ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<__imlib_ItemInList>:
  endbr64
  test   %rdi,%rdi
  je     L38
  push   %r12
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  test   %rdx,%rdx
  je     L15
  test   %esi,%esi
  jle    L15
  lea    -0x1(%rsi),%eax
  mov    %rdi,%rbx
  lea    0x8(%rdi,%rax,8),%r12
  jmp    L59
  cs nopw 0x0(%rax,%rax,1)
L53:
  add    $0x8,%rbx
  cmp    %r12,%rbx
  je     L15
L59:
  mov    (%rbx),%rdi
  mov    %rbp,%rsi
  call   <strcmp@plt>
  test   %eax,%eax
  jne    L53
  pop    %rbx
  mov    $0x1,%eax
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L15:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
  nop
L38:
  xor    %eax,%eax
  ret
```

assistant: ```c
int __imlib_ItemInList(char **list, int size, char *item) {
  int i;
  if (!size)
    return 0;
  if (!list)
    return 0;
  if (!item)
    return 0;
  for (i = 0; i < size; i++) {
    if (!strcmp(list[i], item))
      return 1;
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<__imlib_ItemInList>:
  endbr64
  test   %rdi,%rdi
  je     L47
  push   %r12
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  test   %rdx,%rdx
  je     L19
  test   %esi,%esi
  jle    L19
  lea    -0x1(%rsi),%eax
  mov    %rdi,%rbx
  lea    0x8(%rdi,%rax,8),%r12
  jmp    L48
  cs nopw 0x0(%rax,%rax,1)
L27:
  add    $0x8,%rbx
  cmp    %r12,%rbx
  je     L19
L48:
  mov    (%rbx),%rdi
  mov    %rbp,%rsi
  call   <strcmp@plt>
  test   %eax,%eax
  jne    L27
  pop    %rbx
  mov    $0x1,%eax
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L19:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
  nop
L47:
  xor    %eax,%eax
  ret
```

assistant: ```c
int __imlib_ItemInList(char **list, int size, char *item) {
  int i;
  if (!size)
    return 0;
  if (!list)
    return 0;
  if (!item)
    return 0;
  for (i = 0; i < size; i++) {
    if (!strcmp(list[i], item))
      return 1;
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<foo>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movl   $0x0,-0x4(%rbp)
  jmp    L11
L1:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  lea    D15(%rip),%rax
  mov    (%rdx,%rax,1),%eax
  lea    0x0(,%rax,8),%ecx
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  lea    D15(%rip),%rax
  mov    %ecx,(%rdx,%rax,1)
  addl   $0x1,-0x4(%rbp)
L11:
  cmpl   $0xf,-0x4(%rbp)
  jle    L1
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void foo(void) {
  static unsigned int bm[16];
  int j;
  for (j = 0; j < 16; j++)
    bm[j] = bm[j] * 8;
}
```
user: Translate the below assembly code into C programming language:

```asm
<foo>:
  endbr64
  lea    D4(%rip),%rax
  lea    0x40(%rax),%rdx
L1:
  shll   $0x3,(%rax)
  add    $0x4,%rax
  cmp    %rdx,%rax
  jne    L1
  ret
```

assistant: ```c
void foo(void) {
  static unsigned int bm[16];
  int j;
  for (j = 0; j < 16; j++)
    bm[j] = bm[j] * 8;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<foo>:
  endbr64
  lea    D8(%rip),%rax
  lea    0x40(%rax),%rdx
  nop
L8:
  shll   $0x3,(%rax)
  add    $0x4,%rax
  cmp    %rdx,%rax
  jne    L8
  ret
```

