ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Translate the below assembly code into C programming language:

```asm
<parse_cm_file>:
  endbr64
  push   %r13
  mov    %rdi,%r13
  lea    D28(%rip),%rdi
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x68,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x58(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  mov    %r13,%rdx
  mov    $0x3e8,%esi
  lea    D45(%rip),%rdi
  call   <init_city@plt>
  lea    D24(%rip),%rdi
  mov    %rax,%r12
  call   <puts@plt>
  lea    D32(%rip),%rsi
  mov    %r13,%rdi
  call   <fopen@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  je     L30
L39:
  mov    %rsp,%rbx
  jmp    L32
  nopl   0x0(%rax)
L38:
  mov    %rbx,%rsi
  mov    %r12,%rdi
  call   <decode_line@plt>
L32:
  mov    %rbp,%rdx
  mov    $0x50,%esi
  mov    %rbx,%rdi
  call   <fgets@plt>
  test   %rax,%rax
  jne    L38
  call   <init_fb@plt>
  mov    %r12,%rsi
  mov    %rax,%rdi
  call   <draw_city@plt>
  mov    %rbp,%rdi
  call   <close@plt>
  mov    0x58(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L31
  add    $0x68,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax,%rax,1)
L30:
  mov    %r13,%rsi
  lea    D48(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  call   <abort@plt>
  jmp    L39
L31:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D28
type: string
name: extract_symbolic_data
arguments: tag: D45
type: string
name: extract_symbolic_data
arguments: tag: D24
type: string
name: extract_symbolic_data
arguments: tag: D32
type: string
name: extract_symbolic_data
arguments: tag: D48
type: string

tool: D28: Initializing the city...

tool: D45: Ferrara

tool: D24: Parsing the file

tool: D32: r

tool: D48: Cannot open file %s


assistant: ```c
void parse_cm_file(char *filename) {
  struct city *the_city;
  printf("Initializing the city...\n");
  the_city = init_city("Ferrara", 1000, filename);
  printf("Parsing the file\n");
  char buff[81];
  FILE *fd = fopen(filename, "r");
  if (fd == NULL) {
    printf("Cannot open file %s\n", filename);
    abort();
  }
  while (fgets(buff, 80, fd) != NULL) {
    decode_line(the_city, buff);
  }
  struct fbdev *fb = init_fb();
  draw_city(fb, the_city);
  close(fd);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<get_byte>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  nop
L4:
  mov    $0x40013800,%eax
  mov    (%rax),%rax
  and    $0x20,%eax
  test   %rax,%rax
  je     L4
  mov    $0x40013804,%eax
  mov    (%rax),%rax
  pop    %rbp
  ret
```

assistant: ```c
char get_byte(void) {
  while (!((*(volatile unsigned long *)0x40013800) & 0x20))
    ;
  return (char)*(volatile unsigned long *)0x40013804;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<get_byte>:
  endbr64
L3:
  mov    0x40013800,%rax
  test   $0x20,%al
  je     L3
  mov    0x40013804,%rax
  ret
```

assistant: ```c
char get_byte(void) {
  while (!((*(volatile unsigned long *)0x40013800) & 0x20))
    ;
  return (char)*(volatile unsigned long *)0x40013804;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<get_byte>:
  endbr64
  nopl   0x0(%rax)
L3:
  mov    0x40013800,%rax
  test   $0x20,%al
  je     L3
  mov    0x40013804,%rax
  ret
```

assistant: ```c
char get_byte(void) {
  while (!((*(volatile unsigned long *)0x40013800) & 0x20))
    ;
  return (char)*(volatile unsigned long *)0x40013804;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<get_byte>:
  endbr64
  nopl   0x0(%rax)
L2:
  mov    0x40013800,%rax
  test   $0x20,%al
  je     L2
  mov    0x40013804,%rax
  ret
```

assistant: ```c
char get_byte(void) {
  while (!((*(volatile unsigned long *)0x40013800) & 0x20))
    ;
  return (char)*(volatile unsigned long *)0x40013804;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<ems_begin_>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <emsBegin@plt>
  nop
  leave
  ret
```

