ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Converte this assembly snippet into its equivalent C code:

```asm
<my_strlen>:
  endbr64
  xor    %eax,%eax
  cmpb   $0x0,(%rdi)
  je     L10
  nopl   0x0(%rax,%rax,1)
L2:
  add    $0x1,%rax
  cmpb   $0x0,(%rdi,%rax,1)
  jne    L2
  ret
  nopl   0x0(%rax,%rax,1)
L10:
  ret
```

assistant: ```c
size_t my_strlen(const char *str) {
  size_t len = 0;
  while (str[len]) {
    len++;
  }
  return len;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<uart_puts>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x4(%rbp)
  movw   $0x0,-0x6(%rbp)
  jmp    L12
L7:
  movswq -0x6(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <uart_putc@plt>
  movzwl -0x6(%rbp),%eax
  add    $0x1,%eax
  mov    %ax,-0x6(%rbp)
L12:
  movswl -0x6(%rbp),%eax
  cmp    %eax,-0x4(%rbp)
  jg     L7
  nop
  nop
  leave
  ret
```

assistant: ```c
void uart_puts(const char *str) {
  int len = strlen(str);
  for (short i = 0; i < len; i++) {
    uart_putc(str[i]);
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<uart_puts>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  call   <strlen@plt>
  test   %eax,%eax
  jle    L3
  mov    %eax,%ebp
  mov    $0x0,%ebx
L14:
  movswq %bx,%rax
  movsbl (%r12,%rax,1),%edi
  call   <uart_putc@plt>
  add    $0x1,%ebx
  movswl %bx,%eax
  cmp    %ebp,%eax
  jl     L14
L3:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void uart_puts(const char *str) {
  int len = strlen(str);
  for (short i = 0; i < len; i++) {
    uart_putc(str[i]);
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<uart_puts>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  call   <strlen@plt>
  test   %eax,%eax
  jle    L8
  mov    %eax,%ebp
  xor    %ebx,%ebx
  nopl   0x0(%rax,%rax,1)
L16:
  movswq %bx,%rax
  add    $0x1,%ebx
  movsbl (%r12,%rax,1),%edi
  call   <uart_putc@plt>
  movswl %bx,%eax
  cmp    %ebp,%eax
  jl     L16
L8:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void uart_puts(const char *str) {
  int len = strlen(str);
  for (short i = 0; i < len; i++) {
    uart_putc(str[i]);
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<uart_puts>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  call   <strlen@plt>
  test   %eax,%eax
  jle    L1
  mov    %eax,%ebp
  xor    %ebx,%ebx
  nopl   0x0(%rax,%rax,1)
L11:
  movswq %bx,%rax
  add    $0x1,%ebx
  movsbl (%r12,%rax,1),%edi
  call   <uart_putc@plt>
  movswl %bx,%eax
  cmp    %ebp,%eax
  jl     L11
L1:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void uart_puts(const char *str) {
  int len = strlen(str);
  for (short i = 0; i < len; i++) {
    uart_putc(str[i]);
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<icmCIE94sq>:
  endbr64
  push   %r14
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  movsd  (%rdi),%xmm0
  subsd  (%rsi),%xmm0
  movapd %xmm0,%xmm5
  mulsd  %xmm0,%xmm5
  movsd  %xmm5,0x8(%rsp)
  movsd  0x8(%rdi),%xmm0
  mov    0x8(%rsi),%rbx
  movapd %xmm0,%xmm1
  movq   %rbx,%xmm6
  subsd  %xmm6,%xmm1
  movsd  0x10(%rdi),%xmm3
  mov    0x10(%rsi),%rbp
  movapd %xmm3,%xmm4
  movq   %rbp,%xmm7
  subsd  %xmm7,%xmm4
  mulsd  %xmm1,%xmm1
  movapd %xmm1,%xmm2
  addsd  %xmm5,%xmm2
  mulsd  %xmm4,%xmm4
  addsd  %xmm4,%xmm2
  movq   %xmm2,%r14
  mulsd  %xmm0,%xmm0
  mulsd  %xmm3,%xmm3
  addsd  %xmm3,%xmm0
  pxor   %xmm1,%xmm1
  ucomisd %xmm0,%xmm1
  ja     L73
  sqrtsd %xmm0,%xmm0
  movsd  %xmm0,0x10(%rsp)
L12:
  movq   %rbx,%xmm0
  mulsd  %xmm0,%xmm0
  movq   %rbp,%xmm1
  mulsd  %xmm1,%xmm1
  addsd  %xmm1,%xmm0
  pxor   %xmm1,%xmm1
  ucomisd %xmm0,%xmm1
  ja     L4
  sqrtsd %xmm0,%xmm0
  movsd  %xmm0,0x18(%rsp)
L34:
  movsd  0x10(%rsp),%xmm0
  mulsd  0x18(%rsp),%xmm0
  pxor   %xmm1,%xmm1
  ucomisd %xmm0,%xmm1
  ja     L32
  sqrtsd %xmm0,%xmm0
L61:
  movsd  0x18(%rsp),%xmm1
  subsd  0x10(%rsp),%xmm1
  mulsd  %xmm1,%xmm1
  movq   %r14,%xmm2
  subsd  0x8(%rsp),%xmm2
  subsd  %xmm1,%xmm2
  pxor   %xmm3,%xmm3
  comisd %xmm2,%xmm3
  ja     L9
L44:
  movapd %xmm0,%xmm3
  mulsd  D7(%rip),%xmm3
  movsd  D15(%rip),%xmm4
  addsd  %xmm4,%xmm3
  mulsd  D9(%rip),%xmm0
  addsd  %xmm4,%xmm0
  mulsd  %xmm3,%xmm3
  divsd  %xmm3,%xmm1
  addsd  0x8(%rsp),%xmm1
  mulsd  %xmm0,%xmm0
  divsd  %xmm0,%xmm2
  addsd  %xmm2,%xmm1
  movapd %xmm1,%xmm0
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r14
  ret
L73:
  call   <sqrt@plt>
  movsd  %xmm0,0x10(%rsp)
  jmp    L12
L4:
  call   <sqrt@plt>
  movsd  %xmm0,0x18(%rsp)
  jmp    L34
L32:
  call   <sqrt@plt>
  jmp    L61
L9:
  movapd %xmm3,%xmm2
  jmp    L44
```

