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

```asm
<roman_to_int>:
  endbr64
  test   %rdi,%rdi
  je     L34
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  mov    %eax,%edx
  sub    $0x1,%edx
  js     L18
  cltq
  movslq %edx,%rcx
  mov    %edx,%edx
  xor    %esi,%esi
  lea    -0x2(%rbx,%rax,1),%rdi
  add    %rbx,%rcx
  xor    %r8d,%r8d
  sub    %rdx,%rdi
  lea    D0(%rip),%r9
  jmp    L0
  nopl   0x0(%rax)
L44:
  mov    %edx,%esi
L0:
  movzbl (%rcx),%eax
  xor    %edx,%edx
  sub    $0x43,%eax
  cmp    $0x15,%al
  ja     L26
  movzbl %al,%eax
  mov    (%r9,%rax,4),%edx
L26:
  mov    %edx,%eax
  neg    %eax
  cmp    %esi,%edx
  cmovge %edx,%eax
  sub    $0x1,%rcx
  add    %eax,%r8d
  cmp    %rdi,%rcx
  jne    L44
  mov    %r8d,%eax
  pop    %rbx
  ret
  nop
L18:
  xor    %r8d,%r8d
  pop    %rbx
  mov    %r8d,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L34:
  xor    %eax,%eax
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D0
type_infered: string

tool: D0: d

assistant: ```c
int roman_to_int(char *s) {
  if (!s) {
    return 0;
  }
  int num = 0, len, i, temp, last = 0;
  len = strlen(s);
  for (i = len - 1; i >= 0; --i) {
    switch (s[i]) {
    case 'I':
      temp = 1;
      break;
    case 'V':
      temp = 5;
      break;
    case 'X':
      temp = 10;
      break;
    case 'L':
      temp = 50;
      break;
    case 'C':
      temp = 100;
      break;
    case 'D':
      temp = 500;
      break;
    case 'M':
      temp = 1000;
      break;
    default:
      temp = 0;
      break;
    }
    num += (temp >= last ? temp : -1 * temp);
    last = temp;
  }
  return num;
}
```
user: Translate the below assembly code into C programming language:

```asm
<roman_to_int>:
  endbr64
  test   %rdi,%rdi
  je     L41
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  mov    %eax,%edx
  sub    $0x1,%edx
  js     L45
  cltq
  movslq %edx,%rcx
  mov    %edx,%edx
  xor    %esi,%esi
  lea    -0x2(%rbx,%rax,1),%rdi
  add    %rbx,%rcx
  xor    %r8d,%r8d
  sub    %rdx,%rdi
  lea    D7(%rip),%r9
  jmp    L23
  nopl   0x0(%rax)
L19:
  mov    %edx,%esi
L23:
  movzbl (%rcx),%eax
  xor    %edx,%edx
  sub    $0x43,%eax
  cmp    $0x15,%al
  ja     L25
  movzbl %al,%eax
  mov    (%r9,%rax,4),%edx
L25:
  mov    %edx,%eax
  neg    %eax
  cmp    %esi,%edx
  cmovge %edx,%eax
  sub    $0x1,%rcx
  add    %eax,%r8d
  cmp    %rdi,%rcx
  jne    L19
  mov    %r8d,%eax
  pop    %rbx
  ret
  nop
L45:
  xor    %r8d,%r8d
  pop    %rbx
  mov    %r8d,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L41:
  xor    %eax,%eax
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D7
type: string

tool: D7: d

assistant: ```c
int roman_to_int(char *s) {
  if (!s) {
    return 0;
  }
  int num = 0, len, i, temp, last = 0;
  len = strlen(s);
  for (i = len - 1; i >= 0; --i) {
    switch (s[i]) {
    case 'I':
      temp = 1;
      break;
    case 'V':
      temp = 5;
      break;
    case 'X':
      temp = 10;
      break;
    case 'L':
      temp = 50;
      break;
    case 'C':
      temp = 100;
      break;
    case 'D':
      temp = 500;
      break;
    case 'M':
      temp = 1000;
      break;
    default:
      temp = 0;
      break;
    }
    num += (temp >= last ? temp : -1 * temp);
    last = temp;
  }
  return num;
}
```
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    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x0,-0x4(%rbp)
  jmp    L5
L11:
  mov    -0x4(%rbp),%edx
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x4(%rbp)
L5:
  cmpl   $0xff,-0x4(%rbp)
  jle    L11
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: read_data
arguments: data_label: D2
data_type: string

