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
<call_after_alloca>:
  endbr64
  sub    $0x8,%rsp
  movswl %si,%esi
  movsbl %dil,%edi
  xor    %eax,%eax
  push   $0xd
  mov    $0xb,%r9d
  mov    $0x62,%r8d
  push   $0xc
  call   <call_after_alloca_subr@plt>
  add    $0x18,%rsp
  ret
```

assistant: ```c
int call_after_alloca(c, s, i, l)
char c;
short s;
int i;
long l;
{
  char *z = __builtin_alloca(s + 50);
  call_after_alloca_subr(c, s, i, l, 'b', 11, 12, (unsigned long)13);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<finalizer_scrub128>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L3
L8:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movl   $0xa5a5a5a5,(%rax)
  addl   $0x1,-0xc(%rbp)
L3:
  cmpl   $0xf,-0xc(%rbp)
  jle    L8
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void finalizer_scrub128(void *ptr) {
  uint64_t *sz_ptr = (uint64_t *)ptr;
  int i;
  for (i = 0; i < 16; i++)
    sz_ptr[i] = 0xa5a5a5a5a5a5a5a5ULL;
}
```
user: What is the c source code of the assembly code below:

```asm
<finalizer_scrub128>:
  endbr64
  lea    0x40(%rdi),%rax
L0:
  movl   $0xa5a5a5a5,(%rdi)
  add    $0x4,%rdi
  cmp    %rax,%rdi
  jne    L0
  ret
```

assistant: ```c
void finalizer_scrub128(void *ptr) {
  uint64_t *sz_ptr = (uint64_t *)ptr;
  int i;
  for (i = 0; i < 16; i++)
    sz_ptr[i] = 0xa5a5a5a5a5a5a5a5ULL;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<finalizer_scrub128>:
  endbr64
  movdqa D7(%rip),%xmm0
  movups %xmm0,(%rdi)
  movups %xmm0,0x10(%rdi)
  movups %xmm0,0x20(%rdi)
  movups %xmm0,0x30(%rdi)
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D7
type_guessed: byte[16]

tool: {"D7": [0xa5,0xa5,0xa5,0xa5,0xa5,0xa5,0xa5,0xa5,0xa5,0xa5,0xa5,0xa5,0xa5,0xa5,0xa5,0xa5]}

assistant: ```c
void finalizer_scrub128(void *ptr) {
  uint64_t *sz_ptr = (uint64_t *)ptr;
  int i;
  for (i = 0; i < 16; i++)
    sz_ptr[i] = 0xa5a5a5a5a5a5a5a5ULL;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<finalizer_scrub128>:
  endbr64
  movdqa D6(%rip),%xmm0
  movups %xmm0,(%rdi)
  movups %xmm0,0x10(%rdi)
  movups %xmm0,0x20(%rdi)
  movups %xmm0,0x30(%rdi)
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D6
type_guessed: byte[16]

tool: {"D6": [0xa5,0xa5,0xa5,0xa5,0xa5,0xa5,0xa5,0xa5,0xa5,0xa5,0xa5,0xa5,0xa5,0xa5,0xa5,0xa5]}

assistant: ```c
void finalizer_scrub128(void *ptr) {
  uint64_t *sz_ptr = (uint64_t *)ptr;
  int i;
  for (i = 0; i < 16; i++)
    sz_ptr[i] = 0xa5a5a5a5a5a5a5a5ULL;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<func2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D5
datatype: string

tool: D5: Func2


assistant: ```c
int func2(int **array, int m) {
  printf("Func2\n");
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<func2>:
  endbr64
  sub    $0x8,%rsp
  lea    D2(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: tag: D2
type_infered: string

tool: D2: Func2


assistant: ```c
int func2(int **array, int m) {
  printf("Func2\n");
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<func2>:
  endbr64
  sub    $0x8,%rsp
  lea    D1(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D1
data_type: string

tool: D1: Func2


assistant: ```c
int func2(int **array, int m) {
  printf("Func2\n");
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<func2>:
  endbr64
  sub    $0x8,%rsp
  lea    D0(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D0
datatype: string

