ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %r15
  xor    %edi,%edi
  lea    D16(%rip),%r15
  push   %r14
  lea    D26(%rip),%r14
  push   %r13
  lea    D25(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  xor    %eax,%eax
  call   <print_version@plt>
  movl   $0x3,0x8(%rsp)
  movl   $0x1,0xc(%rsp)
  nopl   0x0(%rax)
L12:
  mov    0xc(%rsp),%ebp
  mov    0x8(%rsp),%eax
  mov    $0xa,%edi
  mov    %ebp,%esi
  lea    -0x3(%rax),%r12d
  call   <pow@plt>
  lea    -0x1(%rbp),%esi
  mov    $0xa,%edi
  mov    %eax,%ebx
  call   <pow@plt>
  mov    %eax,%r8d
  mov    $0xf4240,%eax
  cltd
  idiv   %r8d
  mov    %eax,%ebp
L15:
  mov    %ebx,%edx
  mov    %r15,%rsi
  mov    %r14,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %ebx,%esi
  mov    %ebp,%edi
  add    $0x1,%r12d
  call   <benchmark@plt>
  mov    %ebp,%edi
  call   <print_results@plt>
  xor    %eax,%eax
  mov    %r12d,%esi
  mov    %r13,%rdi
  call   <printf@plt>
  cmp    0x8(%rsp),%r12d
  jne    L15
  lea    0x3(%r12),%eax
  addl   $0x1,0xc(%rsp)
  mov    %eax,0x8(%rsp)
  cmp    $0x12,%r12d
  jne    L12
  mov    $0x12,%esi
  mov    $0x12,%edi
  call   <print_summary@plt>
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D16
type_infered: string
name: unpack_data
arguments: data_label: D26
type_infered: string
name: unpack_data
arguments: data_label: D25
type_infered: string

tool: D16: dnanmeanwd

tool: D26: # c::%s:len=%d


tool: D25: ok %d benchmark finished


assistant: ```c
int main(void) {
  double elapsed;
  int count;
  int iter;
  int len;
  int i;
  int j;
  srand(time(NULL));
  print_version();
  count = 0;
  for (i = 1; i <= 6; i++) {
    len = pow(10, i);
    iter = 1000000 / pow(10, i - 1);
    for (j = 0; j < 3; j++) {
      count += 1;
      printf("# c::%s:len=%d\n", "dnanmeanwd", len);
      elapsed = benchmark(iter, len);
      print_results(iter, elapsed);
      printf("ok %d benchmark finished\n", count);
    }
  }
  print_summary(count, count);
}
```
user: What is the c source code of the assembly code below:

```asm
<PrintStringReverse>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L6
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L22
  mov    -0x8(%rbp),%rax
  add    $0x1,%rax
  mov    %rax,%rdi
  call   <PrintStringReverse>
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%esi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L4
L6:
  nop
  jmp    L4
L22:
  nop
L4:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: tag: D7
type_guessed: string

