(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L53
  mov    %esi,%eax
  add    $0x90,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L53:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int missingNumber(int *nums, int numsSize) {
  int i, j;
  int bits[32];
  int res = 0;
  memset(bits, 0, sizeof(int) * 32);
  for (j = 0; j < 32; j++)
    bits[j] += (numsSize >> j) & 1;
  for (i = 0; i < numsSize; i++) {
    for (j = 0; j < 32; j++)
      bits[j] += (nums[i] >> j) & 1;
    for (j = 0; j < 32; j++)
      bits[j] += (i >> j) & 1;
  }
  for (j = 0; j < 32; j++)
    res += (bits[j] % 2) << j;
  return res;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<missingNumber>:
  endbr64
  push   %r12
  mov    $0x80,%edx
  movslq %esi,%r12
  xor    %esi,%esi
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x90,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x88(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  mov    %rbx,%rdi
  call   <memset@plt>
  mov    %rbx,%rax
  xor    %ecx,%ecx
  nop
L0:
  mov    %r12d,%edx
  sar    %cl,%edx
  add    $0x1,%ecx
  and    $0x1,%edx
  add    %edx,(%rax)
  add    $0x4,%rax
  cmp    $0x20,%ecx
  jne    L0
  test   %r12d,%r12d
  jle    L20
  xor    %r8d,%r8d
  xchg   %ax,%ax
L63:
  mov    0x0(%rbp,%r8,4),%edi
  mov    %r8d,%esi
  mov    %rbx,%rax
  xor    %ecx,%ecx
  nopl   (%rax)
L47:
  mov    %edi,%edx
  sar    %cl,%edx
  add    $0x1,%ecx
  and    $0x1,%edx
  add    %edx,(%rax)
  add    $0x4,%rax
  cmp    $0x20,%ecx
  jne    L47
  mov    %rbx,%rax
  xor    %ecx,%ecx
  nopw   0x0(%rax,%rax,1)
L41:
  mov    %esi,%edx
  sar    %cl,%edx
  add    $0x1,%ecx
  and    $0x1,%edx
  add    %edx,(%rax)
  add    $0x4,%rax
  cmp    $0x20,%ecx
  jne    L41
  add    $0x1,%r8
  cmp    %r8,%r12
  jne    L63
L20:
  xor    %ecx,%ecx
  xor    %r8d,%r8d
  nopl   0x0(%rax,%rax,1)
L69:
  mov    (%rbx,%rcx,4),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  and    $0x1,%eax
  sub    %edx,%eax
  shl    %cl,%eax
  add    $0x1,%rcx
  add    %eax,%r8d
  cmp    $0x20,%rcx
  jne    L69
  mov    0x88(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L46
  add    $0x90,%rsp
  mov    %r8d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L46:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int missingNumber(int *nums, int numsSize) {
  int i, j;
  int bits[32];
  int res = 0;
  memset(bits, 0, sizeof(int) * 32);
  for (j = 0; j < 32; j++)
    bits[j] += (numsSize >> j) & 1;
  for (i = 0; i < numsSize; i++) {
    for (j = 0; j < 32; j++)
      bits[j] += (nums[i] >> j) & 1;
    for (j = 0; j < 32; j++)
      bits[j] += (i >> j) & 1;
  }
  for (j = 0; j < 32; j++)
    res += (bits[j] % 2) << j;
  return res;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<missingNumber>:
  endbr64
  push   %r12
  mov    $0x80,%edx
  movslq %esi,%r12
  xor    %esi,%esi
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x90,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x88(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  mov    %rbx,%rdi
  call   <memset@plt>
  mov    %rbx,%rax
  xor    %ecx,%ecx
  nop
L30:
  mov    %r12d,%edx
  sar    %cl,%edx
  add    $0x1,%ecx
  and    $0x1,%edx
  add    %edx,(%rax)
  add    $0x4,%rax
  cmp    $0x20,%ecx
  jne    L30
  test   %r12d,%r12d
  jle    L46
  xor    %r8d,%r8d
  xchg   %ax,%ax
L41:
  mov    0x0(%rbp,%r8,4),%edi
