ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What would the C source look like for the assembly code provided here:

```asm
<shred>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    $0x0,%eax
  call   <fork@plt>
  test   %rax,%rax
  jne    L27
  mov    -0x2c(%rbp),%edx
  lea    -0x12(%rbp),%rax
  lea    D25(%rip),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <sprintf@plt>
  lea    -0x12(%rbp),%rdx
  mov    -0x28(%rbp),%rax
  sub    $0x8,%rsp
  push   $0x0
  mov    %rdx,%r9
  lea    D0(%rip),%r8
  lea    D38(%rip),%rdx
  mov    %rdx,%rcx
  mov    %rax,%rdx
  lea    D46(%rip),%rax
  mov    %rax,%rsi
  lea    D46(%rip),%rax
  mov    %rax,%rdi
  call   <execl@plt>
  add    $0x10,%rsp
  jmp    L9
L27:
  mov    $0x0,%edi
  call   <wait@plt>
  mov    $0x0,%eax
L9:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L26
  call   <__stack_chk_fail@plt>
L26:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D25
datatype: string
name: retrieve_data
arguments: tag: D0
datatype: string
name: retrieve_data
arguments: tag: D38
datatype: string
name: retrieve_data
arguments: tag: D46
datatype: string

tool: D25: %u

tool: D0: -s

tool: D38: -z

tool: D46: /usr/bin/shred

assistant: ```c
int shred(char *filename, int upToByte) {
  if (fork() == 0) {
    char bytes[10];
    sprintf(bytes, "%u", upToByte);
    execl("/usr/bin/shred", "/usr/bin/shred", filename, "-z", "-s", bytes, NULL);
  } else {
    wait(NULL);
    return 0;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<shred>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %rdi,%rbp
  mov    %esi,%ebx
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <fork@plt>
  test   %rax,%rax
  jne    L20
  lea    0xe(%rsp),%r12
  mov    %ebx,%edx
  lea    D20(%rip),%rsi
  mov    %r12,%rdi
  call   <sprintf@plt>
  sub    $0x8,%rsp
  push   $0x0
  mov    %r12,%r9
  lea    D10(%rip),%r8
  lea    D24(%rip),%rcx
  mov    %rbp,%rdx
  lea    D2(%rip),%rdi
  mov    %rdi,%rsi
  call   <execl@plt>
  add    $0x10,%rsp
L10:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L17
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L20:
  mov    $0x0,%edi
  call   <wait@plt>
  mov    $0x0,%eax
  jmp    L10
L17:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D20
data_type: string
name: retrieve_data
arguments: data_label: D10
data_type: string
name: retrieve_data
arguments: data_label: D24
data_type: string
name: retrieve_data
arguments: data_label: D2
data_type: string

tool: D20: %u

tool: D10: -s

tool: D24: -z

tool: D2: /usr/bin/shred

assistant: ```c
int shred(char *filename, int upToByte) {
  if (fork() == 0) {
    char bytes[10];
    sprintf(bytes, "%u", upToByte);
    execl("/usr/bin/shred", "/usr/bin/shred", filename, "-z", "-s", bytes, NULL);
  } else {
    wait(NULL);
    return 0;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<shred>:
  endbr64
  push   %r13
  mov    %rdi,%r13
  push   %r12
  mov    %esi,%r12d
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <fork@plt>
  test   %rax,%rax
  jne    L24
  lea    0xe(%rsp),%rbx
  mov    %r12d,%edx
  lea    D38(%rip),%rsi
  mov    %rbx,%rdi
  call   <sprintf@plt>
  sub    $0x8,%rsp
  mov    %r13,%rdx
  mov    %rbx,%r9
  push   $0x0
  lea    D7(%rip),%rdi
  lea    D9(%rip),%r8
  lea    D33(%rip),%rcx
  mov    %rdi,%rsi
  call   <execl@plt>
  pop    %rax
  pop    %rdx
L7:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L27
  add    $0x20,%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L24:
  xor    %edi,%edi
  call   <wait@plt>
  xor    %eax,%eax
  jmp    L7
L27:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D38
type: string
name: read_data
arguments: label: D7
type: string
name: read_data
arguments: label: D9
type: string
name: read_data
arguments: label: D33
type: string

tool: D38: %u

tool: D7: /usr/bin/shred

tool: D9: -s

tool: D33: -z

assistant: ```c
int shred(char *filename, int upToByte) {
  if (fork() == 0) {
    char bytes[10];
    sprintf(bytes, "%u", upToByte);
    execl("/usr/bin/shred", "/usr/bin/shred", filename, "-z", "-s", bytes, NULL);
  } else {
    wait(NULL);
    return 0;
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<shred>:
  endbr64
  push   %r13
  mov    %rdi,%r13
  push   %r12
  mov    %esi,%r12d
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <fork@plt>
  test   %rax,%rax
  jne    L2
  lea    0xe(%rsp),%rbx
  mov    %r12d,%edx
  lea    D4(%rip),%rsi
  mov    %rbx,%rdi
  call   <sprintf@plt>
  sub    $0x8,%rsp
  mov    %r13,%rdx
  mov    %rbx,%r9
  push   $0x0
  lea    D13(%rip),%rdi
  lea    D38(%rip),%r8
  lea    D12(%rip),%rcx
  mov    %rdi,%rsi
  call   <execl@plt>
  pop    %rax
  pop    %rdx
L16:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L17
  add    $0x20,%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L2:
  xor    %edi,%edi
  call   <wait@plt>
  xor    %eax,%eax
  jmp    L16
L17:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D4
data_type: string
name: parse_data
arguments: label: D13
data_type: string
name: parse_data
arguments: label: D38
data_type: string
name: parse_data
arguments: label: D12
data_type: string

tool: D4: %u

tool: D13: /usr/bin/shred

tool: D38: -s

tool: D12: -z

assistant: ```c
int shred(char *filename, int upToByte) {
  if (fork() == 0) {
    char bytes[10];
    sprintf(bytes, "%u", upToByte);
    execl("/usr/bin/shred", "/usr/bin/shred", filename, "-z", "-s", bytes, NULL);
  } else {
    wait(NULL);
    return 0;
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<render_byte>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%edx
  mov    -0x10(%rbp),%rax
  mov    %dl,(%rax)
  mov    -0x10(%rbp),%rax
  add    $0x1,%rax
  pop    %rbp
  ret
```

