ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<copy_2d>:
  endbr64
  imul   %r8,%rcx
  push   %r12
  mov    %rdi,%r12
  mov    (%rsi),%rsi
  mov    (%rdi),%rdi
  imul   %rcx,%rdx
  call   <memcpy@plt>
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
void **copy_2d(void **dst, const void **src, size_t n, size_t m, size_t size) {
  memcpy(dst[0], src[0], n * m * size);
  return dst;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  add    $0x1,%rax
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x10(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    -0x20(%rbp),%rax
  mov    (%rax),%rcx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0x10(%rbp),%rax
  lea    -0x1(%rax),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x31,%al
  jne    L0
  mov    -0x10(%rbp),%rax
  lea    -0x1(%rax),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movb   $0x32,(%rax)
L0:
  mov    -0x8(%rbp),%rcx
  mov    -0x8(%rbp),%rax
  mov    $0x0,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <execl@plt>
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <perror@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: label: D3
type: string

tool: D3: execl failed

assistant: ```c
int main(int argc, char **argv) {
  size_t len;
  char *bin;
  len = strlen(argv[0]);
  bin = malloc(len + 1);
  memcpy(bin, argv[0], len + 1);
  if (bin[len - 1] == '1')
    bin[len - 1] = '2';
  execl(bin, bin, (char *)NULL);
  perror("execl failed");
  exit(1);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    (%rsi),%r12
  mov    %r12,%rdi
  call   <strlen@plt>
  mov    %rax,%rbp
  lea    0x1(%rax),%r13
  mov    %r13,%rdi
  call   <malloc@plt>
  mov    %rax,%rbx
  mov    %r13,%rdx
  mov    %r12,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  lea    -0x1(%rbx,%rbp,1),%rax
  cmpb   $0x31,(%rax)
  je     L5
L15:
  mov    $0x0,%edx
  mov    %rbx,%rsi
  mov    %rbx,%rdi
  call   <execl@plt>
  lea    D2(%rip),%rdi
  call   <perror@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L5:
  movb   $0x32,(%rax)
  jmp    L15
```

tool_calls: name: unpack_data
arguments: data_tag: D2
type_guessed: string

tool: D2: execl failed

assistant: ```c
int main(int argc, char **argv) {
  size_t len;
  char *bin;
  len = strlen(argv[0]);
  bin = malloc(len + 1);
  memcpy(bin, argv[0], len + 1);
  if (bin[len - 1] == '1')
    bin[len - 1] = '2';
  execl(bin, bin, (char *)NULL);
  perror("execl failed");
  exit(1);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    (%rsi),%r12
  mov    %r12,%rdi
  call   <strlen@plt>
  lea    0x1(%rax),%r13
  mov    %rax,%rbx
  mov    %r13,%rdi
  call   <malloc@plt>
  mov    %r13,%rdx
  mov    %r12,%rsi
  mov    %rax,%rbp
  mov    %rax,%rdi
  call   <memcpy@plt>
  lea    -0x1(%rbp,%rbx,1),%rax
  cmpb   $0x31,(%rax)
  jne    L5
  movb   $0x32,(%rax)
L5:
  mov    %rbp,%rsi
  xor    %edx,%edx
  mov    %rbp,%rdi
  call   <execl@plt>
  lea    D9(%rip),%rdi
  call   <perror@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D9
data_type: string

tool: D9: execl failed

assistant: ```c
int main(int argc, char **argv) {
  size_t len;
  char *bin;
  len = strlen(argv[0]);
  bin = malloc(len + 1);
  memcpy(bin, argv[0], len + 1);
  if (bin[len - 1] == '1')
    bin[len - 1] = '2';
  execl(bin, bin, (char *)NULL);
  perror("execl failed");
  exit(1);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    (%rsi),%r12
  mov    %r12,%rdi
  call   <strlen@plt>
