ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  cmpl   $0x0,-0x4(%rbp)
  js     L5
  mov    $0x0,%eax
  jmp    L6
L2:
  addl   $0x1,-0x4(%rbp)
L5:
  cmpl   $0x9,-0x4(%rbp)
  jle    L2
  cmpl   $0xa,-0x4(%rbp)
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  mov    $0x0,%eax
L6:
  leave
  ret
```

assistant: ```c
int main() {
  int x;
  if (!(x < 0))
    return 0;
  while (x < 10) {
    x = x + 1;
  }
  assert(x == 10);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  mov    $0x0,%eax
  ret
```

assistant: ```c
int main() {
  int x;
  if (!(x < 0))
    return 0;
  while (x < 10) {
    x = x + 1;
  }
  assert(x == 10);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int main() {
  int x;
  if (!(x < 0))
    return 0;
  while (x < 10) {
    x = x + 1;
  }
  assert(x == 10);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int main() {
  int x;
  if (!(x < 0))
    return 0;
  while (x < 10) {
    x = x + 1;
  }
  assert(x == 10);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<full_mod31_32>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  cltq
  mov    $0x84210842,%edx
  imul   %rdx,%rax
  shr    $0x24,%rax
  mov    %eax,-0x8(%rbp)
  mov    -0x14(%rbp),%edx
  mov    -0x8(%rbp),%eax
  add    %eax,%edx
  mov    -0x8(%rbp),%eax
  shl    $0x5,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
uint8_t full_mod31_32(uint32_t a) {
  uint32_t t = ((0x84210842ULL * a) >> 36);
  uint8_t r = a + t - (t << 5);
  return r;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<full_mod31_32>:
  endbr64
  movslq %edi,%rdx
  mov    $0x84210842,%eax
  imul   %rax,%rdx
  shr    $0x24,%rdx
  lea    (%rdi,%rdx,1),%eax
  shl    $0x5,%edx
  sub    %edx,%eax
  ret
```

assistant: ```c
uint8_t full_mod31_32(uint32_t a) {
  uint32_t t = ((0x84210842ULL * a) >> 36);
  uint8_t r = a + t - (t << 5);
  return r;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<full_mod31_32>:
  endbr64
  movslq %edi,%rdx
  mov    $0x84210842,%eax
  imul   %rax,%rdx
  shr    $0x24,%rdx
  lea    (%rdi,%rdx,1),%eax
  shl    $0x5,%edx
  sub    %edx,%eax
  ret
```

assistant: ```c
uint8_t full_mod31_32(uint32_t a) {
  uint32_t t = ((0x84210842ULL * a) >> 36);
  uint8_t r = a + t - (t << 5);
  return r;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<full_mod31_32>:
  endbr64
  movslq %edi,%rdx
  mov    $0x84210842,%eax
  imul   %rax,%rdx
  shr    $0x24,%rdx
  lea    (%rdi,%rdx,1),%eax
  shl    $0x5,%edx
  sub    %edx,%eax
  ret
```

assistant: ```c
uint8_t full_mod31_32(uint32_t a) {
  uint32_t t = ((0x84210842ULL * a) >> 36);
  uint8_t r = a + t - (t << 5);
  return r;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<square_matrix_add>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %ecx,-0x2c(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L3
L11:
  movl   $0x0,-0x4(%rbp)
  jmp    L22
L16:
  mov    -0x8(%rbp),%eax
  imul   -0x2c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x4(%rbp),%eax
  cltq
  add    %rdx,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%ecx
  mov    -0x8(%rbp),%eax
  imul   -0x2c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x4(%rbp),%eax
  cltq
  add    %rdx,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x8(%rbp),%eax
  imul   -0x2c(%rbp),%eax
  movslq %eax,%rsi
