ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What would the C source look like for the assembly code provided here:

```asm
<read_asm_line>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r12
  mov    %rsi,%r14
  call   <ignore_whitespaces@plt>
  mov    %eax,%ebp
  cmp    $0xffffffff,%eax
  je     L1
  mov    $0x0,%ebx
  mov    $0x0,%ebp
  mov    $0x1,%r15d
L20:
  mov    %ebx,%r13d
  mov    %r12,%rdi
  call   <mygetc@plt>
  mov    D8(%rip),%rdx
  cmp    %eax,(%rdx)
  je     L6
  cmp    $0x3b,%eax
  je     L6
  cmp    $0xa,%eax
  je     L6
  cmp    $0x3a,%eax
  cmove  %r15d,%ebp
  mov    %al,(%r14,%rbx,1)
  add    $0x1,%rbx
  jmp    L20
L6:
  movslq %r13d,%r13
  movb   $0x0,(%r14,%r13,1)
  mov    D8(%rip),%rdx
  cmp    %eax,(%rdx)
  mov    $0xffffffff,%eax
  cmove  %eax,%ebp
L1:
  mov    %ebp,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
int read_asm_line(FILE *in, char *buf) {
  int c;
  int is_label = 0;
  int li = 0;
  if (ignore_whitespaces(in) == -1)
    return -1;
  while ((c = mygetc(in)) != EOF && c != ';' && c != '\n') {
    if (c == ':')
      is_label = 1;
    buf[li] = c;
    li++;
  }
  buf[li] = '\0';
  if (c == EOF)
    return -1;
  else
    return is_label;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<read_asm_line>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  call   <ignore_whitespaces@plt>
  cmp    $0xffffffff,%eax
  je     L60
  mov    D3(%rip),%r13
  xor    %r15d,%r15d
  xor    %r14d,%r14d
  jmp    L43
  nopl   0x0(%rax)
L19:
  cmp    $0x3b,%eax
  je     L13
  cmp    $0xa,%eax
  je     L13
  cmp    $0x3a,%eax
  mov    $0x1,%edx
  mov    %al,(%r12,%r15,1)
  cmove  %edx,%r14d
  add    $0x1,%r15
L43:
  mov    %rbx,%rdi
  movslq %r15d,%rbp
  call   <mygetc@plt>
  cmp    %eax,0x0(%r13)
  jne    L19
L13:
  movb   $0x0,(%r12,%rbp,1)
  cmp    %eax,0x0(%r13)
  je     L60
L26:
  add    $0x8,%rsp
  mov    %r14d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L60:
  mov    $0xffffffff,%r14d
  jmp    L26
```

assistant: ```c
int read_asm_line(FILE *in, char *buf) {
  int c;
  int is_label = 0;
  int li = 0;
  if (ignore_whitespaces(in) == -1)
    return -1;
  while ((c = mygetc(in)) != EOF && c != ';' && c != '\n') {
    if (c == ':')
      is_label = 1;
    buf[li] = c;
    li++;
  }
  buf[li] = '\0';
  if (c == EOF)
    return -1;
  else
    return is_label;
}
```
user: Translate the below assembly code into C programming language:

```asm
<read_asm_line>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  call   <ignore_whitespaces@plt>
  cmp    $0xffffffff,%eax
  je     L52
  mov    D0(%rip),%r13
  xor    %r15d,%r15d
  xor    %r14d,%r14d
  jmp    L50
  nopl   0x0(%rax)
L35:
  cmp    $0x3b,%eax
  je     L4
  cmp    $0xa,%eax
  je     L4
  cmp    $0x3a,%eax
  mov    $0x1,%edx
  mov    %al,(%r12,%r15,1)
  cmove  %edx,%r14d
  add    $0x1,%r15
L50:
  mov    %rbx,%rdi
  movslq %r15d,%rbp
  call   <mygetc@plt>
  cmp    %eax,0x0(%r13)
  jne    L35
L4:
  movb   $0x0,(%r12,%rbp,1)
  cmp    %eax,0x0(%r13)
  je     L52
L23:
  add    $0x8,%rsp
  mov    %r14d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L52:
  mov    $0xffffffff,%r14d
  jmp    L23
```

assistant: ```c
int read_asm_line(FILE *in, char *buf) {
  int c;
  int is_label = 0;
  int li = 0;
  if (ignore_whitespaces(in) == -1)
    return -1;
  while ((c = mygetc(in)) != EOF && c != ';' && c != '\n') {
    if (c == ':')
      is_label = 1;
    buf[li] = c;
    li++;
  }
  buf[li] = '\0';
  if (c == EOF)
    return -1;
  else
    return is_label;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<peek>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <getc@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x18(%rbp),%rdx
  mov    -0x4(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <ungetc@plt>
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int peek(FILE *in) {
  int c;
  c = getc(in);
  ungetc(c, in);
  return c;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<peek>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  call   <getc@plt>
  mov    %eax,%ebx
  mov    %rbp,%rsi
  mov    %eax,%edi
  call   <ungetc@plt>
  mov    %ebx,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int peek(FILE *in) {
  int c;
  c = getc(in);
  ungetc(c, in);
  return c;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<peek>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <getc@plt>
  mov    %rbp,%rsi
  mov    %eax,%edi
  mov    %eax,%r12d
  call   <ungetc@plt>
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int peek(FILE *in) {
  int c;
  c = getc(in);
  ungetc(c, in);
  return c;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<peek>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <getc@plt>
  mov    %rbp,%rsi
  mov    %eax,%edi
  mov    %eax,%r12d
  call   <ungetc@plt>
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int peek(FILE *in) {
  int c;
  c = getc(in);
  ungetc(c, in);
  return c;
}
```
user: Translate the below assembly code into C programming language:

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

assistant: ```c
int main() {
  loopback();
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

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

assistant: ```c
int main() {
  loopback();
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

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

assistant: ```c
int main() {
  loopback();
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

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

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

```asm
<cmp_test>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %eax,-0x8(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %eax,-0x4(%rbp)
  mov    -0x8(%rbp),%eax
  cmp    -0x4(%rbp),%eax
  jge    L0
  mov    $0xffffffff,%eax
  jmp    L10
L0:
  mov    -0x8(%rbp),%eax
  cmp    -0x4(%rbp),%eax
  jle    L29
  mov    $0x1,%eax
  jmp    L10
L29:
  mov    -0x8(%rbp),%eax
  cmp    -0x4(%rbp),%eax
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  mov    $0x0,%eax
L10:
  leave
  ret
```

