ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What is the c source code of the assembly code below:

```asm
<dosub>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  mov    -0x28(%rbp),%rax
  mov    0x10(%rax),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x28(%rbp),%rax
  add    $0x18,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <atoi@plt>
  movslq %eax,%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x1c(%rbp)
  cmpl   $0x4,-0x2c(%rbp)
  jle    L14
  mov    -0x28(%rbp),%rax
  add    $0x20,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <atoi@plt>
  cmp    %eax,-0x1c(%rbp)
  jl     L10
  mov    -0x28(%rbp),%rax
  add    $0x20,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <atoi@plt>
  jmp    L23
L10:
  mov    -0x1c(%rbp),%eax
L23:
  mov    %eax,-0x1c(%rbp)
L14:
  mov    -0x8(%rbp),%rax
  cmp    -0x10(%rbp),%rax
  jb     L32
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  movslq %eax,%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  cmp    %rax,-0x8(%rbp)
  jae    L32
  mov    -0x1c(%rbp),%eax
  cltq
  lea    -0x1(%rax),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x18(%rbp)
  jmp    L58
L43:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <pushback@plt>
  subq   $0x1,-0x18(%rbp)
L58:
  mov    -0x18(%rbp),%rax
  cmp    -0x8(%rbp),%rax
  jae    L43
L32:
  nop
  leave
  ret
```

assistant: ```c
void dosub(const char *argv[], int argc) {
  const char *ap, *fc, *k;
  int nc;
  ap = argv[2];
  fc = ap + atoi(argv[3]);
  nc = strlen(fc);
  if (argc >= 5)
    nc = ((nc) < (atoi(argv[4])) ? (nc) : (atoi(argv[4])));
  if (fc >= ap && fc < ap + strlen(ap))
    for (k = fc + nc - 1; k >= fc; k--)
      pushback(*k);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<dosub>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    %esi,%r13d
  mov    0x10(%rdi),%r12
  mov    0x18(%rdi),%rdi
  call   <atoi@plt>
  movslq %eax,%rbx
  add    %r12,%rbx
  mov    %rbx,%rdi
  call   <strlen@plt>
  mov    %eax,%r14d
  cmp    $0x4,%r13d
  jg     L23
L15:
  cmp    %rbx,%r12
  jbe    L61
L58:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L23:
  mov    0x20(%rbp),%rdi
  call   <atoi@plt>
  cmp    %r14d,%eax
  jg     L15
  mov    0x20(%rbp),%rdi
  call   <atoi@plt>
  mov    %eax,%r14d
  jmp    L15
L61:
  mov    %r12,%rdi
  call   <strlen@plt>
  cltq
  add    %rax,%r12
  cmp    %r12,%rbx
  jae    L58
  movslq %r14d,%rax
  lea    -0x1(%rbx,%rax,1),%rbp
  cmp    %rbp,%rbx
  ja     L58
L74:
  movsbl 0x0(%rbp),%edi
  call   <pushback@plt>
  sub    $0x1,%rbp
  cmp    %rbp,%rbx
  jbe    L74
  jmp    L58
```

assistant: ```c
void dosub(const char *argv[], int argc) {
  const char *ap, *fc, *k;
  int nc;
  ap = argv[2];
  fc = ap + atoi(argv[3]);
  nc = strlen(fc);
  if (argc >= 5)
    nc = ((nc) < (atoi(argv[4])) ? (nc) : (atoi(argv[4])));
  if (fc >= ap && fc < ap + strlen(ap))
    for (k = fc + nc - 1; k >= fc; k--)
