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

```asm
<compymove>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %rdi,-0x48(%rbp)
  mov    %esi,-0x4c(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x4c(%rbp),%edx
  mov    -0x48(%rbp),%rcx
  lea    -0x30(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <bestmove@plt>
  mov    %eax,-0x38(%rbp)
  mov    $0x0,%eax
  call   <rand@plt>
  cltd
  idivl  -0x38(%rbp)
  mov    %edx,%eax
  sub    %eax,-0x38(%rbp)
  movl   $0x0,-0x34(%rbp)
  jmp    L10
L49:
  mov    -0x34(%rbp),%eax
  cltq
  mov    -0x30(%rbp,%rax,4),%eax
  test   %eax,%eax
  je     L26
  subl   $0x1,-0x38(%rbp)
L26:
  addl   $0x1,-0x34(%rbp)
L10:
  cmpl   $0x9,-0x34(%rbp)
  jg     L28
  cmpl   $0x0,-0x38(%rbp)
  jne    L49
L28:
  subl   $0x1,-0x34(%rbp)
  mov    -0x34(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    -0x34(%rbp),%eax
  cltq
  mov    -0x30(%rbp,%rax,4),%eax
  mov    %eax,%esi
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x34(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x48(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x34(%rbp),%eax
  cltq
  mov    -0x30(%rbp,%rax,4),%ecx
  mov    -0x34(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rsi
  mov    -0x48(%rbp),%rax
  add    %rsi,%rax
  sub    %ecx,%edx
  mov    %edx,(%rax)
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L6
  call   <__stack_chk_fail@plt>
L6:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D6
data_type: string

tool: D6: 
Computer takes %d from pile %d leaving...


assistant: ```c
void compymove(int *p, int nimber) {
  int num, c, bests[10];
  num = bestmove(bests, p, nimber);
  num -= rand() % num;
  for (c = 0; c < 10 && num; c++)
    if (bests[c])
      num--;
  c--;
  printf("\nComputer takes %d from pile %d leaving...\n", bests[c], c + 1);
  p[c] -= bests[c];
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<compymove>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %rdi,%rbp
  mov    %esi,%edx
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  mov    %rbp,%rsi
  call   <bestmove@plt>
  mov    %eax,%ebx
  mov    $0x0,%eax
  call   <rand@plt>
  cltd
  idiv   %ebx
  mov    %ebx,%eax
  sub    %edx,%eax
  je     L23
  mov    $0x1,%edx
  mov    %rsp,%r8
L14:
  cmpl   $0x1,-0x4(%r8,%rdx,4)
  adc    $0xffffffff,%eax
  mov    %edx,%edi
  cmp    $0x9,%edx
  setle  %sil
  test   %eax,%eax
  setne  %cl
  add    $0x1,%rdx
  test   %cl,%sil
  jne    L14
L3:
  lea    -0x1(%rdi),%ebx
  movslq %ebx,%rbx
  mov    (%rsp,%rbx,4),%esi
  mov    %edi,%edx
  lea    D6(%rip),%rdi
  call   <printf@plt>
  mov    (%rsp,%rbx,4),%eax
  sub    %eax,0x0(%rbp,%rbx,4)
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L33
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  ret
L23:
  mov    %eax,%edi
  jmp    L3
L33:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D6
type_infered: string

tool: D6: 
Computer takes %d from pile %d leaving...


