ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x28(%rbp)
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x20(%rbp),%rax
  mov    %rax,%rsi
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x24(%rbp)
  movl   $0x0,-0x2c(%rbp)
  jmp    L69
L37:
  mov    -0x2c(%rbp),%eax
  cltq
  movzbl -0x20(%rbp,%rax,1),%edx
  mov    -0x24(%rbp),%eax
  sub    -0x2c(%rbp),%eax
  sub    $0x1,%eax
  cltq
  movzbl -0x20(%rbp,%rax,1),%eax
  cmp    %al,%dl
  je     L54
  movl   $0x1,-0x28(%rbp)
  jmp    L0
L54:
  addl   $0x1,-0x2c(%rbp)
L69:
  mov    -0x2c(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jl     L37
L0:
  cmpl   $0x1,-0x28(%rbp)
  jne    L60
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L67
L60:
  lea    D17(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L67:
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L61
  call   <__stack_chk_fail@plt>
L61:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D0
type_infered: string
name: extract_symbolic_data
arguments: tag: D19
type_infered: string
name: extract_symbolic_data
arguments: tag: D7
type_infered: string
name: extract_symbolic_data
arguments: tag: D17
type_infered: string

tool: D0: Vavedi simvoli:

tool: D19: %s

tool: D7: 0

tool: D17: 1

assistant: ```c
int main() {
  char s[20];
  int i, j, p = 0;
  printf("Vavedi simvoli:");
  scanf("%s", s);
  j = strlen(s);
  for (i = 0; i < j; i++) {
    if (s[i] != s[j - i - 1]) {
      p = 1;
      break;
    }
  }
  if (p == 1) {
    printf("0");
  } else {
    printf("1");
  }
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    D28(%rip),%rdi
  call   <printf@plt>
  mov    %rsp,%rbx
  mov    %rbx,%rsi
  lea    D30(%rip),%rdi
  call   <scanf@plt>
  mov    %rbx,%rdi
  call   <strlen@plt>
  test   %eax,%eax
  jle    L44
  mov    %rsp,%rdx
  movslq %eax,%rcx
  add    %rbx,%rcx
  lea    -0x1(%rax),%eax
  lea    0x1(%rsp,%rax,1),%rax
L37:
  movzbl -0x1(%rcx),%esi
  cmp    %sil,(%rdx)
  jne    L28
  add    $0x1,%rdx
  sub    $0x1,%rcx
  cmp    %rax,%rdx
  jne    L37
L44:
  lea    D22(%rip),%rdi
  call   <printf@plt>
L0:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  mov    $0x0,%eax
  add    $0x20,%rsp
  pop    %rbx
  ret
L28:
  lea    D14(%rip),%rdi
  call   <printf@plt>
  jmp    L0
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D28
type: string
name: parse_data
arguments: label: D30
type: string
name: parse_data
arguments: label: D22
type: string
name: parse_data
arguments: label: D14
type: string

tool: D28: Vavedi simvoli:

tool: D30: %s

tool: D22: 1

tool: D14: 0

assistant: ```c
int main() {
  char s[20];
  int i, j, p = 0;
  printf("Vavedi simvoli:");
  scanf("%s", s);
  j = strlen(s);
  for (i = 0; i < j; i++) {
    if (s[i] != s[j - i - 1]) {
      p = 1;
      break;
    }
  }
  if (p == 1) {
    printf("0");
  } else {
    printf("1");
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbx
  lea    D16(%rip),%rdi
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  call   <printf@plt>
  lea    D5(%rip),%rdi
  mov    %rbx,%rsi
  call   <scanf@plt>
  mov    %rbx,%rdi
  call   <strlen@plt>
  test   %eax,%eax
  jle    L29
  movslq %eax,%rcx
  sub    $0x1,%eax
  mov    %rbx,%rdx
  add    %rbx,%rcx
  lea    0x1(%rsp,%rax,1),%rax
  jmp    L37
  cs nopw 0x0(%rax,%rax,1)
L50:
  add    $0x1,%rdx
  sub    $0x1,%rcx
  cmp    %rax,%rdx
  je     L29
L37:
  movzbl -0x1(%rcx),%esi
  cmp    %sil,(%rdx)
  je     L50
  lea    D10(%rip),%rdi
  call   <printf@plt>
  jmp    L23
L29:
  lea    D0(%rip),%rdi
  call   <printf@plt>
L23:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L49
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbx
  ret
L49:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D16
type: string
name: extract_symbolic_data
arguments: data_tag: D5
type: string
name: extract_symbolic_data
arguments: data_tag: D10
type: string
name: extract_symbolic_data
arguments: data_tag: D0
type: string

