ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请从给定的汇编指令推导出C源代码:

```asm
<HSMMCSDIsIntClockStable>:
  endbr64
  movl   $0x0,-0x4(%rsp)
  mov    0x22c(%edi),%eax
  lea    0x22c(%rdi),%edx
  shr    %eax
  and    $0x1,%eax
  mov    %eax,-0x4(%rsp)
  mov    -0x4(%rsp),%eax
  test   %esi,%esi
  jne    L25
  jmp    L27
  nopl   0x0(%rax)
L9:
  mov    (%rdx),%eax
  shr    %eax
  and    $0x1,%eax
  mov    %eax,-0x4(%rsp)
  mov    -0x4(%rsp),%eax
  sub    $0x1,%esi
  je     L27
L25:
  cmp    $0x1,%eax
  jne    L9
L27:
  mov    -0x4(%rsp),%eax
  ret
```

assistant: ```c
unsigned int HSMMCSDIsIntClockStable(unsigned int baseAddr, unsigned int retry) {
  volatile unsigned int status = 0;
  do {
    status = ((*((volatile unsigned int *)(baseAddr + (0x22C)))) & (0x00000002u)) >> (0x00000001u);
    if ((status == 1) || (retry == 0)) {
      break;
    }
  } while (retry--);
  return status;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<HSMMCSDIsIntClockStable>:
  endbr64
  movl   $0x0,-0x4(%rsp)
  mov    0x22c(%edi),%eax
  lea    0x22c(%rdi),%edx
  shr    %eax
  and    $0x1,%eax
  mov    %eax,-0x4(%rsp)
  mov    -0x4(%rsp),%eax
  test   %esi,%esi
  jne    L0
  jmp    L34
  nopl   0x0(%rax)
L28:
  mov    (%rdx),%eax
  shr    %eax
  and    $0x1,%eax
  mov    %eax,-0x4(%rsp)
  mov    -0x4(%rsp),%eax
  sub    $0x1,%esi
  je     L34
L0:
  cmp    $0x1,%eax
  jne    L28
L34:
  mov    -0x4(%rsp),%eax
  ret
```

assistant: ```c
unsigned int HSMMCSDIsIntClockStable(unsigned int baseAddr, unsigned int retry) {
  volatile unsigned int status = 0;
  do {
    status = ((*((volatile unsigned int *)(baseAddr + (0x22C)))) & (0x00000002u)) >> (0x00000001u);
    if ((status == 1) || (retry == 0)) {
      break;
    }
  } while (retry--);
  return status;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<stpncpy>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    -0x28(%rbp),%rdx
  mov    -0x20(%rbp),%rsi
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rcx
  mov    $0x0,%edx
  mov    %rax,%rdi
  call   <memccpy@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L15
  mov    -0x8(%rbp),%rax
  sub    $0x1,%rax
  jmp    L2
L15:
  mov    -0x18(%rbp),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
L2:
  leave
  ret
```

assistant: ```c
char *stpncpy(char *dest, const char *src, size_t n) {
  char *x = memccpy(dest, src, 0, n);
  if (x) {
    return x - 1;
  } else
    return dest + n;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<stpncpy>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rdx,%rbp
  mov    %rdx,%rcx
  mov    $0x0,%edx
  call   <memccpy@plt>
  lea    -0x1(%rax),%rdx
  add    %rbp,%rbx
  test   %rax,%rax
  mov    %rdx,%rax
  cmove  %rbx,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
char *stpncpy(char *dest, const char *src, size_t n) {
  char *x = memccpy(dest, src, 0, n);
  if (x) {
    return x - 1;
  } else
    return dest + n;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<stpncpy>:
  endbr64
  push   %rbp
  mov    %rdx,%rcx
  mov    %rdx,%rbp
  xor    %edx,%edx
  push   %rbx
  mov    %rdi,%rbx
  add    %rbp,%rbx
  sub    $0x8,%rsp
  call   <memccpy@plt>
  lea    -0x1(%rax),%rdx
  test   %rax,%rax
  mov    %rdx,%rax
  cmove  %rbx,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
char *stpncpy(char *dest, const char *src, size_t n) {
  char *x = memccpy(dest, src, 0, n);
  if (x) {
    return x - 1;
  } else
    return dest + n;
}
```
user: What is the c source code of the assembly code below:

