ylfeng/ReF-Decompile-dataset huggingface dataset:
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    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movb   $0x65,0xf(%rsp)
  lea    D52(%rip),%r12
  lea    D33(%rip),%rbx
  lea    D61(%rip),%rbp
  lea    D13(%rip),%r13
  jmp    L26
L4:
  mov    %r13,%rdi
  call   <puts@plt>
L58:
  lea    D21(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0xf(%rsp),%rsi
  lea    D0(%rip),%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  cmpb   $0x68,0xf(%rsp)
  je     L14
L26:
  movl   $0x0,0x10(%rsp)
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0x10(%rsp),%rsi
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  movl   $0x0,0x14(%rsp)
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0x14(%rsp),%rsi
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    0x10(%rsp),%edi
  test   %edi,%edi
  js     L4
  mov    0x14(%rsp),%esi
  test   %esi,%esi
  js     L4
  call   <kuvvet_alma@plt>
  mov    %rax,%rcx
  mov    0x14(%rsp),%edx
  mov    0x10(%rsp),%esi
  lea    D23(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L58
L14:
  lea    D41(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L35
  mov    $0x0,%eax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L35:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D52
data_type: string
name: read_data
arguments: label: D33
data_type: string
name: read_data
arguments: label: D61
data_type: string
name: read_data
arguments: label: D13
data_type: string
name: read_data
arguments: label: D21
data_type: string
name: read_data
arguments: label: D0
data_type: string
name: read_data
arguments: label: D23
data_type: string
name: read_data
arguments: label: D41
data_type: string

tool: D52: Kuvveti alinacak sayiyi giriniz: 

tool: D33: %d

tool: D61: Kuvveti giriniz: 

tool: D13: Sayiyi veya Kuvveti negatif girdiniz!!

tool: D21: Baska sayi hesaplamak istiyormusunuz?(e veya h): 

tool: D0:  %c

tool: D23: tabani %d kuvveti %d olan sayinin sonucu = %ld


tool: D41: Programdan ciktiniz.

assistant: ```c
int main() {
  char program_cikisi = 'e';
  while (program_cikisi != 'h') {
    int sayi = 0;
    printf("Kuvveti alinacak sayiyi giriniz: ");
    scanf("%d", &sayi);
    int kuvvet = 0;
    printf("Kuvveti giriniz: ");
    scanf("%d", &kuvvet);
    long int sonuc;
    if (sayi >= 0 && kuvvet >= 0) {
      sonuc = kuvvet_alma(sayi, kuvvet);
      printf("tabani %d kuvveti %d olan sayinin sonucu = %ld\n", sayi, kuvvet, sonuc);
    } else {
      printf("Sayiyi veya Kuvveti negatif girdiniz!!\n");
    }
    printf("Baska sayi hesaplamak istiyormusunuz?(e veya h): ");
    scanf(" %c", &program_cikisi);
  }
  printf("Programdan ciktiniz.");
  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   %r15
  lea    D49(%rip),%r15
  push   %r14
  push   %r13
  lea    D32(%rip),%r13
  push   %r12
  push   %rbp
  lea    D51(%rip),%rbp
  push   %rbx
  lea    D20(%rip),%rbx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    0x1f(%rsp),%rax
  movb   $0x65,0x1f(%rsp)
  lea    0x20(%rsp),%r14
  mov    %rax,0x8(%rsp)
  lea    0x24(%rsp),%r12
  jmp    L1
  nopl   0x0(%rax)
L21:
  lea    D31(%rip),%rdi
  call   <puts@plt>
L10:
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rsi
  xor    %eax,%eax
  lea    D78(%rip),%rdi
  call   <scanf@plt>
  cmpb   $0x68,0x1f(%rsp)
  je     L7
L1:
  mov    %r15,%rdi
  xor    %eax,%eax
  movl   $0x0,0x20(%rsp)
  call   <printf@plt>
  mov    %r14,%rsi
  mov    %rbx,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    %r13,%rdi
  xor    %eax,%eax
  movl   $0x0,0x24(%rsp)
  call   <printf@plt>
  mov    %rbx,%rdi
  mov    %r12,%rsi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    0x20(%rsp),%edi
  test   %edi,%edi
  js     L21
  mov    0x24(%rsp),%esi
  test   %esi,%esi
  js     L21
  call   <kuvvet_alma@plt>
  mov    0x24(%rsp),%edx
  mov    0x20(%rsp),%esi
  lea    D30(%rip),%rdi
  mov    %rax,%rcx
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L10
  nopl   0x0(%rax)
L7:
  xor    %eax,%eax
  lea    D29(%rip),%rdi
  call   <printf@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L50
  add    $0x38,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L50:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D49
data_type: string
name: unpack_data
arguments: tag: D32
data_type: string
name: unpack_data
arguments: tag: D51
data_type: string
name: unpack_data
arguments: tag: D20
data_type: string
name: unpack_data
arguments: tag: D31
data_type: string
name: unpack_data
arguments: tag: D78
data_type: string
name: unpack_data
arguments: tag: D30
data_type: string
name: unpack_data
arguments: tag: D29
data_type: string

