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
<attr_cmp>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  jmp    L92
L63:
  lea    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <getcmptok@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x24(%rbp)
  js     L82
  mov    -0x24(%rbp),%eax
  add    $0x100,%eax
  mov    %eax,-0x8(%rbp)
  movl   $0xffffffff,-0x24(%rbp)
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x20,%al
  jne    L31
  mov    -0x18(%rbp),%rax
  add    $0x1,%rax
  mov    %rax,-0x18(%rbp)
  jmp    L31
L82:
  lea    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <getcmptok@plt>
  mov    %eax,-0x8(%rbp)
L31:
  mov    -0x4(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jge    L62
  mov    $0xffffffff,%eax
  jmp    L14
L62:
  mov    -0x4(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jle    L92
  mov    $0x1,%eax
  jmp    L14
L92:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L29
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L63
  cmpl   $0x0,-0x24(%rbp)
  jns    L63
L29:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L64
  mov    $0x1,%eax
  jmp    L14
L64:
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L19
  mov    $0xffffffff,%eax
  jmp    L14
L19:
  mov    $0x0,%eax
L14:
  leave
  ret
```

assistant: ```c
int attr_cmp(const char *a, const char *b, int bnum) {
  while (*a && (*b || bnum >= 0)) {
    int ai, bi;
    ai = getcmptok(&a);
    if (bnum >= 0) {
      bi = bnum + 256;
      bnum = -1;
      if (*a == ' ')
        a++;
    } else
      bi = getcmptok(&b);
    if (ai < bi)
      return -1;
    if (ai > bi)
      return 1;
  }
  if (*a)
    return 1;
  if (*b)
    return -1;
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<attr_cmp>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %rdi,0x8(%rsp)
  mov    %rsi,(%rsp)
  mov    %edx,%ebp
  lea    0x8(%rsp),%r12
  jmp    L33
L94:
  mov    %rsp,%rdi
  call   <getcmptok@plt>
L22:
  cmp    %ebx,%eax
  jg     L39
  jl     L7
L33:
  mov    0x8(%rsp),%rax
  cmpb   $0x0,(%rax)
  je     L35
  mov    (%rsp),%rax
  cmpb   $0x0,(%rax)
  jne    L81
  test   %ebp,%ebp
  js     L74
L81:
  mov    %r12,%rdi
  call   <getcmptok@plt>
  mov    %eax,%ebx
  test   %ebp,%ebp
  js     L94
  lea    0x100(%rbp),%eax
  mov    0x8(%rsp),%rdx
  mov    $0xffffffff,%ebp
  cmpb   $0x20,(%rdx)
  jne    L22
  add    $0x1,%rdx
  mov    %rdx,0x8(%rsp)
  jmp    L22
L74:
  mov    $0x1,%eax
  jmp    L28
L35:
  mov    (%rsp),%rax
  movzbl (%rax),%eax
  neg    %al
  sbb    %eax,%eax
L28:
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L39:
  mov    $0xffffffff,%eax
  jmp    L28
L7:
  mov    $0x1,%eax
  jmp    L28
```

assistant: ```c
int attr_cmp(const char *a, const char *b, int bnum) {
  while (*a && (*b || bnum >= 0)) {
    int ai, bi;
    ai = getcmptok(&a);
    if (bnum >= 0) {
      bi = bnum + 256;
      bnum = -1;
      if (*a == ' ')
        a++;
    } else
      bi = getcmptok(&b);
    if (ai < bi)
      return -1;
    if (ai > bi)
      return 1;
  }
  if (*a)
    return 1;
  if (*b)
    return -1;
  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
<attr_cmp>:
  endbr64
  push   %r12
  push   %rbp
  mov    %edx,%ebp
  push   %rbx
  sub    $0x10,%rsp
  mov    %rdi,0x8(%rsp)