tool: D2: %d	%c		

assistant: ```c
main() {
  int ch;
  for (ch = 0; ch <= 255; ch++)
    printf("%d\t%c\t\t", ch, ch);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    $0x0,%ebx
  lea    D2(%rip),%rbp
L3:
  mov    %ebx,%edx
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  add    $0x1,%ebx
  cmp    $0x100,%ebx
  jne    L3
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D2
type_guessed: string

tool: D2: %d	%c		

assistant: ```c
main() {
  int ch;
  for (ch = 0; ch <= 255; ch++)
    printf("%d\t%c\t\t", ch, ch);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D4(%rip),%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  nopl   0x0(%rax,%rax,1)
L2:
  mov    %ebx,%edx
  mov    %ebx,%esi
  mov    %rbp,%rdi
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    $0x100,%ebx
  jne    L2
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: read_data
arguments: tag: D4
datatype: string

tool: D4: %d	%c		

assistant: ```c
main() {
  int ch;
  for (ch = 0; ch <= 255; ch++)
    printf("%d\t%c\t\t", ch, ch);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D9(%rip),%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  nopl   0x0(%rax,%rax,1)
L0:
  mov    %ebx,%edx
  mov    %ebx,%esi
  mov    %rbp,%rdi
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    $0x100,%ebx
  jne    L0
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D9
data_type: string

tool: D9: %d	%c		

assistant: ```c
main() {
  int ch;
  for (ch = 0; ch <= 255; ch++)
    printf("%d\t%c\t\t", ch, ch);
}
```
user: What is the c source code of the assembly code below?

```asm
<image_normalization>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  mov    %rdx,%r12
  mov    %rcx,%r13
  mov    %r8d,%r14d
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x8(%rsp),%rsi
  lea    0x10(%rsp),%rdi
  mov    %r8d,%ecx
  mov    %rbx,%rdx
  call   <getminmax@plt>
  lea    0xc(%rsp),%rsi
  lea    0x14(%rsp),%rdi
  mov    %r14d,%ecx
  mov    %rbp,%rdx
  call   <getminmax@plt>
  movss  0x8(%rsp),%xmm1
  maxss  0xc(%rsp),%xmm1
  movss  0x10(%rsp),%xmm2
  minss  0x14(%rsp),%xmm2
  subss  %xmm2,%xmm1
  comiss D18(%rip),%xmm1
  jbe    L60
  test   %r14d,%r14d
  jle    L28
  mov    %r14d,%r14d
  mov    $0x0,%eax
  cvtss2sd %xmm1,%xmm1
  movsd  D18(%rip),%xmm3
L58:
  movss  (%rbx,%rax,4),%xmm0
  subss  %xmm2,%xmm0
  cvtss2sd %xmm0,%xmm0
  mulsd  %xmm3,%xmm0
  divsd  %xmm1,%xmm0
  cvtsd2ss %xmm0,%xmm0
  movss  %xmm0,(%r12,%rax,4)
  movss  0x0(%rbp,%rax,4),%xmm0
  subss  %xmm2,%xmm0
  cvtss2sd %xmm0,%xmm0
  mulsd  %xmm3,%xmm0
  divsd  %xmm1,%xmm0
  cvtsd2ss %xmm0,%xmm0
  movss  %xmm0,0x0(%r13,%rax,4)
  add    $0x1,%rax
  cmp    %rax,%r14
  jne    L58
L28:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L19
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L60:
  mov    %r14d,%edx
  mov    $0x0,%eax
  test   %r14d,%r14d
  jle    L28
L17:
  movss  (%rbx,%rax,4),%xmm0
  movss  %xmm0,(%r12,%rax,4)
  movss  0x0(%rbp,%rax,4),%xmm0
  movss  %xmm0,0x0(%r13,%rax,4)
  add    $0x1,%rax
  cmp    %rdx,%rax
  jne    L17
  jmp    L28
L19:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D18
type: f64

