ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请从给定的汇编指令推导出C源代码：

```asm
<cmp>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  call   <strlen@plt>
  mov    %r12,%rdi
  mov    %eax,%ebx
  call   <strlen@plt>
  cmp    %eax,%ebx
  jl     L50
  jg     L4
  lea    0x1(%rbx),%ecx
  mov    $0x1,%edx
  test   %ebx,%ebx
  jg     L41
  jmp    L3
  nopl   0x0(%rax)
L2:
  add    $0x1,%rdx
  cmp    %rcx,%rdx
  je     L3
L41:
  movzbl -0x1(%r12,%rdx,1),%esi
  mov    %edx,%eax
  cmp    %sil,-0x1(%rbp,%rdx,1)
  je     L2
  jmp    L13
  nopw   0x0(%rax,%rax,1)
L4:
  add    $0x1,%eax
L13:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L50:
  lea    0x1(%rbx),%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L3:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int cmp(char *a, char *b) {
  int la = strlen(a), lb = strlen(b);
  int i;
  if (la < lb)
    return la + 1;
  else if (lb < la)
    return lb + 1;
  for (i = 0; i < la; i++)
    if (a[i] != b[i])
      return i + 1;
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<disorder>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L6
L5:
  addq   $0x1,-0x8(%rbp)
L6:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0xa,%al
  jne    L5
  mov    -0x8(%rbp),%rax
  movb   $0x0,(%rax)
  mov    -0x20(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <diag@plt>
  mov    $0x0,%edi
  call   <term@plt>
  nop
  leave
  ret
```

assistant: ```c
void disorder(char *s, char *t) {
  char *u;
  for (u = t; *u != '\n'; u++)
    ;
  *u = 0;
  diag(s, t);
  term(0);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<disorder>:
  endbr64
  sub    $0x8,%rsp
  cmpb   $0xa,(%rsi)
  je     L13
  mov    %rsi,%rax
L0:
  add    $0x1,%rax
  cmpb   $0xa,(%rax)
  jne    L0
L16:
  movb   $0x0,(%rax)
  call   <diag@plt>
  mov    $0x0,%edi
  call   <term@plt>
  add    $0x8,%rsp
  ret
L13:
  mov    %rsi,%rax
  jmp    L16
```

assistant: ```c
void disorder(char *s, char *t) {
  char *u;
  for (u = t; *u != '\n'; u++)
    ;
  *u = 0;
  diag(s, t);
  term(0);
}
```
user: Translate the below assembly code into C programming language:

```asm
<disorder>:
  endbr64
  sub    $0x8,%rsp
  cmpb   $0xa,(%rsi)
  mov    %rsi,%rax
  je     L9
L10:
  add    $0x1,%rax
  cmpb   $0xa,(%rax)
  jne    L10
L9:
  movb   $0x0,(%rax)
  call   <diag@plt>
  xor    %edi,%edi
  add    $0x8,%rsp
  jmp    1070 <term@plt>
```

assistant: ```c
void disorder(char *s, char *t) {
  char *u;
  for (u = t; *u != '\n'; u++)
    ;
  *u = 0;
  diag(s, t);
  term(0);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<disorder>:
  endbr64
  sub    $0x8,%rsp
  cmpb   $0xa,(%rsi)
  mov    %rsi,%rax
  je     L11
L16:
  add    $0x1,%rax
  cmpb   $0xa,(%rax)
  jne    L16
L11:
  movb   $0x0,(%rax)
  call   <diag@plt>
  xor    %edi,%edi
  add    $0x8,%rsp
  jmp    1070 <term@plt>
```

assistant: ```c
void disorder(char *s, char *t) {
  char *u;
  for (u = t; *u != '\n'; u++)
    ;
  *u = 0;
  diag(s, t);
  term(0);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<_genieWriteInhLabelFloat>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  movss  %xmm0,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x10(%rbp),%rcx
  mov    -0x1c(%rbp),%edx
  mov    -0x18(%rbp),%eax
  mov    %rcx,%rsi
  mov    %edx,%edi
  movd   %eax,%xmm0
  call   <gcvt@plt>
  lea    -0x10(%rbp),%rdx
  mov    -0x14(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <_genieWriteInhLabel@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L4
  call   <__stack_chk_fail@plt>
L4:
  leave
  ret
```

assistant: ```c
