ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<getHint>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x90,%rsp
  mov    %rdi,-0x88(%rbp)
  mov    %rsi,-0x90(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0x0,-0x60(%rbp)
  movq   $0x0,-0x58(%rbp)
  movq   $0x0,-0x50(%rbp)
  movq   $0x0,-0x48(%rbp)
  movq   $0x0,-0x40(%rbp)
  movq   $0x0,-0x30(%rbp)
  movq   $0x0,-0x28(%rbp)
  movq   $0x0,-0x20(%rbp)
  movq   $0x0,-0x18(%rbp)
  movq   $0x0,-0x10(%rbp)
  movl   $0x0,-0x74(%rbp)
  movl   $0x0,-0x70(%rbp)
  mov    $0x5,%edi
  call   <malloc@plt>
  mov    %rax,-0x68(%rbp)
  jmp    L21
L44:
  mov    -0x88(%rbp),%rax
  movzbl (%rax),%edx
  mov    -0x90(%rbp),%rax
  movzbl (%rax),%eax
  cmp    %al,%dl
  jne    L34
  addl   $0x1,-0x74(%rbp)
  jmp    L70
L34:
  mov    -0x88(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  sub    $0x30,%eax
  movslq %eax,%rdx
  mov    -0x60(%rbp,%rdx,4),%edx
  add    $0x1,%edx
  cltq
  mov    %edx,-0x60(%rbp,%rax,4)
  mov    -0x90(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  sub    $0x30,%eax
  movslq %eax,%rdx
  mov    -0x30(%rbp,%rdx,4),%edx
  add    $0x1,%edx
  cltq
  mov    %edx,-0x30(%rbp,%rax,4)
L70:
  addq   $0x1,-0x88(%rbp)
  addq   $0x1,-0x90(%rbp)
L21:
  mov    -0x88(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L44
  movl   $0x0,-0x6c(%rbp)
  jmp    L19
L69:
  mov    -0x6c(%rbp),%eax
  cltq
  mov    -0x60(%rbp,%rax,4),%eax
  test   %eax,%eax
  je     L49
  mov    -0x6c(%rbp),%eax
  cltq
  mov    -0x30(%rbp,%rax,4),%eax
  test   %eax,%eax
  je     L49
  mov    -0x6c(%rbp),%eax
  cltq
  mov    -0x30(%rbp,%rax,4),%edx
  mov    -0x6c(%rbp),%eax
  cltq
  mov    -0x60(%rbp,%rax,4),%eax
  cmp    %eax,%edx
  cmovle %edx,%eax
  add    %eax,-0x70(%rbp)
L49:
  addl   $0x1,-0x6c(%rbp)
L19:
  cmpl   $0x9,-0x6c(%rbp)
  jle    L69
  mov    -0x70(%rbp),%ecx
  mov    -0x74(%rbp),%edx
  mov    -0x68(%rbp),%rax
  lea    D6(%rip),%rsi
  mov    %rax,%rdi
  call   <sprintf@plt>
  mov    -0x68(%rbp),%rax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L10
  call   <__stack_chk_fail@plt>
L10:
  leave
  ret
```

tool_calls: name: read_data
arguments: data_label: D6
type: string

tool: D6: %dA%dB

assistant: ```c
char *getHint(char *secret, char *guess) {
  int markBulls[10] = {0};
  int markCows[10] = {0};
  int cntBulls = 0;
  int cntCows = 0;
  char *ret = (char *)malloc(sizeof(char) * 5);
  while (*secret != '\0') {
    if (*secret == *guess) {
      cntBulls++;
    } else {
      markBulls[*secret - '0']++;
      markCows[*guess - '0']++;
    }
    secret++;
    guess++;
  }
  for (int i = 0; i < 10; i++) {
    if (markBulls[i] != 0 && markCows[i] != 0) {
      cntCows += (markBulls[i] < markCows[i] ? markBulls[i] : markCows[i]);
    }
  }
  sprintf(ret, "%dA%dB", cntBulls, cntCows);
  return ret;
}
```
user: Translate the below assembly code into C programming language:

```asm
<getHint>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x60,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  mov    %fs:0x28,%rax
  mov    %rax,0x58(%rsp)
  xor    %eax,%eax
  movq   $0x0,(%rsp)
  movq   $0x0,0x8(%rsp)
  movq   $0x0,0x10(%rsp)
  movq   $0x0,0x18(%rsp)
  movq   $0x0,0x20(%rsp)