assistant: ```c
void ems_begin_(int *const status) {

  emsBegin(status);
  return;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ems_begin_>:
  endbr64
  sub    $0x8,%rsp
  call   <emsBegin@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void ems_begin_(int *const status) {

  emsBegin(status);
  return;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<ems_begin_>:
  endbr64
  jmp    1050 <emsBegin@plt>
```

assistant: ```c
void ems_begin_(int *const status) {

  emsBegin(status);
  return;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<ems_begin_>:
  endbr64
  jmp    1050 <emsBegin@plt>
```

assistant: ```c
void ems_begin_(int *const status) {

  emsBegin(status);
  return;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x10(%rbp)
  lea    D26(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D31(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x18(%rbp),%eax
  mov    %eax,-0x14(%rbp)
  jmp    L17
L32:
  mov    -0x14(%rbp),%edx
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %edx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%ecx
  mov    %ecx,%eax
  shl    $0x2,%eax
  add    %ecx,%eax
  add    %eax,%eax
  sub    %eax,%edx
  mov    %edx,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  imul   %eax,%eax
  imul   -0xc(%rbp),%eax
  add    %eax,-0x10(%rbp)
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x14(%rbp)
L17:
  cmpl   $0x0,-0x14(%rbp)
  jne    L32
  mov    -0x18(%rbp),%eax
  cmp    %eax,-0x10(%rbp)
  jne    L12
  mov    -0x18(%rbp),%eax
  mov    %eax,%esi
  lea    D23(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L40
L12:
  mov    -0x18(%rbp),%eax
  mov    %eax,%esi
  lea    D25(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L40:
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L37
  call   <__stack_chk_fail@plt>
L37:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D26
data_type: string
name: retrieve_data
arguments: data_label: D31
data_type: string
name: retrieve_data
arguments: data_label: D23
data_type: string
name: retrieve_data
arguments: data_label: D25
data_type: string

tool: D26: Enter a positive  integer: 

tool: D31: %d

tool: D23: %d is an Armstrong number.

tool: D25: %d is not an Armstrong number.

assistant: ```c
int main() {
  int n, n1, rem, num = 0;
  printf("Enter a positive  integer: ");
  scanf("%d", &n);
  n1 = n;
  while (n1 != 0) {
    rem = n1 % 10;
    num += rem * rem * rem;
    n1 /= 10;
  }
  if (num == n)
    printf("%d is an Armstrong number.", n);
  else
    printf("%d is not an Armstrong number.", n);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D1(%rip),%rdi
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D25(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%r8d
  test   %r8d,%r8d
  je     L34
  mov    %r8d,%edx
  mov    $0x0,%edi
L5:
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  mov    %eax,%esi
  lea    (%rax,%rax,4),%ecx
  add    %ecx,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,%edx
  imul   %eax,%edx
  imul   %edx,%eax
  add    %eax,%edi
  mov    %esi,%edx
  test   %esi,%esi
  jne    L5
  cmp    %edi,%r8d
  je     L34
  mov    %r8d,%esi
  lea    D11(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L40:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L46
  mov    $0x0,%eax
  add    $0x18,%rsp
  ret
L34:
  mov    %r8d,%esi
  lea    D5(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L40
L46:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D1
data_type: string
name: unpack_data
arguments: label: D25
data_type: string
name: unpack_data
arguments: label: D11
data_type: string
name: unpack_data
arguments: label: D5
data_type: string

tool: D1: Enter a positive  integer: 

tool: D25: %d

tool: D11: %d is not an Armstrong number.

tool: D5: %d is an Armstrong number.

