ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 这段汇编代码用C语言表示是什么样的?

```asm
<aloneBubbleSort>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %esi,0xc(%rsp)
  lea    -0x1(%rsi),%r15d
  test   %r15d,%r15d
  jle    L39
  mov    %rdi,%rbp
  mov    $0x0,%r14d
  mov    $0x0,%ebx
  movl   $0x0,0x8(%rsp)
  jmp    L64
L53:
  cmpl   $0x1,0x8(%rsp)
  je     L1
L76:
  add    $0x1,%ebx
  cmp    %r15d,%ebx
  jge    L39
L64:
  mov    %r14d,%esi
  lea    D16(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0xc(%rsp),%esi
  mov    %rbp,%rdi
  call   <printarray@plt>
  movslq %ebx,%rax
  shl    $0x2,%rax
  lea    0x0(%rbp,%rax,1),%r13
  mov    0x0(%r13),%esi
  lea    0x4(%rbp,%rax,1),%r12
  mov    (%r12),%edx
  cmp    %edx,%esi
  jle    L53
  cmpl   $0x0,0x8(%rsp)
  cmove  %ebx,%r14d
  lea    D12(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x0(%r13),%eax
  mov    (%r12),%edx
  mov    %edx,0x0(%r13)
  mov    %eax,(%r12)
  test   %ebx,%ebx
  jle    L11
  sub    $0x2,%ebx
  movl   $0x1,0x8(%rsp)
  jmp    L76
L1:
  mov    %r14d,%ebx
  movl   $0x0,0x8(%rsp)
  jmp    L76
L11:
  mov    $0x0,%ebx
  movl   $0x1,0x8(%rsp)
  jmp    L76
L39:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: unpack_data
arguments: tag: D16
datatype: string
name: unpack_data
arguments: tag: D12
datatype: string

tool: D16: first reverse place at %d	

tool: D12: exchange(%d,%d)


assistant: ```c
void aloneBubbleSort(int a[], int n) {
  int f = 0;
  int i, d = 0;
  for (i = 0; i < n - 1; i++) {
    printf("first reverse place at %d\t", d);
    printarray(a, n);
    if (a[i] > a[i + 1]) {
      if (f == 0) {
        d = i;
      }
      printf("exchange(%d,%d)\n", a[i], a[i + 1]);
      f = 1;
      int t = a[i];
      a[i] = a[i + 1];
      a[i + 1] = t;
      if (i > 0) {
        i -= 2;
      } else {
        i = 0;
      }
    } else {
      if (f == 1) {
        i = d;
        f = 0;
      }
    }
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<aloneBubbleSort>:
  endbr64
  push   %r15
  lea    -0x1(%rsi),%eax
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %eax,0x1c(%rsp)
  test   %eax,%eax
  jle    L29
  mov    %rdi,%rbx
  mov    %esi,%r13d
  xor    %ebp,%ebp
  xor    %r14d,%r14d
  xor    %r12d,%r12d
  lea    D16(%rip),%r15
  jmp    L55
  nopl   0x0(%rax,%rax,1)
L61:
  test   %r12d,%r12d
  lea    D9(%rip),%rdi
  cmove  %r14d,%ebp
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x10(%rsp),%r9
  mov    0x8(%rsp),%r8
  mov    (%r9),%eax
  mov    (%r8),%edx
  mov    %edx,(%r9)
  mov    %eax,(%r8)
  test   %r14d,%r14d
  je     L23
  sub    $0x1,%r14d
  mov    $0x1,%r12d
L42:
  cmp    %r14d,0x1c(%rsp)
  jle    L29
L55:
  mov    %ebp,%esi
  mov    %r15,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r13d,%esi
  mov    %rbx,%rdi
  call   <printarray@plt>
  movslq %r14d,%rax
  shl    $0x2,%rax
  lea    (%rbx,%rax,1),%r9
  lea    0x4(%rbx,%rax,1),%r8
  mov    (%r9),%esi
  mov    (%r8),%edx
  mov    %r9,0x10(%rsp)
  mov    %r8,0x8(%rsp)
  cmp    %edx,%esi
  jg     L61
  add    $0x1,%r14d
  cmp    $0x1,%r12d
  jne    L42
  lea    0x1(%rbp),%r14d
  xor    %r12d,%r12d
  cmp    %r14d,0x1c(%rsp)
  jg     L55
L29:
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax,%rax,1)
L23:
  mov    $0x1,%r14d
  mov    $0x1,%r12d
  jmp    L42
```

