ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D16(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  mov    0x4(%rsp),%esi
  lea    D21(%rip),%rdi
  imul   %esi,%esi
  call   <printf@plt>
  mov    0x4(%rsp),%eax
  lea    D0(%rip),%rdi
  mov    %eax,%esi
  imul   %eax,%esi
  imul   %eax,%esi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D16
type_infered: string
name: extract_symbolic_data
arguments: tag: D21
type_infered: string
name: extract_symbolic_data
arguments: tag: D0
type_infered: string

tool: D16: %d

tool: D21: 
%d


tool: D0: %d


assistant: ```c
int main() {
  int num;
  scanf("%d", &num);
  printf("\n%d\n", num * num);
  printf("%d\n", num * num * num);
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<readLine>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
L6:
  mov    -0x20(%rbp),%rcx
  mov    -0x14(%rbp),%eax
  mov    $0x1,%edx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <read@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jle    L15
  mov    -0x20(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x20(%rbp)
  movzbl (%rax),%eax
  test   %al,%al
  jne    L6
L15:
  cmpl   $0x0,-0x4(%rbp)
  setg   %al
  movzbl %al,%eax
  leave
  ret
```

assistant: ```c
int readLine(int fd, char *str) {
  int n;
  do {
    n = read(fd, str, 1);
  } while (n > 0 && *str++ != '\0');
  return (n > 0);
}
```
user: Translate the below assembly code into C programming language:

```asm
<readLine>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebp
  mov    %rsi,%rbx
L14:
  mov    $0x1,%edx
  mov    %rbx,%rsi
  mov    %ebp,%edi
  call   <read@plt>
  test   %eax,%eax
  jle    L15
  add    $0x1,%rbx
  cmpb   $0x0,-0x1(%rbx)
  jne    L14
L15:
  test   %eax,%eax
  setg   %al
  movzbl %al,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int readLine(int fd, char *str) {
  int n;
  do {
    n = read(fd, str, 1);
  } while (n > 0 && *str++ != '\0');
  return (n > 0);
}
```
user: Translate the below assembly code into C programming language:

```asm
<readLine>:
  endbr64
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x8,%rsp
  jmp    L21
  nopl   0x0(%rax)
L29:
  add    $0x1,%rbx
  cmpb   $0x0,-0x1(%rbx)
  je     L6
L21:
  mov    $0x1,%edx
  mov    %rbx,%rsi
  mov    %ebp,%edi
  call   <read@plt>
  test   %eax,%eax
  jg     L29
L6:
  test   %eax,%eax
  setg   %al
  add    $0x8,%rsp
  movzbl %al,%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int readLine(int fd, char *str) {
  int n;
  do {
    n = read(fd, str, 1);
  } while (n > 0 && *str++ != '\0');
  return (n > 0);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<readLine>:
  endbr64
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x8,%rsp
  jmp    L7
  nopl   0x0(%rax)
L19:
  add    $0x1,%rbx
  cmpb   $0x0,-0x1(%rbx)
  je     L6
L7:
  mov    $0x1,%edx
  mov    %rbx,%rsi
  mov    %ebp,%edi
  call   <read@plt>
  test   %eax,%eax
  jg     L19
L6:
  test   %eax,%eax
  setg   %al
  add    $0x8,%rsp
  movzbl %al,%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int readLine(int fd, char *str) {
  int n;
  do {
    n = read(fd, str, 1);
  } while (n > 0 && *str++ != '\0');
  return (n > 0);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<qsort>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    %rdx,-0x48(%rbp)
  mov    %rcx,-0x50(%rbp)
  mov    -0x38(%rbp),%rax
  mov    %rax,-0x20(%rbp)
  cmpq   $0x1,-0x40(%rbp)
  jbe    L48
  movl   $0x0,-0x2c(%rbp)
  mov    -0x48(%rbp),%rax
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,-0x18(%rbp)
  movq   $0x0,-0x28(%rbp)
  jmp    L28
L19:
  mov    -0x28(%rbp),%rax
  imul   -0x48(%rbp),%rax
  mov    %rax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rdx
  mov    -0x48(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  mov    -0x50(%rbp),%rcx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   *%rcx
  test   %eax,%eax
  jle    L34
  movl   $0x1,-0x2c(%rbp)
  mov    -0x48(%rbp),%rdx
  mov    -0x10(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0x48(%rbp),%rdx
  mov    -0x8(%rbp),%rcx
  mov    -0x10(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0x48(%rbp),%rdx
  mov    -0x18(%rbp),%rcx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
L34:
  addq   $0x1,-0x28(%rbp)
L28:
  mov    -0x40(%rbp),%rax
  sub    $0x1,%rax
  cmp    %rax,-0x28(%rbp)
  jb     L19
  cmpl   $0x0,-0x2c(%rbp)
  je     L29
  mov    -0x40(%rbp),%rax
  lea    -0x1(%rax),%rsi
  mov    -0x50(%rbp),%rcx
  mov    -0x48(%rbp),%rdx
  mov    -0x38(%rbp),%rax
  mov    %rax,%rdi
  call   <qsort>
  jmp    L29
L48:
  nop
L29:
  leave
  ret
```