tool: D16: Vavedi simvoli:

tool: D5: %s

tool: D10: 0

tool: D0: 1

assistant: ```c
int main() {
  char s[20];
  int i, j, p = 0;
  printf("Vavedi simvoli:");
  scanf("%s", s);
  j = strlen(s);
  for (i = 0; i < j; i++) {
    if (s[i] != s[j - i - 1]) {
      p = 1;
      break;
    }
  }
  if (p == 1) {
    printf("0");
  } else {
    printf("1");
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  push   %rbx
  lea    D20(%rip),%rdi
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  call   <printf@plt>
  lea    D38(%rip),%rdi
  mov    %rbx,%rsi
  call   <scanf@plt>
  mov    %rbx,%rdi
  call   <strlen@plt>
  test   %eax,%eax
  jle    L37
  movslq %eax,%rcx
  sub    $0x1,%eax
  mov    %rbx,%rdx
  add    %rbx,%rcx
  lea    0x1(%rsp,%rax,1),%rax
  jmp    L16
  cs nopw 0x0(%rax,%rax,1)
L4:
  add    $0x1,%rdx
  sub    $0x1,%rcx
  cmp    %rax,%rdx
  je     L37
L16:
  movzbl -0x1(%rcx),%esi
  cmp    %sil,(%rdx)
  je     L4
  lea    D16(%rip),%rdi
  call   <printf@plt>
  jmp    L49
L37:
  lea    D37(%rip),%rdi
  call   <printf@plt>
L49:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L13
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbx
  ret
L13:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D20
type_guessed: string
name: extract_symbolic_data
arguments: tag: D38
type_guessed: string
name: extract_symbolic_data
arguments: tag: D16
type_guessed: string
name: extract_symbolic_data
arguments: tag: D37
type_guessed: string

tool: D20: Vavedi simvoli:

tool: D38: %s

tool: D16: 0

tool: D37: 1

assistant: ```c
int main() {
  char s[20];
  int i, j, p = 0;
  printf("Vavedi simvoli:");
  scanf("%s", s);
  j = strlen(s);
  for (i = 0; i < j; i++) {
    if (s[i] != s[j - i - 1]) {
      p = 1;
      break;
    }
  }
  if (p == 1) {
    printf("0");
  } else {
    printf("1");
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<hyoptenuse>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  movsd  %xmm0,-0x18(%rbp)
  movsd  %xmm1,-0x20(%rbp)
  movsd  -0x18(%rbp),%xmm0
  movapd %xmm0,%xmm1
  mulsd  %xmm0,%xmm1
  movsd  -0x20(%rbp),%xmm0
  mulsd  %xmm0,%xmm0
  addsd  %xmm0,%xmm1
  movq   %xmm1,%rax
  movq   %rax,%xmm0
  call   <sqrt@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x8(%rbp)
  movsd  -0x8(%rbp),%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  leave
  ret
```

assistant: ```c
double hyoptenuse(double n, double m) {
  double d;
  d = sqrt(n * n + m * m);
  return d;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<hyoptenuse>:
  endbr64
  mulsd  %xmm0,%xmm0
  mulsd  %xmm1,%xmm1
  addsd  %xmm1,%xmm0
  pxor   %xmm1,%xmm1
  ucomisd %xmm0,%xmm1
  ja     L6
  sqrtsd %xmm0,%xmm0
  ret
L6:
  sub    $0x8,%rsp
  call   <sqrt@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
double hyoptenuse(double n, double m) {
  double d;
  d = sqrt(n * n + m * m);
  return d;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<hyoptenuse>:
  endbr64
  mulsd  %xmm1,%xmm1
  mulsd  %xmm0,%xmm0
  addsd  %xmm1,%xmm0
  pxor   %xmm1,%xmm1
  ucomisd %xmm0,%xmm1
  ja     L0
  sqrtsd %xmm0,%xmm0
  ret
L0:
  jmp    1050 <sqrt@plt>
```