tool_calls: name: parse_data
arguments: label: D16
type: string
name: parse_data
arguments: label: D9
type: string

tool: D16: first reverse place at %d	

tool: D9: exchange(%d,%d)


assistant: ```c
void aloneBubbleSort(int a[], int n) {
  int f = 0;
  int i, d = 0;
  for (i = 0; i < n - 1; i++) {
    printf("first reverse place at %d\t", d);
    printarray(a, n);
    if (a[i] > a[i + 1]) {
      if (f == 0) {
        d = i;
      }
      printf("exchange(%d,%d)\n", a[i], a[i + 1]);
      f = 1;
      int t = a[i];
      a[i] = a[i + 1];
      a[i + 1] = t;
      if (i > 0) {
        i -= 2;
      } else {
        i = 0;
      }
    } else {
      if (f == 1) {
        i = d;
        f = 0;
      }
    }
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<aloneBubbleSort>:
  endbr64
  push   %r15
  lea    -0x1(%rsi),%eax
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %eax,0xc(%rsp)
  test   %eax,%eax
  jle    L60
  mov    %rdi,%rbx
  mov    %esi,%r13d
  xor    %ebp,%ebp
  xor    %r14d,%r14d
  xor    %r12d,%r12d
  lea    D3(%rip),%r15
  jmp    L52
  nopl   0x0(%rax,%rax,1)
L65:
  test   %r12d,%r12d
  lea    D6(%rip),%rdi
  cmove  %r14d,%ebp
  xor    %eax,%eax
  call   <printf@plt>
  mov    (%rsp),%r8
  rolq   $0x20,(%r8)
  test   %r14d,%r14d
  je     L26
  sub    $0x1,%r14d
  mov    $0x1,%r12d
L1:
  cmp    %r14d,0xc(%rsp)
  jle    L60
L52:
  mov    %ebp,%esi
  mov    %r15,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r13d,%esi
  mov    %rbx,%rdi
  call   <printarray@plt>
  movslq %r14d,%rax
  shl    $0x2,%rax
  lea    (%rbx,%rax,1),%r8
  mov    0x4(%rbx,%rax,1),%edx
  mov    (%r8),%esi
  mov    %r8,(%rsp)
  cmp    %edx,%esi
  jg     L65
  add    $0x1,%r14d
  cmp    $0x1,%r12d
  jne    L1
  lea    0x1(%rbp),%r14d
  xor    %r12d,%r12d
  cmp    %r14d,0xc(%rsp)
  jg     L52
L60:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax,%rax,1)
L26:
  mov    $0x1,%r14d
  mov    $0x1,%r12d
  jmp    L1
```

tool_calls: name: parse_data
arguments: tag: D3
type: string
name: parse_data
arguments: tag: D6
type: string

tool: D3: first reverse place at %d	

tool: D6: exchange(%d,%d)


assistant: ```c
void aloneBubbleSort(int a[], int n) {
  int f = 0;
  int i, d = 0;
  for (i = 0; i < n - 1; i++) {
    printf("first reverse place at %d\t", d);
    printarray(a, n);
    if (a[i] > a[i + 1]) {
      if (f == 0) {
        d = i;
      }
      printf("exchange(%d,%d)\n", a[i], a[i + 1]);
      f = 1;
      int t = a[i];
      a[i] = a[i + 1];
      a[i + 1] = t;
      if (i > 0) {
        i -= 2;
      } else {
        i = 0;
      }
    } else {
      if (f == 1) {
        i = d;
        f = 0;
      }
    }
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<UARTCharGetNonBlocking>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  movl   $0x0,-0x4(%rbp)
  movb   $0xff,-0x5(%rbp)
  mov    -0x14(%rbp),%eax
  mov    $0x7f,%esi
  mov    %eax,%edi
  call   <UARTRegConfigModeEnable@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x14(%rbp),%eax
  add    $0x14,%eax
  mov    %eax,%eax
  mov    (%rax),%eax
  and    $0x1,%eax
  test   %eax,%eax
  je     L2
  mov    -0x14(%rbp),%eax
  mov    (%rax),%eax
  mov    %al,-0x5(%rbp)
L2:
  mov    -0x14(%rbp),%eax
  add    $0xc,%eax
  mov    %eax,%eax
  mov    %rax,%rdx
  mov    -0x4(%rbp),%eax
  mov    %eax,(%rdx)
  movzbl -0x5(%rbp),%eax
  leave
  ret
```

