ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将以下的汇编代码转换成C语言代码:

```asm
<RandomFunc>:
  endbr64
  movzwl (%rdi),%eax
  xor    %edx,%edx
  add    $0x7296,%ax
  jne    L4
L3:
  mov    %dx,(%rsi)
  ret
  nopl   0x0(%rax,%rax,1)
L4:
  mov    %eax,%ecx
  and    $0xe,%ecx
  or     $0x1,%ecx
  ror    %cl,%ax
  mov    %eax,%ecx
  shr    $0x2,%cx
  and    $0xe,%ecx
  or     $0x1,%ecx
  rol    %cl,%ax
  mov    %eax,%edx
  test   %ax,%ax
  je     L3
  movzwl %ax,%eax
  imul   %eax,%eax
  shl    $0x4,%eax
  and    $0x3f0,%ax
  or     %eax,%edx
  imul   $0x274,%dx,%dx
  mov    %dx,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned short input[1], unsigned short output[1]) {
  unsigned short state[1];
  unsigned short local1;
  char copy11;
  {
    state[0UL] = (input[0UL] - 51238316UL) - (unsigned short)47038;
    local1 = 0UL;
    while (local1 < 1UL) {
      if (state[0UL] > local1) {
        state[0UL] = (state[local1] >> ((state[local1] & (unsigned short)15) | 1UL)) | (state[local1] << (16 - ((state[local1] & (unsigned short)15) | 1UL)));
        state[0UL] = (state[local1] << (((state[local1] >> (unsigned short)2) & (unsigned short)15) | 1UL)) | (state[local1] >> (16 - (((state[local1] >> (unsigned short)2) & (unsigned short)15) | 1UL)));
      } else {
        copy11 = *((char *)(&state[local1]) + 0);
        *((char *)(&state[local1]) + 0) = *((char *)(&state[local1]) + 1);
        *((char *)(&state[local1]) + 1) = copy11;
        state[local1] |= (((state[local1] << (((state[0UL] >> (unsigned short)2) & (unsigned short)15) | 1UL)) | (state[local1] >> (16 - (((state[0UL] >> (unsigned short)2) & (unsigned short)15) | 1UL)))) & (unsigned short)63) << 4UL;
      }
      if (state[0UL] == local1) {
        state[local1] |= ((state[local1] + state[local1]) & (unsigned short)15) << 2UL;
      } else {
        state[local1] |= (state[local1] * state[local1] & (unsigned short)63) << 4UL;
      }
      local1++;
    }
    output[0UL] = state[0UL] * 1062470260UL;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<RandomFunc>:
  endbr64
  movzwl (%rdi),%eax
  add    $0x7296,%ax
  jne    L5
L12:
  mov    %ax,(%rsi)
  ret
  nopl   0x0(%rax)
L5:
  mov    %eax,%ecx
  and    $0xe,%ecx
  or     $0x1,%ecx
  ror    %cl,%ax
  mov    %eax,%ecx
  shr    $0x2,%cx
  and    $0xe,%ecx
  or     $0x1,%ecx
  rol    %cl,%ax
  test   %ax,%ax
  je     L12
  movzwl %ax,%edx
  imul   %edx,%edx
  shl    $0x4,%edx
  and    $0x3f0,%dx
  or     %edx,%eax
  imul   $0x274,%ax,%ax
  mov    %ax,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned short input[1], unsigned short output[1]) {
  unsigned short state[1];
  unsigned short local1;
  char copy11;
  {
    state[0UL] = (input[0UL] - 51238316UL) - (unsigned short)47038;
    local1 = 0UL;
    while (local1 < 1UL) {
      if (state[0UL] > local1) {
        state[0UL] = (state[local1] >> ((state[local1] & (unsigned short)15) | 1UL)) | (state[local1] << (16 - ((state[local1] & (unsigned short)15) | 1UL)));
        state[0UL] = (state[local1] << (((state[local1] >> (unsigned short)2) & (unsigned short)15) | 1UL)) | (state[local1] >> (16 - (((state[local1] >> (unsigned short)2) & (unsigned short)15) | 1UL)));
      } else {
        copy11 = *((char *)(&state[local1]) + 0);
        *((char *)(&state[local1]) + 0) = *((char *)(&state[local1]) + 1);
        *((char *)(&state[local1]) + 1) = copy11;
        state[local1] |= (((state[local1] << (((state[0UL] >> (unsigned short)2) & (unsigned short)15) | 1UL)) | (state[local1] >> (16 - (((state[0UL] >> (unsigned short)2) & (unsigned short)15) | 1UL)))) & (unsigned short)63) << 4UL;
      }
      if (state[0UL] == local1) {
        state[local1] |= ((state[local1] + state[local1]) & (unsigned short)15) << 2UL;
      } else {
        state[local1] |= (state[local1] * state[local1] & (unsigned short)63) << 4UL;
      }
      local1++;
    }
    output[0UL] = state[0UL] * 1062470260UL;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<sg_nz_put_unaligned_le64>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L7
  mov    -0x10(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <sg_put_unaligned_le64@plt>
L7:
  nop
  leave
  ret
```

