ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Converte this assembly snippet into its equivalent C code:

```asm
<newstring>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  test   %rdi,%rdi
  je     L6
  mov    %rdi,%rbx
  call   <strlen@plt>
  lea    0x1(%rax),%rdi
  call   <mymalloc@plt>
  mov    %rax,%rbp
  mov    %rbx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
L6:
  mov    %rbp,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
char *newstring(char *s) {
  char *q;
  if (s) {
    q = mymalloc(strlen(s) + 1);
    (void)strcpy(q, s);
    return q;
  } else
    return NULL;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<newstring>:
  endbr64
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  sub    $0x8,%rsp
  test   %rdi,%rdi
  je     L9
  mov    %rdi,%rbp
  call   <strlen@plt>
  lea    0x1(%rax),%rdi
  call   <mymalloc@plt>
  mov    %rbp,%rsi
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <strcpy@plt>
L9:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *newstring(char *s) {
  char *q;
  if (s) {
    q = mymalloc(strlen(s) + 1);
    (void)strcpy(q, s);
    return q;
  } else
    return NULL;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<newstring>:
  endbr64
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  sub    $0x8,%rsp
  test   %rdi,%rdi
  je     L9
  mov    %rdi,%rbp
  call   <strlen@plt>
  lea    0x1(%rax),%rdi
  call   <mymalloc@plt>
  mov    %rbp,%rsi
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <strcpy@plt>
L9:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *newstring(char *s) {
  char *q;
  if (s) {
    q = mymalloc(strlen(s) + 1);
    (void)strcpy(q, s);
    return q;
  } else
    return NULL;
}
```
user: Translate the below assembly code into C programming language:

```asm
<RandomFunc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x28(%rbp),%rax
  mov    (%rax),%rax
  sub    $0x99f4bb,%rax
  mov    %rax,-0x10(%rbp)
  movq   $0x0,-0x18(%rbp)
  nop
  mov    -0x10(%rbp),%rax
  lea    -0x2bbf2c0(%rax),%rdx
  mov    -0x30(%rbp),%rax
  mov    %rdx,(%rax)
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L2
  call   <__stack_chk_fail@plt>
L2:
  leave
  ret
```

assistant: ```c
void RandomFunc(unsigned long input[1], unsigned long output[1]) {
  unsigned long state[1];
  unsigned long local2;
  unsigned long local1;
  unsigned short copy11;
  unsigned short copy12;
  {
    state[0UL] = input[0UL] + 0xffffffffff660b45UL;
    local1 = 0UL;
    while (local1 < 0UL) {
      local2 = 0UL;
      while (local2 < 0UL) {
        if (state[0UL] > local2 + local1) {
          copy11 = *((unsigned short *)(&state[local2]) + 2);
          *((unsigned short *)(&state[local2]) + 2) = *((unsigned short *)(&state[local2]) + 3);
          *((unsigned short *)(&state[local2]) + 3) = copy11;
        } else {
          copy12 = *((unsigned short *)(&state[0UL]) + 1);
          *((unsigned short *)(&state[0UL]) + 1) = *((unsigned short *)(&state[0UL]) + 2);
          *((unsigned short *)(&state[0UL]) + 2) = copy12;
        }
        local2 += 2UL;
      }
      local1 += 2UL;
    }
    output[0UL] = state[0UL] - 45871808UL;
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<RandomFunc>:
  endbr64
  mov    (%rdi),%rax
  sub    $0x355e77b,%rax
  mov    %rax,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned long input[1], unsigned long output[1]) {
  unsigned long state[1];
  unsigned long local2;
  unsigned long local1;
  unsigned short copy11;
  unsigned short copy12;
  {
    state[0UL] = input[0UL] + 0xffffffffff660b45UL;
    local1 = 0UL;
    while (local1 < 0UL) {
      local2 = 0UL;
      while (local2 < 0UL) {
        if (state[0UL] > local2 + local1) {
          copy11 = *((unsigned short *)(&state[local2]) + 2);
          *((unsigned short *)(&state[local2]) + 2) = *((unsigned short *)(&state[local2]) + 3);
          *((unsigned short *)(&state[local2]) + 3) = copy11;
        } else {
          copy12 = *((unsigned short *)(&state[0UL]) + 1);
          *((unsigned short *)(&state[0UL]) + 1) = *((unsigned short *)(&state[0UL]) + 2);
          *((unsigned short *)(&state[0UL]) + 2) = copy12;
        }
        local2 += 2UL;
      }
      local1 += 2UL;
    }
