ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    $0x5,%edx
  mov    $0x3,%esi
  lea    D8(%rip),%rbx
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    $0x7,%edx
  mov    $0x5,%esi
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    $0x6,%ebx
  mov    $0x1,%ebp
  lea    D8(%rip),%r12
  jmp    L27
L70:
  mov    %ecx,%edx
  mov    %ebx,%esi
  mov    %r12,%rdi
  call   <printf@plt>
  add    $0x1,%ebp
L94:
  cmp    $0x8,%ebp
  jg     L120
L27:
  mov    %ebx,%ecx
  add    $0x1,%ebx
  test   $0x1,%bl
  je     L94
  movslq %ebx,%rax
  imul   $0x55555556,%rax,%rax
  shr    $0x20,%rax
  mov    %ebx,%edx
  sar    $0x1f,%edx
  sub    %edx,%eax
  lea    (%rax,%rax,2),%eax
  cmp    %eax,%ebx
  je     L94
  movslq %ebx,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x21,%rax
  mov    %ebx,%edx
  sar    $0x1f,%edx
  sub    %edx,%eax
  lea    (%rax,%rax,4),%eax
  cmp    %eax,%ebx
  je     L94
  movslq %ebx,%rax
  imul   $0xffffffff92492493,%rax,%rax
  shr    $0x20,%rax
  add    %ebx,%eax
  sar    $0x2,%eax
  mov    %ebx,%edx
  sar    $0x1f,%edx
  sub    %edx,%eax
  lea    0x0(,%rax,8),%edx
  sub    %eax,%edx
  cmp    %edx,%ebx
  je     L94
  mov    $0x9,%esi
  mov    %ebx,%eax
  cltd
  idiv   %esi
  test   %edx,%edx
  je     L94
  add    $0x3,%ecx
  test   $0x1,%cl
  je     L94
  mov    $0x3,%edi
  mov    %ecx,%eax
  cltd
  idiv   %edi
  test   %edx,%edx
  je     L94
  mov    $0x5,%edi
  mov    %ecx,%eax
  cltd
  idiv   %edi
  test   %edx,%edx
  je     L94
  mov    $0x7,%edi
  mov    %ecx,%eax
  cltd
  idiv   %edi
  test   %edx,%edx
  je     L94
  mov    $0x9,%edi
  mov    %ecx,%eax
  cltd
  idiv   %edi
  test   %edx,%edx
  je     L94
  jmp    L70
L120:
  mov    $0x0,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: retrieve_data
arguments: label: D8
data_type: string

tool: D8: (%d, %d)


assistant: ```c
int main() {
  int a, b;
  a = 3;
  b = 5;
  printf("(%d, %d)\n", a, b);
  a = 5;
  b = 7;
  printf("(%d, %d)\n", a, b);
  int i = 1;
  while (i < 9) {
    a = a + 1;
    b = a + 2;
    if ((a % 2 != 0) && (a % 3 != 0) && (a % 5 != 0) && (a % 7 != 0) && (a % 9 != 0) && (b % 2 != 0) && (b % 3 != 0) && (b % 5 != 0) && (b % 7 != 0) && (b % 9 != 0)) {
      printf("(%d, %d)\n", a, b);
      i++;
    }
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  push   %r13
  mov    $0x5,%edx
  mov    $0x3,%esi
  mov    $0x9,%r13d
  push   %r12
  mov    $0x6,%r12d
  push   %rbp
  lea    D9(%rip),%rbp
  push   %rbx
  mov    %rbp,%rdi
  mov    $0x1,%ebx
  sub    $0x8,%rsp
  call   <printf@plt>
  mov    $0x7,%edx
  mov    $0x5,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  nopw   0x0(%rax,%rax,1)
L71:
  mov    %r12d,%ecx
  add    $0x1,%r12d
  test   $0x1,%r12b
  je     L90
  imul   $0xaaaaaaab,%r12d,%edx
  cmp    $0x55555555,%edx
  jbe    L90
  imul   $0xcccccccd,%r12d,%eax
  cmp    $0x33333333,%eax
  jbe    L90
  imul   $0xb6db6db7,%r12d,%eax
  cmp    $0x24924924,%eax
  jbe    L90
  mov    %r12d,%eax
  cltd
  idiv   %r13d
  test   %edx,%edx
  je     L90
  add    $0x3,%ecx
  mov    $0x3,%esi
  mov    %ecx,%eax
  cltd
  idiv   %esi
  test   %edx,%edx
  je     L90
  mov    %ecx,%eax
  mov    $0x5,%esi
  cltd
  idiv   %esi
  test   %edx,%edx
  je     L90
  mov    %ecx,%eax
  mov    $0x7,%esi
  cltd
  idiv   %esi
  test   %edx,%edx
  je     L90
  mov    %ecx,%eax
  cltd
  idiv   %r13d
  test   %edx,%edx
  je     L90
  mov    %ecx,%edx
  mov    %r12d,%esi
  mov    %rbp,%rdi
  add    $0x1,%ebx
  call   <printf@plt>
  xchg   %ax,%ax
L90:
  cmp    $0x9,%ebx
  jne    L71
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D9
datatype: string

