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

```asm
<merge>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    %rcx,-0x40(%rbp)
  cmpq   $0x0,-0x30(%rbp)
  jne    L37
  movq   $0xffffffffffffffff,-0x18(%rbp)
  jmp    L57
L37:
  mov    -0x30(%rbp),%rax
  sub    $0x1,%rax
  mov    %rax,-0x18(%rbp)
L57:
  cmpq   $0x0,-0x40(%rbp)
  je     L3
  mov    -0x40(%rbp),%rax
  sub    $0x1,%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x30(%rbp),%rdx
  mov    -0x40(%rbp),%rax
  add    %rdx,%rax
  sub    $0x1,%rax
  mov    %rax,-0x8(%rbp)
L54:
  mov    -0x18(%rbp),%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x10(%rbp),%rax
  lea    0x0(,%rax,4),%rcx
  mov    -0x38(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%eax
  cmp    %eax,%edx
  jle    L15
  mov    -0x18(%rbp),%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    -0x8(%rbp),%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x28(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  subq   $0x1,-0x18(%rbp)
  jmp    L33
L15:
  mov    -0x10(%rbp),%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  mov    -0x8(%rbp),%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x28(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  subq   $0x1,-0x10(%rbp)
L33:
  subq   $0x1,-0x8(%rbp)
  jmp    L54
L3:
  nop
  pop    %rbp
  ret
```

assistant: ```c
void merge(int arrayA[], size_t arrayA_sz, int arrayB[], size_t arrayB_sz) {
  ssize_t i, j, k;
  if (arrayA_sz == 0)
    i = -1;
  else
    i = arrayA_sz - 1;
  if (arrayB_sz == 0)
    return;
  j = arrayB_sz - 1;
  k = arrayA_sz + arrayB_sz - 1;
  while (i >= 0 && j >= 0) {
    if (arrayA[i] > arrayB[j]) {
      arrayA[k] = arrayA[i];
      i--;
    } else {
      arrayA[k] = arrayB[j];
      j--;
    }
    k--;
  }
  while (j >= 0) {
    arrayA[k--] = arrayB[j--];
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<merge>:
  endbr64
  mov    %rdi,%r9
  lea    -0x1(%rsi),%r8
  test   %rsi,%rsi
  mov    $0xffffffffffffffff,%rax
  cmove  %rax,%r8
  test   %rcx,%rcx
  je     L29
  lea    -0x1(%rcx),%rdi
  add    %rdi,%rsi
  lea    (%r9,%rsi,4),%rax
  jmp    L9
L38:
  mov    %esi,(%rax)
  sub    $0x1,%r8
L24:
  sub    $0x4,%rax
L9:
  mov    (%r9,%r8,4),%esi
  mov    (%rdx,%rdi,4),%ecx
  cmp    %ecx,%esi
  jg     L38
  mov    %ecx,(%rax)
  sub    $0x1,%rdi
  jmp    L24
L29:
  ret
```

assistant: ```c
void merge(int arrayA[], size_t arrayA_sz, int arrayB[], size_t arrayB_sz) {
  ssize_t i, j, k;
  if (arrayA_sz == 0)
    i = -1;
  else
    i = arrayA_sz - 1;
  if (arrayB_sz == 0)
    return;
  j = arrayB_sz - 1;
  k = arrayA_sz + arrayB_sz - 1;
  while (i >= 0 && j >= 0) {
    if (arrayA[i] > arrayB[j]) {
      arrayA[k] = arrayA[i];
      i--;
    } else {
      arrayA[k] = arrayB[j];
      j--;
    }
    k--;
  }
  while (j >= 0) {
    arrayA[k--] = arrayB[j--];
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<merge>:
  endbr64
  lea    -0x1(%rsi),%r8
  test   %rsi,%rsi
  mov    $0xffffffffffffffff,%rax
  mov    %rdi,%r9
  cmove  %rax,%r8
  test   %rcx,%rcx
  je     L17
  lea    -0x1(%rcx),%rdi
  add    %rdi,%rsi
  lea    (%r9,%rsi,4),%rcx
  jmp    L8
  nopl   0x0(%rax,%rax,1)
L2:
  sub    $0x1,%r8
L11:
  mov    %eax,(%rcx)
  sub    $0x4,%rcx
L8:
