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

```asm
<simplify_sn>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  add    $0x1,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x18(%rbp)
  movl   $0x0,-0x20(%rbp)
  movl   $0x0,-0x1c(%rbp)
  jmp    L10
L34:
  mov    -0x20(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x20,%al
  je     L37
  mov    -0x20(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    -0x1c(%rbp),%edx
  movslq %edx,%rcx
  mov    -0x18(%rbp),%rdx
  lea    (%rcx,%rdx,1),%rbx
  mov    %eax,%edi
  call   <tolower@plt>
  mov    %al,(%rbx)
  addl   $0x1,-0x1c(%rbp)
  jmp    L20
L37:
  nop
L20:
  addl   $0x1,-0x20(%rbp)
L10:
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cmp    %eax,-0x20(%rbp)
  jl     L34
  mov    -0x1c(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    -0x18(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <realloc@plt>
  mov    %rax,-0x18(%rbp)
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x18(%rbp),%rax
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
char *simplify_sn(const char *sn) {
  char *temp;
  int x, count;
  temp = malloc(strlen(sn) + 1);
  for (x = 0, count = 0; x < strlen(sn); x++) {
    if (sn[x] == ' ')
      continue;
    temp[count] = tolower(sn[x]);
    count++;
  }
  temp = realloc(temp, count + 1);
  temp[count] = 0;
  return temp;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<simplify_sn>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  call   <strlen@plt>
  lea    0x1(%rax),%edi
  call   <malloc@plt>
  mov    %rax,%r14
  mov    $0x0,%ebx
  mov    $0x0,%r13d
  jmp    L23
L34:
  add    $0x1,%rbx
L23:
  mov    %r12,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jle    L26
  movzbl (%r12,%rbx,1),%edi
  cmp    $0x20,%dil
  je     L34
  movslq %r13d,%rbp
  add    %r14,%rbp
  movsbl %dil,%edi
  call   <tolower@plt>
  mov    %al,0x0(%rbp)
  add    $0x1,%r13d
  jmp    L34
L26:
  lea    0x1(%r13),%esi
  mov    %r14,%rdi
  call   <realloc@plt>
  movslq %r13d,%r13
  movb   $0x0,(%rax,%r13,1)
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
char *simplify_sn(const char *sn) {
  char *temp;
  int x, count;
  temp = malloc(strlen(sn) + 1);
  for (x = 0, count = 0; x < strlen(sn); x++) {
    if (sn[x] == ' ')
      continue;
    temp[count] = tolower(sn[x]);
    count++;
  }
  temp = realloc(temp, count + 1);
  temp[count] = 0;
  return temp;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<simplify_sn>:
  endbr64
  push   %r14
  push   %r13
  mov    $0x1,%r13d
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  xor    %ebx,%ebx
  call   <strlen@plt>
  lea    0x1(%rax),%edi
  call   <malloc@plt>
  mov    %rax,%r14
  jmp    L4
  nopl   0x0(%rax,%rax,1)
L3:
  movsbl (%r12,%rbx,1),%edi
  cmp    $0x20,%dil
  je     L7
  call   <tolower@plt>
  add    %r14,%rbp
  mov    %al,0x0(%rbp)
  movslq %r13d,%rbp
  add    $0x1,%r13d
L7:
  add    $0x1,%rbx
L4:
  mov    %r12,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jg     L3
  mov    %r13d,%esi
  mov    %r14,%rdi
  call   <realloc@plt>
  movb   $0x0,(%rax,%rbp,1)
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
char *simplify_sn(const char *sn) {
  char *temp;
  int x, count;
  temp = malloc(strlen(sn) + 1);
