ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What would the C source look like for the assembly code provided here?

```asm
<isPalindrome>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x230,%rsp
  mov    %rdi,-0x228(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x228(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x214(%rbp)
  movl   $0x0,-0x21c(%rbp)
  movl   $0x0,-0x218(%rbp)
  cmpl   $0x0,-0x214(%rbp)
  jg     L71
  mov    $0x0,%eax
  jmp    L14
L20:
  mov    -0x21c(%rbp),%eax
  cltq
  movl   $0x0,-0x210(%rbp,%rax,4)
  addl   $0x1,-0x21c(%rbp)
L71:
  cmpl   $0x7f,-0x21c(%rbp)
  jle    L20
  movl   $0x0,-0x21c(%rbp)
  jmp    L7
L24:
  mov    -0x21c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x228(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  movslq %eax,%rdx
  mov    -0x210(%rbp,%rdx,4),%edx
  add    $0x1,%edx
  cltq
  mov    %edx,-0x210(%rbp,%rax,4)
  mov    -0x21c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x228(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  cltq
  mov    -0x210(%rbp,%rax,4),%eax
  cltd
  shr    $0x1f,%edx
  add    %edx,%eax
  and    $0x1,%eax
  sub    %edx,%eax
  cmp    $0x1,%eax
  jne    L36
  addl   $0x1,-0x218(%rbp)
  jmp    L21
L36:
  subl   $0x1,-0x218(%rbp)
L21:
  addl   $0x1,-0x21c(%rbp)
L7:
  mov    -0x21c(%rbp),%eax
  cmp    -0x214(%rbp),%eax
  jl     L24
  cmpl   $0x1,-0x218(%rbp)
  setle  %al
  movzbl %al,%eax
L14:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L38
  call   <__stack_chk_fail@plt>
L38:
  leave
  ret
```

