ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Demonstrate the C code that generates the assembly output below:

```asm
<cvt_num_field>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x3,-0xc(%rbp)
  movl   $0x0,-0x4(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  jns    L77
  mov    -0x4(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x4(%rbp)
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movb   $0x2d,(%rax)
  negq   -0x18(%rbp)
  jmp    L12
L77:
  mov    -0x4(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x4(%rbp)
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movb   $0x2b,(%rax)
  jmp    L12
L29:
  mov    -0x18(%rbp),%rcx
  movabs $0x6666666666666667,%rdx
  mov    %rcx,%rax
  imul   %rdx
  mov    %rdx,%rax
  sar    $0x2,%rax
  sar    $0x3f,%rcx
  mov    %rcx,%rdx
  sub    %rdx,%rax
  mov    %rax,-0x18(%rbp)
  subl   $0x1,-0xc(%rbp)
L12:
  cmpq   $0x270f,-0x18(%rbp)
  jg     L29
  jmp    L8
L18:
  addl   $0x3,-0xc(%rbp)
L8:
  cmpl   $0x0,-0xc(%rbp)
  jle    L18
  mov    $0x4,%eax
  sub    -0xc(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  movl   $0x3e8,-0x8(%rbp)
  jmp    L83
L89:
  mov    -0xc(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0xc(%rbp)
  test   %eax,%eax
  jne    L48
  mov    -0x4(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x4(%rbp)
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movb   $0x2e,(%rax)
L48:
  mov    -0x8(%rbp),%eax
  movslq %eax,%rsi
  mov    -0x18(%rbp),%rax
  cqto
  idiv   %rsi
  lea    0x30(%rax),%ecx
  mov    -0x4(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x4(%rbp)
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    %ecx,%edx
  mov    %dl,(%rax)
  mov    -0x8(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x18(%rbp),%rax
  cqto
  idiv   %rcx
  mov    %rdx,-0x18(%rbp)
  mov    -0x8(%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,-0x8(%rbp)
L83:
  cmpl   $0x0,-0x8(%rbp)
  jg     L89
  nop
  pop    %rbp
  ret
```

