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
<idct_col>:
  endbr64
  push   %rbp
  push   %rbx
  mov    %rdi,%rax
  movswl (%rdi),%ecx
  shl    $0xb,%ecx
  add    $0x10000,%ecx
  movswl 0x10(%rdi),%edx
  movswl 0x20(%rdi),%r8d
  shl    $0xb,%r8d
  movswl 0x30(%rdi),%esi
  lea    (%rcx,%r8,1),%edi
  sub    %r8d,%ecx
  lea    (%rdx,%rsi,1),%r8d
  imul   $0x454,%r8d,%r8d
  imul   $0x620,%edx,%edx
  add    %r8d,%edx
  imul   $0xfffff138,%esi,%esi
  add    %r8d,%esi
  lea    (%rdi,%rdx,1),%r9d
  lea    (%rcx,%rsi,1),%r10d
  sub    %esi,%ecx
  sub    %edx,%edi
  movswl 0x40(%rax),%edx
  movswl 0x50(%rax),%r8d
  movswl 0x60(%rax),%r11d
  movswl 0x70(%rax),%esi
  lea    (%rdx,%rsi,1),%ebx
  imul   $0x235,%ebx,%ebx
  imul   $0x8e4,%edx,%edx
  add    %ebx,%edx
  imul   $0xfffff2b2,%esi,%esi
  add    %ebx,%esi
  lea    (%r8,%r11,1),%ebx
  imul   $0x968,%ebx,%ebx
  imul   $0xfffffce1,%r11d,%r11d
  add    %ebx,%r11d
  imul   $0xfffff04f,%r8d,%r8d
  add    %ebx,%r8d
  lea    (%rdx,%r11,1),%ebx
  lea    (%rsi,%r8,1),%ebp
  sub    %r11d,%edx
  sub    %r8d,%esi
  lea    (%rdx,%rsi,1),%r8d
  sar    $0x8,%r8d
  imul   $0xb5,%r8d,%r8d
  sub    %esi,%edx
  sar    $0x8,%edx
  imul   $0xb5,%edx,%edx
  lea    (%r9,%rbx,1),%esi
  sar    $0x11,%esi
  mov    %si,(%rax)
  lea    (%r10,%r8,1),%esi
  sar    $0x11,%esi
  mov    %si,0x10(%rax)
  lea    (%rcx,%rdx,1),%esi
  sar    $0x11,%esi
  mov    %si,0x20(%rax)
  lea    (%rdi,%rbp,1),%esi
  sar    $0x11,%esi
  mov    %si,0x30(%rax)
  sub    %ebp,%edi
  sar    $0x11,%edi
  mov    %di,0x40(%rax)
  sub    %edx,%ecx
  sar    $0x11,%ecx
  mov    %cx,0x50(%rax)
  sub    %r8d,%r10d
  sar    $0x11,%r10d
  mov    %r10w,0x60(%rax)
  sub    %ebx,%r9d
  sar    $0x11,%r9d
  mov    %r9w,0x70(%rax)
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void idct_col(signed short *const block) {
  int d0, d1, d2, d3;
  int a0, a1, a2, a3, b0, b1, b2, b3;
  int t0, t1, t2, t3;
  d0 = (block[8 * 0] << 11) + 65536;
  d1 = block[8 * 1];
  d2 = block[8 * 2] << 11;
  d3 = block[8 * 3];
  t0 = d0 + d2;
  t1 = d0 - d2;
  do {
    int tmp = 1108 * (d3 + d1);
    t2 = tmp + (2676 - 1108) * d1;
    t3 = tmp - (2676 + 1108) * d3;
  } while (0);
  a0 = t0 + t2;
  a1 = t1 + t3;
  a2 = t1 - t3;
  a3 = t0 - t2;
  d0 = block[8 * 4];
  d1 = block[8 * 5];
  d2 = block[8 * 6];
  d3 = block[8 * 7];
  do {
    int tmp = 565 * (d3 + d0);
    t0 = tmp + (2841 - 565) * d0;
    t1 = tmp - (2841 + 565) * d3;
  } while (0);
  do {
    int tmp = 2408 * (d1 + d2);
    t2 = tmp + (1609 - 2408) * d2;
    t3 = tmp - (1609 + 2408) * d1;
  } while (0);