  mov    (%r9,%r8,4),%eax
  mov    (%rdx,%rdi,4),%esi
  cmp    %esi,%eax
  jg     L2
  sub    $0x1,%rdi
  mov    %esi,%eax
  jmp    L11
L17:
  ret
```

assistant: ```c
void merge(int arrayA[], size_t arrayA_sz, int arrayB[], size_t arrayB_sz) {
  ssize_t i, j, k;
  if (arrayA_sz == 0)
    i = -1;
  else
    i = arrayA_sz - 1;
  if (arrayB_sz == 0)
    return;
  j = arrayB_sz - 1;
  k = arrayA_sz + arrayB_sz - 1;
  while (i >= 0 && j >= 0) {
    if (arrayA[i] > arrayB[j]) {
      arrayA[k] = arrayA[i];
      i--;
    } else {
      arrayA[k] = arrayB[j];
      j--;
    }
    k--;
  }
  while (j >= 0) {
    arrayA[k--] = arrayB[j--];
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<merge>:
  endbr64
  lea    -0x1(%rsi),%r8
  test   %rsi,%rsi
  mov    $0xffffffffffffffff,%rax
  mov    %rdi,%r9
  cmove  %rax,%r8
  test   %rcx,%rcx
  je     L16
  lea    -0x1(%rcx),%rdi
  add    %rdi,%rsi
  lea    (%r9,%rsi,4),%rcx
  jmp    L39
  nopl   0x0(%rax,%rax,1)
L8:
  sub    $0x1,%r8
L38:
  mov    %eax,(%rcx)
  sub    $0x4,%rcx
L39:
  mov    (%r9,%r8,4),%eax
  mov    (%rdx,%rdi,4),%esi
  cmp    %esi,%eax
  jg     L8
  sub    $0x1,%rdi
  mov    %esi,%eax
  jmp    L38
L16:
  ret
```

assistant: ```c
void merge(int arrayA[], size_t arrayA_sz, int arrayB[], size_t arrayB_sz) {
  ssize_t i, j, k;
  if (arrayA_sz == 0)
    i = -1;
  else
    i = arrayA_sz - 1;
  if (arrayB_sz == 0)
    return;
  j = arrayB_sz - 1;
  k = arrayA_sz + arrayB_sz - 1;
  while (i >= 0 && j >= 0) {
    if (arrayA[i] > arrayB[j]) {
      arrayA[k] = arrayA[i];
      i--;
    } else {
      arrayA[k] = arrayB[j];
      j--;
    }
    k--;
  }
  while (j >= 0) {
    arrayA[k--] = arrayB[j--];
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<extw>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  cmpl   $0xff,-0x14(%rbp)
  setbe  %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  mov    -0x14(%rbp),%eax
  and    $0x80,%eax
  test   %eax,%eax
  je     L17
  movq   $0xffffffffffffff00,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %eax,%edx
  mov    -0x14(%rbp),%eax
  or     %edx,%eax
  jmp    L18
L17:
  mov    -0x14(%rbp),%eax
L18:
  leave
  ret
```

assistant: ```c
int extw(unsigned int b) {
  assert(b < 0x100);
  if (b & 0x80) {
    long v = -256;
    return (v | b);
  }
  return b;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<extw>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  cmp    $0xff,%edi
  setbe  %dil
  movzbl %dil,%edi
  call   <assert@plt>
  mov    %ebx,%eax
  or     $0xffffff00,%eax
  test   $0x80,%bl
  cmove  %ebx,%eax
  pop    %rbx
  ret
```

assistant: ```c
int extw(unsigned int b) {
  assert(b < 0x100);
  if (b & 0x80) {
    long v = -256;
    return (v | b);
  }
  return b;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<extw>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  xor    %edi,%edi
  cmp    $0xff,%ebx
  setbe  %dil
  call   <assert@plt>
  mov    %ebx,%eax
  or     $0xffffff00,%eax
  test   $0x80,%bl
  cmove  %ebx,%eax
  pop    %rbx
  ret
```

assistant: ```c
int extw(unsigned int b) {
  assert(b < 0x100);
  if (b & 0x80) {
    long v = -256;
    return (v | b);
  }
  return b;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<extw>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  xor    %edi,%edi