tool: D18: 255

assistant: ```c
void image_normalization(const float *I0, const float *I1, float *I0n, float *I1n, int size) {
  float max0, max1, min0, min1;
  getminmax(&min0, &max0, I0, size);
  getminmax(&min1, &max1, I1, size);
  const float max = (max0 > max1) ? max0 : max1;
  const float min = (min0 < min1) ? min0 : min1;
  const float den = max - min;
  if (den > 0)
    for (int i = 0; i < size; i++) {
      I0n[i] = 255.0 * (I0[i] - min) / den;
      I1n[i] = 255.0 * (I1[i] - min) / den;
    }
  else
    for (int i = 0; i < size; i++) {
      I0n[i] = I0[i];
      I1n[i] = I1[i];
    }
}
```
user: Translate the below assembly code into C programming language:

```asm
<image_normalization>:
  endbr64
  push   %r14
  mov    %r8d,%r14d
  push   %r13
  mov    %rcx,%r13
  mov    %r8d,%ecx
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %rbx,%rdx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x8(%rsp),%rsi
  lea    0x10(%rsp),%rdi
  call   <getminmax@plt>
  lea    0xc(%rsp),%rsi
  mov    %r14d,%ecx
  mov    %rbp,%rdx
  lea    0x14(%rsp),%rdi
  call   <getminmax@plt>
  movss  0x8(%rsp),%xmm1
  movss  0x10(%rsp),%xmm2
  maxss  0xc(%rsp),%xmm1
  minss  0x14(%rsp),%xmm2
  subss  %xmm2,%xmm1
  comiss D15(%rip),%xmm1
  jbe    L23
  test   %r14d,%r14d
  jle    L25
  movsd  D15(%rip),%xmm3
  movslq %r14d,%r8
  xor    %eax,%eax
  cvtss2sd %xmm1,%xmm1
  nopl   0x0(%rax,%rax,1)
L46:
  movss  (%rbx,%rax,4),%xmm0
  subss  %xmm2,%xmm0
  cvtss2sd %xmm0,%xmm0
  mulsd  %xmm3,%xmm0
  divsd  %xmm1,%xmm0
  cvtsd2ss %xmm0,%xmm0
  movss  %xmm0,(%r12,%rax,4)
  movss  0x0(%rbp,%rax,4),%xmm0
  subss  %xmm2,%xmm0
  cvtss2sd %xmm0,%xmm0
  mulsd  %xmm3,%xmm0
  divsd  %xmm1,%xmm0
  cvtsd2ss %xmm0,%xmm0
  movss  %xmm0,0x0(%r13,%rax,4)
  add    $0x1,%rax
  cmp    %rax,%r8
  jne    L46
L25:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L44
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  cs nopw 0x0(%rax,%rax,1)
L23:
  movslq %r14d,%rdx
  xor    %eax,%eax
  test   %r14d,%r14d
  jle    L25
  nopw   0x0(%rax,%rax,1)
L64:
  movss  (%rbx,%rax,4),%xmm0
  movss  %xmm0,(%r12,%rax,4)
  movss  0x0(%rbp,%rax,4),%xmm0
  movss  %xmm0,0x0(%r13,%rax,4)
  add    $0x1,%rax
  cmp    %rax,%rdx
  jne    L64
  jmp    L25
L44:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D15
type: f64