assistant: ```c
void cvt_num_field(long int n, char *s) {
  int dp = 3;
  int d;
  int i = 0;
  if (n < 0) {
    s[i++] = '-';
    n = -n;
  } else {
    s[i++] = '+';
  }
  while (n > 9999) {
    n /= 10;
    dp--;
  }
  while (dp <= 0)
    dp += 3;
  dp = 4 - dp;
  for (d = 1000; d > 0; d /= 10) {
    if (dp-- == 0)
      s[i++] = '.';
    s[i++] = (n / d) + '0';
    n %= d;
  }
  return;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<cvt_num_field>:
  endbr64
  push   %rbx
  mov    %rdi,%rcx
  mov    %rsi,%r11
  mov    $0x2b,%eax
  test   %rdi,%rdi
  js     L0
L64:
  mov    %al,(%r11)
  cmp    $0x270f,%rcx
  jle    L47
  mov    $0x3,%esi
  movabs $0x6666666666666667,%r8
L10:
  mov    %rcx,%rdi
  mov    %rcx,%rax
  imul   %r8
  sar    $0x2,%rdx
  mov    %rcx,%rax
  sar    $0x3f,%rax
  sub    %rax,%rdx
  mov    %rdx,%rcx
  sub    $0x1,%esi
  cmp    $0x1869f,%rdi
  jg     L10
  test   %esi,%esi
  jg     L52
L87:
  add    $0x3,%esi
  test   %esi,%esi
  jle    L87
L52:
  mov    $0x4,%r9d
  sub    %esi,%r9d
  neg    %esi
  mov    $0x1,%r10d
  mov    $0x3e8,%ebx
  jmp    L56
L0:
  neg    %rcx
  mov    $0x2d,%eax
  jmp    L64
L47:
  mov    $0x3,%esi
  jmp    L52
L106:
  movslq %ebx,%rdi
  lea    0x1(%rax),%r10d
  movslq %eax,%r8
  mov    %rcx,%rax
  cqto
  idiv   %rdi
  add    $0x30,%eax
  mov    %al,(%r11,%r8,1)
  mov    %rdx,%rcx
  imul   $0x66666667,%rdi,%rdi
  sar    $0x22,%rdi
  mov    %ebx,%eax
  sar    $0x1f,%eax
  mov    %edi,%ebx
  sub    %eax,%ebx
  sub    $0x1,%r9d
  cmp    %r9d,%esi
  je     L14
L56:
  mov    %r10d,%eax
  test   %r9d,%r9d
  jne    L106
  lea    0x1(%r10),%eax
  movslq %r10d,%r10
  movb   $0x2e,(%r11,%r10,1)
  jmp    L106
L14:
  pop    %rbx
  ret
```

assistant: ```c
void cvt_num_field(long int n, char *s) {
  int dp = 3;
  int d;
  int i = 0;
  if (n < 0) {
    s[i++] = '-';
    n = -n;
  } else {
    s[i++] = '+';
  }
  while (n > 9999) {
    n /= 10;
    dp--;
  }
  while (dp <= 0)
    dp += 3;
  dp = 4 - dp;
  for (d = 1000; d > 0; d /= 10) {
    if (dp-- == 0)
      s[i++] = '.';
    s[i++] = (n / d) + '0';
    n %= d;
  }
  return;
}
```
user: Translate the below assembly code into C programming language:

```asm
<cvt_num_field>:
  endbr64
  mov    %rdi,%rax
  mov    %rsi,%r9
  mov    $0x2b,%edx
  test   %rdi,%rdi
  jns    L67
  neg    %rax
  mov    $0x2d,%edx
L67:
  mov    %dl,(%r9)
  cmp    $0x270f,%rax
  jle    L63
  movabs $0xcccccccccccccccd,%rdi
  mov    $0x3,%ecx
  nopw   0x0(%rax,%rax,1)
L94:
  mov    %rax,%rsi
  mul    %rdi
  sub    $0x1,%ecx
  mov    %rdx,%rax
  shr    $0x3,%rax
  cmp    $0x1869f,%rsi
  jg     L94
  mov    $0x3,%esi
  sub    %ecx,%esi
  test   %ecx,%ecx
  jg     L16
  nopl   0x0(%rax)
L60:
  mov    %ecx,%esi
  add    $0x3,%ecx
  test   %ecx,%ecx
  jle    L60
  neg    %esi
L16:
  lea    -0x3(%rsi),%r10d
  mov    $0x3e8,%r8d
  mov    $0x1,%edi
L3:
  movslq %r8d,%rcx
  cqto
  movslq %edi,%r11
  idiv   %rcx
  imul   $0x66666667,%rcx,%rcx
  sar    $0x22,%rcx
  add    $0x30,%eax
  mov    %al,(%r9,%r11,1)
  mov    %r8d,%eax
  sar    $0x1f,%eax
  sub    %eax,%ecx
  mov    %ecx,%r8d
  cmp    %esi,%r10d
  je     L96
  lea    0x1(%rdi),%eax
  test   %esi,%esi
  jne    L17
  cltq
  add    $0x2,%edi
  movb   $0x2e,(%r9,%rax,1)
L73:
  mov    %rdx,%rax
  sub    $0x1,%esi
  jmp    L3
  nopl   0x0(%rax,%rax,1)
L96:
  ret
  nopl   0x0(%rax)
L17:
  mov    %eax,%edi
  jmp    L73
  nopl   0x0(%rax)
L63:
  xor    %esi,%esi
  jmp    L16
```

