ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0xa,%edi
  call   <queue_create@plt>
  mov    %rax,-0x8(%rbp)
  movl   $0x1,-0xc(%rbp)
  jmp    L26
L78:
  mov    -0xc(%rbp),%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <queue_enqueue@plt>
  addl   $0x1,-0xc(%rbp)
L26:
  cmpl   $0xb,-0xc(%rbp)
  jle    L78
  movl   $0x1,-0xc(%rbp)
  jmp    L17
L49:
  mov    -0xc(%rbp),%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <queue_lookqueue@plt>
  mov    %eax,%esi
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0xc(%rbp)
L17:
  cmpl   $0xb,-0xc(%rbp)
  jle    L49
  movl   $0x1,-0xc(%rbp)
  jmp    L23
L64:
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <queue_dequeue@plt>
  addl   $0x1,-0xc(%rbp)
L23:
  cmpl   $0x5,-0xc(%rbp)
  jle    L64
  movl   $0x1,-0xc(%rbp)
  jmp    L8
L40:
  mov    -0xc(%rbp),%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <queue_lookqueue@plt>
  mov    %eax,%esi
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0xc(%rbp)
L8:
  cmpl   $0xb,-0xc(%rbp)
  jle    L40
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D12
type_infered: string

tool: D12: %d


assistant: ```c
int main() {
  struct queue *q;
  int i, val;
  q = queue_create(10);
  for (i = 1; i <= 11; i++) {
    queue_enqueue(q, i);
  }
  for (i = 1; i <= 11; i++) {
    printf("%d\n", queue_lookqueue(q, i));
  }
  for (i = 1; i <= 5; i++) {
    queue_dequeue(q);
  }
  for (i = 1; i <= 11; i++) {
    printf("%d\n", queue_lookqueue(q, i));
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    $0xa,%edi
  call   <queue_create@plt>
  mov    %rax,%rbp
  mov    $0x1,%ebx
L27:
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <queue_enqueue@plt>
  add    $0x1,%ebx
  cmp    $0xc,%ebx
  jne    L27
  mov    $0x1,%ebx
  lea    D4(%rip),%r12
L2:
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <queue_lookqueue@plt>
  mov    %eax,%esi
  mov    %r12,%rdi
  call   <printf@plt>
  add    $0x1,%ebx
  cmp    $0xc,%ebx
  jne    L2
  mov    %rbp,%rdi
  call   <queue_dequeue@plt>
  mov    %rbp,%rdi
  call   <queue_dequeue@plt>
  mov    %rbp,%rdi
  call   <queue_dequeue@plt>
  mov    %rbp,%rdi
  call   <queue_dequeue@plt>
  mov    %rbp,%rdi
  call   <queue_dequeue@plt>
  mov    $0x1,%ebx
  lea    D4(%rip),%r12
L28:
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <queue_lookqueue@plt>
  mov    %eax,%esi
  mov    %r12,%rdi
  call   <printf@plt>
  add    $0x1,%ebx
  cmp    $0xc,%ebx
  jne    L28
  mov    $0x0,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D4
type_infered: string

tool: D4: %d


assistant: ```c
int main() {
  struct queue *q;
  int i, val;
  q = queue_create(10);
  for (i = 1; i <= 11; i++) {
    queue_enqueue(q, i);
  }
  for (i = 1; i <= 11; i++) {
    printf("%d\n", queue_lookqueue(q, i));
  }
  for (i = 1; i <= 5; i++) {
    queue_dequeue(q);
  }
  for (i = 1; i <= 11; i++) {
    printf("%d\n", queue_lookqueue(q, i));
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r12
  mov    $0xa,%edi
  push   %rbp
  push   %rbx
  mov    $0x1,%ebx
  call   <queue_create@plt>
  mov    %rax,%rbp
  nopw   0x0(%rax,%rax,1)
L2:
  mov    %ebx,%esi
  mov    %rbp,%rdi
  add    $0x1,%ebx
  call   <queue_enqueue@plt>
  cmp    $0xc,%ebx
  jne    L2
  mov    $0x1,%ebx
  lea    D7(%rip),%r12
  xchg   %ax,%ax
L25:
  mov    %ebx,%esi
  mov    %rbp,%rdi
  add    $0x1,%ebx
  call   <queue_lookqueue@plt>
  mov    %r12,%rdi
  mov    %eax,%esi
  call   <printf@plt>
  cmp    $0xc,%ebx
  jne    L25
  mov    %rbp,%rdi
  mov    $0x1,%ebx
  call   <queue_dequeue@plt>
  mov    %rbp,%rdi
  call   <queue_dequeue@plt>
  mov    %rbp,%rdi
  call   <queue_dequeue@plt>
  mov    %rbp,%rdi
  call   <queue_dequeue@plt>
  mov    %rbp,%rdi
  call   <queue_dequeue@plt>
  nopl   0x0(%rax)
L3:
  mov    %ebx,%esi
  mov    %rbp,%rdi
  add    $0x1,%ebx
  call   <queue_lookqueue@plt>
  mov    %r12,%rdi
  mov    %eax,%esi
  call   <printf@plt>
  cmp    $0xc,%ebx
  jne    L3
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: unpack_data
arguments: label: D7
type_guessed: string