assistant: ```c
void *render_byte(int byte, void *dest) {
  *(unsigned char *)dest = (unsigned char)byte;
  return (unsigned char *)dest + 1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<render_byte>:
  endbr64
  mov    %dil,(%rsi)
  lea    0x1(%rsi),%rax
  ret
```

assistant: ```c
void *render_byte(int byte, void *dest) {
  *(unsigned char *)dest = (unsigned char)byte;
  return (unsigned char *)dest + 1;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<render_byte>:
  endbr64
  mov    %dil,(%rsi)
  lea    0x1(%rsi),%rax
  ret
```

assistant: ```c
void *render_byte(int byte, void *dest) {
  *(unsigned char *)dest = (unsigned char)byte;
  return (unsigned char *)dest + 1;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<render_byte>:
  endbr64
  mov    %dil,(%rsi)
  lea    0x1(%rsi),%rax
  ret
```

assistant: ```c
void *render_byte(int byte, void *dest) {
  *(unsigned char *)dest = (unsigned char)byte;
  return (unsigned char *)dest + 1;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<WriteSword>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x18(%rbp),%eax
  mov    %al,-0xa(%rbp)
  mov    -0x18(%rbp),%eax
  shr    $0x8,%eax
  mov    %al,-0x9(%rbp)
  lea    -0xa(%rbp),%rcx
  mov    -0x14(%rbp),%eax
  mov    $0x2,%edx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <write@plt>
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L0
  call   <__stack_chk_fail@plt>
L0:
  leave
  ret
```

assistant: ```c
int WriteSword(int hFilename, unsigned dValue) {
  unsigned char cBuffer[2];
  cBuffer[0] = dValue & 0xFF;
  cBuffer[1] = (dValue >> 8) & 0xFF;
  return write(hFilename, cBuffer, 2