tool_calls: name: retrieve_data
arguments: data_label: D7
type_infered: f64
name: retrieve_data
arguments: data_label: D15
type_infered: f64
name: retrieve_data
arguments: data_label: D9
type_infered: f64

tool: D7: 0.048

tool: D15: 1

tool: D9: 0.014

assistant: ```c
double icmCIE94sq(double Lab1[3], double Lab2[3]) {
  double desq, dhsq;
  double dlsq, dcsq;
  double c12;
  {
    double dl, da, db;
    dl = Lab1[0] - Lab2[0];
    dlsq = dl * dl;
    da = Lab1[1] - Lab2[1];
    db = Lab1[2] - Lab2[2];
    desq = dlsq + da * da + db * db;
  }
  {
    double c1, c2, dc;
    c1 = sqrt(Lab1[1] * Lab1[1] + Lab1[2] * Lab1[2]);
    c2 = sqrt(Lab2[1] * Lab2[1] + Lab2[2] * Lab2[2]);
    c12 = sqrt(c1 * c2);
    dc = c2 - c1;
    dcsq = dc * dc;
  }
  if ((dhsq = desq - dlsq - dcsq) < 0.0)
    dhsq = 0.0;
  {
    double sc, sh;
    sc = 1.0 + 0.048 * c12;
    sh = 1.0 + 0.014 * c12;
    return dlsq + dcsq / (sc * sc) + dhsq / (sh * sh);
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<icmCIE94sq>:
  endbr64
  sub    $0x28,%rsp
  movsd  0x8(%rdi),%xmm0
  movsd  0x8(%rsi),%xmm1
  movsd  0x10(%rdi),%xmm2
  movsd  (%rdi),%xmm4
  movapd %xmm0,%xmm3
  subsd  (%rsi),%xmm4
  movsd  0x10(%rsi),%xmm6
  subsd  %xmm1,%xmm3
  mulsd  %xmm0,%xmm0
  movapd %xmm2,%xmm5
  subsd  %xmm6,%xmm5
  mulsd  %xmm2,%xmm2
  mulsd  %xmm4,%xmm4
  mulsd  %xmm3,%xmm3
  mulsd  %xmm5,%xmm5
  addsd  %xmm0,%xmm2
  addsd  %xmm4,%xmm3
  addsd  %xmm5,%xmm3
  pxor   %xmm5,%xmm5
  ucomisd %xmm2,%xmm5
  ja     L12
  sqrtsd %xmm2,%xmm2
L47:
  mulsd  %xmm1,%xmm1
  mulsd  %xmm6,%xmm6
  addsd  %xmm6,%xmm1
  ucomisd %xmm1,%xmm5
  ja     L6
  sqrtsd %xmm1,%xmm1
L57:
  movapd %xmm2,%xmm0
  mulsd  %xmm1,%xmm0
  ucomisd %xmm0,%xmm5
  ja     L54
  sqrtsd %xmm0,%xmm0
L53:
  subsd  %xmm2,%xmm1
  subsd  %xmm4,%xmm3
  mulsd  %xmm1,%xmm1
  movapd %xmm3,%xmm2
  movsd  D12(%rip),%xmm3
  mulsd  %xmm0,%xmm3
  mulsd  D1(%rip),%xmm0
  subsd  %xmm1,%xmm2
  maxsd  %xmm2,%xmm5
  movapd %xmm5,%xmm2
  movsd  D9(%rip),%xmm5
  add    $0x28,%rsp