  mov    -0x4(%rbp),%eax
  cltq
  add    %rsi,%rax
  lea    0x0(,%rax,4),%rsi
  mov    -0x28(%rbp),%rax
  add    %rsi,%rax
  add    %ecx,%edx
  mov    %edx,(%rax)
  addl   $0x1,-0x4(%rbp)
L22:
  mov    -0x4(%rbp),%eax
  cmp    -0x2c(%rbp),%eax
  jl     L16
  addl   $0x1,-0x8(%rbp)
L3:
  mov    -0x8(%rbp),%eax
  cmp    -0x2c(%rbp),%eax
  jl     L11
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void square_matrix_add(int *a1, int *b1, int *c, int num) {
  int i, j;
  for (i = 0; i < num; i++)
    for (j = 0; j < num; j++)
      *(c + i * num + j) = *(a1 + i * num + j) + *(b1 + i * num + j);
}
```
user: What is the c source code of the assembly code below:

```asm
<square_matrix_add>:
  endbr64
  test   %ecx,%ecx
  jle    L20
  push   %rbp
  push   %rbx
  mov    %rdi,%r8
  mov    %rsi,%r9
  movslq %ecx,%rbx
  lea    0x0(,%rbx,4),%rbp
  mov    %ecx,%edi
  shl    $0x2,%rdi
  mov    $0x0,%r11d
  mov    $0x0,%r10d
L17:
  lea    0x0(,%r11,4),%rax
L24:
  mov    (%r9,%rax,1),%esi
  add    (%r8,%rax,1),%esi
  mov    %esi,(%rdx,%rax,1)
  add    $0x4,%rax
  cmp    %rdi,%rax
  jne    L24
  add    $0x1,%r10d
  add    %rbx,%r11
  add    %rbp,%rdi
  cmp    %r10d,%ecx
  jne    L17
  pop    %rbx
  pop    %rbp
  ret
L20:
  ret
```

assistant: ```c
void square_matrix_add(int *a1, int *b1, int *c, int num) {
  int i, j;
  for (i = 0; i < num; i++)
    for (j = 0; j < num; j++)
      *(c + i * num + j) = *(a1 + i * num + j) + *(b1 + i * num + j);
}
```
user: Translate the below assembly code into C programming language:

```asm
<square_matrix_add>:
  endbr64
  test   %ecx,%ecx
  jle    L4
  push   %rbp
  movslq %ecx,%rbp
  mov    %rdi,%r8
  mov    %rsi,%r9
  push   %rbx
  lea    0x0(,%rbp,4),%rbx
  xor    %r11d,%r11d
  xor    %r10d,%r10d
  mov    %rbx,%rdi
  nopl   0x0(%rax)
L18:
  lea    0x0(,%r11,4),%rax
L0:
  mov    (%r9,%rax,1),%esi
  add    (%r8,%rax,1),%esi
  mov    %esi,(%rdx,%rax,1)
  add    $0x4,%rax
  cmp    %rax,%rdi
  jne    L0
  add    $0x1,%r10d
  add    %rbp,%r11
  add    %rbx,%rdi
  cmp    %r10d,%ecx
  jne    L18
  pop    %rbx
  pop    %rbp
  ret
L4:
  ret
```

assistant: ```c
void square_matrix_add(int *a1, int *b1, int *c, int num) {
  int i, j;
  for (i = 0; i < num; i++)
    for (j = 0; j < num; j++)
      *(c + i * num + j) = *(a1 + i * num + j) + *(b1 + i * num + j);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<square_matrix_add>:
  endbr64
  test   %ecx,%ecx
  jle    L10
  push   %r15
  movslq %ecx,%rax
  mov    %ecx,%r15d
  xor    %r8d,%r8d
  push   %r14
  lea    0x0(,%rax,4),%r14
  and    $0xfffffffc,%r15d
  push   %r13
  xor    %r13d,%r13d
  push   %r12
  mov    %ecx,%r12d
  push   %rbp
  mov    $0x4,%ebp
  push   %rbx
  lea    -0x1(%rcx),%ebx
  mov    %rax,-0x20(%rsp)
  mov    %ecx,%eax
  shr    $0x2,%eax
  mov    %ebx,-0x18(%rsp)
  lea    -0x1(%rax),%r11d
  mov    %ebx,%eax
  xor    %ebx,%ebx
  shl    $0x2,%rax
  add    $0x1,%r11
  mov    %rax,-0x10(%rsp)
  mov    %ecx,%eax
  shl    $0x4,%r11
  and    $0x3,%eax
  mov    %eax,-0x14(%rsp)
  nopl   0x0(%rax)
L84:
  lea    (%rdx,%rbx,1),%rcx
  lea    (%rdi,%rbp,1),%r9
  mov    %rcx,%rax
  lea    0x0(%rbp,%rsi,1),%r10
  sub    %r9,%rax
  mov    %rcx,%r9
  cmp    $0x8,%rax
  seta   %al
  sub    %r10,%r9
  cmp    $0x8,%r9
  seta   %r9b