  movq   $0x0,0x30(%rsp)
  movq   $0x0,0x38(%rsp)
  movq   $0x0,0x40(%rsp)
  movq   $0x0,0x48(%rsp)
  movq   $0x0,0x50(%rsp)
  mov    $0x5,%edi
  call   <malloc@plt>
  mov    %rax,%r12
  movzbl (%rbx),%eax
  test   %al,%al
  je     L8
  mov    $0x0,%ecx
  mov    $0x0,%esi
  jmp    L2
L31:
  add    $0x1,%esi
L54:
  add    $0x1,%rcx
  movzbl (%rbx,%rcx,1),%eax
  test   %al,%al
  je     L96
L2:
  movzbl 0x0(%rbp,%rcx,1),%edx
  cmp    %al,%dl
  je     L31
  movsbl %al,%eax
  sub    $0x30,%eax
  cltq
  addl   $0x1,(%rsp,%rax,4)
  movsbl %dl,%edx
  sub    $0x30,%edx
  movslq %edx,%rdx
  addl   $0x1,0x30(%rsp,%rdx,4)
  jmp    L54
L8:
  mov    $0x0,%esi
L96:
  mov    $0x0,%edx
  mov    $0x0,%edi
  jmp    L79
L13:
  add    $0x4,%rdx
  cmp    $0x28,%rdx
  je     L43
L79:
  mov    (%rsp,%rdx,1),%eax
  test   %eax,%eax
  je     L13
  mov    0x30(%rsp,%rdx,1),%ecx
  cmp    %ecx,%eax
  cmovg  %ecx,%eax
  add    %edi,%eax
  test   %ecx,%ecx
  cmovne %eax,%edi
  jmp    L13
L43:
  mov    %edi,%ecx
  mov    %esi,%edx
  lea    D8(%rip),%rsi
  mov    %r12,%rdi
  call   <sprintf@plt>
  mov    0x58(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L67
  mov    %r12,%rax
  add    $0x60,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L67:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D8
type_guessed: string

tool: D8: %dA%dB

assistant: ```c
char *getHint(char *secret, char *guess) {
  int markBulls[10] = {0};
  int markCows[10] = {0};
  int cntBulls = 0;
  int cntCows = 0;
  char *ret = (char *)malloc(sizeof(char) * 5);
  while (*secret != '\0') {
    if (*secret == *guess) {
      cntBulls++;
    } else {
      markBulls[*secret - '0']++;
      markCows[*guess - '0']++;
    }
    secret++;
    guess++;
  }
  for (int i = 0; i < 10; i++) {
    if (markBulls[i] != 0 && markCows[i] != 0) {
      cntCows += (markBulls[i] < markCows[i] ? markBulls[i] : markCows[i]);
    }
  }
  sprintf(ret, "%dA%dB", cntBulls, cntCows);
  return ret;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<getHint>:
  endbr64
  push   %r12
  pxor   %xmm0,%xmm0
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    $0x5,%edi
  sub    $0x60,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x58(%rsp)
  xor    %eax,%eax
  movaps %xmm0,(%rsp)
  movq   $0x0,0x20(%rsp)
  movq   $0x0,0x50(%rsp)
  movaps %xmm0,0x10(%rsp)
  movaps %xmm0,0x30(%rsp)
  movaps %xmm0,0x40(%rsp)
  call   <malloc@plt>
  mov    %rax,%r12
  movsbl (%rbx),%eax
  test   %al,%al
  je     L98
  xor    %ecx,%ecx
  xor    %r8d,%r8d
  jmp    L24
  nopw   0x0(%rax,%rax,1)
L26:
  sub    $0x30,%eax
  add    $0x1,%rcx
  sub    $0x30,%edx
  cltq
  movslq %edx,%rdx
  addl   $0x1,(%rsp,%rax,4)
  movsbl (%rbx,%rcx,1),%eax
  addl   $0x1,0x30(%rsp,%rdx,4)
  test   %al,%al
  je     L21
L24:
  movsbl 0x0(%rbp,%rcx,1),%edx
  cmp    %al,%dl
  jne    L26
  add    $0x1,%rcx
  add    $0x1,%r8d
  movsbl (%rbx,%rcx,1),%eax
  test   %al,%al
  jne    L24
L21:
  mov    (%rsp),%eax
L95:
  xor    %edx,%edx
  xor    %r9d,%r9d
  jmp    L39
  nopl   0x0(%rax,%rax,1)
L22:
  mov    (%rsp,%rdx,1),%eax
L39:
  test   %eax,%eax
  je     L34
  mov    0x30(%rsp,%rdx,1),%ecx
  cmp    %eax,%ecx
  cmovle %ecx,%eax
  add    %r9d,%eax
