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
<invert>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    -0x20(%rbp),%eax
  cltq
  shl    $0x3,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0x10(%rbp)
  jmp    L37
L35:
  mov    -0x1c(%rbp),%eax
  cltq
  shl    $0x2,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    -0x10(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,8),%rcx
  mov    -0x8(%rbp),%rdx
  add    %rcx,%rdx
  mov    %rax,(%rdx)
  movl   $0x0,-0xc(%rbp)
  jmp    L3
L11:
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0xc(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  test   %eax,%eax
  sete   %cl
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0xc(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rax,%rdx
  movzbl %cl,%eax
  mov    %eax,(%rdx)
  addl   $0x1,-0xc(%rbp)
L3:
  mov    -0xc(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L11
  addl   $0x1,-0x10(%rbp)
L37:
  mov    -0x10(%rbp),%eax
  cmp    -0x20(%rbp),%eax
  jl     L35
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
int **invert(int **original, int w, int h) {
  int i, j;
  int **temp = (int **)malloc(h * sizeof(int *));
  for (i = 0; i < h; i++) {
    temp[i] = (int *)malloc(w * sizeof(int));
    for (j = 0; j < w; j++)
      temp[i][j] = !original[i][j];
  }
  return temp;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<invert>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%r14
  mov    %esi,%r12d
  mov    %edx,%ebx
  lea    0x0(,%rdx,8),%edi
  call   <malloc@plt>
  mov    %rax,0x8(%rsp)
  test   %ebx,%ebx
  jle    L28
  mov    %rax,%rsi
  lea    0x0(,%r12,4),%r15d
  lea    -0x1(%rbx),%eax
  mov    %r14,%rdi
  lea    0x8(%r14,%rax,8),%r14
  lea    -0x1(%r12),%eax
  lea    0x4(,%rax,4),%rbp
  mov    %rdi,%rbx
  sub    %rdi,%rsi
  mov    %rsi,%r13
L10:
  mov    %r15d,%edi
  call   <malloc@plt>
  lea    0x0(%r13,%rbx,1),%rsi
  mov    %rax,(%rsi)
  test   %r12d,%r12d
  jle    L14
  mov    $0x0,%eax
L6:
  mov    (%rsi),%rdx
  mov    (%rbx),%rcx
  cmpl   $0x0,(%rcx,%rax,1)
  sete   %cl
  movzbl %cl,%ecx
  mov    %ecx,(%rdx,%rax,1)
  add    $0x4,%rax
  cmp    %rbp,%rax
  jne    L6
L14:
  add    $0x8,%rbx
  cmp    %r14,%rbx
  jne    L10
L28:
  mov    0x8(%rsp),%rax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
int **invert(int **original, int w, int h) {
  int i, j;
  int **temp = (int **)malloc(h * sizeof(int *));
  for (i = 0; i < h; i++) {
    temp[i] = (int *)malloc(w * sizeof(int));
    for (j = 0; j < w; j++)
      temp[i][j] = !original[i][j];
  }
  return temp;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<invert>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  mov    %edx,%r12d
  push   %rbp
  mov    %rdi,%rbp
  lea    0x0(,%r12,8),%edi
  push   %rbx
  mov    %esi,%ebx
  sub    $0x18,%rsp
  call   <malloc@plt>
  mov    %rax,0x8(%rsp)
  test   %r12d,%r12d
  jle    L6
  mov    %rax,%rdi
  lea    (%rax,%r12,8),%r12
  lea    -0x1(%rbx),%eax
  lea    0x0(,%rbx,4),%r13d
  mov    %rdi,%r15
  sub    %rdi,%rbp
  lea    0x4(,%rax,4),%r14
  nopl   0x0(%rax)
L15:
  mov    %r13d,%edi
  call   <malloc@plt>
  mov    %rax,(%r15)