  cmp    $0xff,%ebx
  setbe  %dil
  call   <assert@plt>
  mov    %ebx,%eax
  or     $0xffffff00,%eax
  test   $0x80,%bl
  cmove  %ebx,%eax
  pop    %rbx
  ret
```

assistant: ```c
int extw(unsigned int b) {
  assert(b < 0x100);
  if (b & 0x80) {
    long v = -256;
    return (v | b);
  }
  return b;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<puts>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  jmp    L9
L16:
  mov    -0x4(%rbp),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <putchar@plt>
L9:
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jne    L16
  mov    $0xa,%edi
  call   <putchar@plt>
  leave
  ret
```

assistant: ```c
int puts(const char *str) {
  int c;
  while ((c = *str++) != '\0')
    putchar(c);
  return (putchar('\n'));
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<puts>:
  endbr64
  push   %rbx
  lea    0x1(%rdi),%rbx
  movzbl (%rdi),%eax
  movsbl %al,%edi
  test   %al,%al
  je     L1
L19:
  call   <putchar@plt>
  add    $0x1,%rbx
  movzbl -0x1(%rbx),%eax
  movsbl %al,%edi
  test   %al,%al
  jne    L19
L1:
  mov    $0xa,%edi
  call   <putchar@plt>
  pop    %rbx
  ret
```

assistant: ```c
int puts(const char *str) {
  int c;
  while ((c = *str++) != '\0')
    putchar(c);
  return (putchar('\n'));
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<puts>:
  endbr64
  push   %rbx
  lea    0x1(%rdi),%rbx
  movsbl (%rdi),%edi
  test   %dil,%dil
  je     L4
  nopl   0x0(%rax)
L7:
  call   <putchar@plt>
  movsbl (%rbx),%edi
  add    $0x1,%rbx
  test   %dil,%dil
  jne    L7
L4:
  mov    $0xa,%edi
  pop    %rbx
  jmp    1050 <putchar@plt>
```

assistant: ```c
int puts(const char *str) {
  int c;
  while ((c = *str++) != '\0')
    putchar(c);
  return (putchar('\n'));
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<puts>:
  endbr64
  push   %rbx
  lea    0x1(%rdi),%rbx
  movsbl (%rdi),%edi
  test   %dil,%dil
  je     L15
  nopl   0x0(%rax)
L8:
  call   <putchar@plt>
  movsbl (%rbx),%edi
  add    $0x1,%rbx
  test   %dil,%dil
  jne    L8
L15:
  mov    $0xa,%edi
  pop    %rbx
  jmp    1050 <putchar@plt>
```

assistant: ```c
int puts(const char *str) {
  int c;
  while ((c = *str++) != '\0')
    putchar(c);
  return (putchar('\n'));
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<sarprintf>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xd0,%rsp
  mov    %rdi,-0xc8(%rbp)
  mov    %rsi,-0xa8(%rbp)
  mov    %rdx,-0xa0(%rbp)
  mov    %rcx,-0x98(%rbp)
  mov    %r8,-0x90(%rbp)
  mov    %r9,-0x88(%rbp)
  test   %al,%al
  je     L8
  movaps %xmm0,-0x80(%rbp)
  movaps %xmm1,-0x70(%rbp)
  movaps %xmm2,-0x60(%rbp)
  movaps %xmm3,-0x50(%rbp)
  movaps %xmm4,-0x40(%rbp)
  movaps %xmm5,-0x30(%rbp)
  movaps %xmm6,-0x20(%rbp)
  movaps %xmm7,-0x10(%rbp)
L8:
  mov    -0xc8(%rbp),%rdx
  mov    -0xb8(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <va_start@plt>
  mov    -0xb8(%rbp),%edx
  mov    -0xc8(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <sarvprintf_std@plt>
  mov    %eax,-0xb4(%rbp)
  mov    -0xb8(%rbp),%eax
  mov    %eax,%edi
  call   <va_end@plt>
  mov    -0xb4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int sarprintf(const char *format, ...) {
  va_list args;
