ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<sd_switch_clock_rate>:
  endbr64
  push   %r12
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  call   <sd_get_clock_divider@plt>
  cmp    $0xffffffff,%eax
  je     L4
  mov    %eax,%ebx
  jmp    L37
  nopl   0x0(%rax)
L13:
  mov    $0x3e8,%edi
  call   <usleep@plt>
L37:
  mov    $0x20300024,%edi
  call   <mmio_read@plt>
  and    $0x3,%eax
  mov    %eax,%r12d
  jne    L13
  mov    $0x2030002c,%edi
  call   <mmio_read@plt>
  mov    $0x2030002c,%edi
  mov    %eax,%ebp
  mov    %eax,%esi
  and    $0xfffffffb,%esi
  and    $0xffff001b,%ebp
  call   <mmio_write@plt>
  or     %ebp,%ebx
  mov    $0x7d0,%edi
  call   <usleep@plt>
  mov    %ebx,%esi
  mov    $0x2030002c,%edi
  or     $0x4,%ebx
  call   <mmio_write@plt>
  mov    $0x7d0,%edi
  call   <usleep@plt>
  mov    $0x2030002c,%edi
  mov    %ebx,%esi
  call   <mmio_write@plt>
  mov    $0x7d0,%edi
  call   <usleep@plt>
L9:
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L4:
  mov    %ebp,%esi
  lea    D1(%rip),%rdi
  mov    $0xffffffff,%r12d
  call   <printf@plt>
  jmp    L9
```

tool_calls: name: disassemble_data
arguments: tag: D1
type: string

tool: D1: EMMC: couldn't get a valid divider for target rate %lu Hz


assistant: ```c
int sd_switch_clock_rate(uint32_t base_clock, uint32_t target_rate) {
  uint32_t divider = sd_get_clock_divider(base_clock, target_rate);
  if (divider == 0xffffffff) {
    printf("EMMC: couldn't get a valid divider for target rate %lu Hz\n", target_rate);
    return -1;
  }
  while (mmio_read(0x20300000 + 0x24) & 0x3)
    usleep(1000);
  uint32_t control1 = mmio_read(0x20300000 + 0x2C);
  control1 &= ~(1 << 2);
  mmio_write(0x20300000 + 0x2C, control1);
  usleep(2000);
  control1 &= ~0xffe0;
  control1 |= divider;
  mmio_write(0x20300000 + 0x2C, control1);
  usleep(2000);
  control1 |= (1 << 2);
  mmio_write(0x20300000 + 0x2C, control1);
  usleep(2000);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<worker>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  lea    D8(%rip),%rax
  mov    %rax,%rsi
  lea    D17(%rip),%rax
  mov    %rax,%rdi
  call   <fopen@plt>
  mov    %rax,-0x18(%rbp)
  movq   $0x100000,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
L4:
  mov    -0x18(%rbp),%rdx
  mov    -0x10(%rbp),%rsi
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rcx
  mov    $0x1,%edx
  mov    %rax,%rdi
  call   <fread@plt>
  jmp    L4
```

tool_calls: name: read_data
arguments: data_label: D8
type: string
name: read_data
arguments: data_label: D17
type: string