tool: D7: %c

assistant: ```c
void PrintStringReverse(char *sz) {
  if (sz == NULL)
    return;
  if (*sz == 0)
    return;
  PrintStringReverse(sz + 1);
  printf("%c", *sz);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<PrintStringReverse>:
  endbr64
  test   %rdi,%rdi
  je     L19
  push   %rbx
  mov    %rdi,%rbx
  cmpb   $0x0,(%rdi)
  jne    L29
L0:
  pop    %rbx
  ret
L29:
  lea    0x1(%rdi),%rdi
  call   <PrintStringReverse>
  movsbl (%rbx),%esi
  lea    D9(%rip),%rdi
  call   <printf@plt>
  jmp    L0
L19:
  ret
```

tool_calls: name: unpack_data
arguments: tag: D9
data_type: string

tool: D9: %c

assistant: ```c
void PrintStringReverse(char *sz) {
  if (sz == NULL)
    return;
  if (*sz == 0)
    return;
  PrintStringReverse(sz + 1);
  printf("%c", *sz);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<PrintStringReverse>:
  endbr64
  test   %rdi,%rdi
  je     L10
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  movsbl (%rdi),%esi
  test   %sil,%sil
  jne    L9
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax)
L9:
  movzbl 0x1(%rdi),%eax
  lea    D0(%rip),%rbp
  test   %al,%al
  jne    L12
  add    $0x8,%rsp
  mov    %rbp,%rdi
  pop    %rbx
  pop    %rbp
  jmp    1050 <printf@plt>
  nopl   (%rax)
L12:
  cmpb   $0x0,0x2(%rdi)
  je     L28
  lea    0x3(%rdi),%rdi
  call   <PrintStringReverse>
  movsbl 0x2(%rbx),%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  movzbl 0x1(%rbx),%eax
L28:
  mov    %rbp,%rdi
  movsbl %al,%esi
  call   <printf@plt>
  movsbl (%rbx),%esi
  add    $0x8,%rsp
  mov    %rbp,%rdi
  pop    %rbx
  pop    %rbp
  jmp    1050 <printf@plt>
  nopl   0x0(%rax,%rax,1)
L10:
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D0
datatype: string

tool: D0: %c

assistant: ```c
void PrintStringReverse(char *sz) {
  if (sz == NULL)
    return;
  if (*sz == 0)
    return;
  PrintStringReverse(sz + 1);
  printf("%c", *sz);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<PrintStringReverse>:
  endbr64
  test   %rdi,%rdi
  je     L9
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  movsbl (%rdi),%esi
  test   %sil,%sil
  jne    L29
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax)
L29:
  movzbl 0x1(%rdi),%eax
  lea    D6(%rip),%rbp
  test   %al,%al
  jne    L5
  add    $0x8,%rsp
  mov    %rbp,%rdi
  pop    %rbx
  pop    %rbp
  jmp    1050 <printf@plt>
  nopl   (%rax)
L5:
  cmpb   $0x0,0x2(%rdi)
  je     L14
  lea    0x3(%rdi),%rdi
  call   <PrintStringReverse>
  movsbl 0x2(%rbx),%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  movzbl 0x1(%rbx),%eax
L14:
  mov    %rbp,%rdi
  movsbl %al,%esi
  call   <printf@plt>
  movsbl (%rbx),%esi
  add    $0x8,%rsp
  mov    %rbp,%rdi
  pop    %rbx
  pop    %rbp
  jmp    1050 <printf@plt>
  nopl   0x0(%rax,%rax,1)
L9:
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D6
type_guessed: string

tool: D6: %c

assistant: ```c
void PrintStringReverse(char *sz) {
  if (sz == NULL)
    return;
  if (*sz == 0)
    return;
  PrintStringReverse(sz + 1);
  printf("%c", *sz);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    $0xf4240,%edi
  call   <alloc_init@plt>
  mov    $0x0,%eax
  call   <alloc_printstat@plt>
  mov    $0x0,%esi
