ylfeng/ReF-Decompile-dataset huggingface dataset:
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    D39(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x14(%rbp),%rax
  mov    %rax,%rsi
  lea    D25(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  jmp    L13
L28:
  lea    D46(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x14(%rbp),%rax
  mov    %rax,%rsi
  lea    D25(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
L13:
  mov    -0x14(%rbp),%eax
  cmp    $0x1,%eax
  jle    L28
  mov    -0x14(%rbp),%eax
  cmp    $0xa,%eax
  jg     L28
  movl   $0x1,-0x10(%rbp)
  jmp    L38
L60:
  movl   $0x1,-0xc(%rbp)
  jmp    L76
L27:
  lea    D59(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0xc(%rbp)
L76:
  mov    -0xc(%rbp),%eax
  cmp    -0x10(%rbp),%eax
  jle    L27
  lea    D50(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x10(%rbp)
L38:
  mov    -0x14(%rbp),%eax
  cmp    %eax,-0x10(%rbp)
  jle    L60
  mov    $0x0,%eax
  call   <getch@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L20
  call   <__stack_chk_fail@plt>
L20:
  leave
  ret
```

tool_calls: name: parse_data
arguments: tag: D39
data_type: string
name: parse_data
arguments: tag: D25
data_type: string
name: parse_data
arguments: tag: D46
data_type: string
name: parse_data
arguments: tag: D59
data_type: string
name: parse_data
arguments: tag: D50
data_type: string

tool: D39: Ola, me chamo Desenhador, eu gosto de desenhar!
Hoje eu quero desenhar um triangulo, me ajuda?
Digite um numero NATURAL entre 2 e 10
e eu desenharei um triangulo especial para voce!
Brinca comigo!!!



tool: D25: %d

tool: D46: Ei, eu te disse que o numero tem que ser NATUAL
e entre DOIS (2) e DEZ (10)
e nao vale trapacear!
Digite um numero valido desta vez, por favor!


tool: D59: *

tool: D50: 


assistant: ```c
int main() {
  int numero, linhas, colunas;
  printf("Ola, me chamo Desenhador, eu gosto de desenhar!\nHoje eu quero desenhar um triangulo, me ajuda?\nDigite um numero NATURAL entre 2 e 10\ne eu desenharei um triangulo especial para voce!\nBrinca comigo!!!\n\n");
  scanf("%d", &numero);
  while ((numero < 2) || (numero > 10)) {
    printf("Ei, eu te disse que o numero tem que ser NATUAL\ne entre DOIS (2) e DEZ (10)\ne nao vale trapacear!\nDigite um numero valido desta vez, por favor!\n");
    scanf("%d", &numero);
  }
  for (linhas = 1; linhas <= numero; linhas++) {
    for (colunas = 1; colunas <= linhas; colunas++) {
      printf("*");
    }
    printf("\n");
  }
  getch();
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D38(%rip),%rdi
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D26(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  sub    $0x2,%eax
  cmp    $0x8,%eax
  jbe    L12
  lea    D25(%rip),%r12
  lea    0x4(%rsp),%rbp
  lea    D26(%rip),%rbx
L9:
  mov    %r12,%rdi
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  sub    $0x2,%eax
  cmp    $0x8,%eax
  ja     L9
L12:
  mov    $0x1,%ebp
  lea    D22(%rip),%r12
  lea    D49(%rip),%r13
  cmpl   $0x0,0x4(%rsp)
  jg     L20
L61:
  mov    $0x0,%eax
  call   <getch@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L72
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L51:
  mov    %r12,%rdi
  call   <printf@plt>
  add    $0x1,%ebx
  cmp    %ebp,%ebx
  jle    L51
L69:
  mov    %r13,%rdi
  call   <printf@plt>
  add    $0x1,%ebp
  cmp    %ebp,0x4(%rsp)
  jl     L61
L20:
  mov    $0x1,%ebx
  test   %ebp,%ebp
  jg     L51
  jmp    L69
L72:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D38
type_infered: string
name: disassemble_data
arguments: data_label: D26
type_infered: string
name: disassemble_data
arguments: data_label: D25
type_infered: string
name: disassemble_data
arguments: data_label: D22
type_infered: string
name: disassemble_data
arguments: data_label: D49
type_infered: string

tool: D38: Ola, me chamo Desenhador, eu gosto de desenhar!