assistant: ```c
double hyoptenuse(double n, double m) {
  double d;
  d = sqrt(n * n + m * m);
  return d;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<hyoptenuse>:
  endbr64
  mulsd  %xmm1,%xmm1
  mulsd  %xmm0,%xmm0
  addsd  %xmm1,%xmm0
  pxor   %xmm1,%xmm1
  ucomisd %xmm0,%xmm1
  ja     L2
  sqrtsd %xmm0,%xmm0
  ret
L2:
  jmp    1050 <sqrt@plt>
```

assistant: ```c
double hyoptenuse(double n, double m) {
  double d;
  d = sqrt(n * n + m * m);
  return d;
}
```
user: What is the c source code of the assembly code below:

```asm
<start>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L17
L2:
  addl   $0x1,-0x4(%rbp)
L17:
  cmpl   $0x0,-0x4(%rbp)
  jle    L2
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void start(int vetor[], int i) {
  int j;
  for (j = 0; j < 1; j++)
    vetor[j] == i;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<start>:
  endbr64
  ret
```

assistant: ```c
void start(int vetor[], int i) {
  int j;
  for (j = 0; j < 1; j++)
    vetor[j] == i;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<start>:
  endbr64
  ret
```

assistant: ```c
void start(int vetor[], int i) {
  int j;
  for (j = 0; j < 1; j++)
    vetor[j] == i;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<string_split>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  jne    L5
  mov    $0x0,%eax
  jmp    L4
L5:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  mov    %al,-0x1(%rbp)
  jmp    L3
L2:
  movsbl -0x1(%rbp),%eax
  mov    -0x20(%rbp),%rdx
  mov    %eax,%edi
  call   *%rdx
  test   %eax,%eax
  je     L40
  mov    -0x18(%rbp),%rax
  movb   $0x0,(%rax)
  mov    -0x18(%rbp),%rax
  add    $0x1,%rax
  jmp    L4
L40:
  addq   $0x1,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  mov    %al,-0x1(%rbp)
L3:
  cmpb   $0x0,-0x1(%rbp)
  jne    L2
  mov    $0x0,%eax
L4:
  leave
  ret
```

assistant: ```c
char *string_split(char *str, int (*sep_match)(int c)) {
  char c;
  if (!str)
    return NULL;
  for (c = *str; c != '\0'; c = *str) {
    if (sep_match(c)) {
      *str = '\0';
      return str + 1;
    }
    str++;
  }
  return NULL;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<string_split>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  test   %rdi,%rdi
  je     L27
  mov    %rsi,%rbp
  movzbl (%rdi),%edi
  test   %dil,%dil
  je     L66
L51:
  movsbl %dil,%edi
  call   *%rbp
  test   %eax,%eax
  jne    L34
  add    $0x1,%rbx
  movzbl (%rbx),%edi
  test   %dil,%dil
  jne    L51
  mov    $0x0,%eax
  jmp    L29
L34:
  movb   $0x0,(%rbx)
  lea    0x1(%rbx),%rax
L29:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L27:
  mov    %rdi,%rax
  jmp    L29
L66:
  mov    $0x0,%eax
  jmp    L29
```

assistant: ```c
char *string_split(char *str, int (*sep_match)(int c)) {
  char c;
  if (!str)
    return NULL;
  for (c = *str; c != '\0'; c = *str) {
    if (sep_match(c)) {
      *str = '\0';
      return str + 1;
    }
    str++;
  }
  return NULL;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<string_split>:
  endbr64
  test   %rdi,%rdi
  je     L24
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  jmp    L23
  nopw   0x0(%rax,%rax,1)
L36:
  call   *%rbp
  mov    %rbx,%rdx
  add    $0x1,%rbx
  test   %eax,%eax
  jne    L18
L23:
  movsbl (%rbx),%edi
  test   %dil,%dil
  jne    L36
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
  xchg   %ax,%ax
L18:
  movb   $0x0,(%rdx)
  add    $0x8,%rsp
  mov    %rbx,%rax
  pop    %rbx
  pop    %rbp
  ret
L24:
  xor    %eax,%eax
  ret
```