  mov    %r8d,%esi
  mov    %rbx,%rax
  xor    %ecx,%ecx
  nopl   (%rax)
L47:
  mov    %edi,%edx
  sar    %cl,%edx
  add    $0x1,%ecx
  and    $0x1,%edx
  add    %edx,(%rax)
  add    $0x4,%rax
  cmp    $0x20,%ecx
  jne    L47
  mov    %rbx,%rax
  xor    %ecx,%ecx
  nopw   0x0(%rax,%rax,1)
L39:
  mov    %esi,%edx
  sar    %cl,%edx
  add    $0x1,%ecx
  and    $0x1,%edx
  add    %edx,(%rax)
  add    $0x4,%rax
  cmp    $0x20,%ecx
  jne    L39
  add    $0x1,%r8
  cmp    %r8,%r12
  jne    L41
L46:
  xor    %ecx,%ecx
  xor    %r8d,%r8d
  nopl   0x0(%rax,%rax,1)
L45:
  mov    (%rbx,%rcx,4),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  and    $0x1,%eax
  sub    %edx,%eax
  shl    %cl,%eax
  add    $0x1,%rcx
  add    %eax,%r8d
  cmp    $0x20,%rcx
  jne    L45
  mov    0x88(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L61
  add    $0x90,%rsp
  mov    %r8d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L61:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int missingNumber(int *nums, int numsSize) {
  int i, j;
  int bits[32];
  int res = 0;
  memset(bits, 0, sizeof(int) * 32);
  for (j = 0; j < 32; j++)
    bits[j] += (numsSize >> j) & 1;
  for (i = 0; i < numsSize; i++) {
    for (j = 0; j < 32; j++)
      bits[j] += (nums[i] >> j) & 1;
    for (j = 0; j < 32; j++)
      bits[j] += (i >> j) & 1;
  }
  for (j = 0; j < 32; j++)
    res += (bits[j] % 2) << j;
  return res;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  add    $0xffffffffffffff80,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D40(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x60(%rbp),%rax
  mov    $0x50,%esi
  mov    %rax,%rdi
  call   <getLine@plt>
  lea    -0x60(%rbp),%rax
  lea    D63(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strtok@plt>
  mov    %rax,-0x68(%rbp)
  movl   $0x0,-0x7c(%rbp)
  jmp    L28
L23:
  cmpl   $0x0,-0x7c(%rbp)
  jne    L56
  lea    -0x80(%rbp),%rdx
  mov    -0x68(%rbp),%rax
  lea    D45(%rip),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <sscanf@plt>
  jmp    L9
L56:
  cmpl   $0x1,-0x7c(%rbp)
  jne    L62
  lea    -0x78(%rbp),%rdx
  mov    -0x68(%rbp),%rax
  lea    D36(%rip),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <sscanf@plt>
  mov    -0x78(%rbp),%rax
  imul   $0x2710,%rax,%rax
  mov    %rax,-0x78(%rbp)
  jmp    L9
L62:
  lea    -0x70(%rbp),%rdx
  mov    -0x68(%rbp),%rax
  lea    D36(%rip),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <sscanf@plt>
  mov    -0x78(%rbp),%rdx
  mov    -0x70(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x78(%rbp)
L9:
  addl   $0x1,-0x7c(%rbp)
  lea    D63(%rip),%rax
  mov    %rax,%rsi
  mov    $0x0,%edi
  call   <strtok@plt>
  mov    %rax,-0x68(%rbp)
L28:
  cmpq   $0x0,-0x68(%rbp)
  jne    L23
  mov    -0x80(%rbp),%eax
  mov    %eax,%esi
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x78(%rbp),%rax
  mov    %rax,%rsi
  lea    D73(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L26
  call   <__stack_chk_fail@plt>
L26:
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D40
type_infered: string
name: parse_data
arguments: data_tag: D63
type_infered: string
name: parse_data
arguments: data_tag: D45
type_infered: string
name: parse_data
arguments: data_tag: D36
type_infered: string
name: parse_data
arguments: data_tag: D4
type_infered: string
name: parse_data
arguments: data_tag: D73
type_infered: string

