ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请从给定的汇编指令推导出C源代码：

```asm
<intadd_>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %r8,%rbp
  mov    %r9,%r12
  mov    0x48(%rsp),%rbx
  mov    0x40(%rsp),%rax
  lea    -0x4(%rax),%r13
  lea    -0x4(%r9),%r14
  lea    -0x4(%r8),%r15
  mov    (%rdi),%eax
  mov    %eax,D77(%rip)
  mov    (%rsi),%eax
  mov    %eax,D144(%rip)
  mov    (%rdx),%edx
  mov    %edx,D70(%rip)
  mov    (%rcx),%edx
  mov    %edx,D113(%rip)
  cltq
  lea    0x0(%r13,%rax,4),%rdi
  mov    %r9,%rcx
  mov    %r8,%rdx
  lea    D70(%rip),%rsi
  call   <lstptr_@plt>
  mov    %eax,D129(%rip)
  mov    %rbx,%r8
  mov    %r12,%rcx
  mov    %rbp,%rdx
  lea    D129(%rip),%rsi
  lea    D77(%rip),%rdi
  call   <insert_@plt>
  movslq D70(%rip),%rax
  lea    0x0(%r13,%rax,4),%rdi
  mov    %r12,%rcx
  mov    %rbp,%rdx
  lea    D113(%rip),%rsi
  call   <lstptr_@plt>
  mov    %eax,D148(%rip)
  mov    %rbx,%r8
  mov    %r12,%rcx
  mov    %rbp,%rdx
  lea    D148(%rip),%rsi
  lea    D77(%rip),%rdi
  call   <insert_@plt>
  movslq D113(%rip),%rax
  lea    0x0(%r13,%rax,4),%rdi
  mov    %r12,%rcx
  mov    %rbp,%rdx
  lea    D144(%rip),%rsi
  call   <lstptr_@plt>
  mov    %eax,D148(%rip)
  mov    %rbx,%r8
  mov    %r12,%rcx
  mov    %rbp,%rdx
  lea    D148(%rip),%rsi
  lea    D77(%rip),%rdi
  call   <insert_@plt>
  movslq (%rbx),%rax
  mov    D144(%rip),%edx
  mov    %edx,(%r15,%rax,4)
  movslq (%rbx),%rax
  mov    D70(%rip),%edx
  mov    %edx,0x0(%rbp,%rax,4)
  movslq (%rbx),%rax
  mov    D113(%rip),%edx
  mov    %edx,0x4(%rbp,%rax,4)
  mov    (%rbx),%eax
  movslq %eax,%rdx
  add    $0x1,%eax
  mov    %eax,(%r14,%rdx,4)
  mov    (%rbx),%eax
  movslq %eax,%rdx
  add    $0x2,%eax
  mov    %eax,(%r12,%rdx,4)
  mov    (%rbx),%eax
  movslq %eax,%rdx
  mov    %eax,0x4(%r12,%rdx,4)
  movslq D77(%rip),%rdx
  add    $0x2,%eax
  mov    %eax,0x0(%r13,%rdx,4)
  addl   $0x3,(%rbx)
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
int intadd_(int *kk, int *i1, int *i2, int *i3, int *list, int *lptr, int *lend, int *lnew) {
  static int k, n1, n2, n3, lp;
  --lend;
  --lptr;
  --list;
  k = *kk;
  n1 = *i1;
  n2 = *i2;
  n3 = *i3;
  lp = lstptr_(&lend[n1], &n2, &list[1], &lptr[1]);
  insert_(&k, &lp, &list[1], &lptr[1], lnew);
  lp = lstptr_(&lend[n2], &n3, &list[1], &lptr[1]);
  insert_(&k, &lp, &list[1], &lptr[1], lnew);
  lp = lstptr_(&lend[n3], &n1, &list[1], &lptr[1]);
  insert_(&k, &lp, &list[1], &lptr[1], lnew);
  list[*lnew] = n1;
  list[*lnew + 1] = n2;
  list[*lnew + 2] = n3;
  lptr[*lnew] = *lnew + 1;
  lptr[*lnew + 1] = *lnew + 2;
  lptr[*lnew + 2] = *lnew;
  lend[k] = *lnew + 2;
  *lnew += 3;
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<intadd_>:
  endbr64
  push   %r15
  lea    -0x4(%r8),%r15
  push   %r14
  lea    -0x4(%r9),%r14
  push   %r13
  mov    %r9,%r13
  push   %r12
  mov    %r8,%r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    0x40(%rsp),%rax
  mov    0x48(%rsp),%rbx
  lea    -0x4(%rax),%rbp
  mov    (%rdi),%eax
  mov    %eax,D26(%rip)
  movslq (%rsi),%rax
  lea    D215(%rip),%rsi
  mov    %eax,D171(%rip)
  mov    (%rdx),%edx
  lea    0x0(%rbp,%rax,4),%rdi
  mov    %edx,D141(%rip)
  mov    (%rcx),%edx
  mov    %r9,%rcx