assistant: ```c
void compymove(int *p, int nimber) {
  int num, c, bests[10];
  num = bestmove(bests, p, nimber);
  num -= rand() % num;
  for (c = 0; c < 10 && num; c++)
    if (bests[c])
      num--;
  c--;
  printf("\nComputer takes %d from pile %d leaving...\n", bests[c], c + 1);
  p[c] -= bests[c];
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<compymove>:
  endbr64
  push   %r12
  mov    %esi,%edx
  mov    %rdi,%r12
  mov    %rdi,%rsi
  push   %rbp
  push   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%rdi
  call   <bestmove@plt>
  mov    %eax,%ebx
  xor    %eax,%eax
  call   <rand@plt>
  cltd
  idiv   %ebx
  mov    %ebx,%eax
  sub    %edx,%eax
  je     L42
  mov    %rbp,%rdx
  xor    %ecx,%ecx
  jmp    L51
  nopl   0x0(%rax,%rax,1)
L28:
  mov    %r8d,%ecx
L51:
  mov    (%rdx),%esi
  lea    0x1(%rcx),%r8d
  cmp    $0x1,%esi
  adc    $0xffffffff,%eax
  add    $0x4,%rdx
  cmp    $0x9,%r8d
  jg     L20
  test   %eax,%eax
  jne    L28
L20:
  movslq %ecx,%rbx
  lea    0x0(,%rbx,4),%rbp
L5:
  mov    %r8d,%edx
  lea    D2(%rip),%rdi
  call   <printf@plt>
  mov    (%rsp,%rbx,4),%eax
  sub    %eax,(%r12,%rbp,1)
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L48
  add    $0x30,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L42:
  mov    -0x4(%rsp),%esi
  xor    %r8d,%r8d
  mov    $0xfffffffffffffffc,%rbp
  mov    $0xffffffffffffffff,%rbx
  jmp    L5
L48:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D2
data_type: string

tool: D2: 
Computer takes %d from pile %d leaving...


assistant: ```c
void compymove(int *p, int nimber) {
  int num, c, bests[10];
  num = bestmove(bests, p, nimber);
  num -= rand() % num;
  for (c = 0; c < 10 && num; c++)
    if (bests[c])
      num--;
  c--;
  printf("\nComputer takes %d from pile %d leaving...\n", bests[c], c + 1);
  p[c] -= bests[c];
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<compymove>:
  endbr64
  push   %r12
  mov    %esi,%edx
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rbp,%rsi
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  call   <bestmove@plt>
  mov    %eax,%ebx
  xor    %eax,%eax
  call   <rand@plt>
  cltd
  idiv   %ebx
  mov    %ebx,%eax
  sub    %edx,%eax
  je     L196
  mov    (%rsp),%esi
  test   %esi,%esi
  je     L98
  mov    $0x1,%edx
  sub    $0x1,%eax
  je     L119
L98:
  mov    0x4(%rsp),%esi
  test   %esi,%esi
  je     L178
  sub    $0x1,%eax
  je     L277
L178:
  mov    0x8(%rsp),%esi
  test   %esi,%esi
  je     L144
  sub    $0x1,%eax
  je     L157
L144:
  mov    0xc(%rsp),%esi
  test   %esi,%esi
  je     L256
  sub    $0x1,%eax
  je     L205
L256:
  mov    0x10(%rsp),%esi
  test   %esi,%esi
  jne    L71
L251:
  mov    0x14(%rsp),%esi
  test   %esi,%esi
  je     L162
  sub    $0x1,%eax
  je     L56
L162:
  mov    0x18(%rsp),%esi
  test   %esi,%esi
  je     L281
  sub    $0x1,%eax
  je     L19
L281:
  mov    0x1c(%rsp),%esi
  test   %esi,%esi
  je     L131
  sub    $0x1,%eax
  je     L94
L131:
  mov    0x20(%rsp),%esi
  test   %esi,%esi
  jne    L239
L218:
  mov    0x24(%rsp),%esi
  mov    $0x9,%eax
  mov    $0xa,%edx
L119:
  movslq %eax,%rbx
  lea    0x0(,%rbx,4),%r12
L184:
  lea    D0(%rip),%rdi
  call   <printf@plt>
  mov    (%rsp,%rbx,4),%eax
  sub    %eax,0x0(%rbp,%r12,1)
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L261
  add    $0x30,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L71:
  sub    $0x1,%eax
  jne    L251
  mov    $0x4,%eax
  mov    $0x5,%edx
  jmp    L119
  nopl   (%rax)
L239:
  cmp    $0x1,%eax
  jne    L218
  mov    $0x8,%eax
  mov    $0x9,%edx
  jmp    L119
L277:
  mov    $0x1,%eax
  mov    $0x2,%edx
  jmp    L119
L157:
  mov    $0x2,%eax
  mov    $0x3,%edx
  jmp    L119
L205:
  mov    $0x3,%eax
  mov    $0x4,%edx
  jmp    L119
L56:
  mov    $0x5,%eax
  mov    $0x6,%edx
  jmp    L119
  nopw   0x0(%rax,%rax,1)
L196:
  mov    -0x4(%rsp),%esi
  xor    %edx,%edx
  mov    $0xfffffffffffffffc,%r12
  mov    $0xffffffffffffffff,%rbx
  jmp    L184
L19:
  mov    $0x6,%eax
  mov    $0x7,%edx
  jmp    L119
L94:
  mov    $0x7,%eax
  mov    $0x8,%edx
  jmp    L119
L261:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D0
data_type: string

tool: D0: 
Computer takes %d from pile %d leaving...