assistant: ```c
void qsort(void *base_, size_t nmemb, size_t size, int (*compar)(const void *, const void *)) {
  char *base = (char *)base_;
  size_t i;
  if (nmemb < 2) {
    return;
  };
  int doneSwaps = 0;
  char *swapBuffer = (char *)malloc(size);
  for (i = 0; i < (nmemb - 1); i++) {
    char *a = &base[i * size];
    char *b = &a[size];
    if (compar(a, b) > 0) {
      doneSwaps = 1;
      memcpy(swapBuffer, a, size);
      memcpy(a, b, size);
      memcpy(b, swapBuffer, size);
    };
  };
  if (doneSwaps) {
    qsort(base_, nmemb - 1, size, compar);
  };
}
```
user: What is the c source code of the assembly code below:

```asm
<qsort>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,0x18(%rsp)
  cmp    $0x1,%rsi
  jbe    L55
  mov    %rdi,%rbx
  mov    %rsi,%r14
  mov    %rdx,%r12
  mov    %rcx,%r15
  mov    %rdx,%rdi
  call   <malloc@plt>
  mov    %rax,0x8(%rsp)
  sub    $0x1,%r14
  movl   $0x0,0x14(%rsp)
  mov    $0x0,%ebp
  jmp    L17
L66:
  add    $0x1,%rbp
  cmp    %r14,%rbp
  je     L2
L17:
  mov    %rbx,%r13
  add    %r12,%rbx
  mov    %rbx,%rsi
  mov    %r13,%rdi
  call   *%r15
  test   %eax,%eax
  jle    L66
  mov    %r12,%rdx
  mov    %r13,%rsi
  mov    0x8(%rsp),%rdi
  call   <memcpy@plt>
  mov    %r12,%rdx
  mov    %rbx,%rsi
  mov    %r13,%rdi
  call   <memcpy@plt>
  mov    %r12,%rdx
  mov    0x8(%rsp),%rsi
  mov    %rbx,%rdi
  call   <memcpy@plt>
  movl   $0x1,0x14(%rsp)
  jmp    L66
L2:
  cmpl   $0x0,0x14(%rsp)
  jne    L21
L55:
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L21:
  mov    %r15,%rcx
  mov    %r12,%rdx
  mov    %r14,%rsi
  mov    0x18(%rsp),%rdi
  call   <qsort>
  jmp    L55
```

assistant: ```c
void qsort(void *base_, size_t nmemb, size_t size, int (*compar)(const void *, const void *)) {
  char *base = (char *)base_;
  size_t i;
  if (nmemb < 2) {
    return;
  };
  int doneSwaps = 0;
  char *swapBuffer = (char *)malloc(size);
  for (i = 0; i < (nmemb - 1); i++) {
    char *a = &base[i * size];
    char *b = &a[size];
    if (compar(a, b) > 0) {
      doneSwaps = 1;
      memcpy(swapBuffer, a, size);
      memcpy(a, b, size);
      memcpy(b, swapBuffer, size);
    };
  };
  if (doneSwaps) {
    qsort(base_, nmemb - 1, size, compar);
  };
}
```
user: Translate the below assembly code into C programming language:

```asm
<qsort>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,0x18(%rsp)
  mov    %rcx,0x8(%rsp)
  nopw   0x0(%rax,%rax,1)
L25:
  cmp    $0x1,%r12
  jbe    L0
  mov    %rbp,%rdi
  sub    $0x1,%r12
  xor    %ebx,%ebx
  call   <malloc@plt>
  movl   $0x0,0x14(%rsp)
  mov    0x18(%rsp),%r15
  mov    %rax,%r13
L13:
  mov    %r15,%r14
  add    %rbp,%r15
  mov    0x8(%rsp),%rax
  mov    %r15,%rsi
  mov    %r14,%rdi
  call   *%rax
  test   %eax,%eax
  jle    L32
  mov    %rbp,%rdx
  mov    %r14,%rsi
  mov    %r13,%rdi
  call   <memcpy@plt>
  mov    %rbp,%rdx
  mov    %r15,%rsi
  mov    %r14,%rdi
  call   <memcpy@plt>
  mov    %rbp,%rdx
  mov    %r13,%rsi
  mov    %r15,%rdi
  call   <memcpy@plt>
  movl   $0x1,0x14(%rsp)
L32:
  add    $0x1,%rbx
  cmp    %r12,%rbx
  jne    L13
  mov    0x14(%rsp),%eax
  test   %eax,%eax
  jne    L25
L0:
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void qsort(void *base_, size_t nmemb, size_t size, int (*compar)(const void *, const void *)) {
  char *base = (char *)base_;
  size_t i;
  if (nmemb < 2) {
    return;
  };
  int doneSwaps = 0;
  char *swapBuffer = (char *)malloc(size);
  for (i = 0; i < (nmemb - 1); i++) {
    char *a = &base[i * size];
    char *b = &a[size];
