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

```asm
<main>:
  endbr64
  push   %rbp
  sub    $0x30,%rsp
  movdqa D1(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  movl   $0x9,0x20(%rsp)
  movaps %xmm0,(%rsp)
  movdqa D18(%rip),%xmm0
  mov    %rbp,%rdi
  movaps %xmm0,0x10(%rsp)
  call   <sum_row_max@plt>
  lea    D10(%rip),%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rdi
  lea    D41(%rip),%rbp
  call   <sum_matrix@plt>
  lea    D48(%rip),%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <string_length@plt>
  mov    %rbp,%rsi
  lea    D54(%rip),%rdi
  mov    %eax,%edx
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x30,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D1
datatype: byte[16]
name: read_data
arguments: label: D18
datatype: byte[16]
name: read_data
arguments: label: D10
datatype: string
name: read_data
arguments: label: D41
datatype: string
name: read_data
arguments: label: D48
datatype: string
name: read_data
arguments: label: D54
datatype: string

tool: {"D1": [0x01,0x00,0x00,0x00,0x02,0x00,0x00,0x00,0x03,0x00,0x00,0x00,0x04,0x00,0x00,0x00]}

tool: {"D18": [0x05,0x00,0x00,0x00,0x06,0x00,0x00,0x00,0x07,0x00,0x00,0x00,0x08,0x00,0x00,0x00]}

tool: D10: Sum of max from each row is: %d


tool: D41: Hello World

tool: D48: Sum of all elements is: %d


tool: D54: Length of '%s' is %d


assistant: ```c
int main() {
  int x[3][3] = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}};
  int sum_max = sum_row_max(x);
  printf("Sum of max from each row is: %d\n", sum_max);
  int sum_elements = sum_matrix(x);
  printf("Sum of all elements is: %d\n", sum_elements);
  char *s = "Hello World";
  int length = string_length(s);
  printf("Length of '%s' is %d\n", s, length);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<factorial>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  cmpq   $0x1,-0x8(%rbp)
  jne    L10
  mov    $0x1,%eax
  jmp    L2
L10:
  mov    -0x8(%rbp),%rax
  sub    $0x1,%rax
  mov    %rax,%rdi
  call   <factorial>
  imul   -0x8(%rbp),%rax
L2:
  leave
  ret
```

assistant: ```c
unsigned long factorial(unsigned long n) {
  if (n == 1)
    return 1;
  return n * factorial(n - 1);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<factorial>:
  endbr64
  mov    %rdi,%rax
  cmp    $0x1,%rdi
  je     L8
  push   %rbx
  mov    %rdi,%rbx
  lea    -0x1(%rdi),%rdi
  call   <factorial>
  imul   %rbx,%rax
  pop    %rbx
  ret
L8:
  ret
```

assistant: ```c
unsigned long factorial(unsigned long n) {
  if (n == 1)
    return 1;
  return n * factorial(n - 1);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<factorial>:
  endbr64
  mov    $0x1,%eax
  cmp    $0x1,%rdi
  je     L4
  nop
L6:
  mov    %rdi,%rdx
  sub    $0x1,%rdi
  imul   %rdx,%rax
  cmp    $0x1,%rdi
  jne    L6
L4:
  ret
```