  mov    $0x3,%edi
  call   <recursive_test@plt>
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  alloc_init(1 * 1000 * 1000);
  alloc_printstat();
  recursive_test(3, 0);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0xf4240,%edi
  call   <alloc_init@plt>
  mov    $0x0,%eax
  call   <alloc_printstat@plt>
  mov    $0x0,%esi
  mov    $0x3,%edi
  call   <recursive_test@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  alloc_init(1 * 1000 * 1000);
  alloc_printstat();
  recursive_test(3, 0);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0xf4240,%edi
  call   <alloc_init@plt>
  xor    %eax,%eax
  call   <alloc_printstat@plt>
  xor    %esi,%esi
  mov    $0x3,%edi
  call   <recursive_test@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  alloc_init(1 * 1000 * 1000);
  alloc_printstat();
  recursive_test(3, 0);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0xf4240,%edi
  call   <alloc_init@plt>
  xor    %eax,%eax
  call   <alloc_printstat@plt>
  xor    %esi,%esi
  mov    $0x3,%edi
  call   <recursive_test@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  alloc_init(1 * 1000 * 1000);
  alloc_printstat();
  recursive_test(3, 0);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<synth_ntom_mono2stereo>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    -0x28(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x28(%rbp),%rcx
  mov    -0x20(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <synth_ntom@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x8(%rbp),%eax
  cltq
  add    %rax,-0x20(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L13
L0:
  mov    -0x20(%rbp),%rax
  lea    0x2(%rax),%rdx
  mov    -0x20(%rbp),%rax
  movzwl (%rax),%eax
  mov    %ax,(%rdx)
  addq   $0x4,-0x20(%rbp)
  addl   $0x1,-0xc(%rbp)
L13:
  mov    -0x28(%rbp),%rax
  mov    (%rax),%eax
  sub    -0x8(%rbp),%eax
  sar    $0x2,%eax
  cmp    %eax,-0xc(%rbp)
  jl     L0
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int synth_ntom_mono2stereo(double *bandPtr, unsigned char *samples, int *pnt) {
  int i, ret;
  int pnt1 = *pnt;
  ret = synth_ntom(bandPtr, 0, samples, pnt);
  samples += pnt1;
  for (i = 0; i < ((*pnt - pnt1) >> 2); i++) {
    ((short *)samples)[1] = ((short *)samples)[0];
    samples += 4;
  }
  return ret;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<synth_ntom_mono2stereo>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rsi,%r12
  mov    %rdx,%rbx
  mov    (%rdx),%ebp
  mov    %rdx,%rcx
  mov    %rsi,%rdx
  mov    $0x0,%esi
  call   <synth_ntom@plt>
  movslq %ebp,%rdx
  lea    (%r12,%rdx,1),%rsi
  mov    (%rbx),%edx
  sub    %ebp,%edx
  cmp    $0x3,%edx
  jle    L3
  mov    $0x0,%ecx
L12:
  movzwl (%rsi,%rcx,4),%edx
  mov    %dx,0x2(%rsi,%rcx,4)
  add    $0x1,%rcx
  mov    (%rbx),%edx
  sub    %ebp,%edx
  sar    $0x2,%edx
  cmp    %ecx,%edx
  jg     L12
L3:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int synth_ntom_mono2stereo(double *bandPtr, unsigned char *samples, int *pnt) {
  int i, ret;
  int pnt1 = *pnt;
  ret = synth_ntom(bandPtr, 0, samples, pnt);
  samples += pnt1;
  for (i = 0; i < ((*pnt - pnt1) >> 2); i++) {
    ((short *)samples)[1] = ((short *)samples)[0];