  mov    %edx,D141(%rip)
  mov    %r8,%rdx
  call   <lstptr_@plt>
  mov    %rbx,%r8
  mov    %r13,%rcx
  mov    %r12,%rdx
  lea    D207(%rip),%r10
  lea    D26(%rip),%r9
  mov    %eax,D110(%rip)
  mov    %r10,%rsi
  mov    %r9,%rdi
  call   <insert_@plt>
  movslq D141(%rip),%rax
  mov    %r13,%rcx
  mov    %r12,%rdx
  lea    D141(%rip),%rsi
  lea    0x0(%rbp,%rax,4),%rdi
  call   <lstptr_@plt>
  mov    %rbx,%r8
  mov    %r13,%rcx
  mov    %r12,%rdx
  lea    D110(%rip),%r10
  lea    D26(%rip),%r9
  mov    %eax,D110(%rip)
  mov    %r10,%rsi
  mov    %r9,%rdi
  call   <insert_@plt>
  movslq D141(%rip),%rax
  mov    %r13,%rcx
  mov    %r12,%rdx
  lea    D171(%rip),%rsi
  lea    0x0(%rbp,%rax,4),%rdi
  call   <lstptr_@plt>
  mov    %r12,%rdx
  mov    %rbx,%r8
  mov    %r13,%rcx
  lea    D110(%rip),%r10
  lea    D26(%rip),%r9
  mov    %eax,D110(%rip)
  mov    %r10,%rsi
  mov    %r9,%rdi
  call   <insert_@plt>
  movslq (%rbx),%rax
  mov    D171(%rip),%edx
  mov    %edx,(%r15,%rax,4)
  movslq (%rbx),%rax
  mov    D141(%rip),%edx
  mov    %edx,(%r12,%rax,4)
  movslq (%rbx),%rax
  mov    D141(%rip),%edx
  mov    %edx,0x4(%r12,%rax,4)
  movslq (%rbx),%rdx
  mov    %rdx,%rax
  add    $0x1,%eax
  mov    %eax,(%r14,%rdx,4)
  movslq (%rbx),%rdx
  mov    %rdx,%rax
  add    $0x2,%eax
  mov    %eax,0x0(%r13,%rdx,4)
  movslq (%rbx),%rdx
  mov    %edx,0x4(%r13,%rdx,4)
  mov    %rdx,%rax
  movslq D26(%rip),%rdx
  add    $0x2,%eax
  mov    %eax,0x0(%rbp,%rdx,4)
  xor    %eax,%eax
  addl   $0x3,(%rbx)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
int intadd_(int *kk, int *i1, int *i2, int *i3, int *list, int *lptr, int *lend, int *lnew) {
  static int k, n1, n2, n3, lp;
  --lend;
  --lptr;
  --list;
  k = *kk;
  n1 = *i1;
  n2 = *i2;
  n3 = *i3;
  lp = lstptr_(&lend[n1], &n2, &list[1], &lptr[1]);
  insert_(&k, &lp, &list[1], &lptr[1], lnew);
  lp = lstptr_(&lend[n2], &n3, &list[1], &lptr[1]);
  insert_(&k, &lp, &list[1], &lptr[1], lnew);
  lp = lstptr_(&lend[n3], &n1, &list[1], &lptr[1]);
  insert_(&k, &lp, &list[1], &lptr[1], lnew);
  list[*lnew] = n1;
  list[*lnew + 1] = n2;
  list[*lnew + 2] = n3;
  lptr[*lnew] = *lnew + 1;
  lptr[*lnew + 1] = *lnew + 2;
  lptr[*lnew + 2] = *lnew;
  lend[k] = *lnew + 2;
  *lnew += 3;
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<intadd_>:
  endbr64
  push   %r15
  lea    -0x4(%r8),%r15
  push   %r14
  lea    -0x4(%r9),%r14
  push   %r13
  mov    %r9,%r13
  push   %r12
  mov    %r8,%r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    0x40(%rsp),%rax
  mov    0x48(%rsp),%rbx
  lea    -0x4(%rax),%rbp
  mov    (%rdi),%eax
  mov    %eax,D165(%rip)
  movslq (%rsi),%rax
  lea    D194(%rip),%rsi
  mov    %eax,D146(%rip)
  mov    (%rdx),%edx
  lea    0x0(%rbp,%rax,4),%rdi
  mov    %edx,D77(%rip)
  mov    (%rcx),%edx
  mov    %r9,%rcx
  mov    %edx,D77(%rip)
  mov    %r8,%rdx
  call   <lstptr_@plt>
  mov    %rbx,%r8
  mov    %r13,%rcx
  mov    %r12,%rdx
  lea    D72(%rip),%r10
  lea    D165(%rip),%r9
  mov    %eax,D204(%rip)
  mov    %r10,%rsi
  mov    %r9,%rdi
  call   <insert_@plt>
  movslq D77(%rip),%rax
  mov    %r13,%rcx
  mov    %r12,%rdx
  lea    D77(%rip),%rsi
  lea    0x0(%rbp,%rax,4),%rdi
  call   <lstptr_@plt>
