ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<rdsr>:
  endbr64
  push   %rbx
  mov    $0x0,%edi
  call   <usart_cs@plt>
  mov    $0x5,%edi
  call   <usart_out_byte@plt>
  mov    $0x0,%eax
  call   <usart_in_byte@plt>
  mov    %eax,%ebx
  mov    $0x1,%edi
  call   <usart_cs@plt>
  mov    %ebx,%eax
  pop    %rbx
  ret
```

assistant: ```c
uint8_t rdsr(void) {
  uint8_t status;
  usart_cs(0);
  usart_out_byte(0x05);
  status = usart_in_byte();
  usart_cs(1);
  return status;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<rdsr>:
  endbr64
  push   %r12
  xor    %edi,%edi
  call   <usart_cs@plt>
  mov    $0x5,%edi
  call   <usart_out_byte@plt>
  xor    %eax,%eax
  call   <usart_in_byte@plt>
  mov    $0x1,%edi
  mov    %eax,%r12d
  call   <usart_cs@plt>
  mov    %r12d,%eax
  pop    %r12
  ret
```

assistant: ```c
uint8_t rdsr(void) {
  uint8_t status;
  usart_cs(0);
  usart_out_byte(0x05);
  status = usart_in_byte();
  usart_cs(1);
  return status;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<rdsr>:
  endbr64
  push   %r12
  xor    %edi,%edi
  call   <usart_cs@plt>
  mov    $0x5,%edi
  call   <usart_out_byte@plt>
  xor    %eax,%eax
  call   <usart_in_byte@plt>
  mov    $0x1,%edi
  mov    %eax,%r12d
  call   <usart_cs@plt>
  mov    %r12d,%eax
  pop    %r12
  ret
```

assistant: ```c
uint8_t rdsr(void) {
  uint8_t status;
  usart_cs(0);
  usart_out_byte(0x05);
  status = usart_in_byte();
  usart_cs(1);
  return status;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x0,-0x4(%rbp)
  cmpl   $0x2,-0x14(%rbp)
  jne    L18
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  jmp    L61
L18:
  cmpl   $0x2,-0x14(%rbp)
  jle    L58
  cmpl   $0x4,-0x14(%rbp)
  jg     L58
  lea    D34(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  movl   $0x0,-0x4(%rbp)
  jmp    L13
L10:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    %rax,%rsi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x4(%rbp)
L13:
  mov    -0x4(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L10
  mov    $0xa,%edi
  call   <putchar@plt>
  jmp    L61
L58:
  lea    D36(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
L61:
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D3
type_infered: string
name: retrieve_data
arguments: data_tag: D34
type_infered: string
name: retrieve_data
arguments: data_tag: D7
type_infered: string
name: retrieve_data
arguments: data_tag: D36
type_infered: string

tool: D3: You only have one argument, you suck!

tool: D34: Here's your arguments.

tool: D7: %s 

tool: D36: You have too many arguments, you suck!

assistant: ```c
int main(int argc, char *argv[]) {
  int i = 0;
  if (argc == 2) {
    printf("You only have one argument, you suck!\n");
  } else if (argc > 2 && argc < 5) {
    printf("Here's your arguments.\n");
    for (i = 0; i < argc; i++) {
      printf("%s ", argv[i]);
    }
    printf("\n");
  } else {
    printf("You have too many arguments, you suck!\n");
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  cmp    $0x2,%edi
  je     L36
  mov    %edi,%ebp
  mov    %rsi,%r12
  lea    -0x3(%rdi),%eax
  cmp    $0x1,%eax
  jbe    L8
  lea    D25(%rip),%rdi
  call   <puts@plt>
L26:
  mov    $0x0,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L36:
  lea    D7(%rip),%rdi
  call   <puts@plt>
  jmp    L26
L8:
  lea    D13(%rip),%rdi
  call   <puts@plt>
  mov    %r12,%rbx
  lea    -0x1(%rbp),%eax
  lea    0x8(%r12,%rax,8),%r12
  lea    D9(%rip),%rbp
L32:
  mov    (%rbx),%rsi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x8,%rbx
  cmp    %r12,%rbx
  jne    L32
  mov    $0xa,%edi
  call   <putchar@plt>
  jmp    L26
```

tool_calls: name: unpack_data
arguments: data_label: D25
datatype: string
name: unpack_data
arguments: data_label: D7
datatype: string
name: unpack_data
arguments: data_label: D13
datatype: string
name: unpack_data
arguments: data_label: D9
datatype: string

tool: D25: You have too many arguments, you suck!