tool: D15: 255

assistant: ```c
void image_normalization(const float *I0, const float *I1, float *I0n, float *I1n, int size) {
  float max0, max1, min0, min1;
  getminmax(&min0, &max0, I0, size);
  getminmax(&min1, &max1, I1, size);
  const float max = (max0 > max1) ? max0 : max1;
  const float min = (min0 < min1) ? min0 : min1;
  const float den = max - min;
  if (den > 0)
    for (int i = 0; i < size; i++) {
      I0n[i] = 255.0 * (I0[i] - min) / den;
      I1n[i] = 255.0 * (I1[i] - min) / den;
    }
  else
    for (int i = 0; i < size; i++) {
      I0n[i] = I0[i];
      I1n[i] = I1[i];
    }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<is_prime>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  movl   $0x2,-0x4(%rbp)
  jmp    L19
L10:
  addl   $0x1,-0x4(%rbp)
L19:
  mov    -0x4(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jge    L42
  mov    -0x14(%rbp),%eax
  cltd
  idivl  -0x4(%rbp)
  mov    %edx,%eax
  test   %eax,%eax
  jne    L10
L42:
  mov    -0x4(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jge    L48
  mov    $0x0,%eax
  jmp    L12
L48:
  mov    $0x1,%eax
L12:
  pop    %rbp
  ret
```

assistant: ```c
int is_prime(int num) {
  int i;
  for (i = 2; i < num && num % i != 0; i++)
    ;
  if (i < num)
    return 0;
  else
    return 1;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<is_prime>:
  endbr64
  cmp    $0x2,%edi
  jle    L59
  test   $0x1,%dil
  je     L18
  mov    $0x2,%ecx
L51:
  add    $0x1,%ecx
  cmp    %ecx,%edi
  je     L49
  mov    %edi,%eax
  cltd
  idiv   %ecx
  test   %edx,%edx
  jne    L51
  jmp    L4
L59:
  mov    $0x2,%ecx
  jmp    L4
L18:
  mov    $0x2,%ecx
  jmp    L4
L49:
  mov    %edi,%ecx
L4:
  cmp    %ecx,%edi
  setle  %al
  movzbl %al,%eax
  ret
```

assistant: ```c
int is_prime(int num) {
  int i;
  for (i = 2; i < num && num % i != 0; i++)
    ;
  if (i < num)
    return 0;
  else
    return 1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<is_prime>:
  endbr64
  cmp    $0x2,%edi
  jle    L10
  test   $0x1,%dil
  je     L37
  mov    $0x2,%ecx
  jmp    L31
  cs nopw 0x0(%rax,%rax,1)
L22:
  mov    %edi,%eax
  cltd
  idiv   %ecx
  test   %edx,%edx
  je     L47
L31:
  add    $0x1,%ecx
  cmp    %ecx,%edi
  jne    L22
L10:
  mov    $0x1,%edx
L47:
  mov    %edx,%eax
  ret
L37:
  xor    %edx,%edx
  jmp    L47
```

assistant: ```c
int is_prime(int num) {
  int i;
  for (i = 2; i < num && num % i != 0; i++)
    ;
  if (i < num)
    return 0;
  else
    return 1;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<is_prime>:
  endbr64
  cmp    $0x2,%edi
  jle    L36
  test   $0x1,%dil
  je     L18
  mov    $0x2,%ecx
  jmp    L16
  cs nopw 0x0(%rax,%rax,1)
L27:
  mov    %edi,%eax
  cltd
  idiv   %ecx
  test   %edx,%edx
  je     L3
L16:
  add    $0x1,%ecx
  cmp    %ecx,%edi
  jne    L27
L36:
  mov    $0x1,%edx
L3:
  mov    %edx,%eax
  ret
L18:
  xor    %edx,%edx
  jmp    L3
```