  lea    0x8(%rsp),%r12
  mov    %rsi,(%rsp)
  jmp    L72
  xchg   %ax,%ax
L68:
  test   %ebp,%ebp
  js     L88
  mov    %r12,%rdi
  call   <getcmptok@plt>
  mov    %eax,%ebx
L83:
  mov    0x8(%rsp),%rdx
  lea    0x100(%rbp),%eax
  mov    $0xffffffff,%ebp
  cmpb   $0x20,(%rdx)
  jne    L70
  add    $0x1,%rdx
  mov    %rdx,0x8(%rsp)
L70:
  cmp    %ebx,%eax
  jg     L84
L41:
  jl     L88
L72:
  mov    0x8(%rsp),%rdx
  mov    (%rsp),%rax
  cmpb   $0x0,(%rdx)
  movzbl (%rax),%eax
  je     L32
  test   %al,%al
  je     L68
  mov    %r12,%rdi
  call   <getcmptok@plt>
  mov    %eax,%ebx
  test   %ebp,%ebp
  jns    L83
  mov    %rsp,%rdi
  call   <getcmptok@plt>
  cmp    %ebx,%eax
  jle    L41
L84:
  add    $0x10,%rsp
  mov    $0xffffffff,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nop
L88:
  add    $0x10,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  xchg   %ax,%ax
L32:
  neg    %al
  sbb    %eax,%eax
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int attr_cmp(const char *a, const char *b, int bnum) {
  while (*a && (*b || bnum >= 0)) {
    int ai, bi;
    ai = getcmptok(&a);
    if (bnum >= 0) {
      bi = bnum + 256;
      bnum = -1;
      if (*a == ' ')
        a++;
    } else
      bi = getcmptok(&b);
    if (ai < bi)
      return -1;
    if (ai > bi)
      return 1;
  }
  if (*a)
    return 1;
  if (*b)
    return -1;
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<attr_cmp>:
  endbr64
  push   %r12
  push   %rbp
  mov    %edx,%ebp
  push   %rbx
  sub    $0x10,%rsp
  mov    %rdi,0x8(%rsp)
  lea    0x8(%rsp),%r12
  mov    %rsi,(%rsp)
  jmp    L40
  xchg   %ax,%ax
L76:
  test   %ebp,%ebp
  js     L28
  mov    %r12,%rdi
  call   <getcmptok@plt>
  mov    %eax,%ebx
L12:
  mov    0x8(%rsp),%rdx
  lea    0x100(%rbp),%eax
  mov    $0xffffffff,%ebp
  cmpb   $0x20,(%rdx)
  jne    L29
  add    $0x1,%rdx
  mov    %rdx,0x8(%rsp)
L29:
  cmp    %ebx,%eax
  jg     L17
L58:
  jl     L28
L40:
  mov    0x8(%rsp),%rdx
  mov    (%rsp),%rax
  cmpb   $0x0,(%rdx)
  movzbl (%rax),%eax
  je     L35
  test   %al,%al
  je     L76
  mov    %r12,%rdi
  call   <getcmptok@plt>
  mov    %eax,%ebx
  test   %ebp,%ebp
  jns    L12
  mov    %rsp,%rdi
  call   <getcmptok@plt>
  cmp    %ebx,%eax
  jle    L58
L17:
  add    $0x10,%rsp
  mov    $0xffffffff,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nop
L28:
  add    $0x10,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  xchg   %ax,%ax
L35:
  neg    %al
  sbb    %eax,%eax
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int attr_cmp(const char *a, const char *b, int bnum) {
  while (*a && (*b || bnum >= 0)) {
    int ai, bi;
    ai = getcmptok(&a);
    if (bnum >= 0) {
      bi = bnum + 256;
      bnum = -1;
      if (*a == ' ')
        a++;
    } else
      bi = getcmptok(&b);
    if (ai < bi)
      return -1;
    if (ai > bi)
      return 1;
  }
  if (*a)
    return 1;
  if (*b)
    return -1;
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<enkf_wrapper_r4>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  add    $0x30,%rax
  mov    (%rax),%rdi
  mov    -0x20(%rbp),%rax
  add    $0x28,%rax
  mov    (%rax),%r9
  mov    -0x20(%rbp),%rax
  add    $0x20,%rax
  mov    (%rax),%r8
  mov    -0x20(%rbp),%rax
  add    $0x18,%rax