assistant: ```c
void sg_nz_put_unaligned_le64(uint64_t val, void *p) {
  if (val)
    sg_put_unaligned_le64(val, p);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<sg_nz_put_unaligned_le64>:
  endbr64
  test   %rdi,%rdi
  jne    L5
  ret
L5:
  sub    $0x8,%rsp
  call   <sg_put_unaligned_le64@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void sg_nz_put_unaligned_le64(uint64_t val, void *p) {
  if (val)
    sg_put_unaligned_le64(val, p);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<sg_nz_put_unaligned_le64>:
  endbr64
  test   %rdi,%rdi
  jne    L0
  ret
  nopw   0x0(%rax,%rax,1)
L0:
  jmp    1050 <sg_put_unaligned_le64@plt>
```

assistant: ```c
void sg_nz_put_unaligned_le64(uint64_t val, void *p) {
  if (val)
    sg_put_unaligned_le64(val, p);
}
```
user: Translate the below assembly code into C programming language:

```asm
<sg_nz_put_unaligned_le64>:
  endbr64
  test   %rdi,%rdi
  jne    L6
  ret
  nopw   0x0(%rax,%rax,1)
L6:
  jmp    1050 <sg_put_unaligned_le64@plt>
```

assistant: ```c
void sg_nz_put_unaligned_le64(uint64_t val, void *p) {
  if (val)
    sg_put_unaligned_le64(val, p);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<arraySort>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  movl   $0x0,-0x8(%rbp)
  mov    -0x2c(%rbp),%eax
  mov    %eax,-0x18(%rbp)
  movl   $0x0,-0x14(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L21
L20:
  movl   $0x0,-0x1c(%rbp)
  jmp    L31
L12:
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    %eax,-0x14(%rbp)
  jge    L17
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x1c(%rbp),%eax
  mov    %eax,-0x10(%rbp)
L17:
  addl   $0x1,-0x1c(%rbp)
L31:
  mov    -0x1c(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jl     L12
  subl   $0x1,-0x18(%rbp)
  mov    -0x18(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x18(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rax,%rdx
  mov    -0x14(%rbp),%eax
  mov    %eax,(%rdx)
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  mov    %eax,(%rdx)
  movl   $0x0,-0x14(%rbp)
  addl   $0x1,-0xc(%rbp)
L21:
  mov    -0x8(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jl     L20
  mov    -0x28(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
int *arraySort(int numbers[], int numbersCount) {
  int i, k = 0, last = numbersCount, max = 0;
  int current, maxContainer;
  int count = 0;
  while (k < last) {
    for (i = 0; i < last; i++) {
      if (numbers[i] > max) {
        max = numbers[i];
        maxContainer = i;
      }
    }
    last--;
    current = numbers[last];
    numbers[last] = max;
    numbers[maxContainer] = current;
    max = 0;
    count++;
  }
  return numbers;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<arraySort>:
  endbr64
  test   %esi,%esi
  jle    L65
  movslq %esi,%r8
  mov    $0x0,%r10d
  mov    $0x0,%r11d
  jmp    L69
L47:
  lea    0x1(%rax),%rdx
  cmp    %rsi,%rax
  je     L1
  mov    %rdx,%rax
L53:
  mov    (%rdi,%rax,4),%edx
  cmp    %ecx,%edx
  jle    L47
  mov    %eax,%r9d
  mov    %edx,%ecx
  jmp    L47
L1:
  mov    -0x4(%rdi,%r8,4),%edx
  mov    %ecx,-0x4(%rdi,%r8,4)
  movslq %r9d,%rax
  mov    %edx,(%rdi,%rax,4)
  sub    $0x1,%r8
  test   %r8d,%r8d
  jle    L65
L69:
  mov    %r10d,%ecx
  test   %r8d,%r8d
  jle    L1
  lea    -0x1(%r8),%esi
  mov    %r11,%rax
  mov    %r10d,%ecx
  jmp    L53
L65:
  mov    %rdi,%rax
  ret
```

assistant: ```c
int *arraySort(int numbers[], int numbersCount) {
  int i, k = 0, last = numbersCount, max = 0;
  int current, maxContainer;
  int count = 0;
  while (k < last) {
    for (i = 0; i < last; i++) {
      if (numbers[i] > max) {
        max = numbers[i];
        maxContainer = i;
      }
    }
    last--;
    current = numbers[last];
    numbers[last] = max;
    numbers[maxContainer] = current;
    max = 0;
    count++;
  }
  return numbers;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<arraySort>:
  endbr64
  mov    %rdi,%r8
  test   %esi,%esi
  jle    L34
  movslq %esi,%rsi
  xchg   %ax,%ax
L29:
  xor    %eax,%eax
  xor    %ecx,%ecx
  nopl   0x0(%rax)
L18:
  mov    (%r8,%rax,4),%edx
  cmp    %ecx,%edx
  jle    L7
  mov    %eax,%r9d
  mov    %edx,%ecx
L7:
  add    $0x1,%rax
  cmp    %eax,%esi
  jg     L18
  mov    -0x4(%r8,%rsi,4),%edx
  movslq %r9d,%rax
  mov    %ecx,-0x4(%r8,%rsi,4)
  sub    $0x1,%rsi
  mov    %edx,(%r8,%rax,4)
  test   %esi,%esi
  jg     L29
L34:
  mov    %r8,%rax
  ret
```