tool: D9: (%d, %d)


assistant: ```c
int main() {
  int a, b;
  a = 3;
  b = 5;
  printf("(%d, %d)\n", a, b);
  a = 5;
  b = 7;
  printf("(%d, %d)\n", a, b);
  int i = 1;
  while (i < 9) {
    a = a + 1;
    b = a + 2;
    if ((a % 2 != 0) && (a % 3 != 0) && (a % 5 != 0) && (a % 7 != 0) && (a % 9 != 0) && (b % 2 != 0) && (b % 3 != 0) && (b % 5 != 0) && (b % 7 != 0) && (b % 9 != 0)) {
      printf("(%d, %d)\n", a, b);
      i++;
    }
  }
  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   %r13
  mov    $0x5,%edx
  mov    $0x3,%esi
  mov    $0x9,%r13d
  push   %r12
  mov    $0x6,%r12d
  push   %rbp
  lea    D2(%rip),%rbp
  push   %rbx
  mov    %rbp,%rdi
  mov    $0x1,%ebx
  sub    $0x8,%rsp
  call   <printf@plt>
  mov    $0x7,%edx
  mov    $0x5,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  nopw   0x0(%rax,%rax,1)
L79:
  mov    %r12d,%ecx
  add    $0x1,%r12d
  test   $0x1,%r12b
  je     L93
  imul   $0xaaaaaaab,%r12d,%edx
  cmp    $0x55555555,%edx
  jbe    L93
  imul   $0xcccccccd,%r12d,%eax
  cmp    $0x33333333,%eax
  jbe    L93
  imul   $0xb6db6db7,%r12d,%eax
  cmp    $0x24924924,%eax
  jbe    L93
  mov    %r12d,%eax
  cltd
  idiv   %r13d
  test   %edx,%edx
  je     L93
  add    $0x3,%ecx
  mov    $0x3,%esi
  mov    %ecx,%eax
  cltd
  idiv   %esi
  test   %edx,%edx
  je     L93
  mov    %ecx,%eax
  mov    $0x5,%esi
  cltd
  idiv   %esi
  test   %edx,%edx
  je     L93
  mov    %ecx,%eax
  mov    $0x7,%esi
  cltd
  idiv   %esi
  test   %edx,%edx
  je     L93
  mov    %ecx,%eax
  cltd
  idiv   %r13d
  test   %edx,%edx
  je     L93
  mov    %ecx,%edx
  mov    %r12d,%esi
  mov    %rbp,%rdi
  add    $0x1,%ebx
  call   <printf@plt>
  xchg   %ax,%ax
L93:
  cmp    $0x9,%ebx
  jne    L79
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D2
type: string

tool: D2: (%d, %d)


