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

```asm
<makeMultipleDateCriteriaFromTxt>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0x0,-0x28(%rbp)
  mov    -0x38(%rbp),%rax
  mov    %rax,%rdi
  call   <strdupa@plt>
  mov    %rax,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  lea    D11(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strtok@plt>
  mov    %rax,-0x20(%rbp)
  jmp    L38
L24:
  mov    -0x20(%rbp),%rax
  lea    D24(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  je     L33
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <makeSingleDateCriteriaFromTxt@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rdx
  lea    -0x28(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <appendDateCriteria@plt>
L33:
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cltq
  lea    0x1(%rax),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  lea    D11(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strtok@plt>
  mov    %rax,-0x20(%rbp)
L38:
  cmpq   $0x0,-0x20(%rbp)
  jne    L24
  mov    -0x28(%rbp),%rax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L27
  call   <__stack_chk_fail@plt>
L27:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D11
type: string
name: unpack_data
arguments: data_label: D24
type: string

tool: 

tool: D24: ,

assistant: ```c
struct DateCriteria *makeMultipleDateCriteriaFromTxt(char *txt) {
  struct DateCriteria *result = NULL;
  char *txtCopy = strdupa(txt);
  char *tmp = strtok(txtCopy, "[]");
  while (tmp != NULL) {
    if (strcmp(tmp, ",") != 0) {
      struct DateCriteria *item = makeSingleDateCriteriaFromTxt(tmp);
      appendDateCriteria(&result, item);
    }
    tmp = strtok(tmp + strlen(tmp) + 1, "[]");
  }
  return result;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<makeMultipleDateCriteriaFromTxt>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movq   $0x0,(%rsp)
  call   <strdupa@plt>
  mov    %rax,%rdi
  lea    D9(%rip),%rsi
  call   <strtok@plt>
  test   %rax,%rax
  je     L36
  mov    %rax,%rbx
  lea    D13(%rip),%r12
  mov    %rsp,%r13
  lea    D9(%rip),%rbp
  jmp    L51
L13:
  mov    %rbx,%rdi
  call   <strlen@plt>
  cltq
  lea    0x1(%rbx,%rax,1),%rdi
  mov    %rbp,%rsi
  call   <strtok@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L36
L51:
  mov    %r12,%rsi
  mov    %rbx,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  je     L13
  mov    %rbx,%rdi
  call   <makeSingleDateCriteriaFromTxt@plt>
  mov    %rax,%rsi
  mov    %r13,%rdi
  call   <appendDateCriteria@plt>
  jmp    L13
L36:
  mov    (%rsp),%rax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L1
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D9
datatype: string
name: read_data
arguments: label: D13
datatype: string

tool: 

tool: D13: ,

assistant: ```c
struct DateCriteria *makeMultipleDateCriteriaFromTxt(char *txt) {
  struct DateCriteria *result = NULL;
  char *txtCopy = strdupa(txt);
  char *tmp = strtok(txtCopy, "[]");
  while (tmp != NULL) {
    if (strcmp(tmp, ",") != 0) {
      struct DateCriteria *item = makeSingleDateCriteriaFromTxt(tmp);
      appendDateCriteria(&result, item);
    }
    tmp = strtok(tmp + strlen(tmp) + 1, "[]");
  }
  return result;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<makeMultipleDateCriteriaFromTxt>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  lea    D3(%rip),%rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movq   $0x0,(%rsp)
  call   <strdupa@plt>
