ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D7(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  lea    D21(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    D6(%rip),%rdi
  mov    %rsp,%rsi
  call   <scanf@plt>
  movsd  (%rsp),%xmm0
  mov    $0x2,%eax
  movsd  D26(%rip),%xmm1
  lea    D31(%rip),%rdi
  mulsd  %xmm0,%xmm1
  mulsd  D2(%rip),%xmm1
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D7
datatype: string
name: parse_data
arguments: tag: D21
datatype: string
name: parse_data
arguments: tag: D6
datatype: string
name: parse_data
arguments: tag: D26
datatype: f64
name: parse_data
arguments: tag: D31
datatype: string
name: parse_data
arguments: tag: D2
datatype: byte[8]

tool: D7: ***** Liters to Gallons ****

tool: D21: Enter the number of liters: 

tool: D6: %lf

tool: D26: 1.057

tool: D31: %g liters = %g gallons


tool: {"D2": [0x00,0x00,0x00,0x00,0x00,0x00,0xd0,0x3f]}

assistant: ```c
int main(void) {
  double liters, gallons;
  printf("***** Liters to Gallons ****\n");
  printf("Enter the number of liters: ");
  scanf("%lf", &liters);
  gallons = liters * 1.057 / 4.0;
  printf("%g liters = %g gallons\n", liters, gallons);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D56(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  lea    D55(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    D29(%rip),%rdi
  mov    %rsp,%rsi
  call   <scanf@plt>
  movsd  (%rsp),%xmm0
  mov    $0x2,%eax
  movsd  D16(%rip),%xmm1
  lea    D54(%rip),%rdi
  mulsd  %xmm0,%xmm1
  mulsd  D22(%rip),%xmm1
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D56
type: string
name: disassemble_data
arguments: label: D55
type: string
name: disassemble_data
arguments: label: D29
type: string
name: disassemble_data
arguments: label: D16
type: f64
name: disassemble_data
arguments: label: D54
type: string
name: disassemble_data
arguments: label: D22
type: byte[8]

tool: D56: ***** Liters to Gallons ****

tool: D55: Enter the number of liters: 

tool: D29: %lf

tool: D16: 1.057

tool: D54: %g liters = %g gallons


tool: {"D22": [0x00,0x00,0x00,0x00,0x00,0x00,0xd0,0x3f]}

assistant: ```c
int main(void) {
  double liters, gallons;
  printf("***** Liters to Gallons ****\n");
  printf("Enter the number of liters: ");
  scanf("%lf", &liters);
  gallons = liters * 1.057 / 4.0;
  printf("%g liters = %g gallons\n", liters, gallons);
  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   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D21(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x14(%rbp),%rax
  mov    %rax,%rsi
  lea    D41(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    D34(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D41(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x14(%rbp),%edx
  mov    -0x10(%rbp),%eax
  add    %edx,%eax
  pxor   %xmm0,%xmm0
  cvtsi2ss %eax,%xmm0
  movss  D19(%rip),%xmm1
  divss  %xmm1,%xmm0
  movss  %xmm0,-0xc(%rbp)
  pxor   %xmm2,%xmm2
  cvtss2sd -0xc(%rbp),%xmm2
  movq   %xmm2,%rax
  movq   %rax,%xmm0
  lea    D27(%rip),%rax
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L0
  call   <__stack_chk_fail@plt>
L0:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D21
data_type: string
name: disassemble_data
arguments: data_tag: D41
data_type: string
name: disassemble_data
arguments: data_tag: D34
data_type: string
name: disassemble_data
arguments: data_tag: D19
data_type: byte[4]
name: disassemble_data
arguments: data_tag: D27
data_type: string

tool: D21: Entre com a NOTA 1: 

tool: D41: %d

tool: D34: Entre com a NOTA 2: 