```asm
<stpncpy>:
  endbr64
  push   %rbp
  mov    %rdx,%rcx
  mov    %rdx,%rbp
  xor    %edx,%edx
  push   %rbx
  mov    %rdi,%rbx
  add    %rbp,%rbx
  sub    $0x8,%rsp
  call   <memccpy@plt>
  lea    -0x1(%rax),%rdx
  test   %rax,%rax
  mov    %rdx,%rax
  cmove  %rbx,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
char *stpncpy(char *dest, const char *src, size_t n) {
  char *x = memccpy(dest, src, 0, n);
  if (x) {
    return x - 1;
  } else
    return dest + n;
}
```
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
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x0,%edi
  call   <test__popcountsi2@plt>
  test   %rax,%rax
  je     L145
  mov    $0x1,%eax
  jmp    L50
L145:
  mov    $0x1,%edi
  call   <test__popcountsi2@plt>
  test   %rax,%rax
  je     L71
  mov    $0x1,%eax
  jmp    L50
L71:
  mov    $0x2,%edi
  call   <test__popcountsi2@plt>
  test   %rax,%rax
  je     L8
  mov    $0x1,%eax
  jmp    L50
L8:
  mov    $0xfffffffd,%edi
  call   <test__popcountsi2@plt>
  test   %rax,%rax
  je     L90
  mov    $0x1,%eax
  jmp    L50
L90:
  mov    $0xfffffffe,%edi
  call   <test__popcountsi2@plt>
  test   %rax,%rax
  je     L75
  mov    $0x1,%eax
  jmp    L50
L75:
  mov    $0xffffffff,%edi
  call   <test__popcountsi2@plt>
  test   %rax,%rax
  je     L131
  mov    $0x1,%eax
  jmp    L50
L131:
  movl   $0x0,-0x4(%rbp)
  jmp    L22
L58:
  mov    $0x0,%eax
  call   <rand@plt>
  mov    %eax,%edi
  call   <test__popcountsi2@plt>
  test   %rax,%rax
  je     L150
  mov    $0x1,%eax
  jmp    L50
L150:
  addl   $0x1,-0x4(%rbp)
L22:
  cmpl   $0x270f,-0x4(%rbp)
  jle    L58
  mov    $0x0,%eax
L50:
  leave
  ret
```

assistant: ```c
int main() {
  if (test__popcountsi2(0))
    return 1;
  if (test__popcountsi2(1))
    return 1;
  if (test__popcountsi2(2))
    return 1;
  if (test__popcountsi2(0xFFFFFFFD))
    return 1;
  if (test__popcountsi2(0xFFFFFFFE))
    return 1;
  if (test__popcountsi2(0xFFFFFFFF))
    return 1;
  int i;
  for (i = 0; i < 10000; ++i)
    if (test__popcountsi2(rand()))
      return 1;
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %rbx
  mov    $0x0,%edi
  call   <test__popcountsi2@plt>
  mov    %rax,%rdx
  mov    $0x1,%eax
  test   %rdx,%rdx
  je     L21
L27:
  pop    %rbx
  ret
L21:
  mov    $0x1,%edi
  call   <test__popcountsi2@plt>
  mov    %rax,%rdx
  mov    $0x1,%eax
  test   %rdx,%rdx
  jne    L27
  mov    $0x2,%edi
  call   <test__popcountsi2@plt>
  mov    %rax,%rdx
  mov    $0x1,%eax
  test   %rdx,%rdx
  jne    L27
  mov    $0xfffffffd,%edi
  call   <test__popcountsi2@plt>
  mov    %rax,%rdx
  mov    $0x1,%eax
  test   %rdx,%rdx
  jne    L27
  mov    $0xfffffffe,%edi
  call   <test__popcountsi2@plt>
  mov    %rax,%rdx
  mov    $0x1,%eax
  test   %rdx,%rdx
  jne    L27
  mov    $0xffffffff,%edi
  call   <test__popcountsi2@plt>
  test   %rax,%rax
  jne    L0
  mov    $0x2710,%ebx
L23:
  mov    $0x0,%eax
  call   <rand@plt>
  mov    %eax,%edi
  call   <test__popcountsi2@plt>
  test   %rax,%rax
  jne    L102
  sub    $0x1,%ebx
  jne    L23
  jmp    L27
L0:
  mov    $0x1,%eax
  jmp    L27
L102:
  mov    $0x1,%eax
  jmp    L27
```

assistant: ```c
int main() {
  if (test__popcountsi2(0))
    return 1;
  if (test__popcountsi2(1))
    return 1;
  if (test__popcountsi2(2))
    return 1;
  if (test__popcountsi2(0xFFFFFFFD))
    return 1;
  if (test__popcountsi2(0xFFFFFFFE))
    return 1;
  if (test__popcountsi2(0xFFFFFFFF))
    return 1;
  int i;
  for (i = 0; i < 10000; ++i)
    if (test__popcountsi2(rand()))
      return 1;
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbx
  xor    %edi,%edi
  call   <test__popcountsi2@plt>
  test   %rax,%rax
  je     L65
L51:
  mov    $0x1,%eax
  pop    %rbx
  ret
L65:
  mov    $0x1,%edi
  call   <test__popcountsi2@plt>
  test   %rax,%rax
  jne    L51
  mov    $0x2,%edi
  call   <test__popcountsi2@plt>
  test   %rax,%rax
  jne    L51
  mov    $0xfffffffd,%edi
  call   <test__popcountsi2@plt>
  test   %rax,%rax
  jne    L51
  mov    $0xfffffffe,%edi
  call   <test__popcountsi2@plt>
  test   %rax,%rax
  jne    L51
  or     $0xffffffff,%edi
  call   <test__popcountsi2@plt>
  test   %rax,%rax
  jne    L51
  mov    $0x2710,%ebx
L55:
  xor    %eax,%eax
  call   <rand@plt>
  mov    %eax,%edi
  call   <test__popcountsi2@plt>
  test   %rax,%rax
  jne    L51
  sub    $0x1,%ebx
  jne    L55
  pop    %rbx
  ret
```

