ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Translate the below assembly code into C programming language:

```asm
<reverse_a_string>:
  endbr64
  cmpb   $0x0,(%rdi)
  je     L11
  mov    $0x1,%eax
L1:
  lea    -0x1(%rax),%ecx
  mov    %eax,%edx
  mov    %rax,%r9
  add    $0x1,%rax
  cmpb   $0x0,-0x1(%rdi,%rax,1)
  jne    L1
  test   %edx,%edx
  jle    L9
  movslq %ecx,%rax
  add    %rax,%rdi
  mov    %rsi,%rdx
  mov    %rdi,%rax
  sub    $0x1,%edi
L4:
  movzbl (%rax),%r8d
  mov    %r8b,(%rdx)
  sub    $0x1,%rax
  add    $0x1,%rdx
  mov    %edi,%r8d
  sub    %eax,%r8d
  cmp    %ecx,%r8d
  jl     L4
L9:
  movb   $0x0,(%rsi,%r9,1)
  ret
L11:
  mov    $0x0,%r9d
  jmp    L9
```

assistant: ```c
void reverse_a_string(char *src, char *dst) {
  int end;
  int i = 0;
  while (src[i] != '\0') {
    i++;
  }
  end = i;
  i = 0;
  while (i < end) {
    dst[i] = src[end - 1 - i];
    i++;
  }
  dst[end] = '\0';
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<reverse_a_string>:
  endbr64
  cmpb   $0x0,(%rdi)
  je     L12
  mov    $0x1,%eax
  xchg   %ax,%ax
L22:
  mov    %rax,%rdx
  lea    -0x1(%rax),%ecx
  lea    0x1(%rax),%rax
  cmpb   $0x0,(%rdi,%rdx,1)
  jne    L22
  movslq %ecx,%rax
  mov    %ecx,%ecx
  add    %rax,%rdi
  lea    0x1(%rsi,%rcx,1),%r8
  mov    %rsi,%rax
L23:
  movzbl (%rdi),%ecx
  add    $0x1,%rax
  sub    $0x1,%rdi
  mov    %cl,-0x1(%rax)
  cmp    %r8,%rax
  jne    L23
  add    %rdx,%rsi
L12:
  movb   $0x0,(%rsi)
  ret
```

assistant: ```c
void reverse_a_string(char *src, char *dst) {
  int end;
  int i = 0;
  while (src[i] != '\0') {
    i++;
  }
  end = i;
  i = 0;
  while (i < end) {
    dst[i] = src[end - 1 - i];
    i++;
  }
  dst[end] = '\0';
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<reverse_a_string>:
  endbr64
  cmpb   $0x0,(%rdi)
  je     L7
  mov    $0x1,%eax
  xchg   %ax,%ax
L3:
  mov    %rax,%rdx
  lea    -0x1(%rax),%ecx
  lea    0x1(%rax),%rax
  cmpb   $0x0,(%rdi,%rdx,1)
  jne    L3
  movslq %ecx,%rax
  mov    %ecx,%ecx
  add    %rax,%rdi
  lea    0x1(%rsi,%rcx,1),%r8
  mov    %rsi,%rax
L28:
  movzbl (%rdi),%ecx
  add    $0x1,%rax
  sub    $0x1,%rdi
  mov    %cl,-0x1(%rax)
  cmp    %rax,%r8
  jne    L28
  add    %rdx,%rsi
L7:
  movb   $0x0,(%rsi)
  ret
```