  for (x = 0, count = 0; x < strlen(sn); x++) {
    if (sn[x] == ' ')
      continue;
    temp[count] = tolower(sn[x]);
    count++;
  }
  temp = realloc(temp, count + 1);
  temp[count] = 0;
  return temp;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<simplify_sn>:
  endbr64
  push   %r14
  push   %r13
  mov    $0x1,%r13d
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  xor    %ebx,%ebx
  call   <strlen@plt>
  lea    0x1(%rax),%edi
  call   <malloc@plt>
  mov    %rax,%r14
  jmp    L21
  nopl   0x0(%rax,%rax,1)
L10:
  movsbl (%r12,%rbx,1),%edi
  cmp    $0x20,%dil
  je     L18
  call   <tolower@plt>
  add    %r14,%rbp
  mov    %al,0x0(%rbp)
  movslq %r13d,%rbp
  add    $0x1,%r13d
L18:
  add    $0x1,%rbx
L21:
  mov    %r12,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jg     L10
  mov    %r13d,%esi
  mov    %r14,%rdi
  call   <realloc@plt>
  movb   $0x0,(%rax,%rbp,1)
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
char *simplify_sn(const char *sn) {
  char *temp;
  int x, count;
  temp = malloc(strlen(sn) + 1);
  for (x = 0, count = 0; x < strlen(sn); x++) {
    if (sn[x] == ' ')
      continue;
    temp[count] = tolower(sn[x]);
    count++;
  }
  temp = realloc(temp, count + 1);
  temp[count] = 0;
  return temp;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<HashGen>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L2
L18:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  xor    %eax,-0x8(%rbp)
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x1(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  shl    $0x8,%eax
  xor    %eax,-0x8(%rbp)
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x2(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  shl    $0x10,%eax
  xor    %eax,-0x8(%rbp)
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x3(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  shl    $0x18,%eax
  xor    %eax,-0x8(%rbp)
  addl   $0x1,-0x4(%rbp)
L2:
  mov    -0x1c(%rbp),%eax
  lea    -0x4(%rax),%edx
  mov    -0x4(%rbp),%eax
  cmp    %eax,%edx
  ja     L18
  mov    -0x8(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned int HashGen(unsigned char *file, unsigned int size) {
  unsigned int HASH = 0;
  for (int i = 0; i < size - 4; i++) {
    HASH ^= file[i];
    HASH ^= file[i + 1] << 8;
    HASH ^= file[i + 2] << 16;
    HASH ^= file[i + 3] << 24;
  }
  return HASH;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<HashGen>:
  endbr64
  cmp    $0x4,%esi
  je     L5
  mov    %rdi,%rdx
  lea    -0x5(%rsi),%eax
  lea    0x1(%rdi,%rax,1),%r8
  mov    $0x0,%esi
L3:
  movzbl 0x1(%rdx),%eax
  shl    $0x8,%eax
  movzbl 0x2(%rdx),%ecx
  shl    $0x10,%ecx
  xor    %ecx,%eax
  movzbl 0x3(%rdx),%ecx
  shl    $0x18,%ecx
  movzbl (%rdx),%edi
  xor    %edi,%ecx
  xor    %ecx,%eax
  xor    %eax,%esi
  add    $0x1,%rdx
  cmp    %r8,%rdx
  jne    L3
L0:
  mov    %esi,%eax
  ret
L5:
  mov    $0x0,%esi
  jmp    L0
```

assistant: ```c
unsigned int HashGen(unsigned char *file, unsigned int size) {
  unsigned int HASH = 0;
  for (int i = 0; i < size - 4; i++) {
    HASH ^= file[i];
    HASH ^= file[i + 1] << 8;
    HASH ^= file[i + 2] << 16;
    HASH ^= file[i + 3] << 24;
  }
  return HASH;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<HashGen>:
  endbr64