    samples += 4;
  }
  return ret;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<synth_ntom_mono2stereo>:
  endbr64
  push   %r12
  mov    %rdx,%rcx
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  mov    (%rdx),%r12d
  mov    %rsi,%rbx
  mov    %rsi,%rdx
  xor    %esi,%esi
  call   <synth_ntom@plt>
  movslq %r12d,%rdx
  lea    (%rbx,%rdx,1),%rsi
  mov    0x0(%rbp),%edx
  sub    %r12d,%edx
  sar    $0x2,%edx
  test   %edx,%edx
  jle    L17
  movslq %edx,%rdx
  lea    (%rsi,%rdx,4),%rcx
  nopl   0x0(%rax)
L18:
  movzwl (%rsi),%edx
  add    $0x4,%rsi
  mov    %dx,-0x2(%rsi)
  cmp    %rsi,%rcx
  jne    L18
L17:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int synth_ntom_mono2stereo(double *bandPtr, unsigned char *samples, int *pnt) {
  int i, ret;
  int pnt1 = *pnt;
  ret = synth_ntom(bandPtr, 0, samples, pnt);
  samples += pnt1;
  for (i = 0; i < ((*pnt - pnt1) >> 2); i++) {
    ((short *)samples)[1] = ((short *)samples)[0];
    samples += 4;
  }
  return ret;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<synth_ntom_mono2stereo>:
  endbr64
  push   %r12
  mov    %rdx,%rcx
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdx,%rbx
  mov    (%rdx),%r12d
  mov    %rsi,%rdx
  xor    %esi,%esi
  call   <synth_ntom@plt>
  mov    (%rbx),%edi
  mov    %eax,%r9d
  movslq %r12d,%rax
  sub    %r12d,%edi
  lea    0x0(%rbp,%rax,1),%rcx
  sar    $0x2,%edi
  test   %edi,%edi
  jle    L104
  lea    -0x1(%rdi),%esi
  cmp    $0x7,%esi
  jbe    L24
  mov    %esi,%r8d
  mov    %rcx,%rdx
  lea    0x2(%rbp,%rax,1),%rax
  shr    $0x3,%r8d
  sub    $0x1,%r8d
  shl    $0x5,%r8
  lea    0x20(%rcx,%r8,1),%r10
  nopl   0x0(%rax)
L50:
  movdqu (%rdx),%xmm1
  movdqu 0x10(%rdx),%xmm0
  add    $0x20,%rdx
  add    $0x20,%rax
  pextrw $0x0,%xmm1,%r8d
  mov    %r8w,-0x20(%rax)
  pextrw $0x2,%xmm1,%r8d
  mov    %r8w,-0x1c(%rax)
  pextrw $0x4,%xmm1,%r8d
  mov    %r8w,-0x18(%rax)
  pextrw $0x6,%xmm1,%r8d
  mov    %r8w,-0x14(%rax)
  pextrw $0x0,%xmm0,%r8d
  mov    %r8w,-0x10(%rax)
  pextrw $0x2,%xmm0,%r8d
  mov    %r8w,-0xc(%rax)
  pextrw $0x4,%xmm0,%r8d
  mov    %r8w,-0x8(%rax)
  pextrw $0x6,%xmm0,%r8d
  mov    %r8w,-0x4(%rax)
  cmp    %rdx,%r10
  jne    L50
  mov    %esi,%eax
  and    $0xfffffff8,%eax
  mov    %eax,%edx
  lea    (%rcx,%rdx,4),%rcx
L92:
  movzwl (%rcx),%edx
  mov    %dx,0x2(%rcx)
  lea    0x1(%rax),%edx
  cmp    %edi,%edx
  jge    L104
  movzwl 0x4(%rcx),%edx
  mov    %dx,0x6(%rcx)
  lea    0x2(%rax),%edx
  cmp    %edx,%edi
  jle    L104
  movzwl 0x8(%rcx),%edx
  mov    %dx,0xa(%rcx)
  lea    0x3(%rax),%edx
  cmp    %edx,%edi
  jle    L104
  movzwl 0xc(%rcx),%edx
  mov    %dx,0xe(%rcx)
  lea    0x4(%rax),%edx
  cmp    %edx,%edi
  jle    L104
  movzwl 0x10(%rcx),%edx
  mov    %dx,0x12(%rcx)
  lea    0x5(%rax),%edx
  cmp    %edx,%edi
  jle    L104
  movzwl 0x14(%rcx),%edx
  mov    %dx,0x16(%rcx)
  lea    0x6(%rax),%edx
  cmp    %edx,%edi
  jle    L104
  movzwl 0x18(%rcx),%edx
  add    $0x7,%eax
  mov    %dx,0x1a(%rcx)
  cmp    %eax,%edi
  jle    L104
  movzwl 0x1c(%rcx),%eax
  mov    %ax,0x1e(%rcx)
L104:
  pop    %rbx
  mov    %r9d,%eax
  pop    %rbp
  pop    %r12
  ret
L24:
  xor    %eax,%eax
  jmp    L92
```