tool: D7: %d


assistant: ```c
int main() {
  struct queue *q;
  int i, val;
  q = queue_create(10);
  for (i = 1; i <= 11; i++) {
    queue_enqueue(q, i);
  }
  for (i = 1; i <= 11; i++) {
    printf("%d\n", queue_lookqueue(q, i));
  }
  for (i = 1; i <= 5; i++) {
    queue_dequeue(q);
  }
  for (i = 1; i <= 11; i++) {
    printf("%d\n", queue_lookqueue(q, i));
  }
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  push   %r12
  mov    $0xa,%edi
  push   %rbp
  push   %rbx
  mov    $0x1,%ebx
  call   <queue_create@plt>
  mov    %rax,%rbp
  nopw   0x0(%rax,%rax,1)
L4:
  mov    %ebx,%esi
  mov    %rbp,%rdi
  add    $0x1,%ebx
  call   <queue_enqueue@plt>
  cmp    $0xc,%ebx
  jne    L4
  mov    $0x1,%ebx
  lea    D5(%rip),%r12
  xchg   %ax,%ax
L12:
  mov    %ebx,%esi
  mov    %rbp,%rdi
  add    $0x1,%ebx
  call   <queue_lookqueue@plt>
  mov    %r12,%rdi
  mov    %eax,%esi
  call   <printf@plt>
  cmp    $0xc,%ebx
  jne    L12
  mov    $0x5,%ebx
L39:
  mov    %rbp,%rdi
  call   <queue_dequeue@plt>
  sub    $0x1,%ebx
  jne    L39
  mov    $0x1,%ebx
  nopl   0x0(%rax,%rax,1)
L29:
  mov    %ebx,%esi
  mov    %rbp,%rdi
  add    $0x1,%ebx
  call   <queue_lookqueue@plt>
  mov    %r12,%rdi
  mov    %eax,%esi
  call   <printf@plt>
  cmp    $0xc,%ebx
  jne    L29
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: disassemble_data
arguments: label: D5
type_guessed: string

tool: D5: %d


assistant: ```c
int main() {
  struct queue *q;
  int i, val;
  q = queue_create(10);
  for (i = 1; i <= 11; i++) {
    queue_enqueue(q, i);
  }
  for (i = 1; i <= 11; i++) {
    printf("%d\n", queue_lookqueue(q, i));
  }
  for (i = 1; i <= 5; i++) {
    queue_dequeue(q);
  }
  for (i = 1; i <= 11; i++) {
    printf("%d\n", queue_lookqueue(q, i));
  }
  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
<is_prime>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  movl   $0x2,-0x4(%rbp)
  jmp    L4
L25:
  mov    -0x14(%rbp),%eax
  mov    $0x0,%edx
  divl   -0x4(%rbp)
  mov    %edx,%eax
  test   %eax,%eax
  jne    L16
  mov    $0x0,%eax
  jmp    L8
L16:
  addl   $0x1,-0x4(%rbp)
L4:
  mov    -0x4(%rbp),%eax
  imul   %eax,%eax
  mov    %eax,%edx
  mov    -0x14(%rbp),%eax
  cmp    %eax,%edx
  jbe    L25
  mov    $0x1,%eax
L8:
  pop    %rbp
  ret
```

assistant: ```c
unsigned int is_prime(int number) {
  for (unsigned int i = 2; (i * i) <= number; i++) {
    if (number % i == 0) {
      return 0;
    }
  }
  return 1;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<is_prime>:
  endbr64
  cmp    $0x3,%edi
  jbe    L13
  mov    %edi,%esi
  and    $0x1,%esi
  je     L30
  mov    $0x2,%ecx
L35:
  add    $0x1,%ecx
  mov    %ecx,%eax
  imul   %ecx,%eax
  cmp    %edi,%eax
  ja     L30
  mov    %edi,%eax
  mov    $0x0,%edx
  div    %ecx
  test   %edx,%edx
  jne    L35
  mov    %edx,%esi
L30:
  mov    %esi,%eax
  ret
L13:
  mov    $0x1,%esi
  jmp    L30
```