  cmp    $0x4,%esi
  je     L14
  lea    -0x5(%rsi),%eax
  xor    %r8d,%r8d
  lea    0x1(%rdi,%rax,1),%rsi
  nopl   0x0(%rax)
L5:
  movzbl 0x1(%rdi),%eax
  movzbl 0x2(%rdi),%edx
  add    $0x1,%rdi
  movzbl -0x1(%rdi),%ecx
  shl    $0x10,%edx
  shl    $0x8,%eax
  xor    %edx,%eax
  movzbl 0x2(%rdi),%edx
  shl    $0x18,%edx
  xor    %ecx,%edx
  xor    %edx,%eax
  xor    %eax,%r8d
  cmp    %rdi,%rsi
  jne    L5
  mov    %r8d,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L14:
  xor    %r8d,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
unsigned int HashGen(unsigned char *file, unsigned int size) {
  unsigned int HASH = 0;
  for (int i = 0; i < size - 4; i++) {
    HASH ^= file[i];
    HASH ^= file[i + 1] << 8;
    HASH ^= file[i + 2] << 16;
    HASH ^= file[i + 3] << 24;
  }
  return HASH;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x70,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x6c(%rbp)
  movq   $0x0,-0x60(%rbp)
  movq   $0x0,-0x58(%rbp)
  movq   $0x0,-0x50(%rbp)
  movq   $0x0,-0x48(%rbp)
  movq   $0x0,-0x40(%rbp)
  movq   $0x0,-0x38(%rbp)
  movq   $0x0,-0x30(%rbp)
  movq   $0x0,-0x28(%rbp)
  movq   $0x0,-0x20(%rbp)
  movq   $0x0,-0x18(%rbp)
  movl   $0x1,-0x68(%rbp)
  movb   $0x0,-0x6d(%rbp)
  jmp    L77
L1:
  cmpb   $0xa,-0x6d(%rbp)
  jne    L45
  cmpl   $0x50,-0x6c(%rbp)
  jle    L93
  mov    $0xa,%edi
  call   <putchar@plt>
L93:
  movl   $0x0,-0x6c(%rbp)
  movl   $0x1,-0x68(%rbp)
  jmp    L77
L45:
  addl   $0x1,-0x6c(%rbp)
  cmpl   $0x50,-0x6c(%rbp)
  jg     L57
  mov    -0x6c(%rbp),%eax
  sub    $0x1,%eax
  cltq
  movzbl -0x6d(%rbp),%edx
  mov    %dl,-0x60(%rbp,%rax,1)
  jmp    L77
L57:
  cmpl   $0x1,-0x68(%rbp)
  jne    L31
  movl   $0x0,-0x64(%rbp)
  jmp    L41
L75:
  mov    -0x64(%rbp),%eax
  cltq
  movzbl -0x60(%rbp,%rax,1),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <putchar@plt>
  addl   $0x1,-0x64(%rbp)
L41:
  cmpl   $0x4f,-0x64(%rbp)
  jle    L75
  movl   $0x0,-0x68(%rbp)
L31:
  movsbl -0x6d(%rbp),%eax
  mov    %eax,%edi
  call   <putchar@plt>
L77:
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %al,-0x6d(%rbp)
  mov    D17(%rip),%rax
  movzbl (%rax),%eax
  cmp    %al,-0x6d(%rbp)
  jne    L1
  mov    D16(%rip),%rax
  mov    (%rax),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L82
  call   <__stack_chk_fail@plt>
L82:
  leave
  ret
```

assistant: ```c
int main(void) {
  uint32_t current_line_length = 0;
  int32_t buffer[80] = {0};
  bool buffer_in_use = true;
  int32_t character = 0;
  while ((character = getchar()) != EOF) {
    if (character == '\n') {
      if (current_line_length > 80) {
        putchar('\n');
      }
      current_line_length = 0;
      buffer_in_use = true;
    } else {
      current_line_length++;
      if (current_line_length <= 80) {
        buffer[current_line_length - 1] = character;
      } else {
        if (buffer_in_use == true) {
          uint32_t i;
          for (i = 0; i < 80; ++i) {
            putchar(buffer[i]);
          }
          buffer_in_use = false;
        }
        putchar(character);
      }
    }
  }
  return EXIT_SUCCESS;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x60,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x58(%rsp)
  xor    %eax,%eax
  movq   $0x0,(%rsp)
