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

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    $0x1,%ebx
  mov    $0x0,%r12d
  mov    $0x0,%ebp
  jmp    L32
L3:
  add    $0x1,%rbx
  cmp    $0xf4240,%rbx
  je     L26
L32:
  mov    %rbx,%rdi
  call   <calc_length@plt>
  cmp    %ebp,%eax
  jle    L3
  mov    %ebx,%r12d
  mov    %eax,%ebp
  jmp    L3
L26:
  mov    %r12d,%esi
  lea    D8(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D8
datatype: string

tool: D8: %d


assistant: ```c
int main(int argc, char **argv) {
  int longest_length = 0;
  int longest_n = 0;
  for (long i = 1; i < 1000000; ++i) {
    const int len = calc_length(i);
    if (len > longest_length) {
      longest_length = len;
      longest_n = i;
    }
  }
  printf("%d\n", longest_n);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  mov    $0x1,%ebx
  nopw   0x0(%rax,%rax,1)
L18:
  mov    %rbx,%rdi
  call   <calc_length@plt>
  cmp    %ebp,%eax
  jle    L16
  mov    %ebx,%r12d
  mov    %eax,%ebp
L16:
  add    $0x1,%rbx
  cmp    $0xf4240,%rbx
  jne    L18
  mov    %r12d,%esi
  lea    D1(%rip),%rdi
  call   <printf@plt>
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: read_data
arguments: data_label: D1
type_infered: string

tool: D1: %d


assistant: ```c
int main(int argc, char **argv) {
  int longest_length = 0;
  int longest_n = 0;
  for (long i = 1; i < 1000000; ++i) {
    const int len = calc_length(i);
    if (len > longest_length) {
      longest_length = len;
      longest_n = i;
    }
  }
  printf("%d\n", longest_n);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  mov    $0x1,%ebx
  nopw   0x0(%rax,%rax,1)
L1:
  mov    %rbx,%rdi
  call   <calc_length@plt>
  cmp    %ebp,%eax
  jle    L9
  mov    %ebx,%r12d
  mov    %eax,%ebp
L9:
  add    $0x1,%rbx
  cmp    $0xf4240,%rbx
  jne    L1
  mov    %r12d,%esi
  lea    D2(%rip),%rdi
  call   <printf@plt>
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D2
data_type: string

tool: D2: %d


assistant: ```c
int main(int argc, char **argv) {
  int longest_length = 0;
  int longest_n = 0;
  for (long i = 1; i < 1000000; ++i) {
    const int len = calc_length(i);
    if (len > longest_length) {
      longest_length = len;
      longest_n = i;
    }
  }
  printf("%d\n", longest_n);
}
```
user: Translate the below assembly code into C programming language:

```asm
<readLine>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L16
L13:
  addq   $0x1,-0x8(%rbp)
L16:
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    -0x8(%rbp),%rdx
  mov    %al,(%rdx)
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0xa,%al
  jne    L13
  mov    -0x8(%rbp),%rax
  movb   $0x0,(%rax)
  mov    -0x18(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *readLine(char *buffer) {
  char *ch = buffer;
  while ((*ch = getchar()) != '\n')
    ch++;
  *ch = 0;
  return buffer;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<readLine>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    %rdi,%rbx
  jmp    L11
L2:
  add    $0x1,%rbx
L11:
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %al,(%rbx)
  cmp    $0xa,%al
  jne    L2
  movb   $0x0,(%rbx)
  mov    %rbp,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
char *readLine(char *buffer) {
  char *ch = buffer;
  while ((*ch = getchar()) != '\n')
    ch++;
  *ch = 0;
  return buffer;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<readLine>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  jmp    L15
  nopl   0x0(%rax,%rax,1)
L10:
  add    $0x1,%rbx
L15:
  xor    %eax,%eax
  call   <getchar@plt>
  mov    %al,(%rbx)
  cmp    $0xa,%al
  jne    L10
  movb   $0x0,(%rbx)
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  ret
```

assistant: ```c
char *readLine(char *buffer) {
  char *ch = buffer;
  while ((*ch = getchar()) != '\n')
    ch++;
  *ch = 0;
  return buffer;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<readLine>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  jmp    L16
  nopl   0x0(%rax,%rax,1)
L7:
  add    $0x1,%rbx
L16:
  xor    %eax,%eax
  call   <getchar@plt>
  mov    %al,(%rbx)
  cmp    $0xa,%al
  jne    L7
  movb   $0x0,(%rbx)
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  ret
```