assistant: ```c
void foo(void) {
  static unsigned int bm[16];
  int j;
  for (j = 0; j < 16; j++)
    bm[j] = bm[j] * 8;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<foo>:
  endbr64
  movdqa D60(%rip),%xmm0
  pslld  $0x3,%xmm0
  movaps %xmm0,D15(%rip)
  movdqa D15(%rip),%xmm0
  pslld  $0x3,%xmm0
  movaps %xmm0,D20(%rip)
  movdqa D20(%rip),%xmm0
  pslld  $0x3,%xmm0
  movaps %xmm0,D73(%rip)
  movdqa D73(%rip),%xmm0
  pslld  $0x3,%xmm0
  movaps %xmm0,D67(%rip)
  ret
```

assistant: ```c
void foo(void) {
  static unsigned int bm[16];
  int j;
  for (j = 0; j < 16; j++)
    bm[j] = bm[j] * 8;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<maxProduct>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x80000000,-0x10(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L57
L16:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x10(%rbp),%eax
  cmp    -0x4(%rbp),%eax
  jge    L49
  mov    -0x4(%rbp),%eax
  mov    %eax,-0x10(%rbp)
L49:
  mov    -0xc(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0x8(%rbp)
  jmp    L25
L43:
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    -0x4(%rbp),%edx
  imul   %edx,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x10(%rbp),%eax
  cmp    -0x4(%rbp),%eax
  jge    L50
  mov    -0x4(%rbp),%eax
  mov    %eax,-0x10(%rbp)
L50:
  addl   $0x1,-0x8(%rbp)
L25:
  mov    -0x8(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L43
  addl   $0x1,-0xc(%rbp)
L57:
  mov    -0xc(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L16
  mov    -0x10(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int maxProduct(int *nums, int numsSize) {
  int result = -2147483648;
  int i, j, cumprod;
  for (i = 0; i < numsSize; i++) {
    cumprod = nums[i];
    if (result < cumprod)
      result = cumprod;
    for (j = i + 1; j < numsSize; j++) {
      cumprod = cumprod * nums[j];
      if (result < cumprod)
        result = cumprod;
    }
  }
  return result;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<maxProduct>:
  endbr64
  test   %esi,%esi
  jle    L32
  mov    %esi,%r9d
  mov    $0x1,%r8d
  mov    $0x80000000,%ecx
L31:
  mov    -0x4(%rdi,%r8,4),%eax
  cmp    %eax,%ecx
  cmovl  %eax,%ecx
  cmp    %r9,%r8
  je     L5
  mov    %r8,%rdx
L6:
  imul   (%rdi,%rdx,4),%eax
  cmp    %eax,%ecx
  cmovl  %eax,%ecx
  add    $0x1,%rdx
  cmp    %edx,%esi
  jg     L6
  add    $0x1,%r8
  jmp    L31
L32:
  mov    $0x80000000,%ecx
L5:
  mov    %ecx,%eax
  ret
```

