ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What is the c source code of the assembly code below:

```asm
<isnumber>:
  endbr64
  push   %rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  test   %rdi,%rdi
  je     L7
  mov    %rdi,%rbp
  movsbl (%rdi),%edi
  test   %dil,%dil
  jne    L37
L7:
  xor    %eax,%eax
L31:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L10
  add    $0x10,%rsp
  pop    %rbp
  ret
  nopl   0x0(%rax)
L37:
  call   <isspace@plt>
  test   %rax,%rax
  jne    L7
  mov    %rsp,%rsi
  mov    %rbp,%rdi
  call   <strtod@plt>
  mov    (%rsp),%rax
  cmpb   $0x0,(%rax)
  sete   %al
  movzbl %al,%eax
  jmp    L31
L10:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int isnumber(const char *s) {
  if (s == NULL || *s == '\0' || isspace(*s))
    return 0;
  char *p;
  strtod(s, &p);
  return *p == '\0';
}
```
user: Translate the below assembly code into C programming language:

```asm
<ispandigital>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x30,%rsp
  mov    %edi,%ebx
  mov    %esi,%ebp
  mov    %edx,%r12d
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  mov    $0x28,%edx
  mov    $0x0,%esi
  call   <memset@plt>
L105:
  test   %ebx,%ebx
  je     L15
  movslq %ebx,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  mov    %ebx,%edx
  sar    $0x1f,%edx
  sub    %edx,%eax
  mov    %eax,%ecx
  lea    (%rax,%rax,4),%edx
  add    %edx,%edx
  mov    %ebx,%eax
  sub    %edx,%eax
  mov    %ecx,%ebx
  cltq
  mov    (%rsp,%rax,4),%edi
  lea    0x1(%rdi),%edx
  mov    %edx,(%rsp,%rax,4)
  cmp    $0x1,%edx
  jle    L105
  mov    $0x0,%ecx
L74:
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L150
  mov    %ecx,%eax
  add    $0x30,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L15:
  mov    (%rsp),%ecx
  test   %ecx,%ecx
  jne    L136
L146:
  test   %ebp,%ebp
  je     L59
  movslq %ebp,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  mov    %ebp,%edx
  sar    $0x1f,%edx
  sub    %edx,%eax
  mov    %eax,%esi
  lea    (%rax,%rax,4),%edx
  add    %edx,%edx
  mov    %ebp,%eax
  sub    %edx,%eax
  mov    %esi,%ebp
  cltq
  mov    (%rsp,%rax,4),%esi
  lea    0x1(%rsi),%edx
  mov    %edx,(%rsp,%rax,4)
  cmp    $0x1,%edx
  jle    L146
  jmp    L74
L59:
  mov    (%rsp),%ecx
  test   %ecx,%ecx
  jne    L29
L164:
  test   %r12d,%r12d
  je     L162
  movslq %r12d,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  mov    %r12d,%edx
  sar    $0x1f,%edx
  sub    %edx,%eax
  mov    %eax,%esi
  lea    (%rax,%rax,4),%edx
  add    %edx,%edx
  mov    %r12d,%eax
  sub    %edx,%eax
  mov    %esi,%r12d
  cltq
  mov    (%rsp,%rax,4),%esi
  lea    0x1(%rsi),%edx
  mov    %edx,(%rsp,%rax,4)
  cmp    $0x1,%edx
  jle    L164
  jmp    L74
L162:
  mov    (%rsp),%esi
  mov    %r12d,%ecx
  test   %esi,%esi
  jne    L74
  lea    0x4(%rsp),%rax
  lea    0x28(%rsp),%rdx
L134:
  mov    (%rax),%ecx
  cmp    $0x1,%ecx
  jne    L124
  add    $0x4,%rax
  cmp    %rdx,%rax
  jne    L134
  jmp    L74
L136:
  mov    %ebx,%ecx
  jmp    L74
L29:
  mov    %ebp,%ecx
  jmp    L74
L124:
  mov    %esi,%ecx
  jmp    L74
L150:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int ispandigital(int a, int b, int c) {
  int counts[10];
  int i;
  memset(counts, 0, sizeof(counts));
  while (a) {
    int digit = a % 10;
    a /= 10;
    counts[digit]++;
    if (counts[digit] > 1)
      return 0;
  }
  if (counts[0])
    return 0;
  while (b) {
    int digit = b % 10;
    b /= 10;
    counts[digit]++;
    if (counts[digit] > 1)
      return 0;
  }
  if (counts[0])
    return 0;
  while (c) {
    int digit = c % 10;
    c /= 10;
    counts[digit]++;
    if (counts[digit] > 1)
      return 0;
  }
  if (counts[0])
    return 0;
  for (i = 1; i < 10; ++i) {
    if (counts[i] != 1)
      return 0;
  }
  return 1;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<ispandigital>:
  endbr64
  push   %r12
  mov    %edx,%r12d
  mov    $0x28,%edx
  push   %rbp
  mov    %esi,%ebp
  xor    %esi,%esi
  push   %rbx
  mov    %edi,%ebx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  call   <memset@plt>
  jmp    L55
  nopl   0x0(%rax)
L124:
  movslq %ebx,%rax
  mov    %ebx,%edx
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%edx
  sar    $0x22,%rax
  sub    %edx,%eax
  lea    (%rax,%rax,4),%ecx
  mov    %eax,%edx
  mov    %ebx,%eax
  add    %ecx,%ecx
  mov    %edx,%ebx
  sub    %ecx,%eax
  cltq
  mov    (%rsp,%rax,4),%esi
  lea    0x1(%rsi),%edx
  mov    %edx,(%rsp,%rax,4)
  cmp    $0x1,%edx
  jg     L116
L55:
  test   %ebx,%ebx
  jne    L124
  mov    (%rsp),%ecx
  test   %ecx,%ecx
  je     L77
  nopl   (%rax)
L116:
  xor    %eax,%eax
L42:
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L138
  add    $0x30,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L159:
  movslq %ebp,%rax
  mov    %ebp,%edx
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%edx
  sar    $0x22,%rax
  sub    %edx,%eax
  lea    (%rax,%rax,4),%ecx
  mov    %eax,%edx
  mov    %ebp,%eax
  add    %ecx,%ecx
  mov    %edx,%ebp
  sub    %ecx,%eax
  cltq
  mov    (%rsp,%rax,4),%edi
  lea    0x1(%rdi),%edx
  mov    %edx,(%rsp,%rax,4)