int _genieWriteInhLabelFloat(int index, float n, int precision) {
  char str[sizeof(long)];
  gcvt(n, precision, str);
  _genieWriteInhLabel(index, str);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<_genieWriteInhLabelFloat>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %edi,%ebx
  mov    %esi,%edi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%rsi
  call   <gcvt@plt>
  mov    %rbp,%rsi
  mov    %ebx,%edi
  call   <_genieWriteInhLabel@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int _genieWriteInhLabelFloat(int index, float n, int precision) {
  char str[sizeof(long)];
  gcvt(n, precision, str);
  _genieWriteInhLabel(index, str);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<_genieWriteInhLabelFloat>:
  endbr64
  push   %r12
  push   %rbp
  mov    %edi,%ebp
  mov    %esi,%edi
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  mov    %r12,%rsi
  call   <gcvt@plt>
  mov    %r12,%rsi
  mov    %ebp,%edi
  call   <_genieWriteInhLabel@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int _genieWriteInhLabelFloat(int index, float n, int precision) {
  char str[sizeof(long)];
  gcvt(n, precision, str);
  _genieWriteInhLabel(index, str);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<_genieWriteInhLabelFloat>:
  endbr64
  push   %r12
  push   %rbp
  mov    %edi,%ebp
  mov    %esi,%edi
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  mov    %r12,%rsi
  call   <gcvt@plt>
  mov    %r12,%rsi
  mov    %ebp,%edi
  call   <_genieWriteInhLabel@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int _genieWriteInhLabelFloat(int index, float n, int precision) {
  char str[sizeof(long)];
  gcvt(n, precision, str);
  _genieWriteInhLabel(index, str);
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<lookup_normal>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    %ecx,-0x30(%rbp)
  mov    %r8,-0x40(%rbp)
  movl   $0xffffffff,-0x18(%rbp)
  mov    D4(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,-0x14(%rbp)
  movl   $0x0,-0x10(%rbp)
  jmp    L13
L34:
  movl   $0x0,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L16
L47:
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movzbl %al,%edx
  mov    -0x10(%rbp),%eax
  imul   -0x2c(%rbp),%eax
  mov    %eax,%ecx
  mov    -0xc(%rbp),%eax
  add    %ecx,%eax
  movslq %eax,%rcx
  mov    -0x38(%rbp),%rax
  add    %rcx,%rax
  movzbl (%rax),%eax
  movzbl %al,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  imul   %eax,%eax
  add    %eax,-0x8(%rbp)
  addl   $0x1,-0xc(%rbp)
L16:
  mov    -0xc(%rbp),%eax
  cmp    -0x2c(%rbp),%eax
  jl     L47
  mov    -0x8(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jge    L11
  mov    -0x8(%rbp),%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x10(%rbp),%eax
  mov    %eax,-0x18(%rbp)
L11:
  addl   $0x1,-0x10(%rbp)
L13:
  mov    -0x10(%rbp),%eax
  cmp    -0x30(%rbp),%eax
  jl     L34
  mov    -0x18(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int lookup_normal(unsigned char const *const pixel, int const depth, unsigned char const *const palette, int const ncolor, unsigned short *const cachetable) {
  int index;
  int diff;
  int r;
  int i;
  int n;
  int distant;
  index = -1;
  diff = INT_MAX;
  for (i = 0; i < ncolor; i++) {
    distant = 0;
    for (n = 0; n < depth; ++n) {
      r = pixel[n] - palette[i * depth + n];
      distant += r * r;
    }
    if (distant < diff) {
      diff = distant;
      index = i;
    }
  }
  return index;
}
```
user: Translate the below assembly code into C programming language:

```asm
<lookup_normal>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    D6(%rip),%rax
