ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<decrypt>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    %rdx,%r14
  mov    %ecx,%r13d
  mov    %r8d,%ebp
  mov    %r9d,%ebx
  mov    %ecx,%edi
  call   <KeySchedule@plt>
  cmp    $0x2ef00,%r13d
  je     L141
  jg     L65
  cmp    $0x1f500,%r13d
  je     L44
  jle    L42
  cmp    $0x2ee80,%r13d
  jne    L103
  mov    $0xc,%ebx
  mov    $0x4,%ebp
  jmp    L134
L42:
  cmp    $0x1f480,%r13d
  jne    L217
  mov    $0xa,%ebx
  mov    $0x4,%ebp
  jmp    L134
L217:
  cmp    $0x1f4c0,%r13d
  jne    L212
  mov    $0xc,%ebx
  mov    $0x6,%ebp
L134:
  mov    %ebx,%ecx
  mov    %r13d,%edx
  mov    %r14,%rsi
  mov    %r12,%rdi
  call   <AddRoundKey@plt>
  mov    %ebp,%esi
  mov    %r12,%rdi
  call   <InversShiftRow_ByteSub@plt>
  sub    $0x1,%ebx
L74:
  mov    %ebx,%ecx
  mov    %ebp,%edx
  mov    %r14,%rsi
  mov    %r12,%rdi
  call   <AddRoundKey_InversMixColumn@plt>
  mov    %ebp,%esi
  mov    %r12,%rdi
  call   <InversShiftRow_ByteSub@plt>
  sub    $0x1,%ebx
  test   %ebx,%ebx
  jg     L74
L52:
  mov    $0x0,%ecx
  mov    %r13d,%edx
  mov    %r14,%rsi
  mov    %r12,%rdi
  call   <AddRoundKey@plt>
  mov    $0x0,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L103:
  cmp    $0x2eec0,%r13d
  jne    L212
  mov    $0xc,%ebx
  mov    $0x6,%ebp
  jmp    L134
L65:
  cmp    $0x3e8c0,%r13d
  je     L114
  cmp    $0x3e900,%r13d
  jne    L45
  mov    $0xe,%ebx
  mov    $0x8,%ebp
  jmp    L134
L45:
  cmp    $0x3e880,%r13d
  jne    L212
  mov    $0xe,%ebx
  mov    $0x4,%ebp
  jmp    L134
L212:
  mov    %ebx,%ecx
  mov    %r13d,%edx
  mov    %r14,%rsi
  mov    %r12,%rdi
  call   <AddRoundKey@plt>
  mov    %ebp,%esi
  mov    %r12,%rdi
  call   <InversShiftRow_ByteSub@plt>
  sub    $0x1,%ebx
  test   %ebx,%ebx
  jle    L52
  jmp    L74
L141:
  mov    $0xe,%ebx
  mov    $0x8,%ebp
  jmp    L134
L44:
  mov    $0xe,%ebx
  mov    $0x8,%ebp
  jmp    L134
L114:
  mov    $0xe,%ebx
  mov    $0x6,%ebp
  jmp    L134
```

assistant: ```c
int decrypt(int statemt[32], int key[32], int word[4][120], int type, int nb, int round) {
  int i;
  KeySchedule(type, key, word);
  switch (type) {
  case 128128:
    round = 10;
    nb = 4;
    break;
  case 128192:
  case 192192:
    round = 12;
    nb = 6;
    break;
  case 192128:
    round = 12;
    nb = 4;
    break;
  case 128256:
  case 192256:
    round = 14;
    nb = 8;
    break;
  case 256128:
    round = 14;
    nb = 4;
    break;
  case 256192:
    round = 14;
    nb = 6;
    break;
  case 256256:
    round = 14;
    nb = 8;
    break;
  }
  AddRoundKey(statemt, word, type, round);
  InversShiftRow_ByteSub(statemt, nb);
#pragma unroll
  for (i = round - 1; i >= 1; --i) {
    AddRoundKey_InversMixColumn(statemt, word, nb, i);
    InversShiftRow_ByteSub(statemt, nb);
  }
  AddRoundKey(statemt, word, type, 0);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<pokevstr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    %rdx,-0x20(%rbp)
  mov    %ecx,-0x24(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L1
L5:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%edx
  mov    -0x14(%rbp),%ecx
  mov    -0xc(%rbp),%eax
  lea    (%rcx,%rax,1),%edi
  mov    -0x24(%rbp),%ecx
  mov    -0x18(%rbp),%eax
  mov    %eax,%esi
  call   <pokech@plt>
  addl   $0x1,-0xc(%rbp)
  addq   $0x1,-0x8(%rbp)