assistant: ```c
int synth_ntom_mono2stereo(double *bandPtr, unsigned char *samples, int *pnt) {
  int i, ret;
  int pnt1 = *pnt;
  ret = synth_ntom(bandPtr, 0, samples, pnt);
  samples += pnt1;
  for (i = 0; i < ((*pnt - pnt1) >> 2); i++) {
    ((short *)samples)[1] = ((short *)samples)[0];
    samples += 4;
  }
  return ret;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<f1>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  add    $0x4,%rax
  mov    (%rax),%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x8(%rbp),%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  pop    %rbp
  ret
```

assistant: ```c
int f1(volatile int *a) {
  int b = a[1];
  int c = a[0];
  return b + c;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<f1>:
  endbr64
  mov    0x4(%rdi),%eax
  mov    (%rdi),%edx
  add    %edx,%eax
  ret
```

assistant: ```c
int f1(volatile int *a) {
  int b = a[1];
  int c = a[0];
  return b + c;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<f1>:
  endbr64
  mov    0x4(%rdi),%eax
  mov    (%rdi),%edx
  add    %edx,%eax
  ret
```

assistant: ```c
int f1(volatile int *a) {
  int b = a[1];
  int c = a[0];
  return b + c;
}
```
user: Translate the below assembly code into C programming language:

```asm
<f1>:
  endbr64
  mov    0x4(%rdi),%eax
  mov    (%rdi),%edx
  add    %edx,%eax
  ret
```

assistant: ```c
int f1(volatile int *a) {
  int b = a[1];
  int c = a[0];
  return b + c;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<StrTrim>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,-0x18(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  je     L29
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    -0x18(%rbp),%rdx
  lea    (%rax,%rdx,1),%rbx
  jmp    L77
L24:
  movzbl (%rbx),%eax
  test   %al,%al
  je     L89
  movzbl (%rbx),%eax
  movsbl %al,%eax
  mov    %eax,%esi
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <strchr@plt>
  test   %eax,%eax
  je     L48
L89:
  sub    $0x1,%rbx
L77:
  cmp    -0x18(%rbp),%rbx
  jae    L24
  jmp    L61
L48:
  nop
L61:
  cmp    -0x18(%rbp),%rbx
  jb     L3
  movzbl (%rbx),%eax
  test   %al,%al
  je     L2
L3:
  lea    0x1(%rbx),%rax
  movb   $0x0,(%rax)
  jmp    L2
L29:
  nop
L2:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

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

tool: D0:  


assistant: ```c
void StrTrim(char *cmd) {
  register char *ch;
  if (!cmd)
    return;
  for (ch = &cmd[strlen(cmd)]; ch >= cmd; ch--)
    if (*ch && !strchr(" \r\n", *ch))
      break;
  if (ch < cmd || *ch)
    ch[1] = 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<StrTrim>:
  endbr64
  test   %rdi,%rdi
  je     L71
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  call   <strlen@plt>
  add    %rbp,%rax
  mov    %rax,%rbx
  jb     L28
  lea    D1(%rip),%r12
  jmp    L38
L42:
  sub    $0x1,%rbx
  cmp    %rbx,%rbp
  ja     L28
L38:
  movzbl (%rbx),%esi
  test   %sil,%sil
  je     L42
  movsbl %sil,%esi
  mov    %r12,%rdi
  call   <strchr@plt>
  test   %eax,%eax
  jne    L42
  cmp    %rbx,%rbp
  ja     L28
  cmpb   $0x0,(%rbx)
  je     L61
L28:
  movb   $0x0,0x1(%rbx)
L61:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L71:
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D1
data_type: string

tool: D1:  