tool: D0: Func2


assistant: ```c
int func2(int **array, int m) {
  printf("Func2\n");
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<ldtoa_strtold>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    D7(%rip),%rax
  movq   $0x0,(%rax)
  lea    -0x28(%rbp),%rdx
  mov    -0x38(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strtod@plt>
  mov    %rax,-0x48(%rbp)
  fildll -0x48(%rbp)
  fstpt  -0x20(%rbp)
  mov    -0x28(%rbp),%rax
  test   %rax,%rax
  je     L21
  mov    -0x28(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L21
  mov    D7(%rip),%rax
  mov    (%rax),%rax
  test   %rax,%rax
  setne  %al
  movzbl %al,%edx
  mov    -0x40(%rbp),%rax
  mov    %edx,(%rax)
  jmp    L13
L21:
  mov    -0x40(%rbp),%rax
  movl   $0x1,(%rax)
L13:
  fldt   -0x20(%rbp)
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L9
  fstp   %st(0)
  call   <__stack_chk_fail@plt>
L9:
  leave
  ret
```

assistant: ```c
long double ldtoa_strtold(const char *s, int *error) {
  char *endp;
  long double d;
  errno = 0;
  d = (long double)strtod(s, &endp);
  if (endp != NULL && (*endp == '\0'))
    *error = (errno != 0);
  else
    *error = (1 == 1);
  return d;
}
```
user: What is the c source code of the assembly code below?

```asm
<ldtoa_strtold>:
  endbr64
  push   %rbx
  sub    $0x20,%rsp
  mov    %rsi,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    D12(%rip),%rax
  movq   $0x0,(%rax)
  lea    0x10(%rsp),%rsi
  call   <strtod@plt>
  mov    %rax,0x8(%rsp)
  fildll 0x8(%rsp)
  mov    0x10(%rsp),%rdx
  mov    $0x1,%eax
  test   %rdx,%rdx
  je     L29
  cmpb   $0x0,(%rdx)
  jne    L29
  mov    D12(%rip),%rax
  cmpq   $0x0,(%rax)
  setne  %al
  movzbl %al,%eax
L29:
  mov    %eax,(%rbx)
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L24
  add    $0x20,%rsp
  pop    %rbx
  ret
L24:
  fstp   %st(0)
  call   <__stack_chk_fail@plt>
```

assistant: ```c
long double ldtoa_strtold(const char *s, int *error) {
  char *endp;
  long double d;
  errno = 0;
  d = (long double)strtod(s, &endp);
  if (endp != NULL && (*endp == '\0'))
    *error = (errno != 0);
  else
    *error = (1 == 1);
  return d;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ldtoa_strtold>:
  endbr64
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x28,%rsp
  mov    D4(%rip),%rbp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rsi
  movq   $0x0,0x0(%rbp)
  call   <strtod@plt>
  mov    $0x1,%edx
  mov    %rax,0x8(%rsp)
  mov    0x10(%rsp),%rax
  fildll 0x8(%rsp)
  test   %rax,%rax
  je     L24
  cmpb   $0x0,(%rax)
  je     L14
L24:
  mov    %edx,(%rbx)
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L34
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax)
L14:
  xor    %edx,%edx
  cmpq   $0x0,0x0(%rbp)
  setne  %dl
  jmp    L24
L34:
  fstp   %st(0)
  call   <__stack_chk_fail@plt>
```

assistant: ```c
long double ldtoa_strtold(const char *s, int *error) {
  char *endp;
  long double d;
  errno = 0;
  d = (long double)strtod(s, &endp);
  if (endp != NULL && (*endp == '\0'))
    *error = (errno != 0);
  else
    *error = (1 == 1);
  return d;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<ldtoa_strtold>:
  endbr64
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x28,%rsp
  mov    D1(%rip),%rbp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rsi
  movq   $0x0,0x0(%rbp)
  call   <strtod@plt>
  mov    $0x1,%edx
  mov    %rax,0x8(%rsp)
  mov    0x10(%rsp),%rax
  fildll 0x8(%rsp)
  test   %rax,%rax
  je     L16
  cmpb   $0x0,(%rax)
  je     L30
L16:
  mov    %edx,(%rbx)
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax)
L30:
  xor    %edx,%edx
  cmpq   $0x0,0x0(%rbp)
  setne  %dl
  jmp    L16
L2:
  fstp   %st(0)
  call   <__stack_chk_fail@plt>
```

assistant: ```c
long double ldtoa_strtold(const char *s, int *error) {
  char *endp;
  long double d;
  errno = 0;
  d = (long double)strtod(s, &endp);
  if (endp != NULL && (*endp == '\0'))
    *error = (errno != 0);
  else
    *error = (1 == 1);
  return d;
}
```
user: Translate the below assembly code into C programming language:

```asm
<plusOne>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  mov    %rdx,-0x38(%rbp)
  movl   $0x1,-0x18(%rbp)
  mov    -0x2c(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x14(%rbp)
  jmp    L46
L21:
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x18(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  lea    (%rdx,%rax,1),%rsi
  mov    -0xc(%rbp),%ecx
  movslq %ecx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %ecx,%edi
  sar    $0x1f,%edi
  sub    %edi,%eax
  mov    %eax,%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  sub    %eax,%ecx
  mov    %ecx,%edx
  mov    %edx,(%rsi)
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x18(%rbp)
  subl   $0x1,-0x14(%rbp)
L46:
  cmpl   $0x0,-0x14(%rbp)
  jns    L21
  cmpl   $0x1,-0x18(%rbp)
  jne    L11
  mov    -0x2c(%rbp),%eax
  add    $0x1,%eax
  cltq
  shl    $0x2,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    -0x18(%rbp),%edx
  mov    %edx,(%rax)
  movl   $0x0,-0x10(%rbp)
  jmp    L48
L28:
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    -0x10(%rbp),%edx
  movslq %edx,%rdx
  add    $0x1,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x8(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  addl   $0x1,-0x10(%rbp)
L48:
  mov    -0x10(%rbp),%eax
  cmp    -0x2c(%rbp),%eax
  jle    L28
  mov    -0x2c(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    -0x38(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x8(%rbp),%rax
  jmp    L53
L11:
  mov    -0x38(%rbp),%rax
  mov    -0x2c(%rbp),%edx
  mov    %edx,(%rax)
  mov    -0x28(%rbp),%rax
L53:
  leave
  ret
```

assistant: ```c
int *plusOne(int *digits, int digitsSize, int *returnSize) {
  int carry = 1;
  for (int i = digitsSize - 1; i >= 0; i--) {
    int val = *(digits + i) + carry;
    *(digits + i) = val % 10;
    carry = val / 10;
  }
  if (carry == 1) {
    int *tmp = malloc(sizeof(int) * (digitsSize + 1));
    *(tmp + 0) = carry;
    for (int i = 0; i <= digitsSize; i++) {
      *(tmp + i + 1) = *(digits + i);
    }
    *returnSize = digitsSize + 1;
    return tmp;
  } else {
    *returnSize = digitsSize;
    return digits;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<plusOne>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %esi,%ebp
  mov    %rdx,%r12
  mov    %esi,%ecx
  sub    $0x1,%ecx
  js     L31
  movslq %ecx,%rcx
  mov    $0x1,%edx
L30:
  add    (%rbx,%rcx,4),%edx
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  mov    %edx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  lea    (%rax,%rax,4),%esi
  add    %esi,%esi
  sub    %esi,%edx
  mov    %edx,(%rbx,%rcx,4)
  mov    %eax,%edx
  sub    $0x1,%rcx
  test   %ecx,%ecx
  jns    L30
  cmp    $0x1,%eax
  je     L31
L28:
  mov    %ebp,(%r12)
  mov    %rbx,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L31:
  lea    0x1(%rbp),%r13d
  lea    0x0(,%r13,4),%edi
  call   <malloc@plt>
  movl   $0x1,(%rax)
  test   %ebp,%ebp
  js     L23
  lea    0x1(%rbp),%esi
  shl    $0x2,%rsi
  mov    $0x0,%edx
L29:
  mov    (%rbx,%rdx,1),%ecx
  mov    %ecx,0x4(%rax,%rdx,1)
  add    $0x4,%rdx
  cmp    %rdx,%rsi
  jne    L29
L23:
  mov    %r13d,%ebp
  mov    %rax,%rbx
  jmp    L28
```

