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
<main>:
  endbr64
  sub    $0x28,%rsp
  lea    D14(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  lea    0x14(%rsp),%rcx
  lea    0xf(%rsp),%rdx
  lea    0x10(%rsp),%rsi
  lea    D29(%rip),%rdi
  call   <scanf@plt>
  movzbl 0xf(%rsp),%eax
  cmp    $0x2b,%al
  je     L68
  cmp    $0x2d,%al
  je     L32
  cmp    $0x2a,%al
  je     L63
  cmp    $0x2f,%al
  je     L65
  lea    D41(%rip),%rdi
  call   <puts@plt>
L54:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L47
  xor    %eax,%eax
  add    $0x28,%rsp
  ret
L68:
  movss  0x10(%rsp),%xmm0
  addss  0x14(%rsp),%xmm0
  mov    $0x1,%eax
  lea    D21(%rip),%rdi
  cvtss2sd %xmm0,%xmm0
  call   <printf@plt>
  jmp    L54
L65:
  lea    D21(%rip),%rdi
  mov    $0x1,%eax
  movss  0x10(%rsp),%xmm0
  divss  0x14(%rsp),%xmm0
  cvtss2sd %xmm0,%xmm0
  call   <printf@plt>
  jmp    L54
L32:
  movss  0x10(%rsp),%xmm0
  subss  0x14(%rsp),%xmm0
  mov    $0x1,%eax
  lea    D21(%rip),%rdi
  cvtss2sd %xmm0,%xmm0
  call   <printf@plt>
  jmp    L54
L63:
  movss  0x10(%rsp),%xmm0
  mulss  0x14(%rsp),%xmm0
  mov    $0x1,%eax
  lea    D21(%rip),%rdi
  cvtss2sd %xmm0,%xmm0
  call   <printf@plt>
  jmp    L54
L47:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D14
type_guessed: string
name: retrieve_data
arguments: data_label: D29
type_guessed: string
name: retrieve_data
arguments: data_label: D41
type_guessed: string
name: retrieve_data
arguments: data_label: D21
type_guessed: string

tool: D14: Type in your expression.

tool: D29: %f %c %f

tool: D41: Unknown operator.

tool: D21: %.2f


assistant: ```c
int main(void) {
  float value1, value2;
  char operator;
  printf("Type in your expression.\n");
  scanf("%f %c %f", &value1, &operator, & value2);
  if (operator== '+')
    printf("%.2f\n", value1 + value2);
  else if (operator== '-')
    printf("%.2f\n", value1 - value2);
  else if (operator== '*')
    printf("%.2f\n", value1 * value2);
  else if (operator== '/')
    printf("%.2f\n", value1 / value2);
  else
    printf("Unknown operator.\n");
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ft_strmergeflag>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  mov    -0x20(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <ft_strjoin@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x24(%rbp),%eax
  and    $0x1,%eax
  test   %eax,%eax
  je     L17
  lea    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <ft_strdel@plt>
L17:
  mov    -0x24(%rbp),%eax
  and    $0x2,%eax
  test   %eax,%eax
  je     L7
  lea    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <ft_strdel@plt>
L7:
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *ft_strmergeflag(char *a, char *b, int flag) {
  char *ret;
  ret = ft_strjoin(a, b);
  if (flag & 1)
    ft_strdel(&a);
  if (flag & 2)
    ft_strdel(&b);
  return (ret);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<ft_strmergeflag>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,0x8(%rsp)
  mov    %rsi,(%rsp)
  mov    %edx,%ebx
  call   <ft_strjoin@plt>
  mov    %rax,%rbp
  test   $0x1,%bl
  jne    L26
L2:
  test   $0x2,%bl
  jne    L22
L36:
  mov    %rbp,%rax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L26:
  lea    0x8(%rsp),%rdi
  call   <ft_strdel@plt>
  jmp    L2
L22:
  mov    %rsp,%rdi
  call   <ft_strdel@plt>
  jmp    L36
```

assistant: ```c
char *ft_strmergeflag(char *a, char *b, int flag) {
  char *ret;
  ret = ft_strjoin(a, b);
  if (flag & 1)
    ft_strdel(&a);
  if (flag & 2)
    ft_strdel(&b);
  return (ret);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<ft_strmergeflag>:
  endbr64
  push   %r12
  push   %rbx
  mov    %edx,%ebx
  sub    $0x18,%rsp
  mov    %rdi,0x8(%rsp)
  mov    %rsi,(%rsp)
  call   <ft_strjoin@plt>
  mov    %rax,%r12
  test   $0x1,%bl
  jne    L3
  and    $0x2,%ebx
  jne    L15
L28:
  add    $0x18,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L3:
  lea    0x8(%rsp),%rdi
  call   <ft_strdel@plt>
  and    $0x2,%ebx
  je     L28
L15:
  mov    %rsp,%rdi
  call   <ft_strdel@plt>
  add    $0x18,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  ret
```