    if (compar(a, b) > 0) {
      doneSwaps = 1;
      memcpy(swapBuffer, a, size);
      memcpy(a, b, size);
      memcpy(b, swapBuffer, size);
    };
  };
  if (doneSwaps) {
    qsort(base_, nmemb - 1, size, compar);
  };
}
```
user: Translate the below assembly code into C programming language:

```asm
<qsort>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,0x18(%rsp)
  mov    %rcx,0x8(%rsp)
  nopw   0x0(%rax,%rax,1)
L16:
  cmp    $0x1,%r12
  jbe    L10
  mov    %rbp,%rdi
  sub    $0x1,%r12
  xor    %ebx,%ebx
  call   <malloc@plt>
  movl   $0x0,0x14(%rsp)
  mov    0x18(%rsp),%r15
  mov    %rax,%r13
L7:
  mov    %r15,%r14
  add    %rbp,%r15
  mov    0x8(%rsp),%rax
  mov    %r15,%rsi
  mov    %r14,%rdi
  call   *%rax
  test   %eax,%eax
  jle    L3
  mov    %rbp,%rdx
  mov    %r14,%rsi
  mov    %r13,%rdi
  call   <memcpy@plt>
  mov    %rbp,%rdx
  mov    %r15,%rsi
  mov    %r14,%rdi
  call   <memcpy@plt>
  mov    %rbp,%rdx
  mov    %r13,%rsi
  mov    %r15,%rdi
  call   <memcpy@plt>
  movl   $0x1,0x14(%rsp)
L3:
  add    $0x1,%rbx
  cmp    %r12,%rbx
  jne    L7
  mov    0x14(%rsp),%eax
  test   %eax,%eax
  jne    L16
L10:
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void qsort(void *base_, size_t nmemb, size_t size, int (*compar)(const void *, const void *)) {
  char *base = (char *)base_;
  size_t i;
  if (nmemb < 2) {
    return;
  };
  int doneSwaps = 0;
  char *swapBuffer = (char *)malloc(size);
  for (i = 0; i < (nmemb - 1); i++) {
    char *a = &base[i * size];
    char *b = &a[size];
    if (compar(a, b) > 0) {
      doneSwaps = 1;
      memcpy(swapBuffer, a, size);
      memcpy(a, b, size);
      memcpy(b, swapBuffer, size);
    };
  };
  if (doneSwaps) {
    qsort(base_, nmemb - 1, size, compar);
  };
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<CopyArray4>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L14
L18:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x20(%rbp),%rax
  add    %rcx,%rax
  movss  (%rdx),%xmm0
  movss  %xmm0,(%rax)
  mov    -0x4(%rbp),%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,4),%rcx
  mov    -0x20(%rbp),%rax
  add    %rcx,%rax
  movss  (%rdx),%xmm0
  movss  %xmm0,(%rax)
  mov    -0x4(%rbp),%eax
  cltq
  add    $0x2,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  cltq
  add    $0x2,%rax
  lea    0x0(,%rax,4),%rcx
  mov    -0x20(%rbp),%rax
  add    %rcx,%rax
  movss  (%rdx),%xmm0
  movss  %xmm0,(%rax)
  mov    -0x4(%rbp),%eax
  cltq
  add    $0x3,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  cltq
  add    $0x3,%rax
  lea    0x0(,%rax,4),%rcx
  mov    -0x20(%rbp),%rax
  add    %rcx,%rax
  movss  (%rdx),%xmm0
  movss  %xmm0,(%rax)
  addl   $0x4,-0x4(%rbp)
L14:
  mov    -0x4(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jl     L18
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void CopyArray4(float *a, float *b, int len) {
  int i = 0;
  while (i < len) {
    b[i] = a[i];
    b[i + 1] = a[i + 1];
    b[i + 2] = a[i + 2];
    b[i + 3] = a[i + 3];
    i += 4;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<CopyArray4>:
  endbr64
  test   %edx,%edx
  jle    L2
  mov    %rdi,%rax
  sub    $0x1,%edx
  shr    $0x2,%edx
  mov    %edx,%edx
  shl    $0x4,%rdx
  lea    0x10(%rdi,%rdx,1),%rdx
L17:
  movss  (%rax),%xmm0
  movss  %xmm0,(%rsi)
  movss  0x4(%rax),%xmm0
  movss  %xmm0,0x4(%rsi)
  movss  0x8(%rax),%xmm0
  movss  %xmm0,0x8(%rsi)
  movss  0xc(%rax),%xmm0
  movss  %xmm0,0xc(%rsi)
  add    $0x10,%rax
  add    $0x10,%rsi
  cmp    %rdx,%rax
  jne    L17
L2:
  ret
```