  cmp    $0x1,%edx
  jg     L116
L77:
  test   %ebp,%ebp
  jne    L159
  mov    (%rsp),%edx
  test   %edx,%edx
  je     L48
  jmp    L116
  nop
L147:
  movslq %r12d,%rax
  mov    %r12d,%edx
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%edx
  sar    $0x22,%rax
  sub    %edx,%eax
  lea    (%rax,%rax,4),%ecx
  mov    %eax,%edx
  mov    %r12d,%eax
  add    %ecx,%ecx
  mov    %edx,%r12d
  sub    %ecx,%eax
  cltq
  mov    (%rsp,%rax,4),%edi
  lea    0x1(%rdi),%edx
  mov    %edx,(%rsp,%rax,4)
  cmp    $0x1,%edx
  jg     L116
L48:
  test   %r12d,%r12d
  jne    L147
  mov    (%rsp),%eax
  test   %eax,%eax
  jne    L116
  lea    0x4(%rsp),%rdx
  lea    0x28(%rsp),%rcx
  jmp    L80
  nopl   (%rax)
L79:
  add    $0x4,%rdx
  cmp    %rcx,%rdx
  je     L42
L80:
  mov    (%rdx),%eax
  cmp    $0x1,%eax
  je     L79
  jmp    L116
L138:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int ispandigital(int a, int b, int c) {
  int counts[10];
  int i;
  memset(counts, 0, sizeof(counts));
  while (a) {
    int digit = a % 10;
    a /= 10;
    counts[digit]++;
    if (counts[digit] > 1)
      return 0;
  }
  if (counts[0])
    return 0;
  while (b) {
    int digit = b % 10;
    b /= 10;
    counts[digit]++;
    if (counts[digit] > 1)
      return 0;
  }
  if (counts[0])
    return 0;
  while (c) {
    int digit = c % 10;
    c /= 10;
    counts[digit]++;
    if (counts[digit] > 1)
      return 0;
  }
  if (counts[0])
    return 0;
  for (i = 1; i < 10; ++i) {
    if (counts[i] != 1)
      return 0;
  }
  return 1;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<ispandigital>:
  endbr64
  push   %r12
  mov    %edx,%r12d
  mov    $0x28,%edx
  push   %rbp
  mov    %esi,%ebp
  xor    %esi,%esi
  push   %rbx
  mov    %edi,%ebx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  call   <memset@plt>
  jmp    L140
  nopl   0x0(%rax)
L176:
  movslq %ebx,%rax
  mov    %ebx,%edx
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%edx
  sar    $0x22,%rax
  sub    %edx,%eax
  lea    (%rax,%rax,4),%ecx
  mov    %eax,%edx
  mov    %ebx,%eax
  add    %ecx,%ecx
  mov    %edx,%ebx
  sub    %ecx,%eax
  cltq
  mov    (%rsp,%rax,4),%esi
  lea    0x1(%rsi),%edx
  mov    %edx,(%rsp,%rax,4)
  cmp    $0x1,%edx
  jg     L67
L140:
  test   %ebx,%ebx
  jne    L176
  mov    (%rsp),%ecx
  test   %ecx,%ecx
  je     L75
  nopl   (%rax)
L67:
  xor    %eax,%eax
L132:
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L205
  add    $0x30,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L97:
  movslq %ebp,%rax
  mov    %ebp,%edx
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%edx
  sar    $0x22,%rax
  sub    %edx,%eax
  lea    (%rax,%rax,4),%ecx
  mov    %eax,%edx
  mov    %ebp,%eax
  add    %ecx,%ecx
  mov    %edx,%ebp
  sub    %ecx,%eax
  cltq
  mov    (%rsp,%rax,4),%edi
  lea    0x1(%rdi),%edx
  mov    %edx,(%rsp,%rax,4)
  cmp    $0x1,%edx
  jg     L67
L75:
  test   %ebp,%ebp
  jne    L97
  mov    (%rsp),%edx
  test   %edx,%edx
  je     L145
  jmp    L67
  nop
L10:
  movslq %r12d,%rax
  mov    %r12d,%edx
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%edx
  sar    $0x22,%rax
  sub    %edx,%eax
  lea    (%rax,%rax,4),%ecx
  mov    %eax,%edx
  mov    %r12d,%eax
  add    %ecx,%ecx
  mov    %edx,%r12d
  sub    %ecx,%eax
  cltq
  mov    (%rsp,%rax,4),%edi
  lea    0x1(%rdi),%edx
  mov    %edx,(%rsp,%rax,4)
  cmp    $0x1,%edx
  jg     L67
L145:
  test   %r12d,%r12d
  jne    L10
  mov    (%rsp),%eax
  test   %eax,%eax
  jne    L67
  cmpl   $0x1,0x4(%rsp)
  jne    L67
  cmpl   $0x1,0x8(%rsp)
  jne    L67
  cmpl   $0x1,0xc(%rsp)
  jne    L67
  cmpl   $0x1,0x10(%rsp)
  jne    L67
  cmpl   $0x1,0x14(%rsp)
  jne    L67
  cmpl   $0x1,0x18(%rsp)
  jne    L67
  cmpl   $0x1,0x1c(%rsp)
  jne    L67
  cmpl   $0x1,0x20(%rsp)
  jne    L67
  mov    0x24(%rsp),%eax
  cmp    $0x1,%eax
  je     L132
  jmp    L67
L205:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int ispandigital(int a, int b, int c) {
  int counts[10];
  int i;
  memset(counts, 0, sizeof(counts));
  while (a) {
    int digit = a % 10;
    a /= 10;
    counts[digit]++;
    if (counts[digit] > 1)
      return 0;
  }
  if (counts[0])
    return 0;
  while (b) {
    int digit = b % 10;
    b /= 10;
    counts[digit]++;
    if (counts[digit] > 1)
      return 0;
  }
  if (counts[0])
    return 0;
  while (c) {
    int digit = c % 10;
    c /= 10;
    counts[digit]++;
    if (counts[digit] > 1)
      return 0;
  }
  if (counts[0])
    return 0;
  for (i = 1; i < 10; ++i) {
    if (counts[i] != 1)
      return 0;
  }
  return 1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<reverse>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0xc(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  sub    $0x1,%eax
  mov    %eax,-0x8(%rbp)
  jmp    L0
L17:
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0xc(%rbp),%edx
  movslq %edx,%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x4(%rbp),%edx
  mov    %dl,(%rax)
  addl   $0x1,-0xc(%rbp)
  subl   $0x1,-0x8(%rbp)
L0:
  mov    -0xc(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jl     L17
  nop
  nop
  leave
  ret
```