  test   %r9b,%al
  je     L59
  cmp    $0x1,%r12d
  je     L59
  cmpl   $0x2,-0x18(%rsp)
  jbe    L100
  lea    (%rbx,%rdi,1),%r10
  lea    (%rbx,%rsi,1),%r9
  xor    %eax,%eax
L52:
  movdqu (%r10,%rax,1),%xmm0
  movdqu (%r9,%rax,1),%xmm2
  paddd  %xmm2,%xmm0
  movups %xmm0,(%rcx,%rax,1)
  add    $0x10,%rax
  cmp    %r11,%rax
  jne    L52
  cmp    %r15d,%r12d
  je     L120
  mov    -0x14(%rsp),%eax
  mov    %eax,%r9d
  cmp    $0x1,%eax
  je     L29
  mov    %r15d,%ecx
  mov    %r15d,%eax
L40:
  add    %r8,%rcx
  movq   (%rdi,%rcx,4),%xmm0
  movq   (%rsi,%rcx,4),%xmm1
  paddd  %xmm1,%xmm0
  movq   %xmm0,(%rdx,%rcx,4)
  mov    %r9d,%ecx
  and    $0xfffffffe,%ecx
  add    %ecx,%eax
  cmp    %ecx,%r9d
  je     L120
L105:
  cltq
  add    %r8,%rax
  mov    (%rsi,%rax,4),%ecx
  add    (%rdi,%rax,4),%ecx
  mov    %ecx,(%rdx,%rax,4)
L120:
  add    $0x1,%r13d
  add    %r14,%rbx
  add    -0x20(%rsp),%r8
  add    %r14,%rbp
  cmp    %r13d,%r12d
  jne    L84
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L59:
  mov    -0x10(%rsp),%rax
  lea    (%rax,%rbp,1),%r9
  mov    %rbx,%rax
  nopl   0x0(%rax)
L113:
  mov    (%rsi,%rax,1),%ecx
  add    (%rdi,%rax,1),%ecx
  mov    %ecx,(%rdx,%rax,1)
  add    $0x4,%rax
  cmp    %rax,%r9
  jne    L113
  jmp    L120
L100:
  mov    %r12d,%r9d
  xor    %ecx,%ecx
  xor    %eax,%eax
  jmp    L40
L29:
  mov    %r15d,%eax
  jmp    L105
L10:
  ret
```

assistant: ```c
void square_matrix_add(int *a1, int *b1, int *c, int num) {
  int i, j;
  for (i = 0; i < num; i++)
    for (j = 0; j < num; j++)
      *(c + i * num + j) = *(a1 + i * num + j) + *(b1 + i * num + j);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<majorityElement>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,-0xc(%rbp)
  movl   $0x0,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L54
L45:
  cmpl   $0x0,-0x8(%rbp)
  jne    L9
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0xc(%rbp)
  movl   $0x1,-0x8(%rbp)
  jmp    L19
L9:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    %eax,-0xc(%rbp)
  jne    L50
  addl   $0x1,-0x8(%rbp)
  jmp    L19
L50:
  subl   $0x1,-0x8(%rbp)
L19:
  addl   $0x1,-0x4(%rbp)
L54:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L45
  mov    -0xc(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int majorityElement(int *nums, int numsSize) {
  int maj = nums[0], count = 0, i;
  for (i = 0; i < numsSize; ++i) {
    if (!count) {
      maj = nums[i];
      count = 1;
    } else if (nums[i] == maj)
      ++count;
    else
      --count;
  }
  return maj;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<majorityElement>:
  endbr64
  mov    (%rdi),%ecx
  test   %esi,%esi
  jle    L19
  mov    %rdi,%rdx
  lea    -0x1(%rsi),%eax
  lea    0x4(%rdi,%rax,4),%rsi
  mov    $0x0,%eax
  mov    $0x1,%r8d
  jmp    L49
L12:
  lea    0x1(%rax),%edi
  sub    $0x1,%eax
  cmp    %ecx,(%rdx)
  cmove  %edi,%eax
L14:
  add    $0x4,%rdx
  cmp    %rsi,%rdx
  je     L19
L49:
  test   %eax,%eax
  jne    L12
  mov    (%rdx),%ecx
  mov    %r8d,%eax
  jmp    L14
L19:
  mov    %ecx,%eax
  ret
```

assistant: ```c
int majorityElement(int *nums, int numsSize) {
  int maj = nums[0], count = 0, i;
  for (i = 0; i < numsSize; ++i) {
    if (!count) {
      maj = nums[i];
      count = 1;
    } else if (nums[i] == maj)