assistant: ```c
int isPalindrome(char *word) {
  int length = strlen(word);
  int hashTable[128];
  int i = 0, countOdd = 0;
  if (length <= 0)
    return 0;
  for (; i < 128; i++)
    hashTable[i] = 0;
  for (i = 0; i < length; i++) {
    hashTable[(int)word[i]]++;
    if (hashTable[(int)word[i]] % 2 == 1)
      countOdd++;
    else
      countOdd--;
  }
  return countOdd <= 1;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<isPalindrome>:
  endbr64
  push   %rbx
  sub    $0x210,%rsp
  mov    %rdi,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x208(%rsp)
  xor    %eax,%eax
  call   <strlen@plt>
  mov    $0x0,%edx
  test   %eax,%eax
  jle    L37
  mov    %rsp,%rdx
  lea    0x200(%rsp),%rcx
L29:
  movl   $0x0,(%rdx)
  add    $0x4,%rdx
  cmp    %rcx,%rdx
  jne    L29
  mov    %rbx,%rcx
  lea    -0x1(%rax),%eax
  lea    0x1(%rbx,%rax,1),%rdi
  mov    $0x0,%edx
L39:
  movsbq (%rcx),%rsi
  mov    (%rsp,%rsi,4),%eax
  add    $0x1,%eax
  mov    %eax,(%rsp,%rsi,4)
  mov    %eax,%esi
  shr    $0x1f,%esi
  add    %esi,%eax
  and    $0x1,%eax
  sub    %esi,%eax
  lea    0x1(%rdx),%esi
  sub    $0x1,%edx
  cmp    $0x1,%eax
  cmove  %esi,%edx
  add    $0x1,%rcx
  cmp    %rcx,%rdi
  jne    L39
  cmp    $0x1,%edx
  setle  %dl
  movzbl %dl,%edx
L37:
  mov    0x208(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L18
  mov    %edx,%eax
  add    $0x210,%rsp
  pop    %rbx
  ret
L18:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int isPalindrome(char *word) {
  int length = strlen(word);
  int hashTable[128];
  int i = 0, countOdd = 0;
  if (length <= 0)
    return 0;
  for (; i < 128; i++)
    hashTable[i] = 0;
  for (i = 0; i < length; i++) {
    hashTable[(int)word[i]]++;
    if (hashTable[(int)word[i]] % 2 == 1)
      countOdd++;
    else
      countOdd--;
  }
  return countOdd <= 1;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<isPalindrome>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x210,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x208(%rsp)
  xor    %eax,%eax
  call   <strlen@plt>
  mov    %eax,%edx
  xor    %eax,%eax
  test   %edx,%edx
  jle    L2
  mov    %rsp,%rsi
  xor    %eax,%eax
  mov    $0x40,%ecx
  mov    %rsi,%rdi
  rep stos %rax,%es:(%rdi)
  lea    -0x1(%rdx),%eax
  mov    %rbx,%rdi
  xor    %edx,%edx
  lea    0x1(%rbx,%rax,1),%rsi
  nopl   0x0(%rax)
L1:
  movsbq (%rdi),%rcx
  mov    (%rsp,%rcx,4),%eax
  add    $0x1,%eax
  mov    %eax,(%rsp,%rcx,4)
  mov    %eax,%ecx
  shr    $0x1f,%ecx
  add    %ecx,%eax
  and    $0x1,%eax
  sub    %ecx,%eax
  lea    0x1(%rdx),%ecx
  sub    $0x1,%edx
  cmp    $0x1,%eax
  cmove  %ecx,%edx
  add    $0x1,%rdi
  cmp    %rdi,%rsi
  jne    L1
  xor    %eax,%eax
  cmp    $0x1,%edx
  setle  %al
L2:
  mov    0x208(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L0
  add    $0x210,%rsp
  pop    %rbx
  ret
L0:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int isPalindrome(char *word) {
  int length = strlen(word);
  int hashTable[128];
  int i = 0, countOdd = 0;
  if (length <= 0)
    return 0;
  for (; i < 128; i++)
    hashTable[i] = 0;
  for (i = 0; i < length; i++) {
    hashTable[(int)word[i]]++;
    if (hashTable[(int)word[i]] % 2 == 1)
      countOdd++;
    else
      countOdd--;
  }
  return countOdd <= 1;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<isPalindrome>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x210,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x208(%rsp)
  xor    %eax,%eax
  call   <strlen@plt>
  mov    %eax,%edx
  xor    %eax,%eax
  test   %edx,%edx
  jle    L6
  mov    %rsp,%rsi
  xor    %eax,%eax
  mov    $0x40,%ecx
  mov    %rsi,%rdi
  rep stos %rax,%es:(%rdi)
  lea    -0x1(%rdx),%eax
  mov    %rbx,%rdi
  xor    %edx,%edx
  lea    0x1(%rbx,%rax,1),%rsi
  nopl   0x0(%rax)
L1:
  movsbq (%rdi),%rcx
  mov    (%rsp,%rcx,4),%eax
  add    $0x1,%eax
  mov    %eax,(%rsp,%rcx,4)
  mov    %eax,%ecx
  shr    $0x1f,%ecx
  add    %ecx,%eax
  and    $0x1,%eax
  sub    %ecx,%eax
  lea    0x1(%rdx),%ecx
  sub    $0x1,%edx
  cmp    $0x1,%eax
  cmove  %ecx,%edx
  add    $0x1,%rdi
  cmp    %rdi,%rsi
  jne    L1
  xor    %eax,%eax
  cmp    $0x1,%edx
  setle  %al
L6:
  mov    0x208(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L15
  add    $0x210,%rsp
  pop    %rbx
  ret
L15:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int isPalindrome(char *word) {
  int length = strlen(word);
  int hashTable[128];
  int i = 0, countOdd = 0;
  if (length <= 0)
    return 0;
  for (; i < 128; i++)
    hashTable[i] = 0;
  for (i = 0; i < length; i++) {
    hashTable[(int)word[i]]++;
    if (hashTable[(int)word[i]] % 2 == 1)
      countOdd++;
    else
      countOdd--;
  }
  return countOdd <= 1;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  test   %rax,%rax
  je     L32
  mov    -0x10(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L44
L32:
  mov    $0x0,%eax
  jmp    L28
L44:
  cmpl   $0x3,-0x4(%rbp)
  jne    L36
  mov    -0x10(%rbp),%rax
  add    $0x10,%rax
  mov    (%rax),%rdx
  mov    -0x10(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <compile@plt>
  jmp    L28
L36:
  mov    -0x10(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <interpret@plt>
L28:
  leave
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  if (!argv[1] || !*argv[1])
    return 0;
  if (argc == 3)
    return compile(argv[1], argv[2]);
  return interpret(argv[1]);
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  mov    0x8(%rsi),%rdx
  test   %rdx,%rdx
  je     L31
  mov    $0x0,%eax
  cmpb   $0x0,(%rdx)
  je     L11
  sub    $0x8,%rsp
  cmp    $0x3,%edi
  je     L27
  mov    %rdx,%rdi
  call   <interpret@plt>
L36:
  add    $0x8,%rsp
  ret
L27:
  mov    0x10(%rsi),%rsi
  mov    %rdx,%rdi
  call   <compile@plt>
  jmp    L36
L31:
  mov    $0x0,%eax
  ret
L11:
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  if (!argv[1] || !*argv[1])
    return 0;
  if (argc == 3)
    return compile(argv[1], argv[2]);
  return interpret(argv[1]);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  mov    0x8(%rsi),%r8
  test   %r8,%r8
  je     L4
  cmpb   $0x0,(%r8)
  je     L4
  cmp    $0x3,%edi
  je     L20
  mov    %r8,%rdi
  jmp    1060 <interpret@plt>
L4:
  xor    %eax,%eax
  ret
L20:
  mov    0x10(%rsi),%rsi
  mov    %r8,%rdi
  jmp    1070 <compile@plt>
```

assistant: ```c
int main(int argc, char **argv) {
  if (!argv[1] || !*argv[1])
    return 0;
  if (argc == 3)
    return compile(argv[1], argv[2]);
  return interpret(argv[1]);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  mov    0x8(%rsi),%r8
  test   %r8,%r8
  je     L15
  cmpb   $0x0,(%r8)
  je     L15
  cmp    $0x3,%edi
  je     L21
  mov    %r8,%rdi
  jmp    1060 <interpret@plt>
L15:
  xor    %eax,%eax
  ret
L21:
  mov    0x10(%rsi),%rsi
  mov    %r8,%rdi
  jmp    1070 <compile@plt>
```