  addsd  %xmm5,%xmm3
  addsd  %xmm5,%xmm0
  mulsd  %xmm3,%xmm3
  mulsd  %xmm0,%xmm0
  divsd  %xmm3,%xmm1
  divsd  %xmm0,%xmm2
  addsd  %xmm4,%xmm1
  addsd  %xmm2,%xmm1
  movapd %xmm1,%xmm0
  ret
L12:
  movapd %xmm2,%xmm0
  movsd  %xmm3,0x18(%rsp)
  movsd  %xmm4,0x10(%rsp)
  movsd  %xmm6,0x8(%rsp)
  movsd  %xmm1,(%rsp)
  call   <sqrt@plt>
  movsd  0x18(%rsp),%xmm3
  movsd  0x10(%rsp),%xmm4
  pxor   %xmm5,%xmm5
  movsd  0x8(%rsp),%xmm6
  movsd  (%rsp),%xmm1
  movapd %xmm0,%xmm2
  jmp    L47
L54:
  movsd  %xmm1,0x18(%rsp)
  movsd  %xmm2,0x10(%rsp)
  movsd  %xmm3,0x8(%rsp)
  movsd  %xmm4,(%rsp)
  call   <sqrt@plt>
  movsd  0x18(%rsp),%xmm1
  movsd  0x10(%rsp),%xmm2
  pxor   %xmm5,%xmm5
  movsd  0x8(%rsp),%xmm3
  movsd  (%rsp),%xmm4
  jmp    L53
L6:
  movapd %xmm1,%xmm0
  movsd  %xmm2,0x10(%rsp)
  movsd  %xmm3,0x8(%rsp)
  movsd  %xmm4,(%rsp)
  call   <sqrt@plt>
  movsd  0x10(%rsp),%xmm2
  movsd  0x8(%rsp),%xmm3
  pxor   %xmm5,%xmm5
  movsd  (%rsp),%xmm4
  movapd %xmm0,%xmm1
  jmp    L57
```

tool_calls: name: unpack_data
arguments: label: D12
type_infered: f64
name: unpack_data
arguments: label: D1
type_infered: f64
name: unpack_data
arguments: label: D9
type_infered: f64

tool: D12: 0.048

tool: D1: 0.014

tool: D9: 1

assistant: ```c
double icmCIE94sq(double Lab1[3], double Lab2[3]) {
  double desq, dhsq;
  double dlsq, dcsq;
  double c12;
  {
    double dl, da, db;
    dl = Lab1[0] - Lab2[0];
    dlsq = dl * dl;
    da = Lab1[1] - Lab2[1];
    db = Lab1[2] - Lab2[2];
    desq = dlsq + da * da + db * db;
  }
  {
    double c1, c2, dc;
    c1 = sqrt(Lab1[1] * Lab1[1] + Lab1[2] * Lab1[2]);
    c2 = sqrt(Lab2[1] * Lab2[1] + Lab2[2] * Lab2[2]);
    c12 = sqrt(c1 * c2);
    dc = c2 - c1;
    dcsq = dc * dc;
  }
  if ((dhsq = desq - dlsq - dcsq) < 0.0)
    dhsq = 0.0;
  {
    double sc, sh;
    sc = 1.0 + 0.048 * c12;
    sh = 1.0 + 0.014 * c12;
    return dlsq + dcsq / (sc * sc) + dhsq / (sh * sh);
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<icmCIE94sq>:
  endbr64
  sub    $0x28,%rsp
  movsd  0x8(%rdi),%xmm0
  movsd  0x8(%rsi),%xmm1
  movsd  0x10(%rdi),%xmm2
  movsd  (%rdi),%xmm4
  movapd %xmm0,%xmm3
  subsd  (%rsi),%xmm4
  movsd  0x10(%rsi),%xmm6
  subsd  %xmm1,%xmm3
  mulsd  %xmm0,%xmm0
  movapd %xmm2,%xmm5
  subsd  %xmm6,%xmm5
  mulsd  %xmm2,%xmm2
  mulsd  %xmm4,%xmm4
  mulsd  %xmm3,%xmm3
  mulsd  %xmm5,%xmm5
  addsd  %xmm0,%xmm2
  addsd  %xmm4,%xmm3
  addsd  %xmm5,%xmm3
  pxor   %xmm5,%xmm5
  ucomisd %xmm2,%xmm5
  ja     L37
  sqrtsd %xmm2,%xmm2
L9:
  mulsd  %xmm1,%xmm1
  mulsd  %xmm6,%xmm6
  addsd  %xmm6,%xmm1
  ucomisd %xmm1,%xmm5
  ja     L30
  sqrtsd %xmm1,%xmm1
L32:
  movapd %xmm2,%xmm0
  mulsd  %xmm1,%xmm0
  ucomisd %xmm0,%xmm5
  ja     L52
  sqrtsd %xmm0,%xmm0
L3:
  subsd  %xmm2,%xmm1
  subsd  %xmm4,%xmm3
  mulsd  %xmm1,%xmm1
  movapd %xmm3,%xmm2
  movsd  D28(%rip),%xmm3
  mulsd  %xmm0,%xmm3
  mulsd  D7(%rip),%xmm0
  subsd  %xmm1,%xmm2
  maxsd  %xmm2,%xmm5
  movapd %xmm5,%xmm2
  movsd  D17(%rip),%xmm5
  add    $0x28,%rsp
  addsd  %xmm5,%xmm3
  addsd  %xmm5,%xmm0
  mulsd  %xmm3,%xmm3
  mulsd  %xmm0,%xmm0
  divsd  %xmm3,%xmm1
  divsd  %xmm0,%xmm2
  addsd  %xmm4,%xmm1
  addsd  %xmm2,%xmm1
  movapd %xmm1,%xmm0
  ret
L37:
  movapd %xmm2,%xmm0
  movsd  %xmm3,0x18(%rsp)
  movsd  %xmm4,0x10(%rsp)
  movsd  %xmm6,0x8(%rsp)
  movsd  %xmm1,(%rsp)
  call   <sqrt@plt>
  movsd  0x18(%rsp),%xmm3
  movsd  0x10(%rsp),%xmm4
  pxor   %xmm5,%xmm5
  movsd  0x8(%rsp),%xmm6
  movsd  (%rsp),%xmm1
  movapd %xmm0,%xmm2
  jmp    L9
L52:
  movsd  %xmm1,0x18(%rsp)
  movsd  %xmm2,0x10(%rsp)
  movsd  %xmm3,0x8(%rsp)
  movsd  %xmm4,(%rsp)
  call   <sqrt@plt>
  movsd  0x18(%rsp),%xmm1
  movsd  0x10(%rsp),%xmm2
  pxor   %xmm5,%xmm5
  movsd  0x8(%rsp),%xmm3
  movsd  (%rsp),%xmm4
  jmp    L3
L30:
  movapd %xmm1,%xmm0
  movsd  %xmm2,0x10(%rsp)
  movsd  %xmm3,0x8(%rsp)
  movsd  %xmm4,(%rsp)
  call   <sqrt@plt>
  movsd  0x10(%rsp),%xmm2
  movsd  0x8(%rsp),%xmm3
  pxor   %xmm5,%xmm5
  movsd  (%rsp),%xmm4
  movapd %xmm0,%xmm1
  jmp    L32
```