assistant: ```c
signed char UARTCharGetNonBlocking(unsigned int baseAdd) {
  unsigned int lcrRegValue = 0;
  signed char retVal = -1;
  lcrRegValue = UARTRegConfigModeEnable(baseAdd, (0x007F));
  if ((*((volatile unsigned int *)(baseAdd + (0x14)))) & (0x00000001u)) {
    retVal = (signed char)(*((volatile unsigned int *)(baseAdd + (0x0))));
  }
  (*((volatile unsigned int *)(baseAdd + (0xC)))) = lcrRegValue;
  return retVal;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<UARTCharGetNonBlocking>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  mov    $0x7f,%esi
  call   <UARTRegConfigModeEnable@plt>
  mov    %eax,%edx
  mov    0x14(%ebx),%eax
  test   $0x1,%al
  je     L3
  mov    (%ebx),%eax
L5:
  mov    %edx,0xc(%ebx)
  pop    %rbx
  ret
L3:
  mov    $0xffffffff,%eax
  jmp    L5
```

assistant: ```c
signed char UARTCharGetNonBlocking(unsigned int baseAdd) {
  unsigned int lcrRegValue = 0;
  signed char retVal = -1;
  lcrRegValue = UARTRegConfigModeEnable(baseAdd, (0x007F));
  if ((*((volatile unsigned int *)(baseAdd + (0x14)))) & (0x00000001u)) {
    retVal = (signed char)(*((volatile unsigned int *)(baseAdd + (0x0))));
  }
  (*((volatile unsigned int *)(baseAdd + (0xC)))) = lcrRegValue;
  return retVal;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<UARTCharGetNonBlocking>:
  endbr64
  push   %rbx
  mov    $0x7f,%esi
  mov    %edi,%ebx
  call   <UARTRegConfigModeEnable@plt>
  mov    %eax,%edx
  mov    0x14(%ebx),%eax
  test   $0x1,%al
  je     L2
  mov    (%ebx),%eax
L18:
  mov    %edx,0xc(%ebx)
  pop    %rbx
  ret
  nopl   0x0(%rax)
L2:
  mov    $0xffffffff,%eax
  jmp    L18
```