  va_start(args, format);
  int res = sarvprintf_std(format, args);
  va_end(args);
  return res;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<sarprintf>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rdi,%rsi
  mov    $0x0,%ebp
  mov    %ebp,%edi
  call   <va_start@plt>
  mov    %ebp,%esi
  mov    %rbx,%rdi
  call   <sarvprintf_std@plt>
  mov    %eax,%ebx
  mov    %ebp,%edi
  call   <va_end@plt>
  mov    %ebx,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int sarprintf(const char *format, ...) {
  va_list args;
  va_start(args, format);
  int res = sarvprintf_std(format, args);
  va_end(args);
  return res;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<sarprintf>:
  endbr64
  push   %r13
  xor    %r13d,%r13d
  mov    %rdi,%rsi
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  mov    %r13d,%edi
  call   <va_start@plt>
  mov    %r13d,%esi
  mov    %rbp,%rdi
  call   <sarvprintf_std@plt>
  mov    %r13d,%edi
  mov    %eax,%r12d
  call   <va_end@plt>
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int sarprintf(const char *format, ...) {
  va_list args;
  va_start(args, format);
  int res = sarvprintf_std(format, args);
  va_end(args);
  return res;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<sarprintf>:
  endbr64
  push   %r13
  xor    %r13d,%r13d
  mov    %rdi,%rsi
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  mov    %r13d,%edi
  call   <va_start@plt>
  mov    %r13d,%esi
  mov    %rbp,%rdi
  call   <sarvprintf_std@plt>
  mov    %r13d,%edi
  mov    %eax,%r12d
  call   <va_end@plt>
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int sarprintf(const char *format, ...) {
  va_list args;
  va_start(args, format);
  int res = sarvprintf_std(format, args);
  va_end(args);
  return res;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<debug_dump_memory_bytes>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  mov    -0x8(%rbp),%rdx
  mov    $0xf180000c,%eax
  mov    %edx,(%rax)
  mov    $0xf1800010,%edx
  mov    -0xc(%rbp),%eax
  mov    %eax,(%rdx)
  mov    $0xf1800014,%eax
  movl   $0x1,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void debug_dump_memory_bytes(const void *mem, int len) {
  *((volatile unsigned int *)((((((((((0xf0000000)) + (4 * 1024 * 1024)) + (4 * 1024 * 1024)) + (4 * 1024 * 1024)) + (4 * 1024 * 1024)) + (4 * 1024 * 1024)) + (4 * 1024 * 1024)) + 12))) = (unsigned int)mem;
  *((volatile unsigned int *)((((((((((0xf0000000)) + (4 * 1024 * 1024)) + (4 * 1024 * 1024)) + (4 * 1024 * 1024)) + (4 * 1024 * 1024)) + (4 * 1024 * 1024)) + (4 * 1024 * 1024)) + 16))) = len;
  *((volatile unsigned int *)((((((((((0xf0000000)) + (4 * 1024 * 1024)) + (4 * 1024 * 1024)) + (4 * 1024 * 1024)) + (4 * 1024 * 1024)) + (4 * 1024 * 1024)) + (4 * 1024 * 1024)) + 20))) = 1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<debug_dump_memory_bytes>:
  endbr64
  mov    %rdi,%rax
  movabs %eax,0xf180000c
  mov    %esi,%eax
  movabs %eax,0xf1800010
  mov    $0xf1800014,%eax
  movl   $0x1,(%rax)
  ret
```

assistant: ```c
void debug_dump_memory_bytes(const void *mem, int len) {
  *((volatile unsigned int *)((((((((((0xf0000000)) + (4 * 1024 * 1024)) + (4 * 1024 * 1024)) + (4 * 1024 * 1024)) + (4 * 1024 * 1024)) + (4 * 1024 * 1024)) + (4 * 1024 * 1024)) + 12))) = (unsigned int)mem;