Hoje eu quero desenhar um triangulo, me ajuda?
Digite um numero NATURAL entre 2 e 10
e eu desenharei um triangulo especial para voce!
Brinca comigo!!!



tool: D26: %d

tool: D25: Ei, eu te disse que o numero tem que ser NATUAL
e entre DOIS (2) e DEZ (10)
e nao vale trapacear!
Digite um numero valido desta vez, por favor!


tool: D22: *

tool: D49: 


assistant: ```c
int main() {
  int numero, linhas, colunas;
  printf("Ola, me chamo Desenhador, eu gosto de desenhar!\nHoje eu quero desenhar um triangulo, me ajuda?\nDigite um numero NATURAL entre 2 e 10\ne eu desenharei um triangulo especial para voce!\nBrinca comigo!!!\n\n");
  scanf("%d", &numero);
  while ((numero < 2) || (numero > 10)) {
    printf("Ei, eu te disse que o numero tem que ser NATUAL\ne entre DOIS (2) e DEZ (10)\ne nao vale trapacear!\nDigite um numero valido desta vez, por favor!\n");
    scanf("%d", &numero);
  }
  for (linhas = 1; linhas <= numero; linhas++) {
    for (colunas = 1; colunas <= linhas; colunas++) {
      printf("*");
    }
    printf("\n");
  }
  getch();
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %r13
  lea    D32(%rip),%rdi
  push   %r12
  push   %rbp
  push   %rbx
  lea    D30(%rip),%rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rbp
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  sub    $0x2,%eax
  cmp    $0x8,%eax
  jbe    L40
  lea    D26(%rip),%r12
  nopl   0x0(%rax)
L10:
  mov    %r12,%rdi
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  sub    $0x2,%eax
  cmp    $0x8,%eax
  ja     L10
L40:
  mov    $0x1,%ebp
  lea    D39(%rip),%r12
  lea    D35(%rip),%r13
  nopw   0x0(%rax,%rax,1)
L36:
  mov    $0x1,%ebx
  nopl   (%rax)
L7:
  mov    %r12,%rdi
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    %ebp,%ebx
  jle    L7
  mov    %r13,%rdi
  add    $0x1,%ebp
  call   <printf@plt>
  cmp    %ebp,0x4(%rsp)
  jge    L36
  xor    %eax,%eax
  call   <getch@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L29
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L29:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D32
datatype: string
name: disassemble_data
arguments: label: D30
datatype: string
name: disassemble_data
arguments: label: D26
datatype: string
name: disassemble_data
arguments: label: D39
datatype: string
name: disassemble_data
arguments: label: D35
datatype: string

tool: D32: Ola, me chamo Desenhador, eu gosto de desenhar!
Hoje eu quero desenhar um triangulo, me ajuda?
Digite um numero NATURAL entre 2 e 10
e eu desenharei um triangulo especial para voce!
Brinca comigo!!!



tool: D30: %d

tool: D26: Ei, eu te disse que o numero tem que ser NATUAL
e entre DOIS (2) e DEZ (10)
e nao vale trapacear!
Digite um numero valido desta vez, por favor!


