ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    $0x4,%edx
  mov    $0x9,%esi
  xor    %ebp,%ebp
  push   %rbx
  sub    $0x138,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x128(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  lea    0x90(%rsp),%rbx
  call   <fill_roll_counts@plt>
  mov    $0x6,%esi
  mov    $0x6,%edx
  mov    %rbx,%rdi
  call   <fill_roll_counts@plt>
  lea    0x4(%rsp),%rsi
  mov    $0x1,%ecx
  nopl   0x0(%rax,%rax,1)
L11:
  mov    %rsi,%rax
  xor    %edx,%edx
  nopl   (%rax)
L8:
  add    (%rax),%edx
  add    $0x4,%rax
  cmp    %rbx,%rax
  jne    L8
  imul   -0x4(%rax,%rcx,4),%edx
  add    $0x1,%rcx
  add    $0x4,%rsi
  movslq %edx,%rdx
  add    %rdx,%rbp
  cmp    $0x24,%rcx
  jne    L11
  mov    $0x9,%esi
  mov    $0x4,%edi
  call   <power@plt>
  mov    $0x6,%esi
  mov    $0x6,%edi
  mov    %rax,%rbx
  call   <power@plt>
  pxor   %xmm0,%xmm0
  pxor   %xmm1,%xmm1
  lea    D7(%rip),%rdi
  imul   %rax,%rbx
  cvtsi2sd %rbp,%xmm0
  cvtsi2sd %rbx,%xmm1
  divsd  %xmm1,%xmm0
  cvttsd2si %xmm0,%rsi
  call   <printf@plt>
  mov    0x128(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L21
  add    $0x138,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L21:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D7
type_guessed: string

tool: D7: %.7f


assistant: ```c
int main(void) {
  int max_roll_a = 9 * 4;
  int max_roll_b = 6 * 6;
  int roll_counts_a[max_roll_a];
  int roll_counts_b[max_roll_b];
  fill_roll_counts(roll_counts_a, 9, 4);
  fill_roll_counts(roll_counts_b, 6, 6);
  long winning_rolls = 0;
  ;
  for (int i = 0; i < max_roll_b; i++) {
    int sum = 0;
    for (int j = i + 1; j < max_roll_a; j++)
      sum += roll_counts_a[j];
    winning_rolls += roll_counts_b[i] * sum;
  }
  long all_rolls = power(4, 9) * power(6, 6);
  printf("%.7f\n", winning_rolls / (double)all_rolls);
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  push   %rbp
  mov    $0x4,%edx
  mov    $0x9,%esi
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x138,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x128(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  lea    0x90(%rsp),%rbp
  call   <fill_roll_counts@plt>
  mov    $0x6,%esi
  mov    %rbp,%rdi
  mov    $0x6,%edx
  call   <fill_roll_counts@plt>
  lea    0x4(%rsp),%rcx
  mov    $0x1,%edi
  mov    $0x23,%esi
  jmp    L27
  nop
L7:
  movslq %edx,%rsi
  add    (%rsp,%rsi,4),%eax
  lea    0x1(%rdx),%esi
  cmp    $0x23,%edx
  je     L118
  movslq %esi,%rsi
  add    (%rsp,%rsi,4),%eax
  lea    0x2(%rdx),%esi
  cmp    $0x22,%edx
  je     L118
  movslq %esi,%rsi
  add    (%rsp,%rsi,4),%eax
L118:
  imul   -0x4(%rbp,%rdi,4),%eax
  add    $0x4,%rcx
  add    $0x1,%rdi
  cltq
  add    %rax,%rbx
  test   %r8d,%r8d
  je     L83
L89:
  mov    %r8d,%esi
L27:
  lea    -0x1(%rsi),%r8d
  mov    %edi,%edx
  cmp    $0x2,%r8d
  jbe    L0
  mov    %esi,%eax
  movdqu (%rcx),%xmm0
  shr    $0x2,%eax
  cmp    $0x1,%eax
  je     L105
  movdqu 0x10(%rcx),%xmm2
  paddd  %xmm2,%xmm0
  cmp    $0x2,%eax
  je     L105
  movdqu 0x20(%rcx),%xmm3
  paddd  %xmm3,%xmm0
  cmp    $0x3,%eax
  je     L105
  movdqu 0x30(%rcx),%xmm4
  paddd  %xmm4,%xmm0
  cmp    $0x4,%eax
  je     L105
  movdqu 0x40(%rcx),%xmm5
  paddd  %xmm5,%xmm0
  cmp    $0x5,%eax
  je     L105
  movdqu 0x50(%rcx),%xmm6
  paddd  %xmm6,%xmm0
  cmp    $0x6,%eax
  je     L105
  movdqu 0x60(%rcx),%xmm7
  paddd  %xmm7,%xmm0
  cmp    $0x8,%eax
  jne    L105
  movdqu 0x70(%rcx),%xmm7
  paddd  %xmm7,%xmm0
L105:
  movdqa %xmm0,%xmm1
  mov    %esi,%r9d
  psrldq $0x8,%xmm1
  and    $0xfffffffc,%r9d
  paddd  %xmm1,%xmm0
  add    %r9d,%edx
  movdqa %xmm0,%xmm1
  psrldq $0x4,%xmm1
  paddd  %xmm1,%xmm0
  movd   %xmm0,%eax
  cmp    %esi,%r9d
  jne    L7
  imul   -0x4(%rbp,%rdi,4),%eax
  add    $0x4,%rcx
  add    $0x1,%rdi
  cltq
  add    %rax,%rbx
  jmp    L89
L0:
  xor    %eax,%eax
  jmp    L7
  nopw   0x0(%rax,%rax,1)
L83:
  mov    $0x9,%esi
  mov    $0x4,%edi
  call   <power@plt>
  mov    $0x6,%esi
  mov    $0x6,%edi
  mov    %rax,%rbp
  call   <power@plt>
  pxor   %xmm0,%xmm0
  pxor   %xmm1,%xmm1
  lea    D1(%rip),%rdi
  imul   %rax,%rbp
  cvtsi2sd %rbx,%xmm0
  cvtsi2sd %rbp,%xmm1
  divsd  %xmm1,%xmm0
  cvttsd2si %xmm0,%rsi
  call   <printf@plt>
  mov    0x128(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L15
  add    $0x138,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L15:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D1
datatype: string