  mov    (%rax),%rcx
  mov    -0x20(%rbp),%rax
  add    $0x10,%rax
  mov    (%rax),%rdx
  mov    -0x20(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rsi
  mov    -0x20(%rbp),%rax
  mov    (%rax),%rax
  sub    $0x8,%rsp
  push   %rdi
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <enkf_driver_r4_@plt>
  add    $0x10,%rsp
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int enkf_wrapper_r4(int argc, void *argv[]) {
  extern void enkf_driver_r4_();
  int ret;
  enkf_driver_r4_((int *)argv[0], (int *)argv[1], (int *)argv[2], (float *)argv[3], (float *)argv[4], (float *)argv[5], (float *)argv[6]);
  return ret;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<enkf_wrapper_r4>:
  endbr64
  sub    $0x10,%rsp
  mov    %rsi,%rax
  mov    0x18(%rsi),%rcx
  mov    0x10(%rsi),%rdx
  mov    0x8(%rsi),%rsi
  push   0x30(%rax)
  mov    0x28(%rax),%r9
  mov    0x20(%rax),%r8
  mov    (%rax),%rdi
  mov    $0x0,%eax
  call   <enkf_driver_r4_@plt>
  mov    $0x0,%eax
  add    $0x18,%rsp
  ret
```

assistant: ```c
int enkf_wrapper_r4(int argc, void *argv[]) {
  extern void enkf_driver_r4_();
  int ret;
  enkf_driver_r4_((int *)argv[0], (int *)argv[1], (int *)argv[2], (float *)argv[3], (float *)argv[4], (float *)argv[5], (float *)argv[6]);
  return ret;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<enkf_wrapper_r4>:
  endbr64
  mov    %rsi,%rax
  sub    $0x10,%rsp
  mov    0x18(%rsi),%rcx
  mov    0x10(%rsi),%rdx
  mov    0x8(%rsi),%rsi
  push   0x30(%rax)
  mov    0x28(%rax),%r9
  mov    0x20(%rax),%r8
  mov    (%rax),%rdi
  xor    %eax,%eax
  call   <enkf_driver_r4_@plt>
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
```

assistant: ```c
int enkf_wrapper_r4(int argc, void *argv[]) {
  extern void enkf_driver_r4_();
  int ret;
  enkf_driver_r4_((int *)argv[0], (int *)argv[1], (int *)argv[2], (float *)argv[3], (float *)argv[4], (float *)argv[5], (float *)argv[6]);
  return ret;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<enkf_wrapper_r4>:
  endbr64
  mov    %rsi,%rax
  sub    $0x10,%rsp
  mov    0x18(%rsi),%rcx
  mov    0x10(%rsi),%rdx
  mov    0x8(%rsi),%rsi
  push   0x30(%rax)
  mov    0x28(%rax),%r9
  mov    0x20(%rax),%r8
  mov    (%rax),%rdi
  xor    %eax,%eax
  call   <enkf_driver_r4_@plt>
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
```

assistant: ```c
int enkf_wrapper_r4(int argc, void *argv[]) {
  extern void enkf_driver_r4_();
  int ret;
  enkf_driver_r4_((int *)argv[0], (int *)argv[1], (int *)argv[2], (float *)argv[3], (float *)argv[4], (float *)argv[5], (float *)argv[6]);
  return ret;
}
```
user: What is the c source code of the assembly code below:

```asm
<wt_getsym>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r12
  push   %rbx
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,%rax
  mov    -0x28(%rbp),%rcx
  mov    %rcx,-0x18(%rbp)
  mov    %rdx,%rcx
  imul   %rax,%rcx
  mov    %rcx,%rbx
  shr    $0x5,%rbx
  imul   %rax,%rdx
  mov    %rbx,%rcx
  shl    $0x5,%rcx
  sub    %rcx,%rdx
  mov    %rdx,%r12
  lea    (%r12,%rax,1),%rdx
  cmp    $0x20,%rdx
  ja     L12
  lea    0x0(,%rbx,4),%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rdx),%esi
  mov    %r12d,%ecx
  mov    %eax,%edx
  add    %edx,%ecx
  mov    $0x20,%edx
  sub    %ecx,%edx