assistant: ```c
char *ft_strmergeflag(char *a, char *b, int flag) {
  char *ret;
  ret = ft_strjoin(a, b);
  if (flag & 1)
    ft_strdel(&a);
  if (flag & 2)
    ft_strdel(&b);
  return (ret);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<ft_strmergeflag>:
  endbr64
  push   %r12
  push   %rbx
  mov    %edx,%ebx
  sub    $0x18,%rsp
  mov    %rdi,0x8(%rsp)
  mov    %rsi,(%rsp)
  call   <ft_strjoin@plt>
  mov    %rax,%r12
  test   $0x1,%bl
  jne    L3
  and    $0x2,%ebx
  jne    L27
L29:
  add    $0x18,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L3:
  lea    0x8(%rsp),%rdi
  call   <ft_strdel@plt>
  and    $0x2,%ebx
  je     L29
L27:
  mov    %rsp,%rdi
  call   <ft_strdel@plt>
  add    $0x18,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  ret
```

assistant: ```c
char *ft_strmergeflag(char *a, char *b, int flag) {
  char *ret;
  ret = ft_strjoin(a, b);
  if (flag & 1)
    ft_strdel(&a);
  if (flag & 2)
    ft_strdel(&b);
  return (ret);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<aqpipe>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <pipe@plt>
  cmp    $0xffffffff,%eax
  jne    L7
  mov    $0xffffffff,%eax
  jmp    L59
L7:
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,%edi
  call   <__make_fd_async@plt>
  cmp    $0xffffffff,%eax
  jne    L40
  mov    D27(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,%edi
  call   <close@plt>
  mov    -0x18(%rbp),%rax
  add    $0x4,%rax
  mov    (%rax),%eax
  mov    %eax,%edi
  call   <close@plt>
  mov    D27(%rip),%rax
  mov    -0x4(%rbp),%edx
  mov    %edx,(%rax)
  mov    $0xffffffff,%eax
  jmp    L59
L40:
  mov    -0x18(%rbp),%rax
  add    $0x4,%rax
  mov    (%rax),%eax
  mov    %eax,%edi
  call   <__make_fd_async@plt>
  cmp    $0xffffffff,%eax
  jne    L42
  mov    D27(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,%edi
  call   <close@plt>
  mov    -0x18(%rbp),%rax
  add    $0x4,%rax
  mov    (%rax),%eax
  mov    %eax,%edi
  call   <close@plt>
  mov    D27(%rip),%rax
  mov    -0x8(%rbp),%edx
  mov    %edx,(%rax)
  mov    $0xffffffff,%eax
  jmp    L59
L42:
  mov    $0x0,%eax
L59:
  leave
  ret
```