assistant: ```c
int *arraySort(int numbers[], int numbersCount) {
  int i, k = 0, last = numbersCount, max = 0;
  int current, maxContainer;
  int count = 0;
  while (k < last) {
    for (i = 0; i < last; i++) {
      if (numbers[i] > max) {
        max = numbers[i];
        maxContainer = i;
      }
    }
    last--;
    current = numbers[last];
    numbers[last] = max;
    numbers[maxContainer] = current;
    max = 0;
    count++;
  }
  return numbers;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<arraySort>:
  endbr64
  mov    %rdi,%r8
  test   %esi,%esi
  jle    L33
  movslq %esi,%rsi
  xchg   %ax,%ax
L19:
  xor    %eax,%eax
  xor    %ecx,%ecx
  nopl   0x0(%rax)
L1:
  mov    (%r8,%rax,4),%edx
  cmp    %edx,%ecx
  jge    L14
  mov    %eax,%r9d
  mov    %edx,%ecx
L14:
  add    $0x1,%rax
  cmp    %eax,%esi
  jg     L1
  mov    -0x4(%r8,%rsi,4),%edx
  movslq %r9d,%rax
  mov    %ecx,-0x4(%r8,%rsi,4)
  sub    $0x1,%rsi
  mov    %edx,(%r8,%rax,4)
  test   %esi,%esi
  jg     L19
L33:
  mov    %r8,%rax
  ret
```

assistant: ```c
int *arraySort(int numbers[], int numbersCount) {
  int i, k = 0, last = numbersCount, max = 0;
  int current, maxContainer;
  int count = 0;
  while (k < last) {
    for (i = 0; i < last; i++) {
      if (numbers[i] > max) {
        max = numbers[i];
        maxContainer = i;
      }
    }
    last--;
    current = numbers[last];
    numbers[last] = max;
    numbers[maxContainer] = current;
    max = 0;
    count++;
  }
  return numbers;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<str_legacy_to_utf8_internal>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  je     L7
  cmpq   $0x0,-0x20(%rbp)
  jne    L43
L7:
  mov    $0x0,%eax
  jmp    L33
L43:
  mov    -0x18(%rbp),%rax
  mov    $0x0,%edx
  mov    $0xffffffff,%esi
  mov    %rax,%rdi
  call   <g_utf8_validate@plt>
  test   %rax,%rax
  je     L34
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strdup@plt>
  jmp    L33
L34:
  mov    -0x20(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  sub    $0x8,%rsp
  mov    -0x24(%rbp),%ecx
  push   %rcx
  mov    $0x0,%r9d
  mov    $0x0,%r8d
  lea    D6(%rip),%rcx
  mov    $0xffffffff,%esi
  mov    %rax,%rdi
  call   <iconv_string_full@plt>
  add    $0x10,%rsp
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
L33:
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D6
data_type: string

tool: D6: UTF-8

assistant: ```c
char *str_legacy_to_utf8_internal(const char *string, const char *legacy_encoding, int replacement_char) {