  b0 = t0 + t2;
  b3 = t1 + t3;
  t0 -= t2;
  t1 -= t3;
  b1 = ((t0 + t1) >> 8) * 181;
  b2 = ((t0 - t1) >> 8) * 181;
  block[8 * 0] = (a0 + b0) >> 17;
  block[8 * 1] = (a1 + b1) >> 17;
  block[8 * 2] = (a2 + b2) >> 17;
  block[8 * 3] = (a3 + b3) >> 17;
  block[8 * 4] = (a3 - b3) >> 17;
  block[8 * 5] = (a2 - b2) >> 17;
  block[8 * 6] = (a1 - b1) >> 17;
  block[8 * 7] = (a0 - b0) >> 17;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<idct_col>:
  endbr64
  push   %rbp
  mov    %rdi,%rax
  push   %rbx
  movswl (%rdi),%ecx
  movswl 0x20(%rdi),%r8d
  movswl 0x10(%rdi),%edx
  movswl 0x30(%rdi),%esi
  shl    $0xb,%ecx
  movswl 0x60(%rax),%r11d
  shl    $0xb,%r8d
  add    $0x10000,%ecx
  lea    (%rcx,%r8,1),%edi
  sub    %r8d,%ecx
  lea    (%rdx,%rsi,1),%r8d
  imul   $0x454,%r8d,%r8d
  imul   $0x620,%edx,%edx
  imul   $0xfffff138,%esi,%esi
  add    %r8d,%edx
  add    %r8d,%esi
  lea    (%rdi,%rdx,1),%r9d
  sub    %edx,%edi
  movswl 0x40(%rax),%edx
  lea    (%rcx,%rsi,1),%r10d
  sub    %esi,%ecx
  movswl 0x70(%rax),%esi
  movswl 0x50(%rax),%r8d
  lea    (%rdx,%rsi,1),%ebx
  imul   $0x235,%ebx,%ebx
  imul   $0x8e4,%edx,%edx
  imul   $0xfffff2b2,%esi,%esi
  add    %ebx,%edx
  add    %ebx,%esi
  lea    (%r8,%r11,1),%ebx
  imul   $0x968,%ebx,%ebx
  imul   $0xfffffce1,%r11d,%r11d
  imul   $0xfffff04f,%r8d,%r8d
  add    %ebx,%r11d
  add    %ebx,%r8d
  lea    (%rdx,%r11,1),%ebx
  sub    %r11d,%edx
  lea    (%rsi,%r8,1),%ebp
  sub    %r8d,%esi
  lea    (%rdx,%rsi,1),%r8d
  sub    %esi,%edx
  lea    (%r9,%rbx,1),%esi
  sar    $0x8,%r8d
  sar    $0x8,%edx
  imul   $0xb5,%r8d,%r8d
  sar    $0x11,%esi
  imul   $0xb5,%edx,%edx
  mov    %si,(%rax)
  lea    (%r10,%r8,1),%esi
  sar    $0x11,%esi
  mov    %si,0x10(%rax)
  lea    (%rcx,%rdx,1),%esi
  sar    $0x11,%esi
  mov    %si,0x20(%rax)
  lea    (%rdi,%rbp,1),%esi
  sar    $0x11,%esi
  sub    %ebp,%edi
  sub    %ebx,%r9d
  sub    %edx,%ecx
  sub    %r8d,%r10d
  sar    $0x11,%edi
  pop    %rbx
  mov    %si,0x30(%rax)
  sar    $0x11,%ecx
  sar    $0x11,%r10d
  mov    %di,0x40(%rax)
  pop    %rbp
  sar    $0x11,%r9d
  mov    %cx,0x50(%rax)
  mov    %r10w,0x60(%rax)
  mov    %r9w,0x70(%rax)
  ret
```

assistant: ```c
void idct_col(signed short *const block) {
  int d0, d1, d2, d3;
  int a0, a1, a2, a3, b0, b1, b2, b3;
  int t0, t1, t2, t3;
  d0 = (block[8 * 0] << 11) + 65536;
  d1 = block[8 * 1];
  d2 = block[8 * 2] << 11;
  d3 = block[8 * 3];
  t0 = d0 + d2;
  t1 = d0 - d2;
  do {
    int tmp = 1108 * (d3 + d1);
    t2 = tmp + (2676 - 1108) * d1;
    t3 = tmp - (2676 + 1108) * d3;