  test   %ecx,%ecx
  cmovne %eax,%r9d
L34:
  add    $0x4,%rdx
  cmp    $0x28,%rdx
  jne    L22
  mov    %r9d,%ecx
  mov    %r8d,%edx
  lea    D4(%rip),%rsi
  mov    %r12,%rdi
  call   <sprintf@plt>
  mov    0x58(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L75
  add    $0x60,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L98:
  xor    %eax,%eax
  xor    %r8d,%r8d
  jmp    L95
L75:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D4
type: string

tool: D4: %dA%dB

assistant: ```c
char *getHint(char *secret, char *guess) {
  int markBulls[10] = {0};
  int markCows[10] = {0};
  int cntBulls = 0;
  int cntCows = 0;
  char *ret = (char *)malloc(sizeof(char) * 5);
  while (*secret != '\0') {
    if (*secret == *guess) {
      cntBulls++;
    } else {
      markBulls[*secret - '0']++;
      markCows[*guess - '0']++;
    }
    secret++;
    guess++;
  }
  for (int i = 0; i < 10; i++) {
    if (markBulls[i] != 0 && markCows[i] != 0) {
      cntCows += (markBulls[i] < markCows[i] ? markBulls[i] : markCows[i]);
    }
  }
  sprintf(ret, "%dA%dB", cntBulls, cntCows);
  return ret;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<getHint>:
  endbr64
  push   %r12
  pxor   %xmm0,%xmm0
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    $0x5,%edi
  sub    $0x60,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x58(%rsp)
  xor    %eax,%eax
  movaps %xmm0,(%rsp)
  movq   $0x0,0x20(%rsp)
  movq   $0x0,0x50(%rsp)
  movaps %xmm0,0x10(%rsp)
  movaps %xmm0,0x30(%rsp)
  movaps %xmm0,0x40(%rsp)
  call   <malloc@plt>
  mov    %rax,%r12
  movsbl (%rbx),%eax
  test   %al,%al
  je     L17
  xor    %ecx,%ecx
  xor    %r8d,%r8d
  nopl   0x0(%rax,%rax,1)
L55:
  movsbl 0x0(%rbp,%rcx,1),%edx
  cmp    %al,%dl
  je     L70
  sub    $0x30,%eax
  add    $0x1,%rcx
  sub    $0x30,%edx
  cltq
  movslq %edx,%rdx
  addl   $0x1,(%rsp,%rax,4)
  movsbl (%rbx,%rcx,1),%eax
  addl   $0x1,0x30(%rsp,%rdx,4)
  test   %al,%al
  jne    L55
L41:
  movdqa 0x30(%rsp),%xmm4
  movdqa (%rsp),%xmm3
  pxor   %xmm1,%xmm1
  mov    0x20(%rsp),%eax
  movdqa %xmm3,%xmm0
  movdqa %xmm4,%xmm2
  pcmpeqd %xmm1,%xmm0
  pcmpeqd %xmm1,%xmm2
  pcmpeqd %xmm1,%xmm0
  pcmpeqd %xmm1,%xmm2
  pand   %xmm2,%xmm0
  movdqa %xmm4,%xmm2
  pcmpgtd %xmm3,%xmm2
  pand   %xmm2,%xmm3
  pandn  %xmm4,%xmm2
  movdqa 0x10(%rsp),%xmm4
  por    %xmm3,%xmm2
  movdqa 0x40(%rsp),%xmm3
  pand   %xmm2,%xmm0
  movdqa %xmm4,%xmm2
  movdqa %xmm3,%xmm5
  pcmpeqd %xmm1,%xmm2
  pcmpeqd %xmm1,%xmm5
  pcmpeqd %xmm1,%xmm2
  pcmpeqd %xmm5,%xmm1
  pand   %xmm2,%xmm1
  movdqa %xmm4,%xmm2
  pcmpgtd %xmm3,%xmm2
  pand   %xmm2,%xmm3
  pandn  %xmm4,%xmm2
  por    %xmm3,%xmm2
  pand   %xmm2,%xmm1
  paddd  %xmm1,%xmm0
  movdqa %xmm0,%xmm1
  psrldq $0x8,%xmm1
  paddd  %xmm1,%xmm0
  movdqa %xmm0,%xmm1
  psrldq $0x4,%xmm1
  paddd  %xmm1,%xmm0
  movd   %xmm0,%ecx
  test   %eax,%eax
  je     L82
  mov    0x50(%rsp),%edx
  cmp    %eax,%edx
  cmovle %edx,%eax
  add    %ecx,%eax
  test   %edx,%edx
  cmovne %eax,%ecx
L82:
  mov    0x24(%rsp),%eax
  test   %eax,%eax
  je     L47
  mov    0x54(%rsp),%edx
  cmp    %edx,%eax
  cmovg  %edx,%eax
  add    %ecx,%eax
  test   %edx,%edx
  cmovne %eax,%ecx
L47:
  mov    %r8d,%edx