tool_calls: name: read_data
arguments: label: D28
data_type: f64
name: read_data
arguments: label: D7
data_type: f64
name: read_data
arguments: label: D17
data_type: f64

tool: D28: 0.048

tool: D7: 0.014

tool: D17: 1

assistant: ```c
double icmCIE94sq(double Lab1[3], double Lab2[3]) {
  double desq, dhsq;
  double dlsq, dcsq;
  double c12;
  {
    double dl, da, db;
    dl = Lab1[0] - Lab2[0];
    dlsq = dl * dl;
    da = Lab1[1] - Lab2[1];
    db = Lab1[2] - Lab2[2];
    desq = dlsq + da * da + db * db;
  }
  {
    double c1, c2, dc;
    c1 = sqrt(Lab1[1] * Lab1[1] + Lab1[2] * Lab1[2]);
    c2 = sqrt(Lab2[1] * Lab2[1] + Lab2[2] * Lab2[2]);
    c12 = sqrt(c1 * c2);
    dc = c2 - c1;
    dcsq = dc * dc;
  }
  if ((dhsq = desq - dlsq - dcsq) < 0.0)
    dhsq = 0.0;
  {
    double sc, sh;
    sc = 1.0 + 0.048 * c12;
    sh = 1.0 + 0.014 * c12;
    return dlsq + dcsq / (sc * sc) + dhsq / (sh * sh);
  }
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D23(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    -0x14(%rbp),%rax
  mov    %rax,%rsi
  lea    D21(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x14(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  movl   $0x1,-0x10(%rbp)
  jmp    L4
L12:
  mov    -0xc(%rbp),%ecx
  movslq %ecx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %ecx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  sub    %eax,%ecx
  mov    %ecx,%edx
  test   %edx,%edx
  jne    L23
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0xc(%rbp)
  jmp    L4
L23:
  mov    -0xc(%rbp),%ecx
  movslq %ecx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %ecx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  sub    %eax,%ecx
  mov    %ecx,%edx
  mov    -0x10(%rbp),%eax
  imul   %edx,%eax
  mov    %eax,-0x10(%rbp)
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0xc(%rbp)
L4:
  cmpl   $0x0,-0xc(%rbp)
  jne    L12
  mov    -0x14(%rbp),%eax
  mov    -0x10(%rbp),%edx
  mov    %eax,%esi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  call   <getch@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: unpack_data
arguments: data_label: D23
data_type: string
name: unpack_data
arguments: data_label: D21
data_type: string
name: unpack_data
arguments: data_label: D9
data_type: string

tool: D23: Vamos brincar com os numeros!
Me diga um numero NATURAL que eu irei lhe dizer
qual a multiplicacao de todos os seus numeros
Ex: Se voce me dar 123, eu te darei 6, pois 6 = 1 x 2 x 3

tool: D21: %d

tool: D9: O numero dado e o: %d
O multiplo de seus digitos e: %d

assistant: ```c
int main() {
  int numero, mult, clone;
  printf("Vamos brincar com os numeros!\nMe diga um numero NATURAL que eu irei lhe dizer\nqual a multiplicacao de todos os seus numeros\nEx: Se voce me dar 123, eu te darei 6, pois 6 = 1 x 2 x 3\n");
  scanf("%d", &numero);
  clone = numero;
  mult = 1;
  while ((clone) != 0) {
    if ((clone % 10) == 0) {
      mult = mult;
      clone = clone / 10;
    } else {
      mult = mult * (clone % 10);
      clone = clone / 10;
    }
  }
  printf("O numero dado e o: %d\nO multiplo de seus digitos e: %d", numero, mult);
  getch();
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D14(%rip),%rdi
  call   <puts@plt>
  lea    0x4(%rsp),%rsi
  lea    D7(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%esi
  test   %esi,%esi
  je     L37
  mov    %esi,%edx
  mov    $0x1,%edi
  jmp    L28
L47:
  imul   %eax,%edi
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  sar    $0x1f,%edx
  sub    %edx,%eax
  mov    %eax,%edx
L12:
  test   %edx,%edx
  je     L51
L28:
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  lea    (%rax,%rax,4),%ecx
  add    %ecx,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  jne    L47
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  sar    $0x1f,%edx
  sub    %edx,%eax
  mov    %eax,%edx
  jmp    L12
L37:
  mov    $0x1,%edi
L51:
  mov    %edi,%edx
  lea    D23(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  call   <getch@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L31
  mov    $0x0,%eax
  add    $0x18,%rsp
  ret
L31:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D14
type: string
name: extract_symbolic_data
arguments: label: D7
type: string
name: extract_symbolic_data
arguments: label: D23
type: string