  mov    %rbx,%r8
  mov    %r13,%rcx
  mov    %r12,%rdx
  lea    D204(%rip),%r10
  lea    D165(%rip),%r9
  mov    %eax,D204(%rip)
  mov    %r10,%rsi
  mov    %r9,%rdi
  call   <insert_@plt>
  movslq D77(%rip),%rax
  mov    %r13,%rcx
  mov    %r12,%rdx
  lea    D146(%rip),%rsi
  lea    0x0(%rbp,%rax,4),%rdi
  call   <lstptr_@plt>
  mov    %r12,%rdx
  mov    %rbx,%r8
  mov    %r13,%rcx
  lea    D204(%rip),%r10
  lea    D165(%rip),%r9
  mov    %eax,D204(%rip)
  mov    %r10,%rsi
  mov    %r9,%rdi
  call   <insert_@plt>
  movslq (%rbx),%rax
  mov    D146(%rip),%edx
  mov    %edx,(%r15,%rax,4)
  movslq (%rbx),%rax
  mov    D77(%rip),%edx
  mov    %edx,(%r12,%rax,4)
  movslq (%rbx),%rax
  mov    D77(%rip),%edx
  mov    %edx,0x4(%r12,%rax,4)
  movslq (%rbx),%rdx
  mov    %rdx,%rax
  add    $0x1,%eax
  mov    %eax,(%r14,%rdx,4)
  movslq (%rbx),%rdx
  mov    %rdx,%rax
  add    $0x2,%eax
  mov    %eax,0x0(%r13,%rdx,4)
  movslq (%rbx),%rdx
  mov    %edx,0x4(%r13,%rdx,4)
  mov    %rdx,%rax
  movslq D165(%rip),%rdx
  add    $0x2,%eax
  mov    %eax,0x0(%rbp,%rdx,4)
  xor    %eax,%eax
  addl   $0x3,(%rbx)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
int intadd_(int *kk, int *i1, int *i2, int *i3, int *list, int *lptr, int *lend, int *lnew) {
  static int k, n1, n2, n3, lp;
  --lend;
  --lptr;
  --list;
  k = *kk;
  n1 = *i1;
  n2 = *i2;
  n3 = *i3;
  lp = lstptr_(&lend[n1], &n2, &list[1], &lptr[1]);
  insert_(&k, &lp, &list[1], &lptr[1], lnew);
  lp = lstptr_(&lend[n2], &n3, &list[1], &lptr[1]);
  insert_(&k, &lp, &list[1], &lptr[1], lnew);
  lp = lstptr_(&lend[n3], &n1, &list[1], &lptr[1]);
  insert_(&k, &lp, &list[1], &lptr[1], lnew);
  list[*lnew] = n1;
  list[*lnew + 1] = n2;
  list[*lnew + 2] = n3;
  lptr[*lnew] = *lnew + 1;
  lptr[*lnew + 1] = *lnew + 2;
  lptr[*lnew + 2] = *lnew;
  lend[k] = *lnew + 2;
  *lnew += 3;
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<copy_pair_table>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  movzwl (%rax),%eax
  cwtl
  add    $0x2,%eax
  cltq
  add    %eax,%eax
  mov    %eax,%edi
  call   <space@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x18(%rbp),%rax
  movzwl (%rax),%eax
  cwtl
  add    $0x2,%eax
  cltq
  add    %eax,%eax
  mov    %eax,%edx
  mov    -0x18(%rbp),%rcx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
short *copy_pair_table(const short *pt) {
  short *table = (short *)space(sizeof(short) * (pt[0] + 2));
  memcpy(table, pt, sizeof(short) * (pt[0] + 2));
  return table;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<copy_pair_table>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  movswl (%rdi),%eax
  lea    0x4(%rax,%rax,1),%edi
  call   <space@plt>
  mov    %rax,%rbp
  movswl (%rbx),%eax
  lea    0x4(%rax,%rax,1),%edx
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <memcpy@plt>
  mov    %rbp,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
short *copy_pair_table(const short *pt) {
  short *table = (short *)space(sizeof(short) * (pt[0] + 2));
  memcpy(table, pt, sizeof(short) * (pt[0] + 2));
  return table;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<copy_pair_table>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  movswl (%rdi),%eax