tool: D7: You only have one argument, you suck!

tool: D13: Here's your arguments.

tool: D9: %s 

assistant: ```c
int main(int argc, char *argv[]) {
  int i = 0;
  if (argc == 2) {
    printf("You only have one argument, you suck!\n");
  } else if (argc > 2 && argc < 5) {
    printf("Here's your arguments.\n");
    for (i = 0; i < argc; i++) {
      printf("%s ", argv[i]);
    }
    printf("\n");
  } else {
    printf("You have too many arguments, you suck!\n");
  }
  return 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
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  cmp    $0x2,%edi
  je     L39
  lea    -0x3(%rdi),%eax
  mov    %edi,%ebp
  cmp    $0x1,%eax
  jbe    L36
  lea    D28(%rip),%rdi
  call   <puts@plt>
L30:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
L36:
  lea    D3(%rip),%rdi
  mov    %rsi,%rbx
  call   <puts@plt>
  lea    -0x1(%rbp),%eax
  lea    D37(%rip),%rbp
  lea    0x8(%rbx,%rax,8),%r12
L27:
  mov    (%rbx),%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x8,%rbx
  call   <printf@plt>
  cmp    %r12,%rbx
  jne    L27
  mov    $0xa,%edi
  call   <putchar@plt>
  jmp    L30
L39:
  lea    D23(%rip),%rdi
  call   <puts@plt>
  jmp    L30
```

tool_calls: name: retrieve_data
arguments: data_tag: D28
data_type: string
name: retrieve_data
arguments: data_tag: D3
data_type: string
name: retrieve_data
arguments: data_tag: D37
data_type: string
name: retrieve_data
arguments: data_tag: D23
data_type: string

tool: D28: You have too many arguments, you suck!

tool: D3: Here's your arguments.

tool: D37: %s 

tool: D23: You only have one argument, you suck!

assistant: ```c
int main(int argc, char *argv[]) {
  int i = 0;
  if (argc == 2) {
    printf("You only have one argument, you suck!\n");
  } else if (argc > 2 && argc < 5) {
    printf("Here's your arguments.\n");
    for (i = 0; i < argc; i++) {
      printf("%s ", argv[i]);
    }
    printf("\n");
  } else {
    printf("You have too many arguments, you suck!\n");
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  cmp    $0x2,%edi
  je     L19
  lea    -0x3(%rdi),%eax
  mov    %edi,%ebp
  cmp    $0x1,%eax
  jbe    L2
  lea    D3(%rip),%rdi
  call   <puts@plt>
L38:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
L2:
  lea    D6(%rip),%rdi
  mov    %rsi,%rbx
  call   <puts@plt>
  lea    -0x1(%rbp),%eax
  lea    D16(%rip),%rbp
  lea    0x8(%rbx,%rax,8),%r12
L9:
  mov    (%rbx),%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x8,%rbx
  call   <printf@plt>
  cmp    %r12,%rbx
  jne    L9
  mov    $0xa,%edi
  call   <putchar@plt>
  jmp    L38
L19:
  lea    D17(%rip),%rdi
  call   <puts@plt>
  jmp    L38
```

tool_calls: name: parse_data
arguments: data_label: D3
datatype: string
name: parse_data
arguments: data_label: D6
datatype: string
name: parse_data
arguments: data_label: D16
datatype: string
name: parse_data
arguments: data_label: D17
datatype: string

tool: D3: You have too many arguments, you suck!

tool: D6: Here's your arguments.

tool: D16: %s 

tool: D17: You only have one argument, you suck!