assistant: ```c
int main() {
  if (test__popcountsi2(0))
    return 1;
  if (test__popcountsi2(1))
    return 1;
  if (test__popcountsi2(2))
    return 1;
  if (test__popcountsi2(0xFFFFFFFD))
    return 1;
  if (test__popcountsi2(0xFFFFFFFE))
    return 1;
  if (test__popcountsi2(0xFFFFFFFF))
    return 1;
  int i;
  for (i = 0; i < 10000; ++i)
    if (test__popcountsi2(rand()))
      return 1;
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  push   %rbx
  xor    %edi,%edi
  call   <test__popcountsi2@plt>
  test   %rax,%rax
  je     L36
L78:
  mov    $0x1,%eax
  pop    %rbx
  ret
L36:
  mov    $0x1,%edi
  call   <test__popcountsi2@plt>
  test   %rax,%rax
  jne    L78
  mov    $0x2,%edi
  call   <test__popcountsi2@plt>
  test   %rax,%rax
  jne    L78
  mov    $0xfffffffd,%edi
  call   <test__popcountsi2@plt>
  test   %rax,%rax
  jne    L78
  mov    $0xfffffffe,%edi
  call   <test__popcountsi2@plt>
  test   %rax,%rax
  jne    L78
  or     $0xffffffff,%edi
  call   <test__popcountsi2@plt>
  test   %rax,%rax
  jne    L78
  mov    $0x2710,%ebx
L54:
  xor    %eax,%eax
  call   <rand@plt>
  mov    %eax,%edi
  call   <test__popcountsi2@plt>
  test   %rax,%rax
  jne    L78
  sub    $0x1,%ebx
  jne    L54
  pop    %rbx
  ret
```

assistant: ```c
int main() {
  if (test__popcountsi2(0))
    return 1;
  if (test__popcountsi2(1))
    return 1;
  if (test__popcountsi2(2))
    return 1;
  if (test__popcountsi2(0xFFFFFFFD))
    return 1;
  if (test__popcountsi2(0xFFFFFFFE))
    return 1;
  if (test__popcountsi2(0xFFFFFFFF))
    return 1;
  int i;
  for (i = 0; i < 10000; ++i)
    if (test__popcountsi2(rand()))
      return 1;
  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
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x4000,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  cltq
  shl    $0x2,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rdx
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x1000,-0x10(%rbp)
  jmp    L26
L23:
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x8(%rbp),%rax
  add    %rax,%rdx
  mov    -0x10(%rbp),%eax
  mov    %eax,(%rdx)
  mov    -0x10(%rbp),%eax
  and    $0x3ff,%eax
  test   %eax,%eax
  jne    L2
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x8(%rbp),%rax
  lea    (%rcx,%rax,1),%rsi
  mov    -0x10(%rbp),%eax
  mov    %edx,%ecx
  mov    %rsi,%rdx
  mov    %eax,%esi
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L2:
  addl   $0x1,-0x10(%rbp)
L26:
  mov    -0x10(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jl     L23
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D8
datatype: string
name: parse_data
arguments: data_tag: D4
datatype: string

tool: D8: Allocate %d ints at %p


tool: D4: p[%d] at %p is %d


assistant: ```c
int main() {
  int i;
  int size = 4 * 4096;
  int *p = malloc(size * sizeof(int));
  printf("Allocate %d ints at %p\n", size, p);
  for (i = 4096; i < size; i++) {
    p[i] = i;
    if (i % 1024 == 0)
      printf("p[%d] at %p is %d\n", i, &p[i], p[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   %r12
  push   %rbp
  push   %rbx
  mov    $0x10000,%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  mov    %rax,%rdx
  mov    $0x4000,%esi
  lea    D19(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x4000,%rbp
  mov    $0x1000,%ebx
  lea    D18(%rip),%r12
  jmp    L10
L12:
  add    $0x1,%ebx
  add    $0x4,%rbp
  cmp    $0x4000,%ebx
  je     L27
L10:
  mov    %ebx,0x0(%rbp)
  test   $0x3ff,%ebx
  jne    L12
  mov    %ebx,%ecx
  mov    %rbp,%rdx
  mov    %ebx,%esi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L12
L27:
  mov    $0x0,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D19