  } while (0);
  a0 = t0 + t2;
  a1 = t1 + t3;
  a2 = t1 - t3;
  a3 = t0 - t2;
  d0 = block[8 * 4];
  d1 = block[8 * 5];
  d2 = block[8 * 6];
  d3 = block[8 * 7];
  do {
    int tmp = 565 * (d3 + d0);
    t0 = tmp + (2841 - 565) * d0;
    t1 = tmp - (2841 + 565) * d3;
  } while (0);
  do {
    int tmp = 2408 * (d1 + d2);
    t2 = tmp + (1609 - 2408) * d2;
    t3 = tmp - (1609 + 2408) * d1;
  } while (0);
  b0 = t0 + t2;
  b3 = t1 + t3;
  t0 -= t2;
  t1 -= t3;
  b1 = ((t0 + t1) >> 8) * 181;
  b2 = ((t0 - t1) >> 8) * 181;
  block[8 * 0] = (a0 + b0) >> 17;
  block[8 * 1] = (a1 + b1) >> 17;
  block[8 * 2] = (a2 + b2) >> 17;
  block[8 * 3] = (a3 + b3) >> 17;
  block[8 * 4] = (a3 - b3) >> 17;
  block[8 * 5] = (a2 - b2) >> 17;
  block[8 * 6] = (a1 - b1) >> 17;
  block[8 * 7] = (a0 - b0) >> 17;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<idct_col>:
  endbr64
  push   %rbp
  mov    %rdi,%rax
  push   %rbx
  movswl (%rdi),%ecx
  movswl 0x20(%rdi),%r8d
  movswl 0x10(%rdi),%edx
  movswl 0x30(%rdi),%esi
  shl    $0xb,%ecx
  movswl 0x60(%rax),%r11d
  shl    $0xb,%r8d
  add    $0x10000,%ecx
  lea    (%rcx,%r8,1),%edi
  sub    %r8d,%ecx
  lea    (%rdx,%rsi,1),%r8d
  imul   $0x454,%r8d,%r8d
  imul   $0x620,%edx,%edx
  imul   $0xfffff138,%esi,%esi
  add    %r8d,%edx
  add    %r8d,%esi
  lea    (%rdi,%rdx,1),%r9d
  sub    %edx,%edi
  movswl 0x40(%rax),%edx
  lea    (%rcx,%rsi,1),%r10d
  sub    %esi,%ecx
  movswl 0x70(%rax),%esi
  movswl 0x50(%rax),%r8d
  lea    (%rdx,%rsi,1),%ebx
  imul   $0x235,%ebx,%ebx
  imul   $0x8e4,%edx,%edx
  imul   $0xfffff2b2,%esi,%esi
  add    %ebx,%edx
  add    %ebx,%esi
  lea    (%r8,%r11,1),%ebx
  imul   $0x968,%ebx,%ebx
  imul   $0xfffffce1,%r11d,%r11d
  imul   $0xfffff04f,%r8d,%r8d
  add    %ebx,%r11d
  add    %ebx,%r8d
  lea    (%rdx,%r11,1),%ebx
  sub    %r11d,%edx
  lea    (%rsi,%r8,1),%ebp
  sub    %r8d,%esi
  lea    (%rdx,%rsi,1),%r8d
  sub    %esi,%edx
  lea    (%r9,%rbx,1),%esi
  sar    $0x8,%r8d
  sar    $0x8,%edx
  imul   $0xb5,%r8d,%r8d
  sar    $0x11,%esi
  imul   $0xb5,%edx,%edx
  mov    %si,(%rax)
  lea    (%r10,%r8,1),%esi
  sar    $0x11,%esi
  mov    %si,0x10(%rax)
  lea    (%rcx,%rdx,1),%esi
  sar    $0x11,%esi
  mov    %si,0x20(%rax)
  lea    (%rdi,%rbp,1),%esi
  sar    $0x11,%esi
  sub    %ebp,%edi
  sub    %ebx,%r9d
  sub    %edx,%ecx
  sub    %r8d,%r10d
  sar    $0x11,%edi
  pop    %rbx
  mov    %si,0x30(%rax)
  sar    $0x11,%ecx
  sar    $0x11,%r10d
  mov    %di,0x40(%rax)
  pop    %rbp
  sar    $0x11,%r9d
  mov    %cx,0x50(%rax)
  mov    %r10w,0x60(%rax)
  mov    %r9w,0x70(%rax)
  ret
```

