ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    $0xffffffc9,%r8d
  mov    $0x78,%ecx
  movsd  D37(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    $0x6f,%eax
  lea    0x2(%rsp),%rdx
  mov    $0xd,%esi
  mov    %ax,0x6(%rsp)
  lea    D42(%rip),%rdi
  mov    $0x1,%eax
  movl   $0x6c6c6568,0x2(%rsp)
  call   <printf@plt>
  mov    D37(%rip),%rax
  lea    D19(%rip),%rdi
  movq   %rax,%xmm0
  movq   %rax,%xmm2
  movq   %rax,%xmm1
  mov    $0x3,%eax
  call   <printf@plt>
  xor    %eax,%eax
  lea    D11(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D37
type_infered: byte[8]
name: extract_symbolic_data
arguments: data_tag: D42
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D19
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D11
type_infered: string

tool: {"D37": [0x00,0x00,0x00,0xe0,0xfc,0xb0,0x28,0x40]}

tool: D42: Some values: %d, "%s", %f, '%c', and %d.


tool: D19: Variable precision: 1: %1f, 2: %2f, 3: %3f


tool: D11: A percent sign: %%.


assistant: ```c
int main() {
  const char foo[] = "hello";
  const int bar = -55;
  const float zip = 12.3456789;
  printf("Some values: %d, \"%s\", %f, '%c', and %d.\n", 13, foo, zip, 'x', bar);
  printf("Variable precision: 1: %1f, 2: %2f, 3: %3f\n", zip, zip, zip);
  printf("A percent sign: %%.\n");
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    $0xffffffc9,%r8d
  mov    $0x78,%ecx
  movsd  D47(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    $0x6f,%eax
  lea    0x2(%rsp),%rdx
  mov    $0xd,%esi
  mov    %ax,0x6(%rsp)
  lea    D13(%rip),%rdi
  mov    $0x1,%eax
  movl   $0x6c6c6568,0x2(%rsp)
  call   <printf@plt>
  mov    D47(%rip),%rax
  lea    D9(%rip),%rdi
  movq   %rax,%xmm0
  movq   %rax,%xmm2
  movq   %rax,%xmm1
  mov    $0x3,%eax
  call   <printf@plt>
  xor    %eax,%eax
  lea    D10(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D47
datatype: byte[8]
name: read_data
arguments: data_tag: D13
datatype: string
name: read_data
arguments: data_tag: D9
datatype: string
name: read_data
arguments: data_tag: D10
datatype: string

tool: {"D47": [0x00,0x00,0x00,0xe0,0xfc,0xb0,0x28,0x40]}

tool: D13: Some values: %d, "%s", %f, '%c', and %d.


tool: D9: Variable precision: 1: %1f, 2: %2f, 3: %3f


tool: D10: A percent sign: %%.


assistant: ```c
int main() {
  const char foo[] = "hello";
  const int bar = -55;
  const float zip = 12.3456789;
  printf("Some values: %d, \"%s\", %f, '%c', and %d.\n", 13, foo, zip, 'x', bar);
  printf("Variable precision: 1: %1f, 2: %2f, 3: %3f\n", zip, zip, zip);
  printf("A percent sign: %%.\n");
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<insb>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L2
L12:
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <inb@plt>
  mov    %eax,%ecx
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  mov    %ecx,%edx
  mov    %dl,(%rax)
L2:
  mov    -0x28(%rbp),%rax
  lea    -0x1(%rax),%rdx
  mov    %rdx,-0x28(%rbp)
  test   %rax,%rax
  jne    L12
  nop
  nop
  leave
  ret
```

assistant: ```c
void insb(unsigned long port, void *dst, unsigned long count) {
  unsigned char *p = dst;
  while (count--)
    *p++ = inb(port);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<insb>:
  endbr64
  test   %rdx,%rdx
  je     L2
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    %rsi,%rbx
  lea    (%rsi,%rdx,1),%rbp
L15:
  mov    %r12,%rdi
  call   <inb@plt>
  add    $0x1,%rbx
  mov    %al,-0x1(%rbx)
  cmp    %rbx,%rbp
  jne    L15
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L2:
  ret
```

