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

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rdi
  call   <pipe@plt>
  movb   $0x1,0x17(%rsp)
  mov    $0x0,%ebx
  lea    D2(%rip),%r12
L1:
  lea    0x17(%rsp),%rsi
  mov    $0x1,%edx
  mov    0x10(%rsp),%edi
  call   <write@plt>
  lea    0x1(%rbx),%ebp
  mov    %ebx,%esi
  mov    %r12,%rdi
  call   <printf@plt>
  mov    %ebp,%ebx
  jmp    L1
```

tool_calls: name: retrieve_data
arguments: label: D2
datatype: string

tool: D2: %d


assistant: ```c
int main() {
  int fd[2];
  pipe(fd);
  char d[1];
  d[0] = 1;
  int i = 0;
  while (1) {
    write(fd[1], d, 1);
    printf("%d\n", i++);
  }
}
```
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
  lea    D0(%rip),%r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rdi
  lea    0x17(%rsp),%r13
  call   <pipe@plt>
  movb   $0x1,0x17(%rsp)
  nopl   0x0(%rax,%rax,1)
L9:
  mov    0x10(%rsp),%edi
  mov    $0x1,%edx
  mov    %r13,%rsi
  lea    0x1(%rbp),%ebx
  call   <write@plt>
  mov    %ebp,%esi
  mov    %r12,%rdi
  mov    %ebx,%ebp
  call   <printf@plt>
  jmp    L9
```

tool_calls: name: unpack_data
arguments: data_tag: D0
type: string

tool: D0: %d


assistant: ```c
int main() {
  int fd[2];
  pipe(fd);
  char d[1];
  d[0] = 1;
  int i = 0;
  while (1) {
    write(fd[1], d, 1);
    printf("%d\n", i++);
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  lea    D9(%rip),%r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rdi
  lea    0x17(%rsp),%r13
  call   <pipe@plt>
  movb   $0x1,0x17(%rsp)
  nopl   0x0(%rax,%rax,1)
L1:
  mov    0x10(%rsp),%edi
  mov    $0x1,%edx
  mov    %r13,%rsi
  lea    0x1(%rbp),%ebx
  call   <write@plt>
  mov    %ebp,%esi
  mov    %r12,%rdi
  mov    %ebx,%ebp
  call   <printf@plt>
  jmp    L1
```

tool_calls: name: extract_symbolic_data
arguments: label: D9
datatype: string

tool: D9: %d


assistant: ```c
int main() {
  int fd[2];
  pipe(fd);
  char d[1];
  d[0] = 1;
  int i = 0;
  while (1) {
    write(fd[1], d, 1);
    printf("%d\n", i++);
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<csum_partial>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    -0x1c(%rbp),%ecx
  mov    -0x18(%rbp),%rax
  mov    $0x0,%edx
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <do_csum@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x20(%rbp),%eax
  add    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %eax,%edx
  mov    -0x8(%rbp),%rax
  shr    $0x20,%rax
  add    %rdx,%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
unsigned int csum_partial(const unsigned char *buff, int len, unsigned int sum) {
  unsigned long result = do_csum(buff, len, 0);
  result += sum;
  result = (result & 0xffffffff) + (result >> 32);
  return result;
}
```
user: What is the c source code of the assembly code below?

