ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<write_starting_from_block>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x38(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,-0x18(%rbp)
  mov    $0x1000,%edi
  call   <malloc@plt>
  mov    %rax,-0x10(%rbp)
  movl   $0x0,-0x14(%rbp)
  jmp    L11
L17:
  mov    -0x24(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,%edi
  call   <get_first_empty_block_from@plt>
  mov    %eax,-0x1c(%rbp)
  mov    -0x14(%rbp),%eax
  imul   $0xffc,%eax,%eax
  movslq %eax,%rdx
  mov    -0x30(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0x10(%rbp),%rax
  mov    $0xffc,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <memcpy@plt>
  mov    -0x14(%rbp),%eax
  add    $0x1,%eax
  shl    $0xc,%eax
  cltq
  lea    -0x4(%rax),%rdx
  mov    -0x10(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  lea    -0x1c(%rbp),%rax
  mov    $0x4,%edx
  mov    %rax,%rsi
  mov    %rcx,%rdi
  mov    $0x0,%eax
  call   <memcpy@plt>
  mov    -0x10(%rbp),%rdx
  mov    -0x24(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <block_write@plt>
  mov    -0x1c(%rbp),%eax
  mov    %eax,-0x24(%rbp)
  subl   $0xffc,-0x18(%rbp)
  addl   $0x1,-0x14(%rbp)
L11:
  cmpl   $0xffc,-0x18(%rbp)
  jg     L17
  movl   $0xffffffff,-0x1c(%rbp)
  mov    -0x10(%rbp),%rax
  mov    $0x1000,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  mov    -0x14(%rbp),%eax
  imul   $0xffc,%eax,%eax
  movslq %eax,%rdx
  mov    -0x30(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0x18(%rbp),%edx
  mov    -0x10(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <memcpy@plt>
  mov    -0x14(%rbp),%eax
  add    $0x1,%eax
  shl    $0xc,%eax
  cltq
  lea    -0x4(%rax),%rdx
  mov    -0x10(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  lea    -0x1c(%rbp),%rax
  mov    $0x4,%edx
  mov    %rax,%rsi
  mov    %rcx,%rdi
  mov    $0x0,%eax
  call   <memcpy@plt>
  mov    -0x10(%rbp),%rdx
  mov    -0x24(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <block_write@plt>
  mov    -0x24(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L21
  call   <__stack_chk_fail@plt>
L21:
  leave
  ret
```

assistant: ```c
int write_starting_from_block(int last, char *buffer, int *n_byte) {
  int next, nbyte = *n_byte;
  char *buff = (char *)malloc(4096);
  int i = 0;
  while (nbyte > 4096 - 4) {
    next = get_first_empty_block_from(last + 1);
    memcpy(buff, buffer + (i * (4096 - 4)), 4096 - 4);
    memcpy(buff + ((i + 1) * 4096 - 4), &next, 4);
    block_write(last, buff);
    last = next;
    nbyte -= (4096 - 4);
    i++;
  }
  next = -1;
  memset(buff, 0, 4096);
  memcpy(buff, buffer + (i * (4096 - 4)), nbyte);
  memcpy(buff + ((i + 1) * 4096 - 4), &next, 4);
  block_write(last, buff);
  return last;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<write_starting_from_block>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %edi,%ebx
  mov    %rsi,%r14
  mov    %rsi,0x8(%rsp)
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    (%rdx),%ebp
  mov    $0x1000,%edi
  call   <malloc@plt>
  mov    %rax,%r15
  cmp    $0xffc,%ebp
  jle    L12