  if (string == NULL || legacy_encoding == NULL) {
    return NULL;
  }
  if (g_utf8_validate(string, -1, NULL)) {
    return strdup(string);
  }
  char *utf8_string = iconv_string_full(string, -1, legacy_encoding, "UTF-8", NULL, NULL, replacement_char);
  return utf8_string;
}
```
user: What is the c source code of the assembly code below:

```asm
<str_legacy_to_utf8_internal>:
  endbr64
  test   %rdi,%rdi
  je     L14
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  mov    %edx,%r12d
  test   %rsi,%rsi
  je     L10
  mov    $0x0,%edx
  mov    $0xffffffff,%esi
  call   <g_utf8_validate@plt>
  test   %rax,%rax
  jne    L42
  sub    $0x8,%rsp
  push   %r12
  mov    $0x0,%r9d
  mov    $0x0,%r8d
  lea    D7(%rip),%rcx
  mov    %rbp,%rdx
  mov    $0xffffffff,%esi
  mov    %rbx,%rdi
  call   <iconv_string_full@plt>
  add    $0x10,%rsp
L18:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L42:
  mov    %rbx,%rdi
  call   <strdup@plt>
  jmp    L18
L10:
  mov    $0x0,%eax
  jmp    L18
L14:
  mov    $0x0,%eax
  ret
```

tool_calls: name: read_data
arguments: data_label: D7
type: string

tool: D7: UTF-8

assistant: ```c
char *str_legacy_to_utf8_internal(const char *string, const char *legacy_encoding, int replacement_char) {
  if (string == NULL || legacy_encoding == NULL) {
    return NULL;
  }
  if (g_utf8_validate(string, -1, NULL)) {
    return strdup(string);
  }
  char *utf8_string = iconv_string_full(string, -1, legacy_encoding, "UTF-8", NULL, NULL, replacement_char);
  return utf8_string;
}
```
user: What is the c source code of the assembly code below:

```asm
<str_legacy_to_utf8_internal>:
  endbr64
  test   %rdi,%rdi
  je     L2
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  push   %rbx
  test   %rsi,%rsi
  je     L22
  mov    %edx,%ebx
  mov    $0xffffffff,%esi
  xor    %edx,%edx
  mov    %rdi,%rbp
  call   <g_utf8_validate@plt>
  test   %rax,%rax
  jne    L19
  sub    $0x8,%rsp
  mov    %r12,%rdx
  mov    %rbp,%rdi
  xor    %r9d,%r9d
  push   %rbx
  lea    D5(%rip),%rcx
  xor    %r8d,%r8d
  mov    $0xffffffff,%esi
  call   <iconv_string_full@plt>
  pop    %rdx
  pop    %rcx
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L19:
  pop    %rbx
  mov    %rbp,%rdi
  pop    %rbp
  pop    %r12
  jmp    1080 <strdup@plt>
  nopl   0x0(%rax)
L22:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
  nop
L2:
  xor    %eax,%eax
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D5
data_type: string