  *((volatile unsigned int *)((((((((((0xf0000000)) + (4 * 1024 * 1024)) + (4 * 1024 * 1024)) + (4 * 1024 * 1024)) + (4 * 1024 * 1024)) + (4 * 1024 * 1024)) + (4 * 1024 * 1024)) + 16))) = len;
  *((volatile unsigned int *)((((((((((0xf0000000)) + (4 * 1024 * 1024)) + (4 * 1024 * 1024)) + (4 * 1024 * 1024)) + (4 * 1024 * 1024)) + (4 * 1024 * 1024)) + (4 * 1024 * 1024)) + 20))) = 1;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<debug_dump_memory_bytes>:
  endbr64
  mov    %rdi,%rax
  movabs %eax,0xf180000c
  mov    %esi,%eax
  movabs %eax,0xf1800010
  mov    $0xf1800014,%eax
  movl   $0x1,(%rax)
  ret
```

assistant: ```c
void debug_dump_memory_bytes(const void *mem, int len) {
  *((volatile unsigned int *)((((((((((0xf0000000)) + (4 * 1024 * 1024)) + (4 * 1024 * 1024)) + (4 * 1024 * 1024)) + (4 * 1024 * 1024)) + (4 * 1024 * 1024)) + (4 * 1024 * 1024)) + 12))) = (unsigned int)mem;
  *((volatile unsigned int *)((((((((((0xf0000000)) + (4 * 1024 * 1024)) + (4 * 1024 * 1024)) + (4 * 1024 * 1024)) + (4 * 1024 * 1024)) + (4 * 1024 * 1024)) + (4 * 1024 * 1024)) + 16))) = len;
  *((volatile unsigned int *)((((((((((0xf0000000)) + (4 * 1024 * 1024)) + (4 * 1024 * 1024)) + (4 * 1024 * 1024)) + (4 * 1024 * 1024)) + (4 * 1024 * 1024)) + (4 * 1024 * 1024)) + 20))) = 1;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<d>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <a@plt>
  mov    $0x0,%eax
  call   <g@plt>
  nop
  pop    %rbp
  ret
```

assistant: ```c
void d() {
  a();
  g();
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<d>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <a@plt>
  mov    $0x0,%eax
  call   <g@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void d() {
  a();
  g();
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<d>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <a@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1060 <g@plt>
```

assistant: ```c
void d() {
  a();
  g();
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<d>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <a@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1060 <g@plt>
```

assistant: ```c
void d() {
  a();
  g();
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<remove>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    D19(%rip),%rax
  mov    (%rax),%edx
  mov    D4(%rip),%rax
  mov    %edx,(%rax)
  mov    $0x1,%eax
  pop    %rbp
  ret
```

assistant: ```c
int remove(const char *pathname) {
  errno = ENOTSUP;
  return 1;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<remove>:
  endbr64
  mov    D14(%rip),%rax
  mov    (%rax),%edx
  mov    D10(%rip),%rax
  mov    %edx,(%rax)
  mov    $0x1,%eax
  ret
```

assistant: ```c
int remove(const char *pathname) {
  errno = ENOTSUP;
  return 1;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<remove>:
  endbr64
  mov    D10(%rip),%rax
  mov    (%rax),%edx
  mov    D9(%rip),%rax
  mov    %edx,(%rax)
  mov    $0x1,%eax
  ret
```

assistant: ```c
int remove(const char *pathname) {
  errno = ENOTSUP;
  return 1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<remove>:
  endbr64
  mov    D18(%rip),%rax
  mov    (%rax),%edx
  mov    D8(%rip),%rax