  lea    0x1(%rax),%r13
  mov    %rax,%rbx
  mov    %r13,%rdi
  call   <malloc@plt>
  mov    %r13,%rdx
  mov    %r12,%rsi
  mov    %rax,%rbp
  mov    %rax,%rdi
  call   <memcpy@plt>
  lea    -0x1(%rbp,%rbx,1),%rax
  cmpb   $0x31,(%rax)
  jne    L8
  movb   $0x32,(%rax)
L8:
  mov    %rbp,%rsi
  xor    %edx,%edx
  mov    %rbp,%rdi
  call   <execl@plt>
  lea    D4(%rip),%rdi
  call   <perror@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: disassemble_data
arguments: label: D4
data_type: string

tool: D4: execl failed

assistant: ```c
int main(int argc, char **argv) {
  size_t len;
  char *bin;
  len = strlen(argv[0]);
  bin = malloc(len + 1);
  memcpy(bin, argv[0], len + 1);
  if (bin[len - 1] == '1')
    bin[len - 1] = '2';
  execl(bin, bin, (char *)NULL);
  perror("execl failed");
  exit(1);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x42554941,-0xd(%rbp)
  movb   $0x0,-0x9(%rbp)
  lea    -0xd(%rbp),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <rev@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L1
  call   <__stack_chk_fail@plt>
L1:
  leave
  ret
```

assistant: ```c
int main() {
  char b[5] = {"AIUB"};
  rev(b);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0x42554941,0x3(%rsp)
  movb   $0x0,0x7(%rsp)
  lea    0x3(%rsp),%rdi
  call   <rev@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  mov    $0x0,%eax
  add    $0x18,%rsp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main() {
  char b[5] = {"AIUB"};
  rev(b);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x3(%rsp),%rdi
  movl   $0x42554941,0x3(%rsp)
  movb   $0x0,0x7(%rsp)
  call   <rev@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main() {
  char b[5] = {"AIUB"};
  rev(b);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x3(%rsp),%rdi
  movl   $0x42554941,0x3(%rsp)
  movb   $0x0,0x7(%rsp)
  call   <rev@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main() {
  char b[5] = {"AIUB"};
  rev(b);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<unfold_token>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    %rcx,-0x20(%rbp)
  mov    %r8d,%eax
  mov    %al,-0x10(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L22
  cmpq   $0x0,-0x18(%rbp)
  je     L22
  cmpq   $0x0,-0x20(%rbp)
  je     L22
  subl   $0x1,-0xc(%rbp)
  jmp    L76
L20:
  addq   $0x1,-0x18(%rbp)
L76:
  mov    -0x18(%rbp),%rax
  cmp    -0x20(%rbp),%rax
  jae    L172
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L20
  jmp    L172
L115:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  cmp    $0xd,%eax
  je     L206
  cmp    $0xd,%eax
  jg     L87
  test   %eax,%eax
  je     L196
  cmp    $0xa,%eax
  jne    L87
  jmp    L206
L67:
  addq   $0x1,-0x18(%rbp)
L206:
  mov    -0x18(%rbp),%rax
  cmp    -0x20(%rbp),%rax
  jae    L138
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L67
L138:
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  movb   $0x20,(%rax)
  subl   $0x1,-0xc(%rbp)
  jmp    L179
L87:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  cmp    %al,-0x10(%rbp)
  je     L139
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  mov    -0x18(%rbp),%rdx
  movzbl (%rdx),%edx
  mov    %dl,(%rax)
  subl   $0x1,-0xc(%rbp)
  jmp    L179
L139:
  mov    -0x8(%rbp),%rax
  movb   $0x0,(%rax)
  addq   $0x1,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  jmp    L36
L196:
  nop
L179:
  addq   $0x1,-0x18(%rbp)
L172:
  mov    -0x18(%rbp),%rax