tool: D49: Kuvveti alinacak sayiyi giriniz: 

tool: D32: Kuvveti giriniz: 

tool: D51: Baska sayi hesaplamak istiyormusunuz?(e veya h): 

tool: D20: %d

tool: D31: Sayiyi veya Kuvveti negatif girdiniz!!

tool: D78:  %c

tool: D30: tabani %d kuvveti %d olan sayinin sonucu = %ld


tool: D29: Programdan ciktiniz.

assistant: ```c
int main() {
  char program_cikisi = 'e';
  while (program_cikisi != 'h') {
    int sayi = 0;
    printf("Kuvveti alinacak sayiyi giriniz: ");
    scanf("%d", &sayi);
    int kuvvet = 0;
    printf("Kuvveti giriniz: ");
    scanf("%d", &kuvvet);
    long int sonuc;
    if (sayi >= 0 && kuvvet >= 0) {
      sonuc = kuvvet_alma(sayi, kuvvet);
      printf("tabani %d kuvveti %d olan sayinin sonucu = %ld\n", sayi, kuvvet, sonuc);
    } else {
      printf("Sayiyi veya Kuvveti negatif girdiniz!!\n");
    }
    printf("Baska sayi hesaplamak istiyormusunuz?(e veya h): ");
    scanf(" %c", &program_cikisi);
  }
  printf("Programdan ciktiniz.");
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %r15
  lea    D14(%rip),%r15
  push   %r14
  push   %r13
  lea    D63(%rip),%r13
  push   %r12
  push   %rbp
  lea    D0(%rip),%rbp
  push   %rbx
  lea    D45(%rip),%rbx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    0x1f(%rsp),%rax
  movb   $0x65,0x1f(%rsp)
  lea    0x20(%rsp),%r14
  mov    %rax,0x8(%rsp)
  lea    0x24(%rsp),%r12
  jmp    L58
  nopl   0x0(%rax)
L55:
  lea    D76(%rip),%rdi
  call   <puts@plt>
L41:
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rsi
  xor    %eax,%eax
  lea    D66(%rip),%rdi
  call   <scanf@plt>
  cmpb   $0x68,0x1f(%rsp)
  je     L46
L58:
  mov    %r15,%rdi
  xor    %eax,%eax
  movl   $0x0,0x20(%rsp)
  call   <printf@plt>
  mov    %r14,%rsi
  mov    %rbx,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    %r13,%rdi
  xor    %eax,%eax
  movl   $0x0,0x24(%rsp)
  call   <printf@plt>
  mov    %rbx,%rdi
  mov    %r12,%rsi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    0x20(%rsp),%edi
  test   %edi,%edi
  js     L55
  mov    0x24(%rsp),%esi
  test   %esi,%esi
  js     L55
  call   <kuvvet_alma@plt>
  mov    0x24(%rsp),%edx
  mov    0x20(%rsp),%esi
  lea    D73(%rip),%rdi
  mov    %rax,%rcx
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L41
  nopl   0x0(%rax)
L46:
  xor    %eax,%eax
  lea    D62(%rip),%rdi
  call   <printf@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L54
  add    $0x38,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L54:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D14
data_type: string
name: parse_data
arguments: tag: D63
data_type: string
name: parse_data
arguments: tag: D0
data_type: string
name: parse_data
arguments: tag: D45
data_type: string
name: parse_data
arguments: tag: D76
data_type: string
name: parse_data
arguments: tag: D66
data_type: string
name: parse_data
arguments: tag: D73
data_type: string
name: parse_data
arguments: tag: D62
data_type: string