assistant: ```c
char *string_split(char *str, int (*sep_match)(int c)) {
  char c;
  if (!str)
    return NULL;
  for (c = *str; c != '\0'; c = *str) {
    if (sep_match(c)) {
      *str = '\0';
      return str + 1;
    }
    str++;
  }
  return NULL;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<string_split>:
  endbr64
  test   %rdi,%rdi
  je     L29
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  jmp    L37
  nopw   0x0(%rax,%rax,1)
L16:
  call   *%rbp
  mov    %rbx,%rdx
  add    $0x1,%rbx
  test   %eax,%eax
  jne    L15
L37:
  movsbl (%rbx),%edi
  test   %dil,%dil
  jne    L16
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
  xchg   %ax,%ax
L15:
  movb   $0x0,(%rdx)
  add    $0x8,%rsp
  mov    %rbx,%rax
  pop    %rbx
  pop    %rbp
  ret
L29:
  xor    %eax,%eax
  ret
```

assistant: ```c
char *string_split(char *str, int (*sep_match)(int c)) {
  char c;
  if (!str)
    return NULL;
  for (c = *str; c != '\0'; c = *str) {
    if (sep_match(c)) {
      *str = '\0';
      return str + 1;
    }
    str++;
  }
  return NULL;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  add    $0xffffffffffffff80,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x78(%rbp)
  jmp    L44
L36:
  lea    -0x70(%rbp),%rdx
  mov    -0x78(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rdx,%rax
  mov    %rax,%rsi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  addl   $0x1,-0x78(%rbp)
L44:
  cmpl   $0x3,-0x78(%rbp)
  jle    L36
  movl   $0x0,-0x78(%rbp)
  jmp    L11
L32:
  movl   $0x0,-0x74(%rbp)
  jmp    L65
L8:
  lea    -0x50(%rbp),%rdx
  mov    -0x74(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x78(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rcx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  mov    %rax,%rsi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  addl   $0x1,-0x74(%rbp)
L65:
  cmpl   $0x3,-0x74(%rbp)
  jle    L8
  addl   $0x1,-0x78(%rbp)
L11:
  cmpl   $0x3,-0x78(%rbp)
  jle    L32
  lea    -0x60(%rbp),%rdx
  lea    -0x50(%rbp),%rcx
  lea    -0x70(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <multVetorMatriz@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L47
  call   <__stack_chk_fail@plt>
L47:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D5
data_type: string

tool: D5: %d

assistant: ```c
int main() {
  int i, j, A[4][4], v[4], w[4];
  for (i = 0; i < 4; i++)
    scanf("%d", &v[i]);
  for (i = 0; i < 4; i++)
    for (j = 0; j < 4; j++)
      scanf("%d", &A[i][j]);
  multVetorMatriz(v, A, w);
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x78,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  lea    0x10(%rsp),%r12
  lea    D9(%rip),%rbp
L35:
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  add    $0x4,%rbx
  cmp    %r12,%rbx
  jne    L35
  lea    0x30(%rsp),%rbp
  mov    $0x0,%r13d
  lea    D9(%rip),%r12
L28:
  lea    -0x10(%rbp),%rbx
L15:
  mov    %rbx,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  add    $0x4,%rbx
  cmp    %rbp,%rbx
  jne    L15
  add    $0x4,%r13
  add    $0x10,%rbp
  cmp    $0x10,%r13
  jne    L28
  lea    0x10(%rsp),%rdx
  lea    0x20(%rsp),%rsi
  mov    %rsp,%rdi
  call   <multVetorMatriz@plt>
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  mov    $0x0,%eax
  add    $0x78,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D9
data_type: string

tool: D9: %d

assistant: ```c
int main() {
  int i, j, A[4][4], v[4], w[4];
  for (i = 0; i < 4; i++)
    scanf("%d", &v[i]);
  for (i = 0; i < 4; i++)
    for (j = 0; j < 4; j++)
      scanf("%d", &A[i][j]);
  multVetorMatriz(v, A, w);
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  lea    D2(%rip),%r12
  push   %rbp
  push   %rbx
  sub    $0x78,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r14
  lea    0x10(%rsp),%r13
  mov    %r14,%rbx
L15:
  mov    %rbx,%rsi