assistant: ```c
void reverse(char s[]) {
  int i, j, tmp;
  for (i = 0, j = strlen(s) - 1; i < j; i++, j--) {
    tmp = s[i];
    s[i] = s[j];
    s[j] = tmp;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<reverse>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  sub    $0x1,%eax
  test   %eax,%eax
  jle    L5
  cltq
  mov    $0x0,%edx
L13:
  movzbl (%rbx,%rdx,1),%ecx
  movzbl (%rbx,%rax,1),%esi
  mov    %sil,(%rbx,%rdx,1)
  mov    %cl,(%rbx,%rax,1)
  add    $0x1,%rdx
  sub    $0x1,%rax
  cmp    %eax,%edx
  jl     L13
L5:
  pop    %rbx
  ret
```

assistant: ```c
void reverse(char s[]) {
  int i, j, tmp;
  for (i = 0, j = strlen(s) - 1; i < j; i++, j--) {
    tmp = s[i];
    s[i] = s[j];
    s[j] = tmp;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<reverse>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  sub    $0x1,%eax
  test   %eax,%eax
  jle    L8
  cltq
  xor    %edx,%edx
  nopl   0x0(%rax,%rax,1)
L10:
  movzbl (%rbx,%rdx,1),%ecx
  movzbl (%rbx,%rax,1),%esi
  mov    %sil,(%rbx,%rdx,1)
  add    $0x1,%rdx
  mov    %cl,(%rbx,%rax,1)
  sub    $0x1,%rax
  cmp    %eax,%edx
  jl     L10
L8:
  pop    %rbx
  ret
```