assistant: ```c
void insb(unsigned long port, void *dst, unsigned long count) {
  unsigned char *p = dst;
  while (count--)
    *p++ = inb(port);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<insb>:
  endbr64
  test   %rdx,%rdx
  je     L4
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  lea    (%rsi,%rdx,1),%rbp
  push   %rbx
  mov    %rsi,%rbx
  nopw   0x0(%rax,%rax,1)
L5:
  mov    %r12,%rdi
  add    $0x1,%rbx
  call   <inb@plt>
  mov    %al,-0x1(%rbx)
  cmp    %rbp,%rbx
  jne    L5
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L4:
  ret
```

assistant: ```c
void insb(unsigned long port, void *dst, unsigned long count) {
  unsigned char *p = dst;
  while (count--)
    *p++ = inb(port);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<insb>:
  endbr64
  test   %rdx,%rdx
  je     L12
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  lea    (%rsi,%rdx,1),%rbp
  push   %rbx
  mov    %rsi,%rbx
  nopw   0x0(%rax,%rax,1)
L18:
  mov    %r12,%rdi
  add    $0x1,%rbx
  call   <inb@plt>
  mov    %al,-0x1(%rbx)
  cmp    %rbp,%rbx
  jne    L18
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L12:
  ret
```

assistant: ```c
void insb(unsigned long port, void *dst, unsigned long count) {
  unsigned char *p = dst;
  while (count--)
    *p++ = inb(port);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D44(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x24(%rbp),%rax
  mov    %rax,%rsi
  lea    D45(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    D14(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x20(%rbp),%rax
  mov    %rax,%rsi
  lea    D45(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    D30(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x1c(%rbp),%rax
  mov    %rax,%rsi
  lea    D45(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x1c(%rbp),%esi
  mov    -0x24(%rbp),%eax
  lea    -0x14(%rbp),%rcx
  lea    -0x18(%rbp),%rdx
  mov    %eax,%edi
  call   <euclidesExt@plt>
  mov    %eax,-0xc(%rbp)
  mov    -0x20(%rbp),%eax
  cltd
  idivl  -0xc(%rbp)
  mov    %edx,%eax
  test   %eax,%eax
  jne    L31
  mov    -0x20(%rbp),%eax
  cltd
  idivl  -0xc(%rbp)
  mov    %eax,%edx
  mov    -0x18(%rbp),%eax
  imul   %edx,%eax
  mov    -0x1c(%rbp),%ecx
  cltd
  idiv   %ecx
  mov    %edx,%eax
  mov    %eax,-0x18(%rbp)
  movl   $0x0,-0x10(%rbp)
  jmp    L1
L4:
  mov    -0x10(%rbp),%eax
  lea    0x1(%rax),%ecx
  mov    -0x1c(%rbp),%eax
  cltd
  idivl  -0xc(%rbp)
  imul   %eax,%ecx
  mov    %ecx,%edx
  mov    -0x18(%rbp),%eax
  add    %edx,%eax
  mov    -0x1c(%rbp),%ecx
  cltd
  idiv   %ecx
  mov    %edx,%eax
  mov    %eax,%esi
  lea    D38(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x10(%rbp)
L1:
  mov    -0x10(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jl     L4
  jmp    L30
L31:
  lea    D47(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
L30:
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L24
  call   <__stack_chk_fail@plt>
L24:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: tag: D44
type: string
name: unpack_data
arguments: tag: D45
type: string
name: unpack_data
arguments: tag: D14
type: string
name: unpack_data
arguments: tag: D30
type: string
name: unpack_data
arguments: tag: D38
type: string
name: unpack_data
arguments: tag: D47
type: string

tool: D44: Digite o valor de A: 

tool: D45: %i

tool: D14: Digite o valor de B: 

tool: D30: Digite o valor de n: 