assistant: ```c
int main(int argc, char **argv) {
  if (!argv[1] || !*argv[1])
    return 0;
  if (argc == 3)
    return compile(argv[1], argv[2]);
  return interpret(argv[1]);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<lstptr_>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %rcx,-0x30(%rbp)
  subq   $0x4,-0x30(%rbp)
  subq   $0x4,-0x28(%rbp)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,D21(%rip)
L32:
  mov    D21(%rip),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,D23(%rip)
  mov    -0x20(%rbp),%rax
  mov    (%rax),%edx
  mov    D23(%rip),%eax
  cmp    %eax,%edx
  je     L30
  mov    D21(%rip),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,D15(%rip)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%edx
  mov    D15(%rip),%eax
  cmp    %eax,%edx
  je     L2
  jmp    L32
L30:
  nop
  jmp    L14
L2:
  nop
L14:
  mov    D15(%rip),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int lstptr_(int *lpl, int *nb, int *list, int *lptr) {
  int ret_val;
  static int nd, lp;
  --lptr;
  --list;
  lp = lptr[*lpl];
L1:
  nd = list[lp];
  if (nd == *nb) {
    goto L2;
  }
  lp = lptr[lp];
  if (lp != *lpl) {
    goto L1;
  }
L2:
  ret_val = lp;
  return ret_val;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<lstptr_>:
  endbr64
  sub    $0x4,%rcx
  mov    (%rdi),%edi
  movslq %edi,%rax
  mov    (%rcx,%rax,4),%eax
  mov    %eax,D32(%rip)
  mov    (%rsi),%r9d
  mov    $0x0,%r8d
  mov    $0x1,%r10d
L3:
  movslq %eax,%rsi
  cmp    %r9d,-0x4(%rdx,%rsi,4)
  je     L27
  mov    (%rcx,%rsi,4),%eax
  mov    %r10d,%r8d
  cmp    %eax,%edi
  jne    L3
  mov    %eax,D30(%rip)
L28:
  mov    D30(%rip),%eax
  ret
L27:
  test   %r8b,%r8b
  je     L28
  mov    %eax,D30(%rip)
  jmp    L28
```

assistant: ```c
int lstptr_(int *lpl, int *nb, int *list, int *lptr) {
  int ret_val;
  static int nd, lp;
  --lptr;
  --list;
  lp = lptr[*lpl];
L1:
  nd = list[lp];
  if (nd == *nb) {
    goto L2;
  }
  lp = lptr[lp];
  if (lp != *lpl) {
    goto L1;
  }
L2:
  ret_val = lp;
  return ret_val;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<lstptr_>:
  endbr64
  movslq (%rdi),%rax
  sub    $0x4,%rcx
  mov    (%rsi),%r9d
  xor    %r8d,%r8d
  mov    (%rcx,%rax,4),%r10d
  mov    %rax,%rdi
  mov    %r10d,D6(%rip)
  mov    %r10d,%eax
  jmp    L25
  nopl   0x0(%rax)
L20:
  mov    (%rcx,%rsi,4),%eax
  mov    $0x1,%r8d
  cmp    %eax,%edi
  je     L3
L25:
  movslq %eax,%rsi
  cmp    %r9d,-0x4(%rdx,%rsi,4)
  jne    L20
  test   %r8b,%r8b
  je     L19
L3:
  mov    %eax,D11(%rip)
  ret
  nopl   0x0(%rax,%rax,1)
L19:
  mov    %r10d,%eax
  ret
```