assistant: ```c
int main() {
  int n, n1, rem, num = 0;
  printf("Enter a positive  integer: ");
  scanf("%d", &n);
  n1 = n;
  while (n1 != 0) {
    rem = n1 % 10;
    num += rem * rem * rem;
    n1 /= 10;
  }
  if (num == n)
    printf("%d is an Armstrong number.", n);
  else
    printf("%d is not an Armstrong number.", n);
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D26(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D4(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%r8d
  test   %r8d,%r8d
  je     L21
  mov    %r8d,%edx
  xor    %edi,%edi
  nopl   0x0(%rax)
L24:
  movslq %edx,%rax
  mov    %edx,%ecx
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%ecx
  sar    $0x22,%rax
  sub    %ecx,%eax
  lea    (%rax,%rax,4),%esi
  mov    %eax,%ecx
  mov    %edx,%eax
  add    %esi,%esi
  sub    %esi,%eax
  mov    %eax,%edx
  imul   %eax,%edx
  imul   %edx,%eax
  mov    %ecx,%edx
  add    %eax,%edi
  test   %ecx,%ecx
  jne    L24
  cmp    %edi,%r8d
  je     L21
  mov    %r8d,%esi
  lea    D28(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
L49:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L21:
  mov    %r8d,%esi
  lea    D16(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L49
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D26
type_infered: string
name: read_data
arguments: label: D4
type_infered: string
name: read_data
arguments: label: D28
type_infered: string
name: read_data
arguments: label: D16
type_infered: string

tool: D26: Enter a positive  integer: 

tool: D4: %d

tool: D28: %d is not an Armstrong number.

tool: D16: %d is an Armstrong number.

assistant: ```c
int main() {
  int n, n1, rem, num = 0;
  printf("Enter a positive  integer: ");
  scanf("%d", &n);
  n1 = n;
  while (n1 != 0) {
    rem = n1 % 10;
    num += rem * rem * rem;
    n1 /= 10;
  }
  if (num == n)
    printf("%d is an Armstrong number.", n);
  else
    printf("%d is not an Armstrong number.", n);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D27(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D16(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%r8d
  test   %r8d,%r8d
  je     L49
  mov    %r8d,%edx
  xor    %edi,%edi
  nopl   0x0(%rax)
L8:
  movslq %edx,%rax
  mov    %edx,%ecx
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%ecx
  sar    $0x22,%rax
  sub    %ecx,%eax
  lea    (%rax,%rax,4),%esi
  mov    %eax,%ecx
  mov    %edx,%eax
  add    %esi,%esi
  sub    %esi,%eax
  mov    %eax,%edx
  imul   %eax,%edx
  imul   %edx,%eax
  mov    %ecx,%edx
  add    %eax,%edi
  test   %ecx,%ecx
  jne    L8
  cmp    %edi,%r8d
  je     L49
  mov    %r8d,%esi
  lea    D15(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
L36:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L10
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L49:
  mov    %r8d,%esi
  lea    D38(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L36
L10:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D27
type_guessed: string
name: disassemble_data
arguments: label: D16
type_guessed: string
name: disassemble_data
arguments: label: D15
type_guessed: string
name: disassemble_data
arguments: label: D38
type_guessed: string

tool: D27: Enter a positive  integer: 

tool: D16: %d

tool: D15: %d is not an Armstrong number.

tool: D38: %d is an Armstrong number.