tool: {"D19": [0x00,0x00,0x00,0x40]}

tool: D27: 
Media = %f


assistant: ```c
main() {
  int n1, n2;
  float media;
  printf("Entre com a NOTA 1: ");
  scanf("%d", &n1);
  printf("Entre com a NOTA 2: ");
  scanf("%d", &n2);
  media = (float)(n1 + n2) / 2;
  printf("\nMedia = %f\n", media);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D13(%rip),%rdi
  call   <printf@plt>
  mov    %rsp,%rsi
  lea    D20(%rip),%rbx
  mov    %rbx,%rdi
  call   <scanf@plt>
  lea    D0(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  add    (%rsp),%eax
  pxor   %xmm0,%xmm0
  cvtsi2ss %eax,%xmm0
  mulss  D15(%rip),%xmm0
  cvtss2sd %xmm0,%xmm0
  lea    D4(%rip),%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  mov    $0x0,%eax
  add    $0x10,%rsp
  pop    %rbx
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D13
data_type: string
name: unpack_data
arguments: data_label: D20
data_type: string
name: unpack_data
arguments: data_label: D0
data_type: string
name: unpack_data
arguments: data_label: D15
data_type: byte[4]
name: unpack_data
arguments: data_label: D4
data_type: string

tool: D13: Entre com a NOTA 1: 

tool: D20: %d

tool: D0: Entre com a NOTA 2: 

tool: {"D15": [0x00,0x00,0x00,0x3f]}

tool: D4: 
Media = %f


assistant: ```c
main() {
  int n1, n2;
  float media;
  printf("Entre com a NOTA 1: ");
  scanf("%d", &n1);
  printf("Entre com a NOTA 2: ");
  scanf("%d", &n2);
  media = (float)(n1 + n2) / 2;
  printf("\nMedia = %f\n", media);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D35(%rip),%rdi
  lea    D11(%rip),%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rsp,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    D44(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rdi
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  pxor   %xmm0,%xmm0
  add    (%rsp),%eax
  cvtsi2ss %eax,%xmm0
  lea    D36(%rip),%rdi
  mov    $0x1,%eax
  mulss  D33(%rip),%xmm0
  cvtss2sd %xmm0,%xmm0
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D35
type_guessed: string
name: read_data
arguments: data_label: D11
type_guessed: string
name: read_data
arguments: data_label: D44
type_guessed: string
name: read_data
arguments: data_label: D36
type_guessed: string
name: read_data
arguments: data_label: D33
type_guessed: byte[4]

tool: D35: Entre com a NOTA 1: 

tool: D11: %d

tool: D44: Entre com a NOTA 2: 

tool: D36: 
Media = %f


tool: {"D33": [0x00,0x00,0x00,0x3f]}

assistant: ```c
main() {
  int n1, n2;
  float media;
  printf("Entre com a NOTA 1: ");
  scanf("%d", &n1);
  printf("Entre com a NOTA 2: ");
  scanf("%d", &n2);
  media = (float)(n1 + n2) / 2;
  printf("\nMedia = %f\n", media);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D37(%rip),%rdi
  lea    D23(%rip),%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rsp,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    D45(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rdi
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  pxor   %xmm0,%xmm0
  add    (%rsp),%eax
  cvtsi2ss %eax,%xmm0
  lea    D46(%rip),%rdi
  mov    $0x1,%eax
  mulss  D33(%rip),%xmm0
  cvtss2sd %xmm0,%xmm0
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D37
type_infered: string
name: disassemble_data
arguments: tag: D23
type_infered: string
name: disassemble_data
arguments: tag: D45
type_infered: string
name: disassemble_data
arguments: tag: D46
type_infered: string
name: disassemble_data
arguments: tag: D33
type_infered: byte[4]

tool: D37: Entre com a NOTA 1: 

tool: D23: %d

tool: D45: Entre com a NOTA 2: 

tool: D46: 
Media = %f


tool: {"D33": [0x00,0x00,0x00,0x3f]}

assistant: ```c
main() {
  int n1, n2;
  float media;
  printf("Entre com a NOTA 1: ");
  scanf("%d", &n1);
  printf("Entre com a NOTA 2: ");
  scanf("%d", &n2);
  media = (float)(n1 + n2) / 2;
  printf("\nMedia = %f\n", media);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<CRYPTOIntEnable>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    $0x40024780,%eax
  movq   $0x1,(%rax)
  mov    -0x8(%rbp),%rcx
  mov    $0x40024784,%eax
  mov    (%rax),%rax
  mov    $0x40024784,%edx
  or     %rcx,%rax
  mov    %rax,(%rdx)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void CRYPTOIntEnable(uint32_t ui32IntFlags) {