  mov    %rbp,%rsi
  mov    %rax,%rdi
  call   <strtok@plt>
  test   %rax,%rax
  je     L1
  mov    %rax,%rbx
  lea    D15(%rip),%r12
  mov    %rsp,%r13
  jmp    L2
  nopl   0x0(%rax)
L42:
  mov    %rbx,%rdi
  call   <strlen@plt>
  mov    %rbp,%rsi
  cltq
  lea    0x1(%rbx,%rax,1),%rdi
  call   <strtok@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L1
L2:
  mov    %r12,%rsi
  mov    %rbx,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  je     L42
  mov    %rbx,%rdi
  call   <makeSingleDateCriteriaFromTxt@plt>
  mov    %r13,%rdi
  mov    %rax,%rsi
  call   <appendDateCriteria@plt>
  jmp    L42
  nopl   0x0(%rax)
L1:
  mov    (%rsp),%rax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L21
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L21:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D3
type_infered: string
name: unpack_data
arguments: data_tag: D15
type_infered: string

tool: 

tool: D15: ,

assistant: ```c
struct DateCriteria *makeMultipleDateCriteriaFromTxt(char *txt) {
  struct DateCriteria *result = NULL;
  char *txtCopy = strdupa(txt);
  char *tmp = strtok(txtCopy, "[]");
  while (tmp != NULL) {
    if (strcmp(tmp, ",") != 0) {
      struct DateCriteria *item = makeSingleDateCriteriaFromTxt(tmp);
      appendDateCriteria(&result, item);
    }
    tmp = strtok(tmp + strlen(tmp) + 1, "[]");
  }
  return result;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<makeMultipleDateCriteriaFromTxt>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  lea    D2(%rip),%rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movq   $0x0,(%rsp)
  call   <strdupa@plt>
  mov    %rbp,%rsi
  mov    %rax,%rdi
  call   <strtok@plt>
  test   %rax,%rax
  je     L18
  mov    %rax,%rbx
  lea    D13(%rip),%r12
  mov    %rsp,%r13
  jmp    L44
  nopl   0x0(%rax)
L47:
  mov    %rbx,%rdi
  call   <strlen@plt>
  mov    %rbp,%rsi
  cltq
  lea    0x1(%rbx,%rax,1),%rdi
  call   <strtok@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L18
L44:
  mov    %r12,%rsi
  mov    %rbx,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  je     L47
  mov    %rbx,%rdi
  call   <makeSingleDateCriteriaFromTxt@plt>
  mov    %r13,%rdi
  mov    %rax,%rsi
  call   <appendDateCriteria@plt>
  jmp    L47
  nopl   0x0(%rax)
L18:
  mov    (%rsp),%rax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L39
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L39:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D2
datatype: string
name: unpack_data
arguments: data_label: D13
datatype: string

tool: 

tool: D13: ,

assistant: ```c
struct DateCriteria *makeMultipleDateCriteriaFromTxt(char *txt) {
  struct DateCriteria *result = NULL;
  char *txtCopy = strdupa(txt);
  char *tmp = strtok(txtCopy, "[]");
  while (tmp != NULL) {
    if (strcmp(tmp, ",") != 0) {
      struct DateCriteria *item = makeSingleDateCriteriaFromTxt(tmp);
      appendDateCriteria(&result, item);
    }
    tmp = strtok(tmp + strlen(tmp) + 1, "[]");
  }
  return result;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %edi,-0x44(%rbp)
  mov    %rsi,-0x50(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x1a,-0x40(%rbp)
  movl   $0x5,-0x3c(%rbp)
  movl   $0x21,-0x38(%rbp)
  movl   $0x4d,-0x34(%rbp)
  movl   $0x1,-0x30(%rbp)
  movl   $0x3d,-0x2c(%rbp)
  movl   $0xb,-0x28(%rbp)
  movl   $0x3b,-0x24(%rbp)
  movl   $0xf,-0x20(%rbp)
  movl   $0x30,-0x1c(%rbp)
  movl   $0x13,-0x18(%rbp)
  movl   $0x16,-0x14(%rbp)
  lea    -0x40(%rbp),%rax