assistant: ```c
void reverse(char s[]) {
  int i, j, tmp;
  for (i = 0, j = strlen(s) - 1; i < j; i++, j--) {
    tmp = s[i];
    s[i] = s[j];
    s[j] = tmp;
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<reverse>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  sub    $0x1,%eax
  test   %eax,%eax
  jle    L11
  cltq
  xor    %edx,%edx
  nopl   0x0(%rax,%rax,1)
L9:
  movzbl (%rbx,%rdx,1),%ecx
  movzbl (%rbx,%rax,1),%esi
  mov    %sil,(%rbx,%rdx,1)
  add    $0x1,%rdx
  mov    %cl,(%rbx,%rax,1)
  sub    $0x1,%rax
  cmp    %edx,%eax
  jg     L9
L11:
  pop    %rbx
  ret
```

assistant: ```c
void reverse(char s[]) {
  int i, j, tmp;
  for (i = 0, j = strlen(s) - 1; i < j; i++, j--) {
    tmp = s[i];
    s[i] = s[j];
    s[j] = tmp;
  }
}
```
user: Translate the below assembly code into C programming language:

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

assistant: ```c
void stack_free(struct stack *theStack) {
  stack_clear(theStack);
  free(theStack);
  return;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<stack_free>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <stack_clear@plt>
  mov    %rbx,%rdi
  call   <free@plt>
  pop    %rbx
  ret
```

assistant: ```c
void stack_free(struct stack *theStack) {
  stack_clear(theStack);
  free(theStack);
  return;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<stack_free>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  call   <stack_clear@plt>
  mov    %rbp,%rdi
  pop    %rbp
  jmp    1060 <free@plt>
```