  cmp    -0x20(%rbp),%rax
  jae    L75
  cmpl   $0x0,-0xc(%rbp)
  jg     L115
L75:
  mov    -0x8(%rbp),%rax
  movb   $0x0,(%rax)
  mov    -0x18(%rbp),%rax
  jmp    L36
L22:
  mov    $0x0,%eax
L36:
  leave
  ret
```

assistant: ```c
char *unfold_token(char *buf, int buflen, char *tokb, char *toke, char delim) {
  if (buf && tokb && toke) {
    buflen--;
    for (; (tokb < toke) && isspace((int)*tokb); tokb++)
      ;
    while ((tokb < toke) && (buflen > 0)) {
      switch (*tokb) {
      case '\n':
      case '\r':
        for (; (tokb < toke) && isspace((int)*tokb); tokb++)
          ;
        *buf++ = ' ';
        buflen--;
        break;
      case '\0':
        break;
      default:
        if (*tokb != delim) {
          *buf++ = *tokb;
          buflen--;
        } else {
          *buf = '\0';
          return (++tokb);
        }
      }
      tokb++;
    }
    *buf = '\0';
    return tokb;
  }
  return NULL;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<unfold_token>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdx,%rbx
  test   %rdx,%rdx
  setne  %dl
  test   %rcx,%rcx
  setne  %al
  test   %al,%dl
  je     L136
  mov    %rdi,%r13
  mov    %rcx,%rbp
  mov    %r8d,%r14d
  test   %rdi,%rdi
  je     L136
  lea    -0x1(%rsi),%r12d
  cmp    %rcx,%rbx
  jae    L34
L146:
  movsbl (%rbx),%edi
  call   <isspace@plt>
  test   %rax,%rax
  je     L34
  add    $0x1,%rbx
  cmp    %rbx,%rbp
  jne    L146
  mov    %rbp,%rbx
L110:
  movb   $0x0,0x0(%r13)
  mov    %rbx,%rax
L87:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L34:
  cmp    %rbx,%rbp
  jbe    L110
  test   %r12d,%r12d
  jg     L92
  jmp    L110
L83:
  cmp    %rbx,%rbp
  jbe    L169
L56:
  movsbl (%rbx),%edi
  call   <isspace@plt>
  test   %rax,%rax
  je     L169
  add    $0x1,%rbx
  cmp    %rbx,%rbp
  jne    L56
  mov    %rbp,%rbx
L169:
  movb   $0x20,0x0(%r13)
  sub    $0x1,%r12d
  lea    0x1(%r13),%r13
L65:
  add    $0x1,%rbx
  cmp    %rbx,%rbp
  jbe    L110
  test   %r12d,%r12d
  jle    L110
L92:
  movzbl (%rbx),%eax
  cmp    $0xa,%al
  je     L83
  cmp    $0xd,%al
  je     L83
  test   %al,%al
  je     L65
  cmp    %r14b,%al
  je     L188
  mov    %al,0x0(%r13)
  sub    $0x1,%r12d
  lea    0x1(%r13),%r13
  jmp    L65
L188:
  movb   $0x0,0x0(%r13)
  lea    0x1(%rbx),%rax
  jmp    L87
L136:
  mov    $0x0,%eax
  jmp    L87
```

assistant: ```c
char *unfold_token(char *buf, int buflen, char *tokb, char *toke, char delim) {
  if (buf && tokb && toke) {
    buflen--;
    for (; (tokb < toke) && isspace((int)*tokb); tokb++)
      ;
    while ((tokb < toke) && (buflen > 0)) {
      switch (*tokb) {
      case '\n':
      case '\r':
        for (; (tokb < toke) && isspace((int)*tokb); tokb++)
          ;
        *buf++ = ' ';
        buflen--;
        break;
      case '\0':
        break;
      default:
        if (*tokb != delim) {
          *buf++ = *tokb;
          buflen--;
        } else {
          *buf = '\0';
          return (++tokb);
        }
      }
      tokb++;
    }
    *buf = '\0';
    return tokb;
  }
  return NULL;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<unfold_token>:
  endbr64
  push   %r15
  mov    %rdx,%r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  test   %rdx,%rdx
  setne  %dl
  test   %rcx,%rcx
  setne  %al
  test   %al,%dl
  je     L154
  mov    %rdi,%r12
  test   %rdi,%rdi
  je     L154
  mov    %rcx,%rbx
  mov    %r8d,%r13d
  lea    -0x1(%rsi),%ebp