tool: D38: X= %i


tool: D47: Sem solucao

assistant: ```c
void main() {
  int i, a, b, d, n, x, y;
  printf("Digite o valor de A: ");
  scanf("%i", &a);
  printf("Digite o valor de B: ");
  scanf("%i", &b);
  printf("Digite o valor de n: ");
  scanf("%i", &n);
  d = euclidesExt(a, n, &x, &y);
  if (b % d == 0) {
    x = x * (b / d) % n;
    for (i = 0; i < d; i++) {
      printf("X= %i\n", (x + (i + 1) * (n / d)) % n);
    }
  } else {
    printf("Sem solucao\n");
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    D24(%rip),%rdi
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D56(%rip),%rbx
  mov    %rbx,%rdi
  call   <scanf@plt>
  lea    D42(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0x8(%rsp),%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  lea    D58(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0xc(%rsp),%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  lea    0x14(%rsp),%rcx
  lea    0x10(%rsp),%rdx
  mov    0xc(%rsp),%esi
  mov    0x4(%rsp),%edi
  call   <euclidesExt@plt>
  mov    %eax,%ebx
  mov    0x8(%rsp),%ecx
  mov    %ecx,%eax
  cltd
  idiv   %ebx
  test   %edx,%edx
  jne    L46
  mov    %edx,%ebp
  mov    %ecx,%eax
  cltd
  idiv   %ebx
  imul   0x10(%rsp),%eax
  cltd
  idivl  0xc(%rsp)
  mov    %edx,0x10(%rsp)
  test   %ebx,%ebx
  jle    L39
  lea    D34(%rip),%r12
L14:
  add    $0x1,%ebp
  mov    0xc(%rsp),%ecx
  mov    %ecx,%eax
  cltd
  idiv   %ebx
  imul   %ebp,%eax
  add    0x10(%rsp),%eax
  cltd
  idiv   %ecx
  mov    %edx,%esi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  cmp    %ebx,%ebp
  jne    L14
L39:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L27
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L46:
  lea    D14(%rip),%rdi
  call   <puts@plt>
  jmp    L39
L27:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D24
type: string
name: parse_data
arguments: label: D56
type: string
name: parse_data
arguments: label: D42
type: string
name: parse_data
arguments: label: D58
type: string
name: parse_data
arguments: label: D34
type: string
name: parse_data
arguments: label: D14
type: string

tool: D24: Digite o valor de A: 

tool: D56: %i

tool: D42: Digite o valor de B: 

tool: D58: Digite o valor de n: 

tool: D34: X= %i


tool: D14: Sem solucao

assistant: ```c
void main() {
  int i, a, b, d, n, x, y;
  printf("Digite o valor de A: ");
  scanf("%i", &a);
  printf("Digite o valor de B: ");
  scanf("%i", &b);
  printf("Digite o valor de n: ");
  scanf("%i", &n);
  d = euclidesExt(a, n, &x, &y);
  if (b % d == 0) {
    x = x * (b / d) % n;
    for (i = 0; i < d; i++) {
      printf("X= %i\n", (x + (i + 1) * (n / d)) % n);
    }
  } else {
    printf("Sem solucao\n");
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %r12
  lea    D58(%rip),%rdi
  push   %rbp
  lea    D55(%rip),%rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    D13(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x8(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    D45(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0xc(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%esi
  mov    0x4(%rsp),%edi
  lea    0x10(%rsp),%rdx
  lea    0x14(%rsp),%rcx
  call   <euclidesExt@plt>
  mov    %eax,%r12d
  mov    0x8(%rsp),%eax
  cltd
  idiv   %r12d
  test   %edx,%edx
  jne    L25
  imul   0x10(%rsp),%eax
  mov    %edx,%ebp
  lea    D10(%rip),%rbx
  cltd
  idivl  0xc(%rsp)
  mov    %edx,0x10(%rsp)
  test   %r12d,%r12d
  jle    L5
  nopl   0x0(%rax)
L49:
  mov    0xc(%rsp),%ecx
  add    $0x1,%ebp
  mov    %rbx,%rdi
  mov    %ecx,%eax
  cltd
  idiv   %r12d
  imul   %ebp,%eax
  add    0x10(%rsp),%eax
  cltd
  idiv   %ecx
  xor    %eax,%eax
  mov    %edx,%esi
  call   <printf@plt>
  cmp    %r12d,%ebp
  jne    L49
L5:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L10
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L25:
  lea    D34(%rip),%rdi
  call   <puts@plt>
  jmp    L5
L10:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D58
type_infered: string
name: parse_data
arguments: data_label: D55
type_infered: string
name: parse_data
arguments: data_label: D13
type_infered: string
name: parse_data
arguments: data_label: D45
type_infered: string
name: parse_data
arguments: data_label: D10
type_infered: string
name: parse_data
arguments: data_label: D34
type_infered: string

tool: D58: Digite o valor de A: 

tool: D55: %i

tool: D13: Digite o valor de B: 

tool: D45: Digite o valor de n: 