assistant: ```c
int lstptr_(int *lpl, int *nb, int *list, int *lptr) {
  int ret_val;
  static int nd, lp;
  --lptr;
  --list;
  lp = lptr[*lpl];
L1:
  nd = list[lp];
  if (nd == *nb) {
    goto L2;
  }
  lp = lptr[lp];
  if (lp != *lpl) {
    goto L1;
  }
L2:
  ret_val = lp;
  return ret_val;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<lstptr_>:
  endbr64
  movslq (%rdi),%rax
  sub    $0x4,%rcx
  mov    (%rsi),%r9d
  xor    %r8d,%r8d
  mov    (%rcx,%rax,4),%r10d
  mov    %rax,%rdi
  mov    %r10d,D15(%rip)
  mov    %r10d,%eax
  jmp    L1
  nopl   0x0(%rax)
L21:
  mov    (%rcx,%rsi,4),%eax
  mov    $0x1,%r8d
  cmp    %eax,%edi
  je     L28
L1:
  movslq %eax,%rsi
  cmp    %r9d,-0x4(%rdx,%rsi,4)
  jne    L21
  test   %r8b,%r8b
  je     L11
L28:
  mov    %eax,D17(%rip)
  ret
  nopl   0x0(%rax,%rax,1)
L11:
  mov    %r10d,%eax
  ret
```

assistant: ```c
int lstptr_(int *lpl, int *nb, int *list, int *lptr) {
  int ret_val;
  static int nd, lp;
  --lptr;
  --list;
  lp = lptr[*lpl];
L1:
  nd = list[lp];
  if (nd == *nb) {
    goto L2;
  }
  lp = lptr[lp];
  if (lp != *lpl) {
    goto L1;
  }
L2:
  ret_val = lp;
  return ret_val;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<bin_string_to_int>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,-0x28(%rbp)
  movl   $0x0,-0x18(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  sub    $0x1,%eax
  mov    %eax,-0x14(%rbp)
  jmp    L11
L0:
  mov    -0x28(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  lea    -0x30(%rax),%ebx
  mov    -0x14(%rbp),%eax
  test   %rax,%rax
  js     L7
  pxor   %xmm0,%xmm0
  cvtsi2sd %rax,%xmm0
  jmp    L4
L7:
  mov    %rax,%rdx
  shr    %rdx
  and    $0x1,%eax
  or     %rax,%rdx
  pxor   %xmm0,%xmm0
  cvtsi2sd %rdx,%xmm0
  addsd  %xmm0,%xmm0
L4:
  mov    D2(%rip),%rax
  movapd %xmm0,%xmm1
  movq   %rax,%xmm0
  call   <pow@plt>
  movsbl %al,%eax
  imul   %ebx,%eax
  add    %eax,-0x18(%rbp)
  subl   $0x1,-0x14(%rbp)
  addq   $0x1,-0x28(%rbp)
L11:
  mov    -0x28(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L0
  mov    -0x18(%rbp),%eax
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

tool_calls: name: parse_data
arguments: tag: D2
type: byte[8]

tool: {"D2": [0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x40]}

assistant: ```c
unsigned int bin_string_to_int(char *s) {
  unsigned int binary = 0;
  unsigned int index = strlen(s) - 1;
  while (*s != '\0') {
    binary += (*s - 48) * pow((double)2, (double)index);
    index--;
    s++;
  }
  return binary;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<bin_string_to_int>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r13
  call   <strlen@plt>
  movzbl 0x0(%r13),%ebx
  test   %bl,%bl
  je     L5
  mov    %r13,%rbp
  mov    $0x0,%r12d
  lea    -0x1(%rax,%r13,1),%r13d
L6:
  mov    %r13d,%eax
  sub    %ebp,%eax
  pxor   %xmm1,%xmm1
  cvtsi2sd %rax,%xmm1
  mov    D3(%rip),%rax
  movq   %rax,%xmm0
  call   <pow@plt>
  movsbl %bl,%ebx
  sub    $0x30,%ebx
  movsbl %al,%eax
  imul   %eax,%ebx
  add    %ebx,%r12d
  add    $0x1,%rbp
  movzbl 0x0(%rbp),%ebx
  test   %bl,%bl
  jne    L6
L11:
  mov    %r12d,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L5:
  mov    $0x0,%r12d
  jmp    L11
```