assistant: ```c
void compymove(int *p, int nimber) {
  int num, c, bests[10];
  num = bestmove(bests, p, nimber);
  num -= rand() % num;
  for (c = 0; c < 10 && num; c++)
    if (bests[c])
      num--;
  c--;
  printf("\nComputer takes %d from pile %d leaving...\n", bests[c], c + 1);
  p[c] -= bests[c];
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    $0x0,%eax
  call   <createCirListDeque@plt>
  mov    %rax,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    $0x2,%esi
  mov    %rax,%rdi
  call   <addFrontCirListDeque@plt>
  mov    -0x18(%rbp),%rax
  mov    $0x3,%esi
  mov    %rax,%rdi
  call   <addBackCirListDeque@plt>
  mov    -0x18(%rbp),%rax
  mov    $0x1,%esi
  mov    %rax,%rdi
  call   <addFrontCirListDeque@plt>
  mov    -0x18(%rbp),%rax
  mov    $0x4,%esi
  mov    %rax,%rdi
  call   <addBackCirListDeque@plt>
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <printCirListDeque@plt>
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <reverseCirListDeque@plt>
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <printCirListDeque@plt>
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <removeFrontCirListDeque@plt>
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <removeBackCirListDeque@plt>
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <printCirListDeque@plt>
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <frontCirListDeque@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <backCirListDeque@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x8(%rbp)
  movsd  -0x8(%rbp),%xmm0
  mov    -0x10(%rbp),%rax
  movapd %xmm0,%xmm1
  movq   %rax,%xmm0
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <deleteCirListDeque@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D0
type: string

tool: D0: front: %lf	back: %lf


assistant: ```c
int main(int argc, char *argv[]) {
  struct cirListDeque *q = createCirListDeque();
  addFrontCirListDeque(q, 2);
  addBackCirListDeque(q, 3);
  addFrontCirListDeque(q, 1);
  addBackCirListDeque(q, 4);
  printCirListDeque(q);
  reverseCirListDeque(q);
  printCirListDeque(q);
  removeFrontCirListDeque(q);
  removeBackCirListDeque(q);
  printCirListDeque(q);
  double i = frontCirListDeque(q);
  double j = backCirListDeque(q);
  printf("front: %lf\tback: %lf\n", i, j);
  deleteCirListDeque(q);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <createCirListDeque@plt>
  mov    %rax,%rbx
  mov    $0x2,%esi
  mov    %rax,%rdi
  call   <addFrontCirListDeque@plt>
  mov    $0x3,%esi
  mov    %rbx,%rdi
  call   <addBackCirListDeque@plt>
  mov    $0x1,%esi
  mov    %rbx,%rdi
  call   <addFrontCirListDeque@plt>
  mov    $0x4,%esi
  mov    %rbx,%rdi
  call   <addBackCirListDeque@plt>
  mov    %rbx,%rdi
  call   <printCirListDeque@plt>
  mov    %rbx,%rdi
  call   <reverseCirListDeque@plt>
  mov    %rbx,%rdi
  call   <printCirListDeque@plt>
  mov    %rbx,%rdi
  call   <removeFrontCirListDeque@plt>
  mov    %rbx,%rdi
  call   <removeBackCirListDeque@plt>
  mov    %rbx,%rdi
  call   <printCirListDeque@plt>
  mov    %rbx,%rdi
  call   <frontCirListDeque@plt>