tool: D40: Enter phone number: 

tool: D63:  ()-

tool: D45: %d

tool: D36: %ld

tool: D4: Area code = %d


tool: D73: Number = %li


assistant: ```c
int main() {
  char str[80];
  char *tokenPtr;
  int areaCode;
  long number, num2;
  printf("Enter phone number: ");
  getLine(str, 80);
  tokenPtr = strtok(str, " ()-");
  int i = 0;
  while (tokenPtr != NULL) {
    if (i == 0)
      sscanf(tokenPtr, "%d", &areaCode);
    else if (i == 1) {
      sscanf(tokenPtr, "%ld", &number);
      number *= 10000;
    } else {
      sscanf(tokenPtr, "%ld", &num2);
      number += num2;
    }
    i++;
    tokenPtr = strtok(NULL, " ()-");
  }
  printf("Area code = %d\n", areaCode);
  printf("Number = %li\n", number);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  add    $0xffffffffffffff80,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x78(%rsp)
  xor    %eax,%eax
  lea    D15(%rip),%rdi
  call   <printf@plt>
  lea    0x20(%rsp),%rbx
  mov    $0x50,%esi
  mov    %rbx,%rdi
  call   <getLine@plt>
  lea    D10(%rip),%rsi
  mov    %rbx,%rdi
  call   <strtok@plt>
  test   %rax,%rax
  je     L30
  mov    %rax,%rdi
  mov    $0x0,%ebx
  lea    0x18(%rsp),%r13
  lea    D5(%rip),%r12
  lea    D11(%rip),%r14
  lea    D10(%rip),%rbp
  jmp    L62
L31:
  lea    0xc(%rsp),%rdx
  mov    %r14,%rsi
  mov    $0x0,%eax
  call   <sscanf@plt>
L13:
  add    $0x1,%ebx
  mov    %rbp,%rsi
  mov    $0x0,%edi
  call   <strtok@plt>
  mov    %rax,%rdi
  test   %rax,%rax
  je     L30
L62:
  test   %ebx,%ebx
  je     L31
  cmp    $0x1,%ebx
  je     L18
  mov    %r13,%rdx
  mov    %r12,%rsi
  mov    $0x0,%eax
  call   <sscanf@plt>
  mov    0x18(%rsp),%rax
  add    %rax,0x10(%rsp)
  jmp    L13
L18:
  lea    0x10(%rsp),%rdx
  mov    %r12,%rsi
  mov    $0x0,%eax
  call   <sscanf@plt>
  imul   $0x2710,0x10(%rsp),%rax
  mov    %rax,0x10(%rsp)
  jmp    L13
L30:
  mov    0xc(%rsp),%esi
  lea    D50(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x10(%rsp),%rsi
  lea    D45(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x78(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L75
  mov    $0x0,%eax
  sub    $0xffffffffffffff80,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L75:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D15
type_infered: string
name: read_data
arguments: data_label: D10
type_infered: string
name: read_data
arguments: data_label: D5
type_infered: string
name: read_data
arguments: data_label: D11
type_infered: string
name: read_data
arguments: data_label: D50
type_infered: string
name: read_data
arguments: data_label: D45
type_infered: string

tool: D15: Enter phone number: 