  lea    0xffc(%rax),%r13
  mov    $0x0,%r12d
  lea    0x14(%rsp),%rax
  mov    %rax,(%rsp)
L31:
  lea    0x1(%rbx),%edi
  call   <get_first_empty_block_from@plt>
  mov    %eax,0x14(%rsp)
  mov    $0xffc,%edx
  mov    %r14,%rsi
  mov    %r15,%rdi
  mov    $0x0,%eax
  call   <memcpy@plt>
  add    $0x1,%r12d
  mov    $0x4,%edx
  mov    (%rsp),%rsi
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <memcpy@plt>
  mov    %r15,%rsi
  mov    %ebx,%edi
  call   <block_write@plt>
  mov    0x14(%rsp),%ebx
  sub    $0xffc,%ebp
  add    $0xffc,%r14
  add    $0x1000,%r13
  cmp    $0xffc,%ebp
  jg     L31
L23:
  movl   $0xffffffff,0x14(%rsp)
  mov    $0x1000,%edx
  mov    $0x0,%esi
  mov    %r15,%rdi
  call   <memset@plt>
  imul   $0xffc,%r12d,%eax
  cltq
  mov    0x8(%rsp),%rsi
  add    %rax,%rsi
  mov    %ebp,%edx
  mov    %r15,%rdi
  mov    $0x0,%eax
  call   <memcpy@plt>
  lea    0x14(%rsp),%rsi
  add    $0x1,%r12d
  shl    $0xc,%r12d
  movslq %r12d,%r12
  lea    -0x4(%r15,%r12,1),%rdi
  mov    $0x4,%edx
  mov    $0x0,%eax
  call   <memcpy@plt>
  mov    %r15,%rsi
  mov    %ebx,%edi
  call   <block_write@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L17
  mov    %ebx,%eax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L12:
  mov    $0x0,%r12d
  jmp    L23
L17:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int write_starting_from_block(int last, char *buffer, int *n_byte) {
  int next, nbyte = *n_byte;
  char *buff = (char *)malloc(4096);
  int i = 0;
  while (nbyte > 4096 - 4) {
    next = get_first_empty_block_from(last + 1);
    memcpy(buff, buffer + (i * (4096 - 4)), 4096 - 4);
    memcpy(buff + ((i + 1) * 4096 - 4), &next, 4);
    block_write(last, buff);
    last = next;
    nbyte -= (4096 - 4);
    i++;
  }
  next = -1;
  memset(buff, 0, 4096);
  memcpy(buff, buffer + (i * (4096 - 4)), nbyte);
  memcpy(buff + ((i + 1) * 4096 - 4), &next, 4);
  block_write(last, buff);
  return last;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<write_starting_from_block>:
  endbr64
  push   %r15
  push   %r14
  mov    %edi,%r14d
  mov    $0x1000,%edi
  push   %r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    (%rdx),%r15d
  mov    %rsi,0x18(%rsp)
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  call   <malloc@plt>
  mov    %rax,%r13
  cmp    $0xffc,%r15d
  jle    L27
  lea    0xffc(%rax),%rbp
  lea    0x24(%rsp),%rax
  xor    %ebx,%ebx
  mov    %rax,0x8(%rsp)
  nopl   (%rax)
L24:
  lea    0x1(%r14),%edi
  sub    $0xffc,%r15d
  call   <get_first_empty_block_from@plt>
  mov    %r12,%rsi
  mov    $0xffc,%edx
  mov    %r13,%rdi
  mov    %eax,0x24(%rsp)
  xor    %eax,%eax
  add    $0xffc,%r12
  call   <memcpy@plt>
  mov    0x8(%rsp),%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  mov    $0x4,%edx
  mov    %ebx,0x14(%rsp)
  add    $0x1000,%rbp
  add    $0x1,%ebx
  call   <memcpy@plt>
  mov    %r14d,%edi
  mov    %r13,%rsi
  call   <block_write@plt>
  mov    0x24(%rsp),%r14d
  cmp    $0xffc,%r15d
  jg     L24
  mov    0x14(%rsp),%edi
  imul   $0xffc,%ebx,%ebx
  add    $0x2,%edi
  mov    %edi,%eax
  movslq %ebx,%rbx
  add    %rbx,0x18(%rsp)