assistant: ```c
void reverse_a_string(char *src, char *dst) {
  int end;
  int i = 0;
  while (src[i] != '\0') {
    i++;
  }
  end = i;
  i = 0;
  while (i < end) {
    dst[i] = src[end - 1 - i];
    i++;
  }
  dst[end] = '\0';
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<ripemd160_block>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  mov    0x80(%rax),%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x8(%rbp),%rax
  lea    0x4(%rax),%rdx
  mov    -0x10(%rbp),%rax
  mov    0x7c(%rax),%eax
  mov    %eax,(%rdx)
  mov    -0x8(%rbp),%rax
  lea    0x8(%rax),%rdx
  mov    -0x10(%rbp),%rax
  mov    0x78(%rax),%eax
  mov    %eax,(%rdx)
  mov    -0x8(%rbp),%rax
  lea    0xc(%rax),%rdx
  mov    -0x10(%rbp),%rax
  mov    0x74(%rax),%eax
  mov    %eax,(%rdx)
  mov    -0x8(%rbp),%rax
  lea    0x10(%rax),%rdx
  mov    -0x10(%rbp),%rax
  mov    0x70(%rax),%eax
  mov    %eax,(%rdx)
  mov    -0x8(%rbp),%rax
  lea    0x14(%rax),%rdx
  mov    -0x10(%rbp),%rax
  mov    0x6c(%rax),%eax
  mov    %eax,(%rdx)
  mov    -0x8(%rbp),%rax
  lea    0x18(%rax),%rdx
  mov    -0x10(%rbp),%rax
  mov    0x68(%rax),%eax
  mov    %eax,(%rdx)
  mov    -0x8(%rbp),%rax
  lea    0x1c(%rax),%rdx
  mov    -0x10(%rbp),%rax
  mov    0x64(%rax),%eax
  mov    %eax,(%rdx)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void ripemd160_block(unsigned int *out, unsigned int *in) {
  out[0] = in[32 - 0];
  out[1] = in[32 - 1];
  out[2] = in[32 - 2];
  out[3] = in[32 - 3];
  out[4] = in[32 - 4];
  out[5] = in[32 - 5];
  out[6] = in[32 - 6];
  out[7] = in[32 - 7];
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<ripemd160_block>:
  endbr64
  mov    0x80(%rsi),%eax
  mov    %eax,(%rdi)
  mov    0x7c(%rsi),%eax
  mov    %eax,0x4(%rdi)
  mov    0x78(%rsi),%eax
  mov    %eax,0x8(%rdi)
  mov    0x74(%rsi),%eax
  mov    %eax,0xc(%rdi)
  mov    0x70(%rsi),%eax
  mov    %eax,0x10(%rdi)
  mov    0x6c(%rsi),%eax
  mov    %eax,0x14(%rdi)
  mov    0x68(%rsi),%eax
  mov    %eax,0x18(%rdi)
  mov    0x64(%rsi),%eax
  mov    %eax,0x1c(%rdi)
  ret
```

assistant: ```c
void ripemd160_block(unsigned int *out, unsigned int *in) {
  out[0] = in[32 - 0];
  out[1] = in[32 - 1];
  out[2] = in[32 - 2];
  out[3] = in[32 - 3];
  out[4] = in[32 - 4];
  out[5] = in[32 - 5];
  out[6] = in[32 - 6];
  out[7] = in[32 - 7];
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<ripemd160_block>:
  endbr64
  mov    0x80(%rsi),%eax
  mov    %eax,(%rdi)
  mov    0x7c(%rsi),%eax
  mov    %eax,0x4(%rdi)
  mov    0x78(%rsi),%eax
  mov    %eax,0x8(%rdi)
  mov    0x74(%rsi),%eax
  mov    %eax,0xc(%rdi)
  mov    0x70(%rsi),%eax
  mov    %eax,0x10(%rdi)
  mov    0x6c(%rsi),%eax
  mov    %eax,0x14(%rdi)
  mov    0x68(%rsi),%eax
  mov    %eax,0x18(%rdi)
  mov    0x64(%rsi),%eax
  mov    %eax,0x1c(%rdi)
  ret
```