  ;
  (*((volatile unsigned long *)(0x40024000 + 0x00000780))) = 0x00000001;
  (*((volatile unsigned long *)(0x40024000 + 0x00000784))) |= ui32IntFlags;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<CRYPTOIntEnable>:
  endbr64
  mov    %rdi,-0x8(%rsp)
  movq   $0x1,0x40024780
  mov    -0x8(%rsp),%rdx
  mov    0x40024784,%rax
  or     %rdx,%rax
  mov    %rax,0x40024784
  ret
```

assistant: ```c
void CRYPTOIntEnable(uint32_t ui32IntFlags) {

  ;
  (*((volatile unsigned long *)(0x40024000 + 0x00000780))) = 0x00000001;
  (*((volatile unsigned long *)(0x40024000 + 0x00000784))) |= ui32IntFlags;
}
```
user: What is the c source code of the assembly code below:

```asm
<CRYPTOIntEnable>:
  endbr64
  mov    %rdi,-0x8(%rsp)
  movq   $0x1,0x40024780
  mov    -0x8(%rsp),%rdx
  mov    0x40024784,%rax
  or     %rdx,%rax
  mov    %rax,0x40024784
  ret
```

assistant: ```c
void CRYPTOIntEnable(uint32_t ui32IntFlags) {

  ;
  (*((volatile unsigned long *)(0x40024000 + 0x00000780))) = 0x00000001;
  (*((volatile unsigned long *)(0x40024000 + 0x00000784))) |= ui32IntFlags;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<CRYPTOIntEnable>:
  endbr64
  mov    %rdi,-0x8(%rsp)
  movq   $0x1,0x40024780
  mov    -0x8(%rsp),%rdx
  mov    0x40024784,%rax
  or     %rdx,%rax
  mov    %rax,0x40024784
  ret
```

assistant: ```c
void CRYPTOIntEnable(uint32_t ui32IntFlags) {

  ;
  (*((volatile unsigned long *)(0x40024000 + 0x00000780))) = 0x00000001;
  (*((volatile unsigned long *)(0x40024000 + 0x00000784))) |= ui32IntFlags;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<panic>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L3
  mov    -0x8(%rbp),%rax
  mov    %rax,%rsi
  lea    D17(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L7
L3:
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
L7:
  mov    $0x0,%eax
  call   <spin@plt>
  nop
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D17
data_type: string
name: extract_symbolic_data
arguments: data_label: D7
data_type: string

tool: D17: PANIC: %s


tool: D7: PANIC

assistant: ```c
void panic(char *msg) {
  if (msg)
    printf("PANIC: %s\n", msg);
  else
    printf("PANIC\n");
  spin();
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<panic>:
  endbr64
  sub    $0x8,%rsp
  test   %rdi,%rdi
  je     L17
  mov    %rdi,%rsi
  lea    D15(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L0:
  mov    $0x0,%eax
  call   <spin@plt>
  add    $0x8,%rsp
  ret
L17:
  lea    D17(%rip),%rdi
  call   <puts@plt>
  jmp    L0
```

tool_calls: name: unpack_data
arguments: label: D15
data_type: string
name: unpack_data
arguments: label: D17
data_type: string

tool: D15: PANIC: %s


tool: D17: PANIC

assistant: ```c
void panic(char *msg) {
  if (msg)
    printf("PANIC: %s\n", msg);
  else
    printf("PANIC\n");
  spin();
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<panic>:
  endbr64
  sub    $0x8,%rsp
  test   %rdi,%rdi
  je     L1
  mov    %rdi,%rsi
  xor    %eax,%eax
  lea    D4(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1090 <spin@plt>
  nopl   0x0(%rax)
L1:
  lea    D17(%rip),%rdi
  call   <puts@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1090 <spin@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D4
type_guessed: string
name: unpack_data
arguments: data_tag: D17
type_guessed: string