assistant: ```c
int main() {
  int n, n1, rem, num = 0;
  printf("Enter a positive  integer: ");
  scanf("%d", &n);
  n1 = n;
  while (n1 != 0) {
    rem = n1 % 10;
    num += rem * rem * rem;
    n1 /= 10;
  }
  if (num == n)
    printf("%d is an Armstrong number.", n);
  else
    printf("%d is not an Armstrong number.", n);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<insertn_sort_noninc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x1,-0xc(%rbp)
  jmp    L19
L13:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0xc(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x8(%rbp)
  jmp    L49
L10:
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x8(%rbp),%edx
  movslq %edx,%rdx
  add    $0x1,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  subl   $0x1,-0x8(%rbp)
L49:
  cmpl   $0x0,-0x8(%rbp)
  js     L23
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    %eax,-0x4(%rbp)
  jg     L10
L23:
  mov    -0x8(%rbp),%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  mov    %eax,(%rdx)
  addl   $0x1,-0xc(%rbp)
L19:
  mov    -0xc(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L13
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void insertn_sort_noninc(int A[], int n) {
  int i, j, key;
  for (i = 1; i < n; i++) {
    key = A[i];
    for (j = i - 1; j >= 0 && A[j] < key; j--)
      A[j + 1] = A[j];
    A[j + 1] = key;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<insertn_sort_noninc>:
  endbr64
  cmp    $0x1,%esi
  jle    L6
  mov    %rdi,%r9
  lea    -0x1(%rsi),%r10d
  mov    $0x0,%r8d
  jmp    L56
L9:
  movslq %edx,%rdx
  mov    %esi,0x4(%rdi,%rdx,4)
  add    $0x1,%r8
  add    $0x4,%r9
  cmp    %r10,%r8
  je     L6
L56:
  mov    0x4(%r9),%esi
  mov    %r8d,%edx
  mov    %r9,%rax
  test   %r8d,%r8d
  js     L9
L55:
  mov    (%rax),%ecx
  cmp    %esi,%ecx
  jge    L9
  mov    %ecx,0x4(%rax)
  sub    $0x1,%edx
  sub    $0x4,%rax
  cmp    $0xffffffff,%edx
  jne    L55
  jmp    L9
L6:
  ret
```

assistant: ```c
void insertn_sort_noninc(int A[], int n) {
  int i, j, key;
  for (i = 1; i < n; i++) {
    key = A[i];
    for (j = i - 1; j >= 0 && A[j] < key; j--)
      A[j + 1] = A[j];
    A[j + 1] = key;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<insertn_sort_noninc>:
  endbr64
  lea    -0x1(%rsi),%r9d
  xor    %r8d,%r8d
  cmp    $0x1,%esi
  jle    L10
L37:
  mov    0x4(%rdi,%r8,4),%esi
  mov    %r8,%rax
  nopl   0x0(%rax,%rax,1)
L46:
  mov    (%rdi,%rax,4),%edx
  mov    %eax,%ecx
  cmp    %esi,%edx
  jge    L31
  mov    %edx,0x4(%rdi,%rax,4)
  sub    $0x1,%rax
  cmp    $0xffffffff,%eax
  jne    L46
  mov    %rdi,%rax
L45:
  add    $0x1,%r8
  mov    %esi,(%rax)
  cmp    %r8,%r9
  jne    L37
L10:
  ret
  nopl   (%rax)
L31:
  add    $0x1,%ecx
  movslq %ecx,%rcx
  lea    (%rdi,%rcx,4),%rax
  jmp    L45
```