      ++count;
    else
      --count;
  }
  return maj;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<majorityElement>:
  endbr64
  mov    (%rdi),%r8d
  test   %esi,%esi
  jle    L33
  mov    %r8d,%r9d
  xor    %edx,%edx
  xor    %eax,%eax
  jmp    L2
  nopl   0x0(%rax)
L36:
  cmp    %ecx,%esi
  jle    L35
  movslq %ecx,%rax
  mov    %r9d,%r8d
  mov    (%rdi,%rax,4),%r10d
  mov    $0x1,%eax
L4:
  lea    0x1(%rax),%edx
  sub    $0x1,%eax
  cmp    %r10d,%r8d
  cmove  %edx,%eax
  lea    0x1(%rcx),%edx
  cmp    %edx,%esi
  jle    L33
  movslq %edx,%rcx
  mov    (%rdi,%rcx,4),%r9d
L2:
  lea    0x1(%rdx),%ecx
  test   %eax,%eax
  je     L36
  mov    %r9d,%r10d
  mov    %edx,%ecx
  jmp    L4
  nopl   0x0(%rax,%rax,1)
L35:
  mov    %r9d,%r8d
L33:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int majorityElement(int *nums, int numsSize) {
  int maj = nums[0], count = 0, i;
  for (i = 0; i < numsSize; ++i) {
    if (!count) {
      maj = nums[i];
      count = 1;
    } else if (nums[i] == maj)
      ++count;
    else
      --count;
  }
  return maj;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<majorityElement>:
  endbr64
  mov    (%rdi),%r8d
  test   %esi,%esi
  jle    L47
  mov    %r8d,%r9d
  xor    %edx,%edx
  xor    %eax,%eax
  jmp    L9
  nopl   0x0(%rax)
L53:
  cmp    %ecx,%esi
  jle    L39
  movslq %ecx,%rax
  mov    %r9d,%r8d
  mov    (%rdi,%rax,4),%r10d
  mov    $0x1,%eax
L57:
  lea    0x1(%rax),%edx
  sub    $0x1,%eax
  cmp    %r10d,%r8d
  cmove  %edx,%eax
  lea    0x1(%rcx),%edx
  cmp    %edx,%esi
  jle    L47
  movslq %edx,%rcx
  mov    (%rdi,%rcx,4),%r9d
L9:
  lea    0x1(%rdx),%ecx
  test   %eax,%eax
  je     L53
  mov    %r9d,%r10d
  mov    %edx,%ecx
  jmp    L57
  nopl   0x0(%rax,%rax,1)
L39:
  mov    %r9d,%r8d
L47:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int majorityElement(int *nums, int numsSize) {
  int maj = nums[0], count = 0, i;
  for (i = 0; i < numsSize; ++i) {
    if (!count) {
      maj = nums[i];
      count = 1;
    } else if (nums[i] == maj)
      ++count;
    else
      --count;
  }
  return maj;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  cmpl   $0x7,-0x4(%rbp)
  ja     L20
  mov    -0x4(%rbp),%eax
  lea    0x0(,%rax,4),%rdx
  lea    D17(%rip),%rax
  mov    (%rdx,%rax,1),%eax
  cltq
  lea    D17(%rip),%rdx
  add    %rdx,%rax
  notrack jmp *%rax
  mov    $0x0,%eax
  call   <stop_normal@plt>
  jmp    L20
  mov    $0x0,%eax
  call   <stop_symbolic@plt>
  jmp    L20
  mov    $0x0,%eax
  call   <stop_error@plt>
  jmp    L20
  mov    $0x0,%eax
  call   <stop_syscall@plt>
  jmp    L20
  mov    $0x0,%eax
  call   <stop_execnone@plt>
  jmp    L20
  mov    $0x0,%eax
  call   <stop_nostart@plt>
  jmp    L20
  mov    $0x0,%eax
  call   <stop_segfault@plt>
  nop
L20:
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  switch (argc) {
  case 1:
    stop_normal();
    break;
  case 2:
    stop_symbolic();
    break;
  case 3:
    stop_error();
    break;
  case 4:
    stop_syscall();
    break;
  case 5:
    stop_execnone();
    break;
  case 6:
    stop_nostart();
    break;
  case 7:
    stop_segfault();
    break;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  cmp    $0x7,%edi
  ja     L1
  sub    $0x8,%rsp
  mov    %edi,%edi