tool: D8: r

tool: D17: /dev/zero

assistant: ```c
void *worker(void *arg) {
  FILE *fp = fopen("/dev/zero", "r");
  size_t size = 1024 * 1024 * sizeof(char);
  char *buffer = malloc(size);
  while (true) {
    fread(buffer, size, 1, fp);
  }
  free(buffer);
  return NULL;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<worker>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  lea    D12(%rip),%rsi
  lea    D1(%rip),%rdi
  call   <fopen@plt>
  mov    %rax,%rbp
  mov    $0x100000,%edi
  call   <malloc@plt>
  mov    %rax,%rbx
L8:
  mov    %rbp,%rcx
  mov    $0x1,%edx
  mov    $0x100000,%esi
  mov    %rbx,%rdi
  call   <fread@plt>
  jmp    L8
```

tool_calls: name: retrieve_data
arguments: data_tag: D12
datatype: string
name: retrieve_data
arguments: data_tag: D1
datatype: string

tool: D12: r

tool: D1: /dev/zero

assistant: ```c
void *worker(void *arg) {
  FILE *fp = fopen("/dev/zero", "r");
  size_t size = 1024 * 1024 * sizeof(char);
  char *buffer = malloc(size);
  while (true) {
    fread(buffer, size, 1, fp);
  }
  free(buffer);
  return NULL;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<worker>:
  endbr64
  push   %rbp
  lea    D2(%rip),%rsi
  lea    D17(%rip),%rdi
  push   %rbx
  sub    $0x8,%rsp
  call   <fopen@plt>
  mov    $0x100000,%edi
  mov    %rax,%rbp
  call   <malloc@plt>
  mov    %rax,%rbx
  nopl   (%rax)
L5:
  mov    %rbp,%rcx
  mov    $0x1,%edx
  mov    $0x100000,%esi
  mov    %rbx,%rdi
  call   <fread@plt>
  jmp    L5
```

tool_calls: name: disassemble_data
arguments: data_tag: D2
type_infered: string
name: disassemble_data
arguments: data_tag: D17
type_infered: string

tool: D2: r

tool: D17: /dev/zero

assistant: ```c
void *worker(void *arg) {
  FILE *fp = fopen("/dev/zero", "r");
  size_t size = 1024 * 1024 * sizeof(char);
  char *buffer = malloc(size);
  while (true) {
    fread(buffer, size, 1, fp);
  }
  free(buffer);
  return NULL;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<worker>:
  endbr64
  push   %rbp
  lea    D18(%rip),%rsi
  lea    D17(%rip),%rdi
  push   %rbx
  sub    $0x8,%rsp
  call   <fopen@plt>
  mov    $0x100000,%edi
  mov    %rax,%rbp
  call   <malloc@plt>
  mov    %rax,%rbx
  nopl   (%rax)
L2:
  mov    %rbp,%rcx
  mov    $0x1,%edx
  mov    $0x100000,%esi
  mov    %rbx,%rdi
  call   <fread@plt>
  jmp    L2
```

tool_calls: name: extract_symbolic_data
arguments: tag: D18
type_infered: string
name: extract_symbolic_data
arguments: tag: D17
type_infered: string

tool: D18: r

tool: D17: /dev/zero

assistant: ```c
void *worker(void *arg) {
  FILE *fp = fopen("/dev/zero", "r");
  size_t size = 1024 * 1024 * sizeof(char);
  char *buffer = malloc(size);
  while (true) {
    fread(buffer, size, 1, fp);
  }
  free(buffer);
  return NULL;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<doubles>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <string_length@plt>
  cmp    $0x1,%eax
  jle    L59
  lea    0x1(%rbx),%rdx
  lea    -0x1(%rax),%edi
  mov    $0x0,%esi
  jmp    L117
L116:
  movzbl (%rdx),%eax
  cmp    $0x2d,%al
  je     L71
  cmp    $0x2b,%al
  je     L127
L62:
  add    $0x1,%esi
  add    $0x1,%rdx
  cmp    %edi,%esi
  je     L19
L117:
  movzbl -0x1(%rdx),%eax
  cmp    $0x2b,%al
  je     L116
  cmp    $0x2d,%al
  jne    L62
  movzbl (%rdx),%eax
  cmp    $0x2b,%al
  je     L72
  cmp    $0x2d,%al
  jne    L62
  lea    0x1(%rsi),%edx
  lea    D8(%rip),%rcx
  mov    %rbx,%rdi
  call   <replace_sub_string@plt>
  jmp    L129
L71:
  lea    0x1(%rsi),%edx
  lea    D37(%rip),%rcx
  mov    %rbx,%rdi
  call   <replace_sub_string@plt>
  jmp    L129
L127:
  lea    0x1(%rsi),%edx
  lea    D8(%rip),%rcx
  mov    %rbx,%rdi
  call   <replace_sub_string@plt>
  jmp    L129
L72:
  lea    0x1(%rsi),%edx
  lea    D37(%rip),%rcx
  mov    %rbx,%rdi
  call   <replace_sub_string@plt>
  jmp    L129
L19:
  mov    %rbx,%rax
L129:
  pop    %rbx
  ret
L59:
  mov    %rbx,%rax
  jmp    L129
```