assistant: ```c
signed char UARTCharGetNonBlocking(unsigned int baseAdd) {
  unsigned int lcrRegValue = 0;
  signed char retVal = -1;
  lcrRegValue = UARTRegConfigModeEnable(baseAdd, (0x007F));
  if ((*((volatile unsigned int *)(baseAdd + (0x14)))) & (0x00000001u)) {
    retVal = (signed char)(*((volatile unsigned int *)(baseAdd + (0x0))));
  }
  (*((volatile unsigned int *)(baseAdd + (0xC)))) = lcrRegValue;
  return retVal;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<UARTCharGetNonBlocking>:
  endbr64
  push   %rbx
  mov    $0x7f,%esi
  mov    %edi,%ebx
  call   <UARTRegConfigModeEnable@plt>
  mov    %eax,%edx
  mov    0x14(%ebx),%eax
  test   $0x1,%al
  je     L3
  mov    (%ebx),%eax
L7:
  mov    %edx,0xc(%ebx)
  pop    %rbx
  ret
  nopl   0x0(%rax)
L3:
  mov    $0xffffffff,%eax
  jmp    L7
```

assistant: ```c
signed char UARTCharGetNonBlocking(unsigned int baseAdd) {
  unsigned int lcrRegValue = 0;
  signed char retVal = -1;
  lcrRegValue = UARTRegConfigModeEnable(baseAdd, (0x007F));
  if ((*((volatile unsigned int *)(baseAdd + (0x14)))) & (0x00000001u)) {
    retVal = (signed char)(*((volatile unsigned int *)(baseAdd + (0x0))));
  }
  (*((volatile unsigned int *)(baseAdd + (0xC)))) = lcrRegValue;
  return retVal;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<Xascii>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <Xstrlen@plt>
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0x10(%rbp)
  jmp    L48
L25:
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    $0x1f,%eax
  jle    L11
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    $0x7e,%eax
  jg     L11
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x10(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x8(%rbp),%rax
  add    %rcx,%rax
  mov    %dl,(%rax)
  jmp    L19
L11:
  mov    -0x10(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movb   $0x3f,(%rax)
L19:
  addl   $0x1,-0x10(%rbp)
L48:
  mov    -0x10(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jl     L25
  mov    -0x10(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *Xascii(XML_Char *s) {
  char *res;
  int x;
  int len = Xstrlen(s);
  res = malloc(sizeof(char) * (len + 1));
  for (x = 0; x < len; x++) {
    if (s[x] >= 32 && s[x] < 127)
      res[x] = s[x];
    else
      res[x] = '?';
  }
  res[x] = 0;
  return res;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<Xascii>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  call   <Xstrlen@plt>
  mov    %eax,%ebp
  lea    0x1(%rax),%edi
  call   <malloc@plt>
  test   %ebp,%ebp
  jle    L19
  mov    %ebp,%r8d
  mov    $0x0,%edx
  mov    $0x3f,%edi
L23:
  mov    (%rbx,%rdx,4),%ecx
  lea    -0x20(%rcx),%esi
  cmp    $0x5e,%esi
  cmova  %edi,%ecx
  mov    %cl,(%rax,%rdx,1)
  add    $0x1,%rdx
  cmp    %r8,%rdx
  jne    L23
L28:
  movslq %ebp,%rbp
  movb   $0x0,(%rax,%rbp,1)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L19:
  mov    $0x0,%ebp
  jmp    L28
```

assistant: ```c
char *Xascii(XML_Char *s) {
  char *res;
  int x;
  int len = Xstrlen(s);
  res = malloc(sizeof(char) * (len + 1));
  for (x = 0; x < len; x++) {
    if (s[x] >= 32 && s[x] < 127)
      res[x] = s[x];
    else
      res[x] = '?';
  }
  res[x] = 0;
  return res;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<Xascii>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <Xstrlen@plt>
  mov    %eax,%ebx
  lea    0x1(%rax),%edi
  call   <malloc@plt>
  test   %ebx,%ebx
  jle    L5
  movslq %ebx,%rsi
  xor    %edx,%edx
  mov    $0x3f,%r8d
  nopl   0x0(%rax,%rax,1)
L8:
  mov    0x0(%rbp,%rdx,4),%ecx
  lea    -0x20(%rcx),%edi
  cmp    $0x5f,%edi
  cmovae %r8d,%ecx
  mov    %cl,(%rax,%rdx,1)
  add    $0x1,%rdx
  cmp    %rdx,%rsi
  jne    L8
  add    %rax,%rsi
  movb   $0x0,(%rsi)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopw   0x0(%rax,%rax,1)
L5:
  mov    %rax,%rsi
  movb   $0x0,(%rsi)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
char *Xascii(XML_Char *s) {
  char *res;
  int x;
  int len = Xstrlen(s);
  res = malloc(sizeof(char) * (len + 1));
  for (x = 0; x < len; x++) {
    if (s[x] >= 32 && s[x] < 127)
      res[x] = s[x];
    else
      res[x] = '?';
  }
  res[x] = 0;
  return res;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<read_history>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x30,%rsp
  mov    %rdi,-0x1028(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x1028(%rbp),%rax
  lea    D2(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <fopen@plt>
  mov    %rax,-0x1018(%rbp)
  cmpq   $0x0,-0x1018(%rbp)
  jne    L49
  mov    $0xffffffff,%eax
  jmp    L45
L38:
  lea    -0x1010(%rbp),%rax
  mov    $0xd,%esi
  mov    %rax,%rdi
  call   <strchr@plt>
  mov    %rax,-0x1020(%rbp)
  cmpq   $0x0,-0x1020(%rbp)
  jne    L42
  lea    -0x1010(%rbp),%rax
  mov    $0xa,%esi
  mov    %rax,%rdi
  call   <strchr@plt>
  mov    %rax,-0x1020(%rbp)
L42:
  cmpq   $0x0,-0x1020(%rbp)
  je     L18
  mov    -0x1020(%rbp),%rax
  movb   $0x0,(%rax)
L18:
  lea    -0x1010(%rbp),%rax
  mov    %rax,%rdi
  call   <add_history@plt>
L49:
  mov    -0x1018(%rbp),%rdx
  lea    -0x1010(%rbp),%rax
  mov    $0x1000,%esi
  mov    %rax,%rdi
  call   <fgets@plt>
  test   %rax,%rax
  jne    L38
  mov    -0x1018(%rbp),%rax
  mov    %rax,%rdi
  call   <fclose@plt>
  mov    $0x0,%eax
L45:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L59
  call   <__stack_chk_fail@plt>
L59:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: label: D2
datatype: string