tool: D39: *

tool: D35: 


assistant: ```c
int main() {
  int numero, linhas, colunas;
  printf("Ola, me chamo Desenhador, eu gosto de desenhar!\nHoje eu quero desenhar um triangulo, me ajuda?\nDigite um numero NATURAL entre 2 e 10\ne eu desenharei um triangulo especial para voce!\nBrinca comigo!!!\n\n");
  scanf("%d", &numero);
  while ((numero < 2) || (numero > 10)) {
    printf("Ei, eu te disse que o numero tem que ser NATUAL\ne entre DOIS (2) e DEZ (10)\ne nao vale trapacear!\nDigite um numero valido desta vez, por favor!\n");
    scanf("%d", &numero);
  }
  for (linhas = 1; linhas <= numero; linhas++) {
    for (colunas = 1; colunas <= linhas; colunas++) {
      printf("*");
    }
    printf("\n");
  }
  getch();
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %r13
  lea    D38(%rip),%rdi
  push   %r12
  push   %rbp
  push   %rbx
  lea    D16(%rip),%rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rbp
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  sub    $0x2,%eax
  cmp    $0x8,%eax
  jbe    L39
  lea    D31(%rip),%r12
  nopl   0x0(%rax)
L34:
  mov    %r12,%rdi
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  sub    $0x2,%eax
  cmp    $0x8,%eax
  ja     L34
L39:
  mov    $0x1,%ebp
  lea    D8(%rip),%r12
  lea    D0(%rip),%r13
  nopw   0x0(%rax,%rax,1)
L46:
  mov    $0x1,%ebx
  nopl   (%rax)
L48:
  mov    %r12,%rdi
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    %ebp,%ebx
  jle    L48
  mov    %r13,%rdi
  add    $0x1,%ebp
  call   <printf@plt>
  cmp    %ebp,0x4(%rsp)
  jge    L46
  xor    %eax,%eax
  call   <getch@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L26
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L26:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D38
type_guessed: string
name: retrieve_data
arguments: data_tag: D16
type_guessed: string
name: retrieve_data
arguments: data_tag: D31
type_guessed: string
name: retrieve_data
arguments: data_tag: D8
type_guessed: string
name: retrieve_data
arguments: data_tag: D0
type_guessed: string

tool: D38: Ola, me chamo Desenhador, eu gosto de desenhar!
Hoje eu quero desenhar um triangulo, me ajuda?
Digite um numero NATURAL entre 2 e 10
e eu desenharei um triangulo especial para voce!
Brinca comigo!!!



tool: D16: %d

tool: D31: Ei, eu te disse que o numero tem que ser NATUAL
e entre DOIS (2) e DEZ (10)
e nao vale trapacear!
Digite um numero valido desta vez, por favor!


tool: D8: *

tool: D0: 


assistant: ```c
int main() {
  int numero, linhas, colunas;
  printf("Ola, me chamo Desenhador, eu gosto de desenhar!\nHoje eu quero desenhar um triangulo, me ajuda?\nDigite um numero NATURAL entre 2 e 10\ne eu desenharei um triangulo especial para voce!\nBrinca comigo!!!\n\n");
  scanf("%d", &numero);
  while ((numero < 2) || (numero > 10)) {
    printf("Ei, eu te disse que o numero tem que ser NATUAL\ne entre DOIS (2) e DEZ (10)\ne nao vale trapacear!\nDigite um numero valido desta vez, por favor!\n");
    scanf("%d", &numero);
  }
  for (linhas = 1; linhas <= numero; linhas++) {
    for (colunas = 1; colunas <= linhas; colunas++) {
      printf("*");
    }
    printf("\n");
  }
  getch();
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<TimerOutputSrcSelect>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x8(%rbp),%rax
  add    $0x4,%rax
  mov    (%rax),%rax
  mov    -0x8(%rbp),%rdx
  add    $0x4,%rdx
  and    $0xffffffffffffff8f,%rax
  mov    %rax,(%rdx)
  mov    -0x8(%rbp),%rax
  add    $0x4,%rax
