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

```asm
<print_file_info>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    $0x0,%eax
  call   <print_permissions@plt>
  mov    $0x0,%eax
  call   <print_link@plt>
  mov    $0x0,%eax
  call   <print_owner@plt>
  mov    $0x0,%eax
  call   <print_group@plt>
  mov    $0x0,%eax
  call   <print_size@plt>
  mov    $0x0,%eax
  call   <print_date@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <print_file_name@plt>
  mov    $0x0,%eax
  call   <count_blocks@plt>
  nop
  leave
  ret
```

assistant: ```c
void print_file_info(char *dirname, char *filename) {
  print_permissions();
  print_link();
  print_owner();
  print_group();
  print_size();
  print_date();
  print_file_name(filename);
  count_blocks();
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<print_file_info>:
  endbr64
  push   %rbx
  mov    %rsi,%rbx
  mov    $0x0,%eax
  call   <print_permissions@plt>
  mov    $0x0,%eax
  call   <print_link@plt>
  mov    $0x0,%eax
  call   <print_owner@plt>
  mov    $0x0,%eax
  call   <print_group@plt>
  mov    $0x0,%eax
  call   <print_size@plt>
  mov    $0x0,%eax
  call   <print_date@plt>
  mov    %rbx,%rdi
  call   <print_file_name@plt>
  mov    $0x0,%eax
  call   <count_blocks@plt>
  pop    %rbx
  ret
```

assistant: ```c
void print_file_info(char *dirname, char *filename) {
  print_permissions();
  print_link();
  print_owner();
  print_group();
  print_size();
  print_date();
  print_file_name(filename);
  count_blocks();
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<print_file_info>:
  endbr64
  push   %rbp
  xor    %eax,%eax
  mov    %rsi,%rbp
  call   <print_permissions@plt>
  xor    %eax,%eax
  call   <print_link@plt>
  xor    %eax,%eax
  call   <print_owner@plt>
  xor    %eax,%eax
  call   <print_group@plt>
  xor    %eax,%eax
  call   <print_size@plt>
  xor    %eax,%eax
  call   <print_date@plt>
  mov    %rbp,%rdi
  call   <print_file_name@plt>
  xor    %eax,%eax
  pop    %rbp
  jmp    10f0 <count_blocks@plt>
```

assistant: ```c
void print_file_info(char *dirname, char *filename) {
  print_permissions();
  print_link();
  print_owner();
  print_group();
  print_size();
  print_date();
  print_file_name(filename);
  count_blocks();
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<print_file_info>:
  endbr64
  push   %rbp
  xor    %eax,%eax
  mov    %rsi,%rbp
  call   <print_permissions@plt>
  xor    %eax,%eax
  call   <print_link@plt>
  xor    %eax,%eax
  call   <print_owner@plt>
  xor    %eax,%eax
  call   <print_group@plt>
  xor    %eax,%eax
  call   <print_size@plt>
  xor    %eax,%eax
  call   <print_date@plt>
  mov    %rbp,%rdi
  call   <print_file_name@plt>
  xor    %eax,%eax
  pop    %rbp
  jmp    10f0 <count_blocks@plt>
```

assistant: ```c
void print_file_info(char *dirname, char *filename) {
  print_permissions();
  print_link();
  print_owner();
  print_group();
  print_size();
  print_date();
  print_file_name(filename);
  count_blocks();
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  movq   $0xf,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  add    $0x104,%rax
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  size_t size = 15;
  void *p = malloc(size + 260);
  ((char *)p)[size] = 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  mov    $0x0,%eax
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  size_t size = 15;
  void *p = malloc(size + 260);
  ((char *)p)[size] = 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  size_t size = 15;
  void *p = malloc(size + 260);