assistant: ```c
void idct_col(signed short *const block) {
  int d0, d1, d2, d3;
  int a0, a1, a2, a3, b0, b1, b2, b3;
  int t0, t1, t2, t3;
  d0 = (block[8 * 0] << 11) + 65536;
  d1 = block[8 * 1];
  d2 = block[8 * 2] << 11;
  d3 = block[8 * 3];
  t0 = d0 + d2;
  t1 = d0 - d2;
  do {
    int tmp = 1108 * (d3 + d1);
    t2 = tmp + (2676 - 1108) * d1;
    t3 = tmp - (2676 + 1108) * d3;
  } while (0);
  a0 = t0 + t2;
  a1 = t1 + t3;
  a2 = t1 - t3;
  a3 = t0 - t2;
  d0 = block[8 * 4];
  d1 = block[8 * 5];
  d2 = block[8 * 6];
  d3 = block[8 * 7];
  do {
    int tmp = 565 * (d3 + d0);
    t0 = tmp + (2841 - 565) * d0;
    t1 = tmp - (2841 + 565) * d3;
  } while (0);
  do {
    int tmp = 2408 * (d1 + d2);
    t2 = tmp + (1609 - 2408) * d2;
    t3 = tmp - (1609 + 2408) * d1;
  } while (0);
  b0 = t0 + t2;
  b3 = t1 + t3;
  t0 -= t2;
  t1 -= t3;
  b1 = ((t0 + t1) >> 8) * 181;
  b2 = ((t0 - t1) >> 8) * 181;
  block[8 * 0] = (a0 + b0) >> 17;
  block[8 * 1] = (a1 + b1) >> 17;
  block[8 * 2] = (a2 + b2) >> 17;
  block[8 * 3] = (a3 + b3) >> 17;
  block[8 * 4] = (a3 - b3) >> 17;
  block[8 * 5] = (a2 - b2) >> 17;
  block[8 * 6] = (a1 - b1) >> 17;
  block[8 * 7] = (a0 - b0) >> 17;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movsd  D21(%rip),%xmm0
  movsd  %xmm0,-0x10(%rbp)
  lea    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  movl   $0x2,-0x14(%rbp)
  jmp    L16
L27:
  mov    -0x14(%rbp),%eax
  imul   %eax,%eax
  pxor   %xmm2,%xmm2
  cvtsi2sd %eax,%xmm2
  movsd  D21(%rip),%xmm0
  movapd %xmm0,%xmm1
  divsd  %xmm2,%xmm1
  movsd  -0x10(%rbp),%xmm0
  subsd  %xmm1,%xmm0
  movsd  %xmm0,-0x10(%rbp)
  addl   $0x1,-0x14(%rbp)
L16:
  mov    -0x18(%rbp),%eax
  cmp    %eax,-0x14(%rbp)
  jle    L27
  mov    -0x10(%rbp),%rax
  movq   %rax,%xmm0
  lea    D29(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L21
  call   <__stack_chk_fail@plt>
L21:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D21
type_guessed: f64
name: unpack_data
arguments: data_label: D18
type_guessed: string
name: unpack_data
arguments: data_label: D29
type_guessed: string