    output[0UL] = state[0UL] - 45871808UL;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<RandomFunc>:
  endbr64
  mov    (%rdi),%rax
  sub    $0x355e77b,%rax
  mov    %rax,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned long input[1], unsigned long output[1]) {
  unsigned long state[1];
  unsigned long local2;
  unsigned long local1;
  unsigned short copy11;
  unsigned short copy12;
  {
    state[0UL] = input[0UL] + 0xffffffffff660b45UL;
    local1 = 0UL;
    while (local1 < 0UL) {
      local2 = 0UL;
      while (local2 < 0UL) {
        if (state[0UL] > local2 + local1) {
          copy11 = *((unsigned short *)(&state[local2]) + 2);
          *((unsigned short *)(&state[local2]) + 2) = *((unsigned short *)(&state[local2]) + 3);
          *((unsigned short *)(&state[local2]) + 3) = copy11;
        } else {
          copy12 = *((unsigned short *)(&state[0UL]) + 1);
          *((unsigned short *)(&state[0UL]) + 1) = *((unsigned short *)(&state[0UL]) + 2);
          *((unsigned short *)(&state[0UL]) + 2) = copy12;
        }
        local2 += 2UL;
      }
      local1 += 2UL;
    }
    output[0UL] = state[0UL] - 45871808UL;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<RandomFunc>:
  endbr64
  mov    (%rdi),%rax
  sub    $0x355e77b,%rax
  mov    %rax,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned long input[1], unsigned long output[1]) {
  unsigned long state[1];
  unsigned long local2;
  unsigned long local1;
  unsigned short copy11;
  unsigned short copy12;
  {
    state[0UL] = input[0UL] + 0xffffffffff660b45UL;
    local1 = 0UL;
    while (local1 < 0UL) {
      local2 = 0UL;
      while (local2 < 0UL) {
        if (state[0UL] > local2 + local1) {
          copy11 = *((unsigned short *)(&state[local2]) + 2);
          *((unsigned short *)(&state[local2]) + 2) = *((unsigned short *)(&state[local2]) + 3);
          *((unsigned short *)(&state[local2]) + 3) = copy11;
        } else {
          copy12 = *((unsigned short *)(&state[0UL]) + 1);
          *((unsigned short *)(&state[0UL]) + 1) = *((unsigned short *)(&state[0UL]) + 2);
          *((unsigned short *)(&state[0UL]) + 2) = copy12;
        }
        local2 += 2UL;
      }
      local1 += 2UL;
    }
    output[0UL] = state[0UL] - 45871808UL;
  }
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<print1>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L7
L11:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x4(%rbp),%eax
  movq   %rdx,%xmm0
  mov    %eax,%esi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x4(%rbp)
L7:
  mov    -0x4(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L11
  nop
  nop
  leave
  ret
```

tool_calls: name: parse_data
arguments: label: D9
data_type: string

tool: D9: (%d)=%lf


assistant: ```c
void print1(int m, double A[]) {
  for (int i = 0; i < m; i++)
    printf("(%d)=%lf\n", i, A[i]);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<print1>:
  endbr64
  test   %edi,%edi
  jle    L0
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rsi,%r12
  mov    %edi,%ebp
  mov    $0x0,%ebx
  lea    D2(%rip),%r13
L16:
  movsd  (%r12,%rbx,8),%xmm0
  mov    %ebx,%esi
  mov    %r13,%rdi
  call   <printf@plt>
  add    $0x1,%rbx
  cmp    %rbp,%rbx
  jne    L16
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L0:
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D2
type_guessed: string

tool: D2: (%d)=%lf


assistant: ```c
void print1(int m, double A[]) {
  for (int i = 0; i < m; i++)
    printf("(%d)=%lf\n", i, A[i]);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<print1>:
  endbr64
  test   %edi,%edi
  jle    L5
  push   %r13
  lea    D1(%rip),%r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  movslq %edi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  nopl   0x0(%rax)
L1:
  movsd  (%r12,%rbx,8),%xmm0
  mov    %ebx,%esi
  mov    %r13,%rdi
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    %rbp,%rbx
  jne    L1
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L5:
  ret
```

tool_calls: name: unpack_data
arguments: tag: D1
data_type: string

tool: D1: (%d)=%lf


assistant: ```c
void print1(int m, double A[]) {
  for (int i = 0; i < m; i++)
    printf("(%d)=%lf\n", i, A[i]);
}
```
user: What is the c source code of the assembly code below?