  ((char *)p)[size] = 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<merge>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%r9
  mov    %esi,%r10d
  mov    %edx,%r12d
  mov    %ecx,%r8d
  mov    %fs:0x28,%rax
  mov    %rax,-0x28(%rbp)
  xor    %eax,%eax
  mov    %edx,%r13d
  sub    %esi,%r13d
  mov    %ecx,%ebx
  sub    %edx,%ebx
  lea    0x2(%r13),%eax
  cltq
  lea    0xf(,%rax,4),%rax
  mov    %rax,%rcx
  and    $0xfffffffffffffff0,%rcx
  and    $0xfffffffffffff000,%rax
  mov    %rsp,%rdx
  sub    %rax,%rdx
L37:
  cmp    %rdx,%rsp
  je     L161
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L37
L161:
  mov    %rcx,%rax
  and    $0xfff,%eax
  sub    %rax,%rsp
  test   %rax,%rax
  je     L71
  orq    $0x0,-0x8(%rsp,%rax,1)
L71:
  mov    %rsp,%rsi
  lea    0x1(%rbx),%eax
  cltq
  lea    0xf(,%rax,4),%rax
  mov    %rax,%rcx
  and    $0xfffffffffffffff0,%rcx
  and    $0xfffffffffffff000,%rax
  mov    %rsp,%rdx
  sub    %rax,%rdx
L162:
  cmp    %rdx,%rsp
  je     L70
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L162
L70:
  mov    %rcx,%rax
  and    $0xfff,%eax
  sub    %rax,%rsp
  test   %rax,%rax
  je     L69
  orq    $0x0,-0x8(%rsp,%rax,1)
L69:
  mov    %rsp,%rdi
  test   %r13d,%r13d
  js     L174
  lea    0x1(%r13),%ecx
  shl    $0x2,%rcx
  movslq %r10d,%rax
  lea    (%r9,%rax,4),%r11
  mov    $0x0,%eax
L136:
  mov    (%r11,%rax,1),%edx
  mov    %edx,(%rsi,%rax,1)
  add    $0x4,%rax
  cmp    %rcx,%rax
  jne    L136
  add    $0x1,%r13d
L109:
  movslq %r13d,%r13
  movl   $0x10000,(%rsi,%r13,4)
  test   %ebx,%ebx
  jle    L8
  lea    0x1(%rbx),%r11d
  movslq %r12d,%r12
  lea    (%r9,%r12,4),%rcx
  mov    $0x1,%eax
L101:
  mov    (%rcx,%rax,4),%edx
  mov    %edx,-0x4(%rdi,%rax,4)
  add    $0x1,%rax
  cmp    %r11,%rax
  jne    L101
L173:
  movslq %ebx,%rbx
  movl   $0x10000,(%rdi,%rbx,4)
  cmp    %r8d,%r10d
  jg     L33
  movslq %r10d,%rdx
  lea    (%r9,%rdx,4),%rax
  sub    %r10d,%r8d
  add    %rdx,%r8
  lea    0x4(%r9,%r8,4),%r10
  mov    $0x0,%r9d
  mov    $0x0,%r8d
  jmp    L150
L174:
  mov    $0x0,%r13d
  jmp    L109
L8:
  mov    $0x0,%ebx
  jmp    L173
L134:
  mov    %edx,(%rax)
  add    $0x1,%r9d
L53:
  add    $0x4,%rax
  cmp    %r10,%rax
  je     L33
L150:
  movslq %r8d,%rdx
  mov    (%rsi,%rdx,4),%ecx
  movslq %r9d,%rdx
  mov    (%rdi,%rdx,4),%edx
  cmp    %edx,%ecx
  jg     L134
  mov    %ecx,(%rax)
  add    $0x1,%r8d
  jmp    L53
L33:
  mov    -0x28(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L122
  lea    -0x18(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %rbp
  ret
L122:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void merge(int A[], int p, int q, int r) {
  int n1 = q - p + 1;
  int n2 = r - q;
  int i = 0;
  int j = 0;
  int k = 0;
  int L[n1 + 1];
  int R[n2 + 1];
  for (i = 0; i < n1; i++)
    L[i] = A[p + i];
  L[i] = 65536;
  for (j = 0; j < n2; j++)
    R[j] = A[q + j + 1];
  R[j] = 65536;
  i = 0;
  j = 0;
  for (k = p; k <= r; k++) {
    if (L[i] <= R[j]) {
      A[k] = L[i];
      i++;
    } else {
      A[k] = R[j];
      j++;
    }
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<bcmp>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %edx,-0x14(%rbp)
  jmp    L15
L38:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%edx
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  cmp    %al,%dl
  je     L9
  mov    $0x1,%eax
  jmp    L27
L9:
  subl   $0x1,-0x14(%rbp)
  addq   $0x1,-0x8(%rbp)
  addq   $0x1,-0x10(%rbp)
L15:
  cmpl   $0x0,-0x14(%rbp)
  jg     L38
  mov    $0x0,%eax
L27:
  pop    %rbp
  ret
```

assistant: ```c
int bcmp(s1, s2, len)
char *s1, *s2;
int len;
{
  for (; len > 0; len--, s1++, s2++)