  shl    $0xc,%eax
  movslq %eax,%rdi
  lea    -0x4(%rdi),%rbx
L37:
  xor    %esi,%esi
  mov    $0x1000,%edx
  mov    %r13,%rdi
  movl   $0xffffffff,0x24(%rsp)
  call   <memset@plt>
  mov    0x18(%rsp),%rsi
  mov    %r15d,%edx
  mov    %r13,%rdi
  xor    %eax,%eax
  call   <memcpy@plt>
  mov    0x8(%rsp),%rsi
  lea    0x0(%r13,%rbx,1),%rdi
  xor    %eax,%eax
  mov    $0x4,%edx
  call   <memcpy@plt>
  mov    %r13,%rsi
  mov    %r14d,%edi
  call   <block_write@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L39
  add    $0x38,%rsp
  mov    %r14d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   (%rax)
L27:
  lea    0x24(%rsp),%rax
  mov    $0xffc,%ebx
  mov    %rax,0x8(%rsp)
  jmp    L37
L39:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int write_starting_from_block(int last, char *buffer, int *n_byte) {
  int next, nbyte = *n_byte;
  char *buff = (char *)malloc(4096);
  int i = 0;
  while (nbyte > 4096 - 4) {
    next = get_first_empty_block_from(last + 1);
    memcpy(buff, buffer + (i * (4096 - 4)), 4096 - 4);
    memcpy(buff + ((i + 1) * 4096 - 4), &next, 4);
    block_write(last, buff);
    last = next;
    nbyte -= (4096 - 4);
    i++;
  }
  next = -1;
  memset(buff, 0, 4096);
  memcpy(buff, buffer + (i * (4096 - 4)), nbyte);
  memcpy(buff + ((i + 1) * 4096 - 4), &next, 4);
  block_write(last, buff);
  return last;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<write_starting_from_block>:
  endbr64
  push   %r15
  push   %r14
  mov    %edi,%r14d
  mov    $0x1000,%edi
  push   %r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    (%rdx),%r15d
  mov    %rsi,0x18(%rsp)
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  call   <malloc@plt>
  mov    %rax,%r13
  cmp    $0xffc,%r15d
  jle    L35
  lea    0xffc(%rax),%rbp
  lea    0x24(%rsp),%rax
  xor    %ebx,%ebx
  mov    %rax,0x8(%rsp)
  nopl   (%rax)
L9:
  lea    0x1(%r14),%edi
  sub    $0xffc,%r15d
  call   <get_first_empty_block_from@plt>
  mov    %r12,%rsi
  mov    $0xffc,%edx
  mov    %r13,%rdi
  mov    %eax,0x24(%rsp)
  xor    %eax,%eax
  add    $0xffc,%r12
  call   <memcpy@plt>
  mov    0x8(%rsp),%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  mov    $0x4,%edx
  mov    %ebx,0x14(%rsp)
  add    $0x1000,%rbp
  add    $0x1,%ebx
  call   <memcpy@plt>
  mov    %r14d,%edi
  mov    %r13,%rsi
  call   <block_write@plt>
  mov    0x24(%rsp),%r14d
  cmp    $0xffc,%r15d
  jg     L9
  mov    0x14(%rsp),%edi
  imul   $0xffc,%ebx,%ebx
  add    $0x2,%edi
  mov    %edi,%eax
  movslq %ebx,%rbx
  add    %rbx,0x18(%rsp)
  shl    $0xc,%eax
  movslq %eax,%rdi
  lea    -0x4(%rdi),%rbx
L29:
  xor    %esi,%esi
  mov    $0x1000,%edx
  mov    %r13,%rdi
  movl   $0xffffffff,0x24(%rsp)
  call   <memset@plt>
  mov    0x18(%rsp),%rsi
  mov    %r15d,%edx
  mov    %r13,%rdi
  xor    %eax,%eax
  call   <memcpy@plt>
  mov    0x8(%rsp),%rsi
  lea    0x0(%r13,%rbx,1),%rdi
  xor    %eax,%eax
  mov    $0x4,%edx
  call   <memcpy@plt>
  mov    %r13,%rsi
  mov    %r14d,%edi
  call   <block_write@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L30
  add    $0x38,%rsp