assistant: ```c
unsigned long factorial(unsigned long n) {
  if (n == 1)
    return 1;
  return n * factorial(n - 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
<factorial>:
  endbr64
  mov    $0x1,%eax
  cmp    $0x1,%rdi
  je     L17
  nop
L12:
  mov    %rdi,%rdx
  sub    $0x1,%rdi
  imul   %rdx,%rax
  cmp    $0x1,%rdi
  jne    L12
L17:
  ret
```

assistant: ```c
unsigned long factorial(unsigned long n) {
  if (n == 1)
    return 1;
  return n * factorial(n - 1);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<convertimage>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdx,%rax
  shr    $0x2,%rax
  mov    %rax,0x18(%rsp)
  cmp    $0x3,%rdx
  jbe    L13
  mov    %rdi,%r15
  mov    %rsi,%rbp
  mov    $0x0,%r12d
L14:
  movzbl (%r15,%r12,4),%ebx
  shl    $0x8,%ebx
  movzbl 0x1(%r15,%r12,4),%r13d
  add    $0xffffff80,%r13d
  movzbl 0x2(%r15,%r12,4),%eax
  shl    $0x8,%eax
  mov    %eax,0x14(%rsp)
  movzbl 0x3(%r15,%r12,4),%r14d
  add    $0xffffff80,%r14d
  imul   $0x167,%r14d,%edi
  mov    %edi,0xc(%rsp)
  add    %ebx,%edi
  sar    $0x8,%edi
  call   <clamp@plt>
  mov    %al,0x0(%rbp)
  imul   $0xffffffa8,%r13d,%ecx
  imul   $0xffffff49,%r14d,%r14d
  mov    %ecx,0x10(%rsp)
  lea    (%rcx,%rbx,1),%edi
  add    %r14d,%edi
  sar    $0x8,%edi
  call   <clamp@plt>
  mov    %al,0x1(%rbp)
  imul   $0x1c6,%r13d,%r13d
  lea    0x0(%r13,%rbx,1),%edi
  sar    $0x8,%edi
  call   <clamp@plt>
  mov    %al,0x2(%rbp)
  mov    0xc(%rsp),%edi
  mov    0x14(%rsp),%ebx
  add    %ebx,%edi
  sar    $0x8,%edi
  call   <clamp@plt>
  mov    %al,0x3(%rbp)
  mov    0x10(%rsp),%edi
  add    %ebx,%edi
  add    %r14d,%edi
  sar    $0x8,%edi
  call   <clamp@plt>
  mov    %al,0x4(%rbp)
  lea    0x0(%r13,%rbx,1),%edi
  sar    $0x8,%edi
  call   <clamp@plt>
  mov    %al,0x5(%rbp)
  add    $0x1,%r12
  add    $0x6,%rbp
  cmp    0x18(%rsp),%r12
  jb     L14
L13:
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void convertimage(unsigned char *inbuf, unsigned char *outbuf, size_t insize) {
  size_t i;
  for (i = 0; i < insize / 4; i++) {
    int y1 = inbuf[i * 4 + 0] << 8;
    int u = inbuf[i * 4 + 1] - 128;
    int y2 = inbuf[i * 4 + 2] << 8;
    int v = inbuf[i * 4 + 3] - 128;
    outbuf[i * 6 + 0] = clamp((y1 + (359 * v)) >> 8);
    outbuf[i * 6 + 1] = clamp((y1 - (88 * u) - (183 * v)) >> 8);
    outbuf[i * 6 + 2] = clamp((y1 + (454 * u)) >> 8);
    outbuf[i * 6 + 3] = clamp((y2 + (359 * v)) >> 8);
    outbuf[i * 6 + 4] = clamp((y2 - (88 * u) - (183 * v)) >> 8);
    outbuf[i * 6 + 5] = clamp((y2 + (454 * u)) >> 8);
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<convertimage>:
  endbr64
  shr    $0x2,%rdx
  je     L12
  push   %r15
  lea    (%rdi,%rdx,4),%rax
  mov    %rdi,%r15
  push   %r14
  push   %r13
  mov    %rsi,%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rax,0x8(%rsp)
  nopl   0x0(%rax,%rax,1)
L19:
  movzbl 0x3(%r15),%edx
  movzbl (%r15),%ebp
  add    $0x6,%r13
  add    $0x4,%r15
  movzbl -0x3(%r15),%ebx
  movzbl -0x2(%r15),%r14d
  add    $0xffffff80,%edx
  shl    $0x8,%ebp
  imul   $0x167,%edx,%esi
  mov    %edx,(%rsp)
  add    $0xffffff80,%ebx
  shl    $0x8,%r14d
  imul   $0xffffffa8,%ebx,%r12d
  imul   $0x1c6,%ebx,%ebx
  lea    (%rsi,%rbp,1),%edi
  mov    %esi,0x4(%rsp)
  sar    $0x8,%edi
  call   <clamp@plt>
  mov    (%rsp),%edx
  lea    (%r12,%rbp,1),%edi
  mov    %al,-0x6(%r13)
  imul   $0xffffff49,%edx,%edx
  add    %edx,%edi
  mov    %edx,(%rsp)
  sar    $0x8,%edi
  call   <clamp@plt>
  lea    (%rbx,%rbp,1),%edi
  mov    %al,-0x5(%r13)
  sar    $0x8,%edi
  call   <clamp@plt>
  mov    0x4(%rsp),%esi
  mov    %al,-0x4(%r13)
  lea    (%rsi,%r14,1),%edi
  sar    $0x8,%edi
  call   <clamp@plt>
  mov    (%rsp),%edx
  lea    (%r12,%r14,1),%edi
  mov    %al,-0x3(%r13)
  add    %edx,%edi
  sar    $0x8,%edi
  call   <clamp@plt>
  lea    (%rbx,%r14,1),%edi
  mov    %al,-0x2(%r13)
  sar    $0x8,%edi
  call   <clamp@plt>
  mov    %al,-0x1(%r13)
  cmp    %r15,0x8(%rsp)
  jne    L19
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  xchg   %ax,%ax
L12:
  ret
```