tool: D21: 1

tool: D18: %d

tool: D29: %.6lf


assistant: ```c
int main(int argc, const char *argv[]) {
  int i, m;
  double sum = 1;
  scanf("%d", &m);
  for (i = 2; i <= m; i++) {
    sum -= 1.0 / (i * i);
  }
  printf("%.6lf\n", sum);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  lea    D28(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%ecx
  cmp    $0x1,%ecx
  jle    L3
  add    $0x1,%ecx
  movsd  D21(%rip),%xmm0
  mov    $0x2,%eax
  movapd %xmm0,%xmm3
L35:
  mov    %eax,%edx
  imul   %eax,%edx
  pxor   %xmm2,%xmm2
  cvtsi2sd %edx,%xmm2
  movapd %xmm3,%xmm1
  divsd  %xmm2,%xmm1
  subsd  %xmm1,%xmm0
  add    $0x1,%eax
  cmp    %ecx,%eax
  jne    L35
L27:
  lea    D9(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L20
  mov    $0x0,%eax
  add    $0x18,%rsp
  ret
L3:
  movsd  D21(%rip),%xmm0
  jmp    L27
L20:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D28
data_type: string
name: disassemble_data
arguments: data_tag: D21
data_type: f64
name: disassemble_data
arguments: data_tag: D9
data_type: string

tool: D28: %d

tool: D21: 1

tool: D9: %.6lf


assistant: ```c
int main(int argc, const char *argv[]) {
  int i, m;
  double sum = 1;
  scanf("%d", &m);
  for (i = 2; i <= m; i++) {
    sum -= 1.0 / (i * i);
  }
  printf("%.6lf\n", sum);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D34(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  mov    0x4(%rsp),%ecx
  cmp    $0x1,%ecx
  jle    L13
  movsd  D15(%rip),%xmm3
  add    $0x1,%ecx
  mov    $0x2,%eax
  movapd %xmm3,%xmm0
  cs nopw 0x0(%rax,%rax,1)
L31:
  mov    %eax,%edx
  pxor   %xmm2,%xmm2
  movapd %xmm3,%xmm1
  imul   %eax,%edx
  add    $0x1,%eax
  cvtsi2sd %edx,%xmm2
  divsd  %xmm2,%xmm1
  subsd  %xmm1,%xmm0
  cmp    %eax,%ecx
  jne    L31
L8:
  lea    D10(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L34
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L13:
  movsd  D15(%rip),%xmm0
  jmp    L8
L34:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D34
type_infered: string
name: unpack_data
arguments: label: D15
type_infered: f64
name: unpack_data
arguments: label: D10
type_infered: string

tool: D34: %d

tool: D15: 1

tool: D10: %.6lf


assistant: ```c
int main(int argc, const char *argv[]) {
  int i, m;
  double sum = 1;
  scanf("%d", &m);
  for (i = 2; i <= m; i++) {
    sum -= 1.0 / (i * i);
  }
  printf("%.6lf\n", sum);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D22(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  mov    0x4(%rsp),%ecx
  cmp    $0x1,%ecx
  jle    L11
  movsd  D34(%rip),%xmm3
  add    $0x1,%ecx
  mov    $0x2,%eax
  movapd %xmm3,%xmm0
  cs nopw 0x0(%rax,%rax,1)
L3:
  mov    %eax,%edx
  pxor   %xmm2,%xmm2
  movapd %xmm3,%xmm1
  imul   %eax,%edx
  add    $0x1,%eax
  cvtsi2sd %edx,%xmm2
  divsd  %xmm2,%xmm1
  subsd  %xmm1,%xmm0
  cmp    %ecx,%eax
  jne    L3
L21:
  lea    D7(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L11:
  movsd  D34(%rip),%xmm0
  jmp    L21
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D22
type: string
name: retrieve_data
arguments: label: D34
type: f64
name: retrieve_data
arguments: label: D7
type: string