datatype: string
name: extract_symbolic_data
arguments: label: D18
datatype: string

tool: D19: Allocate %d ints at %p


tool: D18: p[%d] at %p is %d


assistant: ```c
int main() {
  int i;
  int size = 4 * 4096;
  int *p = malloc(size * sizeof(int));
  printf("Allocate %d ints at %p\n", size, p);
  for (i = 4096; i < size; i++) {
    p[i] = i;
    if (i % 1024 == 0)
      printf("p[%d] at %p is %d\n", i, &p[i], p[i]);
  }
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  push   %r12
  mov    $0x10000,%edi
  lea    D19(%rip),%r12
  push   %rbp
  push   %rbx
  mov    $0x1000,%ebx
  call   <malloc@plt>
  mov    $0x4000,%esi
  lea    D6(%rip),%rdi
  mov    %rax,%rbp
  mov    %rax,%rdx
  xor    %eax,%eax
  call   <printf@plt>
  add    $0x4000,%rbp
  jmp    L30
L28:
  add    $0x1,%ebx
  add    $0x4,%rbp
  cmp    $0x4000,%ebx
  je     L35
L30:
  mov    %ebx,0x0(%rbp)
  test   $0x3ff,%ebx
  jne    L28
  mov    %ebx,%ecx
  mov    %rbp,%rdx
  mov    %ebx,%esi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%ebx
  add    $0x4,%rbp
  call   <printf@plt>
  cmp    $0x4000,%ebx
  jne    L30
L35:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D19
data_type: string
name: retrieve_data
arguments: data_tag: D6
data_type: string

tool: D19: p[%d] at %p is %d


tool: D6: Allocate %d ints at %p


assistant: ```c
int main() {
  int i;
  int size = 4 * 4096;
  int *p = malloc(size * sizeof(int));
  printf("Allocate %d ints at %p\n", size, p);
  for (i = 4096; i < size; i++) {
    p[i] = i;
    if (i % 1024 == 0)
      printf("p[%d] at %p is %d\n", i, &p[i], p[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   %r12
  mov    $0x10000,%edi
  lea    D2(%rip),%r12
  push   %rbp
  push   %rbx
  mov    $0x1000,%ebx
  call   <malloc@plt>
  mov    $0x4000,%esi
  lea    D3(%rip),%rdi
  mov    %rax,%rbp
  mov    %rax,%rdx
  xor    %eax,%eax
  call   <printf@plt>
  add    $0x4000,%rbp
  jmp    L10
L31:
  add    $0x1,%ebx
  add    $0x4,%rbp
  cmp    $0x4000,%ebx
  je     L28
L10:
  mov    %ebx,0x0(%rbp)
  test   $0x3ff,%ebx
  jne    L31
  mov    %ebx,%ecx
  mov    %rbp,%rdx
  mov    %ebx,%esi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%ebx
  add    $0x4,%rbp
  call   <printf@plt>
  cmp    $0x4000,%ebx
  jne    L10
L28:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D2
type_guessed: string
name: disassemble_data
arguments: tag: D3
type_guessed: string

tool: D2: p[%d] at %p is %d


tool: D3: Allocate %d ints at %p


assistant: ```c
int main() {
  int i;
  int size = 4 * 4096;
  int *p = malloc(size * sizeof(int));
  printf("Allocate %d ints at %p\n", size, p);
  for (i = 4096; i < size; i++) {
    p[i] = i;
    if (i % 1024 == 0)
      printf("p[%d] at %p is %d\n", i, &p[i], p[i]);
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<invTableau>:
  endbr64
  test   %esi,%esi
  jle    L54
  push   %rbx
  mov    %esi,%ebx
  mov    %esi,%r8d
  shr    $0x1f,%r8d
  add    %esi,%r8d
  sar    %r8d
  mov    %rdi,%r9
  lea    -0x1(%rsi),%eax
  lea    (%rax,%rax,8),%rax
  lea    0x24(%rdi,%rax,4),%r11
  movslq %esi,%r10
  shl    $0x2,%r10
  jmp    L5
L38:
  xor    %edi,%eax
  mov    %eax,(%rsi,%rdx,4)
  xor    -0x4(%rcx),%eax
  mov    %eax,-0x4(%rcx)
  xor    %eax,(%rsi,%rdx,4)
L58:
  add    $0x1,%rdx
  sub    $0x4,%rcx
  cmp    %edx,%r8d
  jle    L48
L61:
  mov    (%rsi,%rdx,4),%eax
  mov    -0x4(%rcx),%edi
  cmp    %edi,%eax
  jne    L38
  jmp    L58
L48:
  add    $0x24,%r9
  cmp    %r11,%r9
  je     L27
L5:
  mov    %r9,%rsi
  lea    (%r10,%r9,1),%rcx
  mov    $0x0,%edx
  cmp    $0x1,%ebx
  jg     L61
  jmp    L48
L27:
  pop    %rbx
  ret
L54:
  ret
```