  mov    $0xb,%esi
  mov    %rax,%rdi
  call   <heapsort@plt>
  lea    -0x40(%rbp),%rax
  mov    $0xc,%esi
  mov    %rax,%rdi
  call   <display@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L6
  call   <__stack_chk_fail@plt>
L6:
  leave
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  int a[] = {26, 5, 33, 77, 1, 61, 11, 59, 15, 48, 19, 22};
  heapsort(a, 11);
  display(a, 12);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  movl   $0x1a,(%rsp)
  movl   $0x5,0x4(%rsp)
  movl   $0x21,0x8(%rsp)
  movl   $0x4d,0xc(%rsp)
  movl   $0x1,0x10(%rsp)
  movl   $0x3d,0x14(%rsp)
  movl   $0xb,0x18(%rsp)
  movl   $0x3b,0x1c(%rsp)
  movl   $0xf,0x20(%rsp)
  movl   $0x30,0x24(%rsp)
  movl   $0x13,0x28(%rsp)
  movl   $0x16,0x2c(%rsp)
  mov    %rsp,%rbx
  mov    $0xb,%esi
  mov    %rbx,%rdi
  call   <heapsort@plt>
  mov    $0xc,%esi
  mov    %rbx,%rdi
  call   <display@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  mov    $0x0,%eax
  add    $0x40,%rsp
  pop    %rbx
  ret
L6:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(int argc, char *argv[]) {
  int a[] = {26, 5, 33, 77, 1, 61, 11, 59, 15, 48, 19, 22};
  heapsort(a, 11);
  display(a, 12);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    $0xb,%esi
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  movabs $0x50000001a,%rax
  mov    %rsp,%rbp
  mov    %rbp,%rdi
  mov    %rax,(%rsp)
  movabs $0x4d00000021,%rax
  mov    %rax,0x8(%rsp)
  movabs $0x3d00000001,%rax
  mov    %rax,0x10(%rsp)
  movabs $0x3b0000000b,%rax
  mov    %rax,0x18(%rsp)
  movabs $0x300000000f,%rax
  mov    %rax,0x20(%rsp)
  movabs $0x1600000013,%rax
  mov    %rax,0x28(%rsp)
  call   <heapsort@plt>
  mov    $0xc,%esi
  mov    %rbp,%rdi
  call   <display@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x40,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(int argc, char *argv[]) {
  int a[] = {26, 5, 33, 77, 1, 61, 11, 59, 15, 48, 19, 22};
  heapsort(a, 11);
  display(a, 12);
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    $0xb,%esi
  sub    $0x40,%rsp
  movdqa D7(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  movaps %xmm0,(%rsp)
  movdqa D16(%rip),%xmm0
  mov    %rbp,%rdi
  movaps %xmm0,0x10(%rsp)
  movdqa D20(%rip),%xmm0
  movaps %xmm0,0x20(%rsp)
  call   <heapsort@plt>
  mov    $0xc,%esi
  mov    %rbp,%rdi
  call   <display@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x40,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D7
type: byte[16]
name: extract_symbolic_data
arguments: label: D16
type: byte[16]
name: extract_symbolic_data
arguments: label: D20
type: byte[16]

tool: {"D7": [0x1a,0x00,0x00,0x00,0x05,0x00,0x00,0x00,0x21,0x00,0x00,0x00,0x4d,0x00,0x00,0x00]}

tool: {"D16": [0x01,0x00,0x00,0x00,0x3d,0x00,0x00,0x00,0x0b,0x00,0x00,0x00,0x3b,0x00,0x00,0x00]}

tool: {"D20": [0x0f,0x00,0x00,0x00,0x30,0x00,0x00,0x00,0x13,0x00,0x00,0x00,0x16,0x00,0x00,0x00]}

assistant: ```c
int main(int argc, char *argv[]) {
  int a[] = {26, 5, 33, 77, 1, 61, 11, 59, 15, 48, 19, 22};
  heapsort(a, 11);
  display(a, 12);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<str_char>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,%eax
  mov    %al,-0x1c(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  add    $0x1,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L2
L17:
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0xc(%rbp),%edx
  movslq %edx,%rcx
  mov    -0x8(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  addl   $0x1,-0xc(%rbp)