assistant: ```c
char *readLine(char *buffer) {
  char *ch = buffer;
  while ((*ch = getchar()) != '\n')
    ch++;
  *ch = 0;
  return buffer;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<c_ivector>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  sub    -0x18(%rbp),%rax
  add    $0x2,%rax
  mov    $0x4,%esi
  mov    %rax,%rdi
  call   <calloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x18(%rbp),%rdx
  mov    $0x1,%eax
  sub    %rdx,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  leave
  ret
```

assistant: ```c
int *c_ivector(nl, nh)
long nh, nl;
{
  int *v;
  v = (int *)calloc(nh - nl + 1 + 1, sizeof(int));
  return v - nl + 1;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<c_ivector>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  sub    %rdi,%rsi
  lea    0x2(%rsi),%rdi
  mov    $0x4,%esi
  call   <calloc@plt>
  mov    %rax,%rdx
  mov    $0x1,%eax
  sub    %rbx,%rax
  lea    (%rdx,%rax,4),%rax
  pop    %rbx
  ret
```

assistant: ```c
int *c_ivector(nl, nh)
long nh, nl;
{
  int *v;
  v = (int *)calloc(nh - nl + 1 + 1, sizeof(int));
  return v - nl + 1;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<c_ivector>:
  endbr64
  push   %rbx
  sub    %rdi,%rsi
  mov    %rdi,%rbx
  lea    0x2(%rsi),%rdi
  mov    $0x4,%esi
  call   <calloc@plt>
  mov    %rax,%r8
  mov    $0x1,%eax
  sub    %rbx,%rax
  pop    %rbx
  lea    (%r8,%rax,4),%rax
  ret
```

assistant: ```c
int *c_ivector(nl, nh)
long nh, nl;
{
  int *v;
  v = (int *)calloc(nh - nl + 1 + 1, sizeof(int));
  return v - nl + 1;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<c_ivector>:
  endbr64
  push   %rbx
  sub    %rdi,%rsi
  mov    %rdi,%rbx
  lea    0x2(%rsi),%rdi
  mov    $0x4,%esi
  call   <calloc@plt>
  mov    %rax,%r8
  mov    $0x1,%eax
  sub    %rbx,%rax
  pop    %rbx
  lea    (%r8,%rax,4),%rax
  ret
```

assistant: ```c
int *c_ivector(nl, nh)
long nh, nl;
{
  int *v;
  v = (int *)calloc(nh - nl + 1 + 1, sizeof(int));
  return v - nl + 1;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x1c(%rbp)
  movl   $0x0,-0x18(%rbp)
  jmp    L39
L30:
  mov    $0x0,%eax
  call   <rand@plt>
  movslq %eax,%rdx
  imul   $0x51eb851f,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x5,%edx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  mov    %edx,-0x14(%rbp)
  mov    -0x14(%rbp),%edx
  imul   $0x64,%edx,%edx
  sub    %edx,%eax
  mov    %eax,-0x14(%rbp)
  mov    $0x0,%eax
  call   <rand@plt>
  movslq %eax,%rdx
  imul   $0x51eb851f,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x5,%edx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  mov    %edx,-0x10(%rbp)
  mov    -0x10(%rbp),%edx
  imul   $0x64,%edx,%edx
  sub    %edx,%eax
  mov    %eax,-0x10(%rbp)
  movl   $0x0,-0x20(%rbp)
  mov    -0x14(%rbp),%edx
  mov    -0x10(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x10(%rbp),%edx
  mov    -0x14(%rbp),%eax
  mov    %eax,%esi
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x20(%rbp),%rax
  mov    %rax,%rsi
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x20(%rbp),%eax
  cmp    %eax,-0xc(%rbp)
  jne    L32
  addl   $0x1,-0x18(%rbp)
L32:
  addl   $0x1,-0x1c(%rbp)
L39:
  cmpl   $0x4,-0x1c(%rbp)
  jle    L30
  mov    -0x18(%rbp),%eax
  mov    %eax,%esi
  lea    D29(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L24
  call   <__stack_chk_fail@plt>
L24:
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_label: D2
type: string
name: parse_data
arguments: data_label: D19
type: string
name: parse_data
arguments: data_label: D29
type: string