assistant: ```c
unsigned int is_prime(int number) {
  for (unsigned int i = 2; (i * i) <= number; i++) {
    if (number % i == 0) {
      return 0;
    }
  }
  return 1;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<is_prime>:
  endbr64
  cmp    $0x3,%edi
  jbe    L15
  mov    %edi,%r8d
  and    $0x1,%r8d
  je     L35
  mov    $0x2,%ecx
  jmp    L52
  nopl   0x0(%rax)
L33:
  mov    %edi,%eax
  xor    %edx,%edx
  div    %ecx
  test   %edx,%edx
  je     L30
L52:
  add    $0x1,%ecx
  mov    %ecx,%eax
  imul   %ecx,%eax
  cmp    %edi,%eax
  jbe    L33
L35:
  mov    %r8d,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L30:
  xor    %r8d,%r8d
  mov    %r8d,%eax
  ret
L15:
  mov    $0x1,%r8d
  jmp    L35
```

assistant: ```c
unsigned int is_prime(int number) {
  for (unsigned int i = 2; (i * i) <= number; i++) {
    if (number % i == 0) {
      return 0;
    }
  }
  return 1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<is_prime>:
  endbr64
  cmp    $0x3,%edi
  jbe    L51
  mov    %edi,%r8d
  and    $0x1,%r8d
  je     L16
  mov    $0x2,%ecx
  jmp    L24
  nopl   0x0(%rax)
L19:
  mov    %edi,%eax
  xor    %edx,%edx
  div    %ecx
  test   %edx,%edx
  je     L22
L24:
  add    $0x1,%ecx
  mov    %ecx,%eax
  imul   %ecx,%eax
  cmp    %edi,%eax
  jbe    L19
L16:
  mov    %r8d,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L22:
  xor    %r8d,%r8d
  mov    %r8d,%eax
  ret
L51:
  mov    $0x1,%r8d
  jmp    L16
```

assistant: ```c
unsigned int is_prime(int number) {
  for (unsigned int i = 2; (i * i) <= number; i++) {
    if (number % i == 0) {
      return 0;
    }
  }
  return 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
<rt_mutex_lock>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x4(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <pthread_mutex_lock@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
signed int rt_mutex_lock(union anon$40 *mp) {
  signed int status = 0;
  status = pthread_mutex_lock(mp);
  return status;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<rt_mutex_lock>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <pthread_mutex_lock@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
signed int rt_mutex_lock(union anon$40 *mp) {
  signed int status = 0;
  status = pthread_mutex_lock(mp);
  return status;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<rt_mutex_lock>:
  endbr64
  xor    %eax,%eax
  jmp    1050 <pthread_mutex_lock@plt>
```

assistant: ```c
signed int rt_mutex_lock(union anon$40 *mp) {
  signed int status = 0;
  status = pthread_mutex_lock(mp);
  return status;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<rt_mutex_lock>:
  endbr64
  xor    %eax,%eax
  jmp    1050 <pthread_mutex_lock@plt>
```

assistant: ```c
signed int rt_mutex_lock(union anon$40 *mp) {
  signed int status = 0;
  status = pthread_mutex_lock(mp);
  return status;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ROL64_44>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  rol    $0x2c,%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
uint64_t ROL64_44(uint64_t x) {
  __asm__("rolq $44, %0" : "=r"(x) : "0"(x));
  return x;
}
```
user: Translate the below assembly code into C programming language:

```asm
<ROL64_44>:
  endbr64
  mov    %rdi,%rax
  rol    $0x2c,%rax
  ret
```

assistant: ```c
uint64_t ROL64_44(uint64_t x) {
  __asm__("rolq $44, %0" : "=r"(x) : "0"(x));
  return x;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<ROL64_44>:
  endbr64
  mov    %rdi,%rax
  rol    $0x2c,%rax
  ret
```