L1:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L5
  nop
  nop
  leave
  ret
```

assistant: ```c
void pokevstr(int row, int column, const char *str, int color) {
  const char *p = str;
  int i = 0;
  while (*p) {
    pokech(row + i, column, *p, color);
    i++;
    p++;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<pokevstr>:
  endbr64
  movzbl (%rdx),%eax
  test   %al,%al
  je     L12
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %esi,%r12d
  mov    %ecx,%r13d
  mov    %rdx,%rbx
  sub    %edx,%edi
  mov    %edi,%ebp
L2:
  movsbl %al,%edx
  lea    0x0(%rbp,%rbx,1),%edi
  mov    %r13d,%ecx
  mov    %r12d,%esi
  call   <pokech@plt>
  add    $0x1,%rbx
  movzbl (%rbx),%eax
  test   %al,%al
  jne    L2
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L12:
  ret
```

assistant: ```c
void pokevstr(int row, int column, const char *str, int color) {
  const char *p = str;
  int i = 0;
  while (*p) {
    pokech(row + i, column, *p, color);
    i++;
    p++;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<pokevstr>:
  endbr64
  movzbl (%rdx),%eax
  test   %al,%al
  je     L18
  push   %r13
  sub    %edx,%edi
  push   %r12
  mov    %edi,%r13d
  mov    %ecx,%r12d
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %rdx,%rbx
  sub    $0x8,%rsp
  nopw   0x0(%rax,%rax,1)
L6:
  lea    0x0(%r13,%rbx,1),%edi
  movsbl %al,%edx
  mov    %r12d,%ecx
  mov    %ebp,%esi
  call   <pokech@plt>
  movzbl 0x1(%rbx),%eax
  add    $0x1,%rbx
  test   %al,%al
  jne    L6
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L18:
  ret
```

assistant: ```c
void pokevstr(int row, int column, const char *str, int color) {
  const char *p = str;
  int i = 0;
  while (*p) {
    pokech(row + i, column, *p, color);
    i++;
    p++;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<pokevstr>:
  endbr64
  movzbl (%rdx),%eax
  test   %al,%al
  je     L14
  push   %r13
  sub    %edx,%edi
  push   %r12
  mov    %edi,%r13d
  mov    %ecx,%r12d
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %rdx,%rbx
  sub    $0x8,%rsp
  nopw   0x0(%rax,%rax,1)
L17:
  lea    0x0(%r13,%rbx,1),%edi
  movsbl %al,%edx
  mov    %r12d,%ecx
  mov    %ebp,%esi
  call   <pokech@plt>
  movzbl 0x1(%rbx),%eax
  add    $0x1,%rbx
  test   %al,%al
  jne    L17
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L14:
  ret
```

assistant: ```c
void pokevstr(int row, int column, const char *str, int color) {
  const char *p = str;
  int i = 0;
  while (*p) {
    pokech(row + i, column, *p, color);
    i++;
    p++;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<TriPyramid>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  movsd  %xmm0,-0x18(%rbp)
  mov    D2(%rip),%rax
  mov    $0x0,%edi
  movq   %rax,%xmm0
  call   <pow@plt>
  movsd  %xmm0,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  mov    $0x2,%edi
  movq   %rax,%xmm0
  call   <pow@plt>
  mulsd  -0x20(%rbp),%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movsd  -0x8(%rbp),%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  leave
  ret
```

tool_calls: name: unpack_data
arguments: tag: D2
type_guessed: byte[8]

tool: {"D2": [0x00,0x00,0x00,0x00,0x00,0x00,0x08,0x40]}

assistant: ```c
double TriPyramid(double EdgeLength) {
  double SArea;
  SArea = pow(3, .5) * pow(EdgeLength, 2);
  return SArea;
}
```
user: What is the c source code of the assembly code below?