tool: D1: %.7f


assistant: ```c
int main(void) {
  int max_roll_a = 9 * 4;
  int max_roll_b = 6 * 6;
  int roll_counts_a[max_roll_a];
  int roll_counts_b[max_roll_b];
  fill_roll_counts(roll_counts_a, 9, 4);
  fill_roll_counts(roll_counts_b, 6, 6);
  long winning_rolls = 0;
  ;
  for (int i = 0; i < max_roll_b; i++) {
    int sum = 0;
    for (int j = i + 1; j < max_roll_a; j++)
      sum += roll_counts_a[j];
    winning_rolls += roll_counts_b[i] * sum;
  }
  long all_rolls = power(4, 9) * power(6, 6);
  printf("%.7f\n", winning_rolls / (double)all_rolls);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <curve25519_test@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main(void) {
  curve25519_test();
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <curve25519_test@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  curve25519_test();
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <curve25519_test@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  curve25519_test();
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <curve25519_test@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  curve25519_test();
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<g2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x2a,%al
  je     L5
  mov    $0x0,%eax
  call   <abort@plt>
L5:
  nop
  leave
  ret
```

assistant: ```c
void g2(char *x) {
  if (x[0] != 42)
    abort();
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<g2>:
  endbr64
  cmpb   $0x2a,(%rdi)
  jne    L0
  ret
L0:
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <abort@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void g2(char *x) {
  if (x[0] != 42)
    abort();
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<g2>:
  endbr64
  cmpb   $0x2a,(%rdi)
  jne    L5
  ret
  nopw   0x0(%rax,%rax,1)
L5:
  xor    %eax,%eax
  jmp    1050 <abort@plt>
```

assistant: ```c
void g2(char *x) {
  if (x[0] != 42)
    abort();
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<g2>:
  endbr64
  cmpb   $0x2a,(%rdi)
  jne    L3
  ret
  nopw   0x0(%rax,%rax,1)
L3:
  xor    %eax,%eax
  jmp    1050 <abort@plt>
```

assistant: ```c
void g2(char *x) {
  if (x[0] != 42)
    abort();
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<sanitise>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  jmp    L22
L14:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  mov    %eax,%edi
  call   <isprint@plt>
  test   %rax,%rax
  je     L19
  mov    -0x10(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x10(%rbp)
  mov    -0x8(%rbp),%rdx
  movzbl (%rdx),%edx
  mov    %dl,(%rax)
L19:
  addq   $0x1,-0x8(%rbp)
L22:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L14
  mov    -0x10(%rbp),%rax
  movb   $0x0,(%rax)
  nop
  leave
  ret
```