  test   %ebx,%ebx
  jle    L33
  mov    0x0(%rbp,%r15,1),%rcx
  lea    (%r14,%rax,1),%rsi
  sub    %rax,%rcx
  nopl   0x0(%rax,%rax,1)
L26:
  mov    (%rcx,%rax,1),%edi
  xor    %edx,%edx
  test   %edi,%edi
  sete   %dl
  add    $0x4,%rax
  mov    %edx,-0x4(%rax)
  cmp    %rsi,%rax
  jne    L26
L33:
  add    $0x8,%r15
  cmp    %r12,%r15
  jne    L15
L6:
  mov    0x8(%rsp),%rax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
int **invert(int **original, int w, int h) {
  int i, j;
  int **temp = (int **)malloc(h * sizeof(int *));
  for (i = 0; i < h; i++) {
    temp[i] = (int *)malloc(w * sizeof(int));
    for (j = 0; j < w; j++)
      temp[i][j] = !original[i][j];
  }
  return temp;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<invert>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  mov    %esi,%r13d
  push   %r12
  push   %rbp
  mov    %edx,%ebp
  push   %rbx
  mov    %rdi,%rbx
  lea    0x0(,%rbp,8),%edi
  sub    $0x48,%rsp
  call   <malloc@plt>
  mov    %rax,0x38(%rsp)
  test   %ebp,%ebp
  jle    L61
  mov    %rax,%rsi
  lea    (%rax,%rbp,8),%rax
  lea    0x0(,%r13,4),%r15d
  mov    %r13d,%r14d
  mov    %rax,0x8(%rsp)
  mov    %r13d,%eax
  sub    %rsi,%rbx
  shr    $0x2,%eax
  sub    $0x1,%eax
  mov    %rax,0x18(%rsp)
  mov    %r13d,%eax
  and    $0xfffffffc,%eax
  lea    0x1(%rax),%edi
  movslq %eax,%rbp
  mov    %eax,0x14(%rsp)
  add    $0x2,%eax
  mov    %eax,0x24(%rsp)
  cltq
  movslq %edi,%r12
  shl    $0x2,%rbp
  shl    $0x2,%rax
  mov    %edi,0x20(%rsp)
  shl    $0x2,%r12
  mov    %rax,0x28(%rsp)
  lea    -0x1(%r13),%eax
  mov    %eax,0x10(%rsp)
  lea    0x4(,%rax,4),%rax
  mov    %rax,0x30(%rsp)
  mov    %rbx,%rax
  mov    %rsi,%rbx
  mov    %rax,%r13
  nopw   0x0(%rax,%rax,1)
L82:
  mov    %r15d,%edi
  call   <malloc@plt>
  test   %r14d,%r14d
  movdqa D3(%rip),%xmm2
  mov    %rax,(%rbx)
  mov    %rax,%rdx
  jle    L95
  mov    0x0(%r13,%rbx,1),%rdi
  lea    0x4(%rdi),%rcx
  sub    %rcx,%rax
  cmp    $0x8,%rax
  jbe    L67
  cmpl   $0x2,0x10(%rsp)
  jbe    L67
  mov    0x18(%rsp),%rax
  mov    %rdi,%r9
  pxor   %xmm1,%xmm1
  sub    %rdx,%r9
  lea    0x1(%rax),%rcx
  mov    %rdx,%rax
  shl    $0x4,%rcx
  add    %rdx,%rcx
  nopl   (%rax)
L54:
  movdqu (%rax,%r9,1),%xmm0
  add    $0x10,%rax
  pcmpeqd %xmm1,%xmm0
  pand   %xmm2,%xmm0
  movups %xmm0,-0x10(%rax)
  cmp    %rcx,%rax
  jne    L54
  cmp    0x14(%rsp),%r14d
  je     L95
  mov    (%rdi,%rbp,1),%r9d
  xor    %eax,%eax
  test   %r9d,%r9d
  sete   %al
  mov    %eax,(%rdx,%rbp,1)
  cmp    0x20(%rsp),%r14d
  jle    L95
  mov    (%rdi,%r12,1),%r8d
  xor    %eax,%eax
  test   %r8d,%r8d
  sete   %al
  mov    %eax,(%rdx,%r12,1)
  cmp    0x24(%rsp),%r14d
  jle    L95
  mov    0x28(%rsp),%rsi
  xor    %eax,%eax
  mov    (%rdi,%rsi,1),%edi
  test   %edi,%edi
  sete   %al
  mov    %eax,(%rdx,%rsi,1)
L95:
  add    $0x8,%rbx
  cmp    0x8(%rsp),%rbx
  jne    L82
L61:
  mov    0x38(%rsp),%rax
  add    $0x48,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L67:
  mov    0x30(%rsp),%rax
  sub    %rdx,%rdi
  lea    (%rdx,%rax,1),%rcx
  mov    %rdx,%rax
  nopw   0x0(%rax,%rax,1)
L57:
  mov    (%rdi,%rax,1),%esi