```asm
<csum_partial>:
  endbr64
  push   %rbx
  mov    %edx,%ebx
  mov    $0x0,%edx
  call   <do_csum@plt>
  mov    %ebx,%edx
  add    %rax,%rdx
  mov    %rdx,%rax
  shr    $0x20,%rax
  lea    (%rdx,%rax,1),%eax
  pop    %rbx
  ret
```

assistant: ```c
unsigned int csum_partial(const unsigned char *buff, int len, unsigned int sum) {
  unsigned long result = do_csum(buff, len, 0);
  result += sum;
  result = (result & 0xffffffff) + (result >> 32);
  return result;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<csum_partial>:
  endbr64
  push   %rbx
  mov    %edx,%ebx
  xor    %edx,%edx
  call   <do_csum@plt>
  mov    %ebx,%edx
  pop    %rbx
  add    %rax,%rdx
  mov    %rdx,%rax
  shr    $0x20,%rax
  add    %edx,%eax
  ret
```

assistant: ```c
unsigned int csum_partial(const unsigned char *buff, int len, unsigned int sum) {
  unsigned long result = do_csum(buff, len, 0);
  result += sum;
  result = (result & 0xffffffff) + (result >> 32);
  return result;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<csum_partial>:
  endbr64
  push   %rbx
  mov    %edx,%ebx
  xor    %edx,%edx
  call   <do_csum@plt>
  mov    %ebx,%edx
  pop    %rbx
  add    %rax,%rdx
  mov    %rdx,%rax
  shr    $0x20,%rax
  add    %edx,%eax
  ret
```

assistant: ```c
unsigned int csum_partial(const unsigned char *buff, int len, unsigned int sum) {
  unsigned long result = do_csum(buff, len, 0);
  result += sum;
  result = (result & 0xffffffff) + (result >> 32);
  return result;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  movl   $0x0,-0x14(%rbp)
  movl   $0x0,-0x10(%rbp)
  movl   $0x1,-0xc(%rbp)
  jmp    L19
L4:
  mov    -0xc(%rbp),%eax
  imul   %eax,%eax
  add    %eax,-0x14(%rbp)
  mov    -0xc(%rbp),%eax
  add    %eax,-0x10(%rbp)
  addl   $0x1,-0xc(%rbp)
L19:
  cmpl   $0x64,-0xc(%rbp)
  jle    L4
  mov    -0x10(%rbp),%eax
  imul   %eax,%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x8(%rbp),%eax
  sub    -0x14(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D0
type_guessed: string

tool: D0: result: %d

assistant: ```c
int main() {
  int squaresOfHundred = 0;
  int sumOfHundred = 0;
  int i;
  for (i = 1; i <= 100; i++) {
    squaresOfHundred += i * i;
    sumOfHundred += i;
  }
  int squaresOfSumHundred = sumOfHundred * sumOfHundred;
  int difference = squaresOfSumHundred - squaresOfHundred;
  printf("result: %d", difference);
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x64,%eax
L6:
  sub    $0x1,%eax
  jne    L6
  mov    $0x17ff976,%esi
  lea    D1(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: tag: D1
type_infered: string

tool: D1: result: %d

assistant: ```c
int main() {
  int squaresOfHundred = 0;
  int sumOfHundred = 0;
  int i;
  for (i = 1; i <= 100; i++) {
    squaresOfHundred += i * i;
    sumOfHundred += i;
  }
  int squaresOfSumHundred = sumOfHundred * sumOfHundred;
  int difference = squaresOfSumHundred - squaresOfHundred;
  printf("result: %d", difference);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x17ff976,%esi
  lea    D2(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D2
type: string

