ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    0x8(%rsi),%r13
  test   %r13,%r13
  je     L98
L46:
  mov    %r13,%rdi
  call   <strlen@plt>
  mov    %rax,%r12
  mov    %eax,%esi
  movabs $0xaaaaaaaaaaaaaaab,%rdx
  mul    %rdx
  mov    %rdx,%rbx
  shr    %rbx
  and    $0xfffffffffffffffe,%rdx
  add    %rbx,%rdx
  mov    %r12,%rbx
  sub    %rdx,%rbx
  mov    %ebx,%edx
  mov    %ebx,%r14d
  movl   $0x0,0x4(%rsp)
  movw   $0x0,0x1(%rsp)
  movb   $0x0,0x3(%rsp)
  test   %ebx,%ebx
  jle    L74
  movzbl 0x0(%r13),%eax
  mov    %al,0x2(%rsp)
  cmp    $0x1,%ebx
  jle    L74
  movzbl 0x1(%r13),%eax
  mov    %al,0x3(%rsp)
L74:
  lea    0x1(%rsp),%rdi
  call   <num_to_word@plt>
  cmp    %r12,%rbx
  ja     L97
  movzbl 0x0(%r13,%rbx,1),%eax
  mov    %al,0x4(%rsp)
  movb   $0x0,0x5(%rsp)
  mov    %r14d,%ebp
  mov    $0x1,%ebx
  lea    0x4(%rsp),%r15
L0:
  movslq %ebx,%rax
  imul   $0x55555556,%rax,%rax
  shr    $0x20,%rax
  mov    %ebx,%edx
  sar    $0x1f,%edx
  sub    %edx,%eax
  lea    (%rax,%rax,2),%edx
  mov    %ebx,%eax
  sub    %edx,%eax
  add    $0x1,%ebp
  lea    0x1(%rax),%ebx
  mov    %r12d,%esi
  sub    %ebp,%esi
  add    %r14d,%esi
  mov    %ebp,%edx
  cmp    %r12,%rdx
  ja     L97
  movzbl 0x0(%r13,%rdx,1),%edx
  cltq
  mov    %dl,0x4(%rsp,%rax,1)
  movslq %ebx,%rax
  movb   $0x0,0x4(%rsp,%rax,1)
  cmp    $0x3,%ebx
  jne    L0
  mov    $0x3,%edx
  mov    %r15,%rdi
  call   <num_to_word@plt>
  jmp    L0
L98:
  mov    %rsi,%rbx
  call   <num_to_word_err@plt>
  mov    0x8(%rbx),%r13
  test   %r13,%r13
  jne    L46
  mov    $0x0,%eax
  call   <num_to_word_err@plt>
  jmp    L46
L97:
  lea    D4(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L35
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L35:
  call   <__stack_chk_fail@plt>
```

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

tool: D4: 


assistant: ```c
int main(int argc, char *argv[]) {
  if (!argv[1])
    num_to_word_err();
  char *array = argv[1];
  if (!array)
    num_to_word_err();
  size_t len = strlen(array);
  int tricount = 1;
  int place = len;
  int count = 0;
  int remainder = len % 3;
  unsigned int lena = remainder;
  char triple[4] = {0, 0, 0, 0};
  char remnant[3] = {0, 0, 0};
  while (count < remainder)
    remnant[count] = array[count++];
  num_to_word(remnant, place, remainder);
  while (lena <= len) {
    triple[tricount - 1] = array[lena];
    triple[tricount] = '\0';
    if (tricount == 3)
      num_to_word(triple, place, tricount);
    tricount %= 3;
    for (count = 0; tricount == 3 && count < 4;)
      triple[count++] = 0;
    ++lena;
    ++tricount;
    --place;
  }
  printf("\n");
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    0x8(%rsi),%r13
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  test   %r13,%r13
  je     L66
L59:
  mov    %r13,%rdi
  call   <strlen@plt>
  movl   $0x0,0x14(%rsp)
  movabs $0xaaaaaaaaaaaaaaab,%rdx
  mov    %rax,%r12
  mov    %eax,%esi
  movb   $0x0,0x13(%rsp)
  mul    %rdx
  mov    $0x0,%eax
  mov    %ax,0x11(%rsp)
  mov    %rdx,%rcx
  and    $0xfffffffffffffffe,%rdx
  shr    %rcx
  add    %rcx,%rdx
  mov    %r12,%rcx
  sub    %rdx,%rcx
  mov    %ecx,%edx
  mov    %ecx,%ebp
  je     L14
  movzbl 0x0(%r13),%eax
  mov    %al,0x12(%rsp)
  cmp    $0x2,%rcx
  jne    L25
  movzbl 0x1(%r13),%eax