  movq   $0x0,0x8(%rsp)
  movq   $0x0,0x10(%rsp)
  movq   $0x0,0x18(%rsp)
  movq   $0x0,0x20(%rsp)
  movq   $0x0,0x28(%rsp)
  movq   $0x0,0x30(%rsp)
  movq   $0x0,0x38(%rsp)
  movq   $0x0,0x40(%rsp)
  movq   $0x0,0x48(%rsp)
  mov    $0x1,%r12d
  mov    $0x0,%ebx
  mov    $0x0,%r14d
  jmp    L59
L112:
  cmp    $0x50,%ebx
  jg     L31
  mov    $0x1,%r12d
  mov    %r14d,%ebx
  jmp    L59
L31:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0x1,%r12d
  mov    %r14d,%ebx
  jmp    L59
L97:
  cmp    $0x1,%r12d
  je     L87
L75:
  movsbl %bpl,%edi
  call   <putchar@plt>
  mov    %r13d,%ebx
  mov    %r14d,%r12d
L59:
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %eax,%ebp
  mov    D0(%rip),%rax
  cmp    %bpl,(%rax)
  je     L78
  cmp    $0xa,%bpl
  je     L112
  lea    0x1(%rbx),%r13d
  cmp    $0x50,%r13d
  jg     L97
  movslq %ebx,%rbx
  mov    %bpl,(%rsp,%rbx,1)
  mov    %r13d,%ebx
  jmp    L59
L87:
  mov    %rsp,%rbx
  lea    0x50(%rsp),%r12
L29:
  movsbl (%rbx),%edi
  call   <putchar@plt>
  add    $0x1,%rbx
  cmp    %r12,%rbx
  jne    L29
  jmp    L75
L78:
  mov    D13(%rip),%rax
  mov    (%rax),%eax
  mov    0x58(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L72
  add    $0x60,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L72:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(void) {
  uint32_t current_line_length = 0;
  int32_t buffer[80] = {0};
  bool buffer_in_use = true;
  int32_t character = 0;
  while ((character = getchar()) != EOF) {
    if (character == '\n') {
      if (current_line_length > 80) {
        putchar('\n');
      }
      current_line_length = 0;
      buffer_in_use = true;
    } else {
      current_line_length++;
      if (current_line_length <= 80) {
        buffer[current_line_length - 1] = character;
      } else {
        if (buffer_in_use == true) {
          uint32_t i;
          for (i = 0; i < 80; ++i) {
            putchar(buffer[i]);
          }
          buffer_in_use = false;
        }
        putchar(character);
      }
    }
  }
  return EXIT_SUCCESS;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %r14
  pxor   %xmm0,%xmm0
  push   %r13
  push   %r12
  mov    $0x1,%r12d
  push   %rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x60,%rsp
  mov    D1(%rip),%r14
  mov    %fs:0x28,%rax
  mov    %rax,0x58(%rsp)
  xor    %eax,%eax
  movaps %xmm0,(%rsp)
  movaps %xmm0,0x10(%rsp)
  movaps %xmm0,0x20(%rsp)
  movaps %xmm0,0x30(%rsp)
  movaps %xmm0,0x40(%rsp)
  nopl   0x0(%rax,%rax,1)
L32:
  xor    %eax,%eax
  call   <getchar@plt>
  mov    %eax,%ebp
  cmp    %al,(%r14)
  je     L65
L105:
  cmp    $0xa,%bpl
  je     L83
  lea    0x1(%rbx),%r13d
  cmp    $0x50,%r13d
  jg     L104
  xor    %eax,%eax
  mov    %bpl,(%rsp,%rbx,1)
  movslq %r13d,%rbx
  call   <getchar@plt>
  mov    %eax,%ebp
  cmp    %al,(%r14)
  jne    L105
L65:
  mov    D18(%rip),%rax
  mov    (%rax),%eax
  mov    0x58(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L17
  add    $0x60,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
L83:
  cmp    $0x50,%ebx
  jg     L42
  mov    $0x1,%r12d
  xor    %ebx,%ebx
  jmp    L32
  nop
L104:
  cmp    $0x1,%r12d
  je     L20
L70:
  movsbl %bpl,%edi
  movslq %r13d,%rbx
  xor    %r12d,%r12d
  call   <putchar@plt>
  jmp    L32
  nopw   0x0(%rax,%rax,1)