assistant: ```c
int aqpipe(int fd[2]) {
  if ((pipe(fd)) == -1) {
    return -1;
  };
  if ((__make_fd_async(fd[0])) == -1) {
    int tmp_errno = errno;
    do {
      {
        close(fd[0]);
        close(fd[1]);
      }
    } while (0);
    errno = tmp_errno;
    return -1;
  };
  if ((__make_fd_async(fd[1])) == -1) {
    int tmp_errno = errno;
    do {
      {
        close(fd[0]);
        close(fd[1]);
      }
    } while (0);
    errno = tmp_errno;
    return -1;
  };
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<aqpipe>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  call   <pipe@plt>
  mov    %eax,%ebx
  cmp    $0xffffffff,%eax
  je     L13
  mov    0x0(%rbp),%edi
  call   <__make_fd_async@plt>
  mov    %eax,%ebx
  cmp    $0xffffffff,%eax
  je     L23
  mov    0x4(%rbp),%edi
  call   <__make_fd_async@plt>
  mov    %eax,%ebx
  cmp    $0xffffffff,%eax
  je     L30
  mov    $0x0,%ebx
L13:
  mov    %ebx,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L23:
  mov    D15(%rip),%r12
  mov    (%r12),%r13d
  mov    0x0(%rbp),%edi
  call   <close@plt>
  mov    0x4(%rbp),%edi
  call   <close@plt>
  mov    %r13d,(%r12)
  jmp    L13
L30:
  mov    D15(%rip),%r12
  mov    (%r12),%r13d
  mov    0x0(%rbp),%edi
  call   <close@plt>
  mov    0x4(%rbp),%edi
  call   <close@plt>
  mov    %r13d,(%r12)
  jmp    L13
```

assistant: ```c
int aqpipe(int fd[2]) {
  if ((pipe(fd)) == -1) {
    return -1;
  };
  if ((__make_fd_async(fd[0])) == -1) {
    int tmp_errno = errno;
    do {
      {
        close(fd[0]);
        close(fd[1]);
      }
    } while (0);
    errno = tmp_errno;
    return -1;
  };
  if ((__make_fd_async(fd[1])) == -1) {
    int tmp_errno = errno;
    do {
      {
        close(fd[0]);
        close(fd[1]);
      }
    } while (0);
    errno = tmp_errno;
    return -1;
  };
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<aqpipe>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  call   <pipe@plt>
  mov    %eax,%r12d
  cmp    $0xffffffff,%eax
  je     L7
  mov    (%rbx),%edi
  call   <__make_fd_async@plt>
  mov    %eax,%r12d
  cmp    $0xffffffff,%eax
  je     L30
  mov    0x4(%rbx),%edi
  call   <__make_fd_async@plt>
  mov    %eax,%r12d
  cmp    $0xffffffff,%eax
  je     L30
  xor    %r12d,%r12d
L7:
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L30:
  mov    D1(%rip),%rbp
  mov    (%rbx),%edi
  mov    0x0(%rbp),%r13d
  call   <close@plt>
  mov    0x4(%rbx),%edi
  call   <close@plt>
  mov    %r13d,0x0(%rbp)
  jmp    L7
```

assistant: ```c
int aqpipe(int fd[2]) {
  if ((pipe(fd)) == -1) {
    return -1;
  };
  if ((__make_fd_async(fd[0])) == -1) {
    int tmp_errno = errno;
    do {
      {
        close(fd[0]);
        close(fd[1]);
      }
    } while (0);
    errno = tmp_errno;
    return -1;
  };
  if ((__make_fd_async(fd[1])) == -1) {
    int tmp_errno = errno;
    do {
      {
        close(fd[0]);
        close(fd[1]);
      }
    } while (0);
    errno = tmp_errno;
    return -1;
  };
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<aqpipe>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  call   <pipe@plt>
  mov    %eax,%r12d
  cmp    $0xffffffff,%eax
  je     L1
  mov    (%rbx),%edi
  call   <__make_fd_async@plt>
  mov    %eax,%r12d
  cmp    $0xffffffff,%eax
  je     L19
  mov    0x4(%rbx),%edi
  call   <__make_fd_async@plt>
  mov    %eax,%r12d
  cmp    $0xffffffff,%eax
  je     L19
  xor    %r12d,%r12d
L1:
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L19:
  mov    D3(%rip),%rbp
  mov    (%rbx),%edi
  mov    0x0(%rbp),%r13d
  call   <close@plt>
  mov    0x4(%rbx),%edi
  call   <close@plt>
  mov    %r13d,0x0(%rbp)
  jmp    L1
```

assistant: ```c
int aqpipe(int fd[2]) {
  if ((pipe(fd)) == -1) {
    return -1;
  };
  if ((__make_fd_async(fd[0])) == -1) {
    int tmp_errno = errno;
    do {
      {
        close(fd[0]);
        close(fd[1]);
      }
    } while (0);
    errno = tmp_errno;
    return -1;
  };
  if ((__make_fd_async(fd[1])) == -1) {
    int tmp_errno = errno;
    do {
      {
        close(fd[0]);
        close(fd[1]);
      }
    } while (0);
    errno = tmp_errno;
    return -1;
  };
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<decrypte>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L2
L17:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  mov    %eax,%edx
  mov    -0x20(%rbp),%eax
  mov    %edx,%ecx
  sub    %eax,%ecx
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    %ecx,%edx
  mov    %dl,(%rax)
  addl   $0x1,-0x4(%rbp)
L2:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L17
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void decrypte(char texte[], int longueur, int cle) {
  for (int i = 0; i < longueur; i++) {
    texte[i] = texte[i] - cle;
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<decrypte>:
  endbr64
  test   %esi,%esi
  jle    L5
  mov    %rdi,%rax
  lea    -0x1(%rsi),%ecx
  lea    0x1(%rdi,%rcx,1),%rcx
L16:
  sub    %dl,(%rax)
  add    $0x1,%rax
  cmp    %rcx,%rax
  jne    L16
L5:
  ret
```