  mov    (%rax),%ebx
  test   %ecx,%ecx
  jle    L42
  mov    %esi,%r11d
  mov    %rdx,%r12
  mov    %ecx,%ebp
  mov    $0x0,%r10d
  mov    $0x0,%r9d
  mov    $0xffffffff,%r13d
  mov    %esi,%r8d
  jmp    L61
L55:
  add    $0x1,%r9d
  add    %r11d,%r10d
  cmp    %r9d,%ebp
  je     L44
L61:
  mov    $0x0,%ecx
  test   %r11d,%r11d
  jle    L5
  mov    $0x0,%edx
  mov    $0x0,%ecx
  movslq %r10d,%rsi
  add    %r12,%rsi
L11:
  movzbl (%rdi,%rdx,1),%eax
  movzbl (%rsi,%rdx,1),%r14d
  sub    %r14d,%eax
  imul   %eax,%eax
  add    %eax,%ecx
  add    $0x1,%rdx
  cmp    %r8,%rdx
  jne    L11
L5:
  cmp    %ecx,%ebx
  jle    L55
  mov    %ecx,%ebx
  mov    %r9d,%r13d
  jmp    L55
L42:
  mov    $0xffffffff,%r13d
L44:
  mov    %r13d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int lookup_normal(unsigned char const *const pixel, int const depth, unsigned char const *const palette, int const ncolor, unsigned short *const cachetable) {
  int index;
  int diff;
  int r;
  int i;
  int n;
  int distant;
  index = -1;
  diff = INT_MAX;
  for (i = 0; i < ncolor; i++) {
    distant = 0;
    for (n = 0; n < depth; ++n) {
      r = pixel[n] - palette[i * depth + n];
      distant += r * r;
    }
    if (distant < diff) {
      diff = distant;
      index = i;
    }
  }
  return index;
}
```
user: Translate the below assembly code into C programming language:

```asm
<lookup_normal>:
  endbr64
  mov    D5(%rip),%rax
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    (%rax),%ebx
  test   %ecx,%ecx
  jle    L46
  mov    %esi,%r11d
  mov    %rdx,%r12
  mov    %ecx,%ebp
  xor    %r10d,%r10d
  xor    %r9d,%r9d
  mov    $0xffffffff,%r13d
  lea    -0x1(%rsi),%r8d
  nopl   0x0(%rax)
L2:
  xor    %ecx,%ecx
  test   %r11d,%r11d
  jle    L26
  movslq %r10d,%rsi
  xor    %edx,%edx
  xor    %ecx,%ecx
  add    %r12,%rsi
  nopl   0x0(%rax)
L30:
  movzbl (%rdi,%rdx,1),%eax
  movzbl (%rsi,%rdx,1),%r14d
  sub    %r14d,%eax
  imul   %eax,%eax
  add    %eax,%ecx
  mov    %rdx,%rax
  add    $0x1,%rdx
  cmp    %rax,%r8
  jne    L30
L26:
  cmp    %ecx,%ebx
  jle    L49
  mov    %ecx,%ebx
  mov    %r9d,%r13d
L49:
  add    $0x1,%r9d
  add    %r11d,%r10d
  cmp    %r9d,%ebp
  jne    L2
  pop    %rbx
  mov    %r13d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L46:
  mov    $0xffffffff,%r13d
  pop    %rbx
  pop    %rbp
  mov    %r13d,%eax
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int lookup_normal(unsigned char const *const pixel, int const depth, unsigned char const *const palette, int const ncolor, unsigned short *const cachetable) {
  int index;
  int diff;
  int r;
  int i;
  int n;
  int distant;
  index = -1;
  diff = INT_MAX;
  for (i = 0; i < ncolor; i++) {
    distant = 0;
    for (n = 0; n < depth; ++n) {
      r = pixel[n] - palette[i * depth + n];
      distant += r * r;
    }
    if (distant < diff) {
      diff = distant;
      index = i;
    }
  }
  return index;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<RandomFunc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x18(%rbp),%rax
  movzwl (%rax),%eax
  add    $0xb45,%ax
  mov    %ax,-0xa(%rbp)
  movw   $0x0,-0xc(%rbp)
  nop
  movzwl -0xa(%rbp),%eax
  lea    0x1838(%rax),%edx
  mov    -0x20(%rbp),%rax
  mov    %dx,(%rax)
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L3
  call   <__stack_chk_fail@plt>
L3:
  leave
  ret
```

assistant: ```c
void RandomFunc(unsigned short input[1], unsigned short output[1]) {
  unsigned short state[1];
  unsigned short local1;
  {
    state[0UL] = input[0UL] + (unsigned short)2885;