  mov    %al,0x13(%rsp)
L25:
  lea    0x11(%rsp),%rdi
  mov    %rcx,0x8(%rsp)
  call   <num_to_word@plt>
  mov    0x8(%rsp),%rcx
  cmp    %r12,%rcx
  ja     L81
L46:
  mov    $0x1,%ebx
  lea    0x0(%rbp,%r12,1),%r15d
  mov    $0xaaaaaaab,%r14d
  jmp    L17
L99:
  mov    %ebx,%edx
  lea    0x1(%rbp),%ecx
  imul   %r14,%rdx
  mov    %rcx,%rbp
  shr    $0x21,%rdx
  lea    (%rdx,%rdx,2),%eax
  sub    %eax,%ebx
  lea    0x1(%rbx),%ebx
  cmp    %r12,%rcx
  ja     L81
L17:
  movzbl 0x0(%r13,%rcx,1),%eax
  lea    -0x1(%rbx),%edx
  mov    %r15d,%esi
  movslq %edx,%rdx
  sub    %ebp,%esi
  mov    %al,0x14(%rsp,%rdx,1)
  movslq %ebx,%rax
  movb   $0x0,0x14(%rsp,%rax,1)
  cmp    $0x3,%ebx
  jne    L99
  lea    0x14(%rsp),%rdi
  mov    $0x3,%edx
  call   <num_to_word@plt>
  jmp    L99
L81:
  lea    D8(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L35
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L14:
  lea    0x11(%rsp),%rdi
  xor    %edx,%edx
  mov    %rcx,0x8(%rsp)
  call   <num_to_word@plt>
  mov    0x8(%rsp),%rcx
  jmp    L46
L66:
  mov    %rsi,%rbx
  call   <num_to_word_err@plt>
  mov    0x8(%rbx),%r13
  test   %r13,%r13
  jne    L59
  xor    %eax,%eax
  call   <num_to_word_err@plt>
  jmp    L59
L35:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D8
type_guessed: string

tool: D8: 


assistant: ```c
int main(int argc, char *argv[]) {
  if (!argv[1])
    num_to_word_err();
  char *array = argv[1];
  if (!array)
    num_to_word_err();
  size_t len = strlen(array);
  int tricount = 1;
  int place = len;
  int count = 0;
  int remainder = len % 3;
  unsigned int lena = remainder;
  char triple[4] = {0, 0, 0, 0};
  char remnant[3] = {0, 0, 0};
  while (count < remainder)
    remnant[count] = array[count++];
  num_to_word(remnant, place, remainder);
  while (lena <= len) {
    triple[tricount - 1] = array[lena];
    triple[tricount] = '\0';
    if (tricount == 3)
      num_to_word(triple, place, tricount);
    tricount %= 3;
    for (count = 0; tricount == 3 && count < 4;)
      triple[count++] = 0;
    ++lena;
    ++tricount;
    --place;
  }
  printf("\n");
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    0x8(%rsi),%r12
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  test   %r12,%r12
  je     L105
L116:
  mov    %r12,%rdi
  call   <strlen@plt>
  movl   $0x0,0x4(%rsp)
  movabs $0xaaaaaaaaaaaaaaab,%rdx
  mov    %rax,%rbp
  mov    %eax,%esi
  movb   $0x0,0x3(%rsp)
  mul    %rdx
  mov    $0x0,%eax
  mov    %ax,0x1(%rsp)
  mov    %rdx,%r15
  and    $0xfffffffffffffffe,%rdx
  shr    %r15
  add    %r15,%rdx
  mov    %rbp,%r15
  sub    %rdx,%r15
  mov    %r15d,%edx
  mov    %r15d,%ebx
  je     L23
  movzbl (%r12),%eax
  mov    %al,0x2(%rsp)
  cmp    $0x2,%r15
  jne    L49
  movzbl 0x1(%r12),%eax
  mov    %al,0x3(%rsp)
L49:
  lea    0x1(%rsp),%rdi
  call   <num_to_word@plt>
  cmp    %rbp,%r15
  ja     L31
L76:
  mov    $0x1,%ecx
  lea    (%rbx,%rbp,1),%r13d
  mov    $0xaaaaaaab,%r14d
  cs nopw 0x0(%rax,%rax,1)
L63:
  movzbl (%r12,%r15,1),%eax
  lea    -0x1(%rcx),%edx
  mov    %r13d,%esi
  movslq %edx,%rdx
  sub    %ebx,%esi
  mov    %al,0x4(%rsp,%rdx,1)
  movslq %ecx,%rax
  movb   $0x0,0x4(%rsp,%rax,1)
  cmp    $0x3,%ecx
  je     L58
  mov    %ecx,%edx
  lea    0x1(%rbx),%r15d
  imul   %r14,%rdx
  mov    %r15,%rbx