tool: D14: Kuvveti alinacak sayiyi giriniz: 

tool: D63: Kuvveti giriniz: 

tool: D0: Baska sayi hesaplamak istiyormusunuz?(e veya h): 

tool: D45: %d

tool: D76: Sayiyi veya Kuvveti negatif girdiniz!!

tool: D66:  %c

tool: D73: tabani %d kuvveti %d olan sayinin sonucu = %ld


tool: D62: Programdan ciktiniz.

assistant: ```c
int main() {
  char program_cikisi = 'e';
  while (program_cikisi != 'h') {
    int sayi = 0;
    printf("Kuvveti alinacak sayiyi giriniz: ");
    scanf("%d", &sayi);
    int kuvvet = 0;
    printf("Kuvveti giriniz: ");
    scanf("%d", &kuvvet);
    long int sonuc;
    if (sayi >= 0 && kuvvet >= 0) {
      sonuc = kuvvet_alma(sayi, kuvvet);
      printf("tabani %d kuvveti %d olan sayinin sonucu = %ld\n", sayi, kuvvet, sonuc);
    } else {
      printf("Sayiyi veya Kuvveti negatif girdiniz!!\n");
    }
    printf("Baska sayi hesaplamak istiyormusunuz?(e veya h): ");
    scanf(" %c", &program_cikisi);
  }
  printf("Programdan ciktiniz.");
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<mystrncpy>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %edx,-0x14(%rbp)
  jmp    L32
L0:
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L34
  mov    -0x10(%rbp),%rdx
  lea    0x1(%rdx),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rcx
  mov    %rcx,-0x8(%rbp)
  movzbl (%rdx),%edx
  mov    %dl,(%rax)
  subl   $0x1,-0x14(%rbp)
L32:
  cmpl   $0x0,-0x14(%rbp)
  jg     L0
  mov    -0x8(%rbp),%rax
  movb   $0x0,(%rax)
  jmp    L11
L34:
  nop
L11:
  pop    %rbp
  ret
```

assistant: ```c
void mystrncpy(char *s, char *t, int n) {
  while (n > 0) {
    if (*t == '\0')
      return;
    *s++ = *t++;
    n--;
  }
  *s = '\0';
  return;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<mystrncpy>:
  endbr64
  test   %edx,%edx
  jle    L37
  mov    %edx,%edx
  add    %rdi,%rdx
L23:
  movzbl (%rsi),%eax
  test   %al,%al
  je     L20
  add    $0x1,%rsi
  add    $0x1,%rdi
  mov    %al,-0x1(%rdi)
  cmp    %rdx,%rdi
  jne    L23
L0:
  movb   $0x0,(%rdx)
L20:
  ret
L37:
  mov    %rdi,%rdx
  jmp    L0
```