tool: D4: PANIC: %s


tool: D17: PANIC

assistant: ```c
void panic(char *msg) {
  if (msg)
    printf("PANIC: %s\n", msg);
  else
    printf("PANIC\n");
  spin();
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<panic>:
  endbr64
  sub    $0x8,%rsp
  test   %rdi,%rdi
  je     L4
  mov    %rdi,%rsi
  xor    %eax,%eax
  lea    D13(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1090 <spin@plt>
  nopl   0x0(%rax)
L4:
  lea    D5(%rip),%rdi
  call   <puts@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1090 <spin@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D13
type: string
name: retrieve_data
arguments: data_tag: D5
type: string

tool: D13: PANIC: %s


tool: D5: PANIC

assistant: ```c
void panic(char *msg) {
  if (msg)
    printf("PANIC: %s\n", msg);
  else
    printf("PANIC\n");
  spin();
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movl   $0x0,-0x8(%rbp)
  jmp    L29
L2:
  movl   $0x0,-0x4(%rbp)
  jmp    L1
L3:
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x4(%rbp),%eax
  cltq
  imul   $0x1388,%rax,%rax
  add    %rdx,%rax
  lea    0x0(,%rax,4),%rdx
  lea    D11(%rip),%rax
  mov    (%rdx,%rax,1),%eax
  lea    0x1(%rax),%ecx
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x4(%rbp),%eax
  cltq
  imul   $0x1388,%rax,%rax
  add    %rdx,%rax
  lea    0x0(,%rax,4),%rdx
  lea    D11(%rip),%rax
  mov    %ecx,(%rdx,%rax,1)
  addl   $0x1,-0x4(%rbp)
L1:
  cmpl   $0x1387,-0x4(%rbp)
  jle    L3
  addl   $0x1,-0x8(%rbp)
L29:
  cmpl   $0x1387,-0x8(%rbp)
  jle    L2
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main() {
  static int x[5000][5000];
  for (int j = 0; j < 5000; j++) {
    for (int i = 0; i < 5000; i++) {
      x[i][j]++;
    }
  }
  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
  lea    D5(%rip),%rcx
  lea    0x5f5e100(%rcx),%rdx
  add    $0x5f62f20,%rcx
L3:
  lea    -0x5f5e100(%rdx),%rax
L8:
  addl   $0x1,(%rax)
  add    $0x4e20,%rax
  cmp    %rdx,%rax
  jne    L8
  add    $0x4,%rdx
  cmp    %rcx,%rdx
  jne    L3
  mov    $0x0,%eax
  ret
```

assistant: ```c
int main() {
  static int x[5000][5000];
  for (int j = 0; j < 5000; j++) {
    for (int i = 0; i < 5000; i++) {
      x[i][j]++;
    }
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  lea    D5(%rip),%rcx
  lea    0x5f5e100(%rcx),%rdx
  add    $0x5f62f20,%rcx
L15:
  lea    -0x5f5e100(%rdx),%rax
L12:
  addl   $0x1,(%rax)
  add    $0x4e20,%rax
  cmp    %rdx,%rax
  jne    L12
  lea    0x4(%rax),%rdx
  cmp    %rcx,%rdx
  jne    L15
  xor    %eax,%eax
  ret
```

assistant: ```c
int main() {
  static int x[5000][5000];
  for (int j = 0; j < 5000; j++) {
    for (int i = 0; i < 5000; i++) {
      x[i][j]++;
    }
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  lea    D11(%rip),%rdx
  movdqa D0(%rip),%xmm1
  lea    0x5f5e100(%rdx),%rcx
L18:
  xor    %eax,%eax
  nopl   0x0(%rax)
L3:
  movdqa (%rdx,%rax,1),%xmm0
  paddd  %xmm1,%xmm0
  movaps %xmm0,(%rdx,%rax,1)
  add    $0x10,%rax
  cmp    $0x4e20,%rax
  jne    L3
  add    $0x4e20,%rdx
  cmp    %rcx,%rdx
  jne    L18
  xor    %eax,%eax
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D0
data_type: byte[16]