  lea    D9(%rip),%rsi
  mov    %r12,%rdi
  call   <sprintf@plt>
  mov    0x58(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  add    $0x60,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L70:
  add    $0x1,%rcx
  add    $0x1,%r8d
  movsbl (%rbx,%rcx,1),%eax
  test   %al,%al
  jne    L55
  jmp    L41
L17:
  xor    %r8d,%r8d
  jmp    L41
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D9
type_infered: string

tool: D9: %dA%dB

assistant: ```c
char *getHint(char *secret, char *guess) {
  int markBulls[10] = {0};
  int markCows[10] = {0};
  int cntBulls = 0;
  int cntCows = 0;
  char *ret = (char *)malloc(sizeof(char) * 5);
  while (*secret != '\0') {
    if (*secret == *guess) {
      cntBulls++;
    } else {
      markBulls[*secret - '0']++;
      markCows[*guess - '0']++;
    }
    secret++;
    guess++;
  }
  for (int i = 0; i < 10; i++) {
    if (markBulls[i] != 0 && markCows[i] != 0) {
      cntCows += (markBulls[i] < markCows[i] ? markBulls[i] : markCows[i]);
    }
  }
  sprintf(ret, "%dA%dB", cntBulls, cntCows);
  return ret;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<fast_memfill32_x86>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L5
L12:
  mov    -0x8(%rbp),%rax
  lea    0x4(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  mov    -0x1c(%rbp),%edx
  mov    %edx,(%rax)
L5:
  mov    -0x20(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0x20(%rbp)
  test   %eax,%eax
  jne    L12
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void fast_memfill32_x86(void *v_to, uint32_t val, int words) {
  uint32_t *to = (uint32_t *)(v_to);
  while (words--) {
    *to++ = val;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<fast_memfill32_x86>:
  endbr64
  test   %edx,%edx
  je     L7
  mov    %edx,%edx
  lea    (%rdi,%rdx,4),%rax
L11:
  add    $0x4,%rdi
  mov    %esi,-0x4(%rdi)
  cmp    %rax,%rdi
  jne    L11
L7:
  ret
```

assistant: ```c
void fast_memfill32_x86(void *v_to, uint32_t val, int words) {
  uint32_t *to = (uint32_t *)(v_to);
  while (words--) {
    *to++ = val;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<fast_memfill32_x86>:
  endbr64
  test   %edx,%edx
  je     L0
  mov    %edx,%edx
  lea    (%rdi,%rdx,4),%rax
  xchg   %ax,%ax
L19:
  add    $0x4,%rdi
  mov    %esi,-0x4(%rdi)
  cmp    %rax,%rdi
  jne    L19
L0:
  ret
```

assistant: ```c
void fast_memfill32_x86(void *v_to, uint32_t val, int words) {
  uint32_t *to = (uint32_t *)(v_to);
  while (words--) {
    *to++ = val;
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<fast_memfill32_x86>:
  endbr64
  lea    -0x1(%rdx),%r8d
  test   %edx,%edx
  je     L22
  cmp    $0x2,%r8d
  jbe    L48
  mov    %edx,%ecx
  movd   %esi,%xmm1
  mov    %rdi,%rax
  shr    $0x2,%ecx
  pshufd $0x0,%xmm1,%xmm0
  sub    $0x1,%ecx
  shl    $0x4,%rcx
  lea    0x10(%rdi,%rcx,1),%rcx
  nop
L20:
  movups %xmm0,(%rax)
  add    $0x10,%rax
  cmp    %rcx,%rax
  jne    L20
  mov    %edx,%eax
  and    $0xfffffffc,%eax
  mov    %eax,%ecx
  sub    %eax,%r8d
  lea    (%rdi,%rcx,4),%rdi
  cmp    %eax,%edx
  je     L22
L48:
  mov    %esi,(%rdi)
  test   %r8d,%r8d
  je     L22
  mov    %esi,0x4(%rdi)
  cmp    $0x1,%r8d
  je     L22
  mov    %esi,0x8(%rdi)