assistant: ```c
void stack_free(struct stack *theStack) {
  stack_clear(theStack);
  free(theStack);
  return;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<stack_free>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  call   <stack_clear@plt>
  mov    %rbp,%rdi
  pop    %rbp
  jmp    1060 <free@plt>
```

assistant: ```c
void stack_free(struct stack *theStack) {
  stack_clear(theStack);
  free(theStack);
  return;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  movq   $0x0,-0x10(%rbp)
  mov    $0x8,%edi
  call   <malloc@plt>
  mov    %rax,-0x10(%rbp)
  cmpq   $0x0,-0x10(%rbp)
  je     L27
  mov    -0x10(%rbp),%rax
  movl   $0x4,(%rax)
  mov    -0x10(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L27:
  mov    $0x28,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L25
  movl   $0x0,-0x14(%rbp)
  jmp    L59
L41:
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x8(%rbp),%rax
  add    %rax,%rdx
  mov    $0xa,%eax
  sub    -0x14(%rbp),%eax
  mov    %eax,(%rdx)
  addl   $0x1,-0x14(%rbp)
L59:
  cmpl   $0x9,-0x14(%rbp)
  jle    L41
  movl   $0x0,-0x14(%rbp)
  jmp    L37
L48:
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x8(%rbp),%rax
  lea    (%rcx,%rax,1),%rsi
  mov    -0x14(%rbp),%eax
  mov    %edx,%ecx
  mov    %eax,%edx
  lea    D17(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x14(%rbp)
L37:
  cmpl   $0x9,-0x14(%rbp)
  jle    L48
  mov    $0xa,%edi
  call   <putchar@plt>
L25:
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D3
type: string
name: extract_symbolic_data
arguments: data_tag: D17
type: string

tool: D3: %p	%d


tool: D17: %p	q[%d]=%3d


assistant: ```c
int main(void) {
  int *p = NULL;
  if ((p = malloc(8)) != NULL) {
    *p = 4;
    printf("%p\t%d\n", p, *p);
  }
  int *q;
  int i;
  q = malloc(10 * sizeof(int));
  if (NULL != q) {
    for (i = 0; i < 10; i++) {
      q[i] = 10 - i;
    }
    for (i = 0; i < 10; i++) {
      printf("%p\tq[%d]=%3d\n", &q[i], i, q[i]);
    }
    printf("\n");
  }
  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
  push   %r12
  push   %rbp
  push   %rbx
  mov    $0x8,%edi
  call   <malloc@plt>
  test   %rax,%rax
  je     L21
  movl   $0x4,(%rax)
  mov    $0x4,%edx
  mov    %rax,%rsi
  lea    D16(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L21:
  mov    $0x28,%edi
  call   <malloc@plt>
  test   %rax,%rax
  je     L2
  mov    %rax,%rbx
  mov    $0xa,%edx
L11:
  mov    %edx,(%rax)
  add    $0x4,%rax
  sub    $0x1,%edx
  jne    L11
  mov    $0x0,%ebp
  lea    D1(%rip),%r12
L29:
  mov    (%rbx),%ecx
  mov    %ebp,%edx
  mov    %rbx,%rsi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x1,%ebp
  add    $0x4,%rbx
  cmp    $0xa,%ebp
  jne    L29
  mov    $0xa,%edi
  call   <putchar@plt>
L2:
  mov    $0x0,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: read_data
arguments: data_label: D16
type_infered: string
name: read_data
arguments: data_label: D1
type_infered: string

tool: D16: %p	%d


tool: D1: %p	q[%d]=%3d


assistant: ```c
int main(void) {
  int *p = NULL;
  if ((p = malloc(8)) != NULL) {
    *p = 4;
    printf("%p\t%d\n", p, *p);
  }
  int *q;
  int i;
  q = malloc(10 * sizeof(int));
  if (NULL != q) {
    for (i = 0; i < 10; i++) {
      q[i] = 10 - i;
    }
    for (i = 0; i < 10; i++) {
      printf("%p\tq[%d]=%3d\n", &q[i], i, q[i]);
    }
    printf("\n");
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r12
  mov    $0x8,%edi
  push   %rbp
  push   %rbx
  call   <malloc@plt>
  test   %rax,%rax
  je     L18
  movl   $0x4,(%rax)
  mov    %rax,%rsi
  xor    %eax,%eax
  mov    $0x4,%edx
  lea    D11(%rip),%rdi
  call   <printf@plt>
L18:
  mov    $0x28,%edi
  call   <malloc@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L31
  mov    %rax,%rdx
  mov    $0xa,%eax
  nopl   (%rax)
L23:
  mov    %eax,(%rdx)
  add    $0x4,%rdx