tool: D10:  ()-

tool: D5: %ld

tool: D11: %d

tool: D50: Area code = %d


tool: D45: Number = %li


assistant: ```c
int main() {
  char str[80];
  char *tokenPtr;
  int areaCode;
  long number, num2;
  printf("Enter phone number: ");
  getLine(str, 80);
  tokenPtr = strtok(str, " ()-");
  int i = 0;
  while (tokenPtr != NULL) {
    if (i == 0)
      sscanf(tokenPtr, "%d", &areaCode);
    else if (i == 1) {
      sscanf(tokenPtr, "%ld", &number);
      number *= 10000;
    } else {
      sscanf(tokenPtr, "%ld", &num2);
      number += num2;
    }
    i++;
    tokenPtr = strtok(NULL, " ()-");
  }
  printf("Area code = %d\n", areaCode);
  printf("Number = %li\n", number);
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %r15
  lea    D44(%rip),%rdi
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  lea    D33(%rip),%rbp
  push   %rbx
  sub    $0x88,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x78(%rsp)
  xor    %eax,%eax
  lea    0x20(%rsp),%r12
  call   <printf@plt>
  mov    $0x50,%esi
  mov    %r12,%rdi
  call   <getLine@plt>
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <strtok@plt>
  test   %rax,%rax
  je     L78
  mov    %rax,%rdi
  lea    0x18(%rsp),%r13
  lea    0x10(%rsp),%r15
  xor    %ebx,%ebx
  lea    D36(%rip),%r12
  lea    D32(%rip),%r14
  jmp    L55
  nopl   (%rax)
L37:
  cmp    $0x1,%ebx
  je     L70
  mov    %r13,%rdx
  mov    %r12,%rsi
  xor    %eax,%eax
  call   <sscanf@plt>
  mov    0x18(%rsp),%rax
  add    %rax,0x10(%rsp)
L10:
  xor    %edi,%edi
  mov    %rbp,%rsi
  add    $0x1,%ebx
  call   <strtok@plt>
  mov    %rax,%rdi
  test   %rax,%rax
  je     L78
L55:
  test   %ebx,%ebx
  jne    L37
  lea    0xc(%rsp),%rdx
  mov    %r14,%rsi
  xor    %eax,%eax
  call   <sscanf@plt>
  jmp    L10
  cs nopw 0x0(%rax,%rax,1)
L70:
  mov    %r15,%rdx
  mov    %r12,%rsi
  xor    %eax,%eax
  call   <sscanf@plt>
  imul   $0x2710,0x10(%rsp),%rax
  mov    %rax,0x10(%rsp)
  jmp    L10
  nopl   (%rax)
L78:
  mov    0xc(%rsp),%esi
  lea    D1(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x10(%rsp),%rsi
  xor    %eax,%eax
  lea    D23(%rip),%rdi
  call   <printf@plt>
  mov    0x78(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L54
  add    $0x88,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L54:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D44
datatype: string
name: disassemble_data
arguments: tag: D33
datatype: string
name: disassemble_data
arguments: tag: D36
datatype: string
name: disassemble_data
arguments: tag: D32
datatype: string
name: disassemble_data
arguments: tag: D1
datatype: string
name: disassemble_data
arguments: tag: D23
datatype: string

tool: D44: Enter phone number: 