assistant: ```c
void ripemd160_block(unsigned int *out, unsigned int *in) {
  out[0] = in[32 - 0];
  out[1] = in[32 - 1];
  out[2] = in[32 - 2];
  out[3] = in[32 - 3];
  out[4] = in[32 - 4];
  out[5] = in[32 - 5];
  out[6] = in[32 - 6];
  out[7] = in[32 - 7];
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ripemd160_block>:
  endbr64
  mov    0x80(%rsi),%eax
  mov    %eax,(%rdi)
  mov    0x7c(%rsi),%eax
  mov    %eax,0x4(%rdi)
  mov    0x78(%rsi),%eax
  mov    %eax,0x8(%rdi)
  mov    0x74(%rsi),%eax
  mov    %eax,0xc(%rdi)
  mov    0x70(%rsi),%eax
  mov    %eax,0x10(%rdi)
  mov    0x6c(%rsi),%eax
  mov    %eax,0x14(%rdi)
  mov    0x68(%rsi),%eax
  mov    %eax,0x18(%rdi)
  mov    0x64(%rsi),%eax
  mov    %eax,0x1c(%rdi)
  ret
```

assistant: ```c
void ripemd160_block(unsigned int *out, unsigned int *in) {
  out[0] = in[32 - 0];
  out[1] = in[32 - 1];
  out[2] = in[32 - 2];
  out[3] = in[32 - 3];
  out[4] = in[32 - 4];
  out[5] = in[32 - 5];
  out[6] = in[32 - 6];
  out[7] = in[32 - 7];
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<pathSep>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movq   $0x0,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L44
L1:
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rax
  lea    0x1(%rax),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,(%rax)
L44:
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rax
  movzbl (%rax),%eax
  cmp    $0x2f,%al
  je     L1
  jmp    L2
L28:
  addl   $0x1,-0xc(%rbp)
L2:
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rdx
  mov    -0xc(%rbp),%eax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x2f,%al
  je     L56
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rdx
  mov    -0xc(%rbp),%eax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L28
L56:
  cmpl   $0x0,-0xc(%rbp)
  je     L37
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rdx
  mov    -0xc(%rbp),%eax
  add    %rax,%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,(%rax)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L37
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rax
  movb   $0x0,(%rax)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rax
  lea    0x1(%rax),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,(%rax)
L37:
  mov    -0x8(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
char *pathSep(char **path) {
  char *part = NULL;
  unsigned int length = 0;
  while (**path == '/')
    (*path)++;
  while (((*path)[length] != '/') && ((*path)[length] != '\0'))
    length++;
  if (length > 0) {
    part = *path;
    *path += length;
    if (**path != '\0') {
      **path = '\0';
      (*path)++;
    }
  }
  return part;
}
```
user: What is the c source code of the assembly code below:

```asm
<pathSep>:
  endbr64
  mov    (%rdi),%rsi
  cmpb   $0x2f,(%rsi)
  jne    L21
  lea    0x1(%rsi),%rax
L28:
  mov    %rax,%rsi
  mov    %rax,(%rdi)
  add    $0x1,%rax
  cmpb   $0x2f,-0x1(%rax)
  je     L28
L21:
  movzbl (%rsi),%eax
  cmp    $0x2f,%al
  je     L58
  test   %al,%al
  je     L58
  mov    $0x0,%edx
L6:
  add    $0x1,%edx
  mov    %edx,%eax
  add    %rsi,%rax
  movzbl (%rax),%ecx
  cmp    $0x2f,%cl
  je     L32
  test   %cl,%cl
  jne    L6
L32:
  test   %edx,%edx
  je     L7
  mov    %rax,(%rdi)
  cmpb   $0x0,(%rax)
  je     L65
  movb   $0x0,(%rax)
  addq   $0x1,(%rdi)
L65:
  mov    %rsi,%rax
  ret
L58:
  mov    $0x0,%esi
  jmp    L65
L7:
  mov    $0x0,%esi
  jmp    L65
```