      pushback(*k);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<dosub>:
  endbr64
  push   %r14
  mov    %esi,%r14d
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    0x10(%rdi),%r12
  mov    0x18(%rdi),%rdi
  call   <atoi@plt>
  movslq %eax,%rbx
  add    %r12,%rbx
  mov    %rbx,%rdi
  call   <strlen@plt>
  mov    %eax,%r13d
  cmp    $0x4,%r14d
  jg     L15
L41:
  cmp    %rbx,%r12
  jbe    L25
L13:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
L15:
  mov    0x20(%rbp),%rdi
  call   <atoi@plt>
  cmp    %r13d,%eax
  jg     L41
  mov    0x20(%rbp),%rdi
  call   <atoi@plt>
  mov    %eax,%r13d
  cmp    %rbx,%r12
  ja     L13
  nopw   0x0(%rax,%rax,1)
L25:
  mov    %r12,%rdi
  call   <strlen@plt>
  cltq
  add    %rax,%r12
  cmp    %r12,%rbx
  jae    L13
  movslq %r13d,%rax
  lea    -0x1(%rbx,%rax,1),%rbp
  cmp    %rbp,%rbx
  ja     L13
  nop
L26:
  movsbl 0x0(%rbp),%edi
  sub    $0x1,%rbp
  call   <pushback@plt>
  cmp    %rbp,%rbx
  jbe    L26
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
void dosub(const char *argv[], int argc) {
  const char *ap, *fc, *k;
  int nc;
  ap = argv[2];
  fc = ap + atoi(argv[3]);
  nc = strlen(fc);
  if (argc >= 5)
    nc = ((nc) < (atoi(argv[4])) ? (nc) : (atoi(argv[4])));
  if (fc >= ap && fc < ap + strlen(ap))
    for (k = fc + nc - 1; k >= fc; k--)
      pushback(*k);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<dosub>:
  endbr64
  push   %r14
  mov    %esi,%r14d
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    0x10(%rdi),%r12
  mov    0x18(%rdi),%rdi
  call   <atoi@plt>
  movslq %eax,%rbx
  add    %r12,%rbx
  mov    %rbx,%rdi
  call   <strlen@plt>
  mov    %eax,%r13d
  cmp    $0x4,%r14d
  jg     L24
L61:
  cmp    %rbx,%r12
  jbe    L63
L39:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
L24:
  mov    0x20(%rbp),%rdi
  call   <atoi@plt>
  cmp    %r13d,%eax
  jg     L61
  mov    0x20(%rbp),%rdi
  call   <atoi@plt>
  mov    %eax,%r13d
  cmp    %rbx,%r12
  ja     L39
  nopw   0x0(%rax,%rax,1)
L63:
  mov    %r12,%rdi
  call   <strlen@plt>
  cltq
  add    %rax,%r12
  cmp    %r12,%rbx
  jae    L39
  movslq %r13d,%rax
  lea    -0x1(%rbx,%rax,1),%rbp
  cmp    %rbp,%rbx
  ja     L39
  nop
L54:
  movsbl 0x0(%rbp),%edi
  sub    $0x1,%rbp
  call   <pushback@plt>
  cmp    %rbp,%rbx
  jbe    L54
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
void dosub(const char *argv[], int argc) {
  const char *ap, *fc, *k;
  int nc;
  ap = argv[2];
  fc = ap + atoi(argv[3]);
  nc = strlen(fc);
  if (argc >= 5)
    nc = ((nc) < (atoi(argv[4])) ? (nc) : (atoi(argv[4])));
  if (fc >= ap && fc < ap + strlen(ap))
    for (k = fc + nc - 1; k >= fc; k--)
      pushback(*k);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<dissect_opt_hop>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  mov    (%rax),%rax
  test   %rax,%rax
  je     L9
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  call   <tprintf@plt>
L9:
  nop
  leave
  ret
```