    if (*s1 != *s2)
      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
<bcmp>:
  endbr64
  test   %edx,%edx
  jle    L18
  mov    %edx,%edx
  mov    $0x0,%eax
L9:
  movzbl (%rsi,%rax,1),%ecx
  cmp    %cl,(%rdi,%rax,1)
  jne    L22
  add    $0x1,%rax
  cmp    %rdx,%rax
  jne    L9
  mov    $0x0,%eax
  ret
L18:
  mov    $0x0,%eax
  ret
L22:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int bcmp(s1, s2, len)
char *s1, *s2;
int len;
{
  for (; len > 0; len--, s1++, s2++)
    if (*s1 != *s2)
      return 1;
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<bcmp>:
  endbr64
  test   %edx,%edx
  jle    L27
  movslq %edx,%rdx
  xor    %eax,%eax
  jmp    L13
  nop
L2:
  add    $0x1,%rax
  cmp    %rdx,%rax
  je     L27
L13:
  movzbl (%rsi,%rax,1),%ecx
  cmp    %cl,(%rdi,%rax,1)
  je     L2
  mov    $0x1,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L27:
  xor    %eax,%eax
  ret
```

assistant: ```c
int bcmp(s1, s2, len)
char *s1, *s2;
int len;
{
  for (; len > 0; len--, s1++, s2++)
    if (*s1 != *s2)
      return 1;
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<bcmp>:
  endbr64
  test   %edx,%edx
  jle    L0
  movslq %edx,%rdx
  xor    %eax,%eax
  jmp    L19
  nop
L7:
  add    $0x1,%rax
  cmp    %rdx,%rax
  je     L0
L19:
  movzbl (%rsi,%rax,1),%ecx
  cmp    %cl,(%rdi,%rax,1)
  je     L7
  mov    $0x1,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L0:
  xor    %eax,%eax
  ret
```

assistant: ```c
int bcmp(s1, s2, len)
char *s1, *s2;
int len;
{
  for (; len > 0; len--, s1++, s2++)
    if (*s1 != *s2)
      return 1;
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<my_putnbr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jns    L16
  negl   -0x4(%rbp)
  mov    $0x2d,%edi
  call   <my_putchar@plt>
L16:
  cmpl   $0x9,-0x4(%rbp)
  jle    L10
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,%edi
  call   <my_putnbr>
L10:
  mov    -0x4(%rbp),%ecx
  movslq %ecx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %ecx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  sub    %eax,%ecx
  mov    %ecx,%edx
  lea    0x30(%rdx),%eax
  mov    %eax,%edi
  call   <my_putchar@plt>
  nop
  leave
  ret
```

assistant: ```c
void my_putnbr(int nbr) {
  if (nbr < 0) {
    nbr *= -1;
    my_putchar('-');
  }
  if (nbr / 10 > 0)
    my_putnbr(nbr / 10);
  my_putchar(nbr % 10 + 48);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<my_putnbr>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  test   %edi,%edi
  js     L0
L5:
  cmp    $0x9,%ebx
  jg     L9
L11:
  movslq %ebx,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  mov    %ebx,%edx
  sar    $0x1f,%edx
  sub    %edx,%eax
  lea    (%rax,%rax,4),%eax
  add    %eax,%eax
  sub    %eax,%ebx
  lea    0x30(%rbx),%edi
  call   <my_putchar@plt>
  pop    %rbx
  ret
L0:
  neg    %ebx
  mov    $0x2d,%edi
  call   <my_putchar@plt>
  jmp    L5
L9:
  movslq %ebx,%rdi
  imul   $0x66666667,%rdi,%rdi
  sar    $0x22,%rdi
  mov    %ebx,%eax
  sar    $0x1f,%eax
  sub    %eax,%edi
  call   <my_putnbr>
  jmp    L11
```

assistant: ```c
void my_putnbr(int nbr) {
  if (nbr < 0) {
    nbr *= -1;
    my_putchar('-');
  }
  if (nbr / 10 > 0)
    my_putnbr(nbr / 10);
  my_putchar(nbr % 10 + 48);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<my_putnbr>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%ebx
  sub    $0x8,%rsp
  test   %edi,%edi
  js     L57
L36:
  mov    %ebx,%r12d
  cmp    $0x9,%ebx
  jg     L40
L7:
  mov    $0xcccccccd,%eax
  add    $0x8,%rsp
  imul   %r12,%rax
  shr    $0x23,%rax
  lea    (%rax,%rax,4),%eax
  add    %eax,%eax
  sub    %eax,%ebx