  shr    $0x21,%rdx
  lea    (%rdx,%rdx,2),%eax
  sub    %eax,%ecx
  lea    0x1(%rcx),%ecx
  cmp    %rbp,%r15
  jbe    L63
L31:
  lea    D3(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L12
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L58:
  lea    0x4(%rsp),%rdi
  mov    $0x3,%edx
  lea    0x1(%rbx),%r15d
  call   <num_to_word@plt>
  mov    %r15,%rbx
  cmp    %r15,%rbp
  jb     L31
  mov    $0x1,%ecx
  jmp    L63
L23:
  lea    0x1(%rsp),%rdi
  xor    %edx,%edx
  call   <num_to_word@plt>
  jmp    L76
L105:
  mov    %rsi,%rbx
  call   <num_to_word_err@plt>
  mov    0x8(%rbx),%r12
  test   %r12,%r12
  jne    L116
  xor    %eax,%eax
  call   <num_to_word_err@plt>
  jmp    L116
L12:
  call   <__stack_chk_fail@plt>
```

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

tool: D3: 


assistant: ```c
int main(int argc, char *argv[]) {
  if (!argv[1])
    num_to_word_err();
  char *array = argv[1];
  if (!array)
    num_to_word_err();
  size_t len = strlen(array);
  int tricount = 1;
  int place = len;
  int count = 0;
  int remainder = len % 3;
  unsigned int lena = remainder;
  char triple[4] = {0, 0, 0, 0};
  char remnant[3] = {0, 0, 0};
  while (count < remainder)
    remnant[count] = array[count++];
  num_to_word(remnant, place, remainder);
  while (lena <= len) {
    triple[tricount - 1] = array[lena];
    triple[tricount] = '\0';
    if (tricount == 3)
      num_to_word(triple, place, tricount);
    tricount %= 3;
    for (count = 0; tricount == 3 && count < 4;)
      triple[count++] = 0;
    ++lena;
    ++tricount;
    --place;
  }
  printf("\n");
  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    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x1,-0x8(%rbp)
  jmp    L27
L23:
  movl   $0x0,-0x4(%rbp)
  mov    -0x8(%rbp),%edx
  movslq %edx,%rax
  imul   $0x55555556,%rax,%rax
  shr    $0x20,%rax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  mov    %eax,%ecx
  add    %ecx,%ecx
  add    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  test   %eax,%eax
  jne    L40
  mov    $0x46,%edi
  call   <putchar@plt>
  mov    $0x69,%edi
  call   <putchar@plt>
  mov    $0x7a,%edi
  call   <putchar@plt>
  mov    $0x7a,%edi
  call   <putchar@plt>
  movl   $0x1,-0x4(%rbp)
L40:
  mov    -0x8(%rbp),%edx
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    %eax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  mov    %eax,%ecx
  shl    $0x2,%ecx
  add    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  test   %eax,%eax
  jne    L1
  mov    $0x42,%edi
  call   <putchar@plt>
  mov    $0x75,%edi
  call   <putchar@plt>
  mov    $0x7a,%edi
  call   <putchar@plt>
  mov    $0x7a,%edi
  call   <putchar@plt>
  movl   $0x1,-0x4(%rbp)
L1:
  cmpl   $0x0,-0x4(%rbp)
  jne    L14
  mov    -0x8(%rbp),%eax
  mov    %eax,%edi
  call   <print_int@plt>
L14:
  mov    $0xa,%edi
  call   <putchar@plt>
  addl   $0x1,-0x8(%rbp)
L27:
  cmpl   $0x64,-0x8(%rbp)
  jle    L23
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main() {
  int i;
  for (i = 1; i <= 100; i++) {
    int done = 0;
    if (i % 3 == 0) {
      putchar('F');
      putchar('i');
      putchar('z');
      putchar('z');
      done = 1;
    }
    if (i % 5 == 0) {
      putchar('B');
      putchar('u');
      putchar('z');
      putchar('z');
      done = 1;