assistant: ```c
void cvt_num_field(long int n, char *s) {
  int dp = 3;
  int d;
  int i = 0;
  if (n < 0) {
    s[i++] = '-';
    n = -n;
  } else {
    s[i++] = '+';
  }
  while (n > 9999) {
    n /= 10;
    dp--;
  }
  while (dp <= 0)
    dp += 3;
  dp = 4 - dp;
  for (d = 1000; d > 0; d /= 10) {
    if (dp-- == 0)
      s[i++] = '.';
    s[i++] = (n / d) + '0';
    n %= d;
  }
  return;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D13(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    -0x20(%rbp),%rax
  mov    %rax,%rsi
  lea    D69(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    D39(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    -0x1c(%rbp),%rax
  mov    %rax,%rsi
  lea    D69(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  movss  -0x20(%rbp),%xmm1
  movss  -0x1c(%rbp),%xmm0
  addss  %xmm1,%xmm0
  movss  %xmm0,-0x18(%rbp)
  pxor   %xmm2,%xmm2
  cvtss2sd -0x18(%rbp),%xmm2
  movq   %xmm2,%rax
  movq   %rax,%xmm0
  lea    D68(%rip),%rax
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  movss  -0x20(%rbp),%xmm0
  movss  -0x1c(%rbp),%xmm1
  subss  %xmm1,%xmm0
  movss  %xmm0,-0x14(%rbp)
  pxor   %xmm3,%xmm3
  cvtss2sd -0x14(%rbp),%xmm3
  movq   %xmm3,%rax
  movq   %rax,%xmm0
  lea    D42(%rip),%rax
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  movss  -0x20(%rbp),%xmm1
  movss  -0x1c(%rbp),%xmm0
  mulss  %xmm1,%xmm0
  movss  %xmm0,-0x10(%rbp)
  pxor   %xmm4,%xmm4
  cvtss2sd -0x10(%rbp),%xmm4
  movq   %xmm4,%rax
  movq   %rax,%xmm0
  lea    D21(%rip),%rax
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  movss  -0x20(%rbp),%xmm0
  movss  -0x1c(%rbp),%xmm1
  divss  %xmm1,%xmm0
  movss  %xmm0,-0xc(%rbp)
  pxor   %xmm5,%xmm5
  cvtss2sd -0xc(%rbp),%xmm5
  movq   %xmm5,%rax
  movq   %rax,%xmm0
  lea    D64(%rip),%rax
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L2
  call   <__stack_chk_fail@plt>
L2:
  leave
  ret
```

tool_calls: name: parse_data
arguments: label: D13
type_infered: string
name: parse_data
arguments: label: D69
type_infered: string
name: parse_data
arguments: label: D39
type_infered: string
name: parse_data
arguments: label: D68
type_infered: string
name: parse_data
arguments: label: D42
type_infered: string
name: parse_data
arguments: label: D21
type_infered: string
name: parse_data
arguments: label: D64
type_infered: string

tool: D13: what is the first integer

tool: D69: %f

tool: D39: what is the second integer

tool: D68: the sum for the two integers is:%f 


tool: D42: the difference for the two integers is:%f


tool: D21: the product for the two integers is: %f


tool: D64: the quotient for the two integers is:%f


assistant: ```c
int main() {
  float first;
  printf("what is the first integer\n");
  scanf("%f", &first);
  float second;
  printf("what is the second integer\n");
  scanf("%f", &second);
  float sum = (float)first + second;
  printf("the sum for the two integers is:%f \n", sum);
  float difference = (float)first - second;
  printf("the difference for the two integers is:%f\n", difference);
  float product = (float)first * second;
  printf("the product for the two integers is: %f\n", product);
  float quotient = (float)first / second;
  printf("the quotient for the two integers is:%f\n", quotient);
  return 0;
}
```
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   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D29(%rip),%rdi
  call   <puts@plt>
  mov    %rsp,%rsi
  lea    D13(%rip),%rbx
  mov    %rbx,%rdi
  call   <scanf@plt>
  lea    D0(%rip),%rdi
  call   <puts@plt>
  lea    0x4(%rsp),%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  movss  (%rsp),%xmm0
  addss  0x4(%rsp),%xmm0
  cvtss2sd %xmm0,%xmm0
  lea    D39(%rip),%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  movss  (%rsp),%xmm0
  subss  0x4(%rsp),%xmm0
  cvtss2sd %xmm0,%xmm0
  lea    D50(%rip),%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  movss  (%rsp),%xmm0
  mulss  0x4(%rsp),%xmm0
  cvtss2sd %xmm0,%xmm0
  lea    D2(%rip),%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  movss  (%rsp),%xmm0
  divss  0x4(%rsp),%xmm0
  cvtss2sd %xmm0,%xmm0
  lea    D27(%rip),%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  mov    $0x0,%eax
  add    $0x10,%rsp
  pop    %rbx
  ret
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D29
datatype: string
name: parse_data
arguments: tag: D13
datatype: string
name: parse_data
arguments: tag: D0
datatype: string
name: parse_data
arguments: tag: D39
datatype: string
name: parse_data
arguments: tag: D50
datatype: string
name: parse_data
arguments: tag: D2
datatype: string
name: parse_data
arguments: tag: D27
datatype: string