```asm
<print1>:
  endbr64
  test   %edi,%edi
  jle    L8
  push   %r13
  lea    D0(%rip),%r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  movslq %edi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  nopl   0x0(%rax)
L16:
  movsd  (%r12,%rbx,8),%xmm0
  mov    %ebx,%esi
  mov    %r13,%rdi
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    %rbx,%rbp
  jne    L16
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L8:
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D0
type: string

tool: D0: (%d)=%lf


assistant: ```c
void print1(int m, double A[]) {
  for (int i = 0; i < m; i++)
    printf("(%d)=%lf\n", i, A[i]);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<read_paragraph>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  movl   $0x0,-0x14(%rbp)
  movl   $0x0,-0x10(%rbp)
  movq   $0x0,-0x8(%rbp)
  jmp    L11
L6:
  addl   $0x1,-0x10(%rbp)
  mov    -0x10(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jle    L36
  addl   $0x64,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <realloc@plt>
  mov    %rax,-0x8(%rbp)
L36:
  mov    -0x10(%rbp),%eax
  cltq
  lea    -0x1(%rax),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    -0xc(%rbp),%edx
  mov    %dl,(%rax)
  movzbl (%rax),%eax
  cmp    $0xa,%al
  jne    L4
  subl   $0x1,-0x10(%rbp)
  mov    -0x10(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  jmp    L15
L4:
  mov    -0x10(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
L11:
  mov    -0x30(%rbp),%rax
  mov    -0x28(%rbp),%rdx
  mov    %rax,%rdi
  call   *%rdx
  mov    %eax,-0xc(%rbp)
  mov    D1(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,-0xc(%rbp)
  jne    L6
L15:
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *read_paragraph(int (*read)(void *), void *stream) {
  int buffer_size = 0, length = 0;
  char *buffer = NULL;
  int ch;
  while ((ch = (*read)(stream)) != EOF) {
    if (++length > buffer_size) {
      buffer_size += 100;
      buffer = (char *)realloc(buffer, buffer_size + 1);
    }
    if ((buffer[length - 1] = ch) == '\n') {
      buffer[--length] = '\0';
      break;
    }
    buffer[length] = '\0';
  }
  return buffer;
}
```
user: Translate the below assembly code into C programming language:

```asm
<read_paragraph>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%r15
  mov    %rsi,%r14
  mov    $0x0,%ebp
  mov    $0x0,%r12d
  mov    $0x0,%r13d
  jmp    L28
L49:
  lea    0x1(%rbp),%rax
  mov    %bl,(%r12,%rbp,1)
  cmp    $0xa,%bl
  je     L2
  movb   $0x0,0x1(%r12,%rbp,1)
  mov    %rax,%rbp
L28:
  mov    %ebp,0x8(%rsp)
  mov    %r14,%rdi
  call   *%r15
  mov    %eax,%ebx
  mov    D3(%rip),%rax
  cmp    %ebx,(%rax)
  je     L27
  lea    0x1(%rbp),%eax
  cmp    %eax,%r13d
  jge    L49
  lea    0x64(%r13),%eax
  mov    %eax,0xc(%rsp)
  lea    0x65(%r13),%esi