assistant: ```c
void sanitise(char *buf) {
  unsigned char *q, *r;
  for (q = r = (unsigned char *)buf; *r; r++)
    if (isprint((int)*r))
      *(q++) = *r;
  *q = 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<sanitise>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  movzbl (%rdi),%edi
  test   %dil,%dil
  je     L28
  mov    %rbp,%rbx
  jmp    L25
L47:
  add    $0x1,%rbx
  movzbl (%rbx),%edi
  test   %dil,%dil
  je     L28
L25:
  movzbl %dil,%edi
  call   <isprint@plt>
  test   %rax,%rax
  je     L47
  movzbl (%rbx),%eax
  mov    %al,0x0(%rbp)
  lea    0x1(%rbp),%rbp
  jmp    L47
L28:
  movb   $0x0,0x0(%rbp)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void sanitise(char *buf) {
  unsigned char *q, *r;
  for (q = r = (unsigned char *)buf; *r; r++)
    if (isprint((int)*r))
      *(q++) = *r;
  *q = 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<sanitise>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  movzbl (%rdi),%edi
  test   %dil,%dil
  je     L14
  mov    %rbp,%rbx
  nopl   0x0(%rax,%rax,1)
L11:
  call   <isprint@plt>
  test   %rax,%rax
  je     L29
  movzbl (%rbx),%eax
  add    $0x1,%rbp
  mov    %al,-0x1(%rbp)
L29:
  movzbl 0x1(%rbx),%edi
  add    $0x1,%rbx
  test   %dil,%dil
  jne    L11
L14:
  movb   $0x0,0x0(%rbp)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void sanitise(char *buf) {
  unsigned char *q, *r;
  for (q = r = (unsigned char *)buf; *r; r++)
    if (isprint((int)*r))
      *(q++) = *r;
  *q = 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<sanitise>:
  endbr64
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %rbx,%rbp
  sub    $0x8,%rsp
  movzbl (%rdi),%edi
  test   %dil,%dil
  je     L29
  nopl   0x0(%rax,%rax,1)
L4:
  call   <isprint@plt>
  test   %rax,%rax
  je     L31
  movzbl (%rbx),%edx
  add    $0x1,%rbx
  lea    0x1(%rbp),%rax
  mov    %dl,0x0(%rbp)
  movzbl (%rbx),%edi
  mov    %rax,%rbp
  test   %dil,%dil
  jne    L4
L29:
  movb   $0x0,0x0(%rbp)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  xchg   %ax,%ax
L31:
  movzbl 0x1(%rbx),%edi
  add    $0x1,%rbx
  test   %dil,%dil
  jne    L4
  movb   $0x0,0x0(%rbp)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void sanitise(char *buf) {
  unsigned char *q, *r;
  for (q = r = (unsigned char *)buf; *r; r++)
    if (isprint((int)*r))
      *(q++) = *r;
  *q = 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<atomic_cmpxchg>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    -0x28(%rbp),%rdx
  mov    -0x18(%rbp),%rcx
  mov    -0x20(%rbp),%rax
  lock cmpxchg %rdx,(%rcx)
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
unsigned long atomic_cmpxchg(volatile void *ptr, unsigned long old, unsigned long new) {
  unsigned long prev;
  asm volatile("lock; "
               "cmpxchgq %1,%2"
               : "=a"(prev)
               : "r"(new), "m"(*(volatile long *)ptr), "0"(old)
               : "memory");
  return prev;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<atomic_cmpxchg>:
  endbr64
  mov    %rsi,%rax
  lock cmpxchg %rdx,(%rdi)
  ret
```

assistant: ```c
unsigned long atomic_cmpxchg(volatile void *ptr, unsigned long old, unsigned long new) {
  unsigned long prev;
  asm volatile("lock; "
               "cmpxchgq %1,%2"
               : "=a"(prev)
               : "r"(new), "m"(*(volatile long *)ptr), "0"(old)
               : "memory");
  return prev;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<atomic_cmpxchg>:
  endbr64
  mov    %rsi,%rax
  lock cmpxchg %rdx,(%rdi)
  ret
```

assistant: ```c
unsigned long atomic_cmpxchg(volatile void *ptr, unsigned long old, unsigned long new) {
  unsigned long prev;
  asm volatile("lock; "
               "cmpxchgq %1,%2"
               : "=a"(prev)
               : "r"(new), "m"(*(volatile long *)ptr), "0"(old)
               : "memory");
  return prev;
}
```
user: What is the c source code of the assembly code below?