tool: D33:  ()-

tool: D36: %ld

tool: D32: %d

tool: D1: Area code = %d


tool: D23: Number = %li


assistant: ```c
int main() {
  char str[80];
  char *tokenPtr;
  int areaCode;
  long number, num2;
  printf("Enter phone number: ");
  getLine(str, 80);
  tokenPtr = strtok(str, " ()-");
  int i = 0;
  while (tokenPtr != NULL) {
    if (i == 0)
      sscanf(tokenPtr, "%d", &areaCode);
    else if (i == 1) {
      sscanf(tokenPtr, "%ld", &number);
      number *= 10000;
    } else {
      sscanf(tokenPtr, "%ld", &num2);
      number += num2;
    }
    i++;
    tokenPtr = strtok(NULL, " ()-");
  }
  printf("Area code = %d\n", areaCode);
  printf("Number = %li\n", number);
  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
  lea    D17(%rip),%rdi
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  lea    D43(%rip),%rbp
  push   %rbx
  sub    $0x88,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x78(%rsp)
  xor    %eax,%eax
  lea    0x20(%rsp),%r12
  call   <printf@plt>
  mov    $0x50,%esi
  mov    %r12,%rdi
  call   <getLine@plt>
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <strtok@plt>
  test   %rax,%rax
  je     L66
  mov    %rax,%rdi
  lea    0x18(%rsp),%r13
  lea    0x10(%rsp),%r15
  xor    %ebx,%ebx
  lea    D20(%rip),%r12
  lea    D10(%rip),%r14
  jmp    L58
  nopl   (%rax)
L33:
  cmp    $0x1,%ebx
  je     L16
  mov    %r13,%rdx
  mov    %r12,%rsi
  xor    %eax,%eax
  call   <sscanf@plt>
  mov    0x18(%rsp),%rax
  add    %rax,0x10(%rsp)
L28:
  xor    %edi,%edi
  mov    %rbp,%rsi
  add    $0x1,%ebx
  call   <strtok@plt>
  mov    %rax,%rdi
  test   %rax,%rax
  je     L66
L58:
  test   %ebx,%ebx
  jne    L33
  lea    0xc(%rsp),%rdx
  mov    %r14,%rsi
  xor    %eax,%eax
  call   <sscanf@plt>
  jmp    L28
  cs nopw 0x0(%rax,%rax,1)
L16:
  mov    %r15,%rdx
  mov    %r12,%rsi
  xor    %eax,%eax
  call   <sscanf@plt>
  imul   $0x2710,0x10(%rsp),%rax
  mov    %rax,0x10(%rsp)
  jmp    L28
  nopl   (%rax)
L66:
  mov    0xc(%rsp),%esi
  lea    D15(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x10(%rsp),%rsi
  xor    %eax,%eax
  lea    D40(%rip),%rdi
  call   <printf@plt>
  mov    0x78(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L61
  add    $0x88,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L61:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D17
type_infered: string
name: parse_data
arguments: tag: D43
type_infered: string
name: parse_data
arguments: tag: D20
type_infered: string
name: parse_data
arguments: tag: D10
type_infered: string
name: parse_data
arguments: tag: D15
type_infered: string
name: parse_data
arguments: tag: D40
type_infered: string

tool: D17: Enter phone number: 

tool: D43:  ()-

tool: D20: %ld

tool: D10: %d

tool: D15: Area code = %d


tool: D40: Number = %li


assistant: ```c
int main() {
  char str[80];
  char *tokenPtr;
  int areaCode;
  long number, num2;
  printf("Enter phone number: ");
  getLine(str, 80);
  tokenPtr = strtok(str, " ()-");
  int i = 0;
  while (tokenPtr != NULL) {
    if (i == 0)
      sscanf(tokenPtr, "%d", &areaCode);
    else if (i == 1) {
      sscanf(tokenPtr, "%ld", &number);
      number *= 10000;
    } else {
      sscanf(tokenPtr, "%ld", &num2);
      number += num2;
    }
    i++;
    tokenPtr = strtok(NULL, " ()-");
  }
  printf("Area code = %d\n", areaCode);
  printf("Number = %li\n", number);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0xd(%rbp),%rax
  mov    %rax,%rsi
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  movzbl -0xd(%rbp),%eax
  movsbl %al,%edx
  movzbl -0xa(%rbp),%eax
  movsbl %al,%eax
  add    %edx,%eax
  lea    -0x60(%rax),%ecx
  movzbl -0xa(%rbp),%eax
  movsbl %al,%edx
  movzbl -0xd(%rbp),%eax
  movsbl %al,%eax
  mov    %eax,%esi
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L8
  call   <__stack_chk_fail@plt>
L8:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D0
type: string
name: disassemble_data
arguments: tag: D10
type: string

tool: D0: %s

tool: D10: %c + %c = %d

assistant: ```c
int main() {
  char number[5];
  scanf("%s", &number);
  printf("%c + %c = %d", number[0], number[3], (int)number[0] + (int)number[3] - 2 * (int)'0');
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x3(%rsp),%rsi
  lea    D13(%rip),%rdi
  call   <scanf@plt>
  movsbl 0x3(%rsp),%esi
  movsbl 0x6(%rsp),%edx
  lea    -0x60(%rsi,%rdx,1),%ecx
  lea    D9(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  mov    $0x0,%eax
  add    $0x18,%rsp
  ret
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D13
datatype: string
name: extract_symbolic_data
arguments: tag: D9
datatype: string