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

tool: D6: , Option(s) recognized 

assistant: ```c
void dissect_opt_hop(struct pkt_buff *pkt, size_t *opt_len) {
  if (*opt_len)
    tprintf(", Option(s) recognized ");
}
```
user: What is the c source code of the assembly code below?

```asm
<dissect_opt_hop>:
  endbr64
  cmpq   $0x0,(%rsi)
  jne    L9
  ret
L9:
  sub    $0x8,%rsp
  lea    D3(%rip),%rdi
  call   <tprintf@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: tag: D3
data_type: string

tool: D3: , Option(s) recognized 

assistant: ```c
void dissect_opt_hop(struct pkt_buff *pkt, size_t *opt_len) {
  if (*opt_len)
    tprintf(", Option(s) recognized ");
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<dissect_opt_hop>:
  endbr64
  cmpq   $0x0,(%rsi)
  jne    L0
  ret
  nopl   0x0(%rax,%rax,1)
L0:
  lea    D5(%rip),%rdi
  jmp    1050 <tprintf@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D5
type_infered: string

tool: D5: , Option(s) recognized 

assistant: ```c
void dissect_opt_hop(struct pkt_buff *pkt, size_t *opt_len) {
  if (*opt_len)
    tprintf(", Option(s) recognized ");
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<dissect_opt_hop>:
  endbr64
  cmpq   $0x0,(%rsi)
  jne    L9
  ret
  nopl   0x0(%rax,%rax,1)
L9:
  lea    D9(%rip),%rdi
  jmp    1050 <tprintf@plt>
```

tool_calls: name: parse_data
arguments: tag: D9
type_guessed: string

tool: D9: , Option(s) recognized 

assistant: ```c
void dissect_opt_hop(struct pkt_buff *pkt, size_t *opt_len) {
  if (*opt_len)
    tprintf(", Option(s) recognized ");
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<print_list>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    $0x7b,%edi
  call   <putchar@plt>
  cmpl   $0x0,-0x1c(%rbp)
  jle    L30
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,%esi
  lea    D14(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L30:
  movl   $0x1,-0x4(%rbp)
  jmp    L31
L25:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,%esi
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x4(%rbp)
L31:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L25
  mov    $0x7d,%edi
  call   <putchar@plt>
  cmpl   $0x0,-0x20(%rbp)
  je     L26
  mov    $0xa,%edi
  call   <putchar@plt>
L26:
  nop
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D14
type_infered: string
name: retrieve_data
arguments: tag: D12
type_infered: string

tool: D14: %d

tool: D12: ,%d

assistant: ```c
void print_list(int *list, int n, int nl) {
  int i;
  printf("{");
  if (n > 0)
    printf("%d", list[0]);
  for (i = 1; i < n; i++)
    printf(",%d", list[i]);
  printf("}");
  if (nl)
    printf("\n");
}
```
user: What is the c source code of the assembly code below?

```asm
<print_list>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r12
  mov    %esi,%ebp
  mov    %edx,%r13d
  mov    $0x7b,%edi
  call   <putchar@plt>
  test   %ebp,%ebp
  jg     L26
L36:
  cmp    $0x1,%ebp
  jle    L0
  lea    0x4(%r12),%rbx
  lea    -0x2(%rbp),%eax
  lea    0x8(%r12,%rax,4),%r12
  lea    D12(%rip),%rbp
L38:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x4,%rbx
  cmp    %r12,%rbx
  jne    L38
L0:
  mov    $0x7d,%edi
  call   <putchar@plt>
  test   %r13d,%r13d
  jne    L32
L13:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L26:
  mov    (%r12),%esi
  lea    D1(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L36
L32:
  mov    $0xa,%edi
  call   <putchar@plt>
  jmp    L13
```

tool_calls: name: extract_symbolic_data
arguments: label: D12
datatype: string
name: extract_symbolic_data
arguments: label: D1
datatype: string

tool: D12: ,%d

tool: D1: %d

assistant: ```c
void print_list(int *list, int n, int nl) {
  int i;
  printf("{");
  if (n > 0)
    printf("%d", list[0]);
  for (i = 1; i < n; i++)
    printf(",%d", list[i]);
  printf("}");
  if (nl)
    printf("\n");
}
```
user: What is the c source code of the assembly code below:

```asm
<print_list>:
  endbr64
  push   %r13
  mov    %edx,%r13d
  push   %r12
  mov    %rdi,%r12
  mov    $0x7b,%edi
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  sub    $0x8,%rsp
  call   <putchar@plt>
  test   %ebp,%ebp
  jg     L10
L8:
  mov    $0x7d,%edi
  call   <putchar@plt>
  test   %r13d,%r13d
  jne    L16
L24:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  xchg   %ax,%ax
L10:
  mov    (%r12),%esi
  xor    %eax,%eax
  lea    D4(%rip),%rdi
  call   <printf@plt>
  cmp    $0x1,%ebp
  jle    L8
  lea    -0x2(%rbp),%eax
  lea    0x4(%r12),%rbx
  lea    D0(%rip),%rbp
  lea    0x8(%r12,%rax,4),%r12
  nopl   0x0(%rax,%rax,1)
L43:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %r12,%rbx
  jne    L43
  mov    $0x7d,%edi
  call   <putchar@plt>
  test   %r13d,%r13d
  je     L24
L16:
  add    $0x8,%rsp
  mov    $0xa,%edi
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    1060 <putchar@plt>
```

tool_calls: name: parse_data
arguments: label: D4
type_infered: string
name: parse_data
arguments: label: D0
type_infered: string

tool: D4: %d

tool: D0: ,%d

assistant: ```c
void print_list(int *list, int n, int nl) {
  int i;
  printf("{");
  if (n > 0)
    printf("%d", list[0]);
  for (i = 1; i < n; i++)
    printf(",%d", list[i]);
  printf("}");
  if (nl)
    printf("\n");
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<print_list>:
  endbr64
  push   %r13
  mov    %edx,%r13d
  push   %r12
  mov    %rdi,%r12
  mov    $0x7b,%edi
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  sub    $0x8,%rsp
  call   <putchar@plt>
  test   %ebp,%ebp
  jg     L31
L0:
  mov    $0x7d,%edi
  call   <putchar@plt>
  test   %r13d,%r13d
  jne    L19
L17:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  xchg   %ax,%ax
L31:
  mov    (%r12),%esi
  xor    %eax,%eax
  lea    D6(%rip),%rdi
  call   <printf@plt>
  cmp    $0x1,%ebp
  jle    L0
  lea    -0x2(%rbp),%eax
  lea    0x4(%r12),%rbx
  lea    D11(%rip),%rbp
  lea    0x8(%r12,%rax,4),%r12
  nopl   0x0(%rax,%rax,1)
L1:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %r12,%rbx
  jne    L1
  mov    $0x7d,%edi
  call   <putchar@plt>
  test   %r13d,%r13d
  je     L17
L19:
  add    $0x8,%rsp
  mov    $0xa,%edi
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    1060 <putchar@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D6
datatype: string
name: unpack_data
arguments: data_tag: D11
datatype: string

tool: D6: %d

tool: D11: ,%d

assistant: ```c
void print_list(int *list, int n, int nl) {
  int i;
  printf("{");
  if (n > 0)
    printf("%d", list[0]);
  for (i = 1; i < n; i++)
    printf(",%d", list[i]);
  printf("}");
  if (nl)
    printf("\n");
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<strcasecmp>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  jmp    L10
L16:
  addq   $0x1,-0x18(%rbp)
  addq   $0x1,-0x20(%rbp)
L10:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L25
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  mov    %eax,%edi
  call   <tolower@plt>
  mov    %eax,%ebx
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  mov    %eax,%edi
  call   <tolower@plt>
  cmp    %eax,%ebx
  je     L16
L25:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  mov    %eax,%edi
  call   <tolower@plt>
  mov    %eax,%ebx
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  mov    %eax,%edi
  call   <tolower@plt>
  mov    %eax,%edx
  mov    %ebx,%eax
  sub    %edx,%eax
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int strcasecmp(const char *s1, const char *s2) {
  while ((*s1 != '\0') && (tolower(*(unsigned char *)s1) == tolower(*(unsigned char *)s2))) {
    s1++;
    s2++;
  }
  return tolower(*(unsigned char *)s1) - tolower(*(unsigned char *)s2);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<strcasecmp>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    %rsi,%r12
  movzbl (%rdi),%edi
  test   %dil,%dil
  je     L21
L13:
  movzbl %dil,%edi
  call   <tolower@plt>
  mov    %eax,%ebx
  movzbl (%r12),%edi
  call   <tolower@plt>
  cmp    %eax,%ebx
  jne    L21
  add    $0x1,%rbp
  add    $0x1,%r12
  movzbl 0x0(%rbp),%edi
  test   %dil,%dil
  jne    L13
L21:
  movzbl 0x0(%rbp),%edi
  call   <tolower@plt>
  mov    %eax,%ebx
  movzbl (%r12),%edi
  call   <tolower@plt>
  mov    %eax,%edx
  mov    %ebx,%eax
  sub    %edx,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int strcasecmp(const char *s1, const char *s2) {
  while ((*s1 != '\0') && (tolower(*(unsigned char *)s1) == tolower(*(unsigned char *)s2))) {
    s1++;
    s2++;
  }
  return tolower(*(unsigned char *)s1) - tolower(*(unsigned char *)s2);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<strcasecmp>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  movzbl (%rdi),%edi
  test   %dil,%dil
  jne    L15
  jmp    L18
  nopl   0x0(%rax,%rax,1)
L25:
  movzbl 0x1(%rbp),%edi
  add    $0x1,%rbp
  add    $0x1,%r12
  test   %dil,%dil
  je     L18
L15:
  call   <tolower@plt>
  movzbl (%r12),%edi
  mov    %eax,%ebx
  call   <tolower@plt>
  cmp    %eax,%ebx
  je     L25
L18:
  movzbl 0x0(%rbp),%edi
  call   <tolower@plt>
  movzbl (%r12),%edi
  mov    %eax,%ebx
  call   <tolower@plt>
  mov    %eax,%r8d
  mov    %ebx,%eax
  pop    %rbx
  pop    %rbp
  sub    %r8d,%eax
  pop    %r12
  ret
```