assistant: ```c
uint64_t ROL64_44(uint64_t x) {
  __asm__("rolq $44, %0" : "=r"(x) : "0"(x));
  return x;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<foo>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,-0x4(%rbp)
  jmp    L5
L15:
  movl   $0x2,-0x1c(%rbp)
L5:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jg     L15
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int foo(int *p1, int p2) {
  int z = *p1;
  while (z > p2)
    p2 = 2;
  return z;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<foo>:
  endbr64
  mov    (%rdi),%eax
  mov    $0x2,%edx
  jmp    L10
L7:
  mov    %edx,%esi
L10:
  cmp    %eax,%esi
  jl     L7
  ret
```

assistant: ```c
int foo(int *p1, int p2) {
  int z = *p1;
  while (z > p2)
    p2 = 2;
  return z;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<foo>:
  endbr64
  mov    (%rdi),%eax
  jmp    L15
  nopl   0x0(%rax,%rax,1)
L17:
  mov    $0x2,%esi
L15:
  cmp    %eax,%esi
  jl     L17
  ret
```

assistant: ```c
int foo(int *p1, int p2) {
  int z = *p1;
  while (z > p2)
    p2 = 2;
  return z;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<foo>:
  endbr64
  mov    (%rdi),%eax
  jmp    L17
  nopl   0x0(%rax,%rax,1)
L12:
  mov    $0x2,%esi
L17:
  cmp    %eax,%esi
  jl     L12
  ret
```

assistant: ```c
int foo(int *p1, int p2) {
  int z = *p1;
  while (z > p2)
    p2 = 2;
  return z;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<lubksb>:
  endbr64
  test   %edx,%edx
  jle    L123
  push   %rbx
  mov    %rdi,%r8
  lea    0x1(%rdx),%ebx
  mov    $0x1,%edi
  mov    $0x0,%r11d
  pxor   %xmm2,%xmm2
  movapd %xmm2,%xmm3
  jmp    L99
L68:
  cmp    %edi,%r11d
  jge    L89
  mov    (%r8,%rdi,8),%r10
  movslq %r11d,%rax
L2:
  movsd  (%r10,%rax,8),%xmm0
  mulsd  (%rsi,%rax,8),%xmm0
  subsd  %xmm0,%xmm1
  add    $0x1,%rax
  cmp    %eax,%r9d
  jg     L2
  jmp    L89
L70:
  mov    %r9d,%r11d
L89:
  movsd  %xmm1,(%rsi,%rdi,8)
  add    $0x1,%rdi
  cmp    %rbx,%rdi
  je     L112
L99:
  mov    %edi,%r9d
  movslq (%rcx,%rdi,4),%rax
  lea    (%rsi,%rax,8),%rax
  movsd  (%rax),%xmm1
  movsd  (%rsi,%rdi,8),%xmm0
  movsd  %xmm0,(%rax)
  test   %r11d,%r11d
  jne    L68
  ucomisd %xmm2,%xmm1
  jp     L70
  ucomisd %xmm3,%xmm1
  je     L89
  jmp    L70
L112:
  lea    0x1(%rdx),%r9d
  movslq %edx,%rdi
  shl    $0x3,%rdi
L28:
  movsd  (%rsi,%rdi,1),%xmm1
  cmp    %r9d,%edx
  jl     L92
  mov    (%r8,%rdi,1),%rcx
  movslq %r9d,%rax
L1:
  movsd  (%rcx,%rax,8),%xmm0
  mulsd  (%rsi,%rax,8),%xmm0
  subsd  %xmm0,%xmm1
  add    $0x1,%rax
  cmp    %eax,%edx
  jge    L1
L92:
  mov    (%r8,%rdi,1),%rax
  divsd  (%rax,%rdi,1),%xmm1
  movsd  %xmm1,(%rsi,%rdi,1)
  sub    $0x1,%r9d
  sub    $0x8,%rdi
  cmp    $0x1,%r9d
  jne    L28
  pop    %rbx
  ret
L123:
  ret
```