tool: D2: result: %d

assistant: ```c
int main() {
  int squaresOfHundred = 0;
  int sumOfHundred = 0;
  int i;
  for (i = 1; i <= 100; i++) {
    squaresOfHundred += i * i;
    sumOfHundred += i;
  }
  int squaresOfSumHundred = sumOfHundred * sumOfHundred;
  int difference = squaresOfSumHundred - squaresOfHundred;
  printf("result: %d", difference);
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x17ff976,%esi
  lea    D3(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

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

tool: D3: result: %d

assistant: ```c
int main() {
  int squaresOfHundred = 0;
  int sumOfHundred = 0;
  int i;
  for (i = 1; i <= 100; i++) {
    squaresOfHundred += i * i;
    sumOfHundred += i;
  }
  int squaresOfSumHundred = sumOfHundred * sumOfHundred;
  int difference = squaresOfSumHundred - squaresOfHundred;
  printf("result: %d", difference);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<GenMesMobile>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,%eax
  mov    %esi,%edx
  mov    %ax,-0x14(%rbp)
  mov    %edx,%eax
  mov    %al,-0x18(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movzwl -0x14(%rbp),%eax
  movzbl %al,%eax
  mov    %eax,%esi
  mov    $0x194,%edi
  call   <RegWriteA@plt>
  movzwl -0x14(%rbp),%eax
  shr    $0x8,%ax
  movzbl %al,%eax
  and    $0x1,%eax
  mov    %eax,%esi
  mov    $0x195,%edi
  call   <RegWriteA@plt>
  mov    $0x0,%esi
  mov    $0x1041,%edi
  call   <RamWrite32A@plt>
  cmpb   $0x0,-0x18(%rbp)
  jne    L26
  mov    $0x4,%esi
  mov    $0x193,%edi
  call   <RegWriteA@plt>
  mov    $0x0,%esi
  mov    $0x192,%edi
  call   <RegWriteA@plt>
  mov    $0x3a7ffff7,%esi
  mov    $0x1230,%edi
  call   <RamWrite32A@plt>
  jmp    L33
L26:
  mov    $0x1,%esi
  mov    $0x193,%edi
  call   <RegWriteA@plt>
  mov    $0x0,%esi
  mov    $0x192,%edi
  call   <RegWriteA@plt>
  mov    $0x3f800000,%esi
  mov    $0x1230,%edi
  call   <RamWrite32A@plt>
L33:
  mov    $0x0,%esi
  mov    $0x198,%edi
  call   <RegWriteA@plt>
  mov    $0x1,%esi
  mov    $0x190,%edi
  call   <RegWriteA@plt>
  mov    $0x64,%edi
  call   <WitTim@plt>
  lea    -0xb(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x190,%edi
  call   <RegReadA@plt>
  movzbl -0xb(%rbp),%eax
  test   %al,%al
  jne    L8
  mov    $0x64,%edi
  call   <WitTim@plt>
L8:
  mov    $0x1,%edi
  call   <RamAccFixMod@plt>
  lea    -0xa(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x1041,%edi
  call   <RamReadA@plt>
  mov    $0x0,%edi
  call   <RamAccFixMod@plt>
  movzwl -0xa(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L21
  call   <__stack_chk_fail@plt>
L21:
  leave
  ret
```

assistant: ```c
short GenMesMobile(unsigned short UsRamAdd, unsigned char UcMesMod) {
  short SsMesRlt;
  unsigned char UcMesFin;
  RegWriteA(0x0194, (unsigned char)UsRamAdd);
  RegWriteA(0x0195, (unsigned char)((UsRamAdd >> 8) & 0x0001));
  RamWrite32A(0x1041, 0x00000000);
  if (!UcMesMod) {
    RegWriteA(0x0193, 0x04);
    RegWriteA(0x0192, 0x00);
    RamWrite32A(0x1230, 0x3A7FFFF7);
  } else {
    RegWriteA(0x0193, 0x01);
    RegWriteA(0x0192, 0x00);