tool: {"D0": [0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00]}

assistant: ```c
int main() {
  static int x[5000][5000];
  for (int j = 0; j < 5000; j++) {
    for (int i = 0; i < 5000; i++) {
      x[i][j]++;
    }
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<unitTest>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x121eb25,%esi
  mov    $0x0,%edi
  call   <check@plt>
  mov    $0x121eb26,%esi
  mov    $0x1,%edi
  call   <check@plt>
  mov    $0x121eb89,%esi
  mov    $0x1f,%edi
  call   <check@plt>
  mov    $0x121ebed,%esi
  mov    $0x3c,%edi
  call   <check@plt>
  mov    $0x121ef8f,%esi
  mov    $0x16d,%edi
  call   <check@plt>
  mov    $0x1221235,%esi
  mov    $0x16e,%edi
  call   <check@plt>
  nop
  pop    %rbp
  ret
```

assistant: ```c
void unitTest() {
  check(0, 19000101);
  check(1, 19000102);
  check(31, 19000201);
  check(60, 19000301);
  check(365, 19001231);
  check(366, 19010101);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<unitTest>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x121eb25,%esi
  mov    $0x0,%edi
  call   <check@plt>
  mov    $0x121eb26,%esi
  mov    $0x1,%edi
  call   <check@plt>
  mov    $0x121eb89,%esi
  mov    $0x1f,%edi
  call   <check@plt>
  mov    $0x121ebed,%esi
  mov    $0x3c,%edi
  call   <check@plt>
  mov    $0x121ef8f,%esi
  mov    $0x16d,%edi
  call   <check@plt>
  mov    $0x1221235,%esi
  mov    $0x16e,%edi
  call   <check@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void unitTest() {
  check(0, 19000101);
  check(1, 19000102);
  check(31, 19000201);
  check(60, 19000301);
  check(365, 19001231);
  check(366, 19010101);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<unitTest>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x121eb25,%esi
  xor    %edi,%edi
  call   <check@plt>
  mov    $0x121eb26,%esi
  mov    $0x1,%edi
  call   <check@plt>
  mov    $0x121eb89,%esi
  mov    $0x1f,%edi
  call   <check@plt>
  mov    $0x121ebed,%esi
  mov    $0x3c,%edi
  call   <check@plt>
  mov    $0x121ef8f,%esi
  mov    $0x16d,%edi
  call   <check@plt>
  mov    $0x1221235,%esi
  mov    $0x16e,%edi
  add    $0x8,%rsp
  jmp    1050 <check@plt>
```

assistant: ```c
void unitTest() {
  check(0, 19000101);
  check(1, 19000102);
  check(31, 19000201);
  check(60, 19000301);
  check(365, 19001231);
  check(366, 19010101);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<unitTest>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x121eb25,%esi
  xor    %edi,%edi
  call   <check@plt>
  mov    $0x121eb26,%esi
  mov    $0x1,%edi
  call   <check@plt>
  mov    $0x121eb89,%esi
  mov    $0x1f,%edi
  call   <check@plt>
  mov    $0x121ebed,%esi
  mov    $0x3c,%edi
  call   <check@plt>
  mov    $0x121ef8f,%esi
  mov    $0x16d,%edi
  call   <check@plt>
  mov    $0x1221235,%esi
  mov    $0x16e,%edi
  add    $0x8,%rsp
  jmp    1050 <check@plt>
```

assistant: ```c
void unitTest() {
  check(0, 19000101);
  check(1, 19000102);
  check(31, 19000201);
  check(60, 19000301);
  check(365, 19001231);
  check(366, 19010101);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<my_strpos>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,%eax
  mov    %al,-0x1c(%rbp)
  movl   $0xffffffff,-0x4(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  je     L48
  movl   $0x0,-0x4(%rbp)
  jmp    L1
L40:
  addl   $0x1,-0x4(%rbp)
L1:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    %al,-0x1c(%rbp)
  je     L10
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L40
L10:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L48
  movl   $0xffffffff,-0x4(%rbp)
L48:
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int my_strpos(char *str, char s) {
  int i = -1;
  if (str) {
    for (i = 0; str[i] != s && str[i] != '\0'; i++)
      ;
    if (str[i] == '\0')
      i = -1;
  }
  return i;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<my_strpos>:
  endbr64
  mov    %esi,%r8d
  test   %rdi,%rdi
  je     L11
  movzbl (%rdi),%edx
  test   %dl,%dl
  je     L19
  cmp    %sil,%dl
  je     L19
  mov    $0x1,%eax
L48:
  mov    %eax,%ecx
  movzbl (%rdi,%rax,1),%edx
  add    $0x1,%rax
  test   %dl,%dl
  je     L2
  cmp    %r8b,%dl
  jne    L48
L2:
  test   %dl,%dl
  mov    $0xffffffff,%eax
  cmove  %eax,%ecx
L31:
  mov    %ecx,%eax
  ret
L19:
  mov    $0x0,%ecx
  jmp    L2
L11:
  mov    $0xffffffff,%ecx
  jmp    L31
```