assistant: ```c
int main(int argc, char *argv[]) {
  int i = 0;
  if (argc == 2) {
    printf("You only have one argument, you suck!\n");
  } else if (argc > 2 && argc < 5) {
    printf("Here's your arguments.\n");
    for (i = 0; i < argc; i++) {
      printf("%s ", argv[i]);
    }
    printf("\n");
  } else {
    printf("You have too many arguments, you suck!\n");
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x48,%rsp
  mov    %edi,-0x44(%rbp)
  mov    %rsi,-0x50(%rbp)
  mov    -0x50(%rbp),%rdx
  mov    -0x44(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <checkArguments@plt>
  cmp    $0x1,%eax
  jne    L23
  mov    $0x1,%edi
  call   <exit@plt>
L23:
  movq   $0x4,-0x30(%rbp)
  mov    -0x50(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <atoi@plt>
  mov    %eax,-0x3c(%rbp)
  mov    -0x50(%rbp),%rax
  add    $0x10,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <atol@plt>
  mov    %rax,-0x28(%rbp)
  mov    $0x0,%edi
  call   <time@plt>
  mov    %rax,-0x20(%rbp)
  mov    D9(%rip),%rax
  mov    $0x3,%edi
  movq   %rax,%xmm0
  call   <pow@plt>
  mov    %eax,-0x38(%rbp)
  mov    -0x38(%rbp),%eax
  cltq
  mov    $0x0,%edx
  divq   -0x30(%rbp)
  mov    %eax,%edx
  mov    -0x3c(%rbp),%eax
  cltq
  imul   %edx,%eax
  mov    %eax,-0x34(%rbp)
  mov    -0x34(%rbp),%eax
  cltq
  shl    $0x2,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x18(%rbp)
  jmp    L25
L9:
  mov    -0x34(%rbp),%ecx
  mov    -0x18(%rbp),%rax
  mov    $0xfc,%edx
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <writeBits@plt>
  mov    -0x34(%rbp),%ecx
  mov    -0x18(%rbp),%rax
  mov    $0x7d,%edx
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <writeBits@plt>
L25:
  mov    -0x20(%rbp),%rdx
  mov    -0x28(%rbp),%rax
  lea    (%rdx,%rax,1),%rbx
  mov    $0x0,%edi
  call   <time@plt>
  cmp    %rax,%rbx
  jg     L9
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D9
datatype: byte[8]

tool: {"D9": [0x00,0x00,0x00,0x00,0x00,0x00,0x90,0x40]}

assistant: ```c
int main(int argc, char **argv) {
  if (checkArguments(argc, argv) == 1) {
    exit(1);
  }
  long unsigned intSize = sizeof(int);
  int gbsToAllocate = atoi(argv[1]);
  long duration = atol(argv[2]);
  time_t startTime = time(NULL);
  int gbFactor = (int)pow((double)1024, 3);
  int arrayLength = (gbFactor / intSize) * gbsToAllocate;
  volatile int *array = malloc(sizeof(int) * arrayLength);
  while (startTime + duration > time(NULL)) {
    writeBits(array, arrayLength, 252);
    writeBits(array, arrayLength, 125);
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rsi,%rbx
  call   <checkArguments@plt>
  cmp    $0x1,%eax
  je     L2
L20:
  mov    0x8(%rbx),%rdi
  call   <atoi@plt>
  mov    %eax,%ebp
  mov    0x10(%rbx),%rdi
  call   <atol@plt>
  mov    %rax,%r12
  mov    $0x0,%edi
  call   <time@plt>
  mov    %rax,%r13
  mov    $0x3,%edi
  movsd  D6(%rip),%xmm0
  call   <pow@plt>
  mov    %eax,%ebx
  sar    $0x2,%ebx
  imul   %ebp,%ebx
  lea    0x0(,%rbx,4),%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  add    %r13,%r12
  jmp    L37
L2:
  mov    $0x1,%edi
  call   <exit@plt>
  jmp    L20
L10:
  mov    $0xfc,%edx
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <writeBits@plt>
  mov    $0x7d,%edx
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <writeBits@plt>
L37:
  mov    $0x0,%edi
  call   <time@plt>
  cmp    %rax,%r12
  jg     L10
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: parse_data
arguments: tag: D6
type_infered: byte[8]