    local1 = 0UL;
    while (local1 < (unsigned short)0) {
      if (state[0UL] > local1) {
      }
      if (state[0UL] == local1) {
      }
      local1 += 2UL;
    }
    output[0UL] = state[0UL] + 316086328UL;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<RandomFunc>:
  endbr64
  movzwl (%rdi),%eax
  add    $0x237d,%ax
  mov    %ax,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned short input[1], unsigned short output[1]) {
  unsigned short state[1];
  unsigned short local1;
  {
    state[0UL] = input[0UL] + (unsigned short)2885;
    local1 = 0UL;
    while (local1 < (unsigned short)0) {
      if (state[0UL] > local1) {
      }
      if (state[0UL] == local1) {
      }
      local1 += 2UL;
    }
    output[0UL] = state[0UL] + 316086328UL;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<RandomFunc>:
  endbr64
  movzwl (%rdi),%eax
  add    $0x237d,%ax
  mov    %ax,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned short input[1], unsigned short output[1]) {
  unsigned short state[1];
  unsigned short local1;
  {
    state[0UL] = input[0UL] + (unsigned short)2885;
    local1 = 0UL;
    while (local1 < (unsigned short)0) {
      if (state[0UL] > local1) {
      }
      if (state[0UL] == local1) {
      }
      local1 += 2UL;
    }
    output[0UL] = state[0UL] + 316086328UL;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<getpasswd>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,%eax
  mov    %al,-0x24(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <getpass@plt>
  mov    %rax,-0x8(%rbp)
  movzbl -0x24(%rbp),%eax
  sub    $0x1,%eax
  movzbl %al,%edx
  mov    -0x8(%rbp),%rcx
  mov    -0x20(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strncpy@plt>
  movzbl -0x24(%rbp),%eax
  lea    -0x1(%rax),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <validated_or_get_len@plt>
  leave
  ret
```

assistant: ```c
int getpasswd(const char *prompt, char *password, unsigned char psize) {
  char *inp_pass;
  inp_pass = getpass(prompt);
  strncpy(password, inp_pass, psize - 1);
  password[psize - 1] = '\0';
  return validated_or_get_len(password);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<getpasswd>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rsi,%rbp
  mov    %edx,%ebx
  call   <getpass@plt>
  mov    %rax,%rsi
  lea    -0x1(%rbx),%edx
  movzbl %dl,%edx
  mov    %rbp,%rdi
  call   <strncpy@plt>
  movzbl %bl,%ebx
  movb   $0x0,-0x1(%rbp,%rbx,1)
  mov    %rbp,%rdi
  call   <validated_or_get_len@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int getpasswd(const char *prompt, char *password, unsigned char psize) {
  char *inp_pass;
  inp_pass = getpass(prompt);
  strncpy(password, inp_pass, psize - 1);
  password[psize - 1] = '\0';
  return validated_or_get_len(password);
}
```
user: What is the c source code of the assembly code below:

```asm
<getpasswd>:
  endbr64
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %edx,%ebx
  sub    $0x8,%rsp
  call   <getpass@plt>
  lea    -0x1(%rbx),%edx
  mov    %rbp,%rdi
  movzbl %bl,%ebx
  mov    %rax,%rsi
  movzbl %dl,%edx
  call   <strncpy@plt>
  movb   $0x0,-0x1(%rbp,%rbx,1)
  add    $0x8,%rsp
  mov    %rbp,%rdi
  pop    %rbx
  pop    %rbp
  jmp    1080 <validated_or_get_len@plt>
```

assistant: ```c
int getpasswd(const char *prompt, char *password, unsigned char psize) {
  char *inp_pass;
  inp_pass = getpass(prompt);
  strncpy(password, inp_pass, psize - 1);
  password[psize - 1] = '\0';
  return validated_or_get_len(password);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<getpasswd>:
  endbr64