tool_calls: name: extract_symbolic_data
arguments: label: D8
datatype: string
name: extract_symbolic_data
arguments: label: D37
datatype: string

tool: D8: +

tool: D37: -

assistant: ```c
char *doubles(char buffer[]) {
  int length = string_length(buffer);
  int count;
  for (count = 0; count < length - 1; count++) {
    if ((buffer[count] == '+') && (buffer[count + 1] == '-')) {
      buffer = replace_sub_string(buffer, count, count + 1, "-");
      break;
    }
    if ((buffer[count] == '+') && (buffer[count + 1] == '+')) {
      buffer = replace_sub_string(buffer, count, count + 1, "+");
      break;
    }
    if ((buffer[count] == '-') && (buffer[count + 1] == '+')) {
      buffer = replace_sub_string(buffer, count, count + 1, "-");
      break;
    }
    if ((buffer[count] == '-') && (buffer[count + 1] == '-')) {
      buffer = replace_sub_string(buffer, count, count + 1, "+");
      break;
    }
  }
  return buffer;
}
```
user: Translate the below assembly code into C programming language:

```asm
<doubles>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  call   <string_length@plt>
  cmp    $0x1,%eax
  jle    L45
  lea    -0x2(%rax),%ecx
  xor    %eax,%eax
  jmp    L71
  nopl   0x0(%rax)
L75:
  cmp    $0x2d,%dl
  jne    L43
  movzbl 0x1(%rbp,%rax,1),%edx
  cmp    $0x2b,%dl
  je     L44
  cmp    $0x2d,%dl
  je     L40
L43:
  lea    0x1(%rax),%rdx
  cmp    %rax,%rcx
  je     L45
  mov    %rdx,%rax
L71:
  movzbl 0x0(%rbp,%rax,1),%edx
  mov    %eax,%esi
  cmp    $0x2b,%dl
  jne    L75
  movzbl 0x1(%rbp,%rax,1),%edx
  cmp    $0x2d,%dl
  je     L44
  cmp    $0x2b,%dl
  jne    L43
L40:
  mov    %rbp,%rdi
  lea    0x1(%rsi),%edx
  lea    D2(%rip),%rcx
  pop    %rbp
  jmp    1060 <replace_sub_string@plt>
  xchg   %ax,%ax
L45:
  mov    %rbp,%rax
  pop    %rbp
  ret
  nopl   (%rax)
L44:
  mov    %rbp,%rdi
  lea    0x1(%rsi),%edx
  lea    D14(%rip),%rcx
  pop    %rbp
  jmp    1060 <replace_sub_string@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D2
data_type: string
name: disassemble_data
arguments: data_tag: D14
data_type: string

tool: D2: +

tool: D14: -

assistant: ```c
char *doubles(char buffer[]) {
  int length = string_length(buffer);
  int count;
  for (count = 0; count < length - 1; count++) {
    if ((buffer[count] == '+') && (buffer[count + 1] == '-')) {
      buffer = replace_sub_string(buffer, count, count + 1, "-");
      break;
    }
    if ((buffer[count] == '+') && (buffer[count + 1] == '+')) {
      buffer = replace_sub_string(buffer, count, count + 1, "+");
      break;
    }
    if ((buffer[count] == '-') && (buffer[count + 1] == '+')) {
      buffer = replace_sub_string(buffer, count, count + 1, "-");
      break;
    }
    if ((buffer[count] == '-') && (buffer[count + 1] == '-')) {
      buffer = replace_sub_string(buffer, count, count + 1, "+");
      break;
    }
  }
  return buffer;
}
```
user: Translate the below assembly code into C programming language:

```asm
<doubles>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  call   <string_length@plt>
  cmp    $0x1,%eax
  jle    L12
  lea    -0x2(%rax),%ecx
  xor    %eax,%eax
  jmp    L73
  nopl   0x0(%rax)
L57:
  cmp    $0x2d,%dl
  jne    L38
  movzbl 0x1(%rbp,%rax,1),%edx
  cmp    $0x2b,%dl
  je     L29
  cmp    $0x2d,%dl
  je     L83
L38:
  lea    0x1(%rax),%rdx
  cmp    %rax,%rcx
  je     L12
  mov    %rdx,%rax
L73:
  movzbl 0x0(%rbp,%rax,1),%edx
  mov    %eax,%esi
  cmp    $0x2b,%dl
  jne    L57
  movzbl 0x1(%rbp,%rax,1),%edx
  cmp    $0x2d,%dl
  je     L29
  cmp    $0x2b,%dl
  jne    L38
L83:
  mov    %rbp,%rdi
  lea    0x1(%rsi),%edx
  lea    D17(%rip),%rcx
  pop    %rbp
  jmp    1060 <replace_sub_string@plt>
  xchg   %ax,%ax
L12:
  mov    %rbp,%rax
  pop    %rbp
  ret
  nopl   (%rax)
L29:
  mov    %rbp,%rdi
  lea    0x1(%rsi),%edx
  lea    D1(%rip),%rcx
  pop    %rbp
  jmp    1060 <replace_sub_string@plt>
```

tool_calls: name: unpack_data
arguments: tag: D17
type_infered: string
name: unpack_data
arguments: tag: D1
type_infered: string

tool: D17: +

tool: D1: -

assistant: ```c
char *doubles(char buffer[]) {
  int length = string_length(buffer);
  int count;
  for (count = 0; count < length - 1; count++) {
    if ((buffer[count] == '+') && (buffer[count + 1] == '-')) {
      buffer = replace_sub_string(buffer, count, count + 1, "-");
      break;
    }
    if ((buffer[count] == '+') && (buffer[count + 1] == '+')) {
      buffer = replace_sub_string(buffer, count, count + 1, "+");
      break;
    }
    if ((buffer[count] == '-') && (buffer[count + 1] == '+')) {
      buffer = replace_sub_string(buffer, count, count + 1, "-");
      break;
    }
    if ((buffer[count] == '-') && (buffer[count + 1] == '-')) {
      buffer = replace_sub_string(buffer, count, count + 1, "+");
      break;
    }
  }
  return buffer;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<kaprekar>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,%eax
  mov    %ax,-0x14(%rbp)
  cmpw   $0x181e,-0x14(%rbp)
  jne    L12
  mov    $0x0,%eax
  jmp    L35
L12:
  movswl -0x14(%rbp),%eax
  mov    %eax,%edi
  call   <desc_digits@plt>
  mov    %ax,-0x4(%rbp)
  movswl -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <asc_digits@plt>
  mov    %ax,-0x2(%rbp)
  movb   $0x1,-0x5(%rbp)
  jmp    L31
L8:
  movzwl -0x4(%rbp),%eax
  movzwl -0x2(%rbp),%edx
  sub    %edx,%eax
  cwtl
  mov    %eax,%edi
  call   <desc_digits@plt>
  mov    %ax,-0x4(%rbp)
  movswl -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <asc_digits@plt>
  mov    %ax,-0x2(%rbp)
  movzbl -0x5(%rbp),%eax
  add    $0x1,%eax
  mov    %al,-0x5(%rbp)
L31:
  movswl -0x4(%rbp),%eax
  movswl -0x2(%rbp),%edx
  sub    %edx,%eax
  cmp    $0x181e,%eax
  jne    L8
  movzbl -0x5(%rbp),%eax
L35:
  leave
  ret
```