assistant: ```c
void StrTrim(char *cmd) {
  register char *ch;
  if (!cmd)
    return;
  for (ch = &cmd[strlen(cmd)]; ch >= cmd; ch--)
    if (*ch && !strchr(" \r\n", *ch))
      break;
  if (ch < cmd || *ch)
    ch[1] = 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<StrTrim>:
  endbr64
  test   %rdi,%rdi
  je     L31
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  call   <strlen@plt>
  add    %rbp,%rax
  mov    %rax,%rbx
  jb     L71
  lea    D0(%rip),%r12
  jmp    L2
  cs nopw 0x0(%rax,%rax,1)
L62:
  sub    $0x1,%rbx
  cmp    %rbx,%rbp
  ja     L71
L2:
  movsbl (%rbx),%esi
  test   %sil,%sil
  je     L62
  mov    %r12,%rdi
  call   <strchr@plt>
  test   %eax,%eax
  jne    L62
  cmp    %rbx,%rbp
  ja     L71
  cmpb   $0x0,(%rbx)
  je     L36
L71:
  movb   $0x0,0x1(%rbx)
L36:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L31:
  ret
```

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

tool: D0:  


assistant: ```c
void StrTrim(char *cmd) {
  register char *ch;
  if (!cmd)
    return;
  for (ch = &cmd[strlen(cmd)]; ch >= cmd; ch--)
    if (*ch && !strchr(" \r\n", *ch))
      break;
  if (ch < cmd || *ch)
    ch[1] = 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<StrTrim>:
  endbr64
  test   %rdi,%rdi
  je     L28
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  call   <strlen@plt>
  add    %rbp,%rax
  mov    %rax,%rbx
  jb     L15
  lea    D7(%rip),%r12
  jmp    L78
  cs nopw 0x0(%rax,%rax,1)
L58:
  sub    $0x1,%rbx
  cmp    %rbx,%rbp
  ja     L15
L78:
  movsbl (%rbx),%esi
  test   %sil,%sil
  je     L58
  mov    %r12,%rdi
  call   <strchr@plt>
  test   %eax,%eax
  jne    L58
  cmp    %rbx,%rbp
  ja     L15
  cmpb   $0x0,(%rbx)
  je     L29
L15:
  movb   $0x0,0x1(%rbx)
L29:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L28:
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D7
data_type: string

tool: D7:  


assistant: ```c
void StrTrim(char *cmd) {
  register char *ch;
  if (!cmd)
    return;
  for (ch = &cmd[strlen(cmd)]; ch >= cmd; ch--)
    if (*ch && !strchr(" \r\n", *ch))
      break;
  if (ch < cmd || *ch)
    ch[1] = 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<modinv>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x38(%rbp)
  mov    %rsi,-0x40(%rbp)
  movq   $0x0,-0x30(%rbp)
  movq   $0x1,-0x28(%rbp)
  mov    -0x40(%rbp),%rax
  mov    %rax,-0x20(%rbp)
  mov    -0x38(%rbp),%rax
  cqto
  idivq  -0x40(%rbp)
  mov    %rdx,-0x18(%rbp)
  jmp    L2
L10:
  mov    -0x20(%rbp),%rax
  cqto
  idivq  -0x18(%rbp)
  mov    %rax,-0x10(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x10(%rbp),%rax
  imul   -0x28(%rbp),%rax
  mov    %rax,%rdx
  mov    -0x30(%rbp),%rax
  sub    %rdx,%rax
  mov    %rax,-0x28(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,-0x30(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x10(%rbp),%rax
  imul   -0x18(%rbp),%rax
  mov    %rax,%rdx
  mov    -0x20(%rbp),%rax
  sub    %rdx,%rax
  mov    %rax,-0x18(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,-0x20(%rbp)
L2:
  cmpq   $0x0,-0x18(%rbp)
  jne    L10
  cmpq   $0x0,-0x30(%rbp)
  jns    L32
  mov    -0x30(%rbp),%rdx
  mov    -0x40(%rbp),%rax
  add    %rdx,%rax
  jmp    L12
L32:
  mov    -0x30(%rbp),%rax
L12:
  pop    %rbp
  ret
```