L2:
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L17
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0xc(%rbp),%edx
  movslq %edx,%rcx
  mov    -0x8(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  add    $0x2,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x18(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L5
L28:
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    -0xc(%rbp),%edx
  movslq %edx,%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  addl   $0x1,-0xc(%rbp)
L5:
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L28
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  movzbl -0x1c(%rbp),%eax
  mov    %al,(%rdx)
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x1(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    -0x18(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *str_char(char *str, char c) {
  char *cpy;
  int y;
  cpy = malloc(sizeof(char) * (strlen(str) + 1));
  y = 0;
  while (str[y] != '\0') {
    cpy[y] = str[y];
    y++;
  }
  cpy[y] = str[y];
  free(str);
  str = malloc(sizeof(char) * (strlen(cpy) + 2));
  y = 0;
  while (cpy[y] != '\0') {
    str[y] = cpy[y];
    y++;
  }
  str[y] = c;
  str[y + 1] = '\0';
  free(cpy);
  return (str);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<str_char>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    %esi,%ebp
  call   <strlen@plt>
  lea    0x1(%rax),%edi
  call   <malloc@plt>
  mov    %rax,%rbx
  movzbl (%r12),%edx
  test   %dl,%dl
  je     L27
  mov    $0x0,%eax
L5:
  mov    %dl,(%rbx,%rax,1)
  add    $0x1,%rax
  movzbl (%r12,%rax,1),%edx
  test   %dl,%dl
  jne    L5
L30:
  movb   $0x0,(%rbx,%rax,1)
  mov    %r12,%rdi
  call   <free@plt>
  mov    %rbx,%rdi
  call   <strlen@plt>
  lea    0x2(%rax),%edi
  call   <malloc@plt>
  mov    %rax,%r12
  movzbl (%rbx),%edx
  test   %dl,%dl
  je     L8
  mov    $0x0,%eax
L0:
  mov    %dl,(%r12,%rax,1)
  add    $0x1,%rax
  movzbl (%rbx,%rax,1),%edx
  test   %dl,%dl
  jne    L0
L24:
  mov    %bpl,(%r12,%rax,1)
  movb   $0x0,0x1(%r12,%rax,1)
  mov    %rbx,%rdi
  call   <free@plt>
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L27:
  mov    $0x0,%eax
  jmp    L30
L8:
  mov    $0x0,%eax
  jmp    L24
```

assistant: ```c
char *str_char(char *str, char c) {
  char *cpy;
  int y;
  cpy = malloc(sizeof(char) * (strlen(str) + 1));
  y = 0;
  while (str[y] != '\0') {
    cpy[y] = str[y];
    y++;
  }
  cpy[y] = str[y];
  free(str);
  str = malloc(sizeof(char) * (strlen(cpy) + 2));
  y = 0;
  while (cpy[y] != '\0') {
    str[y] = cpy[y];
    y++;
  }
  str[y] = c;
  str[y + 1] = '\0';
  free(cpy);
  return (str);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<str_char>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  mov    %esi,%ebx
  call   <strlen@plt>
  lea    0x1(%rax),%edi
  call   <malloc@plt>
  movzbl (%r12),%edx
  mov    %rax,%rbp
  test   %dl,%dl
  je     L29
  xor    %eax,%eax
  nopl   0x0(%rax,%rax,1)
L45:
  mov    %dl,0x0(%rbp,%rax,1)
  add    $0x1,%rax
  movzbl (%r12,%rax,1),%edx
  test   %dl,%dl
  jne    L45
  add    %rbp,%rax
L29:
  movb   $0x0,(%rax)
  mov    %r12,%rdi
  call   <free@plt>
  mov    %rbp,%rdi
  call   <strlen@plt>
  lea    0x2(%rax),%edi
  call   <malloc@plt>
  movzbl 0x0(%rbp),%edx
  mov    %rax,%r12
  test   %dl,%dl
  je     L37
  xor    %eax,%eax
  nopl   0x0(%rax)
L7:
  mov    %dl,(%r12,%rax,1)
  mov    %rax,%rcx
  add    $0x1,%rax
  movzbl 0x0(%rbp,%rax,1),%edx