assistant: ```c
int *plusOne(int *digits, int digitsSize, int *returnSize) {
  int carry = 1;
  for (int i = digitsSize - 1; i >= 0; i--) {
    int val = *(digits + i) + carry;
    *(digits + i) = val % 10;
    carry = val / 10;
  }
  if (carry == 1) {
    int *tmp = malloc(sizeof(int) * (digitsSize + 1));
    *(tmp + 0) = carry;
    for (int i = 0; i <= digitsSize; i++) {
      *(tmp + i + 1) = *(digits + i);
    }
    *returnSize = digitsSize + 1;
    return tmp;
  } else {
    *returnSize = digitsSize;
    return digits;
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<plusOne>:
  endbr64
  push   %r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  movslq %esi,%rbx
  mov    %ebx,%ecx
  sub    $0x8,%rsp
  sub    $0x1,%ecx
  js     L42
  movslq %ecx,%rcx
  mov    $0x1,%edx
  cs nopw 0x0(%rax,%rax,1)
L52:
  add    (%r12,%rcx,4),%edx
  movslq %edx,%rax
  mov    %edx,%esi
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%esi
  sar    $0x22,%rax
  sub    %esi,%eax
  lea    (%rax,%rax,4),%esi
  add    %esi,%esi
  sub    %esi,%edx
  mov    %edx,(%r12,%rcx,4)
  sub    $0x1,%rcx
  mov    %eax,%edx
  test   %ecx,%ecx
  jns    L52
  cmp    $0x1,%eax
  je     L42
L26:
  mov    %ebx,0x0(%rbp)
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L42:
  lea    0x1(%rbx),%r13d
  lea    0x0(,%r13,4),%edi
  call   <malloc@plt>
  lea    0x4(,%rbx,4),%rsi
  xor    %edx,%edx
  movl   $0x1,(%rax)
  test   %ebx,%ebx
  js     L29
  nopl   (%rax)
L46:
  mov    (%r12,%rdx,1),%ecx
  mov    %ecx,0x4(%rax,%rdx,1)
  add    $0x4,%rdx
  cmp    %rsi,%rdx
  jne    L46
L29:
  mov    %r13d,%ebx
  mov    %rax,%r12
  jmp    L26
```

assistant: ```c
int *plusOne(int *digits, int digitsSize, int *returnSize) {
  int carry = 1;
  for (int i = digitsSize - 1; i >= 0; i--) {
    int val = *(digits + i) + carry;
    *(digits + i) = val % 10;
    carry = val / 10;
  }
  if (carry == 1) {
    int *tmp = malloc(sizeof(int) * (digitsSize + 1));
    *(tmp + 0) = carry;
    for (int i = 0; i <= digitsSize; i++) {
      *(tmp + i + 1) = *(digits + i);
    }
    *returnSize = digitsSize + 1;
    return tmp;
  } else {
    *returnSize = digitsSize;
    return digits;
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<plusOne>:
  endbr64
  push   %r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  movslq %esi,%rbx
  mov    %ebx,%ecx
  sub    $0x8,%rsp
  sub    $0x1,%ecx
  js     L66
  movslq %ecx,%rcx
  mov    $0x1,%edx
  cs nopw 0x0(%rax,%rax,1)
L71:
  add    (%r12,%rcx,4),%edx
  movslq %edx,%rax
  mov    %edx,%esi
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%esi
  sar    $0x22,%rax
  sub    %esi,%eax
  lea    (%rax,%rax,4),%esi
  add    %esi,%esi
  sub    %esi,%edx
  mov    %edx,(%r12,%rcx,4)
  sub    $0x1,%rcx
  mov    %eax,%edx
  test   %ecx,%ecx
  jns    L71
  cmp    $0x1,%eax
  je     L66
L23:
  mov    %ebx,0x0(%rbp)
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L66:
  lea    0x1(%rbx),%r13d
  lea    0x0(,%r13,4),%edi
  call   <malloc@plt>
  movl   $0x1,(%rax)
  mov    %rax,%rdx
  test   %ebx,%ebx
  js     L128
  sub    %r12,%rax
  cmp    $0x8,%rax
  jbe    L43
  cmp    $0x2,%ebx
  jbe    L43
  mov    %r13d,%ecx
  xor    %eax,%eax
  shr    $0x2,%ecx
  shl    $0x4,%rcx
L24:
  movdqu (%r12,%rax,1),%xmm0
  movups %xmm0,0x4(%rdx,%rax,1)
  add    $0x10,%rax
  cmp    %rcx,%rax
  jne    L24
  mov    %r13d,%eax
  and    $0xfffffffc,%eax
  test   $0x3,%r13b
  je     L128
  movslq %eax,%rsi
  lea    0x0(,%rsi,4),%rcx
  mov    (%r12,%rsi,4),%esi
  mov    %esi,0x4(%rdx,%rcx,1)
  lea    0x1(%rax),%esi
  cmp    %ebx,%esi
  jg     L128
  mov    0x4(%r12,%rcx,1),%esi
  add    $0x2,%eax
  mov    %esi,0x8(%rdx,%rcx,1)
  cmp    %eax,%ebx
  jl     L128
  mov    0x8(%r12,%rcx,1),%eax
  mov    %eax,0xc(%rdx,%rcx,1)
L128:
  mov    %r13d,%ebx
  mov    %rdx,%r12
  jmp    L23
L43:
  lea    0x4(,%rbx,4),%rsi
  xor    %eax,%eax
  nopl   0x0(%rax,%rax,1)
L82:
  mov    (%r12,%rax,1),%ecx
  mov    %ecx,0x4(%rdx,%rax,1)
  add    $0x4,%rax
  cmp    %rsi,%rax
  jne    L82
  jmp    L128
```