  cmp    %rcx,%r15
  jb     L89
  jmp    L54
  nopl   0x0(%rax,%rax,1)
L29:
  add    $0x1,%r15
  cmp    %r15,%rbx
  je     L54
L89:
  movsbl (%r15),%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L29
  test   %ebp,%ebp
  jle    L54
  cmp    %r15,%rbx
  jbe    L54
  lea    0x1(%rbx),%r14
  jmp    L40
  nopw   0x0(%rax,%rax,1)
L5:
  test   %dil,%dil
  je     L79
  cmp    %r13b,%dil
  je     L90
  mov    %dil,(%r12)
  sub    $0x1,%ebp
  add    $0x1,%r12
L79:
  lea    0x1(%r15),%rax
  mov    %rax,%r15
  cmp    %rax,%rbx
  jbe    L54
L158:
  test   %ebp,%ebp
  jle    L54
L40:
  movsbl (%r15),%edi
  cmp    $0xa,%dil
  je     L74
  cmp    $0xd,%dil
  jne    L5
L74:
  lea    0x1(%r15),%rax
  cmp    %r15,%rbx
  ja     L156
  jmp    L109
  nopl   0x0(%rax)
L200:
  cmp    %rax,%rbx
  je     L140
  movsbl (%rax),%edi
  mov    %rax,%r15
L156:
  call   <isspace@plt>
  mov    %rax,%r8
  lea    0x1(%r15),%rax
  test   %r8,%r8
  jne    L200
L109:
  movb   $0x20,(%r12)
  sub    $0x1,%ebp
  add    $0x1,%r12
L202:
  mov    %rax,%r15
  cmp    %rax,%rbx
  ja     L158
L54:
  movb   $0x0,(%r12)
  add    $0x8,%rsp
  mov    %r15,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L140:
  movb   $0x20,(%r12)
  mov    %r14,%rax
  sub    $0x1,%ebp
  add    $0x1,%r12
  jmp    L202
  nopl   0x0(%rax)
L154:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L90:
  movb   $0x0,(%r12)
  add    $0x8,%rsp
  lea    0x1(%r15),%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
char *unfold_token(char *buf, int buflen, char *tokb, char *toke, char delim) {
  if (buf && tokb && toke) {
    buflen--;
    for (; (tokb < toke) && isspace((int)*tokb); tokb++)
      ;
    while ((tokb < toke) && (buflen > 0)) {
      switch (*tokb) {
      case '\n':
      case '\r':
        for (; (tokb < toke) && isspace((int)*tokb); tokb++)
          ;
        *buf++ = ' ';
        buflen--;
        break;
      case '\0':
        break;
      default:
        if (*tokb != delim) {
          *buf++ = *tokb;
          buflen--;
        } else {
          *buf = '\0';
          return (++tokb);
        }
      }
      tokb++;
    }
    *buf = '\0';
    return tokb;
  }
  return NULL;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<unfold_token>:
  endbr64
  test   %rdx,%rdx
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdx,%rbx
  setne  %dl
  test   %rcx,%rcx
  setne  %al
  test   %al,%dl
  je     L84
  mov    %rdi,%r13
  test   %rdi,%rdi
  je     L84
  mov    %rcx,%rbp
  mov    %r8d,%r14d
  lea    -0x1(%rsi),%r12d
  cmp    %rcx,%rbx
  jb     L23
  jmp    L165
  nopl   0x0(%rax,%rax,1)
L195:
  add    $0x1,%rbx
  cmp    %rbx,%rbp
  je     L165
L23:
  movsbl (%rbx),%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L195
  test   %r12d,%r12d
  jle    L165
  cmp    %rbx,%rbp
  ja     L54
  jmp    L165
  xchg   %ax,%ax
L67:
  test   %al,%al
  je     L140
  cmp    %r14b,%al
  je     L100
  mov    %al,0x0(%r13)
  sub    $0x1,%r12d
  add    $0x1,%r13
L140:
  add    $0x1,%rbx
  cmp    %rbx,%rbp
  jbe    L165
L158:
  test   %r12d,%r12d
  jle    L165
L54:
  movzbl (%rbx),%eax
  cmp    $0xa,%al
  je     L15
  cmp    $0xd,%al
  jne    L67
L15:
  cmp    %rbx,%rbp
  ja     L85
  jmp    L6
  nopl   0x0(%rax)
L61:
  cmp    %rbx,%rbp
  je     L6
L85:
  movsbl (%rbx),%edi
  call   <isspace@plt>
  mov    %rbx,%rdx
  add    $0x1,%rbx
  test   %rax,%rax
  jne    L61
  mov    %rdx,%rbx
L6:
  add    $0x1,%rbx