assistant: ```c
int cmp_test(void *left_key, void *right_key) {
  int left = (int)left_key;
  int right = (int)right_key;
  if (left < right) {
    return -1;
  } else if (left > right) {
    return 1;
  } else {
    assert(left == right);
    return 0;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<cmp_test>:
  endbr64
  sub    $0x8,%rsp
  mov    $0xffffffff,%eax
  cmp    %esi,%edi
  jl     L16
  mov    $0x1,%eax
  jle    L0
L16:
  add    $0x8,%rsp
  ret
L0:
  sete   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  mov    $0x0,%eax
  jmp    L16
```

assistant: ```c
int cmp_test(void *left_key, void *right_key) {
  int left = (int)left_key;
  int right = (int)right_key;
  if (left < right) {
    return -1;
  } else if (left > right) {
    return 1;
  } else {
    assert(left == right);
    return 0;
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<cmp_test>:
  endbr64
  sub    $0x8,%rsp
  mov    $0xffffffff,%eax
  cmp    %esi,%edi
  jl     L15
  mov    $0x1,%eax
  jle    L6
L15:
  add    $0x8,%rsp
  ret
  nopl   (%rax)
L6:
  sete   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int cmp_test(void *left_key, void *right_key) {
  int left = (int)left_key;
  int right = (int)right_key;
  if (left < right) {
    return -1;
  } else if (left > right) {
    return 1;
  } else {
    assert(left == right);
    return 0;
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<cmp_test>:
  endbr64
  sub    $0x8,%rsp
  mov    $0xffffffff,%eax
  cmp    %esi,%edi
  jl     L17
  mov    $0x1,%eax
  jle    L7
L17:
  add    $0x8,%rsp
  ret
  nopl   (%rax)
L7:
  sete   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int cmp_test(void *left_key, void *right_key) {
  int left = (int)left_key;
  int right = (int)right_key;
  if (left < right) {
    return -1;
  } else if (left > right) {
    return 1;
  } else {
    assert(left == right);
    return 0;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<yrCalc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    %rcx,-0x20(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rsi
  lea    D24(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x10(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,%esi
  lea    D29(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,%esi
  lea    D29(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x20(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,%esi
  lea    D29(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rdx
  mov    -0x10(%rbp),%rax
  mov    (%rax),%eax
  sub    $0x7d1,%eax
  imul   $0x16d,%eax,%eax
  cltq
  add    %rax,%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,(%rax)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rdx
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  sub    $0x1,%eax
  imul   $0x1e,%eax,%eax
  cltq
  add    %rax,%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,(%rax)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rdx
  mov    -0x20(%rbp),%rax
  mov    (%rax),%eax
  cltq
  add    %rax,%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,(%rax)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rax
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D24
type_guessed: string
name: retrieve_data
arguments: data_label: D29
type_guessed: string

tool: D24: 
%d totalDays in yrCalc

tool: D29: 
%d

assistant: ```c
int yrCalc(long *totalDays, int *year, int *month, int *day) {
  printf("\n%d totalDays in yrCalc", totalDays);
  printf("\n%d", *year);
  printf("\n%d", *month);
  printf("\n%d", *day);
  *totalDays += 365 * (*year - 2001);
  *totalDays += 30 * (*month - 1);
  *totalDays += *day;
  return *totalDays;
}
```
user: Translate the below assembly code into C programming language:

```asm
<yrCalc>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %rsi,%r13
  mov    %rdx,%r12
  mov    %rcx,%rbp
  mov    %rdi,%rsi
  lea    D9(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x0(%r13),%esi
  lea    D11(%rip),%r14
  mov    %r14,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    (%r12),%esi
  mov    %r14,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x0(%rbp),%esi