L22:
  ret
```

assistant: ```c
void fast_memfill32_x86(void *v_to, uint32_t val, int words) {
  uint32_t *to = (uint32_t *)(v_to);
  while (words--) {
    *to++ = val;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<assign>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %eax,%edx
  mov    -0x10(%rbp),%rax
  mov    %edx,(%rax)
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D4
type_guessed: string

tool: D4: inside assign


assistant: ```c
void assign(long a, int *b) {
  *b = (int)a;
  printf("inside assign\n");
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<assign>:
  endbr64
  sub    $0x8,%rsp
  mov    %edi,(%rsi)
  lea    D3(%rip),%rdi
  call   <printf@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: data_label: D3
type_guessed: string

tool: D3: inside assign


assistant: ```c
void assign(long a, int *b) {
  *b = (int)a;
  printf("inside assign\n");
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<assign>:
  endbr64
  mov    %edi,(%rsi)
  lea    D0(%rip),%rdi
  jmp    1050 <printf@plt>
```

tool_calls: name: parse_data
arguments: tag: D0
type_guessed: string

tool: D0: inside assign


assistant: ```c
void assign(long a, int *b) {
  *b = (int)a;
  printf("inside assign\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
<assign>:
  endbr64
  mov    %edi,(%rsi)
  lea    D5(%rip),%rdi
  jmp    1050 <printf@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D5
type_guessed: string

tool: D5: inside assign


assistant: ```c
void assign(long a, int *b) {
  *b = (int)a;
  printf("inside assign\n");
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<multToConstVec3f>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  movss  %xmm0,-0xc(%rbp)
  mov    -0x8(%rbp),%rax
  movss  (%rax),%xmm0
  mulss  -0xc(%rbp),%xmm0
  mov    -0x8(%rbp),%rax
  movss  %xmm0,(%rax)
  mov    -0x8(%rbp),%rax
  add    $0x4,%rax
  movss  (%rax),%xmm0
  mov    -0x8(%rbp),%rax
  add    $0x4,%rax
  mulss  -0xc(%rbp),%xmm0
  movss  %xmm0,(%rax)
  mov    -0x8(%rbp),%rax
  add    $0x8,%rax
  movss  (%rax),%xmm0
  mov    -0x8(%rbp),%rax
  add    $0x8,%rax
  mulss  -0xc(%rbp),%xmm0
  movss  %xmm0,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void multToConstVec3f(float *a, float c) {
  a[0] *= c;
  a[1] *= c;
  a[2] *= c;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<multToConstVec3f>:
  endbr64
  movaps %xmm0,%xmm1
  mulss  (%rdi),%xmm1
  movss  %xmm1,(%rdi)
  movaps %xmm0,%xmm1
  mulss  0x4(%rdi),%xmm1
  movss  %xmm1,0x4(%rdi)
  mulss  0x8(%rdi),%xmm0
  movss  %xmm0,0x8(%rdi)
  ret
```

assistant: ```c
void multToConstVec3f(float *a, float c) {
  a[0] *= c;
  a[1] *= c;
  a[2] *= c;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<multToConstVec3f>:
  endbr64
  movss  (%rdi),%xmm1
  mulss  %xmm0,%xmm1
  movss  %xmm1,(%rdi)
  movss  0x4(%rdi),%xmm1
  mulss  %xmm0,%xmm1
  mulss  0x8(%rdi),%xmm0
  movss  %xmm1,0x4(%rdi)
  movss  %xmm0,0x8(%rdi)
  ret
```

assistant: ```c
void multToConstVec3f(float *a, float c) {
  a[0] *= c;
  a[1] *= c;
  a[2] *= c;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<multToConstVec3f>:
  endbr64
  movq   (%rdi),%xmm1
  movaps %xmm0,%xmm2
  shufps $0xe0,%xmm2,%xmm2