  mov    %edx,%ecx
  shl    %cl,%esi
  mov    %esi,%edx
  mov    %eax,%ecx
  mov    $0x20,%eax
  sub    %ecx,%eax
  mov    %eax,%ecx
  sar    %cl,%edx
  mov    %edx,%eax
  mov    %eax,-0x1c(%rbp)
  jmp    L3
L12:
  lea    0x0(,%rbx,4),%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rdx),%edx
  mov    %r12d,%ecx
  sar    %cl,%edx
  mov    %edx,-0x1c(%rbp)
  lea    0x1(%rbx),%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rdx),%esi
  mov    %r12d,%ecx
  mov    %eax,%edx
  add    %edx,%ecx
  mov    $0x40,%edx
  sub    %ecx,%edx
  mov    %edx,%ecx
  shl    %cl,%esi
  mov    %esi,%edx
  mov    %eax,%ecx
  mov    $0x20,%eax
  sub    %ecx,%eax
  mov    %eax,%ecx
  sar    %cl,%edx
  mov    %edx,%eax
  or     %eax,-0x1c(%rbp)
L3:
  mov    -0x1c(%rbp),%eax
  pop    %rbx
  pop    %r12
  pop    %rbp
  ret
```

assistant: ```c
uint32_t wt_getsym(uint64_t *B, register size_t bits, register size_t pos) {
  uint32_t *A = (uint32_t *)B;
  register size_t i = pos * bits / 32, j = pos * bits - 32 * i;
  uint32_t result;
  if (j + bits <= 32)
    result = (A[i] << (32 - j - bits)) >> (32 - bits);
  else {
    result = A[i] >> j;
    result = result | (A[i + 1] << (64 - j - bits)) >> (32 - bits);
  }
  return result;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<wt_getsym>:
  endbr64
  mov    %rdi,%rax
  imul   %rsi,%rdx
  mov    %rdx,%rdi
  shr    $0x5,%rdi
  and    $0x1f,%edx
  lea    (%rsi,%rdx,1),%rcx
  cmp    $0x20,%rcx
  ja     L7
  mov    $0x20,%r8d
  sub    %esi,%r8d
  mov    %r8d,%ecx
  sub    %edx,%ecx
  mov    (%rax,%rdi,4),%eax
  shl    %cl,%eax
  mov    %r8d,%ecx
  sar    %cl,%eax
  ret
L7:
  mov    (%rax,%rdi,4),%r8d
  mov    %edx,%ecx
  sar    %cl,%r8d
  mov    $0x40,%ecx
  sub    %esi,%ecx
  sub    %edx,%ecx
  mov    0x4(%rax,%rdi,4),%eax
  shl    %cl,%eax
  mov    $0x20,%ecx
  sub    %esi,%ecx
  sar    %cl,%eax
  or     %r8d,%eax
  ret
```

assistant: ```c
uint32_t wt_getsym(uint64_t *B, register size_t bits, register size_t pos) {
  uint32_t *A = (uint32_t *)B;
  register size_t i = pos * bits / 32, j = pos * bits - 32 * i;
  uint32_t result;
  if (j + bits <= 32)
    result = (A[i] << (32 - j - bits)) >> (32 - bits);
  else {
    result = A[i] >> j;
    result = result | (A[i + 1] << (64 - j - bits)) >> (32 - bits);
  }
  return result;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<wt_getsym>:
  endbr64
  imul   %rsi,%rdx
  mov    %rsi,%rax
  mov    %rdx,%r8
  shr    $0x5,%rdx
  and    $0x1f,%r8d
  mov    (%rdi,%rdx,4),%esi
  lea    0x0(,%rdx,4),%r9
  mov    $0x20,%edx
  lea    (%rax,%r8,1),%rcx
  sub    %eax,%edx
  cmp    $0x20,%rcx
  ja     L7
  mov    %edx,%ecx
  sub    %r8d,%ecx
  shl    %cl,%esi
  mov    %edx,%ecx
  mov    %esi,%eax
  sar    %cl,%eax
  ret
L7:
  mov    %r8d,%ecx
  sar    %cl,%esi
  mov    $0x40,%ecx
  sub    %eax,%ecx
  mov    0x4(%rdi,%r9,1),%eax
  sub    %r8d,%ecx
  shl    %cl,%eax
  mov    %edx,%ecx
  sar    %cl,%eax
  or     %esi,%eax
  ret
```

assistant: ```c
uint32_t wt_getsym(uint64_t *B, register size_t bits, register size_t pos) {
  uint32_t *A = (uint32_t *)B;
  register size_t i = pos * bits / 32, j = pos * bits - 32 * i;
  uint32_t result;
  if (j + bits <= 32)
    result = (A[i] << (32 - j - bits)) >> (32 - bits);
  else {
    result = A[i] >> j;