  xor    %edx,%edx
  test   %esi,%esi
  sete   %dl
  add    $0x4,%rax
  mov    %edx,-0x4(%rax)
  cmp    %rcx,%rax
  jne    L57
  jmp    L95
```

tool_calls: name: disassemble_data
arguments: tag: D3
datatype: byte[16]

tool: {"D3": [0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00]}

assistant: ```c
int **invert(int **original, int w, int h) {
  int i, j;
  int **temp = (int **)malloc(h * sizeof(int *));
  for (i = 0; i < h; i++) {
    temp[i] = (int *)malloc(w * sizeof(int));
    for (j = 0; j < w; j++)
      temp[i][j] = !original[i][j];
  }
  return temp;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<digits>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x1869f,-0x4(%rbp)
  jg     L98
  cmpl   $0x3e7,-0x4(%rbp)
  jg     L3
  cmpl   $0x63,-0x4(%rbp)
  jg     L138
  mov    $0x2,%eax
  jmp    L70
L138:
  mov    $0x3,%eax
  jmp    L70
L3:
  cmpl   $0x270f,-0x4(%rbp)
  jg     L88
  mov    $0x4,%eax
  jmp    L70
L88:
  mov    $0x5,%eax
  jmp    L70
L98:
  cmpl   $0x98967f,-0x4(%rbp)
  jg     L117
  cmpl   $0xf423f,-0x4(%rbp)
  jg     L11
  mov    $0x6,%eax
  jmp    L70
L11:
  mov    $0x7,%eax
  jmp    L70
L117:
  cmpl   $0x5f5e0ff,-0x4(%rbp)
  jg     L78
  mov    $0x8,%eax
  jmp    L70
L78:
  mov    $0x9,%eax
L70:
  pop    %rbp
  ret
```

assistant: ```c
int digits(ssize_t n) {
  if (n < 100000) {
    if (n < 1000) {
      if (n < 100)
        return 2;
      else
        return 3;
    } else {
      if (n < 10000)
        return 4;
      else
        return 5;
    }
  } else {
    if (n < 10000000) {
      if (n < 1000000)
        return 6;
      else
        return 7;
    } else {
      if (n < 100000000)
        return 8;
      else
        return 9;
    }
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<digits>:
  endbr64
  cmp    $0x1869f,%edi
  jg     L9
  cmp    $0x3e7,%edi
  jg     L23
  cmp    $0x63,%edi
  setg   %al
  movzbl %al,%eax
  add    $0x2,%eax
  ret
L23:
  cmp    $0x270f,%edi
  setg   %al
  movzbl %al,%eax
  add    $0x4,%eax
  ret
L9:
  cmp    $0x98967f,%edi
  jg     L24
  cmp    $0xf423f,%edi
  setg   %al
  movzbl %al,%eax
  add    $0x6,%eax
  ret
L24:
  cmp    $0x5f5e0ff,%edi
  setg   %al
  movzbl %al,%eax
  add    $0x8,%eax
  ret
```

assistant: ```c
int digits(ssize_t n) {
  if (n < 100000) {
    if (n < 1000) {
      if (n < 100)
        return 2;
      else
        return 3;
    } else {
      if (n < 10000)
        return 4;
      else
        return 5;
    }
  } else {
    if (n < 10000000) {
      if (n < 1000000)
        return 6;
      else
        return 7;
    } else {
      if (n < 100000000)
        return 8;
      else
        return 9;
    }
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<digits>:
  endbr64
  cmp    $0x1869f,%edi
  jg     L9
  cmp    $0x3e7,%edi
  jg     L10
  xor    %eax,%eax
  cmp    $0x63,%edi
  setg   %al
  add    $0x2,%eax
  ret
L9:
  cmp    $0x98967f,%edi
  jg     L13
  xor    %eax,%eax
  cmp    $0xf423f,%edi
  setg   %al
  add    $0x6,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L13:
  xor    %eax,%eax
  cmp    $0x5f5e0ff,%edi
  setg   %al
  add    $0x8,%eax
  ret
  nop
L10:
  xor    %eax,%eax
  cmp    $0x270f,%edi
  setg   %al
  add    $0x4,%eax
  ret
```

assistant: ```c
int digits(ssize_t n) {
  if (n < 100000) {
    if (n < 1000) {
      if (n < 100)
        return 2;
      else
        return 3;
    } else {
      if (n < 10000)
        return 4;
      else
        return 5;
    }
  } else {
    if (n < 10000000) {
      if (n < 1000000)
        return 6;
      else
        return 7;
    } else {
      if (n < 100000000)
        return 8;
      else
        return 9;
    }
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<digits>:
  endbr64
  cmp    $0x1869f,%edi
  jg     L14
  cmp    $0x3e7,%edi
  jg     L7
  xor    %eax,%eax
  cmp    $0x63,%edi
  setg   %al
  add    $0x2,%eax
  ret
L14:
  cmp    $0x98967f,%edi
  jg     L16
  xor    %eax,%eax
  cmp    $0xf423f,%edi
  setg   %al
  add    $0x6,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L16:
  xor    %eax,%eax
  cmp    $0x5f5e0ff,%edi
  setg   %al
  add    $0x8,%eax
  ret
  nop
L7:
  xor    %eax,%eax
  cmp    $0x270f,%edi
  setg   %al
  add    $0x4,%eax
  ret
```

assistant: ```c
int digits(ssize_t n) {
  if (n < 100000) {
    if (n < 1000) {
      if (n < 100)
        return 2;
      else
        return 3;
    } else {
      if (n < 10000)
        return 4;
      else
        return 5;
    }
  } else {
    if (n < 10000000) {
      if (n < 1000000)
        return 6;
      else
        return 7;
    } else {
      if (n < 100000000)
        return 8;
      else
        return 9;
    }
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0xc(%rbp),%rcx
  lea    -0x10(%rbp),%rdx
  lea    -0x14(%rbp),%rax
  mov    %rax,%rsi
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x14(%rbp),%ecx
  mov    -0x10(%rbp),%edx
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L6
  call   <__stack_chk_fail@plt>
L6:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D1
type: string
name: extract_symbolic_data
arguments: data_tag: D8
type: string
name: extract_symbolic_data
arguments: data_tag: D9
type: string

tool: D1: What's the 3 digit number you wish reversed? 

tool: D8: %1d%1d%1d

tool: D9: Your number reversed is %1d%1d%1d.


assistant: ```c
int main(void) {
  int first_digit, second_digit, third_digit;
  printf("What's the 3 digit number you wish reversed? ");
  scanf("%1d%1d%1d", &first_digit, &second_digit, &third_digit);
  printf("Your number reversed is %1d%1d%1d.\n", third_digit, second_digit, first_digit);
  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
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    D8(%rip),%rdi
  call   <printf@plt>
  lea    0x14(%rsp),%rcx
  lea    0x10(%rsp),%rdx
  lea    0xc(%rsp),%rsi
  lea    D16(%rip),%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%ecx
  mov    0x10(%rsp),%edx
  mov    0x14(%rsp),%esi
  lea    D4(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  mov    $0x0,%eax
  add    $0x28,%rsp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D8
datatype: string
name: read_data
arguments: label: D16
datatype: string
name: read_data
arguments: label: D4
datatype: string

tool: D8: What's the 3 digit number you wish reversed? 