    }
    if (!done)
      print_int(i);
    putchar('\n');
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbx
  mov    $0x1,%ebx
  jmp    L40
L13:
  mov    $0x46,%edi
  call   <putchar@plt>
  mov    $0x69,%edi
  call   <putchar@plt>
  mov    $0x7a,%edi
  call   <putchar@plt>
  mov    $0x7a,%edi
  call   <putchar@plt>
  movslq %ebx,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x21,%rax
  mov    %ebx,%edx
  sar    $0x1f,%edx
  sub    %edx,%eax
  lea    (%rax,%rax,4),%eax
  cmp    %eax,%ebx
  je     L47
L36:
  mov    $0xa,%edi
  call   <putchar@plt>
  add    $0x1,%ebx
  cmp    $0x65,%ebx
  je     L2
L40:
  movslq %ebx,%rax
  imul   $0x55555556,%rax,%rax
  shr    $0x20,%rax
  mov    %ebx,%edx
  sar    $0x1f,%edx
  sub    %edx,%eax
  lea    (%rax,%rax,2),%eax
  cmp    %eax,%ebx
  je     L13
  movslq %ebx,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x21,%rax
  mov    %ebx,%edx
  sar    $0x1f,%edx
  sub    %edx,%eax
  lea    (%rax,%rax,4),%eax
  cmp    %eax,%ebx
  jne    L41
L47:
  mov    $0x42,%edi
  call   <putchar@plt>
  mov    $0x75,%edi
  call   <putchar@plt>
  mov    $0x7a,%edi
  call   <putchar@plt>
  mov    $0x7a,%edi
  call   <putchar@plt>
  jmp    L36
L41:
  mov    %ebx,%edi
  call   <print_int@plt>
  jmp    L36
L2:
  mov    $0x0,%eax
  pop    %rbx
  ret
```

assistant: ```c
int main() {
  int i;
  for (i = 1; i <= 100; i++) {
    int done = 0;
    if (i % 3 == 0) {
      putchar('F');
      putchar('i');
      putchar('z');
      putchar('z');
      done = 1;
    }
    if (i % 5 == 0) {
      putchar('B');
      putchar('u');
      putchar('z');
      putchar('z');
      done = 1;
    }
    if (!done)
      print_int(i);
    putchar('\n');
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %r12
  mov    $0xcccccccd,%r12d
  push   %rbp
  mov    $0x1,%ebp
  push   %rbx
  jmp    L31
  nopl   (%rax)
L40:
  test   %ebx,%ebx
  je     L45
  mov    %ebp,%edi
  call   <print_int@plt>
L30:
  mov    $0xa,%edi
  add    $0x1,%ebp
  call   <putchar@plt>
  cmp    $0x65,%ebp
  je     L18
L31:
  mov    %ebp,%ebx
  imul   %r12,%rbx
  shr    $0x22,%rbx
  lea    (%rbx,%rbx,4),%eax
  mov    %ebp,%ebx
  sub    %eax,%ebx
  imul   $0xaaaaaaab,%ebp,%eax
  cmp    $0x55555555,%eax
  ja     L40
  mov    $0x46,%edi
  call   <putchar@plt>
  mov    $0x69,%edi
  call   <putchar@plt>
  mov    $0x7a,%edi
  call   <putchar@plt>
  mov    $0x7a,%edi
  call   <putchar@plt>
  test   %ebx,%ebx
  jne    L30
L45:
  mov    $0x42,%edi
  call   <putchar@plt>
  mov    $0x75,%edi
  call   <putchar@plt>
  mov    $0x7a,%edi
  call   <putchar@plt>
  mov    $0x7a,%edi
  call   <putchar@plt>
  jmp    L30
L18:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int main() {
  int i;
  for (i = 1; i <= 100; i++) {
    int done = 0;
    if (i % 3 == 0) {
      putchar('F');
      putchar('i');
      putchar('z');
      putchar('z');
      done = 1;
    }
    if (i % 5 == 0) {
      putchar('B');
      putchar('u');
      putchar('z');
      putchar('z');
      done = 1;
    }
    if (!done)
      print_int(i);
    putchar('\n');
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r12
  mov    $0xcccccccd,%r12d
  push   %rbp
  mov    $0x1,%ebp
  push   %rbx
  jmp    L37
  nopl   (%rax)
L29:
  test   %ebx,%ebx
  je     L8
  mov    %ebp,%edi
  call   <print_int@plt>