    RamWrite32A(0x1230, 0x3F800000);
  }
  RegWriteA(0x0198, 0x00);
  RegWriteA(0x0190, 0x01);
  WitTim(100);
  RegReadA(0x0190, &UcMesFin);
  if (0x00 == UcMesFin) {
    WitTim(100);
  }
  RamAccFixMod(0x01);
  RamReadA(0x1041, (unsigned short *)&SsMesRlt);
  RamAccFixMod(0x00);
  return (SsMesRlt);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<GenMesMobile>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %edi,%ebx
  mov    %esi,%ebp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movzbl %dil,%esi
  mov    $0x194,%edi
  call   <RegWriteA@plt>
  mov    %ebx,%esi
  shr    $0x8,%si
  and    $0x1,%esi
  mov    $0x195,%edi
  call   <RegWriteA@plt>
  mov    $0x0,%esi
  mov    $0x1041,%edi
  call   <RamWrite32A@plt>
  test   %bpl,%bpl
  jne    L31
  mov    $0x4,%esi
  mov    $0x193,%edi
  call   <RegWriteA@plt>
  mov    $0x0,%esi
  mov    $0x192,%edi
  call   <RegWriteA@plt>
  mov    $0x3a7ffff7,%esi
  mov    $0x1230,%edi
  call   <RamWrite32A@plt>
L5:
  mov    $0x0,%esi
  mov    $0x198,%edi
  call   <RegWriteA@plt>
  mov    $0x1,%esi
  mov    $0x190,%edi
  call   <RegWriteA@plt>
  mov    $0x64,%edi
  call   <WitTim@plt>
  lea    0x5(%rsp),%rsi
  mov    $0x190,%edi
  call   <RegReadA@plt>
  cmpb   $0x0,0x5(%rsp)
  je     L23
L3:
  mov    $0x1,%edi
  call   <RamAccFixMod@plt>
  lea    0x6(%rsp),%rsi
  mov    $0x1041,%edi
  call   <RamReadA@plt>
  mov    $0x0,%edi
  call   <RamAccFixMod@plt>
  movzwl 0x6(%rsp),%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L17
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L31:
  mov    $0x1,%esi
  mov    $0x193,%edi
  call   <RegWriteA@plt>
  mov    $0x0,%esi
  mov    $0x192,%edi
  call   <RegWriteA@plt>
  mov    $0x3f800000,%esi
  mov    $0x1230,%edi
  call   <RamWrite32A@plt>
  jmp    L5
L23:
  mov    $0x64,%edi
  call   <WitTim@plt>
  jmp    L3
L17:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
short GenMesMobile(unsigned short UsRamAdd, unsigned char UcMesMod) {
  short SsMesRlt;
  unsigned char UcMesFin;
  RegWriteA(0x0194, (unsigned char)UsRamAdd);
  RegWriteA(0x0195, (unsigned char)((UsRamAdd >> 8) & 0x0001));
  RamWrite32A(0x1041, 0x00000000);
  if (!UcMesMod) {
    RegWriteA(0x0193, 0x04);
    RegWriteA(0x0192, 0x00);
    RamWrite32A(0x1230, 0x3A7FFFF7);
  } else {
    RegWriteA(0x0193, 0x01);
    RegWriteA(0x0192, 0x00);
    RamWrite32A(0x1230, 0x3F800000);
  }
  RegWriteA(0x0198, 0x00);
  RegWriteA(0x0190, 0x01);
  WitTim(100);
  RegReadA(0x0190, &UcMesFin);
  if (0x00 == UcMesFin) {
    WitTim(100);
  }
  RamAccFixMod(0x01);
  RamReadA(0x1041, (unsigned short *)&SsMesRlt);
  RamAccFixMod(0x00);
  return (SsMesRlt);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<GenMesMobile>:
  endbr64
  push   %rbp
  mov    %esi,%ebp
  movzbl %dil,%esi
  push   %rbx
  mov    %edi,%ebx
  mov    $0x194,%edi
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <RegWriteA@plt>
  mov    %ebx,%esi
  mov    $0x195,%edi
  shr    $0x8,%si
  and    $0x1,%esi
  call   <RegWriteA@plt>
  xor    %esi,%esi
  mov    $0x1041,%edi
  call   <RamWrite32A@plt>
  test   %bpl,%bpl