tool: D14: Vamos brincar com os numeros!
Me diga um numero NATURAL que eu irei lhe dizer
qual a multiplicacao de todos os seus numeros
Ex: Se voce me dar 123, eu te darei 6, pois 6 = 1 x 2 x 3

tool: D7: %d

tool: D23: O numero dado e o: %d
O multiplo de seus digitos e: %d

assistant: ```c
int main() {
  int numero, mult, clone;
  printf("Vamos brincar com os numeros!\nMe diga um numero NATURAL que eu irei lhe dizer\nqual a multiplicacao de todos os seus numeros\nEx: Se voce me dar 123, eu te darei 6, pois 6 = 1 x 2 x 3\n");
  scanf("%d", &numero);
  clone = numero;
  mult = 1;
  while ((clone) != 0) {
    if ((clone % 10) == 0) {
      mult = mult;
      clone = clone / 10;
    } else {
      mult = mult * (clone % 10);
      clone = clone / 10;
    }
  }
  printf("O numero dado e o: %d\nO multiplo de seus digitos e: %d", numero, mult);
  getch();
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D12(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  lea    0x4(%rsp),%rsi
  lea    D22(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%r9d
  test   %r9d,%r9d
  je     L47
  mov    %r9d,%edx
  mov    $0x1,%r8d
  nopl   0x0(%rax,%rax,1)
L4:
  movslq %edx,%rax
  mov    %edx,%ecx
  mov    %edx,%esi
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%ecx
  sar    $0x22,%rax
  sub    %ecx,%eax
  lea    (%rax,%rax,4),%ecx
  mov    %eax,%edx
  add    %ecx,%ecx
  sub    %ecx,%esi
  je     L35
  imul   %esi,%r8d
L35:
  test   %eax,%eax
  jne    L4
L6:
  xor    %eax,%eax
  mov    %r8d,%edx
  lea    D20(%rip),%rdi
  mov    %r9d,%esi
  call   <printf@plt>
  xor    %eax,%eax
  call   <getch@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L38
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L47:
  mov    $0x1,%r8d
  jmp    L6
L38:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D12
data_type: string
name: disassemble_data
arguments: data_label: D22
data_type: string
name: disassemble_data
arguments: data_label: D20
data_type: string

tool: D12: Vamos brincar com os numeros!
Me diga um numero NATURAL que eu irei lhe dizer
qual a multiplicacao de todos os seus numeros
Ex: Se voce me dar 123, eu te darei 6, pois 6 = 1 x 2 x 3

tool: D22: %d

tool: D20: O numero dado e o: %d
O multiplo de seus digitos e: %d

assistant: ```c
int main() {
  int numero, mult, clone;
  printf("Vamos brincar com os numeros!\nMe diga um numero NATURAL que eu irei lhe dizer\nqual a multiplicacao de todos os seus numeros\nEx: Se voce me dar 123, eu te darei 6, pois 6 = 1 x 2 x 3\n");
  scanf("%d", &numero);
  clone = numero;
  mult = 1;
  while ((clone) != 0) {
    if ((clone % 10) == 0) {
      mult = mult;
      clone = clone / 10;
    } else {
      mult = mult * (clone % 10);
      clone = clone / 10;
    }
  }
  printf("O numero dado e o: %d\nO multiplo de seus digitos e: %d", numero, mult);
  getch();
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D15(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  lea    0x4(%rsp),%rsi
  lea    D12(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%r9d
  test   %r9d,%r9d
  je     L36
  mov    %r9d,%edx
  mov    $0x1,%r8d
  nopl   0x0(%rax,%rax,1)
L13:
  movslq %edx,%rax
  mov    %edx,%ecx
  mov    %edx,%esi
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%ecx
  sar    $0x22,%rax
  sub    %ecx,%eax
  lea    (%rax,%rax,4),%ecx
  mov    %eax,%edx
  add    %ecx,%ecx
  sub    %ecx,%esi
  je     L40
  imul   %esi,%r8d
L40:
  test   %eax,%eax
  jne    L13
L38:
  xor    %eax,%eax
  mov    %r8d,%edx
  lea    D5(%rip),%rdi
  mov    %r9d,%esi
  call   <printf@plt>
  xor    %eax,%eax
  call   <getch@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L28
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L36:
  mov    $0x1,%r8d
  jmp    L38
L28:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D15
type_guessed: string
name: extract_symbolic_data
arguments: tag: D12
type_guessed: string
name: extract_symbolic_data
arguments: tag: D5
type_guessed: string