  mov    %r12,%rdi
  add    $0x4,%rbx
  call   <scanf@plt>
  cmp    %r13,%rbx
  jne    L15
  lea    0x30(%rsp),%rbp
  xor    %r15d,%r15d
L13:
  lea    -0x10(%rbp),%rbx
L29:
  mov    %rbx,%rsi
  mov    %r12,%rdi
  add    $0x4,%rbx
  call   <scanf@plt>
  cmp    %rbp,%rbx
  jne    L29
  add    $0x4,%r15
  lea    0x10(%rbx),%rbp
  cmp    $0x10,%r15
  jne    L13
  lea    0x20(%rsp),%rsi
  mov    %r13,%rdx
  mov    %r14,%rdi
  call   <multVetorMatriz@plt>
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L12
  add    $0x78,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L12:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D2
type_infered: string

tool: D2: %d

assistant: ```c
int main() {
  int i, j, A[4][4], v[4], w[4];
  for (i = 0; i < 4; i++)
    scanf("%d", &v[i]);
  for (i = 0; i < 4; i++)
    for (j = 0; j < 4; j++)
      scanf("%d", &A[i][j]);
  multVetorMatriz(v, A, w);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  lea    D4(%rip),%r12
  push   %rbp
  push   %rbx
  sub    $0x78,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r14
  lea    0x10(%rsp),%r13
  mov    %r14,%rbx
L15:
  mov    %rbx,%rsi
  mov    %r12,%rdi
  add    $0x4,%rbx
  call   <scanf@plt>
  cmp    %r13,%rbx
  jne    L15
  lea    0x30(%rsp),%rbp
  xor    %r15d,%r15d
L38:
  lea    -0x10(%rbp),%rbx
L37:
  mov    %rbx,%rsi
  mov    %r12,%rdi
  add    $0x4,%rbx
  call   <scanf@plt>
  cmp    %rbp,%rbx
  jne    L37
  add    $0x4,%r15
  lea    0x10(%rbx),%rbp
  cmp    $0x10,%r15
  jne    L38
  lea    0x20(%rsp),%rsi
  mov    %r13,%rdx
  mov    %r14,%rdi
  call   <multVetorMatriz@plt>
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L35
  add    $0x78,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L35:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D4
type: string

tool: D4: %d

assistant: ```c
int main() {
  int i, j, A[4][4], v[4], w[4];
  for (i = 0; i < 4; i++)
    scanf("%d", &v[i]);
  for (i = 0; i < 4; i++)
    for (j = 0; j < 4; j++)
      scanf("%d", &A[i][j]);
  multVetorMatriz(v, A, w);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<sub_const_r_1>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movsd  %xmm0,-0x20(%rbp)
  mov    %rsi,-0x28(%rbp)
  mov    %edx,-0x2c(%rbp)
  mov    %ecx,-0x30(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L3
L10:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm1
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movsd  -0x20(%rbp),%xmm0
  subsd  %xmm1,%xmm0
  movsd  %xmm0,(%rax)
  addl   $0x1,-0x4(%rbp)
L3:
  mov    -0x30(%rbp),%eax
  sub    -0x2c(%rbp),%eax
  cmp    %eax,-0x4(%rbp)
  jle    L10
  mov    -0x30(%rbp),%eax
  sub    -0x2c(%rbp),%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  mov    -0x30(%rbp),%eax
  sub    -0x2c(%rbp),%eax
  add    $0x1,%eax
  pxor   %xmm1,%xmm1
  cvtsi2sd %eax,%xmm1
  mulsd  -0x20(%rbp),%xmm1
  mov    -0x30(%rbp),%eax
  sub    -0x2c(%rbp),%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  subsd  %xmm1,%xmm0
  movsd  %xmm0,(%rax)
  mov    -0x18(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
double *sub_const_r_1(double *res, double cn, double *vec, int lo1, int hi1) {
  int i;
  for (i = 0; i <= hi1 - lo1; i++)
    res[i] = cn - vec[i];
  res[hi1 - lo1 + 1] -= cn * (hi1 - lo1 + 1);
  return res;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<sub_const_r_1>:
  endbr64
  sub    %edx,%ecx
  js     L7
  lea    0x1(%rcx),%edx
  shl    $0x3,%rdx
  mov    $0x0,%eax
L4:
  movapd %xmm0,%xmm1
  subsd  (%rsi,%rax,1),%xmm1
  movsd  %xmm1,(%rdi,%rax,1)
  add    $0x8,%rax
  cmp    %rax,%rdx
  jne    L4
L7:
  movslq %ecx,%rax
  lea    0x8(%rdi,%rax,8),%rax
  add    $0x1,%ecx
  pxor   %xmm1,%xmm1
  cvtsi2sd %ecx,%xmm1
  mulsd  %xmm0,%xmm1
  movsd  (%rax),%xmm0
  subsd  %xmm1,%xmm0
  movsd  %xmm0,(%rax)
  mov    %rdi,%rax
  ret
```