  mov    %r14,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x0(%r13),%eax
  sub    $0x7d1,%eax
  imul   $0x16d,%eax,%eax
  cltq
  add    (%rbx),%rax
  mov    %rax,(%rbx)
  mov    (%r12),%edx
  sub    $0x1,%edx
  imul   $0x1e,%edx,%edx
  movslq %edx,%rdx
  add    %rax,%rdx
  mov    %rdx,(%rbx)
  movslq 0x0(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,(%rbx)
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

tool_calls: name: parse_data
arguments: tag: D9
type_infered: string
name: parse_data
arguments: tag: D11
type_infered: string

tool: D9: 
%d totalDays in yrCalc

tool: D11: 
%d

assistant: ```c
int yrCalc(long *totalDays, int *year, int *month, int *day) {
  printf("\n%d totalDays in yrCalc", totalDays);
  printf("\n%d", *year);
  printf("\n%d", *month);
  printf("\n%d", *day);
  *totalDays += 365 * (*year - 2001);
  *totalDays += 30 * (*month - 1);
  *totalDays += *day;
  return *totalDays;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<yrCalc>:
  endbr64
  push   %r14
  xor    %eax,%eax
  lea    D8(%rip),%r14
  push   %r13
  mov    %rsi,%r13
  mov    %rdi,%rsi
  push   %r12
  mov    %rdi,%r12
  lea    D3(%rip),%rdi
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  mov    %rcx,%rbx
  call   <printf@plt>
  mov    0x0(%r13),%esi
  mov    %r14,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x0(%rbp),%esi
  mov    %r14,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    (%rbx),%esi
  mov    %r14,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x0(%r13),%eax
  mov    0x0(%rbp),%edx
  sub    $0x7d1,%eax
  sub    $0x1,%edx
  imul   $0x16d,%eax,%eax
  imul   $0x1e,%edx,%edx
  cltq
  add    (%r12),%rax
  movslq %edx,%rdx
  add    %rdx,%rax
  movslq (%rbx),%rdx
  pop    %rbx
  pop    %rbp
  add    %rdx,%rax
  mov    %rax,(%r12)
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D8
datatype: string
name: disassemble_data
arguments: tag: D3
datatype: string

tool: D8: 
%d

tool: D3: 
%d totalDays in yrCalc

assistant: ```c
int yrCalc(long *totalDays, int *year, int *month, int *day) {
  printf("\n%d totalDays in yrCalc", totalDays);
  printf("\n%d", *year);
  printf("\n%d", *month);
  printf("\n%d", *day);
  *totalDays += 365 * (*year - 2001);
  *totalDays += 30 * (*month - 1);
  *totalDays += *day;
  return *totalDays;
}
```
user: Translate the below assembly code into C programming language:

```asm
<yrCalc>:
  endbr64
  push   %r14
  xor    %eax,%eax
  lea    D5(%rip),%r14
  push   %r13
  mov    %rsi,%r13
  mov    %rdi,%rsi
  push   %r12
  mov    %rdi,%r12
  lea    D13(%rip),%rdi
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  mov    %rcx,%rbx
  call   <printf@plt>
  mov    0x0(%r13),%esi
  mov    %r14,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x0(%rbp),%esi
  mov    %r14,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    (%rbx),%esi
  mov    %r14,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x0(%r13),%eax
  mov    0x0(%rbp),%edx
  sub    $0x7d1,%eax
  sub    $0x1,%edx
  imul   $0x16d,%eax,%eax
  imul   $0x1e,%edx,%edx
  cltq
  add    (%r12),%rax
  movslq %edx,%rdx
  add    %rdx,%rax
  movslq (%rbx),%rdx
  pop    %rbx
  pop    %rbp
  add    %rdx,%rax
  mov    %rax,(%r12)
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D5
datatype: string
name: disassemble_data
arguments: data_label: D13
datatype: string

tool: D5: 
%d

tool: D13: 
%d totalDays in yrCalc

assistant: ```c
int yrCalc(long *totalDays, int *year, int *month, int *day) {
  printf("\n%d totalDays in yrCalc", totalDays);
  printf("\n%d", *year);
  printf("\n%d", *month);
  printf("\n%d", *day);
  *totalDays += 365 * (*year - 2001);
  *totalDays += 30 * (*month - 1);
  *totalDays += *day;
  return *totalDays;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<sqrt>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  cmpq   $0xff,-0x8(%rbp)
  jg     L7
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <sqrt8@plt>
  jmp    L31
L7:
  cmpq   $0xffff,-0x8(%rbp)
  jg     L33
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <sqrt16@plt>
  jmp    L31
L33:
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <sqrt32@plt>
L31:
  leave
  ret
```