assistant: ```c
int my_strpos(char *str, char s) {
  int i = -1;
  if (str) {
    for (i = 0; str[i] != s && str[i] != '\0'; i++)
      ;
    if (str[i] == '\0')
      i = -1;
  }
  return i;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<my_strpos>:
  endbr64
  mov    %esi,%ecx
  test   %rdi,%rdi
  je     L48
  movzbl (%rdi),%edx
  mov    $0x1,%eax
  xor    %r8d,%r8d
  cmp    %dl,%sil
  jne    L4
  jmp    L12
  nopl   (%rax)
L26:
  mov    %eax,%r8d
  add    $0x1,%rax
  movzbl -0x1(%rdi,%rax,1),%edx
  cmp    %cl,%dl
  je     L12
L4:
  test   %dl,%dl
  jne    L26
L48:
  mov    $0xffffffff,%r8d
  mov    %r8d,%eax
  ret
  xchg   %ax,%ax
L12:
  test   %sil,%sil
  mov    $0xffffffff,%eax
  cmove  %eax,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
int my_strpos(char *str, char s) {
  int i = -1;
  if (str) {
    for (i = 0; str[i] != s && str[i] != '\0'; i++)
      ;
    if (str[i] == '\0')
      i = -1;
  }
  return i;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<my_strpos>:
  endbr64
  mov    %esi,%ecx
  test   %rdi,%rdi
  je     L33
  movzbl (%rdi),%edx
  mov    $0x1,%eax
  xor    %r8d,%r8d
  cmp    %dl,%sil
  jne    L21
  jmp    L46
  nopl   (%rax)
L32:
  mov    %eax,%r8d
  add    $0x1,%rax
  movzbl -0x1(%rdi,%rax,1),%edx
  cmp    %cl,%dl
  je     L46
L21:
  test   %dl,%dl
  jne    L32
L33:
  mov    $0xffffffff,%r8d
  mov    %r8d,%eax
  ret
  xchg   %ax,%ax
L46:
  test   %sil,%sil
  mov    $0xffffffff,%eax
  cmove  %eax,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
int my_strpos(char *str, char s) {
  int i = -1;
  if (str) {
    for (i = 0; str[i] != s && str[i] != '\0'; i++)
      ;
    if (str[i] == '\0')
      i = -1;
  }
  return i;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<process_http_head_request>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  mov    %rdx,-0x10(%rbp)
  mov    %ecx,-0x14(%rbp)
  mov    %r8d,-0x18(%rbp)
  mov    %r9,-0x20(%rbp)
  mov    -0x20(%rbp),%r8
  mov    -0x18(%rbp),%edi
  mov    -0x14(%rbp),%ecx
  mov    -0x10(%rbp),%rdx
  mov    -0x8(%rbp),%esi
  mov    -0x4(%rbp),%eax
  mov    %r8,%r9
  mov    %edi,%r8d
  mov    %eax,%edi
  call   <send_http_response_header@plt>
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <close@plt>
  nop
  leave
  ret
```

assistant: ```c
int process_http_head_request(int soc, int version, char *file, int size, int isdir, char *dirhtml) {
  send_http_response_header(soc, version, file, size, isdir, dirhtml);
  close(soc);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<process_http_head_request>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  call   <send_http_response_header@plt>
  mov    %ebx,%edi
  call   <close@plt>
  pop    %rbx
  ret
```

assistant: ```c
int process_http_head_request(int soc, int version, char *file, int size, int isdir, char *dirhtml) {
  send_http_response_header(soc, version, file, size, isdir, dirhtml);
  close(soc);
}
```