assistant: ```c
void convertimage(unsigned char *inbuf, unsigned char *outbuf, size_t insize) {
  size_t i;
  for (i = 0; i < insize / 4; i++) {
    int y1 = inbuf[i * 4 + 0] << 8;
    int u = inbuf[i * 4 + 1] - 128;
    int y2 = inbuf[i * 4 + 2] << 8;
    int v = inbuf[i * 4 + 3] - 128;
    outbuf[i * 6 + 0] = clamp((y1 + (359 * v)) >> 8);
    outbuf[i * 6 + 1] = clamp((y1 - (88 * u) - (183 * v)) >> 8);
    outbuf[i * 6 + 2] = clamp((y1 + (454 * u)) >> 8);
    outbuf[i * 6 + 3] = clamp((y2 + (359 * v)) >> 8);
    outbuf[i * 6 + 4] = clamp((y2 - (88 * u) - (183 * v)) >> 8);
    outbuf[i * 6 + 5] = clamp((y2 + (454 * u)) >> 8);
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<convertimage>:
  endbr64
  shr    $0x2,%rdx
  je     L11
  push   %r15
  lea    (%rdi,%rdx,4),%rax
  mov    %rdi,%r15
  push   %r14
  push   %r13
  mov    %rsi,%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rax,0x8(%rsp)
  nopl   0x0(%rax,%rax,1)
L8:
  movzbl 0x3(%r15),%edx
  movzbl (%r15),%ebp
  add    $0x6,%r13
  add    $0x4,%r15
  movzbl -0x3(%r15),%ebx
  movzbl -0x2(%r15),%r14d
  add    $0xffffff80,%edx
  shl    $0x8,%ebp
  imul   $0x167,%edx,%esi
  mov    %edx,(%rsp)
  add    $0xffffff80,%ebx
  shl    $0x8,%r14d
  imul   $0xffffffa8,%ebx,%r12d
  imul   $0x1c6,%ebx,%ebx
  lea    (%rsi,%rbp,1),%edi
  mov    %esi,0x4(%rsp)
  sar    $0x8,%edi
  call   <clamp@plt>
  mov    (%rsp),%edx
  lea    (%r12,%rbp,1),%edi
  mov    %al,-0x6(%r13)
  imul   $0xffffff49,%edx,%edx
  add    %edx,%edi
  mov    %edx,(%rsp)
  sar    $0x8,%edi
  call   <clamp@plt>
  lea    (%rbx,%rbp,1),%edi
  mov    %al,-0x5(%r13)
  sar    $0x8,%edi
  call   <clamp@plt>
  mov    0x4(%rsp),%esi
  mov    %al,-0x4(%r13)
  lea    (%rsi,%r14,1),%edi
  sar    $0x8,%edi
  call   <clamp@plt>
  mov    (%rsp),%edx
  lea    (%r12,%r14,1),%edi
  mov    %al,-0x3(%r13)
  add    %edx,%edi
  sar    $0x8,%edi
  call   <clamp@plt>
  lea    (%rbx,%r14,1),%edi
  mov    %al,-0x2(%r13)
  sar    $0x8,%edi
  call   <clamp@plt>
  mov    %al,-0x1(%r13)
  cmp    %r15,0x8(%rsp)
  jne    L8
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  xchg   %ax,%ax
L11:
  ret
```

assistant: ```c
void convertimage(unsigned char *inbuf, unsigned char *outbuf, size_t insize) {
  size_t i;
  for (i = 0; i < insize / 4; i++) {
    int y1 = inbuf[i * 4 + 0] << 8;
    int u = inbuf[i * 4 + 1] - 128;
    int y2 = inbuf[i * 4 + 2] << 8;
    int v = inbuf[i * 4 + 3] - 128;
    outbuf[i * 6 + 0] = clamp((y1 + (359 * v)) >> 8);
    outbuf[i * 6 + 1] = clamp((y1 - (88 * u) - (183 * v)) >> 8);
    outbuf[i * 6 + 2] = clamp((y1 + (454 * u)) >> 8);
    outbuf[i * 6 + 3] = clamp((y2 + (359 * v)) >> 8);
    outbuf[i * 6 + 4] = clamp((y2 - (88 * u) - (183 * v)) >> 8);
    outbuf[i * 6 + 5] = clamp((y2 + (454 * u)) >> 8);
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    0x10(%rbp),%rax
  mov    %rax,-0x18(%rbp)
  mov    $0xbc614e,%eax
  mov    %eax,-0x20(%rbp)
  lea    -0x20(%rbp),%rax
  mov    %rax,%r10
  call   <foo.0>
  mov    $0x0,%eax
  call   <bar@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L8
  call   <__stack_chk_fail@plt>
L8:
  leave
  ret
```