tool: D2: Qual eh resultado da soma de %d + %d = 

tool: D19: %d

tool: D29: Total de acerto(s): %d

assistant: ```c
int main() {
  int i = 0, num1, num2, resultado, resposta, acerto = 0;
  while (i <= 4) {
    num1 = rand() % 100;
    num2 = rand() % 100;
    resposta = 0;
    resultado = num1 + num2;
    printf("Qual eh resultado da soma de %d + %d = ", num1, num2);
    scanf("%d", &resposta);
    if (resultado == resposta) {
      acerto += 1;
    }
    i++;
  }
  printf("Total de acerto(s): %d", acerto);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    $0x5,%ebp
  mov    $0x0,%r12d
  lea    D1(%rip),%r15
  lea    0x4(%rsp),%r14
L4:
  mov    $0x0,%eax
  call   <rand@plt>
  movslq %eax,%rbx
  imul   $0x51eb851f,%rbx,%rbx
  sar    $0x25,%rbx
  cltd
  sub    %edx,%ebx
  imul   $0x64,%ebx,%edx
  sub    %edx,%eax
  mov    %eax,%ebx
  mov    $0x0,%eax
  call   <rand@plt>
  movslq %eax,%rdx
  imul   $0x51eb851f,%rdx,%rdx
  sar    $0x25,%rdx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  imul   $0x64,%edx,%ecx
  sub    %ecx,%eax
  mov    %eax,%edx
  movl   $0x0,0x4(%rsp)
  lea    (%rbx,%rax,1),%r13d
  mov    %ebx,%esi
  mov    %r15,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %r14,%rsi
  lea    D28(%rip),%rdi
  call   <scanf@plt>
  cmp    %r13d,0x4(%rsp)
  sete   %al
  movzbl %al,%eax
  add    %eax,%r12d
  sub    $0x1,%ebp
  jne    L4
  mov    %r12d,%esi
  lea    D6(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L15
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L15:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D1
data_type: string
name: retrieve_data
arguments: label: D28
data_type: string
name: retrieve_data
arguments: label: D6
data_type: string

tool: D1: Qual eh resultado da soma de %d + %d = 

tool: D28: %d

tool: D6: Total de acerto(s): %d

assistant: ```c
int main() {
  int i = 0, num1, num2, resultado, resposta, acerto = 0;
  while (i <= 4) {
    num1 = rand() % 100;
    num2 = rand() % 100;
    resposta = 0;
    resultado = num1 + num2;
    printf("Qual eh resultado da soma de %d + %d = ", num1, num2);
    scanf("%d", &resposta);
    if (resultado == resposta) {
      acerto += 1;
    }
    i++;
  }
  printf("Total de acerto(s): %d", acerto);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r15
  xor    %r15d,%r15d
  push   %r14
  lea    D10(%rip),%r14
  push   %r13
  push   %r12
  lea    D23(%rip),%r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movl   $0x5,0xc(%rsp)
  lea    0x14(%rsp),%r13
L17:
  xor    %eax,%eax
  call   <rand@plt>
  movslq %eax,%rbp
  cltd
  imul   $0x51eb851f,%rbp,%rbp
  sar    $0x25,%rbp
  sub    %edx,%ebp
  imul   $0x64,%ebp,%edx
  sub    %edx,%eax
  mov    %eax,%ebp
  xor    %eax,%eax
  call   <rand@plt>
  mov    %ebp,%esi
  mov    %r14,%rdi
  movl   $0x0,0x14(%rsp)
  movslq %eax,%rdx
  mov    %eax,%ecx
  imul   $0x51eb851f,%rdx,%rdx
  sar    $0x1f,%ecx
  sar    $0x25,%rdx
  sub    %ecx,%edx
  imul   $0x64,%edx,%ecx
  sub    %ecx,%eax
  mov    %eax,%edx
  lea    0x0(%rbp,%rax,1),%ebx
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r13,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  xor    %eax,%eax
  cmp    %ebx,0x14(%rsp)
  sete   %al
  add    %eax,%r15d
  subl   $0x1,0xc(%rsp)
  jne    L17
  xor    %eax,%eax
  mov    %r15d,%esi
  lea    D3(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D10
data_type: string
name: unpack_data
arguments: label: D23
data_type: string
name: unpack_data
arguments: label: D3
data_type: string