tool: D13: %s

tool: D9: %c + %c = %d

assistant: ```c
int main() {
  char number[5];
  scanf("%s", &number);
  printf("%c + %c = %d", number[0], number[3], (int)number[0] + (int)number[3] - 2 * (int)'0');
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D18(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x3(%rsp),%rsi
  call   <scanf@plt>
  movsbl 0x3(%rsp),%esi
  movsbl 0x6(%rsp),%edx
  lea    D13(%rip),%rdi
  lea    -0x60(%rsi,%rdx,1),%ecx
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D18
datatype: string
name: extract_symbolic_data
arguments: data_tag: D13
datatype: string

tool: D18: %s

tool: D13: %c + %c = %d

assistant: ```c
int main() {
  char number[5];
  scanf("%s", &number);
  printf("%c + %c = %d", number[0], number[3], (int)number[0] + (int)number[3] - 2 * (int)'0');
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D9(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x3(%rsp),%rsi
  call   <scanf@plt>
  movsbl 0x3(%rsp),%esi
  movsbl 0x6(%rsp),%edx
  lea    D8(%rip),%rdi
  lea    -0x60(%rsi,%rdx,1),%ecx
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D9
data_type: string
name: retrieve_data
arguments: tag: D8
data_type: string

tool: D9: %s

tool: D8: %c + %c = %d

assistant: ```c
int main() {
  char number[5];
  scanf("%s", &number);
  printf("%c + %c = %d", number[0], number[3], (int)number[0] + (int)number[3] - 2 * (int)'0');
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<prefix_array>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %edx,-0x34(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x20(%rbp)
  cmpl   $0x1,-0x20(%rbp)
  jne    L60
  mov    -0x28(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x2f,%al
  jne    L60
  movl   $0x0,-0x20(%rbp)
L60:
  mov    $0x0,%ebx
  jmp    L17
L59:
  mov    %ebx,%eax
  lea    0x0(,%rax,8),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  add    $0x1,%eax
  mov    %eax,-0x1c(%rbp)
  mov    -0x20(%rbp),%edx
  mov    -0x1c(%rbp),%eax
  add    %edx,%eax
  add    $0x1,%eax
  mov    %eax,%eax
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,-0x18(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  jne    L68
  jmp    L69
L54:
  sub    $0x1,%ebx
  mov    %ebx,%eax
  lea    0x0(,%rax,8),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <free@plt>
L69:
  test   %ebx,%ebx
  jne    L54
  mov    $0x1,%eax
  jmp    L52
L68:
  mov    -0x20(%rbp),%edx
  mov    -0x28(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memmove@plt>
  mov    -0x20(%rbp),%edx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x2f,(%rax)
  mov    -0x1c(%rbp),%edx
  mov    -0x20(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,%ecx
  mov    -0x18(%rbp),%rax
  add    %rax,%rcx
  mov    %ebx,%eax
  lea    0x0(,%rax,8),%rsi
  mov    -0x30(%rbp),%rax
  add    %rsi,%rax
  mov    (%rax),%rax
  mov    %rax,%rsi
  mov    %rcx,%rdi
  call   <memmove@plt>
  mov    %ebx,%eax
  lea    0x0(,%rax,8),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    %ebx,%eax
  lea    0x0(,%rax,8),%rdx
  mov    -0x30(%rbp),%rax
  add    %rax,%rdx
  mov    -0x18(%rbp),%rax
  mov    %rax,(%rdx)
  add    $0x1,%ebx
L17:
  cmp    -0x34(%rbp),%ebx
  jb     L59
  mov    $0x0,%eax
L52:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int prefix_array(dirname, array, n) const char *dirname;
char **array;
unsigned int n;
{
  register unsigned int i;
  unsigned int dirlen = strlen(dirname);
  if (dirlen == 1 && dirname[0] == '/')
    dirlen = 0;
  for (i = 0; i < n; ++i) {
    unsigned int eltlen = strlen(array[i]) + 1;
    char *new = (char *)malloc(dirlen + 1 + eltlen);
    if (new == 0) {
      while (i > 0)
        free((void *)array[--i]);
      return 1;
    }
    bcopy((dirname), (new), (dirlen));
    new[dirlen] = '/';
    bcopy((array[i]), (&new[dirlen + 1]), (eltlen));
    free((void *)array[i]);
    array[i] = new;
  }
  return 0;
}
```