```asm
<TriPyramid>:
  endbr64
  push   %r14
  push   %rbx
  sub    $0x8,%rsp
  movq   %xmm0,%rbx
  mov    $0x0,%edi
  movsd  D7(%rip),%xmm0
  call   <pow@plt>
  movq   %xmm0,%r14
  mov    $0x2,%edi
  movq   %rbx,%xmm0
  call   <pow@plt>
  movq   %r14,%xmm1
  mulsd  %xmm1,%xmm0
  add    $0x8,%rsp
  pop    %rbx
  pop    %r14
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D7
datatype: byte[8]

tool: {"D7": [0x00,0x00,0x00,0x00,0x00,0x00,0x08,0x40]}

assistant: ```c
double TriPyramid(double EdgeLength) {
  double SArea;
  SArea = pow(3, .5) * pow(EdgeLength, 2);
  return SArea;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<TriPyramid>:
  endbr64
  sub    $0x18,%rsp
  xor    %edi,%edi
  movsd  %xmm0,0x8(%rsp)
  movsd  D8(%rip),%xmm0
  call   <pow@plt>
  movsd  0x8(%rsp),%xmm1
  mov    $0x2,%edi
  movsd  %xmm0,(%rsp)
  movapd %xmm1,%xmm0
  call   <pow@plt>
  mulsd  (%rsp),%xmm0
  add    $0x18,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D8
type_guessed: byte[8]

tool: {"D8": [0x00,0x00,0x00,0x00,0x00,0x00,0x08,0x40]}

assistant: ```c
double TriPyramid(double EdgeLength) {
  double SArea;
  SArea = pow(3, .5) * pow(EdgeLength, 2);
  return SArea;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<TriPyramid>:
  endbr64
  sub    $0x18,%rsp
  xor    %edi,%edi
  movsd  %xmm0,0x8(%rsp)
  movsd  D3(%rip),%xmm0
  call   <pow@plt>
  movsd  0x8(%rsp),%xmm1
  mov    $0x2,%edi
  movsd  %xmm0,(%rsp)
  movapd %xmm1,%xmm0
  call   <pow@plt>
  mulsd  (%rsp),%xmm0
  add    $0x18,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D3
data_type: byte[8]

tool: {"D3": [0x00,0x00,0x00,0x00,0x00,0x00,0x08,0x40]}

assistant: ```c
double TriPyramid(double EdgeLength) {
  double SArea;
  SArea = pow(3, .5) * pow(EdgeLength, 2);
  return SArea;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
L51:
  lea    D91(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x1c(%rbp),%rax
  mov    %rax,%rsi
  lea    D89(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  movss  -0x1c(%rbp),%xmm0
  comiss D45(%rip),%xmm0
  ja     L51
  movss  -0x1c(%rbp),%xmm1
  pxor   %xmm0,%xmm0
  comiss %xmm1,%xmm0
  jae    L51
  movss  D65(%rip),%xmm0
  movss  %xmm0,-0x14(%rbp)
  movss  D65(%rip),%xmm0
  movss  %xmm0,-0xc(%rbp)
  jmp    L57
L26:
  movss  D65(%rip),%xmm0
  movss  %xmm0,-0x18(%rbp)
  movss  -0xc(%rbp),%xmm0
  movss  %xmm0,-0x10(%rbp)
  jmp    L32
L35:
  movss  -0x10(%rbp),%xmm0
  movss  D65(%rip),%xmm2
  movaps %xmm0,%xmm1
  subss  %xmm2,%xmm1
  movss  %xmm1,-0x10(%rbp)
  movss  -0x18(%rbp),%xmm1
  mulss  %xmm1,%xmm0
  movss  %xmm0,-0x18(%rbp)
L32:
  movss  -0x10(%rbp),%xmm0
  pxor   %xmm1,%xmm1
  comiss %xmm1,%xmm0
  ja     L35
  movss  D65(%rip),%xmm0
  divss  -0x18(%rbp),%xmm0
  movss  -0x14(%rbp),%xmm1
  addss  %xmm1,%xmm0
  movss  %xmm0,-0x14(%rbp)
  movss  -0xc(%rbp),%xmm1