assistant: ```c
int strcasecmp(const char *s1, const char *s2) {
  while ((*s1 != '\0') && (tolower(*(unsigned char *)s1) == tolower(*(unsigned char *)s2))) {
    s1++;
    s2++;
  }
  return tolower(*(unsigned char *)s1) - tolower(*(unsigned char *)s2);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<strcasecmp>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  movzbl (%rdi),%edi
  test   %dil,%dil
  jne    L13
  jmp    L31
  nopl   0x0(%rax,%rax,1)
L15:
  movzbl 0x1(%rbp),%edi
  add    $0x1,%rbp
  add    $0x1,%r12
  test   %dil,%dil
  je     L31
L13:
  call   <tolower@plt>
  movzbl (%r12),%edi
  mov    %eax,%ebx
  call   <tolower@plt>
  cmp    %eax,%ebx
  je     L15
L31:
  movzbl 0x0(%rbp),%edi
  call   <tolower@plt>
  movzbl (%r12),%edi
  mov    %eax,%ebx
  call   <tolower@plt>
  mov    %eax,%r8d
  mov    %ebx,%eax
  pop    %rbx
  pop    %rbp
  sub    %r8d,%eax
  pop    %r12
  ret
```

assistant: ```c
int strcasecmp(const char *s1, const char *s2) {
  while ((*s1 != '\0') && (tolower(*(unsigned char *)s1) == tolower(*(unsigned char *)s2))) {
    s1++;
    s2++;
  }
  return tolower(*(unsigned char *)s1) - tolower(*(unsigned char *)s2);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<forkbomb_fn>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  call   <fork@plt>
  call   <fork@plt>
  mov    $0x0,%eax
  call   <getppid@plt>
  mov    %eax,-0x4(%rbp)
  jmp    L3
L14:
  mov    $0x3e8,%edi
  call   <usleep@plt>
L3:
  mov    $0x0,%eax
  call   <getppid@plt>
  cmp    %eax,-0x4(%rbp)
  je     L14
  mov    $0x0,%eax
  call   <getppid@plt>
  cmp    %eax,-0x4(%rbp)
  sete   %al
  movzbl %al,%eax
  leave
  ret
```

assistant: ```c
int forkbomb_fn(const char *cgroup, void *arg) {
  int ppid;
  fork();
  fork();
  ppid = getppid();
  while (getppid() == ppid)
    usleep(1000);
  return getppid() == ppid;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<forkbomb_fn>:
  endbr64
  push   %rbx
  call   <fork@plt>
  call   <fork@plt>
  mov    $0x0,%eax
  call   <getppid@plt>
  mov    %eax,%ebx
  jmp    L8
L6:
  mov    $0x3e8,%edi
  call   <usleep@plt>
L8:
  mov    $0x0,%eax
  call   <getppid@plt>
  cmp    %ebx,%eax
  je     L6
  mov    $0x0,%eax
  call   <getppid@plt>
  cmp    %ebx,%eax
  sete   %al
  movzbl %al,%eax
  pop    %rbx
  ret
```

assistant: ```c
int forkbomb_fn(const char *cgroup, void *arg) {
  int ppid;
  fork();
  fork();
  ppid = getppid();
  while (getppid() == ppid)
    usleep(1000);
  return getppid() == ppid;
}
```
user: Translate the below assembly code into C programming language:

```asm
<forkbomb_fn>:
  endbr64
  push   %rbx
  call   <fork@plt>
  call   <fork@plt>
  xor    %eax,%eax
  call   <getppid@plt>
  mov    %eax,%ebx
  jmp    L3
  nopw   0x0(%rax,%rax,1)
L6:
  mov    $0x3e8,%edi
  call   <usleep@plt>
L3:
  xor    %eax,%eax
  call   <getppid@plt>
  cmp    %ebx,%eax
  je     L6
  xor    %eax,%eax
  call   <getppid@plt>
  cmp    %ebx,%eax
  pop    %rbx
  sete   %al
  movzbl %al,%eax
  ret
```