assistant: ```c
char *pathSep(char **path) {
  char *part = NULL;
  unsigned int length = 0;
  while (**path == '/')
    (*path)++;
  while (((*path)[length] != '/') && ((*path)[length] != '\0'))
    length++;
  if (length > 0) {
    part = *path;
    *path += length;
    if (**path != '\0') {
      **path = '\0';
      (*path)++;
    }
  }
  return part;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<pathSep>:
  endbr64
  mov    (%rdi),%r8
  movzbl (%r8),%edx
  cmp    $0x2f,%dl
  jne    L45
  lea    0x1(%r8),%rax
  nopl   0x0(%rax)
L9:
  mov    %rax,(%rdi)
  mov    %rax,%r8
  movzbl (%rax),%edx
  add    $0x1,%rax
  cmp    $0x2f,%dl
  je     L9
L45:
  test   %dl,%dl
  je     L44
  xor    %edx,%edx
L6:
  lea    0x1(%rdx),%eax
  mov    %rax,%rdx
  add    %r8,%rax
  movzbl (%rax),%ecx
  cmp    $0x2f,%cl
  je     L21
  test   %cl,%cl
  jne    L6
L21:
  test   %edx,%edx
  je     L44
  mov    %rax,(%rdi)
  cmpb   $0x0,(%rax)
  je     L30
  movb   $0x0,(%rax)
  addq   $0x1,(%rdi)
L30:
  mov    %r8,%rax
  ret
  nopl   0x0(%rax)
L44:
  xor    %r8d,%r8d
  mov    %r8,%rax
  ret
```

assistant: ```c
char *pathSep(char **path) {
  char *part = NULL;
  unsigned int length = 0;
  while (**path == '/')
    (*path)++;
  while (((*path)[length] != '/') && ((*path)[length] != '\0'))
    length++;
  if (length > 0) {
    part = *path;
    *path += length;
    if (**path != '\0') {
      **path = '\0';
      (*path)++;
    }
  }
  return part;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<pathSep>:
  endbr64
  mov    (%rdi),%r8
  movzbl (%r8),%edx
  cmp    $0x2f,%dl
  jne    L25
  lea    0x1(%r8),%rax
  nopl   0x0(%rax)
L17:
  mov    %rax,(%rdi)
  mov    %rax,%r8
  movzbl (%rax),%edx
  add    $0x1,%rax
  cmp    $0x2f,%dl
  je     L17
L25:
  test   %dl,%dl
  je     L44
  xor    %edx,%edx
L65:
  lea    0x1(%rdx),%eax
  mov    %rax,%rdx
  add    %r8,%rax
  movzbl (%rax),%ecx
  cmp    $0x2f,%cl
  je     L14
  test   %cl,%cl
  jne    L65
L14:
  test   %edx,%edx
  je     L44
  mov    %rax,(%rdi)
  cmpb   $0x0,(%rax)
  je     L27
  movb   $0x0,(%rax)
  addq   $0x1,(%rdi)
L27:
  mov    %r8,%rax
  ret
  nopl   0x0(%rax)
L44:
  xor    %r8d,%r8d
  mov    %r8,%rax
  ret
```

assistant: ```c
char *pathSep(char **path) {
  char *part = NULL;
  unsigned int length = 0;
  while (**path == '/')
    (*path)++;
  while (((*path)[length] != '/') && ((*path)[length] != '\0'))
    length++;
  if (length > 0) {
    part = *path;
    *path += length;
    if (**path != '\0') {
      **path = '\0';
      (*path)++;
    }
  }
  return part;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<power>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  cmpl   $0x0,-0x18(%rbp)
  jns    L18
  mov    $0xffffffff,%eax
  jmp    L5
L18:
  movl   $0x1,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L17
L14:
  mov    -0x8(%rbp),%eax
  imul   -0x14(%rbp),%eax
  mov    %eax,-0x8(%rbp)
  addl   $0x1,-0x4(%rbp)
L17:
  mov    -0x4(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jl     L14
  mov    -0x8(%rbp),%eax
L5:
  pop    %rbp
  ret
```

assistant: ```c
int power(int base, int exponent) {
  if (exponent < 0)
    return -1;
  int result = 1;
  for (int i = 0; i < exponent; i++)
    result *= base;
  return result;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<power>:
  endbr64
  test   %esi,%esi
  js     L43
  jle    L41
  mov    $0x0,%eax
  mov    $0x1,%edx
L26:
  imul   %edi,%edx
  add    $0x1,%eax
  cmp    %eax,%esi
  jne    L26
L39:
  mov    %edx,%eax
  ret
L43:
  mov    $0xffffffff,%edx
  jmp    L39
L41:
  mov    $0x1,%edx
  jmp    L39
```