tool: D22: %d

tool: D34: 1

tool: D7: %.6lf


assistant: ```c
int main(int argc, const char *argv[]) {
  int i, m;
  double sum = 1;
  scanf("%d", &m);
  for (i = 2; i <= m; i++) {
    sum -= 1.0 / (i * i);
  }
  printf("%.6lf\n", sum);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<utf8strlen>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x4(%rbp)
  movl   $0x0,-0xc(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L4
L6:
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,%rdi
  call   <utf8skip@plt>
  mov    -0xc(%rbp),%edx
  add    %edx,%eax
  add    $0x1,%eax
  mov    %eax,-0xc(%rbp)
  addl   $0x1,-0x8(%rbp)
L4:
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L21
  mov    -0xc(%rbp),%eax
  cmp    -0x4(%rbp),%eax
  jl     L6
L21:
  mov    -0x8(%rbp),%eax
  leave
  ret
```

assistant: ```c
int utf8strlen(char *str) {
  int i, j, len;
  len = strlen(str);
  for (i = 0, j = 0; *(str + i) != '\0' && i < len; j++) {
    i = i + utf8skip(str + i) + 1;
  }
  return j;
}
```
user: Translate the below assembly code into C programming language:

```asm
<utf8strlen>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r12
  call   <strlen@plt>
  cmpb   $0x0,(%r12)
  je     L28
  mov    %eax,%r13d
  test   %eax,%eax
  jle    L28
  mov    %r12,%rdi
  mov    $0x0,%ebp
  mov    $0x0,%ebx
L1:
  call   <utf8skip@plt>
  lea    0x1(%rbx,%rax,1),%ebx
  add    $0x1,%ebp
  movslq %ebx,%rdi
  add    %r12,%rdi
  cmpb   $0x0,(%rdi)
  je     L14
  cmp    %ebx,%r13d
  jg     L1
L14:
  mov    %ebp,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L28:
  mov    $0x0,%ebp
  jmp    L14
```

assistant: ```c
int utf8strlen(char *str) {
  int i, j, len;
  len = strlen(str);
  for (i = 0, j = 0; *(str + i) != '\0' && i < len; j++) {
    i = i + utf8skip(str + i) + 1;
  }
  return j;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<utf8strlen>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  cmpb   $0x0,0x0(%rbp)
  je     L44
  mov    %eax,%r13d
  test   %eax,%eax
  jle    L44
  mov    %rbp,%rdi
  xor    %r12d,%r12d
  xor    %ebx,%ebx
  jmp    L31
  nopl   (%rax)
L8:
  cmp    %ebx,%r13d
  jle    L16
L31:
  call   <utf8skip@plt>
  add    $0x1,%r12d
  lea    0x1(%rbx,%rax,1),%ebx
  movslq %ebx,%rdi
  add    %rbp,%rdi
  cmpb   $0x0,(%rdi)
  jne    L8
L16:
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax,%rax,1)
L44:
  add    $0x8,%rsp
  xor    %r12d,%r12d
  pop    %rbx
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int utf8strlen(char *str) {
  int i, j, len;
  len = strlen(str);
  for (i = 0, j = 0; *(str + i) != '\0' && i < len; j++) {
    i = i + utf8skip(str + i) + 1;
  }
  return j;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<utf8strlen>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  cmpb   $0x0,0x0(%rbp)
  je     L0
  mov    %eax,%r13d
  test   %eax,%eax
  jle    L0
  mov    %rbp,%rdi
  xor    %r12d,%r12d
  xor    %ebx,%ebx
  jmp    L48
  nopl   (%rax)
L27:
  cmp    %ebx,%r13d
  jle    L43
L48:
  call   <utf8skip@plt>
  add    $0x1,%r12d
  lea    0x1(%rbx,%rax,1),%ebx
  movslq %ebx,%rdi
  add    %rbp,%rdi
  cmpb   $0x0,(%rdi)
  jne    L27
L43:
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax,%rax,1)
L0:
  add    $0x8,%rsp
  xor    %r12d,%r12d
  pop    %rbx
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int utf8strlen(char *str) {
  int i, j, len;
  len = strlen(str);
  for (i = 0, j = 0; *(str + i) != '\0' && i < len; j++) {
    i = i + utf8skip(str + i) + 1;
  }
  return j;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0xc(%rbp),%rax
  mov    %rax,%rsi
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0xc(%rbp),%eax
  mov    %eax,%edi
  call   <print_seq@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L1
  call   <__stack_chk_fail@plt>
L1:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D1
type_infered: string