  mov    %r12,%rdi
  call   <realloc@plt>
  mov    %rax,%r12
  mov    0xc(%rsp),%r13d
  jmp    L49
L2:
  movslq 0x8(%rsp),%rax
  movb   $0x0,(%r12,%rax,1)
L27:
  mov    %r12,%rax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
char *read_paragraph(int (*read)(void *), void *stream) {
  int buffer_size = 0, length = 0;
  char *buffer = NULL;
  int ch;
  while ((ch = (*read)(stream)) != EOF) {
    if (++length > buffer_size) {
      buffer_size += 100;
      buffer = (char *)realloc(buffer, buffer_size + 1);
    }
    if ((buffer[length - 1] = ch) == '\n') {
      buffer[--length] = '\0';
      break;
    }
    buffer[length] = '\0';
  }
  return buffer;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<read_paragraph>:
  endbr64
  push   %r15
  xor    %r15d,%r15d
  push   %r14
  xor    %r14d,%r14d
  push   %r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x18,%rsp
  jmp    L2
  nopw   0x0(%rax,%rax,1)
L36:
  mov    %dl,(%r15,%r14,1)
  lea    0x1(%r14),%rax
  cmp    $0xa,%dl
  je     L16
L49:
  movb   $0x0,0x1(%r15,%r14,1)
  mov    %rax,%r14
L2:
  mov    %rbp,%rdi
  movslq %r14d,%r13
  call   *%r12
  mov    %eax,%edx
  mov    D7(%rip),%rax
  cmp    %edx,(%rax)
  je     L29
  lea    0x1(%r14),%eax
  cmp    %eax,%ebx
  jge    L36
  lea    0x64(%rbx),%eax
  lea    0x65(%rbx),%esi
  mov    %r15,%rdi
  mov    %edx,0xc(%rsp)
  mov    %eax,0x8(%rsp)
  call   <realloc@plt>
  mov    0xc(%rsp),%edx
  mov    0x8(%rsp),%ebx
  mov    %rax,%r15
  lea    0x1(%r14),%rax
  mov    %dl,(%r15,%r14,1)
  cmp    $0xa,%dl
  jne    L49
L16:
  movb   $0x0,(%r15,%r13,1)
L29:
  add    $0x18,%rsp
  mov    %r15,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
char *read_paragraph(int (*read)(void *), void *stream) {
  int buffer_size = 0, length = 0;
  char *buffer = NULL;
  int ch;
  while ((ch = (*read)(stream)) != EOF) {
    if (++length > buffer_size) {
      buffer_size += 100;
      buffer = (char *)realloc(buffer, buffer_size + 1);
    }
    if ((buffer[length - 1] = ch) == '\n') {
      buffer[--length] = '\0';
      break;
    }
    buffer[length] = '\0';
  }
  return buffer;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<read_paragraph>:
  endbr64
  push   %r15
  xor    %r15d,%r15d
  push   %r14
  xor    %r14d,%r14d
  push   %r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x18,%rsp
  jmp    L2
  nopw   0x0(%rax,%rax,1)
L31:
  mov    %dl,(%r15,%r14,1)
  lea    0x1(%r14),%rax
  cmp    $0xa,%dl
  je     L23
L25:
  movb   $0x0,(%r15,%rax,1)
  mov    %rax,%r14
L2:
  mov    %rbp,%rdi
  movslq %r14d,%r13
  call   *%r12
  mov    %eax,%edx
  mov    D0(%rip),%rax
  cmp    %edx,(%rax)
  je     L30
  lea    0x1(%r14),%eax
  cmp    %eax,%ebx
  jge    L31
  lea    0x64(%rbx),%eax
  mov    %r15,%rdi
  lea    0x65(%rbx),%esi
  mov    %edx,0x8(%rsp)
  mov    %eax,0xc(%rsp)