  lea    0x30(%rbx),%edi
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  jmp    1050 <my_putchar@plt>
  nopw   0x0(%rax,%rax,1)
L40:
  mov    $0xcccccccd,%edi
  mov    %r12,%rbp
  imul   %rdi,%rbp
  shr    $0x23,%rbp
  mov    %ebp,%r14d
  cmp    $0x63,%ebx
  jg     L14
L5:
  mov    $0xcccccccd,%eax
  imul   %r14,%rax
  shr    $0x23,%rax
  lea    (%rax,%rax,4),%eax
  add    %eax,%eax
  sub    %eax,%ebp
  lea    0x30(%rbp),%edi
  call   <my_putchar@plt>
  jmp    L7
  cs nopw 0x0(%rax,%rax,1)
L57:
  mov    $0x2d,%edi
  neg    %ebx
  call   <my_putchar@plt>
  jmp    L36
  nopl   0x0(%rax)
L14:
  mov    %r14,%r13
  imul   %rdi,%r13
  shr    $0x23,%r13
  mov    %r13d,%r15d
  cmp    $0x3e7,%ebx
  jg     L53
L72:
  mov    $0xcccccccd,%eax
  imul   %r15,%rax
  shr    $0x23,%rax
  lea    (%rax,%rax,4),%eax
  add    %eax,%eax
  sub    %eax,%r13d
  lea    0x30(%r13),%edi
  call   <my_putchar@plt>
  jmp    L5
  xchg   %ax,%ax
L53:
  imul   %r15,%rdi
  shr    $0x23,%rdi
  call   <my_putnbr>
  jmp    L72
```

assistant: ```c
void my_putnbr(int nbr) {
  if (nbr < 0) {
    nbr *= -1;
    my_putchar('-');
  }
  if (nbr / 10 > 0)
    my_putnbr(nbr / 10);
  my_putchar(nbr % 10 + 48);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<my_putnbr>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%ebx
  sub    $0x8,%rsp
  test   %edi,%edi
  js     L18
L70:
  mov    %ebx,%r12d
  cmp    $0x9,%ebx
  jg     L56
L39:
  mov    $0xcccccccd,%eax
  add    $0x8,%rsp
  imul   %r12,%rax
  shr    $0x23,%rax
  lea    (%rax,%rax,4),%eax
  add    %eax,%eax
  sub    %eax,%ebx
  lea    0x30(%rbx),%edi
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  jmp    1050 <my_putchar@plt>
  nopw   0x0(%rax,%rax,1)
L56:
  mov    $0xcccccccd,%edi
  mov    %r12,%rbp
  imul   %rdi,%rbp
  shr    $0x23,%rbp
  mov    %ebp,%r14d
  cmp    $0x63,%ebx
  jg     L25
L51:
  mov    $0xcccccccd,%eax
  imul   %r14,%rax
  shr    $0x23,%rax
  lea    (%rax,%rax,4),%eax
  add    %eax,%eax
  sub    %eax,%ebp
  lea    0x30(%rbp),%edi
  call   <my_putchar@plt>
  jmp    L39
  cs nopw 0x0(%rax,%rax,1)
L18:
  mov    $0x2d,%edi
  neg    %ebx
  call   <my_putchar@plt>
  jmp    L70
  nopl   0x0(%rax)
L25:
  mov    %r14,%r13
  imul   %rdi,%r13
  shr    $0x23,%r13
  mov    %r13d,%r15d
  cmp    $0x3e7,%ebx
  jg     L45
L31:
  mov    $0xcccccccd,%eax
  imul   %r15,%rax
  shr    $0x23,%rax
  lea    (%rax,%rax,4),%eax
  add    %eax,%eax
  sub    %eax,%r13d
  lea    0x30(%r13),%edi
  call   <my_putchar@plt>
  jmp    L51
  xchg   %ax,%ax
L45:
  imul   %r15,%rdi
  shr    $0x23,%rdi
  call   <my_putnbr>
  jmp    L31
```

assistant: ```c
void my_putnbr(int nbr) {
  if (nbr < 0) {
    nbr *= -1;
    my_putchar('-');
  }
  if (nbr / 10 > 0)
    my_putnbr(nbr / 10);
  my_putchar(nbr % 10 + 48);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<longestCommonPrefixMain>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D33(%rip),%rax
  mov    %rax,-0x30(%rbp)
  lea    D46(%rip),%rax
  mov    %rax,-0x28(%rbp)
  lea    D61(%rip),%rax
  mov    %rax,-0x20(%rbp)
  lea    D17(%rip),%rax
  mov    %rax,-0x18(%rbp)
  movl   $0x0,-0x34(%rbp)
  movl   $0x4,-0x34(%rbp)
  mov    -0x34(%rbp),%edx
  lea    -0x30(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <longestCommonPrefix@plt>
  mov    %rax,%rsi
  lea    D30(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x34(%rbp),%eax
  mov    %eax,%esi
  mov    $0x0,%edi
  call   <longestCommonPrefix@plt>
  mov    %rax,%rsi
  lea    D30(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x30(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <longestCommonPrefix@plt>
  mov    %rax,%rsi