assistant: ```c
int power(int base, int exponent) {
  if (exponent < 0)
    return -1;
  int result = 1;
  for (int i = 0; i < exponent; i++)
    result *= base;
  return result;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<power>:
  endbr64
  test   %esi,%esi
  js     L13
  je     L15
  xor    %eax,%eax
  mov    $0x1,%r8d
  nopw   0x0(%rax,%rax,1)
L3:
  add    $0x1,%eax
  imul   %edi,%r8d
  cmp    %eax,%esi
  jne    L3
L1:
  mov    %r8d,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L15:
  mov    $0x1,%r8d
  mov    %r8d,%eax
  ret
L13:
  mov    $0xffffffff,%r8d
  jmp    L1
```

assistant: ```c
int power(int base, int exponent) {
  if (exponent < 0)
    return -1;
  int result = 1;
  for (int i = 0; i < exponent; i++)
    result *= base;
  return result;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<power>:
  endbr64
  test   %esi,%esi
  js     L24
  je     L26
  xor    %eax,%eax
  mov    $0x1,%r8d
  nopw   0x0(%rax,%rax,1)
L28:
  add    $0x1,%eax
  imul   %edi,%r8d
  cmp    %eax,%esi
  jne    L28
L5:
  mov    %r8d,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L26:
  mov    $0x1,%r8d
  mov    %r8d,%eax
  ret
L24:
  mov    $0xffffffff,%r8d
  jmp    L5
```