assistant: ```c
int sqrt(long arg) {
  if (arg < 0x100) {
    return sqrt8(arg);
  } else if (arg < 0x10000) {
    return sqrt16(arg);
  } else {
    return sqrt32(arg);
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<sqrt>:
  endbr64
  sub    $0x8,%rsp
  cmp    $0xff,%rdi
  jle    L24
  cmp    $0xffff,%rdi
  jg     L19
  call   <sqrt16@plt>
L23:
  add    $0x8,%rsp
  ret
L24:
  call   <sqrt8@plt>
  jmp    L23
L19:
  call   <sqrt32@plt>
  jmp    L23
```

assistant: ```c
int sqrt(long arg) {
  if (arg < 0x100) {
    return sqrt8(arg);
  } else if (arg < 0x10000) {
    return sqrt16(arg);
  } else {
    return sqrt32(arg);
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<sqrt>:
  endbr64
  cmp    $0xff,%rdi
  jle    L15
  cmp    $0xffff,%rdi
  jg     L18
  jmp    1080 <sqrt16@plt>
  nopl   0x0(%rax,%rax,1)
L18:
  jmp    1070 <sqrt32@plt>
  nopl   (%rax)
L15:
  jmp    1090 <sqrt8@plt>
```

assistant: ```c
int sqrt(long arg) {
  if (arg < 0x100) {
    return sqrt8(arg);
  } else if (arg < 0x10000) {
    return sqrt16(arg);
  } else {
    return sqrt32(arg);
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<sqrt>:
  endbr64
  cmp    $0xff,%rdi
  jle    L0
  cmp    $0xffff,%rdi
  jg     L2
  jmp    1080 <sqrt16@plt>
  nopl   0x0(%rax,%rax,1)
L2:
  jmp    1070 <sqrt32@plt>
  nopl   (%rax)
L0:
  jmp    1090 <sqrt8@plt>
```

assistant: ```c
int sqrt(long arg) {
  if (arg < 0x100) {
    return sqrt8(arg);
  } else if (arg < 0x10000) {
    return sqrt16(arg);
  } else {
    return sqrt32(arg);
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<__memcpy2__>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    -0x18(%rbp),%rax
  sub    -0x20(%rbp),%rax
  cmp    %rax,-0x28(%rbp)
  jbe    L9
  mov    -0x18(%rbp),%rax
  cmp    -0x20(%rbp),%rax
  jae    L27
L9:
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L32
L46:
  mov    -0x8(%rbp),%rdx
  lea    0x1(%rdx),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x10(%rbp),%rax
  lea    0x1(%rax),%rcx
  mov    %rcx,-0x10(%rbp)
  movzbl (%rdx),%edx
  mov    %dl,(%rax)
L32:
  mov    -0x18(%rbp),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  cmp    %rax,-0x10(%rbp)
  jne    L46
  jmp    L51
L27:
  mov    -0x28(%rbp),%rax
  lea    -0x1(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x28(%rbp),%rax
  lea    -0x1(%rax),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x8(%rbp)
  jmp    L59
L29:
  mov    -0x8(%rbp),%rdx
  lea    -0x1(%rdx),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x10(%rbp),%rax
  lea    -0x1(%rax),%rcx
  mov    %rcx,-0x10(%rbp)
  movzbl (%rdx),%edx
  mov    %dl,(%rax)
L59:
  mov    -0x18(%rbp),%rax
  sub    $0x1,%rax
  cmp    %rax,-0x10(%rbp)
  jne    L29
L51:
  mov    -0x18(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
char *__memcpy2__(char *dst, char *src, size_t len) {
  char *p, *q;
  if (dst - src >= len || dst < src)
    for (p = dst, q = src; p != dst + len;)
      *p++ = *q++;
  else
    for (p = dst + len - 1, q = src + len - 1; p != dst - 1;)
      *p-- = *q--;
  return dst;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<__memcpy2__>:
  endbr64
  mov    %rdi,%rcx
  mov    %rdi,%rax
  sub    %rsi,%rax
  cmp    %rdx,%rax
  jae    L51
  cmp    %rsi,%rdi
  jae    L61
L51:
  lea    (%rcx,%rdx,1),%rax
  cmp    %rax,%rcx
  je     L18
  mov    $0x0,%eax
L0:
  movzbl (%rsi,%rax,1),%edi
  mov    %dil,(%rcx,%rax,1)
  add    $0x1,%rax
  cmp    %rdx,%rax
  jne    L0
L18:
  mov    %rcx,%rax
  ret
L61:
  sub    $0x1,%rdx
  lea    (%rdi,%rdx,1),%rdi
  add    %rdx,%rsi
  lea    -0x1(%rcx),%rax
  cmp    %rax,%rdi
  je     L18
  mov    %rax,%r8
  sub    %rdi,%r8
  mov    $0x0,%eax
L33:
  movzbl (%rsi,%rax,1),%edx
  mov    %dl,(%rdi,%rax,1)
  sub    $0x1,%rax
  cmp    %rax,%r8
  jne    L33
  jmp    L18
```