tool: D10: Qual eh resultado da soma de %d + %d = 

tool: D23: %d

tool: D3: Total de acerto(s): %d

assistant: ```c
int main() {
  int i = 0, num1, num2, resultado, resposta, acerto = 0;
  while (i <= 4) {
    num1 = rand() % 100;
    num2 = rand() % 100;
    resposta = 0;
    resultado = num1 + num2;
    printf("Qual eh resultado da soma de %d + %d = ", num1, num2);
    scanf("%d", &resposta);
    if (resultado == resposta) {
      acerto += 1;
    }
    i++;
  }
  printf("Total de acerto(s): %d", acerto);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %r15
  xor    %r15d,%r15d
  push   %r14
  lea    D20(%rip),%r14
  push   %r13
  push   %r12
  lea    D10(%rip),%r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movl   $0x5,0xc(%rsp)
  lea    0x14(%rsp),%r13
L14:
  xor    %eax,%eax
  call   <rand@plt>
  movslq %eax,%rbp
  cltd
  imul   $0x51eb851f,%rbp,%rbp
  sar    $0x25,%rbp
  sub    %edx,%ebp
  imul   $0x64,%ebp,%edx
  sub    %edx,%eax
  mov    %eax,%ebp
  xor    %eax,%eax
  call   <rand@plt>
  mov    %ebp,%esi
  mov    %r14,%rdi
  movl   $0x0,0x14(%rsp)
  movslq %eax,%rdx
  mov    %eax,%ecx
  imul   $0x51eb851f,%rdx,%rdx
  sar    $0x1f,%ecx
  sar    $0x25,%rdx
  sub    %ecx,%edx
  imul   $0x64,%edx,%ecx
  sub    %ecx,%eax
  mov    %eax,%edx
  lea    0x0(%rbp,%rax,1),%ebx
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r13,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  xor    %eax,%eax
  cmp    %ebx,0x14(%rsp)
  sete   %al
  add    %eax,%r15d
  subl   $0x1,0xc(%rsp)
  jne    L14
  xor    %eax,%eax
  mov    %r15d,%esi
  lea    D29(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D20
datatype: string
name: parse_data
arguments: tag: D10
datatype: string
name: parse_data
arguments: tag: D29
datatype: string

tool: D20: Qual eh resultado da soma de %d + %d = 

tool: D10: %d

tool: D29: Total de acerto(s): %d

assistant: ```c
int main() {
  int i = 0, num1, num2, resultado, resposta, acerto = 0;
  while (i <= 4) {
    num1 = rand() % 100;
    num2 = rand() % 100;
    resposta = 0;
    resultado = num1 + num2;
    printf("Qual eh resultado da soma de %d + %d = ", num1, num2);
    scanf("%d", &resposta);
    if (resultado == resposta) {
      acerto += 1;
    }
    i++;
  }
  printf("Total de acerto(s): %d", acerto);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<string_from_format>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xd0,%rsp
  mov    %rdi,-0xc8(%rbp)
  mov    %rsi,-0xa8(%rbp)
  mov    %rdx,-0xa0(%rbp)
  mov    %rcx,-0x98(%rbp)
  mov    %r8,-0x90(%rbp)
  mov    %r9,-0x88(%rbp)
  test   %al,%al
  je     L5
  movaps %xmm0,-0x80(%rbp)
  movaps %xmm1,-0x70(%rbp)
  movaps %xmm2,-0x60(%rbp)
  movaps %xmm3,-0x50(%rbp)
  movaps %xmm4,-0x40(%rbp)
  movaps %xmm5,-0x30(%rbp)
  movaps %xmm6,-0x20(%rbp)
  movaps %xmm7,-0x10(%rbp)
L5:
  mov    -0xc8(%rbp),%rdx
  mov    -0xbc(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <va_start@plt>
  mov    -0xbc(%rbp),%edx
  mov    -0xc8(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <string_from_vformat@plt>
  mov    %rax,-0xb8(%rbp)
  mov    -0xbc(%rbp),%eax
  mov    %eax,%edi
  call   <va_end@plt>
  mov    -0xb8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *string_from_format(const char *format, ...) {
  char *nuevo;
  va_list arguments;
  va_start(arguments, format);
  nuevo = string_from_vformat(format, arguments);
  va_end(arguments);
  return nuevo;
}
```
user: What is the c source code of the assembly code below?