L42:
  mov    $0xa,%edi
  mov    $0x1,%r12d
  xor    %ebx,%ebx
  call   <putchar@plt>
  jmp    L32
L20:
  mov    %rsp,%rbx
  lea    0x50(%rsp),%r12
  nop
L112:
  movsbl (%rbx),%edi
  add    $0x1,%rbx
  call   <putchar@plt>
  cmp    %r12,%rbx
  jne    L112
  jmp    L70
L17:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(void) {
  uint32_t current_line_length = 0;
  int32_t buffer[80] = {0};
  bool buffer_in_use = true;
  int32_t character = 0;
  while ((character = getchar()) != EOF) {
    if (character == '\n') {
      if (current_line_length > 80) {
        putchar('\n');
      }
      current_line_length = 0;
      buffer_in_use = true;
    } else {
      current_line_length++;
      if (current_line_length <= 80) {
        buffer[current_line_length - 1] = character;
      } else {
        if (buffer_in_use == true) {
          uint32_t i;
          for (i = 0; i < 80; ++i) {
            putchar(buffer[i]);
          }
          buffer_in_use = false;
        }
        putchar(character);
      }
    }
  }
  return EXIT_SUCCESS;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r14
  pxor   %xmm0,%xmm0
  push   %r13
  push   %r12
  mov    $0x1,%r12d
  push   %rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x60,%rsp
  mov    D14(%rip),%r14
  mov    %fs:0x28,%rax
  mov    %rax,0x58(%rsp)
  xor    %eax,%eax
  movaps %xmm0,(%rsp)
  movaps %xmm0,0x10(%rsp)
  movaps %xmm0,0x20(%rsp)
  movaps %xmm0,0x30(%rsp)
  movaps %xmm0,0x40(%rsp)
  nopl   0x0(%rax,%rax,1)
L45:
  xor    %eax,%eax
  call   <getchar@plt>
  mov    %eax,%ebp
  cmp    %al,(%r14)
  je     L6
L110:
  cmp    $0xa,%bpl
  je     L41
  lea    0x1(%rbx),%r13d
  cmp    $0x50,%r13d
  jg     L34
  xor    %eax,%eax
  mov    %bpl,(%rsp,%rbx,1)
  movslq %r13d,%rbx
  call   <getchar@plt>
  mov    %eax,%ebp
  cmp    %al,(%r14)
  jne    L110
L6:
  mov    D10(%rip),%rax
  mov    (%rax),%eax
  mov    0x58(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L36
  add    $0x60,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
L41:
  cmp    $0x50,%ebx
  jg     L65
  mov    $0x1,%r12d
  xor    %ebx,%ebx
  jmp    L45
  nop
L34:
  cmp    $0x1,%r12d
  je     L60
L112:
  movsbl %bpl,%edi
  movslq %r13d,%rbx
  xor    %r12d,%r12d
  call   <putchar@plt>
  jmp    L45
  nopw   0x0(%rax,%rax,1)
L65:
  mov    $0xa,%edi
  mov    $0x1,%r12d
  xor    %ebx,%ebx
  call   <putchar@plt>
  jmp    L45
L60:
  mov    %rsp,%rbx
  lea    0x50(%rsp),%r12
  nop
L101:
  movsbl (%rbx),%edi
  add    $0x1,%rbx
  call   <putchar@plt>
  cmp    %r12,%rbx
  jne    L101
  jmp    L112
L36:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(void) {
  uint32_t current_line_length = 0;
  int32_t buffer[80] = {0};
  bool buffer_in_use = true;
  int32_t character = 0;
  while ((character = getchar()) != EOF) {
    if (character == '\n') {
      if (current_line_length > 80) {
        putchar('\n');
      }
      current_line_length = 0;
      buffer_in_use = true;
    } else {
      current_line_length++;
      if (current_line_length <= 80) {
        buffer[current_line_length - 1] = character;
      } else {
        if (buffer_in_use == true) {
          uint32_t i;
          for (i = 0; i < 80; ++i) {
            putchar(buffer[i]);
          }
          buffer_in_use = false;