assistant: ```c
int is_prime(int num) {
  int i;
  for (i = 2; i < num && num % i != 0; i++)
    ;
  if (i < num)
    return 0;
  else
    return 1;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<buzz>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x10(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <gettimeofday@plt>
  mov    -0x10(%rbp),%edx
  mov    -0x24(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x10(%rbp),%eax
  cmp    $0xf4240,%eax
  jle    L22
  mov    -0x10(%rbp),%eax
  sub    $0xf4240,%eax
  mov    %eax,-0x10(%rbp)
  mov    -0xc(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0xc(%rbp)
L22:
  lea    -0x18(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <gettimeofday@plt>
  lea    -0x18(%rbp),%rdx
  lea    -0x10(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <ftimediff@plt>
  movd   %xmm0,%eax
  mov    %eax,-0x1c(%rbp)
  movss  -0x1c(%rbp),%xmm0
  pxor   %xmm1,%xmm1
  comiss %xmm1,%xmm0
  ja     L22
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L28
  call   <__stack_chk_fail@plt>
L28:
  leave
  ret
```

assistant: ```c
void buzz(int delay) {
  struct timeval now, stop;
  float diff;
  gettimeofday(&stop, 0);
  stop.tv_usec += delay;
  if (stop.tv_usec > 1000000) {
    stop.tv_usec -= 1000000;
    stop.tv_sec += 1;
  }
  do {
    gettimeofday(&now, 0);
    diff = ftimediff(&stop, &now);
  } while (diff > 0.0);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<buzz>:
  endbr64
  push   %rbx
  sub    $0x20,%rsp
  mov    %edi,%ebx
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rdi
  mov    $0x0,%esi
  call   <gettimeofday@plt>
  add    0x10(%rsp),%ebx
  mov    %ebx,%edi
  cmp    $0xf4240,%ebx
  jg     L25
  mov    %ebx,0x10(%rsp)
L11:
  lea    0x8(%rsp),%rbx
  mov    $0x0,%esi
  mov    %rbx,%rdi
  call   <gettimeofday@plt>
  lea    0x10(%rsp),%rdi
  mov    %rbx,%rsi
  call   <ftimediff@plt>
  pxor   %xmm1,%xmm1
  comiss %xmm1,%xmm0
  ja     L11
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L39
  add    $0x20,%rsp
  pop    %rbx
  ret
L25:
  sub    $0xf4240,%edi
  mov    %edi,0x10(%rsp)
  addl   $0x1,0x14(%rsp)
  jmp    L11
L39:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void buzz(int delay) {
  struct timeval now, stop;
  float diff;
  gettimeofday(&stop, 0);
  stop.tv_usec += delay;
  if (stop.tv_usec > 1000000) {
    stop.tv_usec -= 1000000;
    stop.tv_sec += 1;
  }
  do {
    gettimeofday(&now, 0);
    diff = ftimediff(&stop, &now);
  } while (diff > 0.0);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<buzz>:
  endbr64
  push   %rbp
  xor    %esi,%esi
  push   %rbx
  mov    %edi,%ebx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rbp
  mov    %rbp,%rdi
  call   <gettimeofday@plt>
  add    0x10(%rsp),%ebx
  mov    %ebx,%edi
  cmp    $0xf4240,%ebx
  jg     L1
  mov    %ebx,0x10(%rsp)
L10:
  lea    0x8(%rsp),%rbx
  nopw   0x0(%rax,%rax,1)
L2:
  xor    %esi,%esi
  mov    %rbx,%rdi
  call   <gettimeofday@plt>
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <ftimediff@plt>
  pxor   %xmm1,%xmm1
  comiss %xmm1,%xmm0
  ja     L2
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopl   (%rax)
L1:
  sub    $0xf4240,%edi
  addl   $0x1,0x14(%rsp)
  mov    %edi,0x10(%rsp)
  jmp    L10
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void buzz(int delay) {
  struct timeval now, stop;
  float diff;
  gettimeofday(&stop, 0);
  stop.tv_usec += delay;
  if (stop.tv_usec > 1000000) {
    stop.tv_usec -= 1000000;
    stop.tv_sec += 1;
  }
  do {
    gettimeofday(&now, 0);
    diff = ftimediff(&stop, &now);
  } while (diff > 0.0);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<buzz>:
  endbr64
  push   %rbp
  xor    %esi,%esi
  push   %rbx
  mov    %edi,%ebx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rbp
  mov    %rbp,%rdi
  call   <gettimeofday@plt>
  add    0x10(%rsp),%ebx
  mov    %ebx,%edi
  cmp    $0xf4240,%ebx
  jg     L25
  mov    %ebx,0x10(%rsp)
L10:
  lea    0x8(%rsp),%rbx
  nopw   0x0(%rax,%rax,1)
L36:
  xor    %esi,%esi
  mov    %rbx,%rdi
  call   <gettimeofday@plt>
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <ftimediff@plt>
  pxor   %xmm1,%xmm1
  comiss %xmm1,%xmm0
  ja     L36
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L23
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopl   (%rax)
L25:
  sub    $0xf4240,%edi
  addl   $0x1,0x14(%rsp)
  mov    %edi,0x10(%rsp)
  jmp    L10
L23:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void buzz(int delay) {
  struct timeval now, stop;
  float diff;
  gettimeofday(&stop, 0);
  stop.tv_usec += delay;
  if (stop.tv_usec > 1000000) {
    stop.tv_usec -= 1000000;
    stop.tv_sec += 1;
  }
  do {
    gettimeofday(&now, 0);
    diff = ftimediff(&stop, &now);
  } while (diff > 0.0);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x1b0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x1a8(%rbp)
  jmp    L21
L8:
  movl   $0x0,-0x1a4(%rbp)
  jmp    L36
L49:
  mov    -0x1a8(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    -0x1a4(%rbp),%eax
  movslq %eax,%rcx
  movslq %edx,%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  add    %rax,%rax
  add    %rcx,%rax
  mov    -0x1a0(%rbp,%rax,4),%edx
  mov    -0x1a4(%rbp),%eax
  movslq %eax,%rsi
  mov    -0x1a8(%rbp),%eax
  movslq %eax,%rcx
  mov    %rcx,%rax
  shl    $0x2,%rax
  add    %rcx,%rax
  add    %rax,%rax
  add    %rsi,%rax
  mov    %edx,-0x1a0(%rbp,%rax,4)
  addl   $0x1,-0x1a4(%rbp)
L36:
  cmpl   $0x9,-0x1a4(%rbp)
  jle    L49
  addl   $0x1,-0x1a8(%rbp)
L21:
  cmpl   $0x9,-0x1a8(%rbp)
  jle    L8
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L45
  call   <__stack_chk_fail@plt>
L45:
  leave
  ret
```