  mulss  0x8(%rdi),%xmm0
  mulps  %xmm2,%xmm1
  movss  %xmm0,0x8(%rdi)
  movlps %xmm1,(%rdi)
  ret
```

assistant: ```c
void multToConstVec3f(float *a, float c) {
  a[0] *= c;
  a[1] *= c;
  a[2] *= c;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0xffffffff,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    %edx
  mov    %eax,%esi
  sar    $0x1f,%esi
  mov    %edx,%ecx
  sub    %esi,%ecx
  mov    %ecx,%edx
  shl    $0x2,%edx
  add    %ecx,%edx
  sub    %edx,%eax
  mov    %eax,-0x8(%rbp)
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    %edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x4(%rbp)
  cmpl   $0xffffffff,-0x8(%rbp)
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  cmpl   $0x0,-0x4(%rbp)
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  movl   $0x1,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    %edx
  mov    %eax,%esi
  sar    $0x1f,%esi
  mov    %edx,%ecx
  sub    %esi,%ecx
  mov    %ecx,%edx
  shl    $0x2,%edx
  add    %ecx,%edx
  sub    %edx,%eax
  mov    %eax,-0x8(%rbp)
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    %edx
  sar    $0x1f,%eax
  sub    %edx,%eax
  mov    %eax,-0x4(%rbp)
  cmpl   $0x1,-0x8(%rbp)
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  cmpl   $0x0,-0x4(%rbp)
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  movl   $0xffffffff,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    %edx
  mov    %eax,%esi
  sar    $0x1f,%esi
  mov    %edx,%ecx
  sub    %esi,%ecx
  mov    %ecx,%edx
  shl    $0x2,%edx
  add    %ecx,%edx
  sub    %edx,%eax
  mov    %eax,-0x8(%rbp)
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    %edx
  sar    $0x1f,%eax
  sub    %edx,%eax
  mov    %eax,-0x4(%rbp)
  cmpl   $0xffffffff,-0x8(%rbp)
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  cmpl   $0x0,-0x4(%rbp)
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main() {
  int x = -1;
  int m = x % 5;
  int r = x / 5;
  assert(m == -1);
  assert(r == 0);
  x = 1;
  m = x % -5;
  r = x / -5;
  assert(m == 1);
  assert(r == 0);
  x = -1;
  m = x % -5;
  r = x / -5;
  assert(m == -1);
  assert(r == 0);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x1,%edi
  call   <assert@plt>
  mov    $0x1,%edi
  call   <assert@plt>
  mov    $0x1,%edi
  call   <assert@plt>
  mov    $0x1,%edi
  call   <assert@plt>
  mov    $0x1,%edi
  call   <assert@plt>
  mov    $0x1,%edi
  call   <assert@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  int x = -1;
  int m = x % 5;
  int r = x / 5;
  assert(m == -1);
  assert(r == 0);
  x = 1;
  m = x % -5;
  r = x / -5;
  assert(m == 1);
  assert(r == 0);
  x = -1;
  m = x % -5;
  r = x / -5;
  assert(m == -1);
  assert(r == 0);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x1,%edi
  call   <assert@plt>
  mov    $0x1,%edi
  call   <assert@plt>
  mov    $0x1,%edi
  call   <assert@plt>
  mov    $0x1,%edi
  call   <assert@plt>
  mov    $0x1,%edi
  call   <assert@plt>
  mov    $0x1,%edi
  call   <assert@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  int x = -1;
  int m = x % 5;
  int r = x / 5;
  assert(m == -1);
  assert(r == 0);
  x = 1;
  m = x % -5;
  r = x / -5;
  assert(m == 1);
  assert(r == 0);