  lea    0x4(%rax,%rax,1),%edi
  call   <space@plt>
  mov    %rbp,%rsi
  mov    %rax,%r12
  movswl 0x0(%rbp),%eax
  mov    %r12,%rdi
  lea    0x4(%rax,%rax,1),%edx
  call   <memcpy@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
short *copy_pair_table(const short *pt) {
  short *table = (short *)space(sizeof(short) * (pt[0] + 2));
  memcpy(table, pt, sizeof(short) * (pt[0] + 2));
  return table;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<copy_pair_table>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  movswl (%rdi),%eax
  lea    0x4(%rax,%rax,1),%edi
  call   <space@plt>
  mov    %rbp,%rsi
  mov    %rax,%r12
  movswl 0x0(%rbp),%eax
  mov    %r12,%rdi
  lea    0x4(%rax,%rax,1),%edx
  call   <memcpy@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
short *copy_pair_table(const short *pt) {
  short *table = (short *)space(sizeof(short) * (pt[0] + 2));
  memcpy(table, pt, sizeof(short) * (pt[0] + 2));
  return table;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ft_strequ>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L36
L46:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%edx
  mov    -0x4(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x20(%rbp),%rax
  add    %rcx,%rax
  movzbl (%rax),%eax
  cmp    %al,%dl
  je     L58
  mov    $0x0,%eax
  jmp    L34
L58:
  addl   $0x1,-0x4(%rbp)
L36:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L54
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L46
L54:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%edx
  mov    -0x4(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x20(%rbp),%rax
  add    %rcx,%rax
  movzbl (%rax),%eax
  cmp    %al,%dl
  je     L31
  mov    $0x0,%eax
  jmp    L34
L31:
  mov    $0x1,%eax
L34:
  pop    %rbp
  ret
```

assistant: ```c
int ft_strequ(const char *s1, const char *s2) {
  int i;
  i = 0;
  while (s1[i] != '\0' && s2[i] != '\0') {
    if (s1[i] != s2[i])
      return (0);
    i++;
  }
  if (s1[i] != s2[i])
    return (0);
  else
    return (1);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<ft_strequ>:
  endbr64
  movzbl (%rdi),%edx
  mov    $0x0,%eax
  test   %dl,%dl
  je     L30
L32:
  movzbl (%rsi,%rax,1),%ecx
  test   %cl,%cl
  je     L30
  cmp    %dl,%cl
  jne    L23
  add    $0x1,%rax
  movzbl (%rdi,%rax,1),%edx
  test   %dl,%dl
  jne    L32
L30:
  cmp    %dl,(%rsi,%rax,1)
  sete   %al
  movzbl %al,%eax
  ret
L23:
  mov    $0x0,%eax
  ret
```

assistant: ```c
int ft_strequ(const char *s1, const char *s2) {
  int i;
  i = 0;
  while (s1[i] != '\0' && s2[i] != '\0') {
    if (s1[i] != s2[i])
      return (0);
    i++;
  }
  if (s1[i] != s2[i])
    return (0);
  else
    return (1);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ft_strequ>:
  endbr64
  movzbl (%rdi),%edx
  xor    %eax,%eax
  test   %dl,%dl
  jne    L46
  jmp    L18
  nop
L45:
  cmp    %dl,%cl
  jne    L51
  add    $0x1,%rax
  movzbl (%rdi,%rax,1),%edx
  test   %dl,%dl
  je     L16
L46:
  movzbl (%rsi,%rax,1),%ecx
  test   %cl,%cl
  jne    L45
L14:
  xor    %eax,%eax
  cmp    %cl,%dl
  sete   %al
  ret
L51:
  xor    %eax,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L16:
  movzbl (%rsi,%rax,1),%ecx
  xor    %eax,%eax
  cmp    %cl,%dl