  lea    D30(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L9
  call   <__stack_chk_fail@plt>
L9:
  leave
  ret
```

tool_calls: name: read_data
arguments: tag: D33
datatype: string
name: read_data
arguments: tag: D46
datatype: string
name: read_data
arguments: tag: D61
datatype: string
name: read_data
arguments: tag: D17
datatype: string
name: read_data
arguments: tag: D30
datatype: string

tool: D33: aqwerr

tool: D46: aqwert

tool: D61: aqwerty

tool: D17: aqwertyu

tool: D30: %s


assistant: ```c
void longestCommonPrefixMain(void) {
  char *strs[] = {"aqwerr", "aqwert", "aqwerty", "aqwertyu"};
  int strsSize = 0;
  { strsSize = sizeof(strs) / sizeof(strs[0]); };
  printf("%s\n", longestCommonPrefix(strs, strsSize));
  printf("%s\n", longestCommonPrefix(NULL, strsSize));
  printf("%s\n", longestCommonPrefix(strs, 0));
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<longestCommonPrefixMain>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    D15(%rip),%rax
  mov    %rax,(%rsp)
  lea    D5(%rip),%rax
  mov    %rax,0x8(%rsp)
  lea    D7(%rip),%rax
  mov    %rax,0x10(%rsp)
  lea    D3(%rip),%rax
  mov    %rax,0x18(%rsp)
  mov    %rsp,%rbp
  mov    $0x4,%esi
  mov    %rbp,%rdi
  call   <longestCommonPrefix@plt>
  mov    %rax,%rsi
  lea    D14(%rip),%rbx
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    $0x4,%esi
  mov    $0x0,%edi
  call   <longestCommonPrefix@plt>
  mov    %rax,%rsi
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    $0x0,%esi
  mov    %rbp,%rdi
  call   <longestCommonPrefix@plt>
  mov    %rax,%rsi
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D15
type_guessed: string
name: read_data
arguments: data_tag: D5
type_guessed: string
name: read_data
arguments: data_tag: D7
type_guessed: string
name: read_data
arguments: data_tag: D3
type_guessed: string
name: read_data
arguments: data_tag: D14
type_guessed: string

tool: D15: aqwerr

tool: D5: aqwert

tool: D7: aqwerty

tool: D3: aqwertyu

tool: D14: %s


assistant: ```c
void longestCommonPrefixMain(void) {
  char *strs[] = {"aqwerr", "aqwert", "aqwerty", "aqwertyu"};
  int strsSize = 0;
  { strsSize = sizeof(strs) / sizeof(strs[0]); };
  printf("%s\n", longestCommonPrefix(strs, strsSize));
  printf("%s\n", longestCommonPrefix(NULL, strsSize));
  printf("%s\n", longestCommonPrefix(strs, 0));
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<longestCommonPrefixMain>:
  endbr64
  push   %r12
  mov    $0x4,%esi
  push   %rbp
  lea    D27(%rip),%rbp
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  lea    D4(%rip),%rax
  mov    %rsp,%r12
  mov    %r12,%rdi
  mov    %rax,(%rsp)
  lea    D28(%rip),%rax
  mov    %rax,0x8(%rsp)
  lea    D5(%rip),%rax
  mov    %rax,0x10(%rsp)
  lea    D16(%rip),%rax
  mov    %rax,0x18(%rsp)
  call   <longestCommonPrefix@plt>
  mov    %rbp,%rdi
  mov    %rax,%rsi
  call   <printf@plt>
  xor    %edi,%edi
  mov    $0x4,%esi
  call   <longestCommonPrefix@plt>
  mov    %rbp,%rdi
  mov    %rax,%rsi
  call   <printf@plt>
  xor    %esi,%esi
  mov    %r12,%rdi
  call   <longestCommonPrefix@plt>
  mov    %rbp,%rdi
  mov    %rax,%rsi
  call   <printf@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  add    $0x38,%rsp
  pop    %rbp
  pop    %r12
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D27
data_type: string
name: unpack_data
arguments: data_tag: D4
data_type: string
name: unpack_data
arguments: data_tag: D28
data_type: string
name: unpack_data
arguments: data_tag: D5
data_type: string
name: unpack_data
arguments: data_tag: D16
data_type: string

tool: D27: %s


tool: D4: aqwerr

tool: D28: aqwert