L55:
  mov    $0xa,%edi
  add    $0x1,%ebp
  call   <putchar@plt>
  cmp    $0x65,%ebp
  je     L58
L37:
  mov    %ebp,%ebx
  imul   %r12,%rbx
  shr    $0x22,%rbx
  lea    (%rbx,%rbx,4),%eax
  mov    %ebp,%ebx
  sub    %eax,%ebx
  imul   $0xaaaaaaab,%ebp,%eax
  cmp    $0x55555555,%eax
  ja     L29
  mov    $0x46,%edi
  call   <putchar@plt>
  mov    $0x69,%edi
  call   <putchar@plt>
  mov    $0x7a,%edi
  call   <putchar@plt>
  mov    $0x7a,%edi
  call   <putchar@plt>
  test   %ebx,%ebx
  jne    L55
L8:
  mov    $0x42,%edi
  call   <putchar@plt>
  mov    $0x75,%edi
  call   <putchar@plt>
  mov    $0x7a,%edi
  call   <putchar@plt>
  mov    $0x7a,%edi
  call   <putchar@plt>
  jmp    L55
L58:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int main() {
  int i;
  for (i = 1; i <= 100; i++) {
    int done = 0;
    if (i % 3 == 0) {
      putchar('F');
      putchar('i');
      putchar('z');
      putchar('z');
      done = 1;
    }
    if (i % 5 == 0) {
      putchar('B');
      putchar('u');
      putchar('z');
      putchar('z');
      done = 1;
    }
    if (!done)
      print_int(i);
    putchar('\n');
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<hash>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L5
L2:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  add    %eax,-0x4(%rbp)
  addq   $0x1,-0x18(%rbp)
L5:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L2
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x288df0cb,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x4,%edx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  imul   $0x65,%edx,%edx
  sub    %edx,%eax
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jns    L18
  addl   $0x65,-0x4(%rbp)
L18:
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int hash(name)
char *name;
{
  int sum;
  for (sum = 0; *name != '\0'; name++)
    sum += (sum + *name);
  sum %= 101;
  if (sum < 0)
    sum += 101;
  return (sum);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<hash>:
  endbr64
  movzbl (%rdi),%eax
  test   %al,%al
  je     L16
  mov    $0x0,%edx
L6:
  movsbl %al,%eax
  lea    (%rax,%rdx,2),%edx
  add    $0x1,%rdi
  movzbl (%rdi),%eax
  test   %al,%al
  jne    L6
  movslq %edx,%rax
  imul   $0x288df0cb,%rax,%rax
  sar    $0x24,%rax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  imul   $0x65,%eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  lea    0x65(%rax),%edx
  cmovs  %edx,%eax
  ret
L16:
  mov    $0x0,%eax
  ret
```

assistant: ```c
int hash(name)
char *name;
{
  int sum;
  for (sum = 0; *name != '\0'; name++)
    sum += (sum + *name);
  sum %= 101;
  if (sum < 0)
    sum += 101;
  return (sum);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<hash>:
  endbr64
  movsbl (%rdi),%eax
  test   %al,%al
  je     L14
  xor    %edx,%edx
  nopl   (%rax)
L15:
  add    $0x1,%rdi
  lea    (%rax,%rdx,2),%edx
  movsbl (%rdi),%eax
  test   %al,%al
  jne    L15
  movslq %edx,%rax
  mov    %edx,%ecx
  imul   $0x288df0cb,%rax,%rax
  sar    $0x1f,%ecx
  sar    $0x24,%rax
  sub    %ecx,%eax
  imul   $0x65,%eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  lea    0x65(%rax),%edx
  cmovs  %edx,%eax
  ret
  nopl   0x0(%rax)
L14:
  xor    %eax,%eax
  ret
```

assistant: ```c
int hash(name)
char *name;
{
  int sum;