tool: D2: r

assistant: ```c
int read_history(const char *filename) {
  FILE *fp = fopen(filename, "r");
  char buf[4096];
  if (fp == NULL)
    return -1;
  while (fgets(buf, 4096, fp) != NULL) {
    char *p;
    p = strchr(buf, '\r');
    if (!p)
      p = strchr(buf, '\n');
    if (p)
      *p = '\0';
    add_history(buf);
  }
  fclose(fp);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<read_history>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x1008(%rsp)
  xor    %eax,%eax
  lea    D3(%rip),%rsi
  call   <fopen@plt>
  test   %rax,%rax
  je     L57
  mov    %rax,%rbp
  mov    %rsp,%rbx
  jmp    L2
L0:
  mov    $0xa,%esi
  mov    %rbx,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  jne    L1
L11:
  mov    %rbx,%rdi
  call   <add_history@plt>
L2:
  mov    %rbp,%rdx
  mov    $0x1000,%esi
  mov    %rbx,%rdi
  call   <fgets@plt>
  test   %rax,%rax
  je     L17
  mov    $0xd,%esi
  mov    %rbx,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  je     L0
L1:
  movb   $0x0,(%rax)
  jmp    L11
L17:
  mov    %rbp,%rdi
  call   <fclose@plt>
  mov    $0x0,%eax
L41:
  mov    0x1008(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L30
  add    $0x1018,%rsp
  pop    %rbx
  pop    %rbp
  ret
L57:
  mov    $0xffffffff,%eax
  jmp    L41
L30:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D3
type: string

tool: D3: r

assistant: ```c
int read_history(const char *filename) {
  FILE *fp = fopen(filename, "r");
  char buf[4096];
  if (fp == NULL)
    return -1;
  while (fgets(buf, 4096, fp) != NULL) {
    char *p;
    p = strchr(buf, '\r');
    if (!p)
      p = strchr(buf, '\n');
    if (p)
      *p = '\0';
    add_history(buf);
  }
  fclose(fp);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<read_history>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x18,%rsp
  lea    D4(%rip),%rsi
  mov    %fs:0x28,%rax
  mov    %rax,0x1008(%rsp)
  xor    %eax,%eax
  call   <fopen@plt>
  test   %rax,%rax
  je     L59
  mov    %rax,%rbp
  mov    %rsp,%rbx
  jmp    L50
  nopw   0x0(%rax,%rax,1)
L6:
  movb   $0x0,(%rax)
L69:
  mov    %rbx,%rdi
  call   <add_history@plt>
L50:
  mov    %rbp,%rdx
  mov    $0x1000,%esi
  mov    %rbx,%rdi
  call   <fgets@plt>
  test   %rax,%rax
  je     L11
  mov    $0xd,%esi
  mov    %rbx,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  jne    L6
  mov    $0xa,%esi
  mov    %rbx,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  je     L69
  jmp    L6
  xchg   %ax,%ax
L11:
  mov    %rbp,%rdi
  call   <fclose@plt>
  xor    %eax,%eax
L55:
  mov    0x1008(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L52
  add    $0x1018,%rsp
  pop    %rbx
  pop    %rbp
  ret
L59:
  mov    $0xffffffff,%eax
  jmp    L55
L52:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D4
datatype: string