  lea    D9(%rip),%rdx
  movslq (%rdx,%rdi,4),%rax
  add    %rdx,%rax
  notrack jmp *%rax
  mov    $0x0,%eax
  call   <stop_normal@plt>
L43:
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
  mov    $0x0,%eax
  call   <stop_symbolic@plt>
  jmp    L43
  mov    $0x0,%eax
  call   <stop_error@plt>
  jmp    L43
  mov    $0x0,%eax
  call   <stop_syscall@plt>
  jmp    L43
  mov    $0x0,%eax
  call   <stop_execnone@plt>
  jmp    L43
  mov    $0x0,%eax
  call   <stop_nostart@plt>
  jmp    L43
  mov    $0x0,%eax
  call   <stop_segfault@plt>
  jmp    L43
L1:
  mov    $0x0,%eax
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  switch (argc) {
  case 1:
    stop_normal();
    break;
  case 2:
    stop_symbolic();
    break;
  case 3:
    stop_error();
    break;
  case 4:
    stop_syscall();
    break;
  case 5:
    stop_execnone();
    break;
  case 6:
    stop_nostart();
    break;
  case 7:
    stop_segfault();
    break;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  cmp    $0x7,%edi
  ja     L9
  lea    D3(%rip),%rdx
  mov    %edi,%edi
  sub    $0x8,%rsp
  movslq (%rdx,%rdi,4),%rax
  add    %rdx,%rax
  notrack jmp *%rax
  xor    %eax,%eax
  call   <stop_nostart@plt>
L32:
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
  xor    %eax,%eax
  call   <stop_segfault@plt>
  jmp    L32
  xor    %eax,%eax
  call   <stop_normal@plt>
  jmp    L32
  xor    %eax,%eax
  call   <stop_symbolic@plt>
  jmp    L32
  xor    %eax,%eax
  call   <stop_error@plt>
  jmp    L32
  xor    %eax,%eax
  call   <stop_syscall@plt>
  jmp    L32
  xor    %eax,%eax
  call   <stop_execnone@plt>
  jmp    L32
L9:
  xor    %eax,%eax
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  switch (argc) {
  case 1:
    stop_normal();
    break;
  case 2:
    stop_symbolic();
    break;
  case 3:
    stop_error();
    break;
  case 4:
    stop_syscall();
    break;
  case 5:
    stop_execnone();
    break;
  case 6:
    stop_nostart();
    break;
  case 7:
    stop_segfault();
    break;
  }
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  cmp    $0x7,%edi
  ja     L50
  lea    D9(%rip),%rdx
  mov    %edi,%edi
  sub    $0x8,%rsp
  movslq (%rdx,%rdi,4),%rax
  add    %rdx,%rax
  notrack jmp *%rax
  xor    %eax,%eax
  call   <stop_nostart@plt>
L15:
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
  xor    %eax,%eax
  call   <stop_segfault@plt>
  jmp    L15
  xor    %eax,%eax
  call   <stop_normal@plt>
  jmp    L15
  xor    %eax,%eax
  call   <stop_symbolic@plt>
  jmp    L15
  xor    %eax,%eax
  call   <stop_error@plt>
  jmp    L15
  xor    %eax,%eax
  call   <stop_syscall@plt>
  jmp    L15
  xor    %eax,%eax
  call   <stop_execnone@plt>
  jmp    L15
L50:
  xor    %eax,%eax
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  switch (argc) {
  case 1:
    stop_normal();
    break;
  case 2:
    stop_symbolic();
    break;
  case 3:
    stop_error();
    break;
  case 4:
    stop_syscall();
    break;
  case 5:
    stop_execnone();
    break;
  case 6:
    stop_nostart();
    break;
  case 7:
    stop_segfault();
    break;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<_sbrk>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    D53(%rip),%rax
  test   %rax,%rax
  jne    L20
  mov    D9(%rip),%rax
  mov    %rax,D11(%rip)
L20:
  mov    D11(%rip),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x14(%rbp),%eax
  add    $0x3,%eax
  and    $0xfffffffc,%eax
  mov    %eax,-0x14(%rbp)
  mov    D11(%rip),%rdx
  mov    -0x14(%rbp),%eax
  cltq
  add    %rax,%rdx
  mov    D11(%rip),%rax
  cmp    %rax,%rdx
  jbe    L1
  mov    D31(%rip),%rax
  mov    (%rax),%edx
  mov    D51(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xffffffff,%eax
  jmp    L7
L1:
  mov    D11(%rip),%rdx
  mov    -0x14(%rbp),%eax
  cltq
  add    %rdx,%rax
  mov    %rax,D87(%rip)
  mov    -0x8(%rbp),%rax
L7:
  pop    %rbp
  ret
```

assistant: ```c
caddr_t _sbrk(int incr) {
  extern char _Heap_Begin;
  extern char _Heap_Limit;
  static char *current_heap_end;
  char *current_block_address;
  if (current_heap_end == 0) {
    current_heap_end = &_Heap_Begin;
  }
  current_block_address = current_heap_end;
  incr = (incr + 3) & (~3);
  if (current_heap_end + incr > &_Heap_Limit) {
    errno = ENOMEM;
    return (caddr_t)-1;
  }
  current_heap_end += incr;
  return (caddr_t)current_block_address;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<_sbrk>:
  endbr64
  mov    %edi,%edx
  cmpq   $0x0,D16(%rip)
  je     L11
L16:
  mov    D16(%rip),%rax
  add    $0x3,%edx
  and    $0xfffffffc,%edx
  movslq %edx,%rdx
  add    %rax,%rdx
  cmp    D45(%rip),%rdx
  ja     L7
  mov    %rdx,D58(%rip)
  ret
L11:
  mov    D58(%rip),%rax
  mov    %rax,D4(%rip)
  jmp    L16
L7:
  mov    D4(%rip),%rax
  mov    (%rax),%edx
  mov    D32(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
caddr_t _sbrk(int incr) {
  extern char _Heap_Begin;
  extern char _Heap_Limit;
  static char *current_heap_end;
  char *current_block_address;
  if (current_heap_end == 0) {
    current_heap_end = &_Heap_Begin;
  }
  current_block_address = current_heap_end;
  incr = (incr + 3) & (~3);
  if (current_heap_end + incr > &_Heap_Limit) {
    errno = ENOMEM;
    return (caddr_t)-1;
  }
  current_heap_end += incr;
  return (caddr_t)current_block_address;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<_sbrk>:
  endbr64
  mov    D39(%rip),%rax
  mov    %edi,%edx
  test   %rax,%rax
  je     L16
L4:
  add    $0x3,%edx
  and    $0xfffffffc,%edx
  movslq %edx,%rdx
  add    %rax,%rdx
  cmp    D61(%rip),%rdx
  ja     L6
  mov    %rdx,D1(%rip)
  ret
  nop
L16:
  mov    D1(%rip),%rax
  mov    %rax,D11(%rip)
  jmp    L4
L6:
  mov    D11(%rip),%rax
  mov    (%rax),%edx
  mov    D67(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
caddr_t _sbrk(int incr) {
  extern char _Heap_Begin;
  extern char _Heap_Limit;
  static char *current_heap_end;
  char *current_block_address;
  if (current_heap_end == 0) {
    current_heap_end = &_Heap_Begin;
  }
  current_block_address = current_heap_end;
  incr = (incr + 3) & (~3);
  if (current_heap_end + incr > &_Heap_Limit) {
    errno = ENOMEM;
    return (caddr_t)-1;
  }
  current_heap_end += incr;
  return (caddr_t)current_block_address;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<_sbrk>:
  endbr64
  mov    D25(%rip),%rax
  mov    %edi,%edx
  test   %rax,%rax
  je     L1
L4:
  add    $0x3,%edx
  and    $0xfffffffc,%edx
  movslq %edx,%rdx
  add    %rax,%rdx
  cmp    D52(%rip),%rdx
  ja     L14
  mov    %rdx,D66(%rip)
  ret
  nop
L1:
  mov    D66(%rip),%rax