  sete   %al
  ret
L18:
  movzbl (%rsi),%ecx
  jmp    L14
```

assistant: ```c
int ft_strequ(const char *s1, const char *s2) {
  int i;
  i = 0;
  while (s1[i] != '\0' && s2[i] != '\0') {
    if (s1[i] != s2[i])
      return (0);
    i++;
  }
  if (s1[i] != s2[i])
    return (0);
  else
    return (1);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<ft_strequ>:
  endbr64
  movzbl (%rdi),%edx
  xor    %eax,%eax
  test   %dl,%dl
  jne    L37
  jmp    L41
  nop
L27:
  cmp    %dl,%cl
  jne    L47
  add    $0x1,%rax
  movzbl (%rdi,%rax,1),%edx
  test   %dl,%dl
  je     L48
L37:
  movzbl (%rsi,%rax,1),%ecx
  test   %cl,%cl
  jne    L27
L51:
  xor    %eax,%eax
  cmp    %cl,%dl
  sete   %al
  ret
L47:
  xor    %eax,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L48:
  movzbl (%rsi,%rax,1),%ecx
  xor    %eax,%eax
  cmp    %cl,%dl
  sete   %al
  ret
L41:
  movzbl (%rsi),%ecx
  jmp    L51
```

assistant: ```c
int ft_strequ(const char *s1, const char *s2) {
  int i;
  i = 0;
  while (s1[i] != '\0' && s2[i] != '\0') {
    if (s1[i] != s2[i])
      return (0);
    i++;
  }
  if (s1[i] != s2[i])
    return (0);
  else
    return (1);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<Xorriso_option_list_extras>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  mov    -0x20(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  mov    $0x0,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <Xorriso_list_extras@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int Xorriso_option_list_extras(struct XorrisO *xorriso, char *mode, int flag) {
  int ret;
  ret = Xorriso_list_extras(xorriso, mode, 0);
  return (ret);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<Xorriso_option_list_extras>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%edx
  call   <Xorriso_list_extras@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
int Xorriso_option_list_extras(struct XorrisO *xorriso, char *mode, int flag) {
  int ret;
  ret = Xorriso_list_extras(xorriso, mode, 0);
  return (ret);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<Xorriso_option_list_extras>:
  endbr64
  xor    %edx,%edx
  jmp    1050 <Xorriso_list_extras@plt>
```

assistant: ```c
int Xorriso_option_list_extras(struct XorrisO *xorriso, char *mode, int flag) {
  int ret;
  ret = Xorriso_list_extras(xorriso, mode, 0);
  return (ret);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<Xorriso_option_list_extras>:
  endbr64
  xor    %edx,%edx
  jmp    1050 <Xorriso_list_extras@plt>
```

assistant: ```c
int Xorriso_option_list_extras(struct XorrisO *xorriso, char *mode, int flag) {
  int ret;
  ret = Xorriso_list_extras(xorriso, mode, 0);
  return (ret);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<xrop3_95_0_0>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,%edx
  mov    %esi,%eax
  mov    %dl,-0x14(%rbp)
  mov    %al,-0x18(%rbp)
  movl   $0xff,-0xc(%rbp)
  movzbl -0x18(%rbp),%eax
  shl    $0x8,%eax
  mov    %eax,%edx
  movzbl -0x18(%rbp),%eax
  or     %edx,%eax
  mov    %eax,-0x8(%rbp)
  mov    -0xc(%rbp),%eax
  and    -0x8(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  notl   -0x4(%rbp)
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned xrop3_95_0_0(unsigned char s, unsigned char t) {