    result = result | (A[i + 1] << (64 - j - bits)) >> (32 - bits);
  }
  return result;
}
```
user: Translate the below assembly code into C programming language:

```asm
<wt_getsym>:
  endbr64
  imul   %rsi,%rdx
  mov    %rsi,%rax
  mov    %rdx,%r8
  shr    $0x5,%rdx
  and    $0x1f,%r8d
  mov    (%rdi,%rdx,4),%esi
  lea    0x0(,%rdx,4),%r9
  mov    $0x20,%edx
  lea    (%rax,%r8,1),%rcx
  sub    %eax,%edx
  cmp    $0x20,%rcx
  ja     L6
  mov    %edx,%ecx
  sub    %r8d,%ecx
  shl    %cl,%esi
  mov    %edx,%ecx
  mov    %esi,%eax
  sar    %cl,%eax
  ret
L6:
  mov    %r8d,%ecx
  sar    %cl,%esi
  mov    $0x40,%ecx
  sub    %eax,%ecx
  mov    0x4(%rdi,%r9,1),%eax
  sub    %r8d,%ecx
  shl    %cl,%eax
  mov    %edx,%ecx
  sar    %cl,%eax
  or     %esi,%eax
  ret
```

assistant: ```c
uint32_t wt_getsym(uint64_t *B, register size_t bits, register size_t pos) {
  uint32_t *A = (uint32_t *)B;
  register size_t i = pos * bits / 32, j = pos * bits - 32 * i;
  uint32_t result;
  if (j + bits <= 32)
    result = (A[i] << (32 - j - bits)) >> (32 - bits);
  else {
    result = A[i] >> j;
    result = result | (A[i + 1] << (64 - j - bits)) >> (32 - bits);
  }
  return result;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<OCMalloc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L4
  mov    $0x0,%eax
  jmp    L17
L4:
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <malloc@plt>
L17:
  leave
  ret
```

assistant: ```c
void *OCMalloc(size_t size) {
  if (0 == size) {
    return NULL;
  }
  return malloc(size);
}
```
user: Translate the below assembly code into C programming language:

```asm
<OCMalloc>:
  endbr64
  mov    $0x0,%eax
  test   %rdi,%rdi
  je     L7
  sub    $0x8,%rsp
  call   <malloc@plt>
  add    $0x8,%rsp
  ret
L7:
  ret
```

assistant: ```c
void *OCMalloc(size_t size) {
  if (0 == size) {
    return NULL;
  }
  return malloc(size);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<OCMalloc>:
  endbr64
  test   %rdi,%rdi
  je     L9
  jmp    1050 <malloc@plt>
  xchg   %ax,%ax
L9:
  xor    %eax,%eax
  ret
```

assistant: ```c
void *OCMalloc(size_t size) {
  if (0 == size) {
    return NULL;
  }
  return malloc(size);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<OCMalloc>:
  endbr64
  test   %rdi,%rdi
  je     L6
  jmp    1050 <malloc@plt>
  xchg   %ax,%ax
L6:
  xor    %eax,%eax
  ret
```

assistant: ```c
void *OCMalloc(size_t size) {
  if (0 == size) {
    return NULL;
  }
  return malloc(size);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<run_instruction>:
  endbr64
  mov    %edx,%r8d
  cmp    $0xe,%edi
  ja     L12
  mov    %edi,%edi
  lea    D2(%rip),%rcx
  movslq (%rcx,%rdi,4),%rax
  add    %rcx,%rax
  notrack jmp *%rax
  test   %esi,%esi
  setne  %cl
  test   %edx,%edx
  setne  %al
  movzbl %al,%eax
  and    %ecx,%eax
  ret
  or     %edx,%esi
  setne  %al
  movzbl %al,%eax
  ret
  test   %esi,%esi
  sete   %al
  movzbl %al,%eax
  ret
  cmp    %edx,%esi
  sete   %al
  movzbl %al,%eax
  ret
  cmp    %edx,%esi
  setne  %al
  movzbl %al,%eax
  ret
  cmp    %edx,%esi
  setg   %al
  movzbl %al,%eax
  ret
  cmp    %edx,%esi
  setge  %al
  movzbl %al,%eax
  ret
  cmp    %edx,%esi
  setl   %al
  movzbl %al,%eax
  ret
  cmp    %edx,%esi
  setle  %al
  movzbl %al,%eax