assistant: ```c
void CopyArray4(float *a, float *b, int len) {
  int i = 0;
  while (i < len) {
    b[i] = a[i];
    b[i + 1] = a[i + 1];
    b[i + 2] = a[i + 2];
    b[i + 3] = a[i + 3];
    i += 4;
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<CopyArray4>:
  endbr64
  test   %edx,%edx
  jle    L9
  sub    $0x1,%edx
  shr    $0x2,%edx
  shl    $0x4,%rdx
  lea    0x10(%rdi,%rdx,1),%rax
  nopw   0x0(%rax,%rax,1)
L6:
  movss  (%rdi),%xmm0
  add    $0x10,%rdi
  add    $0x10,%rsi
  movss  %xmm0,-0x10(%rsi)
  movss  -0xc(%rdi),%xmm0
  movss  %xmm0,-0xc(%rsi)
  movss  -0x8(%rdi),%xmm0
  movss  %xmm0,-0x8(%rsi)
  movss  -0x4(%rdi),%xmm0
  movss  %xmm0,-0x4(%rsi)
  cmp    %rax,%rdi
  jne    L6
L9:
  ret
```

assistant: ```c
void CopyArray4(float *a, float *b, int len) {
  int i = 0;
  while (i < len) {
    b[i] = a[i];
    b[i + 1] = a[i + 1];
    b[i + 2] = a[i + 2];
    b[i + 3] = a[i + 3];
    i += 4;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<CopyArray4>:
  endbr64
  test   %edx,%edx
  jle    L34
  lea    0xf(%rdi),%rax
  sub    %rsi,%rax
  cmp    $0x1e,%rax
  jbe    L8
  sub    $0x1,%edx
  xor    %eax,%eax
  shr    $0x2,%edx
  lea    0x1(%rdx),%ecx
  xor    %edx,%edx
  nopw   0x0(%rax,%rax,1)
L38:
  movups (%rdi,%rax,1),%xmm1
  add    $0x1,%edx
  movups %xmm1,(%rsi,%rax,1)
  add    $0x10,%rax
  cmp    %ecx,%edx
  jb     L38
  ret
  nopl   0x0(%rax)
L8:
  lea    -0x1(%rdx),%eax
  shr    $0x2,%eax
  shl    $0x4,%rax
  lea    0x10(%rdi,%rax,1),%rax
  nop
L37:
  movss  (%rdi),%xmm0
  add    $0x10,%rdi
  add    $0x10,%rsi
  movss  %xmm0,-0x10(%rsi)
  movss  -0xc(%rdi),%xmm0
  movss  %xmm0,-0xc(%rsi)
  movss  -0x8(%rdi),%xmm0
  movss  %xmm0,-0x8(%rsi)
  movss  -0x4(%rdi),%xmm0
  movss  %xmm0,-0x4(%rsi)
  cmp    %rdi,%rax
  jne    L37
L34:
  ret
```