tool: D10: X= %i


tool: D34: Sem solucao

assistant: ```c
void main() {
  int i, a, b, d, n, x, y;
  printf("Digite o valor de A: ");
  scanf("%i", &a);
  printf("Digite o valor de B: ");
  scanf("%i", &b);
  printf("Digite o valor de n: ");
  scanf("%i", &n);
  d = euclidesExt(a, n, &x, &y);
  if (b % d == 0) {
    x = x * (b / d) % n;
    for (i = 0; i < d; i++) {
      printf("X= %i\n", (x + (i + 1) * (n / d)) % n);
    }
  } else {
    printf("Sem solucao\n");
  }
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  push   %r12
  lea    D28(%rip),%rdi
  push   %rbp
  lea    D26(%rip),%rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    D12(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x8(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    D54(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0xc(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%esi
  mov    0x4(%rsp),%edi
  lea    0x10(%rsp),%rdx
  lea    0x14(%rsp),%rcx
  call   <euclidesExt@plt>
  mov    %eax,%r12d
  mov    0x8(%rsp),%eax
  cltd
  idiv   %r12d
  test   %edx,%edx
  jne    L6
  imul   0x10(%rsp),%eax
  mov    %edx,%ebp
  lea    D21(%rip),%rbx
  cltd
  idivl  0xc(%rsp)
  mov    %edx,0x10(%rsp)
  test   %r12d,%r12d
  jle    L42
  nopl   0x0(%rax)
L43:
  mov    0xc(%rsp),%ecx
  add    $0x1,%ebp
  mov    %rbx,%rdi
  mov    %ecx,%eax
  cltd
  idiv   %r12d
  imul   %ebp,%eax
  add    0x10(%rsp),%eax
  cltd
  idiv   %ecx
  xor    %eax,%eax
  mov    %edx,%esi
  call   <printf@plt>
  cmp    %r12d,%ebp
  jne    L43
L42:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L6:
  lea    D42(%rip),%rdi
  call   <puts@plt>
  jmp    L42
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D28
data_type: string
name: retrieve_data
arguments: data_label: D26
data_type: string
name: retrieve_data
arguments: data_label: D12
data_type: string
name: retrieve_data
arguments: data_label: D54
data_type: string
name: retrieve_data
arguments: data_label: D21
data_type: string
name: retrieve_data
arguments: data_label: D42
data_type: string

tool: D28: Digite o valor de A: 

tool: D26: %i

tool: D12: Digite o valor de B: 

tool: D54: Digite o valor de n: 

tool: D21: X= %i


tool: D42: Sem solucao

assistant: ```c
void main() {
  int i, a, b, d, n, x, y;
  printf("Digite o valor de A: ");
  scanf("%i", &a);
  printf("Digite o valor de B: ");
  scanf("%i", &b);
  printf("Digite o valor de n: ");
  scanf("%i", &n);
  d = euclidesExt(a, n, &x, &y);
  if (b % d == 0) {
    x = x * (b / d) % n;
    for (i = 0; i < d; i++) {
      printf("X= %i\n", (x + (i + 1) * (n / d)) % n);
    }
  } else {
    printf("Sem solucao\n");
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    $0xc,%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  mov    $0xc,%edi
  call   <malloc@plt>
  mov    %rax,%r13
  mov    $0x3,%esi
  lea    D33(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %rbp,%rsi
  lea    D9(%rip),%rbx
  mov    %rbx,%rdi
  call   <scanf@plt>
  lea    0x4(%rbp),%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  lea    0x8(%rbp),%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    $0x0,%eax
  call   <getchar@plt>
  movss  0x0(%rbp),%xmm0
  movss  %xmm0,0x0(%r13)
  movss  0x4(%rbp),%xmm0
  movss  %xmm0,0x4(%r13)
  movss  0x8(%rbp),%xmm0
  movss  %xmm0,0x8(%r13)
  lea    D34(%rip),%rdi
  call   <puts@plt>
  mov    %rbp,%r12
  mov    $0x0,%ebx
  lea    D50(%rip),%r14
L12:
  add    $0x1,%ebx
  pxor   %xmm0,%xmm0
  cvtss2sd (%r12),%xmm0
  mov    %ebx,%esi
  mov    %r14,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  add    $0x4,%r12
  cmp    $0x3,%ebx
  jne    L12
  lea    D48(%rip),%rdi
  call   <puts@plt>
  mov    %r13,%r12
  mov    $0x0,%ebx
  lea    D50(%rip),%r14
L1:
  add    $0x1,%ebx
  pxor   %xmm0,%xmm0
  cvtss2sd (%r12),%xmm0
  mov    %ebx,%esi
  mov    %r14,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  add    $0x4,%r12
  cmp    $0x3,%ebx
  jne    L1
  mov    %rbp,%rdi
  call   <free@plt>
  mov    %r13,%rdi
  call   <free@plt>
  mov    $0x0,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

tool_calls: name: read_data
arguments: data_label: D33
datatype: string
name: read_data
arguments: data_label: D9
datatype: string
name: read_data
arguments: data_label: D34
datatype: string
name: read_data
arguments: data_label: D50
datatype: string
name: read_data
arguments: data_label: D48
datatype: string