  ret
  cmp    %edx,%esi
  sete   %al
  movzbl %al,%eax
  ret
  cmp    %edx,%esi
  setne  %al
  movzbl %al,%eax
  ret
  lea    (%rsi,%rdx,1),%eax
  ret
  mov    %esi,%eax
  sub    %edx,%eax
  ret
  mov    %esi,%eax
  imul   %edx,%eax
  ret
  mov    $0x0,%eax
  test   %edx,%edx
  jle    L3
  mov    %esi,%eax
  cltd
  idiv   %r8d
  ret
L12:
  mov    $0x0,%eax
L3:
  ret
```

assistant: ```c
int run_instruction(int _operator, int _operand1, int _operand2) {
  int result;
  switch (_operator) {
  case 4: {
    result = (_operand1 && _operand2);
    break;
  }
  case 5: {
    result = (_operand1 || _operand2);
    break;
  }
  case 6: {
    result = (!_operand1);
    break;
  }
  case 7: {
    result = (_operand1 == _operand2);
    break;
  }
  case 8: {
    result = (_operand1 != _operand2);
    break;
  }
  case 9: {
    result = (_operand1 > _operand2);
    break;
  }
  case 10: {
    result = (_operand1 >= _operand2);
    break;
  }
  case 11: {
    result = (_operand1 < _operand2);
    break;
  }
  case 12: {
    result = (_operand1 <= _operand2);
    break;
  }
  case 13: {
    result = (_operand1 == _operand2);
    break;
  }
  case 14: {
    result = (_operand1 != _operand2);
    break;
  }
  case 0: {
    result = (_operand1 + _operand2);
    break;
  }
  case 1: {
    result = (_operand1 - _operand2);
    break;
  }
  case 2: {
    result = (_operand1 * _operand2);
    break;
  }
  case 3: {
    result = (_operand2 > 0 ? (_operand1 / _operand2) : 0);
    break;
  }
  }
  return result;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<run_instruction>:
  endbr64
  mov    %edx,%r8d
  cmp    $0xe,%edi
  ja     1040 <run_instruction.cold>
  lea    D1(%rip),%rcx
  mov    %edi,%edi
  movslq (%rcx,%rdi,4),%rax
  add    %rcx,%rax
  notrack jmp *%rax
  nopl   0x0(%rax,%rax,1)
  xor    %eax,%eax
  cmp    %edx,%esi
  setne  %al
  ret
  xor    %eax,%eax
  cmp    %edx,%esi
  sete   %al
  ret
  nopl   0x0(%rax,%rax,1)
  xor    %eax,%eax
  cmp    %edx,%esi
  setg   %al
  ret
  nopl   0x0(%rax,%rax,1)
  xor    %eax,%eax
  cmp    %edx,%esi
  setge  %al
  ret
  nopl   0x0(%rax,%rax,1)
  xor    %eax,%eax
  cmp    %edx,%esi
  setl   %al
  ret
  nopl   0x0(%rax,%rax,1)
  xor    %eax,%eax
  cmp    %edx,%esi
  setle  %al
  ret
  nopl   0x0(%rax,%rax,1)
  xor    %eax,%eax
  test   %esi,%esi
  sete   %al
  ret
  nopl   0x0(%rax,%rax,1)
  test   %esi,%esi
  setne  %cl
  xor    %eax,%eax
  test   %edx,%edx
  setne  %al
  and    %ecx,%eax
  ret
  nop
  xor    %eax,%eax
  or     %edx,%esi
  setne  %al
  ret
  nopl   0x0(%rax,%rax,1)
  lea    (%rsi,%rdx,1),%eax
  ret
  nopl   0x0(%rax)
  mov    %esi,%eax
  sub    %edx,%eax
  ret
  nopl   (%rax)
  mov    %esi,%eax
  imul   %edx,%eax
  ret
  cs nopw 0x0(%rax,%rax,1)
  xor    %eax,%eax
  test   %edx,%edx
  jle    L5
  mov    %esi,%eax
  cltd
  idiv   %r8d
  ret
  nopl   (%rax)
L5:
  ret
```

assistant: ```c
int run_instruction(int _operator, int _operand1, int _operand2) {
  int result;
  switch (_operator) {
  case 4: {
    result = (_operand1 && _operand2);
    break;
  }
  case 5: {
    result = (_operand1 || _operand2);
    break;
  }
  case 6: {
    result = (!_operand1);
    break;
  }
  case 7: {
    result = (_operand1 == _operand2);
    break;
  }
  case 8: {