assistant: ```c
void invTableau(int table[9][9], int size) {
  for (int i = 0; i < size; i++) {
    for (int j = 0; j < size / 2; j++) {
      if (table[i][j] != table[i][size - (j + 1)]) {
        table[i][j] ^= table[i][size - (j + 1)];
        table[i][size - (j + 1)] ^= table[i][j];
        table[i][j] ^= table[i][size - (j + 1)];
      }
    }
  }
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<invTableau>:
  endbr64
  test   %esi,%esi
  jle    L30
  mov    %esi,%r10d
  lea    -0x1(%rsi),%edx
  mov    $0xfffffffffffffffc,%r9
  sar    %r10d
  lea    (%rdx,%rdx,8),%rdx
  lea    -0x1(%r10),%ecx
  lea    0x0(,%rcx,4),%rax
  add    %rcx,%rdx
  lea    0x4(%rdi,%rax,1),%r8
  lea    0x28(%rdi,%rdx,4),%r11
  movslq %esi,%rdi
  sub    %rax,%r9
  shl    $0x2,%rdi
  sub    %rax,%rdi
  sub    $0x8,%rdi
  nopw   0x0(%rax,%rax,1)
L3:
  lea    (%r9,%r8,1),%rdx
  lea    (%rdi,%r8,1),%rcx
  test   %r10d,%r10d
  je     L48
  nopl   (%rax)
L38:
  mov    (%rdx),%eax
  mov    (%rcx),%esi
  cmp    %esi,%eax
  je     L40
  xor    %esi,%eax
  mov    %eax,(%rdx)
  xor    (%rcx),%eax
  mov    %eax,(%rcx)
  xor    %eax,(%rdx)
L40:
  add    $0x4,%rdx
  sub    $0x4,%rcx
  cmp    %r8,%rdx
  jne    L38
L48:
  add    $0x24,%r8
  cmp    %r11,%r8
  jne    L3
L30:
  ret
```

assistant: ```c
void invTableau(int table[9][9], int size) {
  for (int i = 0; i < size; i++) {
    for (int j = 0; j < size / 2; j++) {
      if (table[i][j] != table[i][size - (j + 1)]) {
        table[i][j] ^= table[i][size - (j + 1)];
        table[i][size - (j + 1)] ^= table[i][j];
        table[i][j] ^= table[i][size - (j + 1)];
      }
    }
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<invTableau>:
  endbr64
  test   %esi,%esi
  jle    L26
  mov    %esi,%eax
  sar    %eax
  je     L26
  lea    -0x1(%rax),%ecx
  lea    -0x1(%rsi),%eax
  lea    (%rax,%rax,8),%rax
  lea    0x0(,%rcx,4),%rdx
  add    %rcx,%rax
  lea    0x4(%rdi,%rdx,1),%r8
  lea    0x28(%rdi,%rax,4),%r10
  movslq %esi,%rax
  mov    $0xfffffffffffffffc,%rdi
  shl    $0x2,%rax
  sub    %rdx,%rdi
  sub    %rdx,%rax
  lea    -0x8(%rax),%r9
  nopl   (%rax)
L25:
  lea    (%rdi,%r8,1),%rdx
  lea    (%r9,%r8,1),%rcx
L30:
  mov    (%rdx),%eax
  mov    (%rcx),%esi
  cmp    %esi,%eax
  je     L1
  xor    %esi,%eax
  mov    %eax,(%rdx)
  xor    (%rcx),%eax
  mov    %eax,(%rcx)
  xor    %eax,(%rdx)
L1:
  add    $0x4,%rdx
  sub    $0x4,%rcx
  cmp    %r8,%rdx
  jne    L30
  add    $0x24,%r8
  cmp    %r10,%r8
  jne    L25
L26:
  ret
```