  mov    %r14d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   (%rax)
L35:
  lea    0x24(%rsp),%rax
  mov    $0xffc,%ebx
  mov    %rax,0x8(%rsp)
  jmp    L29
L30:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int write_starting_from_block(int last, char *buffer, int *n_byte) {
  int next, nbyte = *n_byte;
  char *buff = (char *)malloc(4096);
  int i = 0;
  while (nbyte > 4096 - 4) {
    next = get_first_empty_block_from(last + 1);
    memcpy(buff, buffer + (i * (4096 - 4)), 4096 - 4);
    memcpy(buff + ((i + 1) * 4096 - 4), &next, 4);
    block_write(last, buff);
    last = next;
    nbyte -= (4096 - 4);
    i++;
  }
  next = -1;
  memset(buff, 0, 4096);
  memcpy(buff, buffer + (i * (4096 - 4)), nbyte);
  memcpy(buff + ((i + 1) * 4096 - 4), &next, 4);
  block_write(last, buff);
  return last;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<cut_string>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  movb   $0x0,-0x9(%rbp)
  jmp    L2
L48:
  addq   $0x1,-0x8(%rbp)
  movzbl -0x9(%rbp),%eax
  add    $0x1,%eax
  mov    %al,-0x9(%rbp)
L2:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L37
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x2e,%al
  je     L37
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0xa,%al
  je     L37
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0xd,%al
  je     L37
  cmpb   $0x3f,-0x9(%rbp)
  jle    L48
L37:
  mov    -0x8(%rbp),%rax
  movb   $0x0,(%rax)
  mov    -0x18(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
const char *cut_string(char *s) {
  char *p = s, c = 0;
  for (; *p && *p != '.' && *p != 10 && *p != 13 && c < 64; p++, c++)
    ;
  *p = 0;
  return s;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<cut_string>:
  endbr64
  mov    %rdi,%rax
  movzbl (%rdi),%edx
  cmp    $0x2e,%dl
  ja     L56
  movabs $0x400000002401,%rcx
  bt     %rdx,%rcx
  jb     L43
  mov    %rdi,%rsi
  mov    $0x0,%r8d
  jmp    L41
L56:
  mov    %rdi,%rsi
  mov    $0x0,%r8d
L41:
  mov    $0x1,%r9d
  movabs $0x400000002401,%rdi
  jmp    L1
L64:
  cmp    $0x3f,%r8b
  jg     L39
  test   %dl,%dl
  je     L39
L1:
  add    $0x1,%rsi
  add    $0x1,%r8d
  movzbl (%rsi),%ecx
  mov    %r9d,%edx
  cmp    $0x2e,%cl
  ja     L64
  mov    %rdi,%rdx
  shr    %cl,%rdx
  xor    $0x1,%rdx
  and    $0x1,%edx
  jmp    L64
L43:
  mov    %rdi,%rsi
L39:
  movb   $0x0,(%rsi)
  ret
```

assistant: ```c
const char *cut_string(char *s) {
  char *p = s, c = 0;
  for (; *p && *p != '.' && *p != 10 && *p != 13 && c < 64; p++, c++)
    ;
  *p = 0;
  return s;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<cut_string>:
  endbr64
  movzbl (%rdi),%edx
  mov    %rdi,%rax
  mov    %rdi,%rsi
  cmp    $0x2e,%dl
  ja     L45
  movabs $0xffffbfffffffdbfe,%rcx
  bt     %rdx,%rcx
  jae    L7
L45:
  movabs $0xffffbfffffffdbfe,%r9
  xor    %r8d,%r8d
  jmp    L1
  nopl   0x0(%rax)
L33:
  test   %dl,%dl
  je     L7
L1:
  movzbl 0x1(%rsi),%ecx
  mov    %r9,%rdx
  add    $0x1,%rsi
  add    $0x1,%r8d
  sar    %cl,%rdx
  and    $0x1,%edx
  cmp    $0x2f,%cl
  mov    $0x1,%ecx
  cmovae %ecx,%edx
  cmp    $0x40,%r8b
  jne    L33
L7:
  movb   $0x0,(%rsi)