  movss  D65(%rip),%xmm0
  addss  %xmm1,%xmm0
  movss  %xmm0,-0xc(%rbp)
L57:
  movss  -0x1c(%rbp),%xmm0
  comiss -0xc(%rbp),%xmm0
  jae    L26
  pxor   %xmm3,%xmm3
  cvtss2sd -0x14(%rbp),%xmm3
  movq   %xmm3,%rax
  movq   %rax,%xmm0
  lea    D89(%rip),%rax
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  lea    D37(%rip),%rax
  mov    %rax,%rdi
  call   <system@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L53
  call   <__stack_chk_fail@plt>
L53:
  leave
  ret
```

tool_calls: name: read_data
arguments: data_label: D91
type_guessed: string
name: read_data
arguments: data_label: D89
type_guessed: string
name: read_data
arguments: data_label: D45
type_guessed: byte[4]
name: read_data
arguments: data_label: D65
type_guessed: byte[4]
name: read_data
arguments: data_label: D37
type_guessed: string

tool: D91: Inserire la quantita dei numeri

tool: D89: %f

tool: {"D45": [0x00,0x00,0x20,0x41]}

tool: {"D65": [0x00,0x00,0x80,0x3f]}

tool: D37: PAUSE

assistant: ```c
main() {
  float N;
  float Fat;
  float E;
  float K;
  float I;
  do {
    printf("Inserire la quantita dei numeri");
    scanf("%f", &N);
  } while ((N > 10) || (N <= 0));
  E = 1;
  I = 1;
  while (I <= N) {
    Fat = 1;
    K = I;
    while (K > 0) {
      Fat = Fat * (K--);
    }
    E = E + (1 / Fat);
    I++;
  }
  printf("%f", E);
  system("PAUSE");
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D8(%rip),%r12
  lea    0x4(%rsp),%rbp
  lea    D62(%rip),%rbx
L62:
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  movss  0x4(%rsp),%xmm5
  comiss D38(%rip),%xmm5
  ja     L62
  pxor   %xmm7,%xmm7
  comiss %xmm5,%xmm7
  jae    L62
  movss  D15(%rip),%xmm6
  movaps %xmm6,%xmm7
  comiss %xmm6,%xmm5
  jb     L85
  pxor   %xmm9,%xmm9
  movaps %xmm6,%xmm4
  movaps %xmm9,%xmm3
  movaps %xmm6,%xmm8
  jmp    L88
L85:
  movss  D15(%rip),%xmm7
L45:
  pxor   %xmm0,%xmm0
  cvtss2sd %xmm7,%xmm0
  lea    D62(%rip),%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  lea    D41(%rip),%rdi
  call   <system@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L75
  mov    $0x0,%eax
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L19:
  movaps %xmm0,%xmm2
  subss  %xmm4,%xmm0
  mulss  %xmm2,%xmm1
  comiss %xmm3,%xmm0
  ja     L19
L69:
  movaps %xmm8,%xmm0
  divss  %xmm1,%xmm0
  addss  %xmm0,%xmm7
  addss  %xmm8,%xmm6
  comiss %xmm6,%xmm5
  jb     L45
L88:
  movaps %xmm6,%xmm0
  movss  D15(%rip),%xmm1
  comiss %xmm9,%xmm6
  ja     L19
  jmp    L69
L75:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D8
datatype: string
name: read_data
arguments: tag: D62
datatype: string
name: read_data
arguments: tag: D38
datatype: byte[4]
name: read_data
arguments: tag: D15
datatype: byte[4]
name: read_data
arguments: tag: D41
datatype: string

tool: D8: Inserire la quantita dei numeri

tool: D62: %f

tool: {"D38": [0x00,0x00,0x20,0x41]}