assistant: ```c
double *sub_const_r_1(double *res, double cn, double *vec, int lo1, int hi1) {
  int i;
  for (i = 0; i <= hi1 - lo1; i++)
    res[i] = cn - vec[i];
  res[hi1 - lo1 + 1] -= cn * (hi1 - lo1 + 1);
  return res;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<sub_const_r_1>:
  endbr64
  mov    %rdi,%r8
  sub    %edx,%ecx
  js     L4
  movslq %ecx,%rax
  lea    0x8(,%rax,8),%rdx
  xor    %eax,%eax
  nopl   0x0(%rax,%rax,1)
L10:
  movapd %xmm0,%xmm1
  subsd  (%rsi,%rax,1),%xmm1
  movsd  %xmm1,(%r8,%rax,1)
  add    $0x8,%rax
  cmp    %rax,%rdx
  jne    L10
L4:
  add    $0x1,%ecx
  pxor   %xmm1,%xmm1
  cvtsi2sd %ecx,%xmm1
  movslq %ecx,%rax
  lea    (%r8,%rax,8),%rax
  mulsd  %xmm0,%xmm1
  movsd  (%rax),%xmm0
  subsd  %xmm1,%xmm0
  movsd  %xmm0,(%rax)
  mov    %r8,%rax
  ret
```

assistant: ```c
double *sub_const_r_1(double *res, double cn, double *vec, int lo1, int hi1) {
  int i;
  for (i = 0; i <= hi1 - lo1; i++)
    res[i] = cn - vec[i];
  res[hi1 - lo1 + 1] -= cn * (hi1 - lo1 + 1);
  return res;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<sub_const_r_1>:
  endbr64
  mov    %rdi,%r8
  sub    %edx,%ecx
  js     L24
  lea    0x8(%rsi),%rax
  cmp    %rax,%rdi
  je     L53
  cmp    $0x1,%ecx
  jbe    L53
  add    $0x1,%ecx
  movapd %xmm0,%xmm2
  xor    %eax,%eax
  mov    %ecx,%edx
  unpcklpd %xmm2,%xmm2
  shr    %edx
  shl    $0x4,%rdx
  cs nopw 0x0(%rax,%rax,1)
L55:
  movupd (%rsi,%rax,1),%xmm3
  movapd %xmm2,%xmm1
  subpd  %xmm3,%xmm1
  movups %xmm1,(%r8,%rax,1)
  add    $0x10,%rax
  cmp    %rdx,%rax
  jne    L55
  mov    %ecx,%eax
  and    $0xfffffffe,%eax
  test   $0x1,%cl
  je     L62
  cltq
  movapd %xmm0,%xmm1
  subsd  (%rsi,%rax,8),%xmm1
  movsd  %xmm1,(%r8,%rax,8)
L62:
  pxor   %xmm1,%xmm1
  movslq %ecx,%rax
  cvtsi2sd %ecx,%xmm1
  lea    (%r8,%rax,8),%rax
  mulsd  %xmm0,%xmm1
  movsd  (%rax),%xmm0
  subsd  %xmm1,%xmm0
  movsd  %xmm0,(%rax)
  mov    %r8,%rax
  ret
  nopl   0x0(%rax)
L53:
  movslq %ecx,%rax
  lea    0x8(,%rax,8),%rdx
  xor    %eax,%eax
  nopl   (%rax)
L15:
  movapd %xmm0,%xmm1
  subsd  (%rsi,%rax,1),%xmm1
  movsd  %xmm1,(%r8,%rax,1)
  add    $0x8,%rax
  cmp    %rax,%rdx
  jne    L15
L24:
  add    $0x1,%ecx
  jmp    L62
```

assistant: ```c
double *sub_const_r_1(double *res, double cn, double *vec, int lo1, int hi1) {
  int i;
  for (i = 0; i <= hi1 - lo1; i++)
    res[i] = cn - vec[i];
  res[hi1 - lo1 + 1] -= cn * (hi1 - lo1 + 1);
  return res;
}
```