tool: D29: what is the first integer

tool: D13: %f

tool: D0: what is the second integer

tool: D39: the sum for the two integers is:%f 


tool: D50: the difference for the two integers is:%f


tool: D2: the product for the two integers is: %f


tool: D27: the quotient for the two integers is:%f


assistant: ```c
int main() {
  float first;
  printf("what is the first integer\n");
  scanf("%f", &first);
  float second;
  printf("what is the second integer\n");
  scanf("%f", &second);
  float sum = (float)first + second;
  printf("the sum for the two integers is:%f \n", sum);
  float difference = (float)first - second;
  printf("the difference for the two integers is:%f\n", difference);
  float product = (float)first * second;
  printf("the product for the two integers is: %f\n", product);
  float quotient = (float)first / second;
  printf("the quotient for the two integers is:%f\n", quotient);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D66(%rip),%rdi
  lea    D11(%rip),%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  mov    %rsp,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    D58(%rip),%rdi
  call   <puts@plt>
  lea    0x4(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  movss  (%rsp),%xmm0
  addss  0x4(%rsp),%xmm0
  lea    D34(%rip),%rdi
  mov    $0x1,%eax
  cvtss2sd %xmm0,%xmm0
  call   <printf@plt>
  movss  (%rsp),%xmm0
  subss  0x4(%rsp),%xmm0
  lea    D19(%rip),%rdi
  mov    $0x1,%eax
  cvtss2sd %xmm0,%xmm0
  call   <printf@plt>
  movss  (%rsp),%xmm0
  mulss  0x4(%rsp),%xmm0
  lea    D25(%rip),%rdi
  mov    $0x1,%eax
  cvtss2sd %xmm0,%xmm0
  call   <printf@plt>
  lea    D26(%rip),%rdi
  mov    $0x1,%eax
  movss  (%rsp),%xmm0
  divss  0x4(%rsp),%xmm0
  cvtss2sd %xmm0,%xmm0
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D66
datatype: string
name: parse_data
arguments: label: D11
datatype: string
name: parse_data
arguments: label: D58
datatype: string
name: parse_data
arguments: label: D34
datatype: string
name: parse_data
arguments: label: D19
datatype: string
name: parse_data
arguments: label: D25
datatype: string
name: parse_data
arguments: label: D26
datatype: string

tool: D66: what is the first integer

tool: D11: %f

tool: D58: what is the second integer

tool: D34: the sum for the two integers is:%f 


tool: D19: the difference for the two integers is:%f


tool: D25: the product for the two integers is: %f


tool: D26: the quotient for the two integers is:%f


assistant: ```c
int main() {
  float first;
  printf("what is the first integer\n");
  scanf("%f", &first);
  float second;
  printf("what is the second integer\n");
  scanf("%f", &second);
  float sum = (float)first + second;
  printf("the sum for the two integers is:%f \n", sum);
  float difference = (float)first - second;
  printf("the difference for the two integers is:%f\n", difference);
  float product = (float)first * second;
  printf("the product for the two integers is: %f\n", product);
  float quotient = (float)first / second;
  printf("the quotient for the two integers is:%f\n", quotient);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D48(%rip),%rdi
  lea    D8(%rip),%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  mov    %rsp,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    D38(%rip),%rdi
  call   <puts@plt>
  lea    0x4(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  movss  (%rsp),%xmm0
  addss  0x4(%rsp),%xmm0
  lea    D25(%rip),%rdi
  mov    $0x1,%eax
  cvtss2sd %xmm0,%xmm0
  call   <printf@plt>
  movss  (%rsp),%xmm0
  subss  0x4(%rsp),%xmm0
  lea    D51(%rip),%rdi
  mov    $0x1,%eax
  cvtss2sd %xmm0,%xmm0
  call   <printf@plt>
  movss  (%rsp),%xmm0
  mulss  0x4(%rsp),%xmm0
  lea    D27(%rip),%rdi
  mov    $0x1,%eax
  cvtss2sd %xmm0,%xmm0
  call   <printf@plt>
  lea    D19(%rip),%rdi
  mov    $0x1,%eax
  movss  (%rsp),%xmm0
  divss  0x4(%rsp),%xmm0
  cvtss2sd %xmm0,%xmm0
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D48
type: string
name: parse_data
arguments: label: D8
type: string
name: parse_data
arguments: label: D38
type: string
name: parse_data
arguments: label: D25
type: string
name: parse_data
arguments: label: D51
type: string
name: parse_data
arguments: label: D27
type: string
name: parse_data
arguments: label: D19
type: string