tool: {"D6": [0x00,0x00,0x00,0x00,0x00,0x00,0x90,0x40]}

assistant: ```c
int main(int argc, char **argv) {
  if (checkArguments(argc, argv) == 1) {
    exit(1);
  }
  long unsigned intSize = sizeof(int);
  int gbsToAllocate = atoi(argv[1]);
  long duration = atol(argv[2]);
  time_t startTime = time(NULL);
  int gbFactor = (int)pow((double)1024, 3);
  int arrayLength = (gbFactor / intSize) * gbsToAllocate;
  volatile int *array = malloc(sizeof(int) * arrayLength);
  while (startTime + duration > time(NULL)) {
    writeBits(array, arrayLength, 252);
    writeBits(array, arrayLength, 125);
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x8,%rsp
  call   <checkArguments@plt>
  cmp    $0x1,%eax
  je     L2
L18:
  mov    0x8(%rbx),%rdi
  call   <atoi@plt>
  mov    0x10(%rbx),%rdi
  mov    %eax,%ebp
  call   <atol@plt>
  xor    %edi,%edi
  mov    %rax,%r12
  call   <time@plt>
  movsd  D1(%rip),%xmm0
  mov    $0x3,%edi
  mov    %rax,%r13
  call   <pow@plt>
  add    %r13,%r12
  mov    %eax,%ebx
  sar    $0x2,%ebx
  imul   %ebp,%ebx
  lea    0x0(,%rbx,4),%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  jmp    L14
  xchg   %ax,%ax
L10:
  mov    $0xfc,%edx
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <writeBits@plt>
  mov    $0x7d,%edx
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <writeBits@plt>
L14:
  xor    %edi,%edi
  call   <time@plt>
  cmp    %r12,%rax
  jl     L10
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L2:
  mov    $0x1,%edi
  call   <exit@plt>
  jmp    L18
```

tool_calls: name: disassemble_data
arguments: data_tag: D1
data_type: byte[8]

tool: {"D1": [0x00,0x00,0x00,0x00,0x00,0x00,0x90,0x40]}

assistant: ```c
int main(int argc, char **argv) {
  if (checkArguments(argc, argv) == 1) {
    exit(1);
  }
  long unsigned intSize = sizeof(int);
  int gbsToAllocate = atoi(argv[1]);
  long duration = atol(argv[2]);
  time_t startTime = time(NULL);
  int gbFactor = (int)pow((double)1024, 3);
  int arrayLength = (gbFactor / intSize) * gbsToAllocate;
  volatile int *array = malloc(sizeof(int) * arrayLength);
  while (startTime + duration > time(NULL)) {
    writeBits(array, arrayLength, 252);
    writeBits(array, arrayLength, 125);
  }
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x8,%rsp
  call   <checkArguments@plt>
  cmp    $0x1,%eax
  je     L32
L15:
  mov    0x8(%rbx),%rdi
  call   <atoi@plt>
  mov    0x10(%rbx),%rdi
  mov    %eax,%ebp
  call   <atol@plt>
  xor    %edi,%edi
  mov    %rax,%r12
  call   <time@plt>
  movsd  D7(%rip),%xmm0
  mov    $0x3,%edi
  mov    %rax,%r13
  call   <pow@plt>
  add    %r13,%r12
  mov    %eax,%ebx
  sar    $0x2,%ebx
  imul   %ebp,%ebx
  lea    0x0(,%rbx,4),%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  jmp    L3
  xchg   %ax,%ax
L7:
  mov    $0xfc,%edx
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <writeBits@plt>
  mov    $0x7d,%edx
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <writeBits@plt>
L3:
  xor    %edi,%edi
  call   <time@plt>
  cmp    %r12,%rax
  jl     L7
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L32:
  mov    $0x1,%edi
  call   <exit@plt>
  jmp    L15
```

tool_calls: name: parse_data
arguments: data_label: D7
datatype: byte[8]