tool: D15: Vamos brincar com os numeros!
Me diga um numero NATURAL que eu irei lhe dizer
qual a multiplicacao de todos os seus numeros
Ex: Se voce me dar 123, eu te darei 6, pois 6 = 1 x 2 x 3

tool: D12: %d

tool: D5: O numero dado e o: %d
O multiplo de seus digitos e: %d

assistant: ```c
int main() {
  int numero, mult, clone;
  printf("Vamos brincar com os numeros!\nMe diga um numero NATURAL que eu irei lhe dizer\nqual a multiplicacao de todos os seus numeros\nEx: Se voce me dar 123, eu te darei 6, pois 6 = 1 x 2 x 3\n");
  scanf("%d", &numero);
  clone = numero;
  mult = 1;
  while ((clone) != 0) {
    if ((clone % 10) == 0) {
      mult = mult;
      clone = clone / 10;
    } else {
      mult = mult * (clone % 10);
      clone = clone / 10;
    }
  }
  printf("O numero dado e o: %d\nO multiplo de seus digitos e: %d", numero, mult);
  getch();
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<clear_extra>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  add    $0x3,%rax
  and    $0xfffffffffffffffc,%rax
  mov    %rax,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <dlsizeof@plt>
  sub    -0x20(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L0
  mov    -0x18(%rbp),%rdx
  mov    -0x20(%rbp),%rax
  add    %rax,%rdx
  mov    -0x8(%rbp),%rax
  mov    %rax,%rsi
  mov    %rdx,%rdi
  call   <clear@plt>
L0:
  nop
  leave
  ret
```

assistant: ```c
void clear_extra(void *p, size_t n) {
  n = (n + 3) & ~3;
  size_t extra = dlsizeof(p) - n;
  if (extra)
    clear((char *)p + n, extra);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<clear_extra>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  lea    0x3(%rsi),%rbx
  and    $0xfffffffffffffffc,%rbx
  call   <dlsizeof@plt>
  sub    %rbx,%rax
  jne    L4
L16:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L4:
  lea    0x0(%rbp,%rbx,1),%rdi
  mov    %rax,%rsi
  call   <clear@plt>
  jmp    L16
```

assistant: ```c
void clear_extra(void *p, size_t n) {
  n = (n + 3) & ~3;
  size_t extra = dlsizeof(p) - n;
  if (extra)
    clear((char *)p + n, extra);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<clear_extra>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  lea    0x3(%rsi),%rbx
  and    $0xfffffffffffffffc,%rbx
  sub    $0x8,%rsp
  call   <dlsizeof@plt>
  sub    %rbx,%rax
  jne    L0
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  cs nopw 0x0(%rax,%rax,1)
L0:
  add    $0x8,%rsp
  lea    0x0(%rbp,%rbx,1),%rdi
  mov    %rax,%rsi
  pop    %rbx
  pop    %rbp
  jmp    1060 <clear@plt>
```

assistant: ```c
void clear_extra(void *p, size_t n) {
  n = (n + 3) & ~3;
  size_t extra = dlsizeof(p) - n;
  if (extra)
    clear((char *)p + n, extra);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<clear_extra>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  lea    0x3(%rsi),%rbx
  and    $0xfffffffffffffffc,%rbx
  sub    $0x8,%rsp
  call   <dlsizeof@plt>
  sub    %rbx,%rax
  jne    L0
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  cs nopw 0x0(%rax,%rax,1)
L0:
  add    $0x8,%rsp
  lea    0x0(%rbp,%rbx,1),%rdi
  mov    %rax,%rsi
  pop    %rbx
  pop    %rbp
  jmp    1060 <clear@plt>
```

assistant: ```c
void clear_extra(void *p, size_t n) {
  n = (n + 3) & ~3;
  size_t extra = dlsizeof(p) - n;
  if (extra)
    clear((char *)p + n, extra);
}
```