  call   <realloc@plt>
  mov    0x8(%rsp),%edx
  mov    %rax,%r15
  lea    0x1(%r14),%rax
  mov    %dl,(%r15,%r14,1)
  cmp    $0xa,%dl
  je     L23
  mov    0xc(%rsp),%ebx
  jmp    L25
  nopl   0x0(%rax,%rax,1)
L23:
  movb   $0x0,(%r15,%r13,1)
L30:
  add    $0x18,%rsp
  mov    %r15,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
char *read_paragraph(int (*read)(void *), void *stream) {
  int buffer_size = 0, length = 0;
  char *buffer = NULL;
  int ch;
  while ((ch = (*read)(stream)) != EOF) {
    if (++length > buffer_size) {
      buffer_size += 100;
      buffer = (char *)realloc(buffer, buffer_size + 1);
    }
    if ((buffer[length - 1] = ch) == '\n') {
      buffer[--length] = '\0';
      break;
    }
    buffer[length] = '\0';
  }
  return buffer;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<test_indvars>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  add    $0x960,%rax
  movl   $0x3039,0x18(%rax)
  mov    -0x18(%rbp),%rdx
  add    $0x4,%rdx
  mov    0x18(%rax),%eax
  mov    %eax,(%rdx)
  movl   $0x0,-0x8(%rbp)
  jmp    L35
L70:
  mov    -0x8(%rbp),%eax
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x8(%rbp),%eax
  mov    %eax,(%rdx)
  addl   $0x2,-0x8(%rbp)
L35:
  cmpl   $0x63,-0x8(%rbp)
  jbe    L70
  movl   $0x3,-0x8(%rbp)
  jmp    L45
L9:
  mov    -0x8(%rbp),%eax
  lea    0x4(%rax),%edx
  mov    -0x8(%rbp),%eax
  lea    0x0(,%rax,4),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  mov    %edx,(%rax)
  addl   $0x1,-0x8(%rbp)
L45:
  cmpl   $0x66,-0x8(%rbp)
  jbe    L9
  movl   $0xd,-0x8(%rbp)
  jmp    L65
L51:
  movl   $0x0,-0x4(%rbp)
  jmp    L79
L47:
  mov    -0x8(%rbp),%edx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  lea    0x0(,%rax,4),%rdx
  add    %rdx,%rax
  shl    $0x5,%rax
  mov    %rax,%rdx
  mov    -0x20(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0x8(%rbp),%edx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  lea    0x0(,%rax,4),%rdx
  add    %rdx,%rax
  shl    $0x5,%rax
  mov    %rax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  mov    $0xaaaaaaab,%eax
  imul   %rsi,%rax
  shr    $0x20,%rax
  mov    %eax,%esi
  shr    %esi
  mov    -0x8(%rbp),%eax
  mov    (%rcx,%rax,4),%eax
  mov    %esi,%ecx
  mov    %eax,(%rdx,%rcx,4)
  addl   $0x3,-0x4(%rbp)
L79:
  cmpl   $0x63,-0x4(%rbp)
  jbe    L47
  addl   $0x1,-0x8(%rbp)
L65:
  cmpl   $0x63,-0x8(%rbp)
  jbe    L51
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void test_indvars(int *Array1, int Array2[100][200]) {
  unsigned i, j;
  Array1[1] = Array2[3][6] = 12345;
  for (i = 0; i < 100; i += 2)
    Array1[i] = i;
  for (i = 3; i < 103; i++)
    Array1[i] = i + 4;
  for (i = 13; i < 100; i++)
    for (j = 0; j < 100; j += 3)
      Array2[i][j / 3] = Array2[i][i];
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<test_indvars>:
  endbr64
  mov    %rdi,%rcx
  mov    %rsi,%r10
  movl   $0x3039,0x978(%rsi)
  movl   $0x3039,0x4(%rdi)
  mov    $0x0,%eax
L2:
  mov    %eax,(%rcx,%rax,4)
  add    $0x2,%rax
  cmp    $0x64,%rax
  jne    L2
  mov    $0x3,%eax
L53:
  lea    0x4(%rax),%edx
  mov    %edx,(%rcx,%rax,4)
  add    $0x1,%rax
  cmp    $0x67,%rax
  jne    L53
  lea    0x28d4(%r10),%rsi
  lea    0x28a0(%r10),%r9
  add    $0x13a10,%r10
  mov    $0xaaaaaaab,%r8d
  jmp    L57
L34:
  add    $0x324,%rsi
  add    $0x320,%r9
  cmp    %r10,%rsi
  je     L8
L57:
  mov    %r9,%rdi
  mov    $0x0,%edx
L21:
  mov    (%rsi),%ecx
  mov    %edx,%eax
  imul   %r8,%rax
  shr    $0x21,%rax
  mov    %ecx,(%rdi,%rax,4)
  add    $0x3,%edx
  cmp    $0x66,%edx
  jne    L21
  jmp    L34
L8:
  ret
```

assistant: ```c
void test_indvars(int *Array1, int Array2[100][200]) {
  unsigned i, j;
  Array1[1] = Array2[3][6] = 12345;
  for (i = 0; i < 100; i += 2)
    Array1[i] = i;
  for (i = 3; i < 103; i++)
    Array1[i] = i + 4;
  for (i = 13; i < 100; i++)
    for (j = 0; j < 100; j += 3)
      Array2[i][j / 3] = Array2[i][i];
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<test_indvars>:
  endbr64
  movl   $0x3039,0x978(%rsi)
  mov    %rdi,%rcx
  mov    %rsi,%r9
  xor    %eax,%eax
  movl   $0x3039,0x4(%rdi)
  nopl   (%rax)
L35:
  mov    %eax,(%rcx,%rax,4)
  add    $0x2,%rax
  cmp    $0x64,%rax
  jne    L35
  mov    $0x3,%eax
  nopw   0x0(%rax,%rax,1)
L7:
  lea    0x4(%rax),%edx
  mov    %edx,(%rcx,%rax,4)
  add    $0x1,%rax
  cmp    $0x67,%rax
  jne    L7
  lea    0x28d4(%r9),%rsi
  lea    0x28a0(%r9),%rdi
  mov    $0xaaaaaaab,%r8d
  add    $0x13a10,%r9
  nopl   0x0(%rax,%rax,1)
L12:
  xor    %edx,%edx
  nopw   0x0(%rax,%rax,1)
L20:
  mov    %edx,%eax
  mov    (%rsi),%ecx
  add    $0x3,%edx
  imul   %r8,%rax
  shr    $0x21,%rax
  mov    %ecx,(%rdi,%rax,4)
  cmp    $0x66,%edx
  jne    L20
  add    $0x324,%rsi
  add    $0x320,%rdi
  cmp    %rsi,%r9
  jne    L12
  ret
```

assistant: ```c
void test_indvars(int *Array1, int Array2[100][200]) {
  unsigned i, j;
  Array1[1] = Array2[3][6] = 12345;
  for (i = 0; i < 100; i += 2)
    Array1[i] = i;
  for (i = 3; i < 103; i++)
    Array1[i] = i + 4;
  for (i = 13; i < 100; i++)
    for (j = 0; j < 100; j += 3)
      Array2[i][j / 3] = Array2[i][i];
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<test_indvars>:
  endbr64
  movl   $0x3039,0x978(%rsi)
  mov    %rdi,%rdx
  mov    %rsi,%r9
  xor    %eax,%eax
  movl   $0x3039,0x4(%rdi)
  nopl   (%rax)
L16:
  mov    %eax,(%rdx,%rax,4)
  add    $0x2,%rax
  cmp    $0x64,%rax
  jne    L16
  movdqa D13(%rip),%xmm0
  movdqa D15(%rip),%xmm1
  lea    0xc(%rdx),%rax
  add    $0x19c,%rdx
L17:
  paddd  %xmm1,%xmm0
  add    $0x10,%rax
  movups %xmm0,-0x10(%rax)
  cmp    %rdx,%rax
  jne    L17
  lea    0x28d4(%r9),%rsi