tool: {"D15": [0x00,0x00,0x80,0x3f]}

tool: D41: PAUSE

assistant: ```c
main() {
  float N;
  float Fat;
  float E;
  float K;
  float I;
  do {
    printf("Inserire la quantita dei numeri");
    scanf("%f", &N);
  } while ((N > 10) || (N <= 0));
  E = 1;
  I = 1;
  while (I <= N) {
    Fat = 1;
    K = I;
    while (K > 0) {
      Fat = Fat * (K--);
    }
    E = E + (1 / Fat);
    I++;
  }
  printf("%f", E);
  system("PAUSE");
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r12
  lea    D3(%rip),%r12
  push   %rbp
  lea    D5(%rip),%rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rbx
  nop
L70:
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  movss  0x4(%rsp),%xmm5
  pxor   %xmm4,%xmm4
  comiss D28(%rip),%xmm5
  ja     L70
  comiss %xmm5,%xmm4
  jae    L70
  movss  D43(%rip),%xmm3
  comiss %xmm3,%xmm5
  movaps %xmm3,%xmm6
  movaps %xmm3,%xmm7
  jb     L58
L27:
  comiss %xmm4,%xmm6
  movaps %xmm6,%xmm0
  movaps %xmm3,%xmm1
  jbe    L63
  nopl   0x0(%rax,%rax,1)
L8:
  movaps %xmm0,%xmm2
  subss  %xmm3,%xmm0
  mulss  %xmm2,%xmm1
  comiss %xmm4,%xmm0
  ja     L8
  movaps %xmm3,%xmm0
  divss  %xmm1,%xmm0
L6:
  addss  %xmm3,%xmm6
  addss  %xmm0,%xmm7
  comiss %xmm6,%xmm5
  jae    L27
L58:
  mov    %rbp,%rdi
  pxor   %xmm0,%xmm0
  mov    $0x1,%eax
  cvtss2sd %xmm7,%xmm0
  call   <printf@plt>
  lea    D16(%rip),%rdi
  call   <system@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L69
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L63:
  movaps %xmm3,%xmm0
  jmp    L6
L69:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D3
type_infered: string
name: unpack_data
arguments: data_label: D5
type_infered: string
name: unpack_data
arguments: data_label: D28
type_infered: byte[4]
name: unpack_data
arguments: data_label: D43
type_infered: byte[4]
name: unpack_data
arguments: data_label: D16
type_infered: string

tool: D3: Inserire la quantita dei numeri

tool: D5: %f

tool: {"D28": [0x00,0x00,0x20,0x41]}

tool: {"D43": [0x00,0x00,0x80,0x3f]}

tool: D16: PAUSE

assistant: ```c
main() {
  float N;
  float Fat;
  float E;
  float K;
  float I;
  do {
    printf("Inserire la quantita dei numeri");
    scanf("%f", &N);
  } while ((N > 10) || (N <= 0));
  E = 1;
  I = 1;
  while (I <= N) {
    Fat = 1;
    K = I;
    while (K > 0) {
      Fat = Fat * (K--);
    }
    E = E + (1 / Fat);
    I++;
  }
  printf("%f", E);
  system("PAUSE");
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  push   %r12
  lea    D39(%rip),%r12
  push   %rbp
  lea    D37(%rip),%rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rbx
  nop
L45:
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  movss  0x4(%rsp),%xmm5
  pxor   %xmm4,%xmm4
  comiss D30(%rip),%xmm5
  ja     L45
  comiss %xmm5,%xmm4
  jae    L45
  movss  D20(%rip),%xmm3
  comiss %xmm3,%xmm5
  movaps %xmm3,%xmm6