assistant: ```c
void decrypte(char texte[], int longueur, int cle) {
  for (int i = 0; i < longueur; i++) {
    texte[i] = texte[i] - cle;
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<decrypte>:
  endbr64
  test   %esi,%esi
  jle    L7
  lea    -0x1(%rsi),%eax
  lea    0x1(%rdi,%rax,1),%rax
L16:
  sub    %dl,(%rdi)
  add    $0x1,%rdi
  cmp    %rax,%rdi
  jne    L16
L7:
  ret
```

assistant: ```c
void decrypte(char texte[], int longueur, int cle) {
  for (int i = 0; i < longueur; i++) {
    texte[i] = texte[i] - cle;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<decrypte>:
  endbr64
  mov    %rdi,%rcx
  test   %esi,%esi
  jle    L44
  lea    -0x1(%rsi),%eax
  cmp    $0xe,%eax
  jbe    L3
  mov    %rdi,%rax
  mov    %esi,%edi
  movd   %edx,%xmm1
  shr    $0x4,%edi
  punpcklbw %xmm1,%xmm1
  sub    $0x1,%edi
  punpcklwd %xmm1,%xmm1
  shl    $0x4,%rdi
  pshufd $0x0,%xmm1,%xmm1
  lea    0x10(%rcx,%rdi,1),%rdi
L74:
  movdqu (%rax),%xmm0
  add    $0x10,%rax
  psubb  %xmm1,%xmm0
  movups %xmm0,-0x10(%rax)
  cmp    %rax,%rdi
  jne    L74
  mov    %esi,%eax
  and    $0xfffffff0,%eax
  mov    %eax,%edi
  cmp    %eax,%esi
  je     L116
L101:
  mov    %esi,%r10d
  sub    %eax,%r10d
  lea    -0x1(%r10),%r8d
  cmp    $0x6,%r8d
  jbe    L123
  mov    %edx,%r8d
  movzbl %dl,%r9d
  add    %rcx,%rax
  shl    $0x8,%r8d
  movq   (%rax),%xmm0
  or     %r9d,%r8d
  movd   %r8d,%xmm2
  pshuflw $0x0,%xmm2,%xmm1
  psubb  %xmm1,%xmm0
  movq   %xmm0,(%rax)
  mov    %r10d,%eax
  and    $0xfffffff8,%eax
  add    %eax,%edi
  cmp    %eax,%r10d
  je     L44
L123:
  movslq %edi,%rax
  sub    %dl,(%rcx,%rax,1)
  lea    0x1(%rdi),%eax
  cmp    %eax,%esi
  jle    L44
  cltq
  sub    %dl,(%rcx,%rax,1)
  lea    0x2(%rdi),%eax
  cmp    %eax,%esi
  jle    L44
  cltq
  sub    %dl,(%rcx,%rax,1)
  lea    0x3(%rdi),%eax
  cmp    %eax,%esi
  jle    L44
  cltq
  sub    %dl,(%rcx,%rax,1)
  lea    0x4(%rdi),%eax
  cmp    %eax,%esi
  jle    L44
  cltq
  sub    %dl,(%rcx,%rax,1)
  lea    0x5(%rdi),%eax
  cmp    %eax,%esi
  jle    L44
  cltq
  add    $0x6,%edi
  sub    %dl,(%rcx,%rax,1)
  cmp    %edi,%esi
  jle    L44
  movslq %edi,%rdi
  sub    %dl,(%rcx,%rdi,1)
L44:
  ret
L3:
  xor    %eax,%eax
  xor    %edi,%edi
  jmp    L101
L116:
  ret
```

assistant: ```c
void decrypte(char texte[], int longueur, int cle) {
  for (int i = 0; i < longueur; i++) {
    texte[i] = texte[i] - cle;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<tgoto>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  mov    %edx,-0x10(%rbp)
  mov    -0xc(%rbp),%eax
  lea    0x1(%rax),%ecx
  mov    -0x10(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    -0x8(%rbp),%rax
  mov    %ecx,%r8d
  mov    %edx,%ecx
  mov    %rax,%rdx
  mov    $0x1e,%esi
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <snprintf@plt>
  lea    D2(%rip),%rax
  leave
  ret
```

assistant: ```c
char *tgoto(const char *cap, int col, int row) {
  static char buf[30];
  snprintf(buf, sizeof(buf), cap, row + 1, col + 1);
  return (buf);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<tgoto>:
  endbr64
  push   %rbx
  lea    0x1(%rdx),%ecx
  lea    0x1(%rsi),%r8d
  mov    %rdi,%rdx
  mov    $0x1e,%esi
  lea    D7(%rip),%rbx
  mov    %rbx,%rdi
  call   <snprintf@plt>
  mov    %rbx,%rax
  pop    %rbx
  ret
```

assistant: ```c
char *tgoto(const char *cap, int col, int row) {
  static char buf[30];
  snprintf(buf, sizeof(buf), cap, row + 1, col + 1);
  return (buf);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<tgoto>:
  endbr64
  push   %r12
  lea    D3(%rip),%r12
  lea    0x1(%rdx),%ecx
  mov    %rdi,%rdx
  lea    0x1(%rsi),%r8d
  mov    %r12,%rdi
  mov    $0x1e,%esi
  call   <snprintf@plt>
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
char *tgoto(const char *cap, int col, int row) {
  static char buf[30];
  snprintf(buf, sizeof(buf), cap, row + 1, col + 1);
  return (buf);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<tgoto>:
  endbr64
  push   %r12
  lea    D0(%rip),%r12
  lea    0x1(%rdx),%ecx
  mov    %rdi,%rdx
  lea    0x1(%rsi),%r8d
  mov    %r12,%rdi
  mov    $0x1e,%esi
  call   <snprintf@plt>
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
char *tgoto(const char *cap, int col, int row) {
  static char buf[30];
  snprintf(buf, sizeof(buf), cap, row + 1, col + 1);
  return (buf);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<fn_qtg_1_4294967295_cl_48_57>:
  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    L17
L1:
  mov    -0x4(%rbp),%eax
  add    %eax,-0xc(%rbp)
  addl   $0x1,-0x8(%rbp)
L17:
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,%rdi
  call   <fn_cl_48_57@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0xffffffff,-0x4(%rbp)
  jne    L1
  cmpl   $0x0,-0x8(%rbp)
  jle    L6
  mov    -0xc(%rbp),%eax
  jmp    L38
L6:
  mov    $0xffffffff,%eax
L38:
  leave
  ret
```

assistant: ```c
int fn_qtg_1_4294967295_cl_48_57(const char *text) {
  int pos = 0, p = 0, count = 0;
  while ((p = fn_cl_48_57(text + pos)) != -1) {
    pos += p;
    count++;
  }
  if (count >= 1)
    return pos;
  return -1;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<fn_qtg_1_4294967295_cl_48_57>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    $0x0,%ebp
  mov    $0x0,%ebx
  jmp    L19
L4:
  add    %eax,%ebx
  add    $0x1,%ebp
L19:
  movslq %ebx,%rdi
  add    %r12,%rdi
  call   <fn_cl_48_57@plt>
  cmp    $0xffffffff,%eax
  jne    L4
  test   %ebp,%ebp
  cmovle %eax,%ebx
  mov    %ebx,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int fn_qtg_1_4294967295_cl_48_57(const char *text) {
  int pos = 0, p = 0, count = 0;
  while ((p = fn_cl_48_57(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_cl_48_57>:
  endbr64
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  jmp    L17
  nopw   0x0(%rax,%rax,1)
L18:
  add    %eax,%r12d
  add    $0x1,%ebx
L17:
  movslq %r12d,%rdi
  add    %rbp,%rdi
  call   <fn_cl_48_57@plt>
  cmp    $0xffffffff,%eax
  jne    L18
  test   %ebx,%ebx
  pop    %rbx
  pop    %rbp
  cmove  %eax,%r12d
  mov    %r12d,%eax
  pop    %r12
  ret
```

assistant: ```c
int fn_qtg_1_4294967295_cl_48_57(const char *text) {
  int pos = 0, p = 0, count = 0;
  while ((p = fn_cl_48_57(text + pos)) != -1) {
    pos += p;
    count++;
  }
  if (count >= 1)
    return pos;
  return -1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<fn_qtg_1_4294967295_cl_48_57>:
  endbr64
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  jmp    L7
  nopw   0x0(%rax,%rax,1)
L1:
  add    %eax,%r12d
  add    $0x1,%ebx
L7:
  movslq %r12d,%rdi
  add    %rbp,%rdi
  call   <fn_cl_48_57@plt>
  cmp    $0xffffffff,%eax
  jne    L1
  test   %ebx,%ebx
  pop    %rbx
  pop    %rbp
  cmove  %eax,%r12d
  mov    %r12d,%eax
  pop    %r12
  ret
```

assistant: ```c
int fn_qtg_1_4294967295_cl_48_57(const char *text) {
  int pos = 0, p = 0, count = 0;
  while ((p = fn_cl_48_57(text + pos)) != -1) {
    pos += p;
    count++;
  }
  if (count >= 1)
    return pos;
  return -1;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<ffgsdt>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    %rcx,-0x40(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    $0x0,%edi
  call   <time@plt>
  mov    %eax,-0x14(%rbp)
  lea    -0x14(%rbp),%rax
  mov    %rax,%rdi
  call   <gmtime@plt>
  mov    %rax,-0x10(%rbp)
  cmpq   $0x0,-0x10(%rbp)
  jne    L19
  lea    -0x14(%rbp),%rax
  mov    %rax,%rdi
  call   <localtime@plt>
  mov    %rax,-0x10(%rbp)
L19:
  mov    -0x10(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x28(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x10(%rbp),%rax
  mov    0x4(%rax),%eax
  lea    0x1(%rax),%edx
  mov    -0x30(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x10(%rbp),%rax
  mov    0x8(%rax),%eax
  lea    0x76c(%rax),%edx
  mov    -0x38(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x40(%rbp),%rax
  mov    (%rax),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L15
  call   <__stack_chk_fail@plt>
L15:
  leave
  ret
```

assistant: ```c
int ffgsdt(int *day, int *month, int *year, int *status) {
  time_t now;
  struct tm *date;
  now = time(NULL);
  date = gmtime(&now);
  if (!date) {
    date = localtime(&now);
  }
  *day = date->tm_mday;
  *month = date->tm_mon + 1;
  *year = date->tm_year + 1900;
  return (*status);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<ffgsdt>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%r13
  mov    %rsi,%r12
  mov    %rdx,%rbp
  mov    %rcx,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    $0x0,%edi
  call   <time@plt>
  mov    %eax,0x4(%rsp)
  lea    0x4(%rsp),%rdi
  call   <gmtime@plt>
  test   %rax,%rax
  je     L21
L4:
  mov    (%rax),%edx
  mov    %edx,0x0(%r13)
  mov    0x4(%rax),%ecx
  lea    0x1(%rcx),%edx
  mov    %edx,(%r12)
  mov    0x8(%rax),%eax
  add    $0x76c,%eax
  mov    %eax,0x0(%rbp)
  mov    (%rbx),%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L5
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L21:
  lea    0x4(%rsp),%rdi
  call   <localtime@plt>
  jmp    L4
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int ffgsdt(int *day, int *month, int *year, int *status) {
  time_t now;
  struct tm *date;
  now = time(NULL);
  date = gmtime(&now);
  if (!date) {
    date = localtime(&now);
  }
  *day = date->tm_mday;
  *month = date->tm_mon + 1;
  *year = date->tm_year + 1900;
  return (*status);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<ffgsdt>:
  endbr64
  push   %r14
  push   %r13
  mov    %rdi,%r13
  xor    %edi,%edi
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  mov    %rcx,%rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%r14
  call   <time@plt>
  mov    %r14,%rdi
  mov    %eax,0x4(%rsp)
  call   <gmtime@plt>
  test   %rax,%rax
  je     L24
L29:
  mov    (%rax),%edx
  mov    %edx,0x0(%r13)
  mov    0x4(%rax),%ecx
  lea    0x1(%rcx),%edx
  mov    %edx,(%r12)
  mov    0x8(%rax),%eax
  add    $0x76c,%eax
  mov    %eax,0x0(%rbp)
  mov    (%rbx),%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L4
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax,%rax,1)
L24:
  mov    %r14,%rdi
  call   <localtime@plt>
  jmp    L29
L4:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int ffgsdt(int *day, int *month, int *year, int *status) {
  time_t now;
  struct tm *date;
  now = time(NULL);
  date = gmtime(&now);
  if (!date) {
    date = localtime(&now);
  }
  *day = date->tm_mday;
  *month = date->tm_mon + 1;
  *year = date->tm_year + 1900;
  return (*status);
}
```