tool: D5: aqwerty

tool: D16: aqwertyu

assistant: ```c
void longestCommonPrefixMain(void) {
  char *strs[] = {"aqwerr", "aqwert", "aqwerty", "aqwertyu"};
  int strsSize = 0;
  { strsSize = sizeof(strs) / sizeof(strs[0]); };
  printf("%s\n", longestCommonPrefix(strs, strsSize));
  printf("%s\n", longestCommonPrefix(NULL, strsSize));
  printf("%s\n", longestCommonPrefix(strs, 0));
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<longestCommonPrefixMain>:
  endbr64
  push   %r12
  lea    D42(%rip),%rdx
  mov    $0x4,%esi
  lea    D23(%rip),%rcx
  push   %rbp
  movq   %rdx,%xmm0
  lea    D37(%rip),%rbp
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  lea    D7(%rip),%rax
  mov    %rsp,%r12
  mov    %r12,%rdi
  movq   %rax,%xmm1
  lea    D43(%rip),%rax
  punpcklqdq %xmm1,%xmm0
  movq   %rax,%xmm2
  movaps %xmm0,(%rsp)
  movq   %rcx,%xmm0
  punpcklqdq %xmm2,%xmm0
  movaps %xmm0,0x10(%rsp)
  call   <longestCommonPrefix@plt>
  mov    %rbp,%rdi
  mov    %rax,%rsi
  call   <printf@plt>
  xor    %edi,%edi
  mov    $0x4,%esi
  call   <longestCommonPrefix@plt>
  mov    %rbp,%rdi
  mov    %rax,%rsi
  call   <printf@plt>
  xor    %esi,%esi
  mov    %r12,%rdi
  call   <longestCommonPrefix@plt>
  mov    %rbp,%rdi
  mov    %rax,%rsi
  call   <printf@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x38,%rsp
  pop    %rbp
  pop    %r12
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D42
datatype: string
name: retrieve_data
arguments: label: D23
datatype: string
name: retrieve_data
arguments: label: D37
datatype: string
name: retrieve_data
arguments: label: D7
datatype: string
name: retrieve_data
arguments: label: D43
datatype: string

tool: D42: aqwerr

tool: D23: aqwerty

tool: D37: %s


tool: D7: aqwert

tool: D43: aqwertyu

assistant: ```c
void longestCommonPrefixMain(void) {
  char *strs[] = {"aqwerr", "aqwert", "aqwerty", "aqwertyu"};
  int strsSize = 0;
  { strsSize = sizeof(strs) / sizeof(strs[0]); };
  printf("%s\n", longestCommonPrefix(strs, strsSize));
  printf("%s\n", longestCommonPrefix(NULL, strsSize));
  printf("%s\n", longestCommonPrefix(strs, 0));
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<normalize_array>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    -0x1c(%rbp),%edx
  mov    -0x18(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <mean_array@plt>
  movd   %xmm0,%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x1c(%rbp),%edx
  mov    -0x18(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <variance_array@plt>
  mov    %eax,%edi
  call   <sqrt@plt>
  movd   %xmm0,%eax
  mov    %eax,-0x4(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L8
L0:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movss  (%rax),%xmm0
  subss  -0x8(%rbp),%xmm0
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  divss  -0x4(%rbp),%xmm0
  movss  %xmm0,(%rax)
  addl   $0x1,-0xc(%rbp)
L8:
  mov    -0xc(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L0
  mov    -0x1c(%rbp),%edx
  mov    -0x18(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <mean_array@plt>
  movd   %xmm0,%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x1c(%rbp),%edx
  mov    -0x18(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <variance_array@plt>
  mov    %eax,%edi
  call   <sqrt@plt>
  movd   %xmm0,%eax
  mov    %eax,-0x4(%rbp)
  nop
  leave
  ret
```

assistant: ```c
void normalize_array(float *a, int n) {
  int i;
  float mu = mean_array(a, n);
  float sigma = sqrt(variance_array(a, n));
  for (i = 0; i < n; ++i) {
    a[i] = (a[i] - mu) / sigma;
  }
  mu = mean_array(a, n);
  sigma = sqrt(variance_array(a, n));
}
```