tool_calls: name: read_data
arguments: label: D3
datatype: byte[8]

tool: {"D3": [0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x40]}

assistant: ```c
unsigned int bin_string_to_int(char *s) {
  unsigned int binary = 0;
  unsigned int index = strlen(s) - 1;
  while (*s != '\0') {
    binary += (*s - 48) * pow((double)2, (double)index);
    index--;
    s++;
  }
  return binary;
}
```
user: Translate the below assembly code into C programming language:

```asm
<bin_string_to_int>:
  endbr64
  push   %r13
  mov    %rdi,%r13
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  movsbl 0x0(%r13),%ebx
  test   %bl,%bl
  je     L3
  lea    -0x1(%rax,%r13,1),%ebp
  nopw   0x0(%rax,%rax,1)
L13:
  mov    %ebp,%eax
  pxor   %xmm1,%xmm1
  sub    $0x30,%ebx
  sub    %r13d,%eax
  add    $0x1,%r13
  cvtsi2sd %rax,%xmm1
  mov    D9(%rip),%rax
  movq   %rax,%xmm0
  call   <pow@plt>
  movsbl %al,%eax
  imul   %eax,%ebx
  add    %ebx,%r12d
  movsbl 0x0(%r13),%ebx
  test   %bl,%bl
  jne    L13
L3:
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: unpack_data
arguments: tag: D9
type_infered: byte[8]

tool: {"D9": [0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x40]}

assistant: ```c
unsigned int bin_string_to_int(char *s) {
  unsigned int binary = 0;
  unsigned int index = strlen(s) - 1;
  while (*s != '\0') {
    binary += (*s - 48) * pow((double)2, (double)index);
    index--;
    s++;
  }
  return binary;
}
```
user: Translate the below assembly code into C programming language:

```asm
<bin_string_to_int>:
  endbr64
  push   %r13
  mov    %rdi,%r13
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  movsbl 0x0(%r13),%ebx
  test   %bl,%bl
  je     L17
  lea    -0x1(%rax,%r13,1),%ebp
  nopw   0x0(%rax,%rax,1)
L19:
  mov    %ebp,%eax
  pxor   %xmm1,%xmm1
  sub    $0x30,%ebx
  sub    %r13d,%eax
  add    $0x1,%r13
  cvtsi2sd %rax,%xmm1
  mov    D0(%rip),%rax
  movq   %rax,%xmm0
  call   <pow@plt>
  movsbl %al,%eax
  imul   %eax,%ebx
  add    %ebx,%r12d
  movsbl 0x0(%r13),%ebx
  test   %bl,%bl
  jne    L19
L17:
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: parse_data
arguments: tag: D0
type_infered: byte[8]