assistant: ```c
int main() {
  int i, j, x[10][10], n1, n2, n3, n4, n5;
  for (i = 0; i < 10; i++) {
    for (j = 0; j < 10; j++) {
      x[i][j] = x[i - 1][j];
    }
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  mov    $0x0,%eax
  ret
```

assistant: ```c
int main() {
  int i, j, x[10][10], n1, n2, n3, n4, n5;
  for (i = 0; i < 10; i++) {
    for (j = 0; j < 10; j++) {
      x[i][j] = x[i - 1][j];
    }
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int main() {
  int i, j, x[10][10], n1, n2, n3, n4, n5;
  for (i = 0; i < 10; i++) {
    for (j = 0; j < 10; j++) {
      x[i][j] = x[i - 1][j];
    }
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int main() {
  int i, j, x[10][10], n1, n2, n3, n4, n5;
  for (i = 0; i < 10; i++) {
    for (j = 0; j < 10; j++) {
      x[i][j] = x[i - 1][j];
    }
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<memmove>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    -0x8(%rbp),%rax
  cmp    -0x10(%rbp),%rax
  jae    L1
  mov    -0x18(%rbp),%rdx
  mov    -0x10(%rbp),%rcx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <__memcpy_forward@plt>
  jmp    L0
L1:
  mov    -0x18(%rbp),%rdx
  mov    -0x10(%rbp),%rcx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <__memcpy_backward@plt>
L0:
  leave
  ret
```