  mov    (%rax),%rax
  mov    -0x8(%rbp),%rdx
  add    $0x4,%rdx
  or     -0x10(%rbp),%rax
  mov    %rax,(%rdx)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void TimerOutputSrcSelect(unsigned long ulBase, unsigned long ulSelect) {

  ;

  ;
  (*((volatile unsigned long *)(ulBase + 0x00000004))) &= ~0x00000070;
  (*((volatile unsigned long *)(ulBase + 0x00000004))) |= ulSelect;
}
```
user: What is the c source code of the assembly code below:

```asm
<TimerOutputSrcSelect>:
  endbr64
  mov    0x4(%rdi),%rax
  and    $0xffffffffffffff8f,%rax
  mov    %rax,0x4(%rdi)
  mov    0x4(%rdi),%rax
  or     %rsi,%rax
  mov    %rax,0x4(%rdi)
  ret
```

assistant: ```c
void TimerOutputSrcSelect(unsigned long ulBase, unsigned long ulSelect) {

  ;

  ;
  (*((volatile unsigned long *)(ulBase + 0x00000004))) &= ~0x00000070;
  (*((volatile unsigned long *)(ulBase + 0x00000004))) |= ulSelect;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<TimerOutputSrcSelect>:
  endbr64
  mov    0x4(%rdi),%rax
  and    $0xffffffffffffff8f,%rax
  mov    %rax,0x4(%rdi)
  mov    0x4(%rdi),%rax
  or     %rsi,%rax
  mov    %rax,0x4(%rdi)
  ret
```

assistant: ```c
void TimerOutputSrcSelect(unsigned long ulBase, unsigned long ulSelect) {

  ;

  ;
  (*((volatile unsigned long *)(ulBase + 0x00000004))) &= ~0x00000070;
  (*((volatile unsigned long *)(ulBase + 0x00000004))) |= ulSelect;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<TimerOutputSrcSelect>:
  endbr64
  mov    0x4(%rdi),%rax
  and    $0xffffffffffffff8f,%rax
  mov    %rax,0x4(%rdi)
  mov    0x4(%rdi),%rax
  or     %rsi,%rax
  mov    %rax,0x4(%rdi)
  ret
```

assistant: ```c
void TimerOutputSrcSelect(unsigned long ulBase, unsigned long ulSelect) {

  ;

  ;
  (*((volatile unsigned long *)(ulBase + 0x00000004))) &= ~0x00000070;
  (*((volatile unsigned long *)(ulBase + 0x00000004))) |= ulSelect;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<shiftLeft>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  mov    %edx,-0xc(%rbp)
  cmpl   $0x0,-0xc(%rbp)
  jne    L6
  mov    -0x4(%rbp),%eax
  jmp    L1
L6:
  cmpl   $0x0,-0xc(%rbp)
  jns    L5
  mov    -0xc(%rbp),%eax
  neg    %eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%ecx
  mov    -0x4(%rbp),%eax
  mov    %ecx,%esi
  mov    %eax,%edi
  call   <shiftRight@plt>
  jmp    L1
L5:
  mov    -0x4(%rbp),%eax
L1:
  leave
  ret
```

assistant: ```c
int shiftLeft(int data, int width, int count) {
  if (count == 0)
    return data;
  if (count < 0)
    return shiftRight(data, width, -count);
  return data;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<shiftLeft>:
  endbr64
  mov    %edi,%eax
  test   %edx,%edx
  js     L2
  ret
L2:
  sub    $0x8,%rsp
  neg    %edx
  call   <shiftRight@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
int shiftLeft(int data, int width, int count) {
  if (count == 0)
    return data;
  if (count < 0)
    return shiftRight(data, width, -count);
  return data;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<shiftLeft>:
  endbr64
  mov    %edi,%eax
  test   %edx,%edx
  js     L4
  ret
  nopl   0x0(%rax,%rax,1)
L4:
  neg    %edx
  jmp    1050 <shiftRight@plt>
```

assistant: ```c
int shiftLeft(int data, int width, int count) {
  if (count == 0)
    return data;
  if (count < 0)
    return shiftRight(data, width, -count);
  return data;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<shiftLeft>:
  endbr64
  mov    %edi,%eax
  test   %edx,%edx
  js     L7
  ret