assistant: ```c
int main() {
  int a, b;
  a = 3;
  b = 5;
  printf("(%d, %d)\n", a, b);
  a = 5;
  b = 7;
  printf("(%d, %d)\n", a, b);
  int i = 1;
  while (i < 9) {
    a = a + 1;
    b = a + 2;
    if ((a % 2 != 0) && (a % 3 != 0) && (a % 5 != 0) && (a % 7 != 0) && (a % 9 != 0) && (b % 2 != 0) && (b % 3 != 0) && (b % 5 != 0) && (b % 7 != 0) && (b % 9 != 0)) {
      printf("(%d, %d)\n", a, b);
      i++;
    }
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<gridstats>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x130,%rsp
  mov    %edi,-0x124(%rbp)
  mov    %esi,-0x128(%rbp)
  mov    %edx,-0x12c(%rbp)
  mov    %ecx,-0x130(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x124(%rbp),%eax
  imul   -0x128(%rbp),%eax
  mov    -0x12c(%rbp),%edx
  imul   %edx,%eax
  mov    %eax,-0x114(%rbp)
  pxor   %xmm0,%xmm0
  cvtsi2ssl -0x130(%rbp),%xmm0
  pxor   %xmm1,%xmm1
  cvtsi2ssl -0x114(%rbp),%xmm1
  divss  %xmm1,%xmm0
  movd   %xmm0,%esi
  mov    -0x114(%rbp),%r9d
  mov    -0x12c(%rbp),%r8d
  mov    -0x128(%rbp),%ecx
  mov    -0x124(%rbp),%edx
  lea    -0x110(%rbp),%rax
  sub    $0x8,%rsp
  mov    -0x130(%rbp),%edi
  push   %rdi
  movd   %esi,%xmm0
  lea    D3(%rip),%rsi
  mov    %rax,%rdi
  call   <sprintf@plt>
  add    $0x10,%rsp
  lea    -0x110(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x64,%edi
  call   <rt_ui_message@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L2
  call   <__stack_chk_fail@plt>
L2:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D3
type_infered: string

tool: D3: Grid:  X:%3d  Y:%3d  Z:%3d  Cells:%9d  Obj:%9d  Obj/Cell: %7.3f

assistant: ```c
void gridstats(signed int xs, signed int ys, signed int zs, signed int numobj) {
  char t[256l];
  signed int numcells = xs * ys * zs;
  sprintf(t, "Grid:  X:%3d  Y:%3d  Z:%3d  Cells:%9d  Obj:%9d  Obj/Cell: %7.3f", xs, ys, zs, numcells, numobj, (float)numobj / (float)numcells);
  rt_ui_message(100, t);
}
```
user: Translate the below assembly code into C programming language:

```asm
<gridstats>:
  endbr64
  push   %rbx
  sub    $0x110,%rsp
  mov    %edx,%r8d
  mov    %fs:0x28,%rdx
  mov    %rdx,0x108(%rsp)
  xor    %edx,%edx
  mov    %edi,%r9d
  imul   %esi,%r9d
  imul   %r8d,%r9d
  pxor   %xmm0,%xmm0
  cvtsi2ss %ecx,%xmm0
  pxor   %xmm1,%xmm1
  cvtsi2ss %r9d,%xmm1
  divss  %xmm1,%xmm0
  mov    %rsp,%rbx
  sub    $0x8,%rsp
  push   %rcx
  mov    %esi,%ecx
  mov    %edi,%edx
  lea    D1(%rip),%rsi
  mov    %rbx,%rdi
  call   <sprintf@plt>
  mov    %rbx,%rsi
  mov    $0x64,%edi
  call   <rt_ui_message@plt>
  mov    %rbx,%rsp
  mov    0x108(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x110,%rsp
  pop    %rbx
  ret
L9:
  call   <__stack_chk_fail@plt>
```

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

tool: D1: Grid:  X:%3d  Y:%3d  Z:%3d  Cells:%9d  Obj:%9d  Obj/Cell: %7.3f

assistant: ```c
void gridstats(signed int xs, signed int ys, signed int zs, signed int numobj) {
  char t[256l];
  signed int numcells = xs * ys * zs;
  sprintf(t, "Grid:  X:%3d  Y:%3d  Z:%3d  Cells:%9d  Obj:%9d  Obj/Cell: %7.3f", xs, ys, zs, numcells, numobj, (float)numobj / (float)numcells);
  rt_ui_message(100, t);
}
```
user: What is the c source code of the assembly code below?

```asm
<gridstats>:
  endbr64
  mov    %edi,%r9d
  pxor   %xmm0,%xmm0
  pxor   %xmm1,%xmm1
  push   %rbp
  imul   %esi,%r9d
  mov    %edx,%r8d
  cvtsi2ss %ecx,%xmm0
  sub    $0x110,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x108(%rsp)
  xor    %eax,%eax
  imul   %edx,%r9d
  mov    %rsp,%rbp
  sub    $0x8,%rsp
  push   %rcx
  mov    %edi,%edx
  mov    %esi,%ecx
  mov    %rbp,%rdi
  lea    D6(%rip),%rsi
  cvtsi2ss %r9d,%xmm1
  divss  %xmm1,%xmm0
  call   <sprintf@plt>
  mov    %rbp,%rsi
  mov    $0x64,%edi
  call   <rt_ui_message@plt>
  mov    %rbp,%rsp
  mov    0x108(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  add    $0x110,%rsp
  pop    %rbp
  ret
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D6
datatype: string

tool: D6: Grid:  X:%3d  Y:%3d  Z:%3d  Cells:%9d  Obj:%9d  Obj/Cell: %7.3f

assistant: ```c
void gridstats(signed int xs, signed int ys, signed int zs, signed int numobj) {
  char t[256l];
  signed int numcells = xs * ys * zs;
  sprintf(t, "Grid:  X:%3d  Y:%3d  Z:%3d  Cells:%9d  Obj:%9d  Obj/Cell: %7.3f", xs, ys, zs, numcells, numobj, (float)numobj / (float)numcells);
  rt_ui_message(100, t);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<gridstats>:
  endbr64
  mov    %edi,%r9d
  pxor   %xmm0,%xmm0
  pxor   %xmm1,%xmm1
  push   %rbp
  imul   %esi,%r9d
  mov    %edx,%r8d
  cvtsi2ss %ecx,%xmm0
  sub    $0x110,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x108(%rsp)
  xor    %eax,%eax
  imul   %edx,%r9d
  mov    %rsp,%rbp
  sub    $0x8,%rsp
  push   %rcx
  mov    %edi,%edx
  mov    %esi,%ecx
  mov    %rbp,%rdi
  lea    D5(%rip),%rsi
  cvtsi2ss %r9d,%xmm1
  divss  %xmm1,%xmm0
  call   <sprintf@plt>
  mov    %rbp,%rsi
  mov    $0x64,%edi
  call   <rt_ui_message@plt>
  mov    %rbp,%rsp
  mov    0x108(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x110,%rsp
  pop    %rbp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D5
type_infered: string

tool: D5: Grid:  X:%3d  Y:%3d  Z:%3d  Cells:%9d  Obj:%9d  Obj/Cell: %7.3f

assistant: ```c
void gridstats(signed int xs, signed int ys, signed int zs, signed int numobj) {
  char t[256l];
  signed int numcells = xs * ys * zs;
  sprintf(t, "Grid:  X:%3d  Y:%3d  Z:%3d  Cells:%9d  Obj:%9d  Obj/Cell: %7.3f", xs, ys, zs, numcells, numobj, (float)numobj / (float)numcells);
  rt_ui_message(100, t);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<is_primitive_array>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x5b,%al
  je     L3
  mov    $0x0,%eax
  jmp    L7
L8:
  addq   $0x1,-0x8(%rbp)
L3:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x5b,%al
  je     L8
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x4c,%al
  setne  %al
  movzbl %al,%eax
L7:
  pop    %rbp
  ret
```

assistant: ```c
bool is_primitive_array(const char *name) {
  if (*name != '[')
    return false;
  while (*name == '[')
    name++;
  return *name != 'L';
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<is_primitive_array>:
  endbr64
  mov    $0x0,%eax
  cmpb   $0x5b,(%rdi)
  jne    L9
L10:
  add    $0x1,%rdi
  movzbl (%rdi),%eax
  cmp    $0x5b,%al
  je     L10
  cmp    $0x4c,%al
  setne  %al
  movzbl %al,%eax
L9:
  ret
```

assistant: ```c
bool is_primitive_array(const char *name) {
  if (*name != '[')
    return false;
  while (*name == '[')
    name++;
  return *name != 'L';
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<is_primitive_array>:
  endbr64
  xor    %eax,%eax
  cmpb   $0x5b,(%rdi)
  je     L8
  ret
  nopl   0x0(%rax)
L8:
  movzbl 0x1(%rdi),%eax
  add    $0x1,%rdi
  cmp    $0x5b,%al
  je     L8
  cmp    $0x4c,%al
  setne  %al
  movzbl %al,%eax
  ret
```