assistant: ```c
void *memmove(void *s, const void *ct, size_t n) {
  if (s < ct) {
    return __memcpy_forward(s, ct, n);
  } else {
    return __memcpy_backward(s, ct, n);
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<memmove>:
  endbr64
  sub    $0x8,%rsp
  cmp    %rsi,%rdi
  jae    L3
  call   <__memcpy_forward@plt>
L10:
  add    $0x8,%rsp
  ret
L3:
  call   <__memcpy_backward@plt>
  jmp    L10
```

assistant: ```c
void *memmove(void *s, const void *ct, size_t n) {
  if (s < ct) {
    return __memcpy_forward(s, ct, n);
  } else {
    return __memcpy_backward(s, ct, n);
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<memmove>:
  endbr64
  cmp    %rsi,%rdi
  jae    L3
  jmp    1070 <__memcpy_forward@plt>
  xchg   %ax,%ax
L3:
  jmp    1060 <__memcpy_backward@plt>
```

assistant: ```c
void *memmove(void *s, const void *ct, size_t n) {
  if (s < ct) {
    return __memcpy_forward(s, ct, n);
  } else {
    return __memcpy_backward(s, ct, n);
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<memmove>:
  endbr64
  cmp    %rsi,%rdi
  jae    L9
  jmp    1070 <__memcpy_forward@plt>
  xchg   %ax,%ax
L9:
  jmp    1060 <__memcpy_backward@plt>
```

assistant: ```c
void *memmove(void *s, const void *ct, size_t n) {
  if (s < ct) {
    return __memcpy_forward(s, ct, n);
  } else {
    return __memcpy_backward(s, ct, n);
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<strength_test5>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x4(%rbp)
L18:
  addl   $0x1,-0x4(%rbp)
  mov    -0x18(%rbp),%rax
  add    $0x4,%rax
  mov    (%rax),%eax
  test   %eax,%eax
  je     L27
  mov    -0x4(%rbp),%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  shl    $0x2,%eax
  add    %eax,%edx
  mov    -0x18(%rbp),%rax
  add    $0xc,%rax
  mov    (%rax),%eax
  cmp    %eax,%edx
  jg     L10
L27:
  mov    -0x18(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movl   $0x2,(%rax)
  jmp    L18
L10:
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void strength_test5(int *data) {
  int i = 0;
  while (1) {
    i = i + 1;
    if (data[1] && 21 * i > data[3])
      break;
    data[data[2]] = 2;
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<strength_test5>:
  endbr64
  mov    $0x15,%eax
  jmp    L23
L7:
  movslq 0x8(%rdi),%rdx
  movl   $0x2,(%rdi,%rdx,4)
  add    $0x15,%eax
L23:
  cmpl   $0x0,0x4(%rdi)
  je     L7
  cmp    %eax,0xc(%rdi)
  jge    L7
  ret
```

assistant: ```c
void strength_test5(int *data) {
  int i = 0;
  while (1) {
    i = i + 1;
    if (data[1] && 21 * i > data[3])
      break;
    data[data[2]] = 2;
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<strength_test5>:
  endbr64
  mov    $0x15,%eax
  jmp    L0
  nopl   0x0(%rax,%rax,1)
L18:
  movslq 0x8(%rdi),%rdx
  add    $0x15,%eax
  movl   $0x2,(%rdi,%rdx,4)
L0:
  mov    0x4(%rdi),%edx
  test   %edx,%edx
  je     L18
  cmp    %eax,0xc(%rdi)
  jge    L18
  ret
```

assistant: ```c
void strength_test5(int *data) {
  int i = 0;
  while (1) {
    i = i + 1;
    if (data[1] && 21 * i > data[3])
      break;
    data[data[2]] = 2;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<strength_test5>:
  endbr64
  mov    $0x15,%eax
  jmp    L21
  nopl   0x0(%rax,%rax,1)
L25:
  movslq 0x8(%rdi),%rdx
  add    $0x15,%eax
  movl   $0x2,(%rdi,%rdx,4)
L21:
  mov    0x4(%rdi),%edx
  test   %edx,%edx
  je     L25
  cmp    %eax,0xc(%rdi)
  jge    L25
  ret
```

assistant: ```c
void strength_test5(int *data) {
  int i = 0;
  while (1) {
    i = i + 1;
    if (data[1] && 21 * i > data[3])
      break;
    data[data[2]] = 2;
  }
}
```