tool: D4: r

assistant: ```c
int read_history(const char *filename) {
  FILE *fp = fopen(filename, "r");
  char buf[4096];
  if (fp == NULL)
    return -1;
  while (fgets(buf, 4096, fp) != NULL) {
    char *p;
    p = strchr(buf, '\r');
    if (!p)
      p = strchr(buf, '\n');
    if (p)
      *p = '\0';
    add_history(buf);
  }
  fclose(fp);
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<read_history>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x18,%rsp
  lea    D3(%rip),%rsi
  mov    %fs:0x28,%rax
  mov    %rax,0x1008(%rsp)
  xor    %eax,%eax
  call   <fopen@plt>
  test   %rax,%rax
  je     L78
  mov    %rax,%rbp
  mov    %rsp,%rbx
  jmp    L46
  nopw   0x0(%rax,%rax,1)
L28:
  movb   $0x0,(%rax)
L52:
  mov    %rbx,%rdi
  call   <add_history@plt>
L46:
  mov    %rbp,%rdx
  mov    $0x1000,%esi
  mov    %rbx,%rdi
  call   <fgets@plt>
  test   %rax,%rax
  je     L40
  mov    $0xd,%esi
  mov    %rbx,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  jne    L28
  mov    $0xa,%esi
  mov    %rbx,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  je     L52
  jmp    L28
  xchg   %ax,%ax
L40:
  mov    %rbp,%rdi
  call   <fclose@plt>
  xor    %eax,%eax
L74:
  mov    0x1008(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L30
  add    $0x1018,%rsp
  pop    %rbx
  pop    %rbp
  ret
L78:
  mov    $0xffffffff,%eax
  jmp    L74
L30:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D3
type_guessed: string

tool: D3: r

assistant: ```c
int read_history(const char *filename) {
  FILE *fp = fopen(filename, "r");
  char buf[4096];
  if (fp == NULL)
    return -1;
  while (fgets(buf, 4096, fp) != NULL) {
    char *p;
    p = strchr(buf, '\r');
    if (!p)
      p = strchr(buf, '\n');
    if (p)
      *p = '\0';
    add_history(buf);
  }
  fclose(fp);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<CheckBrand>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x6,-0x18(%rbp)
  jmp    L50
L30:
  mov    -0x18(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    -0x18(%rbp),%edx
  lea    -0x6(%rdx),%ecx
  movzbl (%rax),%edx
  movslq %ecx,%rax
  mov    %dl,-0x12(%rbp,%rax,1)
  addl   $0x1,-0x18(%rbp)
L50:
  mov    -0x18(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x2d,%al
  jne    L30
  lea    -0x12(%rbp),%rax
  lea    D9(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jne    L21
  lea    D58(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L21:
  lea    -0x12(%rbp),%rax
  lea    D69(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jne    L64
  lea    D50(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L64:
  lea    -0x12(%rbp),%rax
  lea    D8(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jne    L9
  lea    D62(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L9:
  lea    -0x12(%rbp),%rax
  lea    D30(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jne    L61
  lea    D34(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L61:
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L55
  call   <__stack_chk_fail@plt>
L55:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D9
datatype: string
name: retrieve_data
arguments: data_tag: D58
datatype: string
name: retrieve_data
arguments: data_tag: D69
datatype: string
name: retrieve_data
arguments: data_tag: D50
datatype: string
name: retrieve_data
arguments: data_tag: D8
datatype: string
name: retrieve_data
arguments: data_tag: D62
datatype: string
name: retrieve_data
arguments: data_tag: D30
datatype: string
name: retrieve_data
arguments: data_tag: D34
datatype: string

tool: D9: 933346

tool: D58: 
Bluebridge

tool: D69: 55583

tool: D50: 
Alyson Books

tool: D8: 4210

tool: D62: 
FUNimation Productions, Ltd.

tool: D30: 01

tool: D34: 
Pyramid Books

assistant: ```c
void CheckBrand(char *digits) {
  int i = 6;
  char a[10];
  for (; digits[i] != '-'; i++)
    a[i - 6] = digits[i];
  if (strcmp(a, "933346") == 0)
    printf("\nBluebridge");
  if (strcmp(a, "55583") == 0)
    printf("\nAlyson Books");
  if (strcmp(a, "4210") == 0)
    printf("\nFUNimation Productions, Ltd.");
  if (strcmp(a, "01") == 0)
    printf("\nPyramid Books");
}
```