  unsigned D = 0x00ff;
  unsigned T = ((unsigned)t << 8) | t;
  unsigned stk1;
  stk1 = D & T;
  stk1 = ~stk1;
  return stk1;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<xrop3_95_0_0>:
  endbr64
  movzbl %sil,%eax
  not    %eax
  ret
```

assistant: ```c
unsigned xrop3_95_0_0(unsigned char s, unsigned char t) {
  unsigned D = 0x00ff;
  unsigned T = ((unsigned)t << 8) | t;
  unsigned stk1;
  stk1 = D & T;
  stk1 = ~stk1;
  return stk1;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<xrop3_95_0_0>:
  endbr64
  movzbl %sil,%eax
  not    %eax
  ret
```

assistant: ```c
unsigned xrop3_95_0_0(unsigned char s, unsigned char t) {
  unsigned D = 0x00ff;
  unsigned T = ((unsigned)t << 8) | t;
  unsigned stk1;
  stk1 = D & T;
  stk1 = ~stk1;
  return stk1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<xrop3_95_0_0>:
  endbr64
  movzbl %sil,%eax
  not    %eax
  ret
```

assistant: ```c
unsigned xrop3_95_0_0(unsigned char s, unsigned char t) {
  unsigned D = 0x00ff;
  unsigned T = ((unsigned)t << 8) | t;
  unsigned stk1;
  stk1 = D & T;
  stk1 = ~stk1;
  return stk1;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<flip>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <random@plt>
  movslq %eax,%rdx
  imul   $0x10624dd3,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x6,%edx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  imul   $0x3e8,%edx,%ecx
  sub    %ecx,%eax
  mov    %eax,%edx
  cmp    $0x1f3,%edx
  jg     L19
  mov    $0x0,%eax
  jmp    L5
L19:
  mov    $0x1,%eax
L5:
  pop    %rbp
  ret
```

assistant: ```c
int flip() {
  if (random() % 1000 < 500)
    return 0;
  return 1;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<flip>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <random@plt>
  movslq %eax,%rdx
  imul   $0x10624dd3,%rdx,%rdx
  sar    $0x26,%rdx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  imul   $0x3e8,%edx,%edx
  sub    %edx,%eax
  cmp    $0x1f3,%eax
  setg   %al
  movzbl %al,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int flip() {
  if (random() % 1000 < 500)
    return 0;
  return 1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<flip>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <random@plt>
  movslq %eax,%rdx
  mov    %eax,%ecx
  imul   $0x10624dd3,%rdx,%rdx
  sar    $0x1f,%ecx
  sar    $0x26,%rdx
  sub    %ecx,%edx
  imul   $0x3e8,%edx,%edx
  sub    %edx,%eax
  cmp    $0x1f3,%eax
  setg   %al
  add    $0x8,%rsp
  movzbl %al,%eax
  ret
```

assistant: ```c
int flip() {
  if (random() % 1000 < 500)
    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
<flip>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <random@plt>
  movslq %eax,%rdx
  mov    %eax,%ecx
  imul   $0x10624dd3,%rdx,%rdx
  sar    $0x1f,%ecx
  sar    $0x26,%rdx
  sub    %ecx,%edx
  imul   $0x3e8,%edx,%edx
  sub    %edx,%eax
  cmp    $0x1f3,%eax
  setg   %al
  add    $0x8,%rsp
  movzbl %al,%eax
  ret
```

assistant: ```c
int flip() {
  if (random() % 1000 < 500)
    return 0;
  return 1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<lowest_of>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  mov    -0x4(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jg     L11