assistant: ```c
char kaprekar(short num) {
  if (num == 6174)
    return 0;
  short big = desc_digits(num);
  short small = asc_digits(big);
  char count = 1;
  while (big - small != 6174) {
    big = desc_digits(big - small);
    small = asc_digits(big);
    count++;
  }
  return count;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<kaprekar>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    $0x0,%r12d
  cmp    $0x181e,%di
  je     L20
  movswl %di,%edi
  call   <desc_digits@plt>
  mov    %eax,%ebx
  movswl %ax,%ebp
  mov    %ebp,%edi
  call   <asc_digits@plt>
  movswl %ax,%edx
  sub    %edx,%ebp
  cmp    $0x181e,%ebp
  je     L3
  mov    $0x1,%r12d
L25:
  sub    %eax,%ebx
  movswl %bx,%edi
  call   <desc_digits@plt>
  mov    %eax,%ebx
  movswl %ax,%ebp
  mov    %ebp,%edi
  call   <asc_digits@plt>
  add    $0x1,%r12d
  movswl %ax,%edx
  sub    %edx,%ebp
  cmp    $0x181e,%ebp
  jne    L25
L20:
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L3:
  mov    $0x1,%r12d
  jmp    L20
```

assistant: ```c
char kaprekar(short num) {
  if (num == 6174)
    return 0;
  short big = desc_digits(num);
  short small = asc_digits(big);
  char count = 1;
  while (big - small != 6174) {
    big = desc_digits(big - small);
    small = asc_digits(big);
    count++;
  }
  return count;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<kaprekar>:
  endbr64
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  push   %rbx
  cmp    $0x181e,%di
  je     L19
  movswl %di,%edi
  mov    $0x1,%r12d
  call   <desc_digits@plt>
  movswl %ax,%ebp
  mov    %ebp,%edi
  mov    %ebp,%ebx
  call   <asc_digits@plt>
  movswl %ax,%edx
  sub    %edx,%ebp
  cmp    $0x181e,%ebp
  je     L19
  nopl   0x0(%rax)
L22:
  sub    %eax,%ebx
  add    $0x1,%r12d
  movswl %bx,%edi
  call   <desc_digits@plt>
  movswl %ax,%ebp
  mov    %ebp,%edi
  mov    %ebp,%ebx
  call   <asc_digits@plt>
  movswl %ax,%edx
  sub    %edx,%ebp
  cmp    $0x181e,%ebp
  jne    L22
L19:
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char kaprekar(short num) {
  if (num == 6174)
    return 0;
  short big = desc_digits(num);
  short small = asc_digits(big);
  char count = 1;
  while (big - small != 6174) {
    big = desc_digits(big - small);
    small = asc_digits(big);
    count++;
  }
  return count;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<kaprekar>:
  endbr64
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  push   %rbx
  cmp    $0x181e,%di
  je     L19
  movswl %di,%edi
  mov    $0x1,%r12d
  call   <desc_digits@plt>
  movswl %ax,%ebp
  mov    %ebp,%edi
  mov    %ebp,%ebx
  call   <asc_digits@plt>
  movswl %ax,%edx
  sub    %edx,%ebp
  cmp    $0x181e,%ebp
  je     L19
  nopl   0x0(%rax)
L28:
  sub    %eax,%ebx
  add    $0x1,%r12d
  movswl %bx,%edi
  call   <desc_digits@plt>
  movswl %ax,%ebp
  mov    %ebp,%edi
  mov    %ebp,%ebx
  call   <asc_digits@plt>
  movswl %ax,%edx
  sub    %edx,%ebp
  cmp    $0x181e,%ebp
  jne    L28
L19:
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char kaprekar(short num) {
  if (num == 6174)
    return 0;
  short big = desc_digits(num);
  short small = asc_digits(big);
  char count = 1;
  while (big - small != 6174) {
    big = desc_digits(big - small);
    small = asc_digits(big);
    count++;
  }
  return count;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<VerifSomaLados>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x4(%rbp)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x18(%rbp),%rax
  add    $0x14,%rax
  mov    (%rax),%eax
  add    %edx,%eax
  cmp    $0x7,%eax
  je     L41
  movl   $0x1,-0x4(%rbp)
  jmp    L29
L41:
  mov    -0x18(%rbp),%rax
  add    $0x4,%rax
  mov    (%rax),%edx
  mov    -0x18(%rbp),%rax
  add    $0xc,%rax
  mov    (%rax),%eax
  add    %edx,%eax
  cmp    $0x7,%eax
  je     L11
  movl   $0x1,-0x4(%rbp)
  jmp    L29
L11:
  mov    -0x18(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%edx
  mov    -0x18(%rbp),%rax
  add    $0x10,%rax
  mov    (%rax),%eax
  add    %edx,%eax
  cmp    $0x7,%eax
  je     L29
  movl   $0x1,-0x4(%rbp)
L29:
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int VerifSomaLados(int *dados) {
  int bolean = 0;
  if (dados[0] + dados[5] != 7)
    bolean = 1;
  else if (dados[1] + dados[3] != 7)
    bolean = 1;
  else if (dados[2] + dados[4] != 7)
    bolean = 1;
  return bolean;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<VerifSomaLados>:
  endbr64
  mov    0x14(%rdi),%eax
  add    (%rdi),%eax
  mov    $0x1,%edx
  cmp    $0x7,%eax
  je     L5
L20:
  mov    %edx,%eax
  ret
L5:
  mov    0xc(%rdi),%eax
  add    0x4(%rdi),%eax
  cmp    $0x7,%eax
  jne    L20
  mov    0x10(%rdi),%eax
  add    0x8(%rdi),%eax
  cmp    $0x7,%eax
  setne  %dl
  movzbl %dl,%edx
  jmp    L20
```