assistant: ```c
void mystrncpy(char *s, char *t, int n) {
  while (n > 0) {
    if (*t == '\0')
      return;
    *s++ = *t++;
    n--;
  }
  *s = '\0';
  return;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<mystrncpy>:
  endbr64
  test   %edx,%edx
  jle    L5
  movslq %edx,%rdx
  add    %rdi,%rdx
  jmp    L32
L3:
  add    $0x1,%rdi
  add    $0x1,%rsi
  mov    %al,-0x1(%rdi)
  cmp    %rdi,%rdx
  je     L9
L32:
  movzbl (%rsi),%eax
  test   %al,%al
  jne    L3
  ret
L5:
  mov    %rdi,%rdx
L9:
  movb   $0x0,(%rdx)
  ret
```

assistant: ```c
void mystrncpy(char *s, char *t, int n) {
  while (n > 0) {
    if (*t == '\0')
      return;
    *s++ = *t++;
    n--;
  }
  *s = '\0';
  return;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<mystrncpy>:
  endbr64
  test   %edx,%edx
  jle    L30
  movslq %edx,%rdx
  add    %rdi,%rdx
  jmp    L26
L1:
  add    $0x1,%rdi
  add    $0x1,%rsi
  mov    %al,-0x1(%rdi)
  cmp    %rdi,%rdx
  je     L3
L26:
  movzbl (%rsi),%eax
  test   %al,%al
  jne    L1
  ret
L30:
  mov    %rdi,%rdx
L3:
  movb   $0x0,(%rdx)
  ret
```

assistant: ```c
void mystrncpy(char *s, char *t, int n) {
  while (n > 0) {
    if (*t == '\0')
      return;
    *s++ = *t++;
    n--;
  }
  *s = '\0';
  return;
}
```
user: Translate the below assembly code into C programming language:

```asm
<process_vm_write>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %edi,-0x34(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    %rdx,-0x48(%rbp)
  mov    %rcx,-0x50(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x48(%rbp),%rax
  mov    %rax,-0x30(%rbp)
  mov    -0x50(%rbp),%rax
  mov    %rax,-0x28(%rbp)
  mov    -0x40(%rbp),%rax
  mov    %rax,-0x20(%rbp)
  mov    -0x50(%rbp),%rax
  mov    %rax,-0x18(%rbp)
  lea    -0x20(%rbp),%rdx
  lea    -0x30(%rbp),%rsi
  mov    -0x34(%rbp),%eax
  mov    $0x0,%r9d
  mov    $0x1,%r8d
  mov    %rdx,%rcx
  mov    $0x1,%edx
  mov    %eax,%edi
  call   <process_vm_writev@plt>
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L6
  call   <__stack_chk_fail@plt>
L6:
  leave
  ret
```

assistant: ```c
ssize_t process_vm_write(pid_t pid, void *dest, void *src, size_t n) {
  struct iovec local = {src, n};
  struct iovec remote = {dest, n};
  return process_vm_writev(pid, &local, 1, &remote, 1, 0);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<process_vm_write>:
  endbr64
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rdx,(%rsp)
  mov    %rcx,0x8(%rsp)
  mov    %rsi,0x10(%rsp)
  mov    %rcx,0x18(%rsp)
  lea    0x10(%rsp),%rcx
  mov    %rsp,%rsi
  mov    $0x0,%r9d
  mov    $0x1,%r8d
  mov    $0x1,%edx
  call   <process_vm_writev@plt>
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L8
  add    $0x38,%rsp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
ssize_t process_vm_write(pid_t pid, void *dest, void *src, size_t n) {
  struct iovec local = {src, n};
  struct iovec remote = {dest, n};
  return process_vm_writev(pid, &local, 1, &remote, 1, 0);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<process_vm_write>:
  endbr64
  sub    $0x38,%rsp
  xor    %r9d,%r9d
  mov    $0x1,%r8d
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rdx,(%rsp)
  mov    $0x1,%edx
  mov    %rcx,0x8(%rsp)
  mov    %rsi,0x10(%rsp)
  mov    %rsp,%rsi
  mov    %rcx,0x18(%rsp)
  lea    0x10(%rsp),%rcx
  call   <process_vm_writev@plt>
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L7
  add    $0x38,%rsp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
ssize_t process_vm_write(pid_t pid, void *dest, void *src, size_t n) {
  struct iovec local = {src, n};
  struct iovec remote = {dest, n};
  return process_vm_writev(pid, &local, 1, &remote, 1, 0);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<process_vm_write>:
  endbr64
  sub    $0x38,%rsp
  xor    %r9d,%r9d
  mov    $0x1,%r8d
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rdx,(%rsp)
  mov    $0x1,%edx
  mov    %rcx,0x8(%rsp)
  mov    %rsi,0x10(%rsp)
  mov    %rsp,%rsi
  mov    %rcx,0x18(%rsp)
  lea    0x10(%rsp),%rcx
  call   <process_vm_writev@plt>
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L9
  add    $0x38,%rsp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
ssize_t process_vm_write(pid_t pid, void *dest, void *src, size_t n) {
  struct iovec local = {src, n};
  struct iovec remote = {dest, n};
  return process_vm_writev(pid, &local, 1, &remote, 1, 0);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<out_hour_12H>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    -0x28(%rbp),%rax
  mov    (%rax),%edx
  movslq %edx,%rax
  imul   $0x2aaaaaab,%rax,%rax
  shr    $0x20,%rax
  sar    %eax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%ecx
  mov    %ecx,%eax
  add    %eax,%eax
  add    %ecx,%eax
  shl    $0x2,%eax
  sub    %eax,%edx
  mov    %edx,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  je     L2
  mov    -0x4(%rbp),%eax
  jmp    L1
L2:
  mov    $0xc,%eax
L1:
  mov    -0x20(%rbp),%rsi
  mov    -0x18(%rbp),%rdi
  mov    $0x30,%r8d
  mov    %eax,%ecx
  mov    $0x2,%edx
  call   <out_dec@plt>
  leave
  ret
```