  movb   $0x20,0x0(%r13)
  sub    $0x1,%r12d
  add    $0x1,%r13
  cmp    %rbx,%rbp
  ja     L158
L165:
  movb   $0x0,0x0(%r13)
  mov    %rbx,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax,%rax,1)
L84:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L100:
  movb   $0x0,0x0(%r13)
  lea    0x1(%rbx),%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
char *unfold_token(char *buf, int buflen, char *tokb, char *toke, char delim) {
  if (buf && tokb && toke) {
    buflen--;
    for (; (tokb < toke) && isspace((int)*tokb); tokb++)
      ;
    while ((tokb < toke) && (buflen > 0)) {
      switch (*tokb) {
      case '\n':
      case '\r':
        for (; (tokb < toke) && isspace((int)*tokb); tokb++)
          ;
        *buf++ = ' ';
        buflen--;
        break;
      case '\0':
        break;
      default:
        if (*tokb != delim) {
          *buf++ = *tokb;
          buflen--;
        } else {
          *buf = '\0';
          return (++tokb);
        }
      }
      tokb++;
    }
    *buf = '\0';
    return tokb;
  }
  return NULL;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<count_year>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  movl   $0x0,-0x8(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  jmp    L16
L12:
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <isleap@plt>
  test   %rax,%rax
  je     L24
  addl   $0x1,-0x8(%rbp)
L24:
  addl   $0x16d,-0x8(%rbp)
  addl   $0x1,-0x4(%rbp)
L16:
  mov    -0x4(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jl     L12
  mov    -0x8(%rbp),%eax
  leave
  ret
```

assistant: ```c
int count_year(int a, int b) {
  int total, i;
  for (total = 0, i = a; i < b; i++) {
    if (isleap(i)) {
      total += 1;
    }
    total += 365;
  }
  return total;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<count_year>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  cmp    %esi,%edi
  jge    L27
  mov    %edi,%ebx
  mov    %esi,%r12d
  mov    $0x0,%ebp
L14:
  mov    %ebx,%edi
  call   <isleap@plt>
  cmp    $0x1,%rax
  sbb    $0xfffffe92,%ebp
  add    $0x1,%ebx
  cmp    %ebx,%r12d
  jne    L14
L12:
  mov    %ebp,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L27:
  mov    $0x0,%ebp
  jmp    L12
```

assistant: ```c
int count_year(int a, int b) {
  int total, i;
  for (total = 0, i = a; i < b; i++) {
    if (isleap(i)) {
      total += 1;
    }
    total += 365;
  }
  return total;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<count_year>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  cmp    %esi,%edi
  jge    L8
  mov    %edi,%ebx
  mov    %esi,%ebp
  xor    %r12d,%r12d
  nopl   0x0(%rax,%rax,1)
L10:
  mov    %ebx,%edi
  call   <isleap@plt>
  cmp    $0x1,%rax
  sbb    $0xfffffe92,%r12d
  add    $0x1,%ebx
  cmp    %ebx,%ebp
  jne    L10
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L8:
  xor    %r12d,%r12d
  pop    %rbx
  pop    %rbp
  mov    %r12d,%eax
  pop    %r12
  ret
```

assistant: ```c
int count_year(int a, int b) {
  int total, i;
  for (total = 0, i = a; i < b; i++) {
    if (isleap(i)) {
      total += 1;
    }
    total += 365;
  }
  return total;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<count_year>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  cmp    %esi,%edi
  jge    L2
  mov    %edi,%ebx
  mov    %esi,%ebp
  xor    %r12d,%r12d
  nopl   0x0(%rax,%rax,1)
L4:
  mov    %ebx,%edi
  call   <isleap@plt>
  cmp    $0x1,%rax