assistant: ```c
void lubksb(double **A, double *b, int n, int *indx) {
  int i, ii = 0, ip, j;
  double sum;
  for (i = 1; i <= n; i++) {
    ip = indx[i];
    sum = b[ip];
    b[ip] = b[i];
    if (ii)
      for (j = ii; j <= i - 1; j++)
        sum -= A[i][j] * b[j];
    else if (sum)
      ii = i;
    b[i] = sum;
  }
  for (i = n; i >= 1; i--) {
    sum = b[i];
    for (j = i + 1; j <= n; j++)
      sum -= A[i][j] * b[j];
    b[i] = sum / A[i][i];
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<lubksb>:
  endbr64
  test   %edx,%edx
  jle    L64
  push   %rbp
  lea    0x1(%rdx),%ebp
  mov    %rdi,%r8
  pxor   %xmm2,%xmm2
  push   %rbx
  mov    %edx,%edi
  mov    %rcx,%rbx
  mov    %rbp,%r9
  mov    $0x1,%edx
  xor    %ecx,%ecx
  jmp    L115
  nopl   0x0(%rax)
L68:
  ucomisd %xmm2,%xmm1
  jp     L86
  je     L35
L86:
  mov    %r10d,%ecx
L35:
  movsd  %xmm1,(%rsi,%rdx,8)
  add    $0x1,%rdx
  cmp    %rdx,%rbp
  je     L1
L115:
  movslq (%rbx,%rdx,4),%rax
  movsd  (%rsi,%rdx,8),%xmm0
  mov    %edx,%r10d
  lea    (%rsi,%rax,8),%rax
  movsd  (%rax),%xmm1
  movsd  %xmm0,(%rax)
  test   %ecx,%ecx
  je     L68
  cmp    %edx,%ecx
  jge    L35
  mov    (%r8,%rdx,8),%r11
  movslq %ecx,%rax
L93:
  movsd  (%r11,%rax,8),%xmm0
  mulsd  (%rsi,%rax,8),%xmm0
  add    $0x1,%rax
  subsd  %xmm0,%xmm1
  cmp    %eax,%r10d
  jg     L93
  movsd  %xmm1,(%rsi,%rdx,8)
  add    $0x1,%rdx
  cmp    %rdx,%rbp
  jne    L115
L1:
  movslq %edi,%rcx
  shl    $0x3,%rcx
  nopl   (%rax)
L14:
  movsd  (%rsi,%rcx,1),%xmm1
  mov    (%r8,%rcx,1),%rdx
  movslq %r9d,%rax
  cmp    %r9d,%edi
  jl     L100
  nopl   0x0(%rax)
L103:
  movsd  (%rdx,%rax,8),%xmm0
  mulsd  (%rsi,%rax,8),%xmm0
  add    $0x1,%rax
  subsd  %xmm0,%xmm1
  cmp    %eax,%edi
  jge    L103
L100:
  sub    $0x1,%r9d
  divsd  (%rdx,%rcx,1),%xmm1
  movsd  %xmm1,(%rsi,%rcx,1)
  sub    $0x8,%rcx
  cmp    $0x1,%r9d
  jne    L14
  pop    %rbx
  pop    %rbp
  ret
L64:
  ret
```

assistant: ```c
void lubksb(double **A, double *b, int n, int *indx) {
  int i, ii = 0, ip, j;
  double sum;
  for (i = 1; i <= n; i++) {
    ip = indx[i];
    sum = b[ip];
    b[ip] = b[i];
    if (ii)
      for (j = ii; j <= i - 1; j++)
        sum -= A[i][j] * b[j];
    else if (sum)
      ii = i;
    b[i] = sum;
  }
  for (i = n; i >= 1; i--) {
    sum = b[i];
    for (j = i + 1; j <= n; j++)
      sum -= A[i][j] * b[j];
    b[i] = sum / A[i][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
<sw_sync_timeline_is_valid>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  cmpl   $0xffffffff,-0x14(%rbp)
  jne    L12
  mov    $0x0,%eax
  jmp    L11
L12:
  mov    D7(%rip),%rax
  mov    (%rax),%ecx
  mov    -0x14(%rbp),%eax
  mov    $0x0,%edx
  mov    %ecx,%esi
  mov    %eax,%edi
  call   <fcntl@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  not    %eax
  shr    $0x1f,%eax
  movzbl %al,%eax
L11:
  leave
  ret
```

assistant: ```c
int sw_sync_timeline_is_valid(int fd) {
  int status;
  if (fd == -1)
    return 0;
  status = fcntl(fd, F_GETFD, 0);
  return (status >= 0);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<sw_sync_timeline_is_valid>:
  endbr64
  mov    $0x0,%eax
  cmp    $0xffffffff,%edi
  je     L9
  sub    $0x8,%rsp
  mov    $0x0,%edx
  mov    D0(%rip),%rax
  mov    (%rax),%esi
  call   <fcntl@plt>
  not    %eax
  shr    $0x1f,%eax
  add    $0x8,%rsp
  ret
L9:
  ret
```

assistant: ```c
int sw_sync_timeline_is_valid(int fd) {
  int status;
  if (fd == -1)
    return 0;
  status = fcntl(fd, F_GETFD, 0);
  return (status >= 0);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<sw_sync_timeline_is_valid>:
  endbr64
  cmp    $0xffffffff,%edi
  je     L4
  sub    $0x8,%rsp
  mov    D3(%rip),%rax
  xor    %edx,%edx
  mov    (%rax),%esi
  call   <fcntl@plt>
  add    $0x8,%rsp
  not    %eax
  shr    $0x1f,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L4:
  xor    %eax,%eax
  ret
```