  jne    L15
  mov    $0x4,%esi
  mov    $0x193,%edi
  call   <RegWriteA@plt>
  xor    %esi,%esi
  mov    $0x192,%edi
  call   <RegWriteA@plt>
  mov    $0x3a7ffff7,%esi
  mov    $0x1230,%edi
  call   <RamWrite32A@plt>
L10:
  xor    %esi,%esi
  mov    $0x198,%edi
  call   <RegWriteA@plt>
  mov    $0x1,%esi
  mov    $0x190,%edi
  call   <RegWriteA@plt>
  mov    $0x64,%edi
  call   <WitTim@plt>
  lea    0x5(%rsp),%rsi
  mov    $0x190,%edi
  call   <RegReadA@plt>
  cmpb   $0x0,0x5(%rsp)
  je     L25
L31:
  mov    $0x1,%edi
  call   <RamAccFixMod@plt>
  lea    0x6(%rsp),%rsi
  mov    $0x1041,%edi
  call   <RamReadA@plt>
  xor    %edi,%edi
  call   <RamAccFixMod@plt>
  movzwl 0x6(%rsp),%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L9
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopl   (%rax)
L15:
  mov    $0x1,%esi
  mov    $0x193,%edi
  call   <RegWriteA@plt>
  xor    %esi,%esi
  mov    $0x192,%edi
  call   <RegWriteA@plt>
  mov    $0x3f800000,%esi
  mov    $0x1230,%edi
  call   <RamWrite32A@plt>
  jmp    L10
  nopw   0x0(%rax,%rax,1)
L25:
  mov    $0x64,%edi
  call   <WitTim@plt>
  jmp    L31
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
short GenMesMobile(unsigned short UsRamAdd, unsigned char UcMesMod) {
  short SsMesRlt;
  unsigned char UcMesFin;
  RegWriteA(0x0194, (unsigned char)UsRamAdd);
  RegWriteA(0x0195, (unsigned char)((UsRamAdd >> 8) & 0x0001));
  RamWrite32A(0x1041, 0x00000000);
  if (!UcMesMod) {
    RegWriteA(0x0193, 0x04);
    RegWriteA(0x0192, 0x00);
    RamWrite32A(0x1230, 0x3A7FFFF7);
  } else {
    RegWriteA(0x0193, 0x01);
    RegWriteA(0x0192, 0x00);
    RamWrite32A(0x1230, 0x3F800000);
  }
  RegWriteA(0x0198, 0x00);
  RegWriteA(0x0190, 0x01);
  WitTim(100);
  RegReadA(0x0190, &UcMesFin);
  if (0x00 == UcMesFin) {
    WitTim(100);
  }
  RamAccFixMod(0x01);
  RamReadA(0x1041, (unsigned short *)&SsMesRlt);
  RamAccFixMod(0x00);
  return (SsMesRlt);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<GenMesMobile>:
  endbr64
  push   %rbp
  mov    %esi,%ebp
  movzbl %dil,%esi
  push   %rbx
  mov    %edi,%ebx
  mov    $0x194,%edi
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <RegWriteA@plt>
  mov    %ebx,%esi
  mov    $0x195,%edi
  shr    $0x8,%si
  and    $0x1,%esi
  call   <RegWriteA@plt>
  xor    %esi,%esi
  mov    $0x1041,%edi
  call   <RamWrite32A@plt>
  test   %bpl,%bpl
  jne    L19
  mov    $0x4,%esi
  mov    $0x193,%edi
  call   <RegWriteA@plt>
  xor    %esi,%esi
  mov    $0x192,%edi
  call   <RegWriteA@plt>
  mov    $0x3a7ffff7,%esi
  mov    $0x1230,%edi
  call   <RamWrite32A@plt>
L23:
  xor    %esi,%esi
  mov    $0x198,%edi
  call   <RegWriteA@plt>
  mov    $0x1,%esi
  mov    $0x190,%edi
  call   <RegWriteA@plt>
  mov    $0x64,%edi
  call   <WitTim@plt>
  lea    0x5(%rsp),%rsi
  mov    $0x190,%edi
  call   <RegReadA@plt>
  cmpb   $0x0,0x5(%rsp)
  je     L16
L34:
  mov    $0x1,%edi
  call   <RamAccFixMod@plt>
  lea    0x6(%rsp),%rsi
  mov    $0x1041,%edi
  call   <RamReadA@plt>
  xor    %edi,%edi
  call   <RamAccFixMod@plt>
  movzwl 0x6(%rsp),%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L24