assistant: ```c
void CopyArray4(float *a, float *b, int len) {
  int i = 0;
  while (i < len) {
    b[i] = a[i];
    b[i + 1] = a[i + 1];
    b[i + 2] = a[i + 2];
    b[i + 3] = a[i + 3];
    i += 4;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<cell>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jne    L30
  lea    D0(%rip),%rax
  jmp    L14
L30:
  cmpl   $0x0,-0x4(%rbp)
  jns    L24
  mov    -0x4(%rbp),%eax
  neg    %eax
  mov    %eax,%edx
  lea    D49(%rip),%rax
  mov    %rax,%rsi
  lea    D34(%rip),%rax
  mov    %rax,%rdi
  call   <sprintf@plt>
  jmp    L7
L24:
  mov    -0x4(%rbp),%eax
  mov    %eax,%edx
  lea    D48(%rip),%rax
  mov    %rax,%rsi
  lea    D34(%rip),%rax
  mov    %rax,%rdi
  call   <sprintf@plt>
L7:
  lea    D34(%rip),%rax
L14:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: label: D0
type: string
name: unpack_data
arguments: label: D49
type: string
name: unpack_data
arguments: label: D48
type: string

tool: D0: $m[$p]

tool: D49: $m[$p-%d]

tool: D48: $m[$p+%d]

assistant: ```c
const char *cell(int mov) {
  static char buf[9 + 3 + sizeof(mov) * 3];
  if (mov == 0)
    return "$m[$p]";
  if (mov < 0)
    sprintf(buf, "$m[$p-%d]", -mov);
  else
    sprintf(buf, "$m[$p+%d]", mov);
  return buf;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<cell>:
  endbr64
  lea    D28(%rip),%rax
  test   %edi,%edi
  je     L6
  push   %rbx
  js     L23
  mov    %edi,%edx
  lea    D47(%rip),%rsi
  lea    D12(%rip),%rbx
  mov    %rbx,%rdi
  call   <sprintf@plt>
  mov    %rbx,%rax
L1:
  pop    %rbx
  ret
L23:
  mov    %edi,%edx
  neg    %edx
  lea    D41(%rip),%rsi
  lea    D12(%rip),%rbx
  mov    %rbx,%rdi
  call   <sprintf@plt>
  mov    %rbx,%rax
  jmp    L1
L6:
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D28
type_guessed: string
name: retrieve_data
arguments: data_tag: D47
type_guessed: string
name: retrieve_data
arguments: data_tag: D41
type_guessed: string