```asm
<string_from_format>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rdi,%rsi
  mov    $0x0,%ebp
  mov    %ebp,%edi
  call   <va_start@plt>
  mov    %ebp,%esi
  mov    %rbx,%rdi
  call   <string_from_vformat@plt>
  mov    %rax,%rbx
  mov    %ebp,%edi
  call   <va_end@plt>
  mov    %rbx,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
char *string_from_format(const char *format, ...) {
  char *nuevo;
  va_list arguments;
  va_start(arguments, format);
  nuevo = string_from_vformat(format, arguments);
  va_end(arguments);
  return nuevo;
}
```
user: Translate the below assembly code into C programming language:

```asm
<string_from_format>:
  endbr64
  push   %r13
  xor    %r13d,%r13d
  mov    %rdi,%rsi
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  mov    %r13d,%edi
  call   <va_start@plt>
  mov    %r13d,%esi
  mov    %rbp,%rdi
  call   <string_from_vformat@plt>
  mov    %r13d,%edi
  mov    %rax,%r12
  call   <va_end@plt>
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *string_from_format(const char *format, ...) {
  char *nuevo;
  va_list arguments;
  va_start(arguments, format);
  nuevo = string_from_vformat(format, arguments);
  va_end(arguments);
  return nuevo;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<string_from_format>:
  endbr64
  push   %r13
  xor    %r13d,%r13d
  mov    %rdi,%rsi
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  mov    %r13d,%edi
  call   <va_start@plt>
  mov    %r13d,%esi
  mov    %rbp,%rdi
  call   <string_from_vformat@plt>
  mov    %r13d,%edi
  mov    %rax,%r12
  call   <va_end@plt>
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *string_from_format(const char *format, ...) {
  char *nuevo;
  va_list arguments;
  va_start(arguments, format);
  nuevo = string_from_vformat(format, arguments);
  va_end(arguments);
  return nuevo;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<cpuid_eax>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  mov    %edi,-0x1c(%rbp)
  mov    -0x1c(%rbp),%eax
  cpuid
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
unsigned int cpuid_eax(unsigned int op) {
  unsigned int eax;
  __asm__("cpuid" : "=a"(eax) : "0"(op) : "bx", "cx", "dx");
  return eax;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<cpuid_eax>:
  endbr64
  push   %rbx
  mov    %edi,%eax
  cpuid
  pop    %rbx
  ret
```

assistant: ```c
unsigned int cpuid_eax(unsigned int op) {
  unsigned int eax;
  __asm__("cpuid" : "=a"(eax) : "0"(op) : "bx", "cx", "dx");
  return eax;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<cpuid_eax>:
  endbr64
  push   %rbx
  mov    %edi,%eax
  cpuid
  pop    %rbx
  ret
```

assistant: ```c
unsigned int cpuid_eax(unsigned int op) {
  unsigned int eax;
  __asm__("cpuid" : "=a"(eax) : "0"(op) : "bx", "cx", "dx");
  return eax;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<cpuid_eax>:
  endbr64
  push   %rbx
  mov    %edi,%eax
  cpuid
  pop    %rbx
  ret
```

assistant: ```c
unsigned int cpuid_eax(unsigned int op) {
  unsigned int eax;
  __asm__("cpuid" : "=a"(eax) : "0"(op) : "bx", "cx", "dx");
  return eax;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<fn_qtg_1_4294967295_sq_ch_45_qtg_2_8_cl_48_57_65_91_97_122>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0xc(%rbp)
  movl   $0x0,-0x4(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L29
L20:
  mov    -0x4(%rbp),%eax
  add    %eax,-0xc(%rbp)
  addl   $0x1,-0x8(%rbp)
L29:
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,%rdi
  call   <fn_sq_ch_45_qtg_2_8_cl_48_57_65_91_97_122@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0xffffffff,-0x4(%rbp)
  jne    L20
  cmpl   $0x0,-0x8(%rbp)
  jle    L32
  mov    -0xc(%rbp),%eax
  jmp    L34
L32:
  mov    $0xffffffff,%eax
L34:
  leave
  ret
```