tool: D1: %d

assistant: ```c
int main() {
  int x;
  scanf("%d", &x);
  print_seq(x);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  lea    D1(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%edi
  call   <print_seq@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  mov    $0x0,%eax
  add    $0x18,%rsp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D1
type_infered: string

tool: D1: %d

assistant: ```c
int main() {
  int x;
  scanf("%d", &x);
  print_seq(x);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D3(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  mov    0x4(%rsp),%edi
  call   <print_seq@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D3
type_infered: string

tool: D3: %d

assistant: ```c
int main() {
  int x;
  scanf("%d", &x);
  print_seq(x);
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D9(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  mov    0x4(%rsp),%edi
  call   <print_seq@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D9
type_infered: string

tool: D9: %d

assistant: ```c
int main() {
  int x;
  scanf("%d", &x);
  print_seq(x);
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<minInOrder>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x1c(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0x4(%rbp)
  jmp    L8
L10:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    %eax,-0x8(%rbp)
  jle    L21
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x8(%rbp)
L21:
  addl   $0x1,-0x4(%rbp)
L8:
  mov    -0x4(%rbp),%eax
  cmp    -0x20(%rbp),%eax
  jle    L10
  mov    -0x8(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int minInOrder(int *numbers, int index1, int index2) {
  int result = numbers[index1];
  for (int i = index1 + 1; i <= index2; i++) {
    if (result > numbers[i]) {
      result = numbers[i];
    }
  }
  return result;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<minInOrder>:
  endbr64
  movslq %esi,%rax
  mov    (%rdi,%rax,4),%ecx
  add    $0x1,%esi
  cmp    %edx,%esi
  jg     L0
  movslq %esi,%rax
L12:
  mov    (%rdi,%rax,4),%esi
  cmp    %esi,%ecx
  cmovg  %esi,%ecx
  add    $0x1,%rax
  cmp    %eax,%edx
  jge    L12
L0:
  mov    %ecx,%eax
  ret
```

assistant: ```c
int minInOrder(int *numbers, int index1, int index2) {
  int result = numbers[index1];
  for (int i = index1 + 1; i <= index2; i++) {
    if (result > numbers[i]) {
      result = numbers[i];
    }
  }
  return result;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<minInOrder>:
  endbr64
  movslq %esi,%rax
  add    $0x1,%esi
  mov    (%rdi,%rax,4),%r8d
  cmp    %edx,%esi
  jg     L13
  movslq %esi,%rax
  nopl   (%rax)
L1:
  mov    (%rdi,%rax,4),%ecx
  cmp    %ecx,%r8d
  cmovg  %ecx,%r8d
  add    $0x1,%rax
  cmp    %eax,%edx
  jge    L1
L13:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int minInOrder(int *numbers, int index1, int index2) {
  int result = numbers[index1];
  for (int i = index1 + 1; i <= index2; i++) {
    if (result > numbers[i]) {
      result = numbers[i];
    }
  }
  return result;
}
```