  mov    %rax,D31(%rip)
  jmp    L4
L14:
  mov    D31(%rip),%rax
  mov    (%rax),%edx
  mov    D32(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
caddr_t _sbrk(int incr) {
  extern char _Heap_Begin;
  extern char _Heap_Limit;
  static char *current_heap_end;
  char *current_block_address;
  if (current_heap_end == 0) {
    current_heap_end = &_Heap_Begin;
  }
  current_block_address = current_heap_end;
  incr = (incr + 3) & (~3);
  if (current_heap_end + incr > &_Heap_Limit) {
    errno = ENOMEM;
    return (caddr_t)-1;
  }
  current_heap_end += incr;
  return (caddr_t)current_block_address;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<fatal>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xd0,%rsp
  mov    %rdi,-0xc8(%rbp)
  mov    %rsi,-0xa8(%rbp)
  mov    %rdx,-0xa0(%rbp)
  mov    %rcx,-0x98(%rbp)
  mov    %r8,-0x90(%rbp)
  mov    %r9,-0x88(%rbp)
  test   %al,%al
  je     L8
  movaps %xmm0,-0x80(%rbp)
  movaps %xmm1,-0x70(%rbp)
  movaps %xmm2,-0x60(%rbp)
  movaps %xmm3,-0x50(%rbp)
  movaps %xmm4,-0x40(%rbp)
  movaps %xmm5,-0x30(%rbp)
  movaps %xmm6,-0x20(%rbp)
  movaps %xmm7,-0x10(%rbp)
L8:
  mov    -0xc8(%rbp),%rdx
  mov    -0xb4(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <va_start@plt>
  mov    -0xb4(%rbp),%edx
  mov    -0xc8(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <vprintf@plt>
  mov    -0xb4(%rbp),%eax
  mov    %eax,%edi
  call   <va_end@plt>
  mov    $0xffffffff,%edi
  call   <exit@plt>
  nop
  leave
  ret
```

assistant: ```c
void fatal(const char *ctrl, ...) {
  va_list ap;
  va_start(ap, ctrl);
  vprintf(ctrl, ap);
  va_end(ap);
  exit(-1);
}
```
user: Translate the below assembly code into C programming language:

```asm
<fatal>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rdi,%rsi
  mov    $0x0,%ebp
  mov    %ebp,%edi
  call   <va_start@plt>
  mov    %ebp,%esi
  mov    %rbx,%rdi
  call   <vprintf@plt>
  mov    %ebp,%edi
  call   <va_end@plt>
  mov    $0xffffffff,%edi
  call   <exit@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void fatal(const char *ctrl, ...) {
  va_list ap;
  va_start(ap, ctrl);
  vprintf(ctrl, ap);
  va_end(ap);
  exit(-1);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<fatal>:
  endbr64
  push   %r12
  xor    %r12d,%r12d
  mov    %rdi,%rsi
  push   %rbp
  mov    %rdi,%rbp
  mov    %r12d,%edi
  sub    $0x8,%rsp
  call   <va_start@plt>
  mov    %r12d,%esi
  mov    %rbp,%rdi
  call   <vprintf@plt>
  mov    %r12d,%edi
  call   <va_end@plt>
  mov    $0xffffffff,%edi
  call   <exit@plt>
  add    $0x8,%rsp
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void fatal(const char *ctrl, ...) {
  va_list ap;
  va_start(ap, ctrl);
  vprintf(ctrl, ap);
  va_end(ap);
  exit(-1);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<fatal>:
  endbr64
  push   %r12
  xor    %r12d,%r12d
  mov    %rdi,%rsi
  push   %rbp
  mov    %rdi,%rbp
  mov    %r12d,%edi
  sub    $0x8,%rsp
  call   <va_start@plt>
  mov    %r12d,%esi
  mov    %rbp,%rdi
  call   <vprintf@plt>
  mov    %r12d,%edi
  call   <va_end@plt>
  mov    $0xffffffff,%edi
  call   <exit@plt>
  add    $0x8,%rsp
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void fatal(const char *ctrl, ...) {
  va_list ap;
  va_start(ap, ctrl);
  vprintf(ctrl, ap);
  va_end(ap);
  exit(-1);
}
```