assistant: ```c
long long modinv(long a, long b) {
  long long t = 0, nt = 1, r = b, nr = a % b, q, tmp;
  while (nr) {
    q = r / nr;
    tmp = nt;
    nt = t - q * nt;
    t = tmp;
    tmp = nr;
    nr = r - q * nr;
    r = tmp;
  }
  return (t < 0) ? t + b : t;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<modinv>:
  endbr64
  mov    %rdi,%rax
  cqto
  idiv   %rsi
  mov    %rdx,%rcx
  test   %rdx,%rdx
  je     L16
  mov    %rsi,%rax
  mov    $0x1,%edi
  mov    $0x0,%r9d
  jmp    L5
L23:
  mov    %rdx,%rcx
L5:
  cqto
  idiv   %rcx
  imul   %rdi,%rax
  mov    %rdi,%r8
  mov    %r9,%rdi
  sub    %rax,%rdi
  mov    %r8,%r9
  mov    %rcx,%rax
  test   %rdx,%rdx
  jne    L23
  add    %r8,%rsi
  test   %r8,%r8
  cmovs  %rsi,%r8
L8:
  mov    %r8,%rax
  ret
L16:
  mov    %rdx,%r8
  jmp    L8
```

assistant: ```c
long long modinv(long a, long b) {
  long long t = 0, nt = 1, r = b, nr = a % b, q, tmp;
  while (nr) {
    q = r / nr;
    tmp = nt;
    nt = t - q * nt;
    t = tmp;
    tmp = nr;
    nr = r - q * nr;
    r = tmp;
  }
  return (t < 0) ? t + b : t;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<modinv>:
  endbr64
  mov    %rdi,%rax
  cqto
  idiv   %rsi
  mov    %rdx,%rcx
  test   %rdx,%rdx
  je     L2
  mov    %rsi,%rax
  mov    $0x1,%edi
  xor    %r8d,%r8d
  jmp    L3
  nopl   0x0(%rax)
L1:
  mov    %rdx,%rcx
L3:
  cqto
  mov    %rdi,%r9
  idiv   %rcx
  imul   %rdi,%rax
  mov    %r8,%rdi
  mov    %r9,%r8
  sub    %rax,%rdi
  mov    %rcx,%rax
  test   %rdx,%rdx
  jne    L1
  add    %r9,%rsi
  test   %r9,%r9
  cmovs  %rsi,%r9
  mov    %r9,%rax
  ret
  cs nopw 0x0(%rax,%rax,1)
L2:
  xor    %r9d,%r9d
  mov    %r9,%rax
  ret
```

assistant: ```c
long long modinv(long a, long b) {
  long long t = 0, nt = 1, r = b, nr = a % b, q, tmp;
  while (nr) {
    q = r / nr;
    tmp = nt;
    nt = t - q * nt;
    t = tmp;
    tmp = nr;
    nr = r - q * nr;
    r = tmp;
  }
  return (t < 0) ? t + b : t;
}
```
user: What is the c source code of the assembly code below:

```asm
<modinv>:
  endbr64
  mov    %rdi,%rax
  cqto
  idiv   %rsi
  mov    %rdx,%rcx
  test   %rdx,%rdx
  je     L15
  mov    %rsi,%rax
  mov    $0x1,%edi
  xor    %r8d,%r8d
  jmp    L17
  nopl   0x0(%rax)
L11:
  mov    %rdx,%rcx
L17:
  cqto
  mov    %rdi,%r9
  idiv   %rcx
  imul   %rdi,%rax
  mov    %r8,%rdi
  mov    %r9,%r8
  sub    %rax,%rdi
  mov    %rcx,%rax
  test   %rdx,%rdx
  jne    L11
  add    %r9,%rsi
  test   %r9,%r9
  cmovs  %rsi,%r9
  mov    %r9,%rax
  ret
  cs nopw 0x0(%rax,%rax,1)
L15:
  xor    %r9d,%r9d
  mov    %r9,%rax
  ret
```

assistant: ```c
long long modinv(long a, long b) {
  long long t = 0, nt = 1, r = b, nr = a % b, q, tmp;
  while (nr) {
    q = r / nr;
    tmp = nt;
    nt = t - q * nt;
    t = tmp;
    tmp = nr;
    nr = r - q * nr;
    r = tmp;
  }
  return (t < 0) ? t + b : t;
}
```