tool: D5: UTF-8

assistant: ```c
char *str_legacy_to_utf8_internal(const char *string, const char *legacy_encoding, int replacement_char) {
  if (string == NULL || legacy_encoding == NULL) {
    return NULL;
  }
  if (g_utf8_validate(string, -1, NULL)) {
    return strdup(string);
  }
  char *utf8_string = iconv_string_full(string, -1, legacy_encoding, "UTF-8", NULL, NULL, replacement_char);
  return utf8_string;
}
```
user: What is the c source code of the assembly code below:

```asm
<str_legacy_to_utf8_internal>:
  endbr64
  test   %rdi,%rdi
  je     L20
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  push   %rbx
  test   %rsi,%rsi
  je     L23
  mov    %edx,%ebx
  mov    $0xffffffff,%esi
  xor    %edx,%edx
  mov    %rdi,%rbp
  call   <g_utf8_validate@plt>
  test   %rax,%rax
  jne    L8
  sub    $0x8,%rsp
  mov    %r12,%rdx
  mov    %rbp,%rdi
  xor    %r9d,%r9d
  push   %rbx
  lea    D4(%rip),%rcx
  xor    %r8d,%r8d
  mov    $0xffffffff,%esi
  call   <iconv_string_full@plt>
  pop    %rdx
  pop    %rcx
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L8:
  pop    %rbx
  mov    %rbp,%rdi
  pop    %rbp
  pop    %r12
  jmp    1080 <strdup@plt>
  nopl   0x0(%rax)
L23:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
  nop
L20:
  xor    %eax,%eax
  ret
```

tool_calls: name: read_data
arguments: data_tag: D4
data_type: string

tool: D4: UTF-8

assistant: ```c
char *str_legacy_to_utf8_internal(const char *string, const char *legacy_encoding, int replacement_char) {
  if (string == NULL || legacy_encoding == NULL) {
    return NULL;
  }
  if (g_utf8_validate(string, -1, NULL)) {
    return strdup(string);
  }
  char *utf8_string = iconv_string_full(string, -1, legacy_encoding, "UTF-8", NULL, NULL, replacement_char);
  return utf8_string;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<strcaseeq>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
L74:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%edx
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%eax
  cmp    %al,%dl
  je     L100
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  mov    %al,-0x2(%rbp)
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%eax
  mov    %al,-0x1(%rbp)
  cmpb   $0x0,-0x2(%rbp)
  je     L51
  cmpb   $0x0,-0x1(%rbp)
  jne    L78
L51:
  mov    $0x1,%eax
  jmp    L81
L78:
  cmpb   $0x40,-0x2(%rbp)
  jbe    L79
  cmpb   $0x5a,-0x2(%rbp)
  ja     L79
  orb    $0x20,-0x2(%rbp)
L79:
  cmpb   $0x40,-0x1(%rbp)
  jbe    L77
  cmpb   $0x5a,-0x1(%rbp)
  ja     L77
  orb    $0x20,-0x1(%rbp)
L77:
  movzbl -0x2(%rbp),%eax
  cmp    -0x1(%rbp),%al
  je     L40
  mov    $0x1,%eax
  jmp    L81
L100:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L40
  mov    $0x0,%eax
  jmp    L81
L40:
  addq   $0x1,-0x18(%rbp)
  addq   $0x1,-0x20(%rbp)
  jmp    L74
L81:
  pop    %rbp
  ret
```