tool: {"D0": [0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x40]}

assistant: ```c
unsigned int bin_string_to_int(char *s) {
  unsigned int binary = 0;
  unsigned int index = strlen(s) - 1;
  while (*s != '\0') {
    binary += (*s - 48) * pow((double)2, (double)index);
    index--;
    s++;
  }
  return binary;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<sync_sbf>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,%eax
  mov    %al,-0xc(%rbp)
  mov    -0x8(%rbp),%rax
  movzbl 0x1(%rax),%edx
  mov    -0x8(%rbp),%rax
  mov    %dl,(%rax)
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rdx
  movzbl -0xc(%rbp),%eax
  mov    %al,(%rdx)
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x24,%al
  jne    L27
  mov    -0x8(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  cmp    $0x40,%al
  jne    L27
  mov    $0x1,%eax
  jmp    L3
L27:
  mov    $0x0,%eax
L3:
  pop    %rbp
  ret
```

assistant: ```c
int sync_sbf(unsigned char *buff, unsigned char data) {
  buff[0] = buff[1];
  buff[1] = data;
  return buff[0] == 0x24 && buff[1] == 0x40;
}
```
user: What is the c source code of the assembly code below?

```asm
<sync_sbf>:
  endbr64
  movzbl 0x1(%rdi),%eax
  mov    %al,(%rdi)
  mov    %sil,0x1(%rdi)
  cmp    $0x40,%sil
  sete   %dl
  cmp    $0x24,%al
  sete   %al
  movzbl %al,%eax
  and    %edx,%eax
  ret
```

assistant: ```c
int sync_sbf(unsigned char *buff, unsigned char data) {
  buff[0] = buff[1];
  buff[1] = data;
  return buff[0] == 0x24 && buff[1] == 0x40;
}
```
user: What is the c source code of the assembly code below?

```asm
<sync_sbf>:
  endbr64
  movzbl 0x1(%rdi),%eax
  cmp    $0x40,%sil
  mov    %sil,0x1(%rdi)
  sete   %dl
  cmp    $0x24,%al
  mov    %al,(%rdi)
  sete   %al
  movzbl %al,%eax
  and    %edx,%eax
  ret
```

assistant: ```c
int sync_sbf(unsigned char *buff, unsigned char data) {
  buff[0] = buff[1];
  buff[1] = data;
  return buff[0] == 0x24 && buff[1] == 0x40;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<sync_sbf>:
  endbr64
  movzbl 0x1(%rdi),%eax
  cmp    $0x40,%sil
  mov    %sil,0x1(%rdi)
  sete   %dl
  cmp    $0x24,%al
  mov    %al,(%rdi)
  sete   %al
  movzbl %al,%eax
  and    %edx,%eax
  ret
```