assistant: ```c
int power(int base, int exponent) {
  if (exponent < 0)
    return -1;
  int result = 1;
  for (int i = 0; i < exponent; i++)
    result *= base;
  return result;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<getline0>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L38
L22:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x8(%rbp),%edx
  mov    %dl,(%rax)
  addl   $0x1,-0x4(%rbp)
L38:
  mov    -0x1c(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0x4(%rbp)
  jge    L43
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %eax,-0x8(%rbp)
  mov    D2(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,-0x8(%rbp)
  je     L43
  cmpl   $0xa,-0x8(%rbp)
  jne    L22
L43:
  cmpl   $0xa,-0x8(%rbp)
  jne    L31
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x8(%rbp),%edx
  mov    %dl,(%rax)
  addl   $0x1,-0x4(%rbp)
L31:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int getline0(char s[], int lim) {
  int c, i;
  for (i = 0; i < lim - 1 && (c = getchar()) != EOF && c != '\n'; ++i)
    s[i] = c;
  if (c == '\n') {
    s[i] = c;
    ++i;
  }
  s[i] = '\0';
  return i;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<getline0>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r13
  cmp    $0x1,%esi
  jle    L58
  mov    %rdi,%rbp
  lea    -0x1(%rsi),%r12d
  mov    $0x0,%ebx
L20:
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    D1(%rip),%rdx
  cmp    %eax,(%rdx)
  je     L50
  cmp    $0xa,%eax
  je     L50
  mov    %al,0x0(%rbp)
  add    $0x1,%ebx
  add    $0x1,%rbp
  cmp    %r12d,%ebx
  jne    L20
  mov    %r12d,%ebx
  jmp    L18
L58:
  mov    $0x0,%ebx
L50:
  cmp    $0xa,%eax
  je     L13
L18:
  movslq %ebx,%rax
  movb   $0x0,0x0(%r13,%rax,1)
  mov    %ebx,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L13:
  movslq %ebx,%rax
  movb   $0xa,0x0(%r13,%rax,1)
  add    $0x1,%ebx
  jmp    L18
```

assistant: ```c
int getline0(char s[], int lim) {
  int c, i;
  for (i = 0; i < lim - 1 && (c = getchar()) != EOF && c != '\n'; ++i)
    s[i] = c;
  if (c == '\n') {
    s[i] = c;
    ++i;
  }
  s[i] = '\0';
  return i;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<getline0>:
  endbr64
  push   %r14
  mov    %rdi,%rcx
  xor    %r14d,%r14d
  push   %r13
  mov    %rdi,%r13
  push   %r12
  push   %rbp
  lea    -0x1(%rsi),%ebp
  push   %rbx
  test   %ebp,%ebp
  jle    L4
  mov    D7(%rip),%r12
  mov    %rdi,%rbx
  xor    %r14d,%r14d
  jmp    L0
  nopl   0x0(%rax,%rax,1)
L6:
  lea    0x1(%r14),%edx
  cmp    $0xa,%eax
  je     L26
  mov    %al,(%rcx)
  lea    0x1(%rcx),%rbx
  cmp    %edx,%ebp
  je     L14
  mov    %edx,%r14d
L0:
  xor    %eax,%eax
  call   <getchar@plt>
  mov    %rbx,%rcx
  cmp    %eax,(%r12)
  jne    L6
  cmp    $0xa,%eax
  je     L27
L4:
  movb   $0x0,(%rcx)
  mov    %r14d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L27:
  lea    0x1(%r14),%edx
L26:
  movslq %edx,%rcx
  movb   $0xa,(%rbx)
  mov    %rcx,%r14
  add    %r13,%rcx
  movb   $0x0,(%rcx)
  mov    %r14d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
L14:
  movslq %ebp,%rcx
  mov    %ebp,%r14d
  add    %r13,%rcx
  mov    %r14d,%eax
  movb   $0x0,(%rcx)
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int getline0(char s[], int lim) {
  int c, i;
  for (i = 0; i < lim - 1 && (c = getchar()) != EOF && c != '\n'; ++i)
    s[i] = c;
  if (c == '\n') {
    s[i] = c;
    ++i;
  }
  s[i] = '\0';
  return i;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<getline0>:
  endbr64
  push   %r14
  mov    %rdi,%rcx
  xor    %r14d,%r14d
  push   %r13
  mov    %rdi,%r13
  push   %r12
  push   %rbp
  lea    -0x1(%rsi),%ebp
  push   %rbx
  test   %ebp,%ebp
  jle    L49
  mov    D3(%rip),%r12
  mov    %rdi,%rbx
  xor    %r14d,%r14d
  jmp    L7
  nopl   0x0(%rax,%rax,1)
L17:
  lea    0x1(%r14),%edx
  cmp    $0xa,%eax
  je     L18
  mov    %al,(%rcx)
  lea    0x1(%rcx),%rbx
  cmp    %edx,%ebp
  je     L3
  mov    %edx,%r14d
L7:
  xor    %eax,%eax
  call   <getchar@plt>
  mov    %rbx,%rcx
  cmp    %eax,(%r12)
  jne    L17
  cmp    $0xa,%eax
  je     L47
L49:
  movb   $0x0,(%rcx)
  mov    %r14d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L47:
  lea    0x1(%r14),%edx
L18:
  movslq %edx,%rcx
  movb   $0xa,(%rbx)
  mov    %rcx,%r14
  add    %r13,%rcx
  movb   $0x0,(%rcx)
  mov    %r14d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
L3:
  movslq %ebp,%rcx
  mov    %ebp,%r14d
  add    %r13,%rcx
  mov    %r14d,%eax
  movb   $0x0,(%rcx)
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int getline0(char s[], int lim) {
  int c, i;
  for (i = 0; i < lim - 1 && (c = getchar()) != EOF && c != '\n'; ++i)
    s[i] = c;
  if (c == '\n') {
    s[i] = c;
    ++i;
  }
  s[i] = '\0';
  return i;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<emsSetd>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xf0,%rsp
  mov    %rdi,-0xe8(%rbp)
  movsd  %xmm0,-0xf0(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0xf0(%rbp),%rdx
  lea    -0xe0(%rbp),%rax
  movq   %rdx,%xmm0
  mov    $0xf,%edx
  lea    D9(%rip),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <sprintf@plt>
  lea    -0xe0(%rbp),%rdx
  mov    -0xe8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <ems1Stok@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L1
  call   <__stack_chk_fail@plt>
L1:
  leave
  ret
```