  for (sum = 0; *name != '\0'; name++)
    sum += (sum + *name);
  sum %= 101;
  if (sum < 0)
    sum += 101;
  return (sum);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<hash>:
  endbr64
  movsbl (%rdi),%eax
  test   %al,%al
  je     L0
  xor    %edx,%edx
  nopl   (%rax)
L11:
  add    $0x1,%rdi
  lea    (%rax,%rdx,2),%edx
  movsbl (%rdi),%eax
  test   %al,%al
  jne    L11
  movslq %edx,%rax
  mov    %edx,%ecx
  imul   $0x288df0cb,%rax,%rax
  sar    $0x1f,%ecx
  sar    $0x24,%rax
  sub    %ecx,%eax
  imul   $0x65,%eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  lea    0x65(%rax),%edx
  cmovs  %edx,%eax
  ret
  nopl   0x0(%rax)
L0:
  xor    %eax,%eax
  ret
```

assistant: ```c
int hash(name)
char *name;
{
  int sum;
  for (sum = 0; *name != '\0'; name++)
    sum += (sum + *name);
  sum %= 101;
  if (sum < 0)
    sum += 101;
  return (sum);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<fiat_p384_nonzero>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r13
  push   %r12
  push   %rbx
  mov    %rdi,-0x30(%rbp)
  mov    %rsi,-0x38(%rbp)
  mov    -0x38(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x38(%rbp),%rax
  add    $0x4,%rax
  mov    (%rax),%ecx
  mov    -0x38(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%esi
  mov    -0x38(%rbp),%rax
  add    $0xc,%rax
  mov    (%rax),%edi
  mov    -0x38(%rbp),%rax
  add    $0x10,%rax
  mov    (%rax),%r8d
  mov    -0x38(%rbp),%rax
  add    $0x14,%rax
  mov    (%rax),%r9d
  mov    -0x38(%rbp),%rax
  add    $0x18,%rax
  mov    (%rax),%r10d
  mov    -0x38(%rbp),%rax
  add    $0x1c,%rax
  mov    (%rax),%r11d
  mov    -0x38(%rbp),%rax
  add    $0x20,%rax
  mov    (%rax),%ebx
  mov    -0x38(%rbp),%rax
  add    $0x24,%rax
  mov    (%rax),%r12d
  mov    -0x38(%rbp),%rax
  add    $0x28,%rax
  mov    (%rax),%r13d
  mov    -0x38(%rbp),%rax
  add    $0x2c,%rax
  mov    (%rax),%eax
  or     %r13d,%eax
  or     %r12d,%eax
  or     %ebx,%eax
  or     %r11d,%eax
  or     %r10d,%eax
  or     %r9d,%eax
  or     %r8d,%eax
  or     %edi,%eax
  or     %esi,%eax
  or     %ecx,%eax
  or     %edx,%eax
  mov    %eax,-0x1c(%rbp)
  mov    -0x30(%rbp),%rax
  mov    -0x1c(%rbp),%edx
  mov    %edx,(%rax)
  nop
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %rbp
  ret
```

assistant: ```c
void fiat_p384_nonzero(uint32_t *out1, const uint32_t arg1[12]) {
  uint32_t x1;
  x1 = ((arg1[0]) | ((arg1[1]) | ((arg1[2]) | ((arg1[3]) | ((arg1[4]) | ((arg1[5]) | ((arg1[6]) | ((arg1[7]) | ((arg1[8]) | ((arg1[9]) | ((arg1[10]) | (arg1[11]))))))))))));
  *out1 = x1;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<fiat_p384_nonzero>:
  endbr64
  mov    (%rsi),%eax
  or     0x4(%rsi),%eax
  or     0x8(%rsi),%eax
  or     0xc(%rsi),%eax
  or     0x10(%rsi),%eax
  or     0x14(%rsi),%eax
  or     0x18(%rsi),%eax
  or     0x1c(%rsi),%eax
  or     0x20(%rsi),%eax
  or     0x24(%rsi),%eax
  or     0x28(%rsi),%eax
  or     0x2c(%rsi),%eax
  mov    %eax,(%rdi)
  ret
```

assistant: ```c
void fiat_p384_nonzero(uint32_t *out1, const uint32_t arg1[12]) {
  uint32_t x1;
  x1 = ((arg1[0]) | ((arg1[1]) | ((arg1[2]) | ((arg1[3]) | ((arg1[4]) | ((arg1[5]) | ((arg1[6]) | ((arg1[7]) | ((arg1[8]) | ((arg1[9]) | ((arg1[10]) | (arg1[11]))))))))))));
  *out1 = x1;
}
```
user: What is the c source code of the assembly code below?