  movq   %xmm0,%rbp
  mov    %rbx,%rdi
  call   <backCirListDeque@plt>
  movapd %xmm0,%xmm1
  movq   %rbp,%xmm0
  lea    D8(%rip),%rdi
  call   <printf@plt>
  mov    %rbx,%rdi
  call   <deleteCirListDeque@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D8
datatype: string

tool: D8: front: %lf	back: %lf


assistant: ```c
int main(int argc, char *argv[]) {
  struct cirListDeque *q = createCirListDeque();
  addFrontCirListDeque(q, 2);
  addBackCirListDeque(q, 3);
  addFrontCirListDeque(q, 1);
  addBackCirListDeque(q, 4);
  printCirListDeque(q);
  reverseCirListDeque(q);
  printCirListDeque(q);
  removeFrontCirListDeque(q);
  removeBackCirListDeque(q);
  printCirListDeque(q);
  double i = frontCirListDeque(q);
  double j = backCirListDeque(q);
  printf("front: %lf\tback: %lf\n", i, j);
  deleteCirListDeque(q);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbp
  xor    %eax,%eax
  sub    $0x10,%rsp
  call   <createCirListDeque@plt>
  mov    $0x2,%esi
  mov    %rax,%rbp
  mov    %rax,%rdi
  call   <addFrontCirListDeque@plt>
  mov    %rbp,%rdi
  mov    $0x3,%esi
  call   <addBackCirListDeque@plt>
  mov    %rbp,%rdi
  mov    $0x1,%esi
  call   <addFrontCirListDeque@plt>
  mov    $0x4,%esi
  mov    %rbp,%rdi
  call   <addBackCirListDeque@plt>
  mov    %rbp,%rdi
  call   <printCirListDeque@plt>
  mov    %rbp,%rdi
  call   <reverseCirListDeque@plt>
  mov    %rbp,%rdi
  call   <printCirListDeque@plt>
  mov    %rbp,%rdi
  call   <removeFrontCirListDeque@plt>
  mov    %rbp,%rdi
  call   <removeBackCirListDeque@plt>
  mov    %rbp,%rdi
  call   <printCirListDeque@plt>
  mov    %rbp,%rdi
  call   <frontCirListDeque@plt>
  mov    %rbp,%rdi
  movsd  %xmm0,0x8(%rsp)
  call   <backCirListDeque@plt>
  movsd  0x8(%rsp),%xmm2
  lea    D2(%rip),%rdi
  movapd %xmm0,%xmm1
  movapd %xmm2,%xmm0
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <deleteCirListDeque@plt>
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: label: D2
data_type: string

tool: D2: front: %lf	back: %lf


assistant: ```c
int main(int argc, char *argv[]) {
  struct cirListDeque *q = createCirListDeque();
  addFrontCirListDeque(q, 2);
  addBackCirListDeque(q, 3);
  addFrontCirListDeque(q, 1);
  addBackCirListDeque(q, 4);
  printCirListDeque(q);
  reverseCirListDeque(q);
  printCirListDeque(q);
  removeFrontCirListDeque(q);
  removeBackCirListDeque(q);
  printCirListDeque(q);
  double i = frontCirListDeque(q);
  double j = backCirListDeque(q);
  printf("front: %lf\tback: %lf\n", i, j);
  deleteCirListDeque(q);
  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
  xor    %eax,%eax
  sub    $0x10,%rsp
  call   <createCirListDeque@plt>
  mov    $0x2,%esi
  mov    %rax,%rbp
  mov    %rax,%rdi
  call   <addFrontCirListDeque@plt>
  mov    %rbp,%rdi
  mov    $0x3,%esi
  call   <addBackCirListDeque@plt>
  mov    %rbp,%rdi
  mov    $0x1,%esi
  call   <addFrontCirListDeque@plt>
  mov    $0x4,%esi
  mov    %rbp,%rdi
  call   <addBackCirListDeque@plt>
  mov    %rbp,%rdi
  call   <printCirListDeque@plt>
  mov    %rbp,%rdi