  test   %dl,%dl
  jne    L7
  add    %r12,%rax
  add    $0x2,%rcx
L1:
  mov    %bl,(%rax)
  mov    %rbp,%rdi
  movb   $0x0,(%r12,%rcx,1)
  call   <free@plt>
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L37:
  mov    $0x1,%ecx
  jmp    L1
```

assistant: ```c
char *str_char(char *str, char c) {
  char *cpy;
  int y;
  cpy = malloc(sizeof(char) * (strlen(str) + 1));
  y = 0;
  while (str[y] != '\0') {
    cpy[y] = str[y];
    y++;
  }
  cpy[y] = str[y];
  free(str);
  str = malloc(sizeof(char) * (strlen(cpy) + 2));
  y = 0;
  while (cpy[y] != '\0') {
    str[y] = cpy[y];
    y++;
  }
  str[y] = c;
  str[y + 1] = '\0';
  free(cpy);
  return (str);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<str_char>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  mov    %esi,%ebx
  call   <strlen@plt>
  lea    0x1(%rax),%edi
  call   <malloc@plt>
  movzbl (%r12),%edx
  mov    %rax,%rbp
  test   %dl,%dl
  je     L34
  xor    %eax,%eax
  nopl   0x0(%rax,%rax,1)
L42:
  mov    %dl,0x0(%rbp,%rax,1)
  add    $0x1,%rax
  movzbl (%r12,%rax,1),%edx
  test   %dl,%dl
  jne    L42
  add    %rbp,%rax
L34:
  movb   $0x0,(%rax)
  mov    %r12,%rdi
  call   <free@plt>
  mov    %rbp,%rdi
  call   <strlen@plt>
  lea    0x2(%rax),%edi
  call   <malloc@plt>
  movzbl 0x0(%rbp),%edx
  mov    %rax,%r12
  test   %dl,%dl
  je     L9
  xor    %eax,%eax
  nopl   0x0(%rax)
L18:
  mov    %dl,(%r12,%rax,1)
  mov    %rax,%rcx
  add    $0x1,%rax
  movzbl 0x0(%rbp,%rax,1),%edx
  test   %dl,%dl
  jne    L18
  add    %r12,%rax
  add    $0x2,%rcx
L0:
  mov    %bl,(%rax)
  mov    %rbp,%rdi
  movb   $0x0,(%r12,%rcx,1)
  call   <free@plt>
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L9:
  mov    $0x1,%ecx
  jmp    L0
```

assistant: ```c
char *str_char(char *str, char c) {
  char *cpy;
  int y;
  cpy = malloc(sizeof(char) * (strlen(str) + 1));
  y = 0;
  while (str[y] != '\0') {
    cpy[y] = str[y];
    y++;
  }
  cpy[y] = str[y];
  free(str);
  str = malloc(sizeof(char) * (strlen(cpy) + 2));
  y = 0;
  while (cpy[y] != '\0') {
    str[y] = cpy[y];
    y++;
  }
  str[y] = c;
  str[y + 1] = '\0';
  free(cpy);
  return (str);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<doPowerstackerSplash>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <eraseDisplay@plt>
  mov    $0x0,%eax
  call   <drawCenteredPowerstacker@plt>
  mov    $0x64,%edi
  call   <wait10Msec@plt>
  mov    $0x0,%eax
  call   <eraseDisplay@plt>
  mov    $0x32,%edi
  call   <wait10Msec@plt>
  mov    $0x0,%ecx
  mov    $0x64,%edx
  mov    $0x40,%esi
  mov    $0x0,%edi
  call   <nxtFillRect@plt>
  mov    $0x0,%eax
  call   <eraseCenteredPowerstacker@plt>
  mov    $0x12c,%edi
  call   <wait10Msec@plt>
  nop
  pop    %rbp
  ret
```

assistant: ```c
void doPowerstackerSplash() {
  eraseDisplay();
  drawCenteredPowerstacker();
  wait10Msec(100);
  eraseDisplay();
  wait10Msec(50);
  nxtFillRect(0, 64, 100, 0);
  eraseCenteredPowerstacker();
  wait10Msec(300);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<doPowerstackerSplash>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <eraseDisplay@plt>
  mov    $0x0,%eax
  call   <drawCenteredPowerstacker@plt>
  mov    $0x64,%edi
  call   <wait10Msec@plt>
  mov    $0x0,%eax
  call   <eraseDisplay@plt>
  mov    $0x32,%edi
  call   <wait10Msec@plt>
  mov    $0x0,%ecx
  mov    $0x64,%edx
  mov    $0x40,%esi
  mov    $0x0,%edi
  call   <nxtFillRect@plt>
  mov    $0x0,%eax
  call   <eraseCenteredPowerstacker@plt>