tool: D16: %1d%1d%1d

tool: D4: Your number reversed is %1d%1d%1d.


assistant: ```c
int main(void) {
  int first_digit, second_digit, third_digit;
  printf("What's the 3 digit number you wish reversed? ");
  scanf("%1d%1d%1d", &first_digit, &second_digit, &third_digit);
  printf("Your number reversed is %1d%1d%1d.\n", third_digit, second_digit, first_digit);
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  lea    D21(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x14(%rsp),%rcx
  lea    0x10(%rsp),%rdx
  lea    0xc(%rsp),%rsi
  lea    D4(%rip),%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%ecx
  mov    0x10(%rsp),%edx
  xor    %eax,%eax
  mov    0x14(%rsp),%esi
  lea    D28(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  xor    %eax,%eax
  add    $0x28,%rsp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D21
type_infered: string
name: retrieve_data
arguments: data_tag: D4
type_infered: string
name: retrieve_data
arguments: data_tag: D28
type_infered: string

tool: D21: What's the 3 digit number you wish reversed? 

tool: D4: %1d%1d%1d

tool: D28: Your number reversed is %1d%1d%1d.


assistant: ```c
int main(void) {
  int first_digit, second_digit, third_digit;
  printf("What's the 3 digit number you wish reversed? ");
  scanf("%1d%1d%1d", &first_digit, &second_digit, &third_digit);
  printf("Your number reversed is %1d%1d%1d.\n", third_digit, second_digit, first_digit);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  lea    D12(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x14(%rsp),%rcx
  lea    0x10(%rsp),%rdx
  lea    0xc(%rsp),%rsi
  lea    D14(%rip),%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%ecx
  mov    0x10(%rsp),%edx
  xor    %eax,%eax
  mov    0x14(%rsp),%esi
  lea    D19(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  xor    %eax,%eax
  add    $0x28,%rsp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D12
data_type: string
name: retrieve_data
arguments: data_tag: D14
data_type: string
name: retrieve_data
arguments: data_tag: D19
data_type: string

tool: D12: What's the 3 digit number you wish reversed? 

tool: D14: %1d%1d%1d

tool: D19: Your number reversed is %1d%1d%1d.


assistant: ```c
int main(void) {
  int first_digit, second_digit, third_digit;
  printf("What's the 3 digit number you wish reversed? ");
  scanf("%1d%1d%1d", &first_digit, &second_digit, &third_digit);
  printf("Your number reversed is %1d%1d%1d.\n", third_digit, second_digit, first_digit);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x88,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x78(%rsp)
  xor    %eax,%eax
  movl   $0x3f800000,(%rsp)
  movl   $0xbfa40000,0x4(%rsp)
  movss  D25(%rip),%xmm1
  movss  %xmm1,0x8(%rsp)
  movss  D49(%rip),%xmm0
  movss  %xmm0,0xc(%rsp)
  movss  %xmm1,0x10(%rsp)
  movl   $0xbff80000,0x14(%rsp)
  movl   $0x3fa00000,0x18(%rsp)
  movl   $0xbf180000,0x1c(%rsp)
  movl   $0x3e800000,0x20(%rsp)
  movl   $0xbd800000,0x24(%rsp)
  movl   $0x3d000000,0x28(%rsp)
  movl   $0x0,0x2c(%rsp)
  movl   $0x0,0x30(%rsp)
  movl   $0x0,0x40(%rsp)
  movss  D8(%rip),%xmm2
  movss  %xmm2,0x44(%rsp)
  movss  D57(%rip),%xmm3
  movss  %xmm3,0x48(%rsp)
  movss  D34(%rip),%xmm4
  movss  %xmm4,0x4c(%rsp)