assistant: ```c
int sw_sync_timeline_is_valid(int fd) {
  int status;
  if (fd == -1)
    return 0;
  status = fcntl(fd, F_GETFD, 0);
  return (status >= 0);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<sw_sync_timeline_is_valid>:
  endbr64
  cmp    $0xffffffff,%edi
  je     L8
  sub    $0x8,%rsp
  mov    D5(%rip),%rax
  xor    %edx,%edx
  mov    (%rax),%esi
  call   <fcntl@plt>
  add    $0x8,%rsp
  not    %eax
  shr    $0x1f,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L8:
  xor    %eax,%eax
  ret
```

assistant: ```c
int sw_sync_timeline_is_valid(int fd) {
  int status;
  if (fd == -1)
    return 0;
  status = fcntl(fd, F_GETFD, 0);
  return (status >= 0);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<all_tests>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  mov    -0x8(%rbp),%edx
  mov    -0x4(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <test1@plt>
  mov    -0x8(%rbp),%edx
  mov    -0x4(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <test2@plt>
  mov    -0x8(%rbp),%edx
  mov    -0x4(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <test3@plt>
  mov    -0x8(%rbp),%edx
  mov    -0x4(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <test4@plt>
  mov    -0x8(%rbp),%edx
  mov    -0x4(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <test5@plt>
  mov    -0x8(%rbp),%edx
  mov    -0x4(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <test6@plt>
  nop
  leave
  ret
```

assistant: ```c
void all_tests(int x, int y) {
  test1(x, y);
  test2(x, y);
  test3(x, y);
  test4(x, y);
  test5(x, y);
  test6(x, y);
}
```
user: Translate the below assembly code into C programming language:

```asm
<all_tests>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebx
  mov    %esi,%ebp
  call   <test1@plt>
  mov    %ebp,%esi
  mov    %ebx,%edi
  call   <test2@plt>
  mov    %ebp,%esi
  mov    %ebx,%edi
  call   <test3@plt>
  mov    %ebp,%esi
  mov    %ebx,%edi
  call   <test4@plt>
  mov    %ebp,%esi
  mov    %ebx,%edi
  call   <test5@plt>
  mov    %ebp,%esi
  mov    %ebx,%edi
  call   <test6@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void all_tests(int x, int y) {
  test1(x, y);
  test2(x, y);
  test3(x, y);
  test4(x, y);
  test5(x, y);
  test6(x, y);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<all_tests>:
  endbr64
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  mov    %edi,%ebp
  sub    $0x8,%rsp
  call   <test1@plt>
  mov    %r12d,%esi
  mov    %ebp,%edi
  call   <test2@plt>
  mov    %r12d,%esi
  mov    %ebp,%edi
  call   <test3@plt>
  mov    %r12d,%esi
  mov    %ebp,%edi
  call   <test4@plt>
  mov    %r12d,%esi
  mov    %ebp,%edi
  call   <test5@plt>
  add    $0x8,%rsp
  mov    %r12d,%esi
  mov    %ebp,%edi
  pop    %rbp
  pop    %r12
  jmp    10b0 <test6@plt>
```

assistant: ```c
void all_tests(int x, int y) {
  test1(x, y);
  test2(x, y);
  test3(x, y);
  test4(x, y);
  test5(x, y);
  test6(x, y);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<all_tests>:
  endbr64
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  mov    %edi,%ebp
  sub    $0x8,%rsp
  call   <test1@plt>
  mov    %r12d,%esi
  mov    %ebp,%edi
  call   <test2@plt>
  mov    %r12d,%esi
  mov    %ebp,%edi
  call   <test3@plt>
  mov    %r12d,%esi
  mov    %ebp,%edi
  call   <test4@plt>
  mov    %r12d,%esi
  mov    %ebp,%edi
  call   <test5@plt>
  add    $0x8,%rsp
  mov    %r12d,%esi
  mov    %ebp,%edi
  pop    %rbp
  pop    %r12
  jmp    10b0 <test6@plt>
```

assistant: ```c
void all_tests(int x, int y) {
  test1(x, y);
  test2(x, y);
  test3(x, y);
  test4(x, y);
  test5(x, y);
  test6(x, y);
}
```