assistant: ```c
int sync_sbf(unsigned char *buff, unsigned char data) {
  buff[0] = buff[1];
  buff[1] = data;
  return buff[0] == 0x24 && buff[1] == 0x40;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<tetris4>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x0,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L25
L41:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  lea    -0x1(%rax),%edx
  mov    -0x4(%rbp),%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,4),%rcx
  mov    -0x20(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%eax
  cmp    %eax,%edx
  jne    L21
  mov    -0x4(%rbp),%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x4(%rbp),%eax
  cltq
  add    $0x2,%rax
  lea    0x0(,%rax,4),%rcx
  mov    -0x20(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%eax
  cmp    %eax,%edx
  jne    L21
  addl   $0x1,-0x8(%rbp)
L21:
  mov    -0x4(%rbp),%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  lea    -0x1(%rax),%edx
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x20(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%eax
  cmp    %eax,%edx
  jne    L52
  addl   $0x1,-0x8(%rbp)
L52:
  addl   $0x1,-0x4(%rbp)
L25:
  mov    -0x14(%rbp),%eax
  sub    $0x2,%eax
  cmp    %eax,-0x4(%rbp)
  jl     L41
  mov    -0x14(%rbp),%eax
  cltq
  shl    $0x2,%rax
  lea    -0x4(%rax),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  lea    -0x1(%rax),%edx
  mov    -0x14(%rbp),%eax
  cltq
  shl    $0x2,%rax
  lea    -0x8(%rax),%rcx
  mov    -0x20(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%eax
  cmp    %eax,%edx
  jne    L46
  addl   $0x1,-0x8(%rbp)
L46:
  mov    -0x8(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int tetris4(int c, int *arr) {
  int counter = 0;
  for (int i = 0; i < c - 2; i++) {
    if (arr[i] - 1 == arr[i + 1] && arr[i + 1] == arr[i + 2])
      counter++;
    if (arr[i + 1] - 1 == arr[i])
      counter++;
  }
  if (arr[c - 1] - 1 == arr[c - 2])
    counter++;
  return counter;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<tetris4>:
  endbr64
  cmp    $0x2,%edi
  jle    L3
  mov    %rsi,%rax
  lea    -0x3(%rdi),%edx
  lea    0x4(%rsi,%rdx,4),%r10
  mov    $0x0,%ecx
  jmp    L20
L24:
  sub    $0x1,%edx
  cmp    %edx,%r8d
  sete   %dl
  movzbl %dl,%edx
  add    %edx,%ecx
  add    $0x4,%rax
  cmp    %r10,%rax
  je     L35
L20:
  mov    (%rax),%r8d
  mov    0x4(%rax),%edx
  lea    -0x1(%r8),%r9d
  cmp    %edx,%r9d
  jne    L24
  cmp    0x8(%rax),%edx
  sete   %r9b
  movzbl %r9b,%r9d
  add    %r9d,%ecx
  jmp    L24
L3:
  mov    $0x0,%ecx
L35:
  movslq %edi,%rdi
  shl    $0x2,%rdi
  mov    -0x4(%rsi,%rdi,1),%eax
  sub    $0x1,%eax
  cmp    -0x8(%rsi,%rdi,1),%eax
  je     L5
L65:
  mov    %ecx,%eax
  ret
L5:
  add    $0x1,%ecx
  jmp    L65
```

assistant: ```c
int tetris4(int c, int *arr) {
  int counter = 0;
  for (int i = 0; i < c - 2; i++) {
    if (arr[i] - 1 == arr[i + 1] && arr[i + 1] == arr[i + 2])
      counter++;
    if (arr[i + 1] - 1 == arr[i])
      counter++;
  }
  if (arr[c - 1] - 1 == arr[c - 2])
    counter++;
  return counter;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<tetris4>:
  endbr64
  cmp    $0x2,%edi
  jle    L46
  lea    -0x3(%rdi),%edx
  mov    %rsi,%rax
  xor    %r8d,%r8d
  lea    0x4(%rsi,%rdx,4),%r10
  nopw   0x0(%rax,%rax,1)
L34:
  mov    (%rax),%ecx
  mov    0x4(%rax),%edx
  lea    -0x1(%rcx),%r9d
  cmp    %edx,%r9d
  jne    L23
  xor    %r9d,%r9d
  cmp    0x8(%rax),%edx
  sete   %r9b
  add    %r9d,%r8d
L23:
  sub    $0x1,%edx
  cmp    %edx,%ecx
  sete   %dl
  add    $0x4,%rax
  movzbl %dl,%edx
  add    %edx,%r8d
  cmp    %r10,%rax
  jne    L34
L40:
  movslq %edi,%rdi
  shl    $0x2,%rdi
  mov    -0x4(%rsi,%rdi,1),%eax
  sub    $0x1,%eax
  cmp    -0x8(%rsi,%rdi,1),%eax
  jne    L24
  add    $0x1,%r8d
L24:
  mov    %r8d,%eax
  ret
  xchg   %ax,%ax
L46:
  xor    %r8d,%r8d
  jmp    L40
```

assistant: ```c
int tetris4(int c, int *arr) {
  int counter = 0;
  for (int i = 0; i < c - 2; i++) {
    if (arr[i] - 1 == arr[i + 1] && arr[i + 1] == arr[i + 2])
      counter++;
    if (arr[i + 1] - 1 == arr[i])
      counter++;
  }
  if (arr[c - 1] - 1 == arr[c - 2])
    counter++;
  return counter;
}
```