assistant: ```c
bool is_primitive_array(const char *name) {
  if (*name != '[')
    return false;
  while (*name == '[')
    name++;
  return *name != 'L';
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<is_primitive_array>:
  endbr64
  xor    %eax,%eax
  cmpb   $0x5b,(%rdi)
  je     L12
  ret
  nopl   0x0(%rax)
L12:
  movzbl 0x1(%rdi),%eax
  add    $0x1,%rdi
  cmp    $0x5b,%al
  je     L12
  cmp    $0x4c,%al
  setne  %al
  movzbl %al,%eax
  ret
```

assistant: ```c
bool is_primitive_array(const char *name) {
  if (*name != '[')
    return false;
  while (*name == '[')
    name++;
  return *name != 'L';
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<my_getnbr_base>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x2d,%al
  jne    L40
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    -0x20(%rbp),%rax
  mov    %rax,%rsi
  mov    %rdx,%rdi
  call   <my_getnbr_base>
  neg    %eax
  jmp    L32
L40:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x2b,%al
  jne    L59
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    -0x20(%rbp),%rax
  mov    %rax,%rsi
  mov    %rdx,%rdi
  call   <my_getnbr_base>
  jmp    L32
L59:
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <my_strlen@plt>
  mov    %eax,-0x4(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L56
L2:
  mov    -0x8(%rbp),%eax
  imul   -0x4(%rbp),%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    -0x20(%rbp),%rdx
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <my_getnbr_baseBIS@plt>
  mov    %eax,%edx
  mov    -0x8(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x8(%rbp)
  addq   $0x1,-0x18(%rbp)
L56:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L2
  mov    -0x8(%rbp),%eax
L32:
  leave
  ret
```