tool: D48: what is the first integer

tool: D8: %f

tool: D38: what is the second integer

tool: D25: the sum for the two integers is:%f 


tool: D51: the difference for the two integers is:%f


tool: D27: the product for the two integers is: %f


tool: D19: the quotient for the two integers is:%f


assistant: ```c
int main() {
  float first;
  printf("what is the first integer\n");
  scanf("%f", &first);
  float second;
  printf("what is the second integer\n");
  scanf("%f", &second);
  float sum = (float)first + second;
  printf("the sum for the two integers is:%f \n", sum);
  float difference = (float)first - second;
  printf("the difference for the two integers is:%f\n", difference);
  float product = (float)first * second;
  printf("the product for the two integers is: %f\n", product);
  float quotient = (float)first / second;
  printf("the quotient for the two integers is:%f\n", quotient);
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<double_unlock_bad>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <pthread_mutex_unlock@plt>
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <pthread_mutex_unlock@plt>
  nop
  leave
  ret
```

assistant: ```c
void double_unlock_bad(pthread_mutex_t *m) {
  pthread_mutex_unlock(m);
  pthread_mutex_unlock(m);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<double_unlock_bad>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <pthread_mutex_unlock@plt>
  mov    %rbx,%rdi
  call   <pthread_mutex_unlock@plt>
  pop    %rbx
  ret
```

assistant: ```c
void double_unlock_bad(pthread_mutex_t *m) {
  pthread_mutex_unlock(m);
  pthread_mutex_unlock(m);
}
```
user: What is the c source code of the assembly code below:

```asm
<double_unlock_bad>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  call   <pthread_mutex_unlock@plt>
  mov    %rbp,%rdi
  pop    %rbp
  jmp    1050 <pthread_mutex_unlock@plt>
```

assistant: ```c
void double_unlock_bad(pthread_mutex_t *m) {
  pthread_mutex_unlock(m);
  pthread_mutex_unlock(m);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<double_unlock_bad>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  call   <pthread_mutex_unlock@plt>
  mov    %rbp,%rdi
  pop    %rbp
  jmp    1050 <pthread_mutex_unlock@plt>
```

assistant: ```c
void double_unlock_bad(pthread_mutex_t *m) {
  pthread_mutex_unlock(m);
  pthread_mutex_unlock(m);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<syscon_store_standard_Laurential>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x68,%rsp
  mov    %edi,-0x54(%rbp)
  mov    %esi,-0x58(%rbp)
  mov    %edx,-0x5c(%rbp)
  mov    %rcx,-0x68(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  mov    %rsp,%rax
  mov    %rax,%rbx
  mov    -0x54(%rbp),%eax
  movslq %eax,%rdx
  sub    $0x1,%rdx
  mov    %rdx,-0x40(%rbp)
  movslq %eax,%rdx
  mov    %rdx,%r10
  mov    $0x0,%r11d
  movslq %eax,%rdx
  mov    %rdx,%r8
  mov    $0x0,%r9d
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    $0x10,%eax
  sub    $0x1,%rax
  add    %rdx,%rax
  mov    $0x10,%esi
  mov    $0x0,%edx
  div    %rsi
  imul   $0x10,%rax,%rax
  mov    %rax,%rcx
  and    $0xfffffffffffff000,%rcx
  mov    %rsp,%rdx
  sub    %rcx,%rdx
L45:
  cmp    %rdx,%rsp
  je     L3
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L45
L3:
  mov    %rax,%rdx
  and    $0xfff,%edx
  sub    %rdx,%rsp
  mov    %rax,%rdx
  and    $0xfff,%edx
  test   %rdx,%rdx
  je     L22
  and    $0xfff,%eax
  sub    $0x8,%rax
  add    %rsp,%rax
  orq    $0x0,(%rax)
L22:
  mov    %rsp,%rax
  add    $0x3,%rax
  shr    $0x2,%rax
  shl    $0x2,%rax
  mov    %rax,-0x38(%rbp)
  mov    -0x54(%rbp),%eax
  mov    %eax,-0x24(%rbp)
  mov    -0x58(%rbp),%eax
  mov    %eax,-0x20(%rbp)
  mov    -0x5c(%rbp),%eax
  mov    %eax,-0x1c(%rbp)
  movl   $0x0,-0x48(%rbp)
  jmp    L14
L20:
  mov    -0x48(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x68(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%ecx
  mov    -0x38(%rbp),%rax
  mov    -0x48(%rbp),%edx
  movslq %edx,%rdx
  mov    %ecx,(%rax,%rdx,4)
  addl   $0x1,-0x48(%rbp)
L14:
  mov    -0x48(%rbp),%eax
  cmp    -0x54(%rbp),%eax
  jl     L20
  mov    -0x30(%rbp),%rcx
  mov    -0x38(%rbp),%rdx
  lea    -0x24(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x4a,%edi
  call   <_ada_use_c2phc@plt>
  mov    %eax,-0x44(%rbp)
  mov    -0x44(%rbp),%eax
  mov    %rbx,%rsp
  mov    -0x18(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L12
  call   <__stack_chk_fail@plt>
L12:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int syscon_store_standard_Laurential(int nc, int n, int k, char *p) {
  int a[3], b[nc], i, fail;
  double *c;
  a[0] = nc;
  a[1] = n;
  a[2] = k;
  for (i = 0; i < nc; i++)
    b[i] = (int)p[i];
  fail = _ada_use_c2phc(74, a, b, c);
  return fail;
}
```
user: Translate the below assembly code into C programming language:

```asm
<syscon_store_standard_Laurential>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,%eax
  mov    %esi,%r9d
  mov    %edx,%r8d
  mov    %rcx,%rsi
  mov    %fs:0x28,%rdx
  mov    %rdx,-0x8(%rbp)
  xor    %edx,%edx
  movslq %edi,%rdx
  lea    0xf(,%rdx,4),%rdx
  mov    %rdx,%rdi
  and    $0xfffffffffffffff0,%rdi
  and    $0xfffffffffffff000,%rdx
  mov    %rsp,%rcx
  sub    %rdx,%rcx
L39:
  cmp    %rcx,%rsp
  je     L15
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L39
L15:
  mov    %rdi,%rdx
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  je     L45
  orq    $0x0,-0x8(%rsp,%rdx,1)
L45:
  mov    %rsp,%rdi
  mov    %eax,-0x14(%rbp)
  mov    %r9d,-0x10(%rbp)
  mov    %r8d,-0xc(%rbp)
  test   %eax,%eax
  jle    L28
  mov    %eax,%ecx
  mov    $0x0,%eax
L30:
  movsbl (%rsi,%rax,1),%edx
  mov    %edx,(%rdi,%rax,4)
  add    $0x1,%rax
  cmp    %rcx,%rax
  jne    L30
L28:
  lea    -0x14(%rbp),%rsi
  mov    $0x0,%ecx
  mov    %rdi,%rdx
  mov    $0x4a,%edi
  call   <_ada_use_c2phc@plt>
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  jne    L49
  leave
  ret
L49:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int syscon_store_standard_Laurential(int nc, int n, int k, char *p) {
  int a[3], b[nc], i, fail;
  double *c;
  a[0] = nc;
  a[1] = n;
  a[2] = k;
  for (i = 0; i < nc; i++)
    b[i] = (int)p[i];
  fail = _ada_use_c2phc(74, a, b, c);
  return fail;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<syscon_store_standard_Laurential>:
  endbr64
  push   %rbp
  mov    %esi,%r9d
  movslq %edi,%rsi
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    0xf(,%rsi,4),%rax
  mov    %rsp,%r10
  mov    %rax,%r8
  and    $0xfffffffffffff000,%rax
  sub    %rax,%r10
  and    $0xfffffffffffffff0,%r8
  cmp    %r10,%rsp
  je     L54
L17:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %r10,%rsp
  jne    L17
L54:
  and    $0xfff,%r8d
  sub    %r8,%rsp
  test   %r8,%r8
  jne    L50
L25:
  mov    %edi,-0x14(%rbp)
  mov    %rsp,%r8
  mov    %r9d,-0x10(%rbp)
  mov    %edx,-0xc(%rbp)
  test   %edi,%edi
  jle    L32
  xor    %eax,%eax
  nopl   0x0(%rax,%rax,1)
L62:
  movsbl (%rcx,%rax,1),%edx
  mov    %edx,(%r8,%rax,4)
  add    $0x1,%rax
  cmp    %rsi,%rax
  jne    L62
L32:
  xor    %ecx,%ecx
  mov    %r8,%rdx
  lea    -0x14(%rbp),%rsi
  mov    $0x4a,%edi
  call   <_ada_use_c2phc@plt>
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  jne    L57
  leave
  ret
  nopl   (%rax)
L50:
  orq    $0x0,-0x8(%rsp,%r8,1)
  jmp    L25
L57:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int syscon_store_standard_Laurential(int nc, int n, int k, char *p) {
  int a[3], b[nc], i, fail;
  double *c;
  a[0] = nc;
  a[1] = n;
  a[2] = k;
  for (i = 0; i < nc; i++)
    b[i] = (int)p[i];
  fail = _ada_use_c2phc(74, a, b, c);
  return fail;
}
```