  sub    $0x1,%eax
  jne    L23
  xor    %ebp,%ebp
  lea    D19(%rip),%r12
  nopl   0x0(%rax)
L39:
  mov    (%rbx),%ecx
  mov    %ebp,%edx
  mov    %rbx,%rsi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%ebp
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    $0xa,%ebp
  jne    L39
  mov    $0xa,%edi
  call   <putchar@plt>
L31:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D11
type_guessed: string
name: extract_symbolic_data
arguments: tag: D19
type_guessed: string

tool: D11: %p	%d


tool: D19: %p	q[%d]=%3d


assistant: ```c
int main(void) {
  int *p = NULL;
  if ((p = malloc(8)) != NULL) {
    *p = 4;
    printf("%p\t%d\n", p, *p);
  }
  int *q;
  int i;
  q = malloc(10 * sizeof(int));
  if (NULL != q) {
    for (i = 0; i < 10; i++) {
      q[i] = 10 - i;
    }
    for (i = 0; i < 10; i++) {
      printf("%p\tq[%d]=%3d\n", &q[i], i, q[i]);
    }
    printf("\n");
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  push   %r12
  mov    $0x8,%edi
  push   %rbp
  push   %rbx
  call   <malloc@plt>
  test   %rax,%rax
  je     L24
  movl   $0x4,(%rax)
  mov    %rax,%rsi
  xor    %eax,%eax
  mov    $0x4,%edx
  lea    D29(%rip),%rdi
  call   <printf@plt>
L24:
  mov    $0x28,%edi
  call   <malloc@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L14
  movdqa D25(%rip),%xmm0
  xor    %ebp,%ebp
  lea    D14(%rip),%r12
  movups %xmm0,(%rax)
  movdqa D17(%rip),%xmm0
  movups %xmm0,0x10(%rax)
  mov    D32(%rip),%rax
  mov    %rax,0x20(%rbx)
L11:
  mov    (%rbx),%ecx
  mov    %ebp,%edx
  mov    %rbx,%rsi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%ebp
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    $0xa,%ebp
  jne    L11
  mov    $0xa,%edi
  call   <putchar@plt>
L14:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D29
type_guessed: string
name: extract_symbolic_data
arguments: tag: D25
type_guessed: byte[16]
name: extract_symbolic_data
arguments: tag: D14
type_guessed: string
name: extract_symbolic_data
arguments: tag: D17
type_guessed: byte[16]
name: extract_symbolic_data
arguments: tag: D32
type_guessed: byte[8]

tool: D29: %p	%d


tool: {"D25": [0x0a,0x00,0x00,0x00,0x09,0x00,0x00,0x00,0x08,0x00,0x00,0x00,0x07,0x00,0x00,0x00]}

tool: D14: %p	q[%d]=%3d


tool: {"D17": [0x06,0x00,0x00,0x00,0x05,0x00,0x00,0x00,0x04,0x00,0x00,0x00,0x03,0x00,0x00,0x00]}

tool: {"D32": [0x02,0x00,0x00,0x00,0x01,0x00,0x00,0x00]}

assistant: ```c
int main(void) {
  int *p = NULL;
  if ((p = malloc(8)) != NULL) {
    *p = 4;
    printf("%p\t%d\n", p, *p);
  }
  int *q;
  int i;
  q = malloc(10 * sizeof(int));
  if (NULL != q) {
    for (i = 0; i < 10; i++) {
      q[i] = 10 - i;
    }
    for (i = 0; i < 10; i++) {
      printf("%p\tq[%d]=%3d\n", &q[i], i, q[i]);
    }
    printf("\n");
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<display>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0x4,0x4(%rsp)
  lea    D78(%rip),%rdi
  call   <printf@plt>
  lea    D28(%rip),%rdi
  call   <printf@plt>
  lea    D80(%rip),%r13
  lea    D4(%rip),%r12
  lea    D45(%rip),%rbp
  lea    D70(%rip),%rbx
  jmp    L32
L30:
  cmp    $0x4,%eax
  jne    L32
  mov    $0x0,%edi
  call   <exit@plt>
  jmp    L32
L5:
  lea    D74(%rip),%rdi
  call   <printf@plt>
  lea    0x2(%rsp),%rdx
  lea    0x3(%rsp),%rsi
  lea    D30(%rip),%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
L32:
  mov    %r13,%rdi
  call   <printf@plt>
  mov    %r12,%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    %rbx,%rdi
  call   <printf@plt>
  lea    D102(%rip),%rdi
  call   <printf@plt>
  lea    D37(%rip),%rdi
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D1(%rip),%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  cmp    $0x3,%eax
  jg     L30
  cmp    $0x1,%eax
  jg     L5
  jne    L32
  mov    $0x0,%eax
  call   <add_node@plt>
  jmp    L5
```