```asm
<atomic_cmpxchg>:
  endbr64
  mov    %rsi,%rax
  lock cmpxchg %rdx,(%rdi)
  ret
```

assistant: ```c
unsigned long atomic_cmpxchg(volatile void *ptr, unsigned long old, unsigned long new) {
  unsigned long prev;
  asm volatile("lock; "
               "cmpxchgq %1,%2"
               : "=a"(prev)
               : "r"(new), "m"(*(volatile long *)ptr), "0"(old)
               : "memory");
  return prev;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<printLong>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0x0,-0x20(%rbp)
  movq   $0x0,-0x18(%rbp)
  movl   $0x0,-0x10(%rbp)
  movl   $0x0,-0x28(%rbp)
  jmp    L18
L28:
  mov    -0x38(%rbp),%rax
  mov    %eax,%esi
  mov    -0x38(%rbp),%rcx
  movabs $0x6666666666666667,%rdx
  mov    %rcx,%rax
  imul   %rdx
  mov    %rdx,%rax
  sar    $0x2,%rax
  sar    $0x3f,%rcx
  mov    %rcx,%rdx
  sub    %rdx,%rax
  mov    %eax,%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  mov    %eax,%edx
  mov    %esi,%eax
  sub    %edx,%eax
  add    $0x30,%eax
  mov    %eax,%edx
  mov    -0x28(%rbp),%eax
  cltq
  mov    %dl,-0x20(%rbp,%rax,1)
  addl   $0x1,-0x28(%rbp)
  mov    -0x38(%rbp),%rcx
  movabs $0x6666666666666667,%rdx
  mov    %rcx,%rax
  imul   %rdx
  mov    %rdx,%rax
  sar    $0x2,%rax
  sar    $0x3f,%rcx
  mov    %rcx,%rdx
  sub    %rdx,%rax
  mov    %rax,-0x38(%rbp)
L18:
  cmpq   $0x0,-0x38(%rbp)
  jne    L28
  mov    -0x28(%rbp),%eax
  mov    %eax,-0x24(%rbp)
  jmp    L39
L9:
  lea    -0x20(%rbp),%rdx
  mov    -0x24(%rbp),%eax
  cltq
  add    %rdx,%rax
  mov    $0x1,%edx
  mov    %rax,%rsi
  mov    $0x2,%edi
  call   <write@plt>
  subl   $0x1,-0x24(%rbp)
L39:
  cmpl   $0x0,-0x24(%rbp)
  jns    L9
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L19
  call   <__stack_chk_fail@plt>
L19:
  leave
  ret
```