assistant: ```c
char *out_hour_12H(char *buf, char *end, const struct tm *t) {
  int hour = t->tm_hour % 12;
  return out_dec(buf, end, 2, hour ? hour : 12, '0');
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<out_hour_12H>:
  endbr64
  sub    $0x8,%rsp
  mov    (%rdx),%edx
  movslq %edx,%rcx
  imul   $0x2aaaaaab,%rcx,%rcx
  sar    $0x21,%rcx
  mov    %edx,%eax
  sar    $0x1f,%eax
  sub    %eax,%ecx
  lea    (%rcx,%rcx,2),%eax
  shl    $0x2,%eax
  sub    %eax,%edx
  mov    %edx,%ecx
  mov    $0xc,%eax
  cmove  %eax,%ecx
  mov    $0x30,%r8d
  mov    $0x2,%edx
  call   <out_dec@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
char *out_hour_12H(char *buf, char *end, const struct tm *t) {
  int hour = t->tm_hour % 12;
  return out_dec(buf, end, 2, hour ? hour : 12, '0');
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<out_hour_12H>:
  endbr64
  movslq (%rdx),%rcx
  mov    $0x30,%r8d
  mov    %rcx,%rdx
  imul   $0x2aaaaaab,%rcx,%rcx
  mov    %edx,%eax
  sar    $0x1f,%eax
  sar    $0x21,%rcx
  sub    %eax,%ecx
  lea    (%rcx,%rcx,2),%eax
  shl    $0x2,%eax
  sub    %eax,%edx
  mov    $0xc,%eax
  mov    %edx,%ecx
  mov    $0x2,%edx
  cmove  %eax,%ecx
  jmp    1050 <out_dec@plt>
```

assistant: ```c
char *out_hour_12H(char *buf, char *end, const struct tm *t) {
  int hour = t->tm_hour % 12;
  return out_dec(buf, end, 2, hour ? hour : 12, '0');
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<out_hour_12H>:
  endbr64
  movslq (%rdx),%rcx
  mov    $0x30,%r8d
  mov    %rcx,%rdx
  imul   $0x2aaaaaab,%rcx,%rcx
  mov    %edx,%eax
  sar    $0x1f,%eax
  sar    $0x21,%rcx
  sub    %eax,%ecx
  lea    (%rcx,%rcx,2),%eax
  shl    $0x2,%eax
  sub    %eax,%edx
  mov    $0xc,%eax
  mov    %edx,%ecx
  mov    $0x2,%edx
  cmove  %eax,%ecx
  jmp    1050 <out_dec@plt>
```