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopl   (%rax)
L19:
  mov    $0x1,%esi
  mov    $0x193,%edi
  call   <RegWriteA@plt>
  xor    %esi,%esi
  mov    $0x192,%edi
  call   <RegWriteA@plt>
  mov    $0x3f800000,%esi
  mov    $0x1230,%edi
  call   <RamWrite32A@plt>
  jmp    L23
  nopw   0x0(%rax,%rax,1)
L16:
  mov    $0x64,%edi
  call   <WitTim@plt>
  jmp    L34
L24:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
short GenMesMobile(unsigned short UsRamAdd, unsigned char UcMesMod) {
  short SsMesRlt;
  unsigned char UcMesFin;
  RegWriteA(0x0194, (unsigned char)UsRamAdd);
  RegWriteA(0x0195, (unsigned char)((UsRamAdd >> 8) & 0x0001));
  RamWrite32A(0x1041, 0x00000000);
  if (!UcMesMod) {
    RegWriteA(0x0193, 0x04);
    RegWriteA(0x0192, 0x00);
    RamWrite32A(0x1230, 0x3A7FFFF7);
  } else {
    RegWriteA(0x0193, 0x01);
    RegWriteA(0x0192, 0x00);
    RamWrite32A(0x1230, 0x3F800000);
  }
  RegWriteA(0x0198, 0x00);
  RegWriteA(0x0190, 0x01);
  WitTim(100);
  RegReadA(0x0190, &UcMesFin);
  if (0x00 == UcMesFin) {
    WitTim(100);
  }
  RamAccFixMod(0x01);
  RamReadA(0x1041, (unsigned short *)&SsMesRlt);
  RamAccFixMod(0x00);
  return (SsMesRlt);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<yudpsocket_client>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    D17(%rip),%rax
  mov    (%rax),%ecx
  mov    D7(%rip),%rax
  mov    (%rax),%eax
  mov    $0x0,%edx
  mov    %ecx,%esi
  mov    %eax,%edi
  call   <socket@plt>
  mov    %eax,-0xc(%rbp)
  movl   $0x1,-0x10(%rbp)
  mov    D24(%rip),%rax
  mov    (%rax),%edx
  mov    D5(%rip),%rax
  mov    (%rax),%esi
  lea    -0x10(%rbp),%rcx
  mov    -0xc(%rbp),%eax
  mov    $0x4,%r8d
  mov    %eax,%edi
  call   <setsockopt@plt>
  mov    -0xc(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L2
  call   <__stack_chk_fail@plt>
L2:
  leave
  ret
```

assistant: ```c
int yudpsocket_client() {
  int socketfd = socket(AF_INET, SOCK_DGRAM, 0);
  int reuseon = 1;
  setsockopt(socketfd, SOL_SOCKET, SO_REUSEADDR, &reuseon, sizeof(reuseon));
  return socketfd;
}
```
user: What is the c source code of the assembly code below?

```asm
<yudpsocket_client>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    $0x0,%edx
  mov    D37(%rip),%rax
  mov    (%rax),%esi
  mov    D38(%rip),%rax
  mov    (%rax),%edi
  call   <socket@plt>
  mov    %eax,%ebx
  movl   $0x1,0x4(%rsp)
  lea    0x4(%rsp),%rcx
  mov    $0x4,%r8d
  mov    D16(%rip),%rax
  mov    (%rax),%edx
  mov    D24(%rip),%rax
  mov    (%rax),%esi
  mov    %ebx,%edi
  call   <setsockopt@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  mov    %ebx,%eax
  add    $0x10,%rsp
  pop    %rbx
  ret
L6:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int yudpsocket_client() {
  int socketfd = socket(AF_INET, SOCK_DGRAM, 0);
  int reuseon = 1;
  setsockopt(socketfd, SOL_SOCKET, SO_REUSEADDR, &reuseon, sizeof(reuseon));
  return socketfd;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<yudpsocket_client>:
  endbr64
  push   %r12
  xor    %edx,%edx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    D35(%rip),%rax