assistant: ```c
extern int strcaseeq(const char *s1, const char *s2) {
  while (1) {
    if (*s1 != *s2) {
      unsigned char c1, c2;
      c1 = *(const unsigned char *)s1;
      c2 = *(const unsigned char *)s2;
      if (c1 == 0 || c2 == 0)
        return 1;
      if (c1 >= 'A' && c1 <= 'Z')
        c1 |= 0x20;
      if (c2 >= 'A' && c2 <= 'Z')
        c2 |= 0x20;
      if (c1 != c2)
        return 1;
    } else {
      if (*s1 == 0)
        return 0;
    }
    s1++;
    s2++;
  }
  return 1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<strcaseeq>:
  endbr64
  mov    $0x0,%ecx
  jmp    L6
L19:
  test   %al,%al
  je     L35
L41:
  add    $0x1,%rcx
L6:
  movzbl (%rdi,%rcx,1),%eax
  movzbl (%rsi,%rcx,1),%edx
  cmp    %dl,%al
  je     L19
  mov    %eax,%r9d
  mov    %edx,%r8d
  test   %al,%al
  je     L46
  test   %dl,%dl
  je     L46
  lea    -0x41(%rax),%r10d
  or     $0x20,%eax
  cmp    $0x1a,%r10b
  cmovb  %eax,%r9d
  lea    -0x41(%rdx),%eax
  or     $0x20,%edx
  cmp    $0x1a,%al
  cmovb  %edx,%r8d
  cmp    %r8b,%r9b
  je     L41
  mov    $0x1,%eax
  ret
L46:
  mov    $0x1,%eax
  ret
L35:
  mov    $0x0,%eax
  ret
```

assistant: ```c
extern int strcaseeq(const char *s1, const char *s2) {
  while (1) {
    if (*s1 != *s2) {
      unsigned char c1, c2;
      c1 = *(const unsigned char *)s1;
      c2 = *(const unsigned char *)s2;
      if (c1 == 0 || c2 == 0)
        return 1;
      if (c1 >= 'A' && c1 <= 'Z')
        c1 |= 0x20;
      if (c2 >= 'A' && c2 <= 'Z')
        c2 |= 0x20;
      if (c1 != c2)
        return 1;
    } else {
      if (*s1 == 0)
        return 0;
    }
    s1++;
    s2++;
  }
  return 1;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<strcaseeq>:
  endbr64
  xor    %ecx,%ecx
  jmp    L3
  nopl   0x0(%rax,%rax,1)
L41:
  test   %al,%al
  je     L20
  test   %dl,%dl
  je     L20
  lea    -0x41(%rax),%r9d
  mov    %eax,%r8d
  or     $0x20,%r8d
  cmp    $0x1a,%r9b
  lea    -0x41(%rdx),%r9d
  cmovae %eax,%r8d
  mov    %edx,%eax
  or     $0x20,%eax
  cmp    $0x1a,%r9b
  cmovae %edx,%eax
  cmp    %al,%r8b
  jne    L20
L8:
  add    $0x1,%rcx
L3:
  movzbl (%rdi,%rcx,1),%eax
  movzbl (%rsi,%rcx,1),%edx
  cmp    %dl,%al
  jne    L41
  test   %al,%al
  jne    L8
  xor    %eax,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L20:
  mov    $0x1,%eax
  ret
```