  nopl   0x0(%rax,%rax,1)
L7:
  neg    %edx
  jmp    1050 <shiftRight@plt>
```

assistant: ```c
int shiftLeft(int data, int width, int count) {
  if (count == 0)
    return data;
  if (count < 0)
    return shiftRight(data, width, -count);
  return data;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<strcpy>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  nop
L2:
  mov    -0x10(%rbp),%rdx
  lea    0x1(%rdx),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rcx
  mov    %rcx,-0x8(%rbp)
  movzbl (%rdx),%edx
  mov    %dl,(%rax)
  movzbl (%rax),%eax
  test   %al,%al
  jne    L2
  mov    -0x18(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
char *strcpy(char *restrict dest, const char *restrict src) {
  const unsigned char *s = (const unsigned char *)src;
  unsigned char *d = (unsigned char *)dest;
  while ((*d++ = *s++))
    ;
  return dest;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<strcpy>:
  endbr64
  mov    %rdi,%rax
  mov    $0x0,%edx
L9:
  movzbl (%rsi,%rdx,1),%ecx
  mov    %cl,(%rax,%rdx,1)
  add    $0x1,%rdx
  test   %cl,%cl
  jne    L9
  ret
```

assistant: ```c
char *strcpy(char *restrict dest, const char *restrict src) {
  const unsigned char *s = (const unsigned char *)src;
  unsigned char *d = (unsigned char *)dest;
  while ((*d++ = *s++))
    ;
  return dest;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<strcpy>:
  endbr64
  mov    %rdi,%rax
  xor    %edx,%edx
  nopl   0x0(%rax)
L9:
  movzbl (%rsi,%rdx,1),%ecx
  mov    %cl,(%rax,%rdx,1)
  add    $0x1,%rdx
  test   %cl,%cl
  jne    L9
  ret
```

assistant: ```c
char *strcpy(char *restrict dest, const char *restrict src) {
  const unsigned char *s = (const unsigned char *)src;
  unsigned char *d = (unsigned char *)dest;
  while ((*d++ = *s++))
    ;
  return dest;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<strcpy>:
  endbr64
  mov    %rdi,%rax
  xor    %edx,%edx
  nopl   0x0(%rax)
L4:
  movzbl (%rsi,%rdx,1),%ecx
  mov    %cl,(%rax,%rdx,1)
  add    $0x1,%rdx
  test   %cl,%cl
  jne    L4
  ret
```

assistant: ```c
char *strcpy(char *restrict dest, const char *restrict src) {
  const unsigned char *s = (const unsigned char *)src;
  unsigned char *d = (unsigned char *)dest;
  while ((*d++ = *s++))
    ;
  return dest;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<wcrossprod>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x38(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    %rdx,-0x48(%rbp)
  mov    %rcx,-0x50(%rbp)
  mov    %r8d,-0x54(%rbp)
  mov    %r9d,-0x58(%rbp)
  mov    -0x38(%rbp),%rax
  mov    %rax,-0x20(%rbp)
  mov    -0x40(%rbp),%rax
  mov    %rax,-0x18(%rbp)
  mov    -0x48(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x50(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0x30(%rbp)
  jmp    L28
L41:
  movl   $0x0,-0x2c(%rbp)
  jmp    L20
L14:
  movl   $0x0,-0x28(%rbp)
  mov    -0x30(%rbp),%eax
  imul   0x10(%rbp),%eax
  mov    %eax,-0x24(%rbp)
  mov    -0x30(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm1
  mov    -0x2c(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  mulsd  %xmm0,%xmm1
  mov    -0x28(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  mov    -0x24(%rbp),%edx
  mov    -0x2c(%rbp),%eax
  add    %edx,%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mulsd  %xmm1,%xmm0
  movsd  %xmm0,(%rax)
  movl   $0x1,-0x28(%rbp)
  jmp    L1
L17:
  mov    -0x24(%rbp),%edx
  mov    -0x2c(%rbp),%eax
  add    %edx,%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm1