tool_calls: name: read_data
arguments: label: D9
data_type: string

tool: D9: %.*g

assistant: ```c
void emsSetd(const char *token, double dvalue) {
  char str[200 + 1];
  ;
  sprintf(str, "%.*g", 15, dvalue);
  ems1Stok(token, str);
  return;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<emsSetd>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0xe8,%rsp
  mov    %rdi,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0xd8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    $0xf,%edx
  lea    D9(%rip),%rsi
  mov    %rbp,%rdi
  call   <sprintf@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <ems1Stok@plt>
  mov    0xd8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0xe8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D9
type: string

tool: D9: %.*g

assistant: ```c
void emsSetd(const char *token, double dvalue) {
  char str[200 + 1];
  ;
  sprintf(str, "%.*g", 15, dvalue);
  ems1Stok(token, str);
  return;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<emsSetd>:
  endbr64
  push   %r12
  lea    D0(%rip),%rsi
  mov    $0xf,%edx
  push   %rbp
  mov    %rdi,%rbp
  sub    $0xe8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xd8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  mov    %r12,%rdi
  call   <sprintf@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <ems1Stok@plt>
  mov    0xd8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  add    $0xe8,%rsp
  pop    %rbp
  pop    %r12
  ret
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D0
type_infered: string

tool: D0: %.*g

assistant: ```c
void emsSetd(const char *token, double dvalue) {
  char str[200 + 1];
  ;
  sprintf(str, "%.*g", 15, dvalue);
  ems1Stok(token, str);
  return;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<emsSetd>:
  endbr64
  push   %r12
  lea    D9(%rip),%rsi
  mov    $0xf,%edx
  push   %rbp
  mov    %rdi,%rbp
  sub    $0xe8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xd8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  mov    %r12,%rdi
  call   <sprintf@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <ems1Stok@plt>
  mov    0xd8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  add    $0xe8,%rsp
  pop    %rbp
  pop    %r12
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D9
type: string

tool: D9: %.*g

assistant: ```c
void emsSetd(const char *token, double dvalue) {
  char str[200 + 1];
  ;
  sprintf(str, "%.*g", 15, dvalue);
  ems1Stok(token, str);
  return;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <welcomeArt@plt>
L85:
  mov    $0x0,%eax
  call   <mainMenu@plt>
  cmp    $0x3,%eax
  je     L9
  cmp    $0x3,%eax
  jg     L85
  cmp    $0x2,%eax
  je     L1
  cmp    $0x2,%eax
  jg     L85
  test   %eax,%eax
  je     L18
  cmp    $0x1,%eax
  je     L5
  jmp    L85
L18:
  mov    $0x0,%eax
  call   <loadGame@plt>
  jmp    L16
L5:
  mov    $0x0,%eax
  call   <displayHighScores@plt>
  jmp    L16
L1:
  mov    $0x0,%eax
  call   <controls@plt>
  jmp    L16
L9:
  mov    $0x0,%eax
  call   <exitYN@plt>
  nop
L16:
  jmp    L85
```

assistant: ```c
int main() {
  welcomeArt();
  do {
    switch (mainMenu()) {
    case 0:
      loadGame();
      break;
    case 1:
      displayHighScores();
      break;
    case 2:
      controls();
      break;
    case 3:
      exitYN();
      break;
    }
  } while (1);
  return (0);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <welcomeArt@plt>
  jmp    L12
L66:
  cmp    $0x3,%eax
  jne    L12
  mov    $0x0,%eax
  call   <exitYN@plt>
  jmp    L12
L70:
  mov    $0x0,%eax
  call   <loadGame@plt>
L12:
  mov    $0x0,%eax
  call   <mainMenu@plt>
  cmp    $0x2,%eax
  je     L69
  jg     L66
  test   %eax,%eax
  je     L70
  cmp    $0x1,%eax
  jne    L12
  mov    $0x0,%eax
  call   <displayHighScores@plt>
  jmp    L12
L69:
  mov    $0x0,%eax
  call   <controls@plt>
  jmp    L12
```

assistant: ```c
int main() {
  welcomeArt();
  do {
    switch (mainMenu()) {
    case 0:
      loadGame();
      break;
    case 1:
      displayHighScores();
      break;
    case 2:
      controls();
      break;
    case 3:
      exitYN();
      break;
    }
  } while (1);
  return (0);
}
```