tool: {"D7": [0x00,0x00,0x00,0x00,0x00,0x00,0x90,0x40]}

assistant: ```c
int main(int argc, char **argv) {
  if (checkArguments(argc, argv) == 1) {
    exit(1);
  }
  long unsigned intSize = sizeof(int);
  int gbsToAllocate = atoi(argv[1]);
  long duration = atol(argv[2]);
  time_t startTime = time(NULL);
  int gbFactor = (int)pow((double)1024, 3);
  int arrayLength = (gbFactor / intSize) * gbsToAllocate;
  volatile int *array = malloc(sizeof(int) * arrayLength);
  while (startTime + duration > time(NULL)) {
    writeBits(array, arrayLength, 252);
    writeBits(array, arrayLength, 125);
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<hpmRotateQuaternion>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xb0,%rsp
  movss  %xmm0,-0x94(%rbp)
  movss  %xmm1,-0x98(%rbp)
  movss  %xmm2,-0x9c(%rbp)
  movss  %xmm3,-0xa0(%rbp)
  mov    %rdi,-0xa8(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x90(%rbp),%rdx
  movss  -0xa0(%rbp),%xmm2
  movss  -0x9c(%rbp),%xmm1
  movss  -0x98(%rbp),%xmm0
  mov    -0x94(%rbp),%eax
  mov    %rdx,%rdi
  movaps %xmm2,%xmm3
  movaps %xmm1,%xmm2
  movaps %xmm0,%xmm1
  movd   %eax,%xmm0
  call   <hpmQuaternionRotation@plt>
  lea    -0x50(%rbp),%rdx
  mov    -0xa8(%rbp),%rcx
  lea    -0x90(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <hpmMultMat4@plt>
  lea    -0x50(%rbp),%rdx
  mov    -0xa8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <copyMat4@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L3
  call   <__stack_chk_fail@plt>
L3:
  leave
  ret
```

assistant: ```c
void hpmRotateQuaternion(float x, float y, float z, float w, float *mat) {
  float rot[16], r[16];
  hpmQuaternionRotation(x, y, z, w, rot);
  hpmMultMat4(rot, mat, r);
  copyMat4(mat, r);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<hpmRotateQuaternion>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x90,%rsp
  mov    %rdi,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x88(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  mov    %r12,%rdi
  call   <hpmQuaternionRotation@plt>
  lea    0x40(%rsp),%rbp
  mov    %rbp,%rdx
  mov    %rbx,%rsi
  mov    %r12,%rdi
  call   <hpmMultMat4@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <copyMat4@plt>
  mov    0x88(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  add    $0x90,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L6:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void hpmRotateQuaternion(float x, float y, float z, float w, float *mat) {
  float rot[16], r[16];
  hpmQuaternionRotation(x, y, z, w, rot);
  hpmMultMat4(rot, mat, r);
  copyMat4(mat, r);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<hpmRotateQuaternion>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x90,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x88(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r13
  lea    0x40(%rsp),%r12
  mov    %r13,%rdi
  call   <hpmQuaternionRotation@plt>
  mov    %rbp,%rsi
  mov    %r13,%rdi
  mov    %r12,%rdx
  call   <hpmMultMat4@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <copyMat4@plt>
  mov    0x88(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  add    $0x90,%rsp
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L1:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void hpmRotateQuaternion(float x, float y, float z, float w, float *mat) {
  float rot[16], r[16];
  hpmQuaternionRotation(x, y, z, w, rot);
  hpmMultMat4(rot, mat, r);
  copyMat4(mat, r);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<hpmRotateQuaternion>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x90,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x88(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r13
  lea    0x40(%rsp),%r12
  mov    %r13,%rdi
  call   <hpmQuaternionRotation@plt>
  mov    %rbp,%rsi
  mov    %r13,%rdi
  mov    %r12,%rdx
  call   <hpmMultMat4@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <copyMat4@plt>
  mov    0x88(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x90,%rsp
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L0:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void hpmRotateQuaternion(float x, float y, float z, float w, float *mat) {
  float rot[16], r[16];
  hpmQuaternionRotation(x, y, z, w, rot);
  hpmMultMat4(rot, mat, r);
  copyMat4(mat, r);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<match_ev_str>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %rcx,-0x30(%rbp)
  mov    -0x18(%rbp),%rsi
  mov    -0x28(%rbp),%rax
  mov    $0x0,%r8d
  mov    $0x0,%ecx
  mov    $0x0,%edx
  mov    %rax,%rdi
  call   <regexec@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jne    L1
  mov    $0x1,%eax
  jmp    L16
L1:
  mov    $0x0,%eax
L16:
  leave
  ret
```