assistant: ```c
char *out_hour_12H(char *buf, char *end, const struct tm *t) {
  int hour = t->tm_hour % 12;
  return out_dec(buf, end, 2, hour ? hour : 12, '0');
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<swap>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    %rcx,-0x28(%rbp)
  mov    D36(%rip),%rax
  cmp    %rax,-0x28(%rbp)
  jbe    L9
  mov    D29(%rip),%rax
  mov    -0x28(%rbp),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <realloc@plt>
  mov    %rax,D48(%rip)
  mov    -0x28(%rbp),%rax
  mov    %rax,D48(%rip)
L9:
  mov    -0x1c(%rbp),%eax
  cltq
  imul   -0x28(%rbp),%rax
  mov    %rax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x20(%rbp),%eax
  cltq
  imul   -0x28(%rbp),%rax
  mov    %rax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x8(%rbp)
  mov    D48(%rip),%rax
  mov    -0x28(%rbp),%rdx
  mov    -0x10(%rbp),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0x28(%rbp),%rdx
  mov    -0x8(%rbp),%rcx
  mov    -0x10(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    D48(%rip),%rcx
  mov    -0x28(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  nop
  leave
  ret
```

assistant: ```c
void swap(void *base, int i, int j, size_t size) {
  static void *temp = NULL;
  static size_t max_size = 0;
  void *elem1, *elem2;
  if (size > max_size) {
    temp = realloc(temp, size);
    max_size = size;
  }
  elem1 = ((char *)base) + (i * size);
  elem2 = ((char *)base) + (j * size);
  memcpy(temp, elem1, size);
  memcpy(elem1, elem2, size);
  memcpy(elem2, temp, size);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<swap>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r13
  mov    %esi,%ebx
  mov    %edx,%ebp
  mov    %rcx,%r12
  cmp    %rcx,D57(%rip)
  jb     L12
L17:
  movslq %ebx,%rbx
  imul   %r12,%rbx
  add    %r13,%rbx
  movslq %ebp,%rbp
  imul   %r12,%rbp
  add    %r13,%rbp
  mov    %r12,%rdx
  mov    %rbx,%rsi
  mov    D14(%rip),%rdi
  call   <memcpy@plt>
  mov    %r12,%rdx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <memcpy@plt>
  mov    %r12,%rdx
  mov    D14(%rip),%rsi
  mov    %rbp,%rdi
  call   <memcpy@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L12:
  mov    %rcx,%rsi
  mov    D14(%rip),%rdi
  call   <realloc@plt>
  mov    %rax,D35(%rip)
  mov    %r12,D35(%rip)
  jmp    L17
```