```asm
<fiat_p384_nonzero>:
  endbr64
  mov    (%rsi),%eax
  or     0x4(%rsi),%eax
  or     0x8(%rsi),%eax
  or     0xc(%rsi),%eax
  or     0x10(%rsi),%eax
  or     0x14(%rsi),%eax
  or     0x18(%rsi),%eax
  or     0x1c(%rsi),%eax
  or     0x20(%rsi),%eax
  or     0x24(%rsi),%eax
  or     0x28(%rsi),%eax
  or     0x2c(%rsi),%eax
  mov    %eax,(%rdi)
  ret
```

assistant: ```c
void fiat_p384_nonzero(uint32_t *out1, const uint32_t arg1[12]) {
  uint32_t x1;
  x1 = ((arg1[0]) | ((arg1[1]) | ((arg1[2]) | ((arg1[3]) | ((arg1[4]) | ((arg1[5]) | ((arg1[6]) | ((arg1[7]) | ((arg1[8]) | ((arg1[9]) | ((arg1[10]) | (arg1[11]))))))))))));
  *out1 = x1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<fiat_p384_nonzero>:
  endbr64
  mov    (%rsi),%eax
  or     0x4(%rsi),%eax
  or     0x8(%rsi),%eax
  or     0xc(%rsi),%eax
  or     0x10(%rsi),%eax
  or     0x14(%rsi),%eax
  or     0x18(%rsi),%eax
  or     0x1c(%rsi),%eax
  or     0x20(%rsi),%eax
  or     0x24(%rsi),%eax
  or     0x28(%rsi),%eax
  or     0x2c(%rsi),%eax
  mov    %eax,(%rdi)
  ret
```

assistant: ```c
void fiat_p384_nonzero(uint32_t *out1, const uint32_t arg1[12]) {
  uint32_t x1;
  x1 = ((arg1[0]) | ((arg1[1]) | ((arg1[2]) | ((arg1[3]) | ((arg1[4]) | ((arg1[5]) | ((arg1[6]) | ((arg1[7]) | ((arg1[8]) | ((arg1[9]) | ((arg1[10]) | (arg1[11]))))))))))));
  *out1 = x1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<read_line>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L28
L14:
  mov    -0x8(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jge    L28
  mov    -0x8(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x8(%rbp)
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x4(%rbp),%edx
  mov    %dl,(%rax)
L28:
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0xa,-0x4(%rbp)
  jne    L14
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x8(%rbp),%eax
  leave
  ret
```

assistant: ```c
int read_line(char str[], int n) {
  int ch, i = 0;
  while ((ch = getchar()) != '\n')
    if (i < n)
      str[i++] = ch;
  str[i] = '\0';
  return i;
}
```
user: What is the c source code of the assembly code below?

```asm
<read_line>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r12
  mov    %esi,%ebp
  mov    $0x0,%ebx
L7:
  mov    %ebx,%r13d
L2:
  mov    $0x0,%eax
  call   <getchar@plt>
  cmp    $0xa,%eax
  je     L29
  cmp    %ebx,%ebp
  jle    L2
  mov    %al,(%r12,%rbx,1)
  add    $0x1,%rbx
  jmp    L7
L29:
  movslq %r13d,%rax
  movb   $0x0,(%r12,%rax,1)
  mov    %r13d,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int read_line(char str[], int n) {
  int ch, i = 0;
  while ((ch = getchar()) != '\n')
    if (i < n)
      str[i++] = ch;
  str[i] = '\0';
  return i;
}
```