assistant: ```c
int match_ev_str(const char *ms, const char *ev, regex_t *regex, const struct servermsg *msg) {
  int ret;
  ret = regexec(regex, ms, 0, NULL, 0);
  if (ret == 0)
    return 1;
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<match_ev_str>:
  endbr64
  sub    $0x8,%rsp
  mov    %rdi,%rsi
  mov    %rdx,%rdi
  mov    $0x0,%r8d
  mov    $0x0,%ecx
  mov    $0x0,%edx
  call   <regexec@plt>
  test   %eax,%eax
  sete   %al
  movzbl %al,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int match_ev_str(const char *ms, const char *ev, regex_t *regex, const struct servermsg *msg) {
  int ret;
  ret = regexec(regex, ms, 0, NULL, 0);
  if (ret == 0)
    return 1;
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<match_ev_str>:
  endbr64
  sub    $0x8,%rsp
  mov    %rdi,%rsi
  xor    %r8d,%r8d
  mov    %rdx,%rdi
  xor    %ecx,%ecx
  xor    %edx,%edx
  call   <regexec@plt>
  test   %eax,%eax
  sete   %al
  add    $0x8,%rsp
  movzbl %al,%eax
  ret
```

assistant: ```c
int match_ev_str(const char *ms, const char *ev, regex_t *regex, const struct servermsg *msg) {
  int ret;
  ret = regexec(regex, ms, 0, NULL, 0);
  if (ret == 0)
    return 1;
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<match_ev_str>:
  endbr64
  sub    $0x8,%rsp
  mov    %rdi,%rsi
  xor    %r8d,%r8d
  mov    %rdx,%rdi
  xor    %ecx,%ecx
  xor    %edx,%edx
  call   <regexec@plt>
  test   %eax,%eax
  sete   %al
  add    $0x8,%rsp
  movzbl %al,%eax
  ret
```

assistant: ```c
int match_ev_str(const char *ms, const char *ev, regex_t *regex, const struct servermsg *msg) {
  int ret;
  ret = regexec(regex, ms, 0, NULL, 0);
  if (ret == 0)
    return 1;
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<buildNext>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0xc(%rbp)
  movl   $0x0,-0x14(%rbp)
  mov    -0xc(%rbp),%eax
  cltq
  shl    $0x2,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movl   $0xffffffff,(%rax)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,-0x10(%rbp)
  jmp    L31
L5:
  cmpl   $0x0,-0x10(%rbp)
  js     L29
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%edx
  mov    -0x10(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x28(%rbp),%rax
  add    %rcx,%rax
  movzbl (%rax),%eax
  cmp    %al,%dl
  jne    L42
L29:
  addl   $0x1,-0x10(%rbp)
  addl   $0x1,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x8(%rbp),%rax
  add    %rax,%rdx
  mov    -0x10(%rbp),%eax
  mov    %eax,(%rdx)
  jmp    L31
L42:
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x10(%rbp)
L31:
  mov    -0xc(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0x14(%rbp)
  jl     L5
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
int *buildNext(char *tar) {
  int m = strlen(tar);
  int j = 0;
  int *next = (int *)malloc(sizeof(int) * m);
  int t = next[0] = -1;
  while (j < m - 1) {
    if (t < 0 || tar[j] == tar[t]) {
      next[++j] = ++t;
    } else {
      t = next[t];
    }
  }
  return next;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<buildNext>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  call   <strlen@plt>
  mov    %eax,%ebx
  lea    0x0(,%rax,4),%edi
  call   <malloc@plt>
  movl   $0xffffffff,(%rax)
  lea    -0x1(%rbx),%esi
  test   %esi,%esi
  jle    L52
  mov    $0xffffffff,%edx
  mov    $0x0,%ecx
  jmp    L45
L47:
  add    $0x1,%edx
  add    $0x1,%ecx
  movslq %ecx,%rdi
  mov    %edx,(%rax,%rdi,4)
L55:
  cmp    %esi,%ecx
  jge    L52
L45:
  test   %edx,%edx
  js     L47
  movslq %edx,%rdi
  movslq %ecx,%r8
  movzbl 0x0(%rbp,%rdi,1),%ebx
  cmp    %bl,0x0(%rbp,%r8,1)
  je     L47
  mov    (%rax,%rdi,4),%edx
  jmp    L55
L52:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int *buildNext(char *tar) {
  int m = strlen(tar);
  int j = 0;
  int *next = (int *)malloc(sizeof(int) * m);
  int t = next[0] = -1;
  while (j < m - 1) {
    if (t < 0 || tar[j] == tar[t]) {
      next[++j] = ++t;
    } else {
      t = next[t];
    }
  }
  return next;
}
```