assistant: ```c
void insertn_sort_noninc(int A[], int n) {
  int i, j, key;
  for (i = 1; i < n; i++) {
    key = A[i];
    for (j = i - 1; j >= 0 && A[j] < key; j--)
      A[j + 1] = A[j];
    A[j + 1] = key;
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<insertn_sort_noninc>:
  endbr64
  lea    -0x1(%rsi),%r9d
  xor    %r8d,%r8d
  cmp    $0x1,%esi
  jle    L2
L1:
  mov    0x4(%rdi,%r8,4),%esi
  mov    %r8,%rax
  nopl   0x0(%rax,%rax,1)
L4:
  mov    (%rdi,%rax,4),%edx
  mov    %eax,%ecx
  cmp    %esi,%edx
  jge    L6
  mov    %edx,0x4(%rdi,%rax,4)
  sub    $0x1,%rax
  cmp    $0xffffffff,%eax
  jne    L4
  mov    %rdi,%rax
L27:
  add    $0x1,%r8
  mov    %esi,(%rax)
  cmp    %r8,%r9
  jne    L1
L2:
  ret
  nopl   (%rax)
L6:
  add    $0x1,%ecx
  movslq %ecx,%rcx
  lea    (%rdi,%rcx,4),%rax
  jmp    L27
```

assistant: ```c
void insertn_sort_noninc(int A[], int n) {
  int i, j, key;
  for (i = 1; i < n; i++) {
    key = A[i];
    for (j = i - 1; j >= 0 && A[j] < key; j--)
      A[j + 1] = A[j];
    A[j + 1] = key;
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<process>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L12
L6:
  mov    $0x0,%eax
  call   <sched_yield@plt>
  test   %rax,%rax
  je     L22
  mov    $0x0,%eax
  call   <abort@plt>
L22:
  addl   $0x1,-0x4(%rbp)
L12:
  cmpl   $0x1f3,-0x4(%rbp)
  jle    L6
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int process(void *arg) {
  int i;
  for (i = 0; i < 500; i++)
    if (sched_yield())
      abort();
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<process>:
  endbr64
  push   %rbx
  mov    $0x1f4,%ebx
  jmp    L31
L33:
  sub    $0x1,%ebx
  je     L4
L31:
  mov    $0x0,%eax
  call   <sched_yield@plt>
  test   %rax,%rax
  je     L33
  mov    $0x0,%eax
  call   <abort@plt>
  jmp    L33
L4:
  mov    $0x0,%eax
  pop    %rbx
  ret
```

assistant: ```c
int process(void *arg) {
  int i;
  for (i = 0; i < 500; i++)
    if (sched_yield())
      abort();
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<process>:
  endbr64
  push   %rbx
  mov    $0x1f4,%ebx
  jmp    L29
  nopl   0x0(%rax)
L31:
  sub    $0x1,%ebx
  je     L36
L29:
  xor    %eax,%eax
  call   <sched_yield@plt>
  test   %rax,%rax
  je     L31
  xor    %eax,%eax
  call   <abort@plt>
  sub    $0x1,%ebx
  jne    L29
L36:
  xor    %eax,%eax
  pop    %rbx
  ret
```

assistant: ```c
int process(void *arg) {
  int i;
  for (i = 0; i < 500; i++)
    if (sched_yield())
      abort();
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<process>:
  endbr64
  push   %rbx
  mov    $0x1f4,%ebx
  jmp    L39
  nopl   0x0(%rax)
L7:
  sub    $0x1,%ebx
  je     L2
L39:
  xor    %eax,%eax
  call   <sched_yield@plt>
  test   %rax,%rax
  je     L7
  xor    %eax,%eax
  call   <abort@plt>
  sub    $0x1,%ebx
  jne    L39
L2:
  xor    %eax,%eax
  pop    %rbx
  ret
```

assistant: ```c
int process(void *arg) {
  int i;
  for (i = 0; i < 500; i++)
    if (sched_yield())
      abort();
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<power>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  cmpl   $0x0,-0xc(%rbp)
  jne    L37
  mov    $0x1,%eax
  jmp    L10
L37:
  cmpl   $0x2,-0xc(%rbp)
  jne    L8
  mov    -0x8(%rbp),%rax
  imul   %rax,%rax
  jmp    L10
L8:
  mov    -0xc(%rbp),%eax
  and    $0x1,%eax
  test   %eax,%eax
  jne    L0
  mov    -0xc(%rbp),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <power>
  mov    $0x2,%esi
  mov    %rax,%rdi
  call   <power>
  jmp    L10
L0:
  mov    -0xc(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <power>
  imul   -0x8(%rbp),%rax
L10:
  leave
  ret
```

assistant: ```c
long int power(long int x, int n) {
  if (n == 0) {
    return 1;
  } else if (n == 2) {
    return x * x;
  } else if (n % 2 == 0) {
    return power(power(x, n / 2), 2);
  } else {
    return x * power(x, n - 1);
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<power>:
  endbr64
  mov    $0x1,%eax
  test   %esi,%esi
  je     L17
  push   %rbx
  mov    %rdi,%rbx
  cmp    $0x2,%esi
  je     L35
  test   $0x1,%sil
  jne    L18
  mov    %esi,%eax
  shr    $0x1f,%eax
  add    %eax,%esi
  sar    %esi
  call   <power>
  mov    %rax,%rdi
  mov    $0x2,%esi
  call   <power>
L44:
  pop    %rbx
  ret
L35:
  mov    %rdi,%rax
  imul   %rdi,%rax
  jmp    L44
L18:
  sub    $0x1,%esi
  call   <power>
  imul   %rbx,%rax
  jmp    L44
L17:
  ret
```