assistant: ```c
int my_getnbr_base(char *str, char *base) {
  int res;
  int len;
  if (*str == '-')
    return (-my_getnbr_base(str + 1, base));
  if (*str == '+')
    return (my_getnbr_base(str + 1, base));
  len = my_strlen(base);
  res = 0;
  while (*str) {
    res *= len;
    res += my_getnbr_baseBIS(*str, base);
    str++;
  }
  return (res);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<my_getnbr_base>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  movzbl (%rdi),%eax
  cmp    $0x2d,%al
  je     L21
  cmp    $0x2b,%al
  je     L42
  mov    %rsi,%rdi
  call   <my_strlen@plt>
  mov    %eax,%r12d
  movzbl (%rbx),%edi
  test   %dil,%dil
  je     L3
  mov    $0x0,%r13d
L65:
  imul   %r12d,%r13d
  movsbl %dil,%edi
  mov    %rbp,%rsi
  call   <my_getnbr_baseBIS@plt>
  add    %eax,%r13d
  add    $0x1,%rbx
  movzbl (%rbx),%edi
  test   %dil,%dil
  jne    L65
L31:
  mov    %r13d,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L21:
  lea    0x1(%rdi),%rdi
  call   <my_getnbr_base>
  neg    %eax
  mov    %eax,%r13d
  jmp    L31
L42:
  lea    0x1(%rdi),%rdi
  call   <my_getnbr_base>
  mov    %eax,%r13d
  jmp    L31
L3:
  mov    $0x0,%r13d
  jmp    L31
```

assistant: ```c
int my_getnbr_base(char *str, char *base) {
  int res;
  int len;
  if (*str == '-')
    return (-my_getnbr_base(str + 1, base));
  if (*str == '+')
    return (my_getnbr_base(str + 1, base));
  len = my_strlen(base);
  res = 0;
  while (*str) {
    res *= len;
    res += my_getnbr_baseBIS(*str, base);
    str++;
  }
  return (res);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<my_getnbr_base>:
  endbr64
  push   %r14
  add    $0x1,%rdi
  push   %r13
  push   %r12
  mov    $0x1,%r12d
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  jmp    L34
  nopl   0x0(%rax,%rax,1)
L54:
  cmp    $0x2b,%al
  jne    L2
L20:
  add    $0x1,%rdi
L34:
  movzbl -0x1(%rdi),%eax
  lea    -0x1(%rdi),%r14
  cmp    $0x2d,%al
  jne    L54
  neg    %r12d
  jmp    L20
  nopl   0x0(%rax)
L2:
  mov    %rbp,%rdi
  call   <my_strlen@plt>
  movsbl (%r14),%edi
  mov    %eax,%r13d
  test   %dil,%dil
  je     L58
  xor    %ebx,%ebx
  cs nopw 0x0(%rax,%rax,1)
L10:
  imul   %r13d,%ebx
  mov    %rbp,%rsi
  add    $0x1,%r14
  call   <my_getnbr_baseBIS@plt>
  movsbl (%r14),%edi
  add    %eax,%ebx
  test   %dil,%dil
  jne    L10
  mov    %r12d,%eax
  imul   %ebx,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L58:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int my_getnbr_base(char *str, char *base) {
  int res;
  int len;
  if (*str == '-')
    return (-my_getnbr_base(str + 1, base));
  if (*str == '+')
    return (my_getnbr_base(str + 1, base));
  len = my_strlen(base);
  res = 0;
  while (*str) {
    res *= len;
    res += my_getnbr_baseBIS(*str, base);
    str++;
  }
  return (res);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<my_getnbr_base>:
  endbr64
  push   %r14
  add    $0x1,%rdi
  push   %r13
  push   %r12
  mov    $0x1,%r12d
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  jmp    L18
  nopl   0x0(%rax,%rax,1)
L19:
  cmp    $0x2b,%al
  jne    L34
L55:
  add    $0x1,%rdi
L18:
  movzbl -0x1(%rdi),%eax
  lea    -0x1(%rdi),%r14
  cmp    $0x2d,%al
  jne    L19
  neg    %r12d
  jmp    L55
  nopl   0x0(%rax)
L34:
  mov    %rbp,%rdi
  call   <my_strlen@plt>
  movsbl (%r14),%edi
  mov    %eax,%r13d
  test   %dil,%dil
  je     L32
  xor    %ebx,%ebx
  cs nopw 0x0(%rax,%rax,1)
L36:
  imul   %r13d,%ebx
  mov    %rbp,%rsi
  add    $0x1,%r14
  call   <my_getnbr_baseBIS@plt>
  movsbl (%r14),%edi
  add    %eax,%ebx
  test   %dil,%dil
  jne    L36
  mov    %r12d,%eax
  imul   %ebx,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L32:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int my_getnbr_base(char *str, char *base) {
  int res;
  int len;
  if (*str == '-')
    return (-my_getnbr_base(str + 1, base));
  if (*str == '+')
    return (my_getnbr_base(str + 1, base));
  len = my_strlen(base);
  res = 0;
  while (*str) {
    res *= len;
    res += my_getnbr_baseBIS(*str, base);
    str++;
  }
  return (res);
}
```
user: Translate the below assembly code into C programming language:

```asm
<zero_array>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L13
L16:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movl   $0x0,(%rax)
  addl   $0x1,-0x4(%rbp)
L13:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L16
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void zero_array(int ar[], int len) {
  for (int i = 0; i < len; i++) {
    ar[i] = 0;
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<zero_array>:
  endbr64
  test   %esi,%esi
  jle    L6
  mov    %rdi,%rax
  lea    -0x1(%rsi),%edx
  lea    0x4(%rdi,%rdx,4),%rdx
L5:
  movl   $0x0,(%rax)
  add    $0x4,%rax
  cmp    %rdx,%rax
  jne    L5
L6:
  ret
```