  call   <reverseCirListDeque@plt>
  mov    %rbp,%rdi
  call   <printCirListDeque@plt>
  mov    %rbp,%rdi
  call   <removeFrontCirListDeque@plt>
  mov    %rbp,%rdi
  call   <removeBackCirListDeque@plt>
  mov    %rbp,%rdi
  call   <printCirListDeque@plt>
  mov    %rbp,%rdi
  call   <frontCirListDeque@plt>
  mov    %rbp,%rdi
  movsd  %xmm0,0x8(%rsp)
  call   <backCirListDeque@plt>
  movsd  0x8(%rsp),%xmm2
  lea    D0(%rip),%rdi
  movapd %xmm0,%xmm1
  movapd %xmm2,%xmm0
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <deleteCirListDeque@plt>
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: label: D0
type: string

tool: D0: front: %lf	back: %lf


assistant: ```c
int main(int argc, char *argv[]) {
  struct cirListDeque *q = createCirListDeque();
  addFrontCirListDeque(q, 2);
  addBackCirListDeque(q, 3);
  addFrontCirListDeque(q, 1);
  addBackCirListDeque(q, 4);
  printCirListDeque(q);
  reverseCirListDeque(q);
  printCirListDeque(q);
  removeFrontCirListDeque(q);
  removeBackCirListDeque(q);
  printCirListDeque(q);
  double i = frontCirListDeque(q);
  double j = backCirListDeque(q);
  printf("front: %lf\tback: %lf\n", i, j);
  deleteCirListDeque(q);
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<debugx32_directive>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void debugx32_directive(const char *directive, const char *params) {
  (void)directive;
  (void)params;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<debugx32_directive>:
  endbr64
  ret
```

assistant: ```c
void debugx32_directive(const char *directive, const char *params) {
  (void)directive;
  (void)params;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<debugx32_directive>:
  endbr64
  ret
```

assistant: ```c
void debugx32_directive(const char *directive, const char *params) {
  (void)directive;
  (void)params;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<debugx32_directive>:
  endbr64
  ret
```

assistant: ```c
void debugx32_directive(const char *directive, const char *params) {
  (void)directive;
  (void)params;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<whitespace>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x8(%rbp)
L26:
  mov    -0x10(%rbp),%rax
  cmp    -0x8(%rbp),%rax
  jne    L24
  mov    $0x1,%eax
  jmp    L14
L24:
  mov    -0x10(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x10(%rbp)
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <xmlc_white_space@plt>
  test   %eax,%eax
  jne    L26
  mov    $0x0,%eax
L14:
  leave
  ret
```

assistant: ```c
int whitespace(char *text, int n_txt) {
  char *s = text, *end = text + n_txt;
  for (;;) {
    if (s == end)
      return 1;
    if (!xmlc_white_space(*(s++)))
      return 0;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<whitespace>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  movslq %esi,%rbp
  add    %rdi,%rbp
L26:
  cmp    %rbp,%rbx
  je     L1
  add    $0x1,%rbx
  movsbl -0x1(%rbx),%edi
  call   <xmlc_white_space@plt>
  test   %eax,%eax
  jne    L26
  jmp    L19
L1:
  mov    $0x1,%eax