tool_calls: name: extract_symbolic_data
arguments: label: D78
type: string
name: extract_symbolic_data
arguments: label: D28
type: string
name: extract_symbolic_data
arguments: label: D80
type: string
name: extract_symbolic_data
arguments: label: D4
type: string
name: extract_symbolic_data
arguments: label: D45
type: string
name: extract_symbolic_data
arguments: label: D70
type: string
name: extract_symbolic_data
arguments: label: D74
type: string
name: extract_symbolic_data
arguments: label: D30
type: string
name: extract_symbolic_data
arguments: label: D102
type: string
name: extract_symbolic_data
arguments: label: D37
type: string
name: extract_symbolic_data
arguments: label: D1
type: string

tool: D78: 

	 Telephone Directory

tool: D28: 
	 ===================

tool: D80: 
	 Menu

tool: D4: 
	 1. Add Sudirriber

tool: D45: 
	 2. Search Suscriber

tool: D70: 
	 3. Remove Suscriber

tool: D74: 

	 Would you like to continue:::(y/n)

tool: D30: %c%c

tool: D102: 
	 4. Exit


tool: D37: 

	 ::Enter Your Choice::

tool: D1: %d

assistant: ```c
void display(void) {
  int choice = 4;
  char cnt, temp;
  printf("\n\n\t Telephone Directory");
  printf("\n\t ===================");
  do {
    printf("\n\t Menu");
    printf("\n\t 1. Add Sudirriber");
    printf("\n\t 2. Search Suscriber");
    printf("\n\t 3. Remove Suscriber");
    printf("\n\t 4. Exit\n");
    printf("\n\n\t ::Enter Your Choice::");
    scanf("%d", &choice);
    switch (choice) {
    case 1:
      add_node();
      break;
    case 2:
      break;
    case 3:
      break;
    case 4:
      exit(0);
    default:
      continue;
    }
    printf("\n\n\t Would you like to continue:::(y/n)");
    scanf("%c%c", &temp, &cnt);
  } while (cnt != 'n' || cnt != 'N');
}
```