assistant: ```c
int main(void) {
  int i = 12345678;
  int foo(void) {
    if (i != 12345678)
      abort();
  }
  foo();
  bar();
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

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

assistant: ```c
int main(void) {
  int i = 12345678;
  int foo(void) {
    if (i != 12345678)
      abort();
  }
  foo();
  bar();
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <bar@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  int i = 12345678;
  int foo(void) {
    if (i != 12345678)
      abort();
  }
  foo();
  bar();
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <bar@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  int i = 12345678;
  int foo(void) {
    if (i != 12345678)
      abort();
  }
  foo();
  bar();
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<hw_wait_for_key>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  nop
L4:
  mov    $0x0,%eax
  call   <hw_read_key@plt>
  test   %al,%al
  je     L4
  mov    $0x0,%eax
  call   <hw_read_key@plt>
  pop    %rbp
  ret
```

assistant: ```c
char hw_wait_for_key() {
  while (hw_read_key() == 0)
    ;
  return hw_read_key();
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<hw_wait_for_key>:
  endbr64
  sub    $0x8,%rsp
L6:
  mov    $0x0,%eax
  call   <hw_read_key@plt>
  test   %al,%al
  je     L6
  mov    $0x0,%eax
  call   <hw_read_key@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
char hw_wait_for_key() {
  while (hw_read_key() == 0)
    ;
  return hw_read_key();
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<hw_wait_for_key>:
  endbr64
  sub    $0x8,%rsp
  nopl   0x0(%rax,%rax,1)
L9:
  xor    %eax,%eax
  call   <hw_read_key@plt>
  test   %al,%al
  je     L9
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1050 <hw_read_key@plt>
```