assistant: ```c
void swap(void *base, int i, int j, size_t size) {
  static void *temp = NULL;
  static size_t max_size = 0;
  void *elem1, *elem2;
  if (size > max_size) {
    temp = realloc(temp, size);
    max_size = size;
  }
  elem1 = ((char *)base) + (i * size);
  elem2 = ((char *)base) + (j * size);
  memcpy(temp, elem1, size);
  memcpy(elem1, elem2, size);
  memcpy(elem2, temp, size);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<swap>:
  endbr64
  push   %r13
  movslq %edx,%r13
  push   %r12
  mov    %rcx,%r12
  push   %rbp
  movslq %esi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  cmp    %rcx,D16(%rip)
  mov    D48(%rip),%rdi
  jb     L13
L18:
  imul   %r12,%rbp
  mov    %r12,%rdx
  imul   %r12,%r13
  add    %rbx,%rbp
  add    %rbx,%r13
  mov    %rbp,%rsi
  call   <memcpy@plt>
  mov    %r12,%rdx
  mov    %r13,%rsi
  mov    %rbp,%rdi
  call   <memcpy@plt>
  mov    D48(%rip),%rsi
  add    $0x8,%rsp
  mov    %r12,%rdx
  pop    %rbx
  mov    %r13,%rdi
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    1060 <memcpy@plt>
  nopl   (%rax)
L13:
  mov    %rcx,%rsi
  call   <realloc@plt>
  mov    %r12,D31(%rip)
  mov    %rax,D31(%rip)
  mov    %rax,%rdi
  jmp    L18
```

assistant: ```c
void swap(void *base, int i, int j, size_t size) {
  static void *temp = NULL;
  static size_t max_size = 0;
  void *elem1, *elem2;
  if (size > max_size) {
    temp = realloc(temp, size);
    max_size = size;
  }
  elem1 = ((char *)base) + (i * size);
  elem2 = ((char *)base) + (j * size);
  memcpy(temp, elem1, size);
  memcpy(elem1, elem2, size);
  memcpy(elem2, temp, size);
}
```
user: What is the c source code of the assembly code below?

```asm
<swap>:
  endbr64
  push   %r13
  movslq %edx,%r13
  push   %r12
  mov    %rcx,%r12
  push   %rbp
  movslq %esi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  cmp    %rcx,D31(%rip)
  mov    D4(%rip),%rdi
  jb     L3
L6:
  imul   %r12,%rbp
  mov    %r12,%rdx
  imul   %r12,%r13
  add    %rbx,%rbp
  add    %rbx,%r13
  mov    %rbp,%rsi
  call   <memcpy@plt>
  mov    %r12,%rdx
  mov    %r13,%rsi
  mov    %rbp,%rdi
  call   <memcpy@plt>
  mov    D4(%rip),%rsi
  add    $0x8,%rsp
  mov    %r12,%rdx
  pop    %rbx
  mov    %r13,%rdi
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    1060 <memcpy@plt>
  nopl   (%rax)
L3:
  mov    %rcx,%rsi
  call   <realloc@plt>
  mov    %r12,D49(%rip)
  mov    %rax,D49(%rip)
  mov    %rax,%rdi
  jmp    L6
```

assistant: ```c
void swap(void *base, int i, int j, size_t size) {
  static void *temp = NULL;
  static size_t max_size = 0;
  void *elem1, *elem2;
  if (size > max_size) {
    temp = realloc(temp, size);
    max_size = size;
  }
  elem1 = ((char *)base) + (i * size);
  elem2 = ((char *)base) + (j * size);
  memcpy(temp, elem1, size);
  memcpy(elem1, elem2, size);
  memcpy(elem2, temp, size);
}
```
user: What is the c source code of the assembly code below?

```asm
<ems_facer_>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %edx,-0x34(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x20(%rbp),%rdx
  mov    -0x34(%rbp),%esi
  mov    -0x28(%rbp),%rax
  mov    %rdx,%rcx
  mov    $0xf,%edx
  mov    %rax,%rdi
  call   <cnfImpn@plt>
  mov    -0x30(%rbp),%rax
  mov    (%rax),%edx
  lea    -0x20(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <emsFacer@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L7
  call   <__stack_chk_fail@plt>
L7:
  leave
  ret
```

assistant: ```c
void ems_facer_(char *const token, int *const fstat, int token_length) {
  char ctok[15 + 1];

  cnfImpn(token, token_length, 15, ctok);
  emsFacer(ctok, *fstat);
  return;
}
```