assistant: ```c
int *plusOne(int *digits, int digitsSize, int *returnSize) {
  int carry = 1;
  for (int i = digitsSize - 1; i >= 0; i--) {
    int val = *(digits + i) + carry;
    *(digits + i) = val % 10;
    carry = val / 10;
  }
  if (carry == 1) {
    int *tmp = malloc(sizeof(int) * (digitsSize + 1));
    *(tmp + 0) = carry;
    for (int i = 0; i <= digitsSize; i++) {
      *(tmp + i + 1) = *(digits + i);
    }
    *returnSize = digitsSize + 1;
    return tmp;
  } else {
    *returnSize = digitsSize;
    return digits;
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<hextorb>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  jmp    L14
L6:
  addq   $0x2,-0x8(%rbp)
  addq   $0x1,-0x10(%rbp)
L14:
  mov    -0x10(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  lea    D1(%rip),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <sscanf@plt>
  cmp    $0x1,%eax
  je     L6
  mov    -0x10(%rbp),%rax
  movb   $0x0,(%rax)
  nop
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D1
datatype: string

tool: D1: %2x

assistant: ```c
void hextorb(uint8_t *hex, uint8_t *rb) {
  while (sscanf(hex, "%2x", rb) == 1) {
    hex += 2;
    rb += 1;
  }
  *rb = '\0';
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<hextorb>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    %rsi,%rbx
  lea    D9(%rip),%r12
  jmp    L0
L5:
  add    $0x2,%rbp
  add    $0x1,%rbx
L0:
  mov    %rbx,%rdx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <sscanf@plt>
  cmp    $0x1,%eax
  je     L5
  movb   $0x0,(%rbx)
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D9
type_guessed: string

tool: D9: %2x

assistant: ```c
void hextorb(uint8_t *hex, uint8_t *rb) {
  while (sscanf(hex, "%2x", rb) == 1) {
    hex += 2;
    rb += 1;
  }
  *rb = '\0';
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<hextorb>:
  endbr64
  push   %r12
  lea    D4(%rip),%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rsi,%rbx
  jmp    L16
  nopw   0x0(%rax,%rax,1)
L1:
  add    $0x2,%rbp
  add    $0x1,%rbx
L16:
  mov    %rbx,%rdx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <sscanf@plt>
  cmp    $0x1,%eax
  je     L1
  movb   $0x0,(%rbx)
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

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

tool: D4: %2x

assistant: ```c
void hextorb(uint8_t *hex, uint8_t *rb) {
  while (sscanf(hex, "%2x", rb) == 1) {
    hex += 2;
    rb += 1;
  }
  *rb = '\0';
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<hextorb>:
  endbr64
  push   %r12
  lea    D8(%rip),%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rsi,%rbx
  jmp    L11
  nopw   0x0(%rax,%rax,1)
L17:
  add    $0x2,%rbp
  add    $0x1,%rbx
L11:
  mov    %rbx,%rdx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <sscanf@plt>
  cmp    $0x1,%eax
  je     L17
  movb   $0x0,(%rbx)
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D8
data_type: string