assistant: ```c
char hw_wait_for_key() {
  while (hw_read_key() == 0)
    ;
  return hw_read_key();
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<hw_wait_for_key>:
  endbr64
  sub    $0x8,%rsp
  nopl   0x0(%rax,%rax,1)
L4:
  xor    %eax,%eax
  call   <hw_read_key@plt>
  test   %al,%al
  je     L4
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1050 <hw_read_key@plt>
```

assistant: ```c
char hw_wait_for_key() {
  while (hw_read_key() == 0)
    ;
  return hw_read_key();
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<err_facer_>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  mov    -0x24(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,%edi
  call   <starMallocAtomic@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rdx
  mov    -0x24(%rbp),%ecx
  mov    -0x18(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <cnfImprt@plt>
  mov    -0x20(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <errFacer@plt>
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <starFree@plt>
  nop
  leave
  ret
```

assistant: ```c
void err_facer_(char *const TOKEN, int *const STATUS, int TOKEN_length) {
  char *token;
  int status;
  ;
  token = starMallocAtomic(TOKEN_length + 1);
  cnfImprt(TOKEN, TOKEN_length, token);
  status = *STATUS;
  errFacer(token, status);
  starFree(token);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<err_facer_>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r13
  mov    %rsi,%r12
  mov    %edx,%ebp
  lea    0x1(%rdx),%edi
  call   <starMallocAtomic@plt>
  mov    %rax,%rbx
  mov    %rax,%rdx
  mov    %ebp,%esi
  mov    %r13,%rdi
  call   <cnfImprt@plt>
  mov    (%r12),%esi
  mov    %rbx,%rdi
  call   <errFacer@plt>
  mov    %rbx,%rdi
  call   <starFree@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
void err_facer_(char *const TOKEN, int *const STATUS, int TOKEN_length) {
  char *token;
  int status;
  ;
  token = starMallocAtomic(TOKEN_length + 1);
  cnfImprt(TOKEN, TOKEN_length, token);
  status = *STATUS;
  errFacer(token, status);
  starFree(token);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<err_facer_>:
  endbr64
  push   %r13
  mov    %rdi,%r13
  lea    0x1(%rdx),%edi
  push   %r12
  mov    %edx,%r12d
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x8,%rsp
  call   <starMallocAtomic@plt>
  mov    %r12d,%esi
  mov    %r13,%rdi
  mov    %rax,%rdx
  mov    %rax,%rbp
  call   <cnfImprt@plt>
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  call   <errFacer@plt>
  add    $0x8,%rsp
  mov    %rbp,%rdi
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    10b0 <starFree@plt>
```

assistant: ```c
void err_facer_(char *const TOKEN, int *const STATUS, int TOKEN_length) {
  char *token;
  int status;
  ;
  token = starMallocAtomic(TOKEN_length + 1);
  cnfImprt(TOKEN, TOKEN_length, token);
  status = *STATUS;
  errFacer(token, status);
  starFree(token);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<err_facer_>:
  endbr64
  push   %r13
  mov    %rdi,%r13
  lea    0x1(%rdx),%edi
  push   %r12
  mov    %edx,%r12d
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x8,%rsp
  call   <starMallocAtomic@plt>
  mov    %r12d,%esi
  mov    %r13,%rdi
  mov    %rax,%rdx
  mov    %rax,%rbp
  call   <cnfImprt@plt>
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  call   <errFacer@plt>
  add    $0x8,%rsp
  mov    %rbp,%rdi
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    10b0 <starFree@plt>
```