  lea    0x28a0(%r9),%rdi
  mov    $0xaaaaaaab,%r8d
  add    $0x13a10,%r9
  nopl   0x0(%rax)
L25:
  xor    %edx,%edx
  nopw   0x0(%rax,%rax,1)
L4:
  mov    %edx,%eax
  mov    (%rsi),%ecx
  add    $0x3,%edx
  imul   %r8,%rax
  shr    $0x21,%rax
  mov    %ecx,(%rdi,%rax,4)
  cmp    $0x66,%edx
  jne    L4
  add    $0x324,%rsi
  add    $0x320,%rdi
  cmp    %rsi,%r9
  jne    L25
  ret
```

tool_calls: name: read_data
arguments: data_label: D13
type: byte[16]
name: read_data
arguments: data_label: D15
type: byte[16]

tool: {"D13": [0x03,0x00,0x00,0x00,0x04,0x00,0x00,0x00,0x05,0x00,0x00,0x00,0x06,0x00,0x00,0x00]}

tool: {"D15": [0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00]}

assistant: ```c
void test_indvars(int *Array1, int Array2[100][200]) {
  unsigned i, j;
  Array1[1] = Array2[3][6] = 12345;
  for (i = 0; i < 100; i += 2)
    Array1[i] = i;
  for (i = 3; i < 103; i++)
    Array1[i] = i + 4;
  for (i = 13; i < 100; i++)
    for (j = 0; j < 100; j += 3)
      Array2[i][j / 3] = Array2[i][i];
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    -0x41000(%rsp),%r11
L23:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L23
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x41020(%rbp)
  movl   $0x2,-0x41014(%rbp)
  movl   $0x5,-0x4101c(%rbp)
  movl   $0x1,-0x41018(%rbp)
  lea    -0x40010(%rbp),%rax
  mov    $0x20000,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  lea    -0x20010(%rbp),%rax
  mov    $0x20000,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  lea    -0x41010(%rbp),%rdx
  mov    $0x0,%eax
  mov    $0x200,%ecx
  mov    %rdx,%rdi
  rep stos %rax,%es:(%rdi)
L9:
  addl   $0x1,-0x41020(%rbp)
  mov    -0x41018(%rbp),%eax
  shl    $0x5,%eax
  mov    %eax,%edx
  mov    -0x41020(%rbp),%eax
  add    $0x2,%eax
  cltq
  mov    -0x41010(%rbp,%rax,4),%eax
  xor    %edx,%eax
  and    $0xffff,%eax
  mov    %eax,-0x41018(%rbp)
  mov    -0x41018(%rbp),%eax
  cltq
  mov    -0x20010(%rbp,%rax,4),%eax
  mov    %eax,-0x41014(%rbp)
  mov    -0x41020(%rbp),%eax
  movzwl %ax,%eax
  cltq
  mov    -0x41014(%rbp),%edx
  mov    %edx,-0x40010(%rbp,%rax,4)
  mov    -0x41018(%rbp),%eax
  cltq
  mov    -0x41020(%rbp),%edx
  mov    %edx,-0x20010(%rbp,%rax,4)
  subl   $0x1,-0x4101c(%rbp)
  cmpl   $0x0,-0x4101c(%rbp)
  jne    L9
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L1
  call   <__stack_chk_fail@plt>
L1:
  leave
  ret
```

assistant: ```c
int main() {
  int strstart = 0;
  int hash_head = 2;
  int prev_length = 5;
  int ins_h = 1;
  int prev[32 << 10] = {0};
  int head[32 << 10] = {0};
  int window[1024] = {0};
  do {
    strstart++;
    ((ins_h = (((ins_h) << ((15 + 3 - 1) / 3)) ^ (window[(strstart) + 3 - 1])) & 0xFFFF), prev[(strstart)&0xFFFF] = hash_head = head[ins_h], head[ins_h] = (strstart));
  } while (--prev_length != 0);
}
```