assistant: ```c
void printLong(long number) {
  char output[20] = {0};
  int length = 0;
  while (number != 0) {
    output[length] = (char)((number - (((long)(number / 10)) * 10)) + 48);
    length++;
    number /= 10;
  }
  for (int x = length; x > -1; x--) {
    write(2, &output[x], 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
<printLong>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movq   $0x0,(%rsp)
  movq   $0x0,0x8(%rsp)
  movl   $0x0,0x10(%rsp)
  test   %rdi,%rdi
  je     L40
  mov    %rdi,%rcx
  mov    $0x1,%edi
  movabs $0x6666666666666667,%r9
  lea    -0x1(%rsp),%r8
L16:
  mov    %ecx,%esi
  mov    %rcx,%rax
  imul   %r9
  sar    $0x2,%rdx
  sar    $0x3f,%rcx
  sub    %rcx,%rdx
  mov    %rdx,%rcx
  add    $0x30,%esi
  lea    (%rdx,%rdx,4),%eax
  add    %eax,%eax
  sub    %eax,%esi
  mov    %sil,(%r8,%rdi,1)
  mov    %edi,%eax
  add    $0x1,%rdi
  test   %rdx,%rdx
  jne    L16
  test   %eax,%eax
  js     L1
L20:
  movslq %eax,%rdx
  lea    (%rsp,%rdx,1),%rdx
  mov    %rdx,%rbx
  sub    %edx,%eax
  mov    %eax,%ebp
L58:
  mov    $0x1,%edx
  mov    %rbx,%rsi
  mov    $0x2,%edi
  call   <write@plt>
  sub    $0x1,%rbx
  mov    %ebp,%eax
  add    %ebx,%eax
  jns    L58
L1:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L59
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  ret
L40:
  mov    $0x0,%eax
  jmp    L20
L59:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void printLong(long number) {
  char output[20] = {0};
  int length = 0;
  while (number != 0) {
    output[length] = (char)((number - (((long)(number / 10)) * 10)) + 48);
    length++;
    number /= 10;
  }
  for (int x = length; x > -1; x--) {
    write(2, &output[x], 1);
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<printLong>:
  endbr64
  push   %rbp
  pxor   %xmm0,%xmm0
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movl   $0x0,0x10(%rsp)
  movaps %xmm0,(%rsp)
  test   %rdi,%rdi
  je     L30
  mov    %rdi,%rcx
  lea    -0x1(%rsp),%r8
  mov    $0x1,%edi
  movabs $0x6666666666666667,%r9
  nopw   0x0(%rax,%rax,1)
L14:
  mov    %rcx,%rax
  mov    %ecx,%esi
  imul   %r9
  mov    %rcx,%rax
  add    $0x30,%esi
  sar    $0x3f,%rax
  sar    $0x2,%rdx
  sub    %rax,%rdx
  lea    (%rdx,%rdx,4),%eax
  mov    %rdx,%rcx
  mov    %rdi,%rdx
  add    %eax,%eax
  sub    %eax,%esi
  movslq %edi,%rax
  mov    %sil,(%r8,%rdi,1)
  add    $0x1,%rdi
  test   %rcx,%rcx
  jne    L14
  lea    (%r8,%rax,1),%rbp
  lea    (%rsp,%rax,1),%rbx
  sub    %rdx,%rbp
  nopl   0x0(%rax)
L11:
  mov    %rbx,%rsi
  mov    $0x1,%edx
  mov    $0x2,%edi
  sub    $0x1,%rbx
  call   <write@plt>
  cmp    %rbx,%rbp
  jne    L11
L40:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L32
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopw   0x0(%rax,%rax,1)
L30:
  mov    %rsp,%rsi
  mov    $0x1,%edx
  mov    $0x2,%edi
  call   <write@plt>
  jmp    L40
L32:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void printLong(long number) {
  char output[20] = {0};
  int length = 0;
  while (number != 0) {
    output[length] = (char)((number - (((long)(number / 10)) * 10)) + 48);
    length++;
    number /= 10;
  }
  for (int x = length; x > -1; x--) {
    write(2, &output[x], 1);
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<printLong>:
  endbr64
  push   %rbp
  pxor   %xmm0,%xmm0
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movl   $0x0,0x10(%rsp)
  movaps %xmm0,(%rsp)
  test   %rdi,%rdi
  je     L7
  mov    %rdi,%rcx
  lea    -0x1(%rsp),%r8
  mov    $0x1,%edi
  movabs $0x6666666666666667,%r9
  nopw   0x0(%rax,%rax,1)
L48:
  mov    %rcx,%rax
  mov    %ecx,%esi
  imul   %r9
  mov    %rcx,%rax
  add    $0x30,%esi
  sar    $0x3f,%rax
  sar    $0x2,%rdx
  sub    %rax,%rdx
  lea    (%rdx,%rdx,4),%eax
  mov    %rdx,%rcx
  mov    %rdi,%rdx
  add    %eax,%eax
  sub    %eax,%esi
  movslq %edi,%rax
  mov    %sil,(%r8,%rdi,1)
  add    $0x1,%rdi
  test   %rcx,%rcx
  jne    L48
  lea    (%r8,%rax,1),%rbp
  lea    (%rsp,%rax,1),%rbx
  sub    %rdx,%rbp
  nopl   0x0(%rax)
L29:
  mov    %rbx,%rsi
  mov    $0x1,%edx
  mov    $0x2,%edi
  sub    $0x1,%rbx
  call   <write@plt>
  cmp    %rbx,%rbp
  jne    L29
L36:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L44
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopw   0x0(%rax,%rax,1)
L7:
  mov    %rsp,%rsi
  mov    $0x1,%edx
  mov    $0x2,%edi
  call   <write@plt>
  jmp    L36
L44:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void printLong(long number) {
  char output[20] = {0};
  int length = 0;
  while (number != 0) {
    output[length] = (char)((number - (((long)(number / 10)) * 10)) + 48);
    length++;
    number /= 10;
  }
  for (int x = length; x > -1; x--) {
    write(2, &output[x], 1);
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<get_action_remorseless>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  cmpl   $0x1,-0x8(%rbp)
  jne    L12
  mov    $0x1,%eax
  jmp    L2
L12:
  mov    $0xffffffff,%eax
L2:
  pop    %rbp
  ret
```

assistant: ```c
int get_action_remorseless(int myRep, int hisRep) {
  if (hisRep == 1) {
    return 1;
  } else {
    return -1;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<get_action_remorseless>:
  endbr64
  cmp    $0x1,%esi
  mov    $0xffffffff,%eax
  cmove  %esi,%eax
  ret
```