  mov    %edx,(%rax)
  mov    $0x1,%eax
  ret
```

assistant: ```c
int remove(const char *pathname) {
  errno = ENOTSUP;
  return 1;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<cmd_read>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x240,%rsp
  mov    %edi,-0x234(%rbp)
  mov    %rsi,-0x240(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x240(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <atoi@plt>
  mov    %eax,-0x220(%rbp)
  mov    D3(%rip),%rax
  mov    (%rax),%edx
  mov    -0x220(%rbp),%eax
  mov    $0x0,%esi
  mov    %eax,%edi
  call   <lseek@plt>
  mov    %eax,-0x21c(%rbp)
  mov    D29(%rip),%rax
  mov    (%rax),%edx
  mov    -0x220(%rbp),%eax
  mov    $0x0,%esi
  mov    %eax,%edi
  call   <lseek@plt>
  mov    %eax,-0x218(%rbp)
  mov    -0x218(%rbp),%eax
  sub    -0x21c(%rbp),%eax
  mov    %eax,-0x228(%rbp)
  mov    D2(%rip),%rax
  mov    (%rax),%edx
  mov    -0x21c(%rbp),%ecx
  mov    -0x220(%rbp),%eax
  mov    %ecx,%esi
  mov    %eax,%edi
  call   <lseek@plt>
  cmpl   $0x2,-0x234(%rbp)
  jle    L7
  mov    -0x240(%rbp),%rax
  add    $0x10,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <atoi@plt>
  mov    %eax,-0x228(%rbp)
L7:
  movl   $0x0,-0x224(%rbp)
  jmp    L54
L38:
  mov    -0x228(%rbp),%eax
  sub    -0x224(%rbp),%eax
  mov    %eax,%edx
  mov    $0x200,%eax
  cmp    %eax,%edx
  cmovbe %edx,%eax
  mov    %eax,%edx
  lea    -0x210(%rbp),%rcx
  mov    -0x220(%rbp),%eax
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <read@plt>
  mov    %eax,-0x214(%rbp)
  cmpl   $0xffffffff,-0x214(%rbp)
  je     L53
  mov    -0x214(%rbp),%edx
  lea    -0x210(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x1,%edi
  call   <write@plt>
  mov    -0x214(%rbp),%eax
  add    %eax,-0x224(%rbp)
L54:
  mov    -0x224(%rbp),%eax
  cmp    -0x228(%rbp),%eax
  jl     L38
  jmp    L10
L53:
  nop
L10:
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L24
  call   <__stack_chk_fail@plt>
L24:
  leave
  ret
```

assistant: ```c
void cmd_read(int argc, char *argv[]) {
  int fd, size;
  int read_out, rsz;
  char buf[512];
  fd = atoi(argv[1]);
  {
    int cur, end;
    cur = lseek(fd, 0, SEEK_CUR);
    end = lseek(fd, 0, SEEK_END);
    size = end - cur;
    lseek(fd, cur, SEEK_SET);
  }
  if (argc > 2)
    size = atoi(argv[2]);
  for (rsz = 0; rsz < size;) {
    read_out = read(fd, buf, ((sizeof(buf)) < (size - rsz) ? (sizeof(buf)) : (size - rsz)));
    if (read_out == -1)
      return;
    write(1, buf, read_out);
    rsz += read_out;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<cmd_read>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x210,%rsp
  mov    %edi,%r13d
  mov    %rsi,%rbp
  mov    %fs:0x28,%rax
  mov    %rax,0x208(%rsp)
  xor    %eax,%eax
  mov    0x8(%rsi),%rdi
  call   <atoi@plt>
  mov    %eax,%r14d
  mov    D23(%rip),%rax
  mov    (%rax),%edx
  mov    $0x0,%esi
  mov    %r14d,%edi
  call   <lseek@plt>
  mov    %eax,%ebx
  mov    D24(%rip),%rax
  mov    (%rax),%edx
  mov    $0x0,%esi
  mov    %r14d,%edi
  call   <lseek@plt>
  mov    %eax,%r12d
  mov    D9(%rip),%rax
  mov    (%rax),%edx
  mov    %ebx,%esi
  mov    %r14d,%edi
  call   <lseek@plt>
  sub    %ebx,%r12d
  cmp    $0x2,%r13d
  jg     L56
L28:
  test   %r12d,%r12d
  jle    L26