assistant: ```c
void err_facer_(char *const TOKEN, int *const STATUS, int TOKEN_length) {
  char *token;
  int status;
  ;
  token = starMallocAtomic(TOKEN_length + 1);
  cnfImprt(TOKEN, TOKEN_length, token);
  status = *STATUS;
  errFacer(token, status);
  starFree(token);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<memcpy_from_eeprom_with_checksum>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  movb   $0x0,-0x2(%rbp)
  jmp    L0
L44:
  mov    -0x1c(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x1c(%rbp)
  mov    %eax,%edi
  call   <eeprom_get_char@plt>
  mov    %al,-0x1(%rbp)
  movzbl -0x2(%rbp),%eax
  add    %eax,%eax
  test   %eax,%eax
  jne    L45
  movzbl -0x2(%rbp),%eax
  test   %al,%al
  jns    L43
L45:
  mov    $0x1,%eax
  jmp    L33
L43:
  mov    $0x0,%eax
L33:
  mov    %al,-0x2(%rbp)
  movzbl -0x1(%rbp),%eax
  add    %al,-0x2(%rbp)
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  movzbl -0x1(%rbp),%edx
  mov    %dl,(%rax)
  subl   $0x1,-0x20(%rbp)
L0:
  cmpl   $0x0,-0x20(%rbp)
  jne    L44
  mov    -0x1c(%rbp),%eax
  mov    %eax,%edi
  call   <eeprom_get_char@plt>
  cmp    %al,-0x2(%rbp)
  sete   %al
  movzbl %al,%eax
  leave
  ret
```

assistant: ```c
int memcpy_from_eeprom_with_checksum(char *destination, unsigned int source, unsigned int size) {
  unsigned char data, checksum = 0;
  for (; size > 0; size--) {
    data = eeprom_get_char(source++);
    checksum = (checksum << 1) || (checksum >> 7);
    checksum += data;
    *(destination++) = data;
  }
  return (checksum == eeprom_get_char(source));
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<memcpy_from_eeprom_with_checksum>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %esi,%r13d
  test   %edx,%edx
  je     L2
  mov    %edx,%r14d
  mov    %edx,%ebp
  add    %rdi,%rbp
  mov    %rdi,%rbx
  mov    $0x0,%r15d
  mov    %esi,%r12d
  sub    %edi,%r12d
L16:
  lea    (%r12,%rbx,1),%edi
  call   <eeprom_get_char@plt>
  movsbl %r15b,%edx
  shr    $0x1f,%edx
  test   %r15b,%r15b
  mov    $0x1,%ecx
  cmovne %ecx,%edx
  lea    (%rax,%rdx,1),%r15d
  add    $0x1,%rbx
  mov    %al,-0x1(%rbx)
  cmp    %rbp,%rbx
  jne    L16
  add    %r14d,%r13d
L14:
  mov    %r13d,%edi
  call   <eeprom_get_char@plt>
  cmp    %r15b,%al
  sete   %al
  movzbl %al,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L2:
  mov    $0x0,%r15d
  jmp    L14
```