  mov    -0x28(%rbp),%eax
  imul   -0x58(%rbp),%eax
  mov    %eax,%edx
  mov    -0x30(%rbp),%eax
  add    %edx,%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm2
  mov    -0x28(%rbp),%eax
  imul   0x10(%rbp),%eax
  mov    %eax,%edx
  mov    -0x2c(%rbp),%eax
  add    %edx,%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  mulsd  %xmm0,%xmm2
  mov    -0x28(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  mulsd  %xmm2,%xmm0
  mov    -0x24(%rbp),%edx
  mov    -0x2c(%rbp),%eax
  add    %edx,%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  addsd  %xmm1,%xmm0
  movsd  %xmm0,(%rax)
  addl   $0x1,-0x28(%rbp)
L1:
  mov    -0x28(%rbp),%eax
  cmp    -0x54(%rbp),%eax
  jl     L17
  addl   $0x1,-0x2c(%rbp)
L20:
  mov    -0x2c(%rbp),%eax
  cmp    0x10(%rbp),%eax
  jl     L14
  addl   $0x1,-0x30(%rbp)
L28:
  mov    -0x30(%rbp),%eax
  cmp    -0x58(%rbp),%eax
  jl     L41
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void wcrossprod(const double *x, const double *y, const double *w, double *z, const int m, const int n, const int p) {
  int i, j, k, ip;
  const double *restrict xr = x, *restrict yr = y, *restrict wr = w;
  double *restrict zr = z;
  for (i = 0; i < n; i++) {
    for (j = 0; j < p; j++) {
      k = 0;
      ip = i * p;
      zr[ip + j] = xr[i] * yr[j] * wr[k];
      for (k = 1; k < m; k++)
        zr[ip + j] += xr[k * n + i] * yr[k * p + j] * wr[k];
    }
  }
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<wcrossprod>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rcx,-0x8(%rsp)
  mov    %r8d,-0x20(%rsp)
  test   %r9d,%r9d
  jle    L8
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  mov    %rdx,%r12
  mov    %r8d,%esi
  movslq %r9d,%r8
  shl    $0x3,%r8
  lea    (%rdi,%r8,1),%r15
  mov    %r9d,%eax
  mov    %rax,-0x18(%rsp)
  movslq 0x38(%rsp),%rdi
  shl    $0x3,%rdi
  lea    0x0(%rbp,%rdi,1),%rax
  mov    %rax,-0x10(%rsp)
  lea    -0x2(%rsi),%eax
  lea    0x10(%rdx,%rax,8),%r9
  movl   $0x0,-0x1c(%rsp)
  mov    $0x0,%r14d
  mov    0x38(%rsp),%eax
  mov    %rax,-0x28(%rsp)
  jmp    L11
L59:
  movslq -0x1c(%rsp),%rax
  mov    -0x8(%rsp),%rcx
  lea    (%rcx,%rax,8),%r11
  mov    -0x10(%rsp),%r13
  mov    $0x0,%r10d
L46:
  mov    %r11,%rsi
  movsd  (%rbx,%r14,8),%xmm0
  mulsd  0x0(%rbp,%r10,8),%xmm0
  mulsd  (%r12),%xmm0
  movsd  %xmm0,(%r11)
  cmpl   $0x1,-0x20(%rsp)
  jle    L14
  lea    0x8(%r12),%rax
  mov    %r13,%rcx
  mov    %r15,%rdx
L23:
  movsd  (%rdx),%xmm0
  mulsd  (%rcx),%xmm0
  mulsd  (%rax),%xmm0
  addsd  (%rsi),%xmm0
  movsd  %xmm0,(%rsi)
  add    %r8,%rdx
  add    %rdi,%rcx
  add    $0x8,%rax
  cmp    %r9,%rax
  jne    L23
L14:
  add    $0x1,%r10
  add    $0x8,%r11
  add    $0x8,%r13
  cmp    -0x28(%rsp),%r10
  jne    L46
L60:
  add    $0x1,%r14
  mov    0x38(%rsp),%edx
  add    %edx,-0x1c(%rsp)
  add    $0x8,%r15
  cmp    -0x18(%rsp),%r14
  je     L8
L11:
  cmpl   $0x0,0x38(%rsp)
  jg     L59
  jmp    L60
L8:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void wcrossprod(const double *x, const double *y, const double *w, double *z, const int m, const int n, const int p) {
  int i, j, k, ip;
  const double *restrict xr = x, *restrict yr = y, *restrict wr = w;
  double *restrict zr = z;
  for (i = 0; i < n; i++) {
    for (j = 0; j < p; j++) {
      k = 0;
      ip = i * p;
      zr[ip + j] = xr[i] * yr[j] * wr[k];
      for (k = 1; k < m; k++)
        zr[ip + j] += xr[k * n + i] * yr[k * p + j] * wr[k];
    }
  }
}
```