  mov    (%rax),%esi
  mov    D4(%rip),%rax
  mov    (%rax),%edi
  call   <socket@plt>
  lea    0x4(%rsp),%rcx
  mov    $0x4,%r8d
  movl   $0x1,0x4(%rsp)
  mov    %eax,%r12d
  mov    D39(%rip),%rax
  mov    %r12d,%edi
  mov    (%rax),%edx
  mov    D29(%rip),%rax
  mov    (%rax),%esi
  call   <setsockopt@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x10,%rsp
  mov    %r12d,%eax
  pop    %r12
  ret
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int yudpsocket_client() {
  int socketfd = socket(AF_INET, SOCK_DGRAM, 0);
  int reuseon = 1;
  setsockopt(socketfd, SOL_SOCKET, SO_REUSEADDR, &reuseon, sizeof(reuseon));
  return socketfd;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<yudpsocket_client>:
  endbr64
  push   %r12
  xor    %edx,%edx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    D5(%rip),%rax
  mov    (%rax),%esi
  mov    D39(%rip),%rax
  mov    (%rax),%edi
  call   <socket@plt>
  lea    0x4(%rsp),%rcx
  mov    $0x4,%r8d
  movl   $0x1,0x4(%rsp)
  mov    %eax,%r12d
  mov    D1(%rip),%rax
  mov    %r12d,%edi
  mov    (%rax),%edx
  mov    D20(%rip),%rax
  mov    (%rax),%esi
  call   <setsockopt@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x10,%rsp
  mov    %r12d,%eax
  pop    %r12
  ret
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int yudpsocket_client() {
  int socketfd = socket(AF_INET, SOCK_DGRAM, 0);
  int reuseon = 1;
  setsockopt(socketfd, SOL_SOCKET, SO_REUSEADDR, &reuseon, sizeof(reuseon));
  return socketfd;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<copy_in>:
  endbr64
  mov    %rdi,%r9
  lea    0x80(%rsi),%r8
  add    $0x200,%rdi
L6:
  lea    -0x80(%r8),%rax
  mov    %r9,%r10
L7:
  mov    (%r10),%edx
  mov    %dl,(%rax)
  movzbl %dh,%ecx
  mov    %cl,0x1(%rax)
  mov    %edx,%esi
  sar    $0x10,%esi
  mov    %sil,0x2(%rax)
  mov    %edx,%esi
  sar    $0x18,%esi
  mov    %sil,0x3(%rax)
  mov    %edx,%esi
  mov    $0x20,%ecx
  sar    %cl,%esi
  mov    %sil,0x4(%rax)
  mov    %edx,%esi
  mov    $0x28,%ecx
  sar    %cl,%esi
  mov    %sil,0x5(%rax)
  mov    %edx,%esi
  mov    $0x30,%ecx
  sar    %cl,%esi
  mov    %sil,0x6(%rax)
  mov    $0x38,%ecx
  sar    %cl,%edx
  mov    %dl,0x7(%rax)
  add    $0x4,%r10
  add    $0x8,%rax
  cmp    %r8,%rax
  jne    L7
  add    $0x40,%r9
  sub    $0xffffffffffffff80,%r8
  cmp    %rdi,%r9
  jne    L6
  ret
```

assistant: ```c
void copy_in(uint64_t *large_A, char *part_buf_A) {
  int i, j;
  for (i = 0; i < ((512) / (64)); i++) {
    for (j = 0; j < 16; j++) {
#pragma HLS PIPELINE
      uint64_t curr = large_A[i * 16 + j];
      part_buf_A[i * 128 + j * 8 + 0] = (uint8_t)(curr >> 0);
      part_buf_A[i * 128 + j * 8 + 1] = (uint8_t)(curr >> 8);
      part_buf_A[i * 128 + j * 8 + 2] = (uint8_t)(curr >> 16);
      part_buf_A[i * 128 + j * 8 + 3] = (uint8_t)(curr >> 24);
      part_buf_A[i * 128 + j * 8 + 4] = (uint8_t)(curr >> 32);
      part_buf_A[i * 128 + j * 8 + 5] = (uint8_t)(curr >> 40);
      part_buf_A[i * 128 + j * 8 + 6] = (uint8_t)(curr >> 48);
      part_buf_A[i * 128 + j * 8 + 7] = (uint8_t)(curr >> 56);
    }
  }
}
```