assistant: ```c
int memcpy_from_eeprom_with_checksum(char *destination, unsigned int source, unsigned int size) {
  unsigned char data, checksum = 0;
  for (; size > 0; size--) {
    data = eeprom_get_char(source++);
    checksum = (checksum << 1) || (checksum >> 7);
    checksum += data;
    *(destination++) = data;
  }
  return (checksum == eeprom_get_char(source));
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<memcpy_from_eeprom_with_checksum>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  mov    %esi,%r13d
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  test   %edx,%edx
  je     L17
  mov    %edx,%r12d
  mov    %esi,%r15d
  mov    %rdi,%rbx
  mov    %edx,%r14d
  add    %rdi,%r12
  xor    %ebp,%ebp
  sub    %edi,%r15d
  nopl   (%rax)
L21:
  lea    (%r15,%rbx,1),%edi
  call   <eeprom_get_char@plt>
  cmp    $0x1,%bpl
  mov    %eax,%ebp
  sbb    $0xff,%bpl
  add    $0x1,%rbx
  mov    %al,-0x1(%rbx)
  cmp    %r12,%rbx
  jne    L21
  add    %r14d,%r13d
L26:
  mov    %r13d,%edi
  call   <eeprom_get_char@plt>
  cmp    %bpl,%al
  sete   %al
  add    $0x8,%rsp
  pop    %rbx
  movzbl %al,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L17:
  xor    %ebp,%ebp
  jmp    L26
```

assistant: ```c
int memcpy_from_eeprom_with_checksum(char *destination, unsigned int source, unsigned int size) {
  unsigned char data, checksum = 0;
  for (; size > 0; size--) {
    data = eeprom_get_char(source++);
    checksum = (checksum << 1) || (checksum >> 7);
    checksum += data;
    *(destination++) = data;
  }
  return (checksum == eeprom_get_char(source));
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<memcpy_from_eeprom_with_checksum>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  mov    %esi,%r13d
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  test   %edx,%edx
  je     L26
  mov    %edx,%r12d
  mov    %esi,%r15d
  mov    %rdi,%rbx
  mov    %edx,%r14d
  add    %rdi,%r12
  xor    %ebp,%ebp
  sub    %edi,%r15d
  nopl   (%rax)
L17:
  lea    (%r15,%rbx,1),%edi
  call   <eeprom_get_char@plt>
  cmp    $0x1,%bpl
  mov    %eax,%ebp
  sbb    $0xff,%bpl
  add    $0x1,%rbx
  mov    %al,-0x1(%rbx)
  cmp    %r12,%rbx
  jne    L17
  add    %r14d,%r13d
L9:
  mov    %r13d,%edi
  call   <eeprom_get_char@plt>
  cmp    %bpl,%al
  sete   %al
  add    $0x8,%rsp
  pop    %rbx
  movzbl %al,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L26:
  xor    %ebp,%ebp
  jmp    L9
```

assistant: ```c
int memcpy_from_eeprom_with_checksum(char *destination, unsigned int source, unsigned int size) {
  unsigned char data, checksum = 0;
  for (; size > 0; size--) {
    data = eeprom_get_char(source++);
    checksum = (checksum << 1) || (checksum >> 7);
    checksum += data;
    *(destination++) = data;
  }
  return (checksum == eeprom_get_char(source));
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<_write>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int _write(SSL_CTX *ctx, int sockfd) {
  int res = 0;
  return res;
}
```
user: What is the c source code of the assembly code below:

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

assistant: ```c
int _write(SSL_CTX *ctx, int sockfd) {
  int res = 0;
  return res;
}
```
user: What is the c source code of the assembly code below:

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

assistant: ```c
int _write(SSL_CTX *ctx, int sockfd) {
  int res = 0;
  return res;
}
```
user: Converte this assembly snippet into its equivalent C code:

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

assistant: ```c
int _write(SSL_CTX *ctx, int sockfd) {
  int res = 0;
  return res;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<alignp>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x1c(%rbp),%eax
  sub    $0x1,%eax
  cltq
  and    -0x10(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L6
  mov    $0x0,%eax
  jmp    L13
L6:
  mov    -0x1c(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %edx,%eax
L13:
  pop    %rbp
  ret
```

assistant: ```c
uint32_t alignp(const void *x, uint32_t pow2) {
  uintptr_t xp = (uintptr_t)x;
  uintptr_t r = xp & (pow2 - 1);
  if (!r)
    return 0;
  return pow2 - r;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<alignp>:
  endbr64
  lea    -0x1(%rsi),%eax
  cltq
  and    %rax,%rdi
  sub    %edi,%esi
  test   %rdi,%rdi
  mov    $0x0,%eax
  cmovne %esi,%eax
  ret
```

assistant: ```c
uint32_t alignp(const void *x, uint32_t pow2) {
  uintptr_t xp = (uintptr_t)x;
  uintptr_t r = xp & (pow2 - 1);
  if (!r)
    return 0;
  return pow2 - r;
}
```