  ret
```

assistant: ```c
const char *cut_string(char *s) {
  char *p = s, c = 0;
  for (; *p && *p != '.' && *p != 10 && *p != 13 && c < 64; p++, c++)
    ;
  *p = 0;
  return s;
}
```
user: What is the c source code of the assembly code below:

```asm
<cut_string>:
  endbr64
  movzbl (%rdi),%ecx
  mov    %rdi,%rax
  mov    %rdi,%rdx
  cmp    $0x2e,%cl
  ja     L13
  movabs $0xffffbfffffffdbfe,%rsi
  bt     %rcx,%rsi
  jae    L2
L13:
  movabs $0xffffbfffffffdbfe,%rdi
  xor    %ecx,%ecx
  xchg   %ax,%ax
L48:
  movzbl 0x1(%rdx),%esi
  add    $0x1,%rdx
  add    $0x1,%ecx
  cmp    $0x2e,%sil
  ja     L43
  cmp    $0x40,%cl
  je     L2
  bt     %rsi,%rdi
  jb     L48
L2:
  movb   $0x0,(%rdx)
  ret
L43:
  cmp    $0x40,%cl
  jne    L48
  movb   $0x0,(%rdx)
  ret
```

assistant: ```c
const char *cut_string(char *s) {
  char *p = s, c = 0;
  for (; *p && *p != '.' && *p != 10 && *p != 13 && c < 64; p++, c++)
    ;
  *p = 0;
  return s;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<hp71_length>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  add    $0x1c,%rax
  movzbl (%rax),%eax
  movzbl %al,%edx
  mov    -0x18(%rbp),%rax
  add    $0x1d,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  shl    $0x8,%eax
  add    %eax,%edx
  mov    -0x18(%rbp),%rax
  add    $0x1e,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  shl    $0x10,%eax
  add    %edx,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int hp71_length(unsigned char *entry) {
  int length;
  length = (*(entry + 28)) + ((*(entry + 29)) << 8) + ((*(entry + 30)) << 16);
  length = (length + 1) / 2;
  return (length);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<hp71_length>:
  endbr64
  movzbl 0x1d(%rdi),%eax
  shl    $0x8,%eax
  movzbl 0x1c(%rdi),%edx
  add    %edx,%eax
  movzbl 0x1e(%rdi),%edx
  shl    $0x10,%edx
  lea    0x1(%rax,%rdx,1),%eax
  sar    %eax
  ret
```

assistant: ```c
int hp71_length(unsigned char *entry) {
  int length;
  length = (*(entry + 28)) + ((*(entry + 29)) << 8) + ((*(entry + 30)) << 16);
  length = (length + 1) / 2;
  return (length);
}
```
user: Translate the below assembly code into C programming language:

```asm
<hp71_length>:
  endbr64
  movzbl 0x1d(%rdi),%eax
  movzbl 0x1c(%rdi),%edx
  shl    $0x8,%eax
  add    %edx,%eax
  movzbl 0x1e(%rdi),%edx
  shl    $0x10,%edx
  lea    0x1(%rax,%rdx,1),%eax
  sar    %eax
  ret
```

assistant: ```c
int hp71_length(unsigned char *entry) {
  int length;
  length = (*(entry + 28)) + ((*(entry + 29)) << 8) + ((*(entry + 30)) << 16);
  length = (length + 1) / 2;
  return (length);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<hp71_length>:
  endbr64
  movzbl 0x1d(%rdi),%eax
  movzbl 0x1c(%rdi),%edx
  shl    $0x8,%eax
  add    %edx,%eax
  movzbl 0x1e(%rdi),%edx
  shl    $0x10,%edx
  lea    0x1(%rax,%rdx,1),%eax
  sar    %eax
  ret
```

assistant: ```c
int hp71_length(unsigned char *entry) {
  int length;
  length = (*(entry + 28)) + ((*(entry + 29)) << 8) + ((*(entry + 30)) << 16);
  length = (length + 1) / 2;
  return (length);
}
```
user: What is the c source code of the assembly code below?