    result = (_operand1 != _operand2);
    break;
  }
  case 9: {
    result = (_operand1 > _operand2);
    break;
  }
  case 10: {
    result = (_operand1 >= _operand2);
    break;
  }
  case 11: {
    result = (_operand1 < _operand2);
    break;
  }
  case 12: {
    result = (_operand1 <= _operand2);
    break;
  }
  case 13: {
    result = (_operand1 == _operand2);
    break;
  }
  case 14: {
    result = (_operand1 != _operand2);
    break;
  }
  case 0: {
    result = (_operand1 + _operand2);
    break;
  }
  case 1: {
    result = (_operand1 - _operand2);
    break;
  }
  case 2: {
    result = (_operand1 * _operand2);
    break;
  }
  case 3: {
    result = (_operand2 > 0 ? (_operand1 / _operand2) : 0);
    break;
  }
  }
  return result;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<run_instruction>:
  endbr64
  mov    %edx,%r8d
  cmp    $0xe,%edi
  ja     1040 <run_instruction.cold>
  lea    D5(%rip),%rcx
  mov    %edi,%edi
  movslq (%rcx,%rdi,4),%rax
  add    %rcx,%rax
  notrack jmp *%rax
  nopl   0x0(%rax,%rax,1)
  xor    %eax,%eax
  cmp    %edx,%esi
  setne  %al
  ret
  xor    %eax,%eax
  cmp    %edx,%esi
  sete   %al
  ret
  nopl   0x0(%rax,%rax,1)
  xor    %eax,%eax
  cmp    %edx,%esi
  setg   %al
  ret
  nopl   0x0(%rax,%rax,1)
  xor    %eax,%eax
  cmp    %edx,%esi
  setge  %al
  ret
  nopl   0x0(%rax,%rax,1)
  xor    %eax,%eax
  cmp    %edx,%esi
  setl   %al
  ret
  nopl   0x0(%rax,%rax,1)
  xor    %eax,%eax
  cmp    %edx,%esi
  setle  %al
  ret
  nopl   0x0(%rax,%rax,1)
  xor    %eax,%eax
  test   %esi,%esi
  sete   %al
  ret
  nopl   0x0(%rax,%rax,1)
  test   %esi,%esi
  setne  %cl
  xor    %eax,%eax
  test   %edx,%edx
  setne  %al
  and    %ecx,%eax
  ret
  nop
  xor    %eax,%eax
  or     %edx,%esi
  setne  %al
  ret
  nopl   0x0(%rax,%rax,1)
  lea    (%rsi,%rdx,1),%eax
  ret
  nopl   0x0(%rax)
  mov    %esi,%eax
  sub    %edx,%eax
  ret
  nopl   (%rax)
  mov    %esi,%eax
  imul   %edx,%eax
  ret
  cs nopw 0x0(%rax,%rax,1)
  xor    %eax,%eax
  test   %edx,%edx
  jle    L6
  mov    %esi,%eax
  cltd
  idiv   %r8d
  ret
  nopl   (%rax)
L6:
  ret
```

assistant: ```c
int run_instruction(int _operator, int _operand1, int _operand2) {
  int result;
  switch (_operator) {
  case 4: {
    result = (_operand1 && _operand2);
    break;
  }
  case 5: {
    result = (_operand1 || _operand2);
    break;
  }
  case 6: {
    result = (!_operand1);
    break;
  }
  case 7: {
    result = (_operand1 == _operand2);
    break;
  }
  case 8: {
    result = (_operand1 != _operand2);
    break;
  }
  case 9: {
    result = (_operand1 > _operand2);
    break;
  }
  case 10: {
    result = (_operand1 >= _operand2);
    break;
  }
  case 11: {
    result = (_operand1 < _operand2);
    break;
  }
  case 12: {
    result = (_operand1 <= _operand2);
    break;
  }
  case 13: {
    result = (_operand1 == _operand2);
    break;
  }
  case 14: {
    result = (_operand1 != _operand2);
    break;
  }
  case 0: {
    result = (_operand1 + _operand2);
    break;
  }
  case 1: {
    result = (_operand1 - _operand2);
    break;
  }
  case 2: {
    result = (_operand1 * _operand2);
    break;
  }
  case 3: {
    result = (_operand2 > 0 ? (_operand1 / _operand2) : 0);
    break;
  }
  }
  return result;
}
```