assistant: ```c
int VerifSomaLados(int *dados) {
  int bolean = 0;
  if (dados[0] + dados[5] != 7)
    bolean = 1;
  else if (dados[1] + dados[3] != 7)
    bolean = 1;
  else if (dados[2] + dados[4] != 7)
    bolean = 1;
  return bolean;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<VerifSomaLados>:
  endbr64
  mov    0x14(%rdi),%eax
  add    (%rdi),%eax
  mov    $0x1,%r8d
  cmp    $0x7,%eax
  je     L3
L5:
  mov    %r8d,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L3:
  mov    0xc(%rdi),%eax
  add    0x4(%rdi),%eax
  cmp    $0x7,%eax
  jne    L5
  mov    0x10(%rdi),%eax
  add    0x8(%rdi),%eax
  xor    %r8d,%r8d
  cmp    $0x7,%eax
  setne  %r8b
  mov    %r8d,%eax
  ret
```

assistant: ```c
int VerifSomaLados(int *dados) {
  int bolean = 0;
  if (dados[0] + dados[5] != 7)
    bolean = 1;
  else if (dados[1] + dados[3] != 7)
    bolean = 1;
  else if (dados[2] + dados[4] != 7)
    bolean = 1;
  return bolean;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<VerifSomaLados>:
  endbr64
  mov    0x14(%rdi),%eax
  add    (%rdi),%eax
  mov    $0x1,%r8d
  cmp    $0x7,%eax
  je     L6
L19:
  mov    %r8d,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L6:
  mov    0xc(%rdi),%eax
  add    0x4(%rdi),%eax
  cmp    $0x7,%eax
  jne    L19
  mov    0x10(%rdi),%eax
  add    0x8(%rdi),%eax
  xor    %r8d,%r8d
  cmp    $0x7,%eax
  setne  %r8b
  mov    %r8d,%eax
  ret
```