assistant: ```c
extern int strcaseeq(const char *s1, const char *s2) {
  while (1) {
    if (*s1 != *s2) {
      unsigned char c1, c2;
      c1 = *(const unsigned char *)s1;
      c2 = *(const unsigned char *)s2;
      if (c1 == 0 || c2 == 0)
        return 1;
      if (c1 >= 'A' && c1 <= 'Z')
        c1 |= 0x20;
      if (c2 >= 'A' && c2 <= 'Z')
        c2 |= 0x20;
      if (c1 != c2)
        return 1;
    } else {
      if (*s1 == 0)
        return 0;
    }
    s1++;
    s2++;
  }
  return 1;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<strcaseeq>:
  endbr64
  xor    %ecx,%ecx
  jmp    L44
  nopl   0x0(%rax,%rax,1)
L4:
  test   %al,%al
  je     L26
  test   %dl,%dl
  je     L26
  lea    -0x41(%rax),%r9d
  mov    %eax,%r8d
  or     $0x20,%r8d
  cmp    $0x1a,%r9b
  lea    -0x41(%rdx),%r9d
  cmovae %eax,%r8d
  mov    %edx,%eax
  or     $0x20,%eax
  cmp    $0x1a,%r9b
  cmovae %edx,%eax
  cmp    %al,%r8b
  jne    L26
L30:
  add    $0x1,%rcx
L44:
  movzbl (%rdi,%rcx,1),%eax
  movzbl (%rsi,%rcx,1),%edx
  cmp    %dl,%al
  jne    L4
  test   %al,%al
  jne    L30
  xor    %eax,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L26:
  mov    $0x1,%eax
  ret
```

assistant: ```c
extern int strcaseeq(const char *s1, const char *s2) {
  while (1) {
    if (*s1 != *s2) {
      unsigned char c1, c2;
      c1 = *(const unsigned char *)s1;
      c2 = *(const unsigned char *)s2;
      if (c1 == 0 || c2 == 0)
        return 1;
      if (c1 >= 'A' && c1 <= 'Z')
        c1 |= 0x20;
      if (c2 >= 'A' && c2 <= 'Z')
        c2 |= 0x20;
      if (c1 != c2)
        return 1;
    } else {
      if (*s1 == 0)
        return 0;
    }
    s1++;
    s2++;
  }
  return 1;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<strlwr_>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L5
L18:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x40,%al
  jle    L15
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x5a,%al
  jg     L15
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  add    $0x20,%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%rax
  mov    %dl,(%rax)
L15:
  addq   $0x1,-0x8(%rbp)
L5:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L18
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void strlwr_(char *string) {
  char *p;
  for (p = string; *p; p++)
    if ('A' <= *p && *p <= 'Z')
      *p += 'a' - 'A';
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<strlwr_>:
  endbr64
  movzbl (%rdi),%eax
  test   %al,%al
  jne    L3
L1:
  ret
L9:
  add    $0x1,%rdi
  movzbl (%rdi),%eax
  test   %al,%al
  je     L1
L3:
  lea    -0x41(%rax),%edx
  cmp    $0x19,%dl
  ja     L9
  add    $0x20,%eax
  mov    %al,(%rdi)
  jmp    L9
```

assistant: ```c
void strlwr_(char *string) {
  char *p;
  for (p = string; *p; p++)
    if ('A' <= *p && *p <= 'Z')
      *p += 'a' - 'A';
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<strlwr_>:
  endbr64
  jmp    L28
  cs nopw 0x0(%rax,%rax,1)
L5:
  lea    -0x41(%rax),%edx
  cmp    $0x19,%dl
  ja     L22
  add    $0x20,%eax
  mov    %al,(%rdi)
L22:
  add    $0x1,%rdi
L28:
  movzbl (%rdi),%eax
  test   %al,%al
  jne    L5
  ret
```

assistant: ```c
void strlwr_(char *string) {
  char *p;
  for (p = string; *p; p++)
    if ('A' <= *p && *p <= 'Z')
      *p += 'a' - 'A';
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<strlwr_>:
  endbr64
  movzbl (%rdi),%eax
  test   %al,%al
  je     L13
  nopl   0x0(%rax,%rax,1)
L20:
  lea    -0x41(%rax),%edx
  cmp    $0x19,%dl
  ja     L31
  add    $0x20,%eax
  add    $0x1,%rdi
  mov    %al,-0x1(%rdi)
  movzbl (%rdi),%eax
  test   %al,%al
  jne    L20
  ret
  nopw   0x0(%rax,%rax,1)
L31:
  movzbl 0x1(%rdi),%eax
  add    $0x1,%rdi
  test   %al,%al
  jne    L20
L13:
  ret
```

assistant: ```c
void strlwr_(char *string) {
  char *p;
  for (p = string; *p; p++)
    if ('A' <= *p && *p <= 'Z')
      *p += 'a' - 'A';
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<print_version>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: tag: D8
datatype: string

tool: D8: Brother scanner driver. Build Mar 17 2022 23:57:03
Copyright 2017 Dariusz Stojaczyk


assistant: ```c
void print_version(void) {
  printf("Brother scanner driver. Build "
         "Mar 17 2022"
         " "
         "23:57:03"
         "\n"
         "Copyright 2017 Dariusz Stojaczyk\n");
}
```