        }
        putchar(character);
      }
    }
  }
  return EXIT_SUCCESS;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<corner_strmm_nt_4x3_lib4>:
  endbr64
  movss  (%rsi),%xmm0
  movss  (%rdi),%xmm14
  pxor   %xmm4,%xmm4
  movss  0x4(%rdi),%xmm13
  movss  0x8(%rdi),%xmm12
  mulss  %xmm0,%xmm14
  movss  0x10(%rsi),%xmm5
  movss  0x14(%rdi),%xmm8
  mulss  %xmm0,%xmm13
  movss  0x18(%rdi),%xmm2
  movss  0x14(%rsi),%xmm7
  mulss  %xmm0,%xmm12
  movss  0x1c(%rdi),%xmm6
  movups 0x20(%rdi),%xmm3
  mulss  0xc(%rdi),%xmm0
  movss  0x24(%rsi),%xmm11
  movss  0x28(%rsi),%xmm10
  addss  %xmm4,%xmm14
  shufps $0x0,%xmm11,%xmm11
  mulps  %xmm3,%xmm11
  addss  %xmm4,%xmm13
  addss  %xmm4,%xmm12
  addss  %xmm4,%xmm0
  movaps %xmm0,%xmm9
  movss  0x10(%rdi),%xmm0
  movaps %xmm0,%xmm1
  mulss  %xmm5,%xmm1
  mulss  %xmm7,%xmm0
  addss  %xmm14,%xmm1
  movaps %xmm8,%xmm14
  mulss  %xmm5,%xmm14
  addss  %xmm4,%xmm0
  mulss  %xmm7,%xmm8
  addss  %xmm13,%xmm14
  movaps %xmm2,%xmm13
  mulss  %xmm5,%xmm13
  addss  %xmm4,%xmm8
  mulss  %xmm6,%xmm5
  mulss  %xmm7,%xmm2
  unpcklps %xmm14,%xmm1
  mulss  %xmm7,%xmm6
  unpcklps %xmm8,%xmm0
  addss  %xmm12,%xmm13
  addss  %xmm9,%xmm5
  addss  %xmm4,%xmm2
  addss  %xmm4,%xmm6
  unpcklps %xmm5,%xmm13
  movss  0x20(%rsi),%xmm5
  movlhps %xmm13,%xmm1
  unpcklps %xmm6,%xmm2
  shufps $0x0,%xmm5,%xmm5
  mulps  %xmm3,%xmm5
  movlhps %xmm2,%xmm0
  addps  %xmm11,%xmm0
  movups %xmm0,0x10(%rdx)
  movaps %xmm10,%xmm0
  shufps $0x0,%xmm0,%xmm0
  mulps  %xmm3,%xmm0
  addps  %xmm5,%xmm1
  movups %xmm1,(%rdx)
  pxor   %xmm1,%xmm1
  addps  %xmm1,%xmm0
  movups %xmm0,0x20(%rdx)
  ret
```

assistant: ```c
void corner_strmm_nt_4x3_lib4(float *A, float *B, float *C) {
  const int bs = 4;
  const int lda = bs;
  const int ldc = bs;
  float a_0, a_1, a_2, a_3, b_0, b_1, b_2, b_3, c_00 = 0, c_01 = 0, c_02 = 0, c_10 = 0, c_11 = 0, c_12 = 0, c_20 = 0, c_21 = 0, c_22 = 0, c_30 = 0, c_31 = 0, c_32 = 0;
  a_0 = A[0 + lda * 0];
  a_1 = A[1 + lda * 0];
  a_2 = A[2 + lda * 0];
  a_3 = A[3 + lda * 0];
  b_0 = B[0 + lda * 0];
  c_00 += a_0 * b_0;
  c_10 += a_1 * b_0;
  c_20 += a_2 * b_0;
  c_30 += a_3 * b_0;
  a_0 = A[0 + lda * 1];
  a_1 = A[1 + lda * 1];
  a_2 = A[2 + lda * 1];
  a_3 = A[3 + lda * 1];
  b_0 = B[0 + lda * 1];
  b_1 = B[1 + lda * 1];
  c_00 += a_0 * b_0;
  c_10 += a_1 * b_0;
  c_20 += a_2 * b_0;
  c_30 += a_3 * b_0;
  c_01 += a_0 * b_1;
  c_11 += a_1 * b_1;
  c_21 += a_2 * b_1;
  c_31 += a_3 * b_1;
  a_0 = A[0 + lda * 2];
  a_1 = A[1 + lda * 2];
  a_2 = A[2 + lda * 2];
  a_3 = A[3 + lda * 2];
  b_0 = B[0 + lda * 2];
  b_1 = B[1 + lda * 2];
  b_2 = B[2 + lda * 2];
  c_00 += a_0 * b_0;
  c_10 += a_1 * b_0;
  c_20 += a_2 * b_0;
  c_30 += a_3 * b_0;
  c_01 += a_0 * b_1;
  c_11 += a_1 * b_1;
  c_21 += a_2 * b_1;
  c_31 += a_3 * b_1;
  c_02 += a_0 * b_2;
  c_12 += a_1 * b_2;
  c_22 += a_2 * b_2;
  c_32 += a_3 * b_2;
  C[0 + ldc * 0] = c_00;
  C[1 + ldc * 0] = c_10;
  C[2 + ldc * 0] = c_20;
  C[3 + ldc * 0] = c_30;
  C[0 + ldc * 1] = c_01;
  C[1 + ldc * 1] = c_11;
  C[2 + ldc * 1] = c_21;
  C[3 + ldc * 1] = c_31;
  C[0 + ldc * 2] = c_02;
  C[1 + ldc * 2] = c_12;
  C[2 + ldc * 2] = c_22;
  C[3 + ldc * 2] = c_32;
}
```
user: What is the c source code of the assembly code below?