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %edx,%ebx
  sub    $0x8,%rsp
  call   <getpass@plt>
  lea    -0x1(%rbx),%edx
  mov    %rbp,%rdi
  movzbl %bl,%ebx
  mov    %rax,%rsi
  movzbl %dl,%edx
  call   <strncpy@plt>
  movb   $0x0,-0x1(%rbp,%rbx,1)
  add    $0x8,%rsp
  mov    %rbp,%rdi
  pop    %rbx
  pop    %rbp
  jmp    1080 <validated_or_get_len@plt>
```

assistant: ```c
int getpasswd(const char *prompt, char *password, unsigned char psize) {
  char *inp_pass;
  inp_pass = getpass(prompt);
  strncpy(password, inp_pass, psize - 1);
  password[psize - 1] = '\0';
  return validated_or_get_len(password);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<uniquePaths>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x24(%rbp)
  mov    %esi,-0x28(%rbp)
  movq   $0x1,-0x10(%rbp)
  movq   $0x1,-0x8(%rbp)
  mov    -0x28(%rbp),%edx
  mov    -0x24(%rbp),%eax
  cmp    %eax,%edx
  cmovle %edx,%eax
  mov    %eax,-0x14(%rbp)
  subl   $0x1,-0x14(%rbp)
  movl   $0x1,-0x18(%rbp)
  jmp    L19
L13:
  mov    -0x28(%rbp),%edx
  mov    -0x24(%rbp),%eax
  add    %edx,%eax
  sub    $0x1,%eax
  sub    -0x18(%rbp),%eax
  cltq
  mov    -0x10(%rbp),%rdx
  imul   %rdx,%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x18(%rbp),%eax
  cltq
  mov    -0x8(%rbp),%rdx
  imul   %rdx,%rax
  mov    %rax,-0x8(%rbp)
  addl   $0x1,-0x18(%rbp)
L19:
  mov    -0x18(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jle    L13
  mov    -0x10(%rbp),%rax
  cqto
  idivq  -0x8(%rbp)
  pop    %rbp
  ret
```

assistant: ```c
int uniquePaths(int m, int n) {
  long p1 = 1, p2 = 1;
  int min = m < n ? m : n;
  min = min - 1;
  for (int i = 1; i <= min; i++) {
    p1 *= (n + m - 1 - i);
    p2 *= i;
  }
  return p1 / p2;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<uniquePaths>:
  endbr64
  cmp    %edi,%esi
  mov    %edi,%r8d
  cmovle %esi,%r8d
  cmp    $0x1,%r8d
  jle    L5
  mov    %r8d,%r8d
  mov    $0x1,%edx
  mov    $0x1,%ecx
  mov    $0x1,%eax
  movslq %edi,%rdi
  movslq %esi,%rsi
  lea    -0x1(%rdi,%rsi,1),%rdi
L12:
  mov    %rdi,%rsi
  sub    %rdx,%rsi
  imul   %rsi,%rax
  imul   %rdx,%rcx
  add    $0x1,%rdx
  cmp    %r8,%rdx
  jne    L12
L15:
  cqto
  idiv   %rcx
  ret
L5:
  mov    $0x1,%ecx
  mov    $0x1,%eax
  jmp    L15
```

assistant: ```c
int uniquePaths(int m, int n) {
  long p1 = 1, p2 = 1;
  int min = m < n ? m : n;
  min = min - 1;
  for (int i = 1; i <= min; i++) {
    p1 *= (n + m - 1 - i);
    p2 *= i;
  }
  return p1 / p2;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<uniquePaths>:
  endbr64
  cmp    %edi,%esi
  mov    %edi,%r8d
  cmovle %esi,%r8d
  cmp    $0x1,%r8d
  jle    L1
  add    %edi,%esi
  movslq %r8d,%r8
  mov    $0x1,%edx
  mov    $0x1,%ecx
  movslq %esi,%rsi
  mov    $0x1,%eax
  sub    $0x1,%rsi
  xchg   %ax,%ax
L0:
  mov    %rsi,%rdi
  imul   %rdx,%rcx
  sub    %rdx,%rdi
  add    $0x1,%rdx
  imul   %rdi,%rax
  cmp    %rdx,%r8
  jne    L0
  cqto
  idiv   %rcx
  ret
  nopl   (%rax)
L1:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int uniquePaths(int m, int n) {
  long p1 = 1, p2 = 1;
  int min = m < n ? m : n;
  min = min - 1;
  for (int i = 1; i <= min; i++) {
    p1 *= (n + m - 1 - i);
    p2 *= i;
  }
  return p1 / p2;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<uniquePaths>:
  endbr64
  cmp    %edi,%esi
  mov    %edi,%r8d
  cmovle %esi,%r8d
  cmp    $0x1,%r8d
  jle    L8
  add    %edi,%esi
  movslq %r8d,%r8
  mov    $0x1,%edx
  mov    $0x1,%ecx
  movslq %esi,%rsi
  mov    $0x1,%eax
  sub    $0x1,%rsi
  xchg   %ax,%ax
L7:
  mov    %rsi,%rdi
  imul   %rdx,%rcx