assistant: ```c
int VerifSomaLados(int *dados) {
  int bolean = 0;
  if (dados[0] + dados[5] != 7)
    bolean = 1;
  else if (dados[1] + dados[3] != 7)
    bolean = 1;
  else if (dados[2] + dados[4] != 7)
    bolean = 1;
  return bolean;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<align64>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  neg    %rax
  mov    %rax,-0x8(%rbp)
  mov    -0x20(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  sub    $0x1,%rax
  and    -0x8(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
unsigned long long align64(unsigned long long offset, unsigned long long alignment) {
  unsigned long long mask = ~(alignment - 1);
  return (offset + (alignment - 1)) & mask;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<align64>:
  endbr64
  lea    -0x1(%rsi,%rdi,1),%rax
  neg    %rsi
  and    %rsi,%rax
  ret
```

assistant: ```c
unsigned long long align64(unsigned long long offset, unsigned long long alignment) {
  unsigned long long mask = ~(alignment - 1);
  return (offset + (alignment - 1)) & mask;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<align64>:
  endbr64
  lea    -0x1(%rsi,%rdi,1),%rax
  neg    %rsi
  and    %rsi,%rax
  ret
```

assistant: ```c
unsigned long long align64(unsigned long long offset, unsigned long long alignment) {
  unsigned long long mask = ~(alignment - 1);
  return (offset + (alignment - 1)) & mask;
}
```
user: Translate the below assembly code into C programming language:

```asm
<print_sys_cfg_rates_usage>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D85(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D87(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D47(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D67(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D81(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D36(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D13(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D46(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D83(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: data_label: D85
type_infered: string
name: parse_data
arguments: data_label: D87
type_infered: string
name: parse_data
arguments: data_label: D47
type_infered: string
name: parse_data
arguments: data_label: D67
type_infered: string
name: parse_data
arguments: data_label: D81
type_infered: string
name: parse_data
arguments: data_label: D36
type_infered: string
name: parse_data
arguments: data_label: D13
type_infered: string
name: parse_data
arguments: data_label: D46
type_infered: string
name: parse_data
arguments: data_label: D83
type_infered: string

tool: D85: 
Usage : sys_cfg_rates [RATES]


tool: D87: 
[RATES] is set of data rates in unit of 500 kbps and each rate can be

tool: D47: 
entered in hexadecimal or decimal format. Rates must be separated by

tool: D67: 
space. Duplicate Rate  fields are not allowed

tool: D81: 
A rate with MSB bit is basic rate, i.e 0x82 is basic rate.

tool: D36: 
Following is the list of supported rates in units of 500 Kbps:

tool: D13: 
Decimal: (2, 4, 11, 22, 12, 18, 24, 36, 48, 72, 96, 108)

tool: D46: 
Hex: (0x02, 0x04, 0x0b, 0x16, 0x0C, 0x12, 0x18, 0x24, 0x30, 0x48, 0x60, 0x6c)

tool: D83: 
Basic rates: (0x82, 0x84, 0x8b, 0x96, 0x8C, 0x92, 0x98, 0xA4, 0xB0, 0xC8, 0xE0, 0xEc)


assistant: ```c
void print_sys_cfg_rates_usage(void) {
  printf("\nUsage : sys_cfg_rates [RATES]\n");
  printf("\n[RATES] is set of data rates in unit of 500 kbps and each rate can be");
  printf("\nentered in hexadecimal or decimal format. Rates must be separated by");
  printf("\nspace. Duplicate Rate  fields are not allowed");
  printf("\nA rate with MSB bit is basic rate, i.e 0x82 is basic rate.");
  printf("\nFollowing is the list of supported rates in units of 500 Kbps:");
  printf("\nDecimal: (2, 4, 11, 22, 12, 18, 24, 36, 48, 72, 96, 108)");
  printf("\nHex: (0x02, 0x04, 0x0b, 0x16, 0x0C, 0x12, 0x18, 0x24, 0x30, 0x48, 0x60, 0x6c)");
  printf("\nBasic rates: (0x82, 0x84, 0x8b, 0x96, 0x8C, 0x92, 0x98, 0xA4, 0xB0, 0xC8, 0xE0, 0xEc)\n");
  return;
}
```