  mov    -0x4(%rbp),%eax
  jmp    L4
L11:
  mov    -0x8(%rbp),%eax
L4:
  pop    %rbp
  ret
```

assistant: ```c
int lowest_of(int arg1, int arg2) {
  if (arg1 <= arg2)
    return arg1;
  else
    return arg2;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<lowest_of>:
  endbr64
  cmp    %esi,%edi
  mov    %esi,%eax
  cmovle %edi,%eax
  ret
```

assistant: ```c
int lowest_of(int arg1, int arg2) {
  if (arg1 <= arg2)
    return arg1;
  else
    return arg2;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<lowest_of>:
  endbr64
  cmp    %esi,%edi
  mov    %esi,%eax
  cmovle %edi,%eax
  ret
```

assistant: ```c
int lowest_of(int arg1, int arg2) {
  if (arg1 <= arg2)
    return arg1;
  else
    return arg2;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<csd2int>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x0,-0x4(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,-0x8(%rbp)
  jmp    L8
L19:
  mov    -0x4(%rbp),%eax
  lea    (%rax,%rax,1),%edx
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x20(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%eax
  add    %edx,%eax
  mov    %eax,-0x4(%rbp)
  subl   $0x1,-0x8(%rbp)
L8:
  cmpl   $0x0,-0x8(%rbp)
  jns    L19
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int csd2int(cprec, code)
int cprec, *code;
{
  int i, a;
  a = 0;
  for (i = cprec; i >= 0; i--)
    a = a * 2 + code[i];
  return (a);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<csd2int>:
  endbr64
  test   %edi,%edi
  js     L1
  movslq %edi,%rdx
  shl    $0x2,%rdx
  lea    (%rsi,%rdx,1),%rax
  lea    -0x4(%rsi,%rdx,1),%rsi
  mov    %edi,%edi
  shl    $0x2,%rdi
  sub    %rdi,%rsi
  mov    $0x0,%edx
L25:
  mov    (%rax),%ecx
  lea    (%rcx,%rdx,2),%edx
  sub    $0x4,%rax
  cmp    %rsi,%rax
  jne    L25
L2:
  mov    %edx,%eax
  ret
L1:
  mov    $0x0,%edx
  jmp    L2
```

assistant: ```c
int csd2int(cprec, code)
int cprec, *code;
{
  int i, a;
  a = 0;
  for (i = cprec; i >= 0; i--)
    a = a * 2 + code[i];
  return (a);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<csd2int>:
  endbr64
  test   %edi,%edi
  js     L9
  movslq %edi,%rdi
  xor    %r8d,%r8d
  lea    (%rsi,%rdi,4),%rax
  sub    $0x4,%rsi
  cs nopw 0x0(%rax,%rax,1)
L1:
  mov    (%rax),%edx
  sub    $0x4,%rax
  lea    (%rdx,%r8,2),%r8d
  cmp    %rsi,%rax
  jne    L1
  mov    %r8d,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L9:
  xor    %r8d,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
int csd2int(cprec, code)
int cprec, *code;
{
  int i, a;
  a = 0;
  for (i = cprec; i >= 0; i--)
    a = a * 2 + code[i];
  return (a);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<csd2int>:
  endbr64
  test   %edi,%edi
  js     L19
  movslq %edi,%rdi
  xor    %r8d,%r8d
  lea    (%rsi,%rdi,4),%rax
  sub    $0x4,%rsi
  cs nopw 0x0(%rax,%rax,1)
L5:
  mov    (%rax),%edx
  sub    $0x4,%rax
  lea    (%rdx,%r8,2),%r8d
  cmp    %rsi,%rax
  jne    L5
  mov    %r8d,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L19:
  xor    %r8d,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
int csd2int(cprec, code)
int cprec, *code;
{
  int i, a;
  a = 0;
  for (i = cprec; i >= 0; i--)
    a = a * 2 + code[i];
  return (a);
}
```