assistant: ```c
int maxProduct(int *nums, int numsSize) {
  int result = -2147483648;
  int i, j, cumprod;
  for (i = 0; i < numsSize; i++) {
    cumprod = nums[i];
    if (result < cumprod)
      result = cumprod;
    for (j = i + 1; j < numsSize; j++) {
      cumprod = cumprod * nums[j];
      if (result < cumprod)
        result = cumprod;
    }
  }
  return result;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<maxProduct>:
  endbr64
  test   %esi,%esi
  jle    L22
  mov    $0x1,%ecx
  mov    $0x80000000,%r8d
  movslq %esi,%r9
  mov    -0x4(%rdi,%rcx,4),%eax
  cmp    %eax,%r8d
  cmovl  %eax,%r8d
  cmp    %rcx,%r9
  je     L14
  cs nopw 0x0(%rax,%rax,1)
L26:
  mov    %rcx,%rdx
  nopl   0x0(%rax,%rax,1)
L31:
  imul   (%rdi,%rdx,4),%eax
  cmp    %eax,%r8d
  cmovl  %eax,%r8d
  add    $0x1,%rdx
  cmp    %edx,%esi
  jg     L31
  add    $0x1,%rcx
  mov    -0x4(%rdi,%rcx,4),%eax
  cmp    %eax,%r8d
  cmovl  %eax,%r8d
  cmp    %rcx,%r9
  jne    L26
L14:
  mov    %r8d,%eax
  ret
L22:
  mov    $0x80000000,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
int maxProduct(int *nums, int numsSize) {
  int result = -2147483648;
  int i, j, cumprod;
  for (i = 0; i < numsSize; i++) {
    cumprod = nums[i];
    if (result < cumprod)
      result = cumprod;
    for (j = i + 1; j < numsSize; j++) {
      cumprod = cumprod * nums[j];
      if (result < cumprod)
        result = cumprod;
    }
  }
  return result;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<maxProduct>:
  endbr64
  test   %esi,%esi
  jle    L37
  mov    (%rdi),%r8d
  movslq %esi,%r9
  mov    $0x1,%ecx
  mov    %r8d,%eax
  cmp    $0x1,%esi
  je     L3
  nopl   0x0(%rax,%rax,1)
L22:
  mov    %rcx,%rdx
  nopl   0x0(%rax,%rax,1)
L28:
  imul   (%rdi,%rdx,4),%eax
  cmp    %eax,%r8d
  cmovl  %eax,%r8d
  add    $0x1,%rdx
  cmp    %edx,%esi
  jg     L28
  mov    (%rdi,%rcx,4),%eax
  cmp    %eax,%r8d
  cmovl  %eax,%r8d
  add    $0x1,%rcx
  cmp    %rcx,%r9
  jne    L22
L3:
  mov    %r8d,%eax
  ret
L37:
  mov    $0x80000000,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
int maxProduct(int *nums, int numsSize) {
  int result = -2147483648;
  int i, j, cumprod;
  for (i = 0; i < numsSize; i++) {
    cumprod = nums[i];
    if (result < cumprod)
      result = cumprod;
    for (j = i + 1; j < numsSize; j++) {
      cumprod = cumprod * nums[j];
      if (result < cumprod)
        result = cumprod;
    }
  }
  return result;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<num_squares>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  cmpl   $0x0,-0x24(%rbp)
  jg     L75
  mov    $0x0,%eax
  jmp    L42
L75:
  mov    -0x24(%rbp),%eax
  mov    %eax,%edi
  call   <sqrt@plt>
  mov    %eax,-0x8(%rbp)
  mov    -0x24(%rbp),%eax
  mov    %eax,-0x10(%rbp)
  movl   $0x1,-0x1c(%rbp)
  jmp    L69
L27:
  mov    -0x24(%rbp),%eax
  mov    %eax,-0x14(%rbp)
  movl   $0x0,-0xc(%rbp)
  mov    -0x1c(%rbp),%eax
  mov    %eax,-0x18(%rbp)
  jmp    L44
L18:
  mov    -0x18(%rbp),%eax
  imul   %eax,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x14(%rbp),%eax
  cltd
  idivl  -0x4(%rbp)
  add    %eax,-0xc(%rbp)
  mov    -0x14(%rbp),%eax
  cltd
  idivl  -0x4(%rbp)
  mov    %edx,-0x14(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  je     L31
  subl   $0x1,-0x18(%rbp)
L44:
  cmpl   $0x0,-0x18(%rbp)
  jg     L18
  jmp    L56
L31:
  nop
L56:
  mov    -0xc(%rbp),%eax
  cmp    -0x10(%rbp),%eax
  jge    L41
  mov    -0xc(%rbp),%eax
  mov    %eax,-0x10(%rbp)
L41:
  addl   $0x1,-0x1c(%rbp)
L69:
  mov    -0x1c(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jle    L27
  mov    -0x10(%rbp),%eax
L42:
  leave
  ret
```

assistant: ```c
int num_squares(int n) {
  if (n <= 0) {
    return 0;
  }
  int end = (int)sqrt(n), square;
  int i, j, temp, least_num = n, current_num;
  for (i = 1; i <= end; ++i) {
    temp = n;
    current_num = 0;
    for (j = i; j >= 1; --j) {
      square = j * j;
      current_num += temp / square;
      temp %= square;
      if (temp == 0) {
        break;
      }
    }
    if (current_num < least_num) {
      least_num = current_num;
    }
  }
  return least_num;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<num_squares>:
  endbr64
  mov    $0x0,%r9d
  test   %edi,%edi
  jle    L52
  push   %rbx
  mov    %edi,%ebx
  call   <sqrt@plt>
  test   %eax,%eax
  jle    L8
  lea    0x1(%rax),%r10d
  mov    %ebx,%r9d
  mov    $0x1,%r8d
  jmp    L53
L9:
  mov    $0x0,%edi
L82:
  cmp    %edi,%r9d
  cmovg  %edi,%r9d
  add    $0x1,%r8d
  cmp    %r8d,%r10d
  je     L0
L53:
  test   %r8d,%r8d
  jle    L9
  mov    %r8d,%ecx
  imul   %r8d,%ecx
  mov    %ebx,%eax
  cltd
  idiv   %ecx
  mov    %eax,%edi
  mov    %edx,%eax
  test   %edx,%edx
  je     L82
  mov    %r8d,%ecx
L88:
  sub    $0x1,%ecx
  je     L82
  mov    %ecx,%esi
  imul   %ecx,%esi
  cltd
  idiv   %esi
  add    %eax,%edi
  mov    %edx,%eax
  test   %edx,%edx
  jne    L88
  jmp    L82
L8:
  mov    %ebx,%r9d
L0:
  mov    %r9d,%eax
  pop    %rbx
  ret
L52:
  mov    %r9d,%eax
  ret
```