assistant: ```c
int get_action_remorseless(int myRep, int hisRep) {
  if (hisRep == 1) {
    return 1;
  } else {
    return -1;
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<get_action_remorseless>:
  endbr64
  xor    %eax,%eax
  cmp    $0x1,%esi
  sete   %al
  lea    -0x1(%rax,%rax,1),%eax
  ret
```

assistant: ```c
int get_action_remorseless(int myRep, int hisRep) {
  if (hisRep == 1) {
    return 1;
  } else {
    return -1;
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<get_action_remorseless>:
  endbr64
  xor    %eax,%eax
  cmp    $0x1,%esi
  sete   %al
  lea    -0x1(%rax,%rax,1),%eax
  ret
```

assistant: ```c
int get_action_remorseless(int myRep, int hisRep) {
  if (hisRep == 1) {
    return 1;
  } else {
    return -1;
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<vmpc_encrypt>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    %rcx,-0x40(%rbp)
  movq   $0x0,-0x18(%rbp)
  cmpq   $0x0,-0x28(%rbp)
  jne    L18
  mov    $0x1,%eax
  jmp    L10
L18:
  movq   $0x0,-0x18(%rbp)
  jmp    L39
L17:
  mov    -0x18(%rbp),%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%ebx
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <vmpc_yield@plt>
  mov    %eax,%edx
  mov    -0x18(%rbp),%rax
  lea    0x0(,%rax,4),%rcx
  mov    -0x38(%rbp),%rax
  add    %rcx,%rax
  xor    %ebx,%edx
  mov    %edx,(%rax)
  addq   $0x1,-0x18(%rbp)
L39:
  mov    -0x18(%rbp),%rax
  cmp    -0x40(%rbp),%rax
  jb     L17
  mov    $0x0,%eax
L10:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int vmpc_encrypt(struct vmpc_ctx *context, uint8_t *input, uint8_t *output, size_t len) {
  size_t i;
  i = 0;
  if (context == NULL)
    return (0x00000001);
  for (i = 0; i < len; ++i)
    output[i] = input[i] ^ vmpc_yield(context);
  return (0x00000000);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<vmpc_encrypt>:
  endbr64
  test   %rdi,%rdi
  je     L29
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r12
  mov    %rsi,%r14
  mov    %rdx,%r15
  mov    %rcx,%r13
  test   %rcx,%rcx
  je     L20
  mov    $0x0,%ebx
L27:
  mov    (%r14,%rbx,4),%ebp
  mov    %r12,%rdi
  call   <vmpc_yield@plt>
  xor    %eax,%ebp
  mov    %ebp,(%r15,%rbx,4)
  add    $0x1,%rbx
  cmp    %rbx,%r13
  jne    L27
  mov    $0x0,%eax
L3:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L29:
  mov    $0x1,%eax
  ret
L20:
  mov    $0x0,%eax
  jmp    L3
```

assistant: ```c
int vmpc_encrypt(struct vmpc_ctx *context, uint8_t *input, uint8_t *output, size_t len) {
  size_t i;
  i = 0;
  if (context == NULL)
    return (0x00000001);
  for (i = 0; i < len; ++i)
    output[i] = input[i] ^ vmpc_yield(context);
  return (0x00000000);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<vmpc_encrypt>:
  endbr64
  test   %rdi,%rdi
  je     L10
  push   %r15
  mov    %rdx,%r15
  push   %r14
  mov    %rsi,%r14
  push   %r13
  mov    %rcx,%r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  test   %rcx,%rcx
  je     L23
  nopw   0x0(%rax,%rax,1)
L28:
  mov    (%r14,%rbx,4),%ebp
  mov    %r12,%rdi
  call   <vmpc_yield@plt>
  xor    %eax,%ebp
  mov    %ebp,(%r15,%rbx,4)
  add    $0x1,%rbx
  cmp    %rbx,%r13
  jne    L28
L23:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L10:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int vmpc_encrypt(struct vmpc_ctx *context, uint8_t *input, uint8_t *output, size_t len) {
  size_t i;
  i = 0;
  if (context == NULL)
    return (0x00000001);
  for (i = 0; i < len; ++i)
    output[i] = input[i] ^ vmpc_yield(context);
  return (0x00000000);
}
```