tool: D33: Please input %d floats for 'arr1':


tool: D9: %f

tool: D34: Arr1:

tool: D50: pos %d: %.3f


tool: D48: Arr2:

assistant: ```c
int main() {
  int i;
  float *arr1, *arr2;
  arr1 = (float *)malloc(3 * sizeof(float));
  arr2 = (float *)malloc(3 * sizeof(float));
  printf("Please input %d floats for 'arr1':\n", 3);
  for (i = 0; i < 3; i++) {
    scanf("%f", arr1 + i);
  }
  getchar();
  for (i = 0; i < 3; i++) {
    *(arr2 + i) = *(arr1 + i);
  }
  printf("Arr1:\n");
  for (i = 0; i < 3; i++) {
    printf("pos %d: %.3f\n", i + 1, arr1[i]);
  }
  printf("Arr2:\n");
  for (i = 0; i < 3; i++) {
    printf("pos %d: %.3f\n", i + 1, arr2[i]);
  }
  free(arr1);
  free(arr2);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  push   %r14
  mov    $0xc,%edi
  lea    D45(%rip),%r14
  push   %r13
  push   %r12
  lea    D11(%rip),%r12
  push   %rbp
  push   %rbx
  xor    %ebx,%ebx
  call   <malloc@plt>
  mov    $0xc,%edi
  mov    %rax,%rbp
  call   <malloc@plt>
  mov    $0x3,%esi
  lea    D22(%rip),%rdi
  mov    %rax,%r13
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r12,%rdi
  mov    %rbp,%rsi
  call   <scanf@plt>
  mov    %r12,%rdi
  lea    0x4(%rbp),%rsi
  call   <scanf@plt>
  mov    %r12,%rdi
  lea    0x8(%rbp),%rsi
  mov    %rbp,%r12
  call   <scanf@plt>
  xor    %eax,%eax
  call   <getchar@plt>
  movss  0x0(%rbp),%xmm0
  lea    D17(%rip),%rdi
  movss  %xmm0,0x0(%r13)
  movss  0x4(%rbp),%xmm0
  movss  %xmm0,0x4(%r13)
  movss  0x8(%rbp),%xmm0
  movss  %xmm0,0x8(%r13)
  call   <puts@plt>
L18:
  add    $0x1,%ebx
  pxor   %xmm0,%xmm0
  mov    %r14,%rdi
  mov    $0x1,%eax
  mov    %ebx,%esi
  cvtss2sd (%r12),%xmm0
  add    $0x4,%r12
  call   <printf@plt>
  cmp    $0x3,%ebx
  jne    L18
  lea    D27(%rip),%rdi
  mov    %r13,%r12
  xor    %ebx,%ebx
  call   <puts@plt>
L4:
  add    $0x1,%ebx
  pxor   %xmm0,%xmm0
  mov    %r14,%rdi
  mov    $0x1,%eax
  mov    %ebx,%esi
  cvtss2sd (%r12),%xmm0
  add    $0x4,%r12
  call   <printf@plt>
  cmp    $0x3,%ebx
  jne    L4
  mov    %rbp,%rdi
  call   <free@plt>
  mov    %r13,%rdi
  call   <free@plt>
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

tool_calls: name: parse_data
arguments: data_label: D45
type_infered: string
name: parse_data
arguments: data_label: D11
type_infered: string
name: parse_data
arguments: data_label: D22
type_infered: string
name: parse_data
arguments: data_label: D17
type_infered: string
name: parse_data
arguments: data_label: D27
type_infered: string

tool: D45: pos %d: %.3f


tool: D11: %f

tool: D22: Please input %d floats for 'arr1':


tool: D17: Arr1:

tool: D27: Arr2:

assistant: ```c
int main() {
  int i;
  float *arr1, *arr2;
  arr1 = (float *)malloc(3 * sizeof(float));
  arr2 = (float *)malloc(3 * sizeof(float));
  printf("Please input %d floats for 'arr1':\n", 3);
  for (i = 0; i < 3; i++) {
    scanf("%f", arr1 + i);
  }
  getchar();
  for (i = 0; i < 3; i++) {
    *(arr2 + i) = *(arr1 + i);
  }
  printf("Arr1:\n");
  for (i = 0; i < 3; i++) {
    printf("pos %d: %.3f\n", i + 1, arr1[i]);
  }
  printf("Arr2:\n");
  for (i = 0; i < 3; i++) {
    printf("pos %d: %.3f\n", i + 1, arr2[i]);
  }
  free(arr1);
  free(arr2);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r14
  mov    $0xc,%edi
  lea    D5(%rip),%r14
  push   %r13
  push   %r12
  lea    D23(%rip),%r12
  push   %rbp
  push   %rbx
  xor    %ebx,%ebx
  call   <malloc@plt>
  mov    $0xc,%edi
  mov    %rax,%rbp
  call   <malloc@plt>
  mov    $0x3,%esi
  lea    D7(%rip),%rdi
  mov    %rax,%r13
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r12,%rdi
  mov    %rbp,%rsi
  call   <scanf@plt>
  mov    %r12,%rdi
  lea    0x4(%rbp),%rsi
  call   <scanf@plt>
  mov    %r12,%rdi
  lea    0x8(%rbp),%rsi
  mov    %rbp,%r12
  call   <scanf@plt>
  xor    %eax,%eax
  call   <getchar@plt>
  movss  0x0(%rbp),%xmm0
  lea    D26(%rip),%rdi
  movss  %xmm0,0x0(%r13)
  movss  0x4(%rbp),%xmm0
  movss  %xmm0,0x4(%r13)
  movss  0x8(%rbp),%xmm0
  movss  %xmm0,0x8(%r13)
  call   <puts@plt>
L11:
  add    $0x1,%ebx
  pxor   %xmm0,%xmm0
  mov    %r14,%rdi
  mov    $0x1,%eax
  mov    %ebx,%esi
  cvtss2sd (%r12),%xmm0
  add    $0x4,%r12
  call   <printf@plt>
  cmp    $0x3,%ebx
  jne    L11
  lea    D49(%rip),%rdi
  mov    %r13,%r12
  xor    %ebx,%ebx
  call   <puts@plt>
L15:
  add    $0x1,%ebx
  pxor   %xmm0,%xmm0
  mov    %r14,%rdi
  mov    $0x1,%eax
  mov    %ebx,%esi
  cvtss2sd (%r12),%xmm0
  add    $0x4,%r12
  call   <printf@plt>
  cmp    $0x3,%ebx
  jne    L15
  mov    %rbp,%rdi
  call   <free@plt>
  mov    %r13,%rdi
  call   <free@plt>
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

tool_calls: name: disassemble_data
arguments: label: D5
type_infered: string
name: disassemble_data
arguments: label: D23
type_infered: string
name: disassemble_data
arguments: label: D7
type_infered: string
name: disassemble_data
arguments: label: D26
type_infered: string
name: disassemble_data
arguments: label: D49
type_infered: string

tool: D5: pos %d: %.3f


tool: D23: %f

tool: D7: Please input %d floats for 'arr1':


tool: D26: Arr1:

tool: D49: Arr2:

assistant: ```c
int main() {
  int i;
  float *arr1, *arr2;
  arr1 = (float *)malloc(3 * sizeof(float));
  arr2 = (float *)malloc(3 * sizeof(float));
  printf("Please input %d floats for 'arr1':\n", 3);
  for (i = 0; i < 3; i++) {
    scanf("%f", arr1 + i);
  }
  getchar();
  for (i = 0; i < 3; i++) {
    *(arr2 + i) = *(arr1 + i);
  }
  printf("Arr1:\n");
  for (i = 0; i < 3; i++) {
    printf("pos %d: %.3f\n", i + 1, arr1[i]);
  }
  printf("Arr2:\n");
  for (i = 0; i < 3; i++) {
    printf("pos %d: %.3f\n", i + 1, arr2[i]);
  }
  free(arr1);
  free(arr2);
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<strlen>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L10
L9:
  addq   $0x1,-0x8(%rbp)
L10:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L9
  mov    -0x8(%rbp),%rax
  sub    -0x18(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
size_t strlen(const char *s) {
  const char *t;
  for (t = s; *t; ++t)
    ;
  return t - s;
}
```
user: Translate the below assembly code into C programming language:

```asm
<strlen>:
  endbr64
  cmpb   $0x0,(%rdi)
  je     L0
  mov    %rdi,%rax
L2:
  add    $0x1,%rax
  cmpb   $0x0,(%rax)
  jne    L2
L22:
  sub    %rdi,%rax
  ret
L0:
  mov    %rdi,%rax
  jmp    L22
```

assistant: ```c
size_t strlen(const char *s) {
  const char *t;
  for (t = s; *t; ++t)
    ;
  return t - s;
}
```