assistant: ```c
int fn_qtg_1_4294967295_sq_ch_45_qtg_2_8_cl_48_57_65_91_97_122(const char *text) {
  int pos = 0, p = 0, count = 0;
  while ((p = fn_sq_ch_45_qtg_2_8_cl_48_57_65_91_97_122(text + pos)) != -1) {
    pos += p;
    count++;
  }
  if (count >= 1)
    return pos;
  return -1;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<fn_qtg_1_4294967295_sq_ch_45_qtg_2_8_cl_48_57_65_91_97_122>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    $0x0,%ebp
  mov    $0x0,%ebx
  jmp    L11
L16:
  add    %eax,%ebx
  add    $0x1,%ebp
L11:
  movslq %ebx,%rdi
  add    %r12,%rdi
  call   <fn_sq_ch_45_qtg_2_8_cl_48_57_65_91_97_122@plt>
  cmp    $0xffffffff,%eax
  jne    L16
  test   %ebp,%ebp
  cmovle %eax,%ebx
  mov    %ebx,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int fn_qtg_1_4294967295_sq_ch_45_qtg_2_8_cl_48_57_65_91_97_122(const char *text) {
  int pos = 0, p = 0, count = 0;
  while ((p = fn_sq_ch_45_qtg_2_8_cl_48_57_65_91_97_122(text + pos)) != -1) {
    pos += p;
    count++;
  }
  if (count >= 1)
    return pos;
  return -1;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<fn_qtg_1_4294967295_sq_ch_45_qtg_2_8_cl_48_57_65_91_97_122>:
  endbr64
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  jmp    L9
  nopw   0x0(%rax,%rax,1)
L10:
  add    %eax,%r12d
  add    $0x1,%ebx
L9:
  movslq %r12d,%rdi
  add    %rbp,%rdi
  call   <fn_sq_ch_45_qtg_2_8_cl_48_57_65_91_97_122@plt>
  cmp    $0xffffffff,%eax
  jne    L10
  test   %ebx,%ebx
  pop    %rbx
  pop    %rbp
  cmove  %eax,%r12d
  mov    %r12d,%eax
  pop    %r12
  ret
```

assistant: ```c
int fn_qtg_1_4294967295_sq_ch_45_qtg_2_8_cl_48_57_65_91_97_122(const char *text) {
  int pos = 0, p = 0, count = 0;
  while ((p = fn_sq_ch_45_qtg_2_8_cl_48_57_65_91_97_122(text + pos)) != -1) {
    pos += p;
    count++;
  }
  if (count >= 1)
    return pos;
  return -1;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<fn_qtg_1_4294967295_sq_ch_45_qtg_2_8_cl_48_57_65_91_97_122>:
  endbr64
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  jmp    L9
  nopw   0x0(%rax,%rax,1)
L0:
  add    %eax,%r12d
  add    $0x1,%ebx
L9:
  movslq %r12d,%rdi
  add    %rbp,%rdi
  call   <fn_sq_ch_45_qtg_2_8_cl_48_57_65_91_97_122@plt>
  cmp    $0xffffffff,%eax
  jne    L0
  test   %ebx,%ebx
  pop    %rbx
  pop    %rbp
  cmove  %eax,%r12d
  mov    %r12d,%eax
  pop    %r12
  ret
```

assistant: ```c
int fn_qtg_1_4294967295_sq_ch_45_qtg_2_8_cl_48_57_65_91_97_122(const char *text) {
  int pos = 0, p = 0, count = 0;
  while ((p = fn_sq_ch_45_qtg_2_8_cl_48_57_65_91_97_122(text + pos)) != -1) {
    pos += p;
    count++;
  }
  if (count >= 1)
    return pos;
  return -1;
}
```