tool: D28: $m[$p]

tool: D47: $m[$p+%d]

tool: D41: $m[$p-%d]

assistant: ```c
const char *cell(int mov) {
  static char buf[9 + 3 + sizeof(mov) * 3];
  if (mov == 0)
    return "$m[$p]";
  if (mov < 0)
    sprintf(buf, "$m[$p-%d]", -mov);
  else
    sprintf(buf, "$m[$p+%d]", mov);
  return buf;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<cell>:
  endbr64
  lea    D27(%rip),%rax
  test   %edi,%edi
  je     L19
  push   %rbx
  mov    %edi,%edx
  js     L11
  lea    D6(%rip),%rbx
  lea    D34(%rip),%rsi
  mov    %rbx,%rdi
  call   <sprintf@plt>
  mov    %rbx,%rax
  pop    %rbx
  ret
  nop
L11:
  lea    D6(%rip),%rbx
  neg    %edx
  lea    D7(%rip),%rsi
  mov    %rbx,%rdi
  call   <sprintf@plt>
  mov    %rbx,%rax
  pop    %rbx
  ret
  nopl   (%rax)
L19:
  ret
```

tool_calls: name: read_data
arguments: label: D27
type_guessed: string
name: read_data
arguments: label: D34
type_guessed: string
name: read_data
arguments: label: D7
type_guessed: string

tool: D27: $m[$p]

tool: D34: $m[$p+%d]

tool: D7: $m[$p-%d]

assistant: ```c
const char *cell(int mov) {
  static char buf[9 + 3 + sizeof(mov) * 3];
  if (mov == 0)
    return "$m[$p]";
  if (mov < 0)
    sprintf(buf, "$m[$p-%d]", -mov);
  else
    sprintf(buf, "$m[$p+%d]", mov);
  return buf;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<cell>:
  endbr64
  lea    D11(%rip),%rax
  test   %edi,%edi
  je     L9
  push   %rbx
  mov    %edi,%edx
  js     L1
  lea    D2(%rip),%rbx
  lea    D8(%rip),%rsi
  mov    %rbx,%rdi
  call   <sprintf@plt>
  mov    %rbx,%rax
  pop    %rbx
  ret
  nop
L1:
  lea    D2(%rip),%rbx
  neg    %edx
  lea    D5(%rip),%rsi
  mov    %rbx,%rdi
  call   <sprintf@plt>
  mov    %rbx,%rax
  pop    %rbx
  ret
  nopl   (%rax)
L9:
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D11
type_infered: string
name: retrieve_data
arguments: data_tag: D8
type_infered: string
name: retrieve_data
arguments: data_tag: D5
type_infered: string

tool: D11: $m[$p]

tool: D8: $m[$p+%d]

tool: D5: $m[$p-%d]

assistant: ```c
const char *cell(int mov) {
  static char buf[9 + 3 + sizeof(mov) * 3];
  if (mov == 0)
    return "$m[$p]";
  if (mov < 0)
    sprintf(buf, "$m[$p-%d]", -mov);
  else
    sprintf(buf, "$m[$p+%d]", mov);
  return buf;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<plm_dPlm>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %edi,-0x34(%rbp)
  mov    %esi,-0x38(%rbp)
  movaps %xmm0,-0x50(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x24(%rbp),%rdx
  movdqa -0x50(%rbp),%xmm0
  mov    -0x38(%rbp),%ecx
  mov    -0x34(%rbp),%eax
  mov    %ecx,%esi
  mov    %eax,%edi
  mov    $0x1,%eax
  call   <plm_lndPlm@plt>
  mov    %eax,%edi
  call   <__floatsitf>
  movaps %xmm0,-0x20(%rbp)
  movdqa -0x20(%rbp),%xmm0
  mov    $0x1,%eax
  call   <expq@plt>
  mov    -0x24(%rbp),%edx
  imul   %edx,%eax
  mov    %eax,%edi
  call   <__floatsitf>
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L9
  call   <__stack_chk_fail@plt>
L9:
  leave
  ret
```