  movaps %xmm3,%xmm7
  jb     L54
L68:
  comiss %xmm4,%xmm6
  movaps %xmm6,%xmm0
  movaps %xmm3,%xmm1
  jbe    L79
  nopl   0x0(%rax,%rax,1)
L77:
  movaps %xmm0,%xmm2
  subss  %xmm3,%xmm0
  mulss  %xmm2,%xmm1
  comiss %xmm4,%xmm0
  ja     L77
  movaps %xmm3,%xmm0
  divss  %xmm1,%xmm0
L26:
  addss  %xmm3,%xmm6
  addss  %xmm0,%xmm7
  comiss %xmm6,%xmm5
  jae    L68
L54:
  mov    %rbp,%rdi
  pxor   %xmm0,%xmm0
  mov    $0x1,%eax
  cvtss2sd %xmm7,%xmm0
  call   <printf@plt>
  lea    D23(%rip),%rdi
  call   <system@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L79:
  movaps %xmm3,%xmm0
  jmp    L26
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D39
datatype: string
name: read_data
arguments: tag: D37
datatype: string
name: read_data
arguments: tag: D30
datatype: byte[4]
name: read_data
arguments: tag: D20
datatype: byte[4]
name: read_data
arguments: tag: D23
datatype: string

tool: D39: Inserire la quantita dei numeri

tool: D37: %f

tool: {"D30": [0x00,0x00,0x20,0x41]}

tool: {"D20": [0x00,0x00,0x80,0x3f]}

tool: D23: PAUSE

assistant: ```c
main() {
  float N;
  float Fat;
  float E;
  float K;
  float I;
  do {
    printf("Inserire la quantita dei numeri");
    scanf("%f", &N);
  } while ((N > 10) || (N <= 0));
  E = 1;
  I = 1;
  while (I <= N) {
    Fat = 1;
    K = I;
    while (K > 0) {
      Fat = Fat * (K--);
    }
    E = E + (1 / Fat);
    I++;
  }
  printf("%f", E);
  system("PAUSE");
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x64,-0x8(%rbp)
  mov    -0x8(%rbp),%eax
  mov    %eax,%edi
  call   <noOfWays@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%edx
  mov    -0x8(%rbp),%eax
  mov    %eax,%esi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x1,%eax
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D7
type_infered: string

tool: D7: Number of ways to reach %d is : %d


assistant: ```c
int main() {
  int sum = 100;
  int n = noOfWays(sum);
  printf("Number of ways to reach %d is : %d\n", sum, n);
  return 1;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x64,%edi
  call   <noOfWays@plt>
  mov    %eax,%edx
  mov    $0x64,%esi
  lea    D0(%rip),%rdi
  call   <printf@plt>
  mov    $0x1,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: label: D0
type: string

tool: D0: Number of ways to reach %d is : %d


assistant: ```c
int main() {
  int sum = 100;
  int n = noOfWays(sum);
  printf("Number of ways to reach %d is : %d\n", sum, n);
  return 1;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x64,%edi
  call   <noOfWays@plt>
  mov    $0x64,%esi
  lea    D0(%rip),%rdi
  mov    %eax,%edx
  call   <printf@plt>
  mov    $0x1,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: label: D0
type: string

tool: D0: Number of ways to reach %d is : %d


assistant: ```c
int main() {
  int sum = 100;
  int n = noOfWays(sum);