  mov    $0x0,%ebp
  mov    %rsp,%r13
L8:
  mov    %r12d,%edx
  sub    %ebp,%edx
  mov    $0x200,%eax
  cmp    %eax,%edx
  cmova  %eax,%edx
  mov    %r13,%rsi
  mov    %r14d,%edi
  call   <read@plt>
  mov    %eax,%ebx
  cmp    $0xffffffff,%eax
  je     L26
  mov    %eax,%edx
  mov    %r13,%rsi
  mov    $0x1,%edi
  call   <write@plt>
  add    %ebx,%ebp
  cmp    %ebp,%r12d
  jg     L8
L26:
  mov    0x208(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L12
  add    $0x210,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L56:
  mov    0x10(%rbp),%rdi
  call   <atoi@plt>
  mov    %eax,%r12d
  jmp    L28
L12:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void cmd_read(int argc, char *argv[]) {
  int fd, size;
  int read_out, rsz;
  char buf[512];
  fd = atoi(argv[1]);
  {
    int cur, end;
    cur = lseek(fd, 0, SEEK_CUR);
    end = lseek(fd, 0, SEEK_END);
    size = end - cur;
    lseek(fd, cur, SEEK_SET);
  }
  if (argc > 2)
    size = atoi(argv[2]);
  for (rsz = 0; rsz < size;) {
    read_out = read(fd, buf, ((sizeof(buf)) < (size - rsz) ? (sizeof(buf)) : (size - rsz)));
    if (read_out == -1)
      return;
    write(1, buf, read_out);
    rsz += read_out;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<cmd_read>:
  endbr64
  push   %r14
  push   %r13
  mov    %edi,%r13d
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  sub    $0x210,%rsp
  mov    0x8(%rsi),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x208(%rsp)
  xor    %eax,%eax
  call   <atoi@plt>
  xor    %esi,%esi
  mov    %eax,%r14d
  mov    D1(%rip),%rax
  mov    %r14d,%edi
  mov    (%rax),%edx
  call   <lseek@plt>
  xor    %esi,%esi
  mov    %r14d,%edi
  mov    %eax,%ebx
  mov    D24(%rip),%rax
  mov    (%rax),%edx
  call   <lseek@plt>
  mov    %ebx,%esi
  mov    %r14d,%edi
  mov    %eax,%r12d
  mov    D25(%rip),%rax
  sub    %ebx,%r12d
  mov    (%rax),%edx
  call   <lseek@plt>
  cmp    $0x2,%r13d
  jg     L52
L28:
  test   %r12d,%r12d
  jle    L68
  xor    %ebp,%ebp
  mov    %rsp,%r13
  jmp    L31
  nopl   0x0(%rax,%rax,1)
L5:
  mov    %eax,%edx
  mov    %r13,%rsi
  mov    $0x1,%edi
  add    %ebx,%ebp
  call   <write@plt>
  cmp    %ebp,%r12d
  jle    L68
L31:
  mov    %r12d,%edx
  mov    $0x200,%eax
  mov    %r13,%rsi
  mov    %r14d,%edi
  sub    %ebp,%edx
  cmp    %eax,%edx
  cmova  %eax,%edx
  call   <read@plt>
  mov    %eax,%ebx
  cmp    $0xffffffff,%eax
  jne    L5
L68:
  mov    0x208(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L57
  add    $0x210,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopw   0x0(%rax,%rax,1)
L52:
  mov    0x10(%rbp),%rdi
  call   <atoi@plt>
  mov    %eax,%r12d
  jmp    L28
L57:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void cmd_read(int argc, char *argv[]) {
  int fd, size;
  int read_out, rsz;
  char buf[512];
  fd = atoi(argv[1]);
  {
    int cur, end;
    cur = lseek(fd, 0, SEEK_CUR);
    end = lseek(fd, 0, SEEK_END);
    size = end - cur;
    lseek(fd, cur, SEEK_SET);
  }
  if (argc > 2)
    size = atoi(argv[2]);
  for (rsz = 0; rsz < size;) {
    read_out = read(fd, buf, ((sizeof(buf)) < (size - rsz) ? (sizeof(buf)) : (size - rsz)));
    if (read_out == -1)
      return;
    write(1, buf, read_out);
    rsz += read_out;
  }
}
```