assistant: ```c
void zero_array(int ar[], int len) {
  for (int i = 0; i < len; i++) {
    ar[i] = 0;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<zero_array>:
  endbr64
  test   %esi,%esi
  jle    L5
  movslq %esi,%rsi
  lea    0x0(,%rsi,4),%rdx
  xor    %esi,%esi
  jmp    1050 <memset@plt>
  nopw   0x0(%rax,%rax,1)
L5:
  ret
```

assistant: ```c
void zero_array(int ar[], int len) {
  for (int i = 0; i < len; i++) {
    ar[i] = 0;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<zero_array>:
  endbr64
  test   %esi,%esi
  jle    L4
  movslq %esi,%rsi
  lea    0x0(,%rsi,4),%rdx
  xor    %esi,%esi
  jmp    1050 <memset@plt>
  nopw   0x0(%rax,%rax,1)
L4:
  ret
```

assistant: ```c
void zero_array(int ar[], int len) {
  for (int i = 0; i < len; i++) {
    ar[i] = 0;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<mystrcpy>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  jmp    L1
L0:
  mov    -0x10(%rbp),%rdx
  lea    0x1(%rdx),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rcx
  mov    %rcx,-0x8(%rbp)
  movzbl (%rdx),%edx
  mov    %dl,(%rax)
L1:
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L0
  mov    -0x8(%rbp),%rax
  movb   $0x0,(%rax)
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int mystrcpy(s1, s2)
char *s1;
char *s2;
{
  while (*s2)
    *s1++ = *s2++;
  *s1 = '\0';
  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
<mystrcpy>:
  endbr64
  movzbl (%rsi),%eax
  test   %al,%al
  je     L4
L12:
  add    $0x1,%rsi
  add    $0x1,%rdi
  mov    %al,-0x1(%rdi)
  movzbl (%rsi),%eax
  test   %al,%al
  jne    L12
L4:
  movb   $0x0,(%rdi)
  mov    $0x0,%eax
  ret
```

assistant: ```c
int mystrcpy(s1, s2)
char *s1;
char *s2;
{
  while (*s2)
    *s1++ = *s2++;
  *s1 = '\0';
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<mystrcpy>:
  endbr64
  jmp    L13
  cs nopw 0x0(%rax,%rax,1)
L12:
  mov    %al,(%rdi)
  add    $0x1,%rdi
  add    $0x1,%rsi
L13:
  movzbl (%rsi),%eax
  test   %al,%al
  jne    L12
  movb   $0x0,(%rdi)
  xor    %eax,%eax
  ret
```

assistant: ```c
int mystrcpy(s1, s2)
char *s1;
char *s2;
{
  while (*s2)
    *s1++ = *s2++;
  *s1 = '\0';
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<mystrcpy>:
  endbr64
  jmp    L11
  cs nopw 0x0(%rax,%rax,1)
L1:
  mov    %al,(%rdi)
  add    $0x1,%rdi
  add    $0x1,%rsi
L11:
  movzbl (%rsi),%eax
  test   %al,%al
  jne    L1
  movb   $0x0,(%rdi)
  xor    %eax,%eax
  ret
```

assistant: ```c
int mystrcpy(s1, s2)
char *s1;
char *s2;
{
  while (*s2)
    *s1++ = *s2++;
  *s1 = '\0';
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<ci20_load_elf_path>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    D6(%rip),%rax
  mov    (%rax),%edx
  mov    -0x20(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <open@plt>
  mov    %eax,-0x8(%rbp)
  cmpl   $0xffffffff,-0x8(%rbp)
  jne    L3
  mov    D1(%rip),%rax
  mov    (%rax),%eax
  jmp    L6
L3:
  mov    -0x28(%rbp),%rdx
  mov    -0x8(%rbp),%ecx
  mov    -0x18(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <ci20_load_elf_fd@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x8(%rbp),%eax
  mov    %eax,%edi
  call   <close@plt>
  mov    -0x4(%rbp),%eax
L6:
  leave
  ret
```

assistant: ```c
int ci20_load_elf_path(struct ci20_dev *dev, const char *path, uint32_t *entry) {
  int filedes, err;
  filedes = open(path, O_RDONLY);
  if (filedes == -1)
    return errno;
  err = ci20_load_elf_fd(dev, filedes, entry);
  close(filedes);
  return err;
}
```