L19:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int whitespace(char *text, int n_txt) {
  char *s = text, *end = text + n_txt;
  for (;;) {
    if (s == end)
      return 1;
    if (!xmlc_white_space(*(s++)))
      return 0;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<whitespace>:
  endbr64
  push   %rbp
  movslq %esi,%rbp
  push   %rbx
  add    %rdi,%rbp
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  jmp    L15
  nopl   (%rax)
L14:
  movsbl (%rbx),%edi
  add    $0x1,%rbx
  call   <xmlc_white_space@plt>
  test   %eax,%eax
  je     L23
L15:
  cmp    %rbp,%rbx
  jne    L14
  mov    $0x1,%eax
L23:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int whitespace(char *text, int n_txt) {
  char *s = text, *end = text + n_txt;
  for (;;) {
    if (s == end)
      return 1;
    if (!xmlc_white_space(*(s++)))
      return 0;
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<whitespace>:
  endbr64
  push   %rbp
  movslq %esi,%rbp
  push   %rbx
  add    %rdi,%rbp
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  jmp    L11
  nopl   (%rax)
L6:
  movsbl (%rbx),%edi
  add    $0x1,%rbx
  call   <xmlc_white_space@plt>
  test   %eax,%eax
  je     L28
L11:
  cmp    %rbp,%rbx
  jne    L6
  mov    $0x1,%eax
L28:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int whitespace(char *text, int n_txt) {
  char *s = text, *end = text + n_txt;
  for (;;) {
    if (s == end)
      return 1;
    if (!xmlc_white_space(*(s++)))
      return 0;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<hash_update>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L5
L3:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  xor    %eax,-0x10(%rbp)
  mov    -0x10(%rbp),%eax
  imul   $0x1000193,%eax,%eax
  mov    %eax,-0x10(%rbp)
  addl   $0x1,-0xc(%rbp)
L5:
  mov    -0xc(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jl     L3
  mov    -0x18(%rbp),%rax
  mov    -0x10(%rbp),%edx
  mov    %edx,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void hash_update(void *hash, void *key, int len) {
  uint32_t seed = *(uint32_t *)hash;
  uint8_t *data = (uint8_t *)key;
  int i;
  for (i = 0; i < len; ++i) {
    seed ^= data[i];
    seed *= 16777619;
  }
  *(uint32_t *)hash = seed;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<hash_update>:
  endbr64
  mov    (%rdi),%eax
  test   %edx,%edx
  jle    L6
  mov    %rsi,%rcx
  lea    -0x1(%rdx),%edx
  lea    0x4(%rsi,%rdx,4),%rdx
L17:
  xor    (%rcx),%eax
  imul   $0x1000193,%eax,%eax
  add    $0x4,%rcx
  cmp    %rdx,%rcx
  jne    L17
L6:
  mov    %eax,(%rdi)
  ret
```

assistant: ```c
void hash_update(void *hash, void *key, int len) {
  uint32_t seed = *(uint32_t *)hash;
  uint8_t *data = (uint8_t *)key;
  int i;
  for (i = 0; i < len; ++i) {
    seed ^= data[i];
    seed *= 16777619;
  }
  *(uint32_t *)hash = seed;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<hash_update>:
  endbr64
  mov    (%rdi),%eax
  test   %edx,%edx
  jle    L9
  sub    $0x1,%edx
  lea    0x4(%rsi,%rdx,4),%rdx
  nopw   0x0(%rax,%rax,1)
L2:
  xor    (%rsi),%eax
  add    $0x4,%rsi