assistant: ```c
int num_squares(int n) {
  if (n <= 0) {
    return 0;
  }
  int end = (int)sqrt(n), square;
  int i, j, temp, least_num = n, current_num;
  for (i = 1; i <= end; ++i) {
    temp = n;
    current_num = 0;
    for (j = i; j >= 1; --j) {
      square = j * j;
      current_num += temp / square;
      temp %= square;
      if (temp == 0) {
        break;
      }
    }
    if (current_num < least_num) {
      least_num = current_num;
    }
  }
  return least_num;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<num_squares>:
  endbr64
  xor    %r10d,%r10d
  test   %edi,%edi
  jle    L49
  push   %rbx
  mov    %edi,%ebx
  call   <sqrt@plt>
  test   %eax,%eax
  jle    L17
  lea    0x1(%rax),%edi
  mov    %ebx,%r10d
  mov    $0x1,%r9d
  nopl   0x0(%rax,%rax,1)
L22:
  mov    %ebx,%eax
  mov    %r9d,%ecx
  xor    %r8d,%r8d
L38:
  mov    %ecx,%esi
  cltd
  imul   %ecx,%esi
  idiv   %esi
  add    %eax,%r8d
  mov    %edx,%eax
  test   %edx,%edx
  je     L10
  sub    $0x1,%ecx
  jne    L38
L10:
  cmp    %r8d,%r10d
  cmovg  %r8d,%r10d
  add    $0x1,%r9d
  cmp    %r9d,%edi
  jne    L22
L8:
  mov    %r10d,%eax
  pop    %rbx
  ret
L49:
  mov    %r10d,%eax
  ret
L17:
  mov    %ebx,%r10d
  jmp    L8
```

assistant: ```c
int num_squares(int n) {
  if (n <= 0) {
    return 0;
  }
  int end = (int)sqrt(n), square;
  int i, j, temp, least_num = n, current_num;
  for (i = 1; i <= end; ++i) {
    temp = n;
    current_num = 0;
    for (j = i; j >= 1; --j) {
      square = j * j;
      current_num += temp / square;
      temp %= square;
      if (temp == 0) {
        break;
      }
    }
    if (current_num < least_num) {
      least_num = current_num;
    }
  }
  return least_num;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<num_squares>:
  endbr64
  xor    %r10d,%r10d
  test   %edi,%edi
  jle    L54
  push   %rbx
  mov    %edi,%ebx
  call   <sqrt@plt>
  test   %eax,%eax
  jle    L58
  lea    0x1(%rax),%edi
  mov    %ebx,%r10d
  mov    $0x1,%r9d
  nopl   0x0(%rax,%rax,1)
L56:
  mov    %ebx,%eax
  mov    %r9d,%ecx
  xor    %r8d,%r8d
L41:
  mov    %ecx,%esi
  cltd
  imul   %ecx,%esi
  idiv   %esi
  add    %eax,%r8d
  mov    %edx,%eax
  test   %edx,%edx
  je     L55
  sub    $0x1,%ecx
  jne    L41
L55:
  cmp    %r8d,%r10d
  cmovg  %r8d,%r10d
  add    $0x1,%r9d
  cmp    %r9d,%edi
  jne    L56
L28:
  mov    %r10d,%eax
  pop    %rbx
  ret
L54:
  mov    %r10d,%eax
  ret
L58:
  mov    %ebx,%r10d
  jmp    L28
```

assistant: ```c
int num_squares(int n) {
  if (n <= 0) {
    return 0;
  }
  int end = (int)sqrt(n), square;
  int i, j, temp, least_num = n, current_num;
  for (i = 1; i <= end; ++i) {
    temp = n;
    current_num = 0;
    for (j = i; j >= 1; --j) {
      square = j * j;
      current_num += temp / square;
      temp %= square;
      if (temp == 0) {
        break;
      }
    }
    if (current_num < least_num) {
      least_num = current_num;
    }
  }
  return least_num;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<user_free>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  mov    %rdi,-0x20(%rbp)
  mov    $0x44,%eax
  mov    $0xd,%esi
  mov    -0x20(%rbp),%rdx
  mov    %esi,%ebx
  mov    %rdx,%rcx
  int    $0x40
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int user_free(void *mem) {
  int val;
  __asm__ __volatile__("int $0x40" : "=a"(val) : "a"(68), "b"(13), "c"(mem));
  return val;
}
```
user: Translate the below assembly code into C programming language:

```asm
<user_free>:
  endbr64
  push   %rbx
  mov    $0x44,%eax
  mov    $0xd,%ebx
  mov    %rdi,%rcx
  int    $0x40
  pop    %rbx
  ret
```

assistant: ```c
int user_free(void *mem) {
  int val;
  __asm__ __volatile__("int $0x40" : "=a"(val) : "a"(68), "b"(13), "c"(mem));
  return val;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<user_free>:
  endbr64
  push   %rbx
  mov    $0x44,%eax
  mov    $0xd,%ebx
  mov    %rdi,%rcx
  int    $0x40
  pop    %rbx
  ret
```

assistant: ```c
int user_free(void *mem) {
  int val;
  __asm__ __volatile__("int $0x40" : "=a"(val) : "a"(68), "b"(13), "c"(mem));
  return val;
}
```
user: Translate the below assembly code into C programming language:

```asm
<errorFn>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
L1:
  jmp    L1
```

assistant: ```c
void errorFn(void) {
  {
  ERROR:
    goto ERROR;
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<errorFn>:
  endbr64
L6:
  jmp    L6
```

assistant: ```c
void errorFn(void) {
  {
  ERROR:
    goto ERROR;
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<errorFn>:
  endbr64
  nopl   0x0(%rax)
L0:
  jmp    L0
```

assistant: ```c
void errorFn(void) {
  {
  ERROR:
    goto ERROR;
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<errorFn>:
  endbr64
  nopl   0x0(%rax)
L3:
  jmp    L3
```

assistant: ```c
void errorFn(void) {
  {
  ERROR:
    goto ERROR;
  }
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xe0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movabs $0x74206120616c6f48,%rax
  mov    %rax,-0xdd(%rbp)
  movl   $0x736f646f,-0xd5(%rbp)
  movb   $0x0,-0xd1(%rbp)
  lea    -0xdd(%rbp),%rax
  mov    %rax,%rsi
  lea    D15(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0xdd(%rbp),%rax
  mov    %rax,%rdi
  call   <longitud_string@plt>
  mov    %eax,%esi
  lea    D31(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0xd0(%rbp),%rax
  lea    D30(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  lea    -0xd0(%rbp),%rax
  mov    %rax,%rsi
  lea    D48(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0xd0(%rbp),%rax
  mov    %rax,%rdi
  call   <longitud_string@plt>
  mov    %eax,%esi
  lea    D31(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L1
  call   <__stack_chk_fail@plt>
L1:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D15
data_type: string
name: extract_symbolic_data
arguments: tag: D31
data_type: string
name: extract_symbolic_data
arguments: tag: D30
data_type: string
name: extract_symbolic_data
arguments: tag: D48
data_type: string

tool: D15: cadena: %s


tool: D31: longitud cadena: %d


tool: D30: leer libros y revistas

tool: D48: cadena2: %s


assistant: ```c
int main(void) {
  char string1[] = "Hola a todos";
  char string2[200];
  printf("cadena: %s\n", string1);
  printf("longitud cadena: %d\n", longitud_string(string1));
  strcpy(string2, "leer libros y revistas");
  printf("cadena2: %s\n", string2);
  printf("longitud cadena: %d\n", longitud_string(string2));
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0xe8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xd8(%rsp)
  xor    %eax,%eax
  movabs $0x74206120616c6f48,%rax
  mov    %rax,0x3(%rsp)
  movl   $0x736f646f,0xb(%rsp)
  movb   $0x0,0xf(%rsp)
  lea    0x3(%rsp),%rbx
  mov    %rbx,%rsi
  lea    D1(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %rbx,%rdi
  call   <longitud_string@plt>
  mov    %eax,%esi
  lea    D2(%rip),%rbp
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0x10(%rsp),%rbx
  lea    D26(%rip),%rsi
  mov    %rbx,%rdi
  call   <strcpy@plt>
  mov    %rbx,%rsi
  lea    D18(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %rbx,%rdi
  call   <longitud_string@plt>
  mov    %eax,%esi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0xd8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  mov    $0x0,%eax
  add    $0xe8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D1
type_guessed: string
name: disassemble_data
arguments: label: D2
type_guessed: string
name: disassemble_data
arguments: label: D26
type_guessed: string
name: disassemble_data
arguments: label: D18
type_guessed: string