  printf("Number of ways to reach %d is : %d\n", sum, n);
  return 1;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x64,%edi
  call   <noOfWays@plt>
  mov    $0x64,%esi
  lea    D0(%rip),%rdi
  mov    %eax,%edx
  call   <printf@plt>
  mov    $0x1,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D0
data_type: string

tool: D0: Number of ways to reach %d is : %d


assistant: ```c
int main() {
  int sum = 100;
  int n = noOfWays(sum);
  printf("Number of ways to reach %d is : %d\n", sum, n);
  return 1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<plot8>:
  endbr64
  sub    $0xd8,%rsp
  mov    %rdx,%r8
  mov    %fs:0x28,%rax
  mov    %rax,0xc8(%rsp)
  xor    %eax,%eax
  movslq %esi,%rsi
  movl   $0x0,(%rsp)
  movl   $0x0,0x4(%rsp)
  movl   $0x0,0x8(%rsp)
  movl   $0x0,0xc(%rsp)
  movl   $0x0,0x10(%rsp)
  movl   $0x1,0x14(%rsp)
  movl   $0x1,0x18(%rsp)
  movl   $0x2,0x1c(%rsp)
  movl   $0x2,0x20(%rsp)
  movl   $0x3,0x24(%rsp)
  movl   $0x3,0x28(%rsp)
  movl   $0x3,0x2c(%rsp)
  movl   $0x3,0x30(%rsp)
  movl   $0x3,0x34(%rsp)
  movl   $0x4,0x38(%rsp)
  movl   $0x4,0x3c(%rsp)
  movl   $0x5,0x40(%rsp)
  movl   $0x5,0x44(%rsp)
  movl   $0x6,0x48(%rsp)
  movl   $0x6,0x4c(%rsp)
  movl   $0x6,0x50(%rsp)
  movl   $0x6,0x54(%rsp)
  movl   $0x6,0x58(%rsp)
  lea    0x1(%rdi),%r11d
  mov    %r11d,0x60(%rsp)
  lea    0x2(%rdi),%r10d
  mov    %r10d,0x64(%rsp)
  lea    0x3(%rdi),%r9d
  mov    %r9d,0x68(%rsp)
  lea    0x4(%rdi),%ecx
  mov    %ecx,0x6c(%rsp)
  lea    0x5(%rdi),%edx
  mov    %edx,0x70(%rsp)
  mov    %edi,0x74(%rsp)
  lea    0x6(%rdi),%eax
  mov    %eax,0x78(%rsp)
  mov    %edi,0x7c(%rsp)
  mov    %eax,0x80(%rsp)
  mov    %r11d,0x84(%rsp)
  mov    %r10d,0x88(%rsp)
  mov    %r9d,0x8c(%rsp)
  mov    %ecx,0x90(%rsp)
  mov    %edx,0x94(%rsp)
  mov    %edi,0x98(%rsp)
  mov    %eax,0x9c(%rsp)
  mov    %edi,0xa0(%rsp)
  mov    %eax,0xa4(%rsp)
  mov    %r11d,0xa8(%rsp)
  mov    %r10d,0xac(%rsp)
  mov    %r9d,0xb0(%rsp)
  mov    %ecx,0xb4(%rsp)
  mov    %edx,0xb8(%rsp)
  mov    $0x0,%edx
L19:
  movslq (%rsp,%rdx,1),%rax
  imul   %rsi,%rax
  movslq 0x60(%rsp,%rdx,1),%rcx
  add    %r8,%rax
  movb   $0x38,(%rax,%rcx,1)
  add    $0x4,%rdx
  cmp    $0x5c,%rdx
  jne    L19
  mov    0xc8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L12
  add    $0xd8,%rsp
  ret
L12:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void plot8(int posicao, int colunas, char matriz[7][colunas]) {
  int cx = 0;
  int cy = 0;
  int p = posicao;
  int x[] = {0, 0, 0, 0, 0, 1, 1, 2, 2, 3, 3, 3, 3, 3, 4, 4, 5, 5, 6, 6, 6, 6, 6};
  int y[] = {p + 1, p + 2, p + 3, p + 4, p + 5, p + 0, p + 6, p + 0, p + 6, p + 1, p + 2, p + 3, p + 4, p + 5, p + 0, p + 6, p + 0, p + 6, p + 1, p + 2, p + 3, p + 4, p + 5};
  int points = sizeof(x) / sizeof(int);
  for (cx = 0; cx < points; ++cx) {
    matriz[x[cx]][y[cx]] = '8';
  }
}
```