assistant: ```c
__float128 plm_dPlm(int l, int m, __float128 x) {
  int sign;
  __float128 value = plm_lndPlm(l, m, x, &sign);
  return sign * expq(value);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<plm_dPlm>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rdx
  mov    $0x1,%eax
  call   <plm_lndPlm@plt>
  mov    %eax,%edi
  call   <__floatsitf>
  mov    $0x1,%eax
  call   <expq@plt>
  imul   0x4(%rsp),%eax
  mov    %eax,%edi
  call   <__floatsitf>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x18,%rsp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
__float128 plm_dPlm(int l, int m, __float128 x) {
  int sign;
  __float128 value = plm_lndPlm(l, m, x, &sign);
  return sign * expq(value);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<plm_dPlm>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rdx
  mov    $0x1,%eax
  call   <plm_lndPlm@plt>
  mov    %eax,%edi
  call   <__floatsitf>
  mov    $0x1,%eax
  call   <expq@plt>
  imul   0x4(%rsp),%eax
  mov    %eax,%edi
  call   <__floatsitf>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  add    $0x18,%rsp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
__float128 plm_dPlm(int l, int m, __float128 x) {
  int sign;
  __float128 value = plm_lndPlm(l, m, x, &sign);
  return sign * expq(value);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<plm_dPlm>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rdx
  mov    $0x1,%eax
  call   <plm_lndPlm@plt>
  mov    %eax,%edi
  call   <__floatsitf>
  mov    $0x1,%eax
  call   <expq@plt>
  imul   0x4(%rsp),%eax
  mov    %eax,%edi
  call   <__floatsitf>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  add    $0x18,%rsp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
__float128 plm_dPlm(int l, int m, __float128 x) {
  int sign;
  __float128 value = plm_lndPlm(l, m, x, &sign);
  return sign * expq(value);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<LucasLehmerTest>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    -0x8(%rbp),%eax
  mov    $0x4,%esi
  mov    %eax,%edi
  call   <mpz_init_set_ui@plt>
  movl   $0x2,-0xc(%rbp)
  jmp    L17
L18:
  mov    -0x8(%rbp),%edx
  mov    -0x8(%rbp),%ecx
  mov    -0x8(%rbp),%eax
  mov    %ecx,%esi
  mov    %eax,%edi
  call   <mpz_mul@plt>
  mov    -0x8(%rbp),%ecx
  mov    -0x8(%rbp),%eax
  mov    $0x2,%edx
  mov    %ecx,%esi
  mov    %eax,%edi
  call   <mpz_sub_ui@plt>
  mov    -0x14(%rbp),%edx
  mov    -0x8(%rbp),%ecx
  mov    -0x8(%rbp),%eax
  mov    %ecx,%esi
  mov    %eax,%edi
  call   <mpz_mod@plt>
  addl   $0x1,-0xc(%rbp)
L17:
  mov    -0x18(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jg     L18
  mov    -0x8(%rbp),%eax
  mov    $0x0,%esi
  mov    %eax,%edi
  call   <mpz_cmp_ui@plt>
  test   %rax,%rax
  sete   %al
  movzbl %al,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x8(%rbp),%eax
  mov    %eax,%edi
  call   <mpz_clear@plt>
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int LucasLehmerTest(mpz_t m, int p) {
  int i, ret;
  mpz_t s;
  mpz_init_set_ui(s, 4);
  for (i = 2; i <= p - 1; i++) {
    mpz_mul(s, s, s);
    mpz_sub_ui(s, s, 2);
    mpz_mod(s, s, m);
  }
  ret = (mpz_cmp_ui(s, 0) == 0);
  mpz_clear(s);
  return (ret);
}
```