  mov    $0x12c,%edi
  call   <wait10Msec@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void doPowerstackerSplash() {
  eraseDisplay();
  drawCenteredPowerstacker();
  wait10Msec(100);
  eraseDisplay();
  wait10Msec(50);
  nxtFillRect(0, 64, 100, 0);
  eraseCenteredPowerstacker();
  wait10Msec(300);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<doPowerstackerSplash>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <eraseDisplay@plt>
  xor    %eax,%eax
  call   <drawCenteredPowerstacker@plt>
  mov    $0x64,%edi
  call   <wait10Msec@plt>
  xor    %eax,%eax
  call   <eraseDisplay@plt>
  mov    $0x32,%edi
  call   <wait10Msec@plt>
  xor    %edi,%edi
  xor    %ecx,%ecx
  mov    $0x64,%edx
  mov    $0x40,%esi
  call   <nxtFillRect@plt>
  xor    %eax,%eax
  call   <eraseCenteredPowerstacker@plt>
  mov    $0x12c,%edi
  add    $0x8,%rsp
  jmp    10a0 <wait10Msec@plt>
```

assistant: ```c
void doPowerstackerSplash() {
  eraseDisplay();
  drawCenteredPowerstacker();
  wait10Msec(100);
  eraseDisplay();
  wait10Msec(50);
  nxtFillRect(0, 64, 100, 0);
  eraseCenteredPowerstacker();
  wait10Msec(300);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<doPowerstackerSplash>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <eraseDisplay@plt>
  xor    %eax,%eax
  call   <drawCenteredPowerstacker@plt>
  mov    $0x64,%edi
  call   <wait10Msec@plt>
  xor    %eax,%eax
  call   <eraseDisplay@plt>
  mov    $0x32,%edi
  call   <wait10Msec@plt>
  xor    %edi,%edi
  xor    %ecx,%ecx
  mov    $0x64,%edx
  mov    $0x40,%esi
  call   <nxtFillRect@plt>
  xor    %eax,%eax
  call   <eraseCenteredPowerstacker@plt>
  mov    $0x12c,%edi
  add    $0x8,%rsp
  jmp    10a0 <wait10Msec@plt>
```

assistant: ```c
void doPowerstackerSplash() {
  eraseDisplay();
  drawCenteredPowerstacker();
  wait10Msec(100);
  eraseDisplay();
  wait10Msec(50);
  nxtFillRect(0, 64, 100, 0);
  eraseCenteredPowerstacker();
  wait10Msec(300);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<remapPIC>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    $0x21,%edi
  call   <inportb@plt>
  mov    %al,-0x2(%rbp)
  mov    $0xa1,%edi
  call   <inportb@plt>
  mov    %al,-0x1(%rbp)
  mov    $0x20,%esi
  mov    $0x20,%edi
  call   <outportb@plt>
  mov    $0x11,%esi
  mov    $0x20,%edi
  call   <outportb@plt>
  mov    $0x11,%esi
  mov    $0xa0,%edi
  call   <outportb@plt>
  mov    -0x14(%rbp),%eax
  movzbl %al,%eax
  mov    %eax,%esi
  mov    $0x21,%edi
  call   <outportb@plt>
  mov    -0x18(%rbp),%eax
  movzbl %al,%eax
  mov    %eax,%esi
  mov    $0xa1,%edi
  call   <outportb@plt>
  mov    $0x4,%esi
  mov    $0x21,%edi
  call   <outportb@plt>
  mov    $0x2,%esi
  mov    $0xa1,%edi
  call   <outportb@plt>
  mov    $0x1,%esi
  mov    $0x21,%edi
  call   <outportb@plt>
  mov    $0x1,%esi
  mov    $0xa1,%edi
  call   <outportb@plt>
  movzbl -0x2(%rbp),%eax
  mov    %eax,%esi
  mov    $0x21,%edi
  call   <outportb@plt>
  movzbl -0x1(%rbp),%eax
  mov    %eax,%esi
  mov    $0xa1,%edi
  call   <outportb@plt>
  nop
  leave
  ret
```

assistant: ```c
void remapPIC(int pic1, int pic2) {
  unsigned char md, sd;
  md = inportb(0x21);
  sd = inportb(0xA1);
  outportb(0x20, 0x20);
  outportb(0x20, 0x10 + 0x01);
  outportb(0xA0, 0x10 + 0x01);
  outportb(0x21, pic1);
  outportb(0xA1, pic2);
  outportb(0x21, 0x04);
  outportb(0xA1, 0x02);
  outportb(0x21, 0x01);
  outportb(0xA1, 0x01);
  outportb(0x21, md);
  outportb(0xA1, sd);
}
```