```asm
<Cmov>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r12
  push   %rbx
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    -0x28(%rbp),%r12
  mov    -0x30(%rbp),%rbx
  movl   $0x8,-0x14(%rbp)
L3:
  mov    %r12,%rdx
  lea    0x2(%rdx),%r12
  mov    %rbx,%rax
  lea    0x2(%rax),%rbx
  movzwl (%rdx),%edx
  mov    %dx,(%rax)
  subl   $0x1,-0x14(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  jne    L3
  nop
  nop
  pop    %rbx
  pop    %r12
  pop    %rbp
  ret
```

assistant: ```c
void Cmov(a, b) void *a, *b;
{
  register short *pa, *pb;
  int i;
  pa = (short *)a;
  pb = (short *)b;
  i = 8;
  do
    *pb++ = *pa++;
  while (--i);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<Cmov>:
  endbr64
  mov    $0x0,%eax
L8:
  movzwl (%rdi,%rax,1),%edx
  mov    %dx,(%rsi,%rax,1)
  add    $0x2,%rax
  cmp    $0x10,%rax
  jne    L8
  ret
```

assistant: ```c
void Cmov(a, b) void *a, *b;
{
  register short *pa, *pb;
  int i;
  pa = (short *)a;
  pb = (short *)b;
  i = 8;
  do
    *pb++ = *pa++;
  while (--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
<Cmov>:
  endbr64
  xor    %eax,%eax
  cs nopw 0x0(%rax,%rax,1)
L3:
  movzwl (%rdi,%rax,1),%edx
  mov    %dx,(%rsi,%rax,1)
  add    $0x2,%rax
  cmp    $0x10,%rax
  jne    L3
  ret
```

assistant: ```c
void Cmov(a, b) void *a, *b;
{
  register short *pa, *pb;
  int i;
  pa = (short *)a;
  pb = (short *)b;
  i = 8;
  do
    *pb++ = *pa++;
  while (--i);
}
```
user: Translate the below assembly code into C programming language:

```asm
<Cmov>:
  endbr64
  lea    0x2(%rdi),%rdx
  mov    %rsi,%rax
  sub    %rdx,%rax
  cmp    $0xc,%rax
  jbe    L6
  movdqu (%rdi),%xmm0
  movups %xmm0,(%rsi)
  ret
  nopl   0x0(%rax)
L6:
  movzwl (%rdi),%eax
  mov    %ax,(%rsi)
  movzwl 0x2(%rdi),%eax
  mov    %ax,0x2(%rsi)
  movzwl 0x4(%rdi),%eax
  mov    %ax,0x4(%rsi)
  movzwl 0x6(%rdi),%eax
  mov    %ax,0x6(%rsi)
  movzwl 0x8(%rdi),%eax
  mov    %ax,0x8(%rsi)
  movzwl 0xa(%rdi),%eax
  mov    %ax,0xa(%rsi)
  movzwl 0xc(%rdi),%eax
  mov    %ax,0xc(%rsi)
  movzwl 0xe(%rdi),%eax
  mov    %ax,0xe(%rsi)
  ret
```

assistant: ```c
void Cmov(a, b) void *a, *b;
{
  register short *pa, *pb;
  int i;
  pa = (short *)a;
  pb = (short *)b;
  i = 8;
  do
    *pb++ = *pa++;
  while (--i);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<int_shifts_are_logical>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movl   $0xffffffff,-0x4(%rbp)
  sarl   -0x4(%rbp)
  cmpl   $0xffffffff,-0x4(%rbp)
  setne  %al
  movzbl %al,%eax
  pop    %rbp
  ret
```

assistant: ```c
int int_shifts_are_logical(void) {
  int x = ~0;
  x >>= 1;
  return !!(~x);
}
```
user: 这段汇编代码用C语言表示是什么样的？

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

assistant: ```c
int int_shifts_are_logical(void) {
  int x = ~0;
  x >>= 1;
  return !!(~x);
}
```
user: Translate the below assembly code into C programming language:

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

assistant: ```c
int int_shifts_are_logical(void) {
  int x = ~0;
  x >>= 1;
  return !!(~x);
}
```
user: 这段汇编代码用C语言表示是什么样的:

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

assistant: ```c
int int_shifts_are_logical(void) {
  int x = ~0;
  x >>= 1;
  return !!(~x);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0xa,-0x4(%rbp)
  nop
  movl   $0x5,-0x4(%rbp)
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x4(%rbp),%eax
  cltq
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x0,-0x8(%rbp)
L18:
  lea    D14(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  cmpl   $0x9,-0x8(%rbp)
  jg     L7
  addl   $0x1,-0x8(%rbp)
  jmp    L18
L7:
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: read_data
arguments: data_label: D7
type_infered: string
name: read_data
arguments: data_label: D14
type_infered: string

tool: D7: Print this: 

tool: D14: This always gets printed!