assistant: ```c
int forkbomb_fn(const char *cgroup, void *arg) {
  int ppid;
  fork();
  fork();
  ppid = getppid();
  while (getppid() == ppid)
    usleep(1000);
  return getppid() == ppid;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<forkbomb_fn>:
  endbr64
  push   %rbx
  call   <fork@plt>
  call   <fork@plt>
  xor    %eax,%eax
  call   <getppid@plt>
  mov    %eax,%ebx
  jmp    L0
  nopw   0x0(%rax,%rax,1)
L18:
  mov    $0x3e8,%edi
  call   <usleep@plt>
L0:
  xor    %eax,%eax
  call   <getppid@plt>
  cmp    %ebx,%eax
  je     L18
  xor    %eax,%eax
  call   <getppid@plt>
  cmp    %ebx,%eax
  pop    %rbx
  sete   %al
  movzbl %al,%eax
  ret
```

assistant: ```c
int forkbomb_fn(const char *cgroup, void *arg) {
  int ppid;
  fork();
  fork();
  ppid = getppid();
  while (getppid() == ppid)
    usleep(1000);
  return getppid() == ppid;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<alloc_LogGrid>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x0,%eax
  call   <alloc_Grid@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <Zero_LogGrid@plt>
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
double *alloc_LogGrid() {
  double *tt;
  tt = alloc_Grid();
  Zero_LogGrid(tt);
  return tt;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<alloc_LogGrid>:
  endbr64
  push   %rbx
  mov    $0x0,%eax
  call   <alloc_Grid@plt>
  mov    %rax,%rbx
  mov    %rax,%rdi
  call   <Zero_LogGrid@plt>
  mov    %rbx,%rax
  pop    %rbx
  ret
```

assistant: ```c
double *alloc_LogGrid() {
  double *tt;
  tt = alloc_Grid();
  Zero_LogGrid(tt);
  return tt;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<alloc_LogGrid>:
  endbr64
  push   %r12
  xor    %eax,%eax
  call   <alloc_Grid@plt>
  mov    %rax,%r12
  mov    %rax,%rdi
  call   <Zero_LogGrid@plt>
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
double *alloc_LogGrid() {
  double *tt;
  tt = alloc_Grid();
  Zero_LogGrid(tt);
  return tt;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<alloc_LogGrid>:
  endbr64
  push   %r12
  xor    %eax,%eax
  call   <alloc_Grid@plt>
  mov    %rax,%r12
  mov    %rax,%rdi
  call   <Zero_LogGrid@plt>
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
double *alloc_LogGrid() {
  double *tt;
  tt = alloc_Grid();
  Zero_LogGrid(tt);
  return tt;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<re_match>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  mov    %ecx,-0x28(%rbp)
  mov    %r8,-0x30(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x28(%rbp),%edi
  mov    -0x24(%rbp),%esi
  mov    -0x20(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    -0x24(%rbp),%ecx
  push   %rcx
  push   -0x30(%rbp)
  mov    %edi,%r9d
  mov    %esi,%r8d
  mov    %rdx,%rcx
  mov    $0x0,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <re_match_2_internal@plt>
  add    $0x10,%rsp
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L3
  call   <__stack_chk_fail@plt>
L3:
  leave
  ret
```

assistant: ```c
int re_match(bufp, string, size, pos, regs)
struct re_pattern_buffer *bufp;
const char *string;
int size, pos;
struct re_registers *regs;
{
  int result = re_match_2_internal(bufp, (void *)0, 0, string, size, pos, regs, size);
  __builtin_alloca(0);
  return result;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<re_match>:
  endbr64
  sub    $0x8,%rsp
  mov    %ecx,%r9d
  push   %rdx
  push   %r8
  mov    %edx,%r8d
  mov    %rsi,%rcx
  mov    $0x0,%edx
  mov    $0x0,%esi
  mov    $0x0,%eax
  call   <re_match_2_internal@plt>
  add    $0x18,%rsp
  ret
```

assistant: ```c
int re_match(bufp, string, size, pos, regs)
struct re_pattern_buffer *bufp;
const char *string;
int size, pos;
struct re_registers *regs;
{
  int result = re_match_2_internal(bufp, (void *)0, 0, string, size, pos, regs, size);
  __builtin_alloca(0);
  return result;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<re_match>:
  endbr64
  sub    $0x8,%rsp
  mov    %ecx,%r9d
  xor    %eax,%eax
  mov    %rsi,%rcx
  push   %rdx
  xor    %esi,%esi
  push   %r8
  mov    %edx,%r8d
  xor    %edx,%edx
  call   <re_match_2_internal@plt>
  add    $0x18,%rsp
  ret
```

assistant: ```c
int re_match(bufp, string, size, pos, regs)
struct re_pattern_buffer *bufp;
const char *string;
int size, pos;
struct re_registers *regs;
{
  int result = re_match_2_internal(bufp, (void *)0, 0, string, size, pos, regs, size);
  __builtin_alloca(0);
  return result;
}
```