```asm
<cpu_is_amd>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <get_cpu_id@plt>
  mov    $0xc,%edx
  mov    %eax,%esi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <memcmp@plt>
  test   %rax,%rax
  sete   %al
  movzbl %al,%eax
  pop    %rbp
  ret
```

assistant: ```c
int cpu_is_amd(void) {
  static char cpustr[] = "AuthenticAMD";
  return memcmp(cpustr, get_cpu_id(), 12) == 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<cpu_is_amd>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <get_cpu_id@plt>
  mov    %eax,%esi
  mov    $0xc,%edx
  lea    D0(%rip),%rdi
  call   <memcmp@plt>
  test   %rax,%rax
  sete   %al
  movzbl %al,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int cpu_is_amd(void) {
  static char cpustr[] = "AuthenticAMD";
  return memcmp(cpustr, get_cpu_id(), 12) == 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<cpu_is_amd>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <get_cpu_id@plt>
  mov    $0xc,%edx
  lea    D3(%rip),%rdi
  mov    %eax,%esi
  call   <memcmp@plt>
  test   %rax,%rax
  sete   %al
  add    $0x8,%rsp
  movzbl %al,%eax
  ret
```

assistant: ```c
int cpu_is_amd(void) {
  static char cpustr[] = "AuthenticAMD";
  return memcmp(cpustr, get_cpu_id(), 12) == 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<cpu_is_amd>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <get_cpu_id@plt>
  mov    $0xc,%edx
  lea    D0(%rip),%rdi
  mov    %eax,%esi
  call   <memcmp@plt>
  test   %rax,%rax
  sete   %al
  add    $0x8,%rsp
  movzbl %al,%eax
  ret
```

assistant: ```c
int cpu_is_amd(void) {
  static char cpustr[] = "AuthenticAMD";
  return memcmp(cpustr, get_cpu_id(), 12) == 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<usage>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rsi
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%edi
  call   <exit@plt>
  nop
  leave
  ret
```

tool_calls: name: parse_data
arguments: label: D1
type_guessed: string

tool: D1: usage :%s ipaddr portnum filename


assistant: ```c
void usage(char *command) {
  printf("usage :%s ipaddr portnum filename\n", command);
  exit(0);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<usage>:
  endbr64
  sub    $0x8,%rsp
  mov    %rdi,%rsi
  lea    D7(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%edi
  call   <exit@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D7
type_guessed: string

tool: D7: usage :%s ipaddr portnum filename


assistant: ```c
void usage(char *command) {
  printf("usage :%s ipaddr portnum filename\n", command);
  exit(0);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<usage>:
  endbr64
  sub    $0x8,%rsp
  mov    %rdi,%rsi
  lea    D2(%rip),%rdi
  call   <printf@plt>
  xor    %edi,%edi
  add    $0x8,%rsp
  jmp    1070 <exit@plt>
```

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

tool: D2: usage :%s ipaddr portnum filename


assistant: ```c
void usage(char *command) {
  printf("usage :%s ipaddr portnum filename\n", command);
  exit(0);
}
```