  imul   $0x1000193,%eax,%eax
  cmp    %rdx,%rsi
  jne    L2
L9:
  mov    %eax,(%rdi)
  ret
```

assistant: ```c
void hash_update(void *hash, void *key, int len) {
  uint32_t seed = *(uint32_t *)hash;
  uint8_t *data = (uint8_t *)key;
  int i;
  for (i = 0; i < len; ++i) {
    seed ^= data[i];
    seed *= 16777619;
  }
  *(uint32_t *)hash = seed;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<hash_update>:
  endbr64
  mov    (%rdi),%eax
  test   %edx,%edx
  jle    L14
  sub    $0x1,%edx
  lea    0x4(%rsi,%rdx,4),%rdx
  nopw   0x0(%rax,%rax,1)
L16:
  xor    (%rsi),%eax
  add    $0x4,%rsi
  imul   $0x1000193,%eax,%eax
  cmp    %rsi,%rdx
  jne    L16
L14:
  mov    %eax,(%rdi)
  ret
```

assistant: ```c
void hash_update(void *hash, void *key, int len) {
  uint32_t seed = *(uint32_t *)hash;
  uint8_t *data = (uint8_t *)key;
  int i;
  for (i = 0; i < len; ++i) {
    seed ^= data[i];
    seed *= 16777619;
  }
  *(uint32_t *)hash = seed;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<adler32>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x1,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L11
L18:
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  movzbl (%rax),%eax
  movzbl %al,%edx
  mov    -0x8(%rbp),%eax
  add    %edx,%eax
  mov    %eax,%ecx
  mov    $0x80078071,%edx
  imul   %rcx,%rdx
  shr    $0x20,%rdx
  shr    $0xf,%edx
  mov    %edx,-0x8(%rbp)
  mov    -0x8(%rbp),%edx
  imul   $0xfff1,%edx,%edx
  sub    %edx,%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x4(%rbp),%edx
  mov    -0x8(%rbp),%eax
  add    %edx,%eax
  mov    %eax,%ecx
  mov    $0x80078071,%edx
  imul   %rcx,%rdx
  shr    $0x20,%rdx
  shr    $0xf,%edx
  mov    %edx,-0x4(%rbp)
  mov    -0x4(%rbp),%edx
  imul   $0xfff1,%edx,%edx
  sub    %edx,%eax
  mov    %eax,-0x4(%rbp)
  subq   $0x1,-0x20(%rbp)
L11:
  cmpq   $0x0,-0x20(%rbp)
  jne    L18
  mov    -0x4(%rbp),%eax
  shl    $0x10,%eax
  or     -0x8(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned int adler32(unsigned char *data, size_t len) {
  unsigned int a = 1, b = 0;
  while (len != 0) {
    a = (a + *data++) % 65521;
    b = (b + a) % 65521;
    len--;
  }
  return (b << 16) | a;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<adler32>:
  endbr64
  test   %rsi,%rsi
  je     L19
  add    %rdi,%rsi
  mov    $0x0,%eax
  mov    $0x1,%edx
  mov    $0x80078071,%r8d
L25:
  add    $0x1,%rdi
  movzbl -0x1(%rdi),%ecx
  add    %edx,%ecx
  mov    %ecx,%edx
  imul   %r8,%rdx
  shr    $0x2f,%rdx
  imul   $0xfff1,%edx,%r9d
  sub    %r9d,%ecx
  mov    %ecx,%edx
  lea    (%rcx,%rax,1),%ecx
  mov    %ecx,%eax
  imul   %r8,%rax
  shr    $0x2f,%rax
  imul   $0xfff1,%eax,%r9d
  mov    %ecx,%eax
  sub    %r9d,%eax
  cmp    %rdi,%rsi
  jne    L25
L26:
  shl    $0x10,%eax
  or     %edx,%eax
  ret
L19:
  mov    $0x0,%eax
  mov    $0x1,%edx
  jmp    L26
```

assistant: ```c
unsigned int adler32(unsigned char *data, size_t len) {
  unsigned int a = 1, b = 0;
  while (len != 0) {
    a = (a + *data++) % 65521;
    b = (b + a) % 65521;
    len--;
  }
  return (b << 16) | a;
}
```