  movss  D58(%rip),%xmm5
  movss  %xmm5,0x50(%rsp)
  movss  %xmm0,0x54(%rsp)
  movss  %xmm1,0x58(%rsp)
  movss  %xmm0,0x5c(%rsp)
  movss  %xmm5,0x60(%rsp)
  movss  %xmm4,0x64(%rsp)
  movss  %xmm3,0x68(%rsp)
  movss  %xmm2,0x6c(%rsp)
  movl   $0x0,0x70(%rsp)
  lea    0x40(%rsp),%rsi
  mov    %rsp,%rdi
  call   <__ESBMC_check_stability@plt>
  mov    %eax,%edi
  call   <assert@plt>
  mov    0x78(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  mov    $0x0,%eax
  add    $0x88,%rsp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D25
datatype: f32
name: retrieve_data
arguments: tag: D49
datatype: byte[4]
name: retrieve_data
arguments: tag: D8
datatype: byte[4]
name: retrieve_data
arguments: tag: D57
datatype: f32
name: retrieve_data
arguments: tag: D34
datatype: byte[4]
name: retrieve_data
arguments: tag: D58
datatype: f32

tool: D25: 1.96875

tool: {"D49": [0x00,0x00,0x00,0xc0]}

tool: {"D8": [0x00,0x00,0xc0,0xbd]}

tool: D57: 0.375

tool: {"D34": [0x00,0x00,0x78,0xbf]}

tool: D58: 1.8125

assistant: ```c
int main() {
  float a[] = {1.000000000000000, -1.281250000000000, 1.968750000000000, -2.000000000000000, 1.968750000000000, -1.937500000000000, 1.250000000000000, -0.593750000000000, 0.250000000000000, -0.062500000000000, 0.031250000000000, 0, 0};
  float b[] = {0, -0.093750000000000, 0.375000000000000, -0.968750000000000, 1.812500000000000, -2.000000000000000, 1.968750000000000, -2.000000000000000, 1.812500000000000, -0.968750000000000, 0.375000000000000, -0.093750000000000, 0};
  assert(__ESBMC_check_stability(a, b));
  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
  sub    $0x88,%rsp
  movabs $0xbff800003ffc0000,%rdx
  movabs $0xbf1800003fa00000,%rcx
  movabs $0xbf7800003ec00000,%rsi
  mov    %fs:0x28,%rax
  mov    %rax,0x78(%rsp)
  movabs $0xbfa400003f800000,%rax
  mov    %rdx,0x10(%rsp)
  mov    %rsp,%rdi
  movabs $0xbd8000003e800000,%rdx
  mov    %rcx,0x18(%rsp)
  movabs $0xbdc0000000000000,%rcx
  mov    %rdx,0x20(%rsp)
  movabs $0xc00000003fe80000,%rdx
  mov    %rsi,0x48(%rsp)
  lea    0x40(%rsp),%rsi
  movq   $0x3d000000,0x28(%rsp)
  movl   $0x0,0x30(%rsp)
  mov    %rax,(%rsp)
  movabs $0xc00000003ffc0000,%rax
  mov    %rax,0x8(%rsp)
  mov    %rax,0x58(%rsp)
  movabs $0xbf7800003fe80000,%rax
  mov    %rax,0x60(%rsp)
  movabs $0xbdc000003ec00000,%rax
  mov    %rcx,0x40(%rsp)
  mov    %rdx,0x50(%rsp)
  mov    %rax,0x68(%rsp)
  movl   $0x0,0x70(%rsp)
  call   <__ESBMC_check_stability@plt>
  mov    %eax,%edi
  call   <assert@plt>
  mov    0x78(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  xor    %eax,%eax
  add    $0x88,%rsp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main() {
  float a[] = {1.000000000000000, -1.281250000000000, 1.968750000000000, -2.000000000000000, 1.968750000000000, -1.937500000000000, 1.250000000000000, -0.593750000000000, 0.250000000000000, -0.062500000000000, 0.031250000000000, 0, 0};
  float b[] = {0, -0.093750000000000, 0.375000000000000, -0.968750000000000, 1.812500000000000, -2.000000000000000, 1.968750000000000, -2.000000000000000, 1.812500000000000, -0.968750000000000, 0.375000000000000, -0.093750000000000, 0};
  assert(__ESBMC_check_stability(a, b));
  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