assistant: ```c
char *__memcpy2__(char *dst, char *src, size_t len) {
  char *p, *q;
  if (dst - src >= len || dst < src)
    for (p = dst, q = src; p != dst + len;)
      *p++ = *q++;
  else
    for (p = dst + len - 1, q = src + len - 1; p != dst - 1;)
      *p-- = *q--;
  return dst;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<__memcpy2__>:
  endbr64
  mov    %rdi,%rax
  mov    %rdi,%r8
  sub    %rsi,%rax
  cmp    %rdx,%rax
  jae    L40
  cmp    %rsi,%rdi
  jae    L12
L40:
  test   %rdx,%rdx
  je     L54
  xor    %eax,%eax
  xchg   %ax,%ax
L36:
  movzbl (%rsi,%rax,1),%ecx
  mov    %cl,(%r8,%rax,1)
  add    $0x1,%rax
  cmp    %rdx,%rax
  jne    L36
L54:
  mov    %r8,%rax
  ret
  nopl   (%rax)
L12:
  sub    $0x1,%rdx
  lea    -0x1(%r8),%rcx
  lea    (%rdi,%rdx,1),%rdi
  add    %rdx,%rsi
  cmp    %rcx,%rdi
  je     L54
  sub    %rdi,%rcx
  xor    %eax,%eax
  nopl   0x0(%rax)
L14:
  movzbl (%rsi,%rax,1),%edx
  mov    %dl,(%rdi,%rax,1)
  sub    $0x1,%rax
  cmp    %rax,%rcx
  jne    L14
  mov    %r8,%rax
  ret
```

assistant: ```c
char *__memcpy2__(char *dst, char *src, size_t len) {
  char *p, *q;
  if (dst - src >= len || dst < src)
    for (p = dst, q = src; p != dst + len;)
      *p++ = *q++;
  else
    for (p = dst + len - 1, q = src + len - 1; p != dst - 1;)
      *p-- = *q--;
  return dst;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<__memcpy2__>:
  endbr64
  mov    %rdi,%rax
  mov    %rdi,%r8
  mov    %rsi,%rcx
  sub    %rsi,%rax
  cmp    %rdx,%rax
  jae    L126
  cmp    %rsi,%rdi
  jae    L70
L126:
  lea    (%r8,%rdx,1),%r9
  cmp    %r9,%r8
  je     L59
  lea    0x1(%rcx),%rdi
  mov    %r8,%rsi
  lea    -0x1(%rdx),%rax
  sub    %rdi,%rsi
  cmp    $0xe,%rsi
  jbe    L160
  cmp    $0x6,%rax
  jbe    L160
  cmp    $0xe,%rax
  jbe    L27
  mov    %rdx,%rsi
  xor    %eax,%eax
  and    $0xfffffffffffffff0,%rsi
L182:
  movdqu (%rcx,%rax,1),%xmm0
  movups %xmm0,(%r8,%rax,1)
  add    $0x10,%rax
  cmp    %rsi,%rax
  jne    L182
  mov    %rdx,%rdi
  and    $0xfffffffffffffff0,%rdi
  lea    (%r8,%rdi,1),%rax
  lea    (%rcx,%rdi,1),%rsi
  cmp    %rdi,%rdx
  je     L59
  sub    %rdi,%rdx
  lea    -0x1(%rdx),%r10
  cmp    $0x6,%r10
  jbe    L130
L64:
  mov    (%rcx,%rdi,1),%rcx
  mov    %rcx,(%r8,%rdi,1)
  mov    %rdx,%rcx
  and    $0xfffffffffffffff8,%rcx
  add    %rcx,%rax
  add    %rcx,%rsi
  cmp    %rcx,%rdx
  je     L59
L130:
  movzbl (%rsi),%edx
  mov    %dl,(%rax)
  lea    0x1(%rax),%rdx
  cmp    %rdx,%r9
  je     L59
  movzbl 0x1(%rsi),%edx
  mov    %dl,0x1(%rax)
  lea    0x2(%rax),%rdx
  cmp    %rdx,%r9
  je     L59
  movzbl 0x2(%rsi),%edx
  mov    %dl,0x2(%rax)
  lea    0x3(%rax),%rdx
  cmp    %rdx,%r9
  je     L59
  movzbl 0x3(%rsi),%edx
  mov    %dl,0x3(%rax)
  lea    0x4(%rax),%rdx
  cmp    %rdx,%r9
  je     L59
  movzbl 0x4(%rsi),%edx
  mov    %dl,0x4(%rax)
  lea    0x5(%rax),%rdx
  cmp    %rdx,%r9
  je     L59
  movzbl 0x5(%rsi),%edx
  mov    %dl,0x5(%rax)
  lea    0x6(%rax),%rdx
  cmp    %rdx,%r9
  je     L59
  movzbl 0x6(%rsi),%edx
  mov    %dl,0x6(%rax)
L59:
  mov    %r8,%rax
  ret
  nopl   0x0(%rax,%rax,1)
L70:
  sub    $0x1,%rdx
  lea    -0x1(%r8),%rsi
  lea    (%rdi,%rdx,1),%rdi
  add    %rdx,%rcx
  cmp    %rsi,%rdi
  je     L59
  sub    %rdi,%rsi
  xor    %eax,%eax
  nopl   0x0(%rax)
L158:
  movzbl (%rcx,%rax,1),%edx
  mov    %dl,(%rdi,%rax,1)
  sub    $0x1,%rax
  cmp    %rsi,%rax
  jne    L158
  mov    %r8,%rax
  ret
L27:
  mov    %rcx,%rsi
  mov    %r8,%rax
  xor    %edi,%edi
  jmp    L64
  nopl   0x0(%rax)
L160:
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L81:
  movzbl (%rcx,%rax,1),%esi
  mov    %sil,(%r8,%rax,1)
  add    $0x1,%rax
  cmp    %rax,%rdx
  jne    L81
  mov    %r8,%rax
  ret
```

assistant: ```c
char *__memcpy2__(char *dst, char *src, size_t len) {
  char *p, *q;
  if (dst - src >= len || dst < src)
    for (p = dst, q = src; p != dst + len;)
      *p++ = *q++;
  else
    for (p = dst + len - 1, q = src + len - 1; p != dst - 1;)
      *p-- = *q--;
  return dst;
}
```