assistant: ```c
int main() {
  int x;
  x = 10;
  goto label2;
  printf("Do not print this");
label2:
  x = 5;
  printf("Print this: ");
  printf(x);
  int counter;
  counter = 0;
label1:
  printf("This always gets printed!");
  if (counter < 10) {
    counter = counter + 1;
    goto label1;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  lea    D28(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x5,%edi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D12(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0xa,%ebx
  lea    D12(%rip),%rbp
L3:
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  sub    $0x1,%ebx
  jne    L3
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: read_data
arguments: data_label: D28
type_guessed: string
name: read_data
arguments: data_label: D12
type_guessed: string

tool: D28: Print this: 

tool: D12: This always gets printed!

assistant: ```c
int main() {
  int x;
  x = 10;
  goto label2;
  printf("Do not print this");
label2:
  x = 5;
  printf("Print this: ");
  printf(x);
  int counter;
  counter = 0;
label1:
  printf("This always gets printed!");
  if (counter < 10) {
    counter = counter + 1;
    goto label1;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D18(%rip),%rdi
  xor    %eax,%eax
  lea    D9(%rip),%rbp
  push   %rbx
  mov    $0xa,%ebx
  sub    $0x8,%rsp
  call   <printf@plt>
  mov    $0x5,%edi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  nopw   0x0(%rax,%rax,1)
L6:
  xor    %eax,%eax
  mov    %rbp,%rdi
  call   <printf@plt>
  sub    $0x1,%ebx
  jne    L6
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D18
type: string
name: parse_data
arguments: data_tag: D9
type: string

tool: D18: Print this: 

tool: D9: This always gets printed!

assistant: ```c
int main() {
  int x;
  x = 10;
  goto label2;
  printf("Do not print this");
label2:
  x = 5;
  printf("Print this: ");
  printf(x);
  int counter;
  counter = 0;
label1:
  printf("This always gets printed!");
  if (counter < 10) {
    counter = counter + 1;
    goto label1;
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<main>:
  endbr64
  push   %rbp
  lea    D7(%rip),%rdi
  xor    %eax,%eax
  lea    D18(%rip),%rbp
  push   %rbx
  mov    $0xa,%ebx
  sub    $0x8,%rsp
  call   <printf@plt>
  mov    $0x5,%edi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  nopw   0x0(%rax,%rax,1)
L3:
  xor    %eax,%eax
  mov    %rbp,%rdi
  call   <printf@plt>
  sub    $0x1,%ebx
  jne    L3
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: read_data
arguments: data_label: D7
type: string
name: read_data
arguments: data_label: D18
type: string

tool: D7: Print this: 

tool: D18: This always gets printed!

assistant: ```c
int main() {
  int x;
  x = 10;
  goto label2;
  printf("Do not print this");
label2:
  x = 5;
  printf("Print this: ");
  printf(x);
  int counter;
  counter = 0;
label1:
  printf("This always gets printed!");
  if (counter < 10) {
    counter = counter + 1;
    goto label1;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <h_world@plt>
  mov    $0x0,%eax
  call   <g_world@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main(void) {
  h_world();
  g_world();
}
```