tool: D8: %2x

assistant: ```c
void hextorb(uint8_t *hex, uint8_t *rb) {
  while (sscanf(hex, "%2x", rb) == 1) {
    hex += 2;
    rb += 1;
  }
  *rb = '\0';
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<puth>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  and    $0xf,%eax
  mov    %eax,%edx
  lea    D7(%rip),%rax
  movzbl (%rdx,%rax,1),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <handleC@plt>
  nop
  leave
  ret
```

assistant: ```c
void puth(unsigned n) {
  static const char hex[16] = {'0', '1', '2', '3', '4', '5', '6', '7', '8', '9', 'A', 'B', 'C', 'D', 'E', 'F'};
  handleC(hex[n & 15]);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<puth>:
  endbr64
  sub    $0x8,%rsp
  and    $0xf,%edi
  lea    D5(%rip),%rax
  movsbl (%rax,%rdi,1),%edi
  call   <handleC@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void puth(unsigned n) {
  static const char hex[16] = {'0', '1', '2', '3', '4', '5', '6', '7', '8', '9', 'A', 'B', 'C', 'D', 'E', 'F'};
  handleC(hex[n & 15]);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<puth>:
  endbr64
  and    $0xf,%edi
  lea    D4(%rip),%rax
  movsbl (%rax,%rdi,1),%edi
  jmp    1050 <handleC@plt>
```

assistant: ```c
void puth(unsigned n) {
  static const char hex[16] = {'0', '1', '2', '3', '4', '5', '6', '7', '8', '9', 'A', 'B', 'C', 'D', 'E', 'F'};
  handleC(hex[n & 15]);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<puth>:
  endbr64
  and    $0xf,%edi
  lea    D2(%rip),%rax
  movsbl (%rax,%rdi,1),%edi
  jmp    1050 <handleC@plt>
```

assistant: ```c
void puth(unsigned n) {
  static const char hex[16] = {'0', '1', '2', '3', '4', '5', '6', '7', '8', '9', 'A', 'B', 'C', 'D', 'E', 'F'};
  handleC(hex[n & 15]);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<insect>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x24(%rbp)
  mov    %esi,-0x28(%rbp)
  mov    %edx,-0x2c(%rbp)
  mov    %ecx,-0x30(%rbp)
  movl   $0x0,-0x4(%rbp)
  mov    -0x24(%rbp),%eax
  cmp    -0x28(%rbp),%eax
  jle    L18
  mov    -0x28(%rbp),%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x24(%rbp),%eax
  mov    %eax,-0x10(%rbp)
  jmp    L17
L18:
  mov    -0x24(%rbp),%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x28(%rbp),%eax
  mov    %eax,-0x10(%rbp)
L17:
  mov    -0x2c(%rbp),%eax
  cmp    -0x30(%rbp),%eax
  jle    L7
  mov    -0x30(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x2c(%rbp),%eax
  mov    %eax,-0x8(%rbp)
  jmp    L9
L7:
  mov    -0x2c(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x30(%rbp),%eax
  mov    %eax,-0x8(%rbp)
L9:
  mov    -0x14(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jg     L16
  mov    -0x10(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jl     L16
  movl   $0x1,-0x4(%rbp)
L16:
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
insect(line_c1, line_c2, object_c1, object_c2) int line_c1, line_c2, object_c1, object_c2;
{
  int l1, l2, o1, o2;
  int status = 0;
  if (line_c1 > line_c2) {
    l1 = line_c2;
    l2 = line_c1;
  } else {
    l1 = line_c1;
    l2 = line_c2;
  }
  if (object_c1 > object_c2) {
    o1 = object_c2;
    o2 = object_c1;
  } else {
    o1 = object_c1;
    o2 = object_c2;
  }
  if ((l1 <= o2) && (l2 >= o1))
    status = 1;
  return (status);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<insect>:
  endbr64
  cmp    %esi,%edi
  jg     L17
  mov    %esi,%eax
  mov    %edi,%esi
  mov    %eax,%edi
L17:
  cmp    %ecx,%edx
  jg     L16
  mov    %ecx,%eax
  mov    %edx,%ecx
  mov    %eax,%edx
L16:
  cmp    %edx,%esi
  setle  %dl
  cmp    %ecx,%edi
  setge  %al
  movzbl %al,%eax
  and    %edx,%eax
  ret
```

assistant: ```c
insect(line_c1, line_c2, object_c1, object_c2) int line_c1, line_c2, object_c1, object_c2;
{
  int l1, l2, o1, o2;
  int status = 0;
  if (line_c1 > line_c2) {
    l1 = line_c2;
    l2 = line_c1;
  } else {
    l1 = line_c1;
    l2 = line_c2;
  }
  if (object_c1 > object_c2) {
    o1 = object_c2;
    o2 = object_c1;
  } else {
    o1 = object_c1;
    o2 = object_c2;
  }
  if ((l1 <= o2) && (l2 >= o1))
    status = 1;
  return (status);
}
```