  sub    %rdx,%rdi
  add    $0x1,%rdx
  imul   %rdi,%rax
  cmp    %rdx,%r8
  jne    L7
  cqto
  idiv   %rcx
  ret
  nopl   (%rax)
L8:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int uniquePaths(int m, int n) {
  long p1 = 1, p2 = 1;
  int min = m < n ? m : n;
  min = min - 1;
  for (int i = 1; i <= min; i++) {
    p1 *= (n + m - 1 - i);
    p2 *= i;
  }
  return p1 / p2;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<hermint>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movsd  %xmm0,-0x28(%rbp)
  mov    %edi,-0x2c(%rbp)
  mov    %rsi,-0x38(%rbp)
  mov    %rdx,-0x40(%rbp)
  mov    %rcx,-0x48(%rbp)
  mov    %r8,-0x50(%rbp)
  mov    %r9,-0x58(%rbp)
  movl   $0x0,-0x1c(%rbp)
  mov    -0x2c(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x18(%rbp)
  jmp    L31
L23:
  mov    -0x18(%rbp),%edx
  mov    -0x1c(%rbp),%eax
  add    %edx,%eax
  sar    %eax
  mov    %eax,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  comisd -0x28(%rbp),%xmm0
  jbe    L14
  mov    -0x14(%rbp),%eax
  mov    %eax,-0x18(%rbp)
  jmp    L31
L14:
  mov    -0x14(%rbp),%eax
  mov    %eax,-0x1c(%rbp)
L31:
  mov    -0x18(%rbp),%eax
  sub    -0x1c(%rbp),%eax
  cmp    $0x1,%eax
  jg     L23
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm1
  movsd  -0x28(%rbp),%xmm0
  subsd  %xmm1,%xmm0
  movsd  %xmm0,-0x10(%rbp)
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x40(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm1
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x48(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm2
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x50(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm3
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x58(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  mulsd  -0x10(%rbp),%xmm0
  addsd  %xmm3,%xmm0
  mulsd  -0x10(%rbp),%xmm0
  addsd  %xmm2,%xmm0
  mulsd  -0x10(%rbp),%xmm0
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movsd  -0x8(%rbp),%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  pop    %rbp
  ret
```

assistant: ```c
double hermint(double xbar, int n, double *x, double *yc1, double *yc2, double *yc3, double *yc4) {
  int klo, khi, k;
  double dx;
  double result;
  klo = 0;
  khi = n - 1;
  while (khi - klo > 1) {
    k = (khi + klo) >> 1;
    if (x[k] > xbar)
      khi = k;
    else
      klo = k;
  }
  dx = xbar - x[klo];
  result = yc1[klo] + dx * (yc2[klo] + dx * (yc3[klo] + dx * yc4[klo]));
  return result;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<hermint>:
  endbr64
  push   %rbx
  mov    %rsi,%r10
  mov    %rdx,%r11
  mov    %rcx,%rbx
  lea    -0x1(%rdi),%ecx
  mov    $0x0,%edx
  jmp    L19
L11:
  lea    (%rdx,%rcx,1),%eax
  sar    %eax
  movslq %eax,%rsi
  movsd  (%r10,%rsi,8),%xmm1
  comisd %xmm0,%xmm1
  cmovbe %eax,%edx
  cmova  %eax,%ecx
L19:
  mov    %ecx,%eax
  sub    %edx,%eax
  cmp    $0x1,%eax
  jg     L11
  movslq %edx,%rdx
  subsd  (%r10,%rdx,8),%xmm0
  movapd %xmm0,%xmm1
  mulsd  (%r9,%rdx,8),%xmm1
  addsd  (%r8,%rdx,8),%xmm1
  mulsd  %xmm0,%xmm1
  addsd  (%rbx,%rdx,8),%xmm1
  mulsd  %xmm1,%xmm0
  addsd  (%r11,%rdx,8),%xmm0
  pop    %rbx
  ret
```

assistant: ```c
double hermint(double xbar, int n, double *x, double *yc1, double *yc2, double *yc3, double *yc4) {
  int klo, khi, k;
  double dx;
  double result;
  klo = 0;
  khi = n - 1;
  while (khi - klo > 1) {
    k = (khi + klo) >> 1;
    if (x[k] > xbar)
      khi = k;
    else
      klo = k;
  }
  dx = xbar - x[klo];
  result = yc1[klo] + dx * (yc2[klo] + dx * (yc3[klo] + dx * yc4[klo]));
  return result;
}
```