tool: D1: cadena: %s


tool: D2: longitud cadena: %d


tool: D26: leer libros y revistas

tool: D18: cadena2: %s


assistant: ```c
int main(void) {
  char string1[] = "Hola a todos";
  char string2[200];
  printf("cadena: %s\n", string1);
  printf("longitud cadena: %d\n", longitud_string(string1));
  strcpy(string2, "leer libros y revistas");
  printf("cadena2: %s\n", string2);
  printf("longitud cadena: %d\n", longitud_string(string2));
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r12
  lea    D38(%rip),%rdi
  lea    D3(%rip),%r12
  push   %rbp
  sub    $0xe8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xd8(%rsp)
  movabs $0x74206120616c6f48,%rax
  lea    0x3(%rsp),%rbp
  movl   $0x736f646f,0xb(%rsp)
  mov    %rbp,%rsi
  movb   $0x0,0xf(%rsp)
  mov    %rax,0x3(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rdi
  lea    0x10(%rsp),%rbp
  call   <longitud_string@plt>
  mov    %r12,%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  lea    D13(%rip),%rsi
  mov    %rbp,%rdi
  call   <strcpy@plt>
  mov    %rbp,%rsi
  lea    D12(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <longitud_string@plt>
  mov    %r12,%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0xd8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  add    $0xe8,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D38
type: string
name: disassemble_data
arguments: label: D3
type: string
name: disassemble_data
arguments: label: D13
type: string
name: disassemble_data
arguments: label: D12
type: string

tool: D38: cadena: %s


tool: D3: longitud cadena: %d


tool: D13: leer libros y revistas

tool: D12: cadena2: %s


assistant: ```c
int main(void) {
  char string1[] = "Hola a todos";
  char string2[200];
  printf("cadena: %s\n", string1);
  printf("longitud cadena: %d\n", longitud_string(string1));
  strcpy(string2, "leer libros y revistas");
  printf("cadena2: %s\n", string2);
  printf("longitud cadena: %d\n", longitud_string(string2));
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %r12
  lea    D32(%rip),%rdi
  lea    D20(%rip),%r12
  push   %rbp
  sub    $0xe8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xd8(%rsp)
  movabs $0x74206120616c6f48,%rax
  lea    0x3(%rsp),%rbp
  movl   $0x736f646f,0xb(%rsp)
  mov    %rbp,%rsi
  movb   $0x0,0xf(%rsp)
  mov    %rax,0x3(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rdi
  lea    0x10(%rsp),%rbp
  call   <longitud_string@plt>
  mov    %r12,%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  lea    D14(%rip),%rsi
  mov    %rbp,%rdi
  call   <strcpy@plt>
  mov    %rbp,%rsi
  lea    D21(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <longitud_string@plt>
  mov    %r12,%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0xd8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0xe8,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D32
type_infered: string
name: unpack_data
arguments: tag: D20
type_infered: string
name: unpack_data
arguments: tag: D14
type_infered: string
name: unpack_data
arguments: tag: D21
type_infered: string

tool: D32: cadena: %s


tool: D20: longitud cadena: %d


tool: D14: leer libros y revistas

tool: D21: cadena2: %s


assistant: ```c
int main(void) {
  char string1[] = "Hola a todos";
  char string2[200];
  printf("cadena: %s\n", string1);
  printf("longitud cadena: %d\n", longitud_string(string1));
  strcpy(string2, "leer libros y revistas");
  printf("cadena2: %s\n", string2);
  printf("longitud cadena: %d\n", longitud_string(string2));
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<underline>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,%eax
  mov    %al,-0x20(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L9
L14:
  movsbl -0x20(%rbp),%eax
  mov    -0x18(%rbp),%rdx
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <fputc@plt>
  addl   $0x1,-0x4(%rbp)
L9:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L14
  mov    -0x18(%rbp),%rax
  mov    %rax,%rsi
  mov    $0xa,%edi
  call   <fputc@plt>
  nop
  leave
  ret
```

assistant: ```c
void underline(FILE *ofp, int count, char character) {
  int i;
  for (i = 0; i < count; i++)
    fprintf(ofp, "%c", character);
  fprintf(ofp, "\n");
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<underline>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r12
  test   %esi,%esi
  jle    L10
  mov    %esi,%r13d
  mov    $0x0,%ebx
  movsbl %dl,%ebp
L13:
  mov    %r12,%rsi
  mov    %ebp,%edi
  call   <fputc@plt>
  add    $0x1,%ebx
  cmp    %ebx,%r13d
  jne    L13
L10:
  mov    %r12,%rsi
  mov    $0xa,%edi
  call   <fputc@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
void underline(FILE *ofp, int count, char character) {
  int i;
  for (i = 0; i < count; i++)
    fprintf(ofp, "%c", character);
  fprintf(ofp, "\n");
}
```