assistant: ```c
void invTableau(int table[9][9], int size) {
  for (int i = 0; i < size; i++) {
    for (int j = 0; j < size / 2; j++) {
      if (table[i][j] != table[i][size - (j + 1)]) {
        table[i][j] ^= table[i][size - (j + 1)];
        table[i][size - (j + 1)] ^= table[i][j];
        table[i][j] ^= table[i][size - (j + 1)];
      }
    }
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<HAL_core_subsystem_version>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  mov    -0x8(%rbp),%rax
  lea    D9(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    $0x5,%eax
  leave
  ret
```

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

tool: D9: 1.00

assistant: ```c
int HAL_core_subsystem_version(char *patchstr, int bufLen) {
  strcpy(patchstr, "1.00");
  return 5;
}
```
user: What is the c source code of the assembly code below?

```asm
<HAL_core_subsystem_version>:
  endbr64
  sub    $0x8,%rsp
  lea    D1(%rip),%rsi
  call   <strcpy@plt>
  mov    $0x5,%eax
  add    $0x8,%rsp
  ret
```

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

tool: D1: 1.00

assistant: ```c
int HAL_core_subsystem_version(char *patchstr, int bufLen) {
  strcpy(patchstr, "1.00");
  return 5;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<HAL_core_subsystem_version>:
  endbr64
  sub    $0x8,%rsp
  lea    D7(%rip),%rsi
  call   <strcpy@plt>
  mov    $0x5,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: label: D7
datatype: string

tool: D7: 1.00

assistant: ```c
int HAL_core_subsystem_version(char *patchstr, int bufLen) {
  strcpy(patchstr, "1.00");
  return 5;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<HAL_core_subsystem_version>:
  endbr64
  sub    $0x8,%rsp
  lea    D2(%rip),%rsi
  call   <strcpy@plt>
  mov    $0x5,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D2
type_infered: string

tool: D2: 1.00

assistant: ```c
int HAL_core_subsystem_version(char *patchstr, int bufLen) {
  strcpy(patchstr, "1.00");
  return 5;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x10(%rbp),%rdx
  lea    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x18(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  cmp    %rax,%rdx
  jne    L18
  mov    -0x18(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  add    %rax,%rax
  mov    %rax,%rsi
  lea    D40(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x18(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  imul   %rdx,%rax
  mov    %rax,%rsi
  lea    D34(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L21
L18:
  mov    -0x18(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  add    %rax,%rax
  mov    %rax,%rsi
  lea    D30(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x18(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  imul   %rdx,%rax
  mov    %rax,%rsi
  lea    D46(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L21:
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L10
  call   <__stack_chk_fail@plt>
L10:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D18
type: string
name: unpack_data
arguments: data_tag: D40
type: string
name: unpack_data
arguments: data_tag: D34
type: string
name: unpack_data
arguments: data_tag: D30
type: string
name: unpack_data
arguments: data_tag: D46
type: string

tool: D18: %ld %ld

tool: D40: The perimeter of the square is %ld m.


tool: D34: The area of the square is %ld m2.


tool: D30: The perimeter of the rectangle is %ld m.


tool: D46: The area of the rectangle is %ld m2.


assistant: ```c
main() {
  long int n1, n2;
  scanf("%ld %ld", &n1, &n2);
  if (n1 == n2) {
    printf("The perimeter of the square is %ld m.\n", ((n1 * 2) + (n2 * 2)));
    printf("The area of the square is %ld m2.\n", (n1 * n2));
  } else {
    printf("The perimeter of the rectangle is %ld m.\n", ((n1 * 2) + (n2 * 2)));
    printf("The area of the rectangle is %ld m2.\n", (n1 * n2));
  }
  return 0;
}
```