assistant: ```c
long int power(long int x, int n) {
  if (n == 0) {
    return 1;
  } else if (n == 2) {
    return x * x;
  } else if (n % 2 == 0) {
    return power(power(x, n / 2), 2);
  } else {
    return x * power(x, n - 1);
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<strncpyz>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  cmpq   $0x0,-0x20(%rbp)
  je     L14
  cmpq   $0x0,-0x18(%rbp)
  je     L14
  cmpl   $0x0,-0x24(%rbp)
  jg     L34
L14:
  mov    $0x0,%eax
  jmp    L26
L34:
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x24(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    -0x4(%rbp),%eax
  cmp    %edx,%eax
  cmovg  %edx,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%edx
  mov    -0x20(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strncpy@plt>
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x4(%rbp),%eax
L26:
  leave
  ret
```

assistant: ```c
int strncpyz(char *dest, const char *src, int max) {
  int len;
  if ((!src) || (!dest) || max < 1)
    return 0;
  len = strlen(src);
  len = (len < max - 1) ? len : max - 1;
  strncpy(dest, src, len);
  dest[len] = '\0';
  return len;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<strncpyz>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edx,%ebx
  test   %rdi,%rdi
  sete   %al
  test   %edx,%edx
  setle  %dl
  or     %dl,%al
  jne    L19
  mov    %rdi,%rbp
  mov    %rsi,%r12
  test   %rsi,%rsi
  je     L19
  mov    %rsi,%rdi
  call   <strlen@plt>
  sub    $0x1,%ebx
  cmp    %eax,%ebx
  cmovg  %eax,%ebx
  mov    %ebx,%edx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <strncpy@plt>
  movslq %ebx,%rax
  movb   $0x0,0x0(%rbp,%rax,1)
  jmp    L25
L19:
  mov    $0x0,%ebx
L25:
  mov    %ebx,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int strncpyz(char *dest, const char *src, int max) {
  int len;
  if ((!src) || (!dest) || max < 1)
    return 0;
  len = strlen(src);
  len = (len < max - 1) ? len : max - 1;
  strncpy(dest, src, len);
  dest[len] = '\0';
  return len;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<strncpyz>:
  endbr64
  test   %rdi,%rdi
  push   %r12
  mov    %edx,%r12d
  sete   %al
  test   %edx,%edx
  push   %rbp
  setle  %dl
  push   %rbx
  or     %dl,%al
  jne    L7
  mov    %rsi,%rbp
  test   %rsi,%rsi
  je     L7
  mov    %rdi,%rbx
  mov    %rsi,%rdi
  sub    $0x1,%r12d
  call   <strlen@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  cmp    %eax,%r12d
  cmovg  %eax,%r12d
  mov    %r12d,%edx
  call   <strncpy@plt>
  movslq %r12d,%rax
  movb   $0x0,(%rbx,%rax,1)
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   (%rax)
L7:
  xor    %r12d,%r12d
  pop    %rbx
  pop    %rbp
  mov    %r12d,%eax
  pop    %r12
  ret
```

assistant: ```c
int strncpyz(char *dest, const char *src, int max) {
  int len;
  if ((!src) || (!dest) || max < 1)
    return 0;
  len = strlen(src);
  len = (len < max - 1) ? len : max - 1;
  strncpy(dest, src, len);
  dest[len] = '\0';
  return len;
}
```