  x = -1;
  m = x % -5;
  r = x / -5;
  assert(m == -1);
  assert(r == 0);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x1,%edi
  call   <assert@plt>
  mov    $0x1,%edi
  call   <assert@plt>
  mov    $0x1,%edi
  call   <assert@plt>
  mov    $0x1,%edi
  call   <assert@plt>
  mov    $0x1,%edi
  call   <assert@plt>
  mov    $0x1,%edi
  call   <assert@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  int x = -1;
  int m = x % 5;
  int r = x / 5;
  assert(m == -1);
  assert(r == 0);
  x = 1;
  m = x % -5;
  r = x / -5;
  assert(m == 1);
  assert(r == 0);
  x = -1;
  m = x % -5;
  r = x / -5;
  assert(m == -1);
  assert(r == 0);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<getNumMarkDistribution>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    %edx,-0x1c(%rbp)
  mov    %rcx,-0x28(%rbp)
  cmpl   $0x0,-0x18(%rbp)
  jns    L15
  mov    $0x0,%eax
  jmp    L99
L15:
  cmpl   $0x0,-0x14(%rbp)
  jne    L20
  cmpl   $0x0,-0x18(%rbp)
  jne    L14
  mov    $0x1,%eax
  jmp    L99
L14:
  mov    $0x0,%eax
  jmp    L99
L20:
  mov    -0x14(%rbp),%eax
  cltq
  imul   $0x4ec4,%rax,%rdx
  mov    -0x28(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rsi
  mov    -0x18(%rbp),%eax
  movslq %eax,%rdx
  mov    %rdx,%rax
  shl    $0x3,%rax
  add    %rdx,%rax
  shl    $0x3,%rax
  sub    %rdx,%rax
  add    %rsi,%rax
  mov    (%rcx,%rax,4),%eax
  cmp    $0xffffffff,%eax
  je     L31
  mov    -0x14(%rbp),%eax
  cltq
  imul   $0x4ec4,%rax,%rdx
  mov    -0x28(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rsi
  mov    -0x18(%rbp),%eax
  movslq %eax,%rdx
  mov    %rdx,%rax
  shl    $0x3,%rax
  add    %rdx,%rax
  shl    $0x3,%rax
  sub    %rdx,%rax
  add    %rsi,%rax
  mov    (%rcx,%rax,4),%eax
  jmp    L99
L31:
  movl   $0x0,-0x8(%rbp)
  mov    -0x1c(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  jmp    L42
L25:
  mov    -0x18(%rbp),%eax
  sub    -0x4(%rbp),%eax
  mov    -0x14(%rbp),%edx
  lea    -0x1(%rdx),%edi
  mov    -0x28(%rbp),%rcx
  mov    -0x1c(%rbp),%edx
  mov    %eax,%esi
  call   <getNumMarkDistribution>
  add    %eax,-0x8(%rbp)
  addl   $0x1,-0x4(%rbp)
L42:
  mov    -0x4(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jle    L25
  mov    -0x14(%rbp),%eax
  cltq
  imul   $0x4ec4,%rax,%rdx
  mov    -0x28(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rsi
  mov    -0x18(%rbp),%eax
  movslq %eax,%rdx
  mov    %rdx,%rax
  shl    $0x3,%rax
  add    %rdx,%rax
  shl    $0x3,%rax
  sub    %rdx,%rax
  lea    (%rax,%rsi,1),%rdx
  mov    -0x8(%rbp),%eax
  mov    %eax,(%rcx,%rdx,4)
  mov    -0x8(%rbp),%eax
L99:
  leave
  ret
```

assistant: ```c
int getNumMarkDistribution(int numSubject, int totalMark, int minToPass, int memoTable[][71][71]) {
  if (totalMark < 0)
    return 0;
  if (numSubject == 0) {
    if (totalMark == 0)
      return 1;
    return 0;
  }
  if (memoTable[numSubject][totalMark][minToPass] != -1)
    return memoTable[numSubject][totalMark][minToPass];
  int numWay = 0, mark;
  for (mark = minToPass; mark <= totalMark; mark++)
    numWay += getNumMarkDistribution(numSubject - 1, totalMark - mark, minToPass, memoTable);
  memoTable[numSubject][totalMark][minToPass] = numWay;
  return numWay;
}
```