  sub    $0x88,%rsp
  movaps D1(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x78(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  lea    0x40(%rsp),%rsi
  movl   $0x0,0x30(%rsp)
  movaps %xmm0,(%rsp)
  movaps D3(%rip),%xmm0
  movl   $0x0,0x70(%rsp)
  movaps %xmm0,0x10(%rsp)
  movaps D27(%rip),%xmm0
  movaps %xmm0,0x20(%rsp)
  movaps D15(%rip),%xmm0
  movaps %xmm0,0x40(%rsp)
  movaps D17(%rip),%xmm0
  movaps %xmm0,0x50(%rsp)
  movaps D42(%rip),%xmm0
  movaps %xmm0,0x60(%rsp)
  call   <__ESBMC_check_stability@plt>
  mov    %eax,%edi
  call   <assert@plt>
  mov    0x78(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  xor    %eax,%eax
  add    $0x88,%rsp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D1
data_type: f32[4]
name: disassemble_data
arguments: data_label: D3
data_type: f32[4]
name: disassemble_data
arguments: data_label: D27
data_type: f32[4]
name: disassemble_data
arguments: data_label: D15
data_type: byte[16]
name: disassemble_data
arguments: data_label: D17
data_type: f32[4]
name: disassemble_data
arguments: data_label: D42
data_type: f32[4]

tool: {"D1": [1.0,-1.28125,1.96875,-2.0]}

tool: {"D3": [1.96875,-1.9375,1.25,-0.59375]}

tool: {"D27": [0.25,-0.0625,0.03125,0.0]}

tool: {"D15": [0x00,0x00,0x00,0x00,0x00,0x00,0xc0,0xbd,0x00,0x00,0xc0,0x3e,0x00,0x00,0x78,0xbf]}

tool: {"D17": [1.8125,-2.0,1.96875,-2.0]}

tool: {"D42": [1.8125,-0.96875,0.375,-0.09375]}

assistant: ```c
int main() {
  float a[] = {1.000000000000000, -1.281250000000000, 1.968750000000000, -2.000000000000000, 1.968750000000000, -1.937500000000000, 1.250000000000000, -0.593750000000000, 0.250000000000000, -0.062500000000000, 0.031250000000000, 0, 0};
  float b[] = {0, -0.093750000000000, 0.375000000000000, -0.968750000000000, 1.812500000000000, -2.000000000000000, 1.968750000000000, -2.000000000000000, 1.812500000000000, -0.968750000000000, 0.375000000000000, -0.093750000000000, 0};
  assert(__ESBMC_check_stability(a, b));
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<printResults>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    $0x1e8480,%edx
  mov    $0x4,%esi
  lea    D30(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x50,%ecx
  mov    $0xa,%edx
  mov    $0xa,%esi
  lea    D24(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x10,%esi
  lea    D26(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <getTimeElapsed@plt>
  mov    %eax,%esi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  nop
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D30
datatype: string
name: disassemble_data
arguments: data_tag: D24
datatype: string
name: disassemble_data
arguments: data_tag: D26
datatype: string
name: disassemble_data
arguments: data_tag: D7
datatype: string

tool: D30: The end.
	Threads: %d, tasks: %d


tool: D24: Tasks breakdown (%%):
	insert: %d, find: %d, remove: %d


tool: D26: Skip list: %d levels.


tool: D7: Time elapsed:
	%lfs


assistant: ```c
void printResults(struct timeval *begTime) {
  printf("The end.\n\tThreads: %d, tasks: %d\n", 4, 500000 * 4);
  printf("Tasks breakdown (%%):\n\tinsert: %d, find: %d, remove: %d\n", 10, 10, 80);
  printf("Skip list: %d levels.\n", 16);
  printf("Time elapsed:\n\t%lfs\n", getTimeElapsed(begTime));
}
```