  sbb    $0xfffffe92,%r12d
  add    $0x1,%ebx
  cmp    %ebx,%ebp
  jne    L4
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L2:
  xor    %r12d,%r12d
  pop    %rbx
  pop    %rbp
  mov    %r12d,%eax
  pop    %r12
  ret
```

assistant: ```c
int count_year(int a, int b) {
  int total, i;
  for (total = 0, i = a; i < b; i++) {
    if (isleap(i)) {
      total += 1;
    }
    total += 365;
  }
  return total;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<latent>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  cmpl   $0x4,-0x14(%rbp)
  jne    L4
  movq   $0x0,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movl   $0x2a,(%rax)
L4:
  nop
  pop    %rbp
  ret
```

assistant: ```c
void latent(int a) {
  if (a == 4) {
    int *p = NULL;
    *p = 42;
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<latent>:
  endbr64
  cmp    $0x4,%edi
  je     L7
L10:
  ret
L7:
  movl   $0x2a,0x0
  jmp    L10
```

assistant: ```c
void latent(int a) {
  if (a == 4) {
    int *p = NULL;
    *p = 42;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<latent>:
  endbr64
  cmp    $0x4,%edi
  je     L5
  ret
  nopw   0x0(%rax,%rax,1)
L5:
  movl   $0x0,0x0
  ud2
```

assistant: ```c
void latent(int a) {
  if (a == 4) {
    int *p = NULL;
    *p = 42;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<latent>:
  endbr64
  cmp    $0x4,%edi
  je     L5
  ret
  nopw   0x0(%rax,%rax,1)
L5:
  movl   $0x0,0x0
  ud2
```

assistant: ```c
void latent(int a) {
  if (a == 4) {
    int *p = NULL;
    *p = 42;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  cmpl   $0x0,-0x1c(%rbp)
  je     L8
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rax
  mov    (%rax),%eax
  mov    %eax,-0x8(%rbp)
L8:
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rax
  mov    (%rax),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x8(%rbp),%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main(int **a, int argc) {
  int b;
  int d, e;
  if (argc) {
    d = *(*a);
  } else {
  }
  e = *(*a);
  return d + e;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  test   %esi,%esi
  je     L2
  mov    (%rdi),%rax
  mov    (%rax),%eax
L2:
  mov    (%rdi),%rdx
  add    (%rdx),%eax
  ret
```

assistant: ```c
int main(int **a, int argc) {
  int b;
  int d, e;
  if (argc) {
    d = *(*a);
  } else {
  }
  e = *(*a);
  return d + e;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  mov    (%rdi),%rax
  mov    (%rax),%edx
  xor    %eax,%eax
  test   %esi,%esi
  cmovne %edx,%eax
  add    %edx,%eax
  ret
```

assistant: ```c
int main(int **a, int argc) {
  int b;
  int d, e;
  if (argc) {
    d = *(*a);
  } else {
  }
  e = *(*a);
  return d + e;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  mov    (%rdi),%rax
  mov    (%rax),%edx
  xor    %eax,%eax
  test   %esi,%esi
  cmovne %edx,%eax
  add    %edx,%eax
  ret
```

assistant: ```c
int main(int **a, int argc) {
  int b;
  int d, e;
  if (argc) {
    d = *(*a);
  } else {
  }
  e = *(*a);
  return d + e;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <tabulate_dim@plt>
  mov    $0x0,%eax
  call   <tabulate_key@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main(void) {
  tabulate_dim();
  tabulate_key();
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <tabulate_dim@plt>
  mov    $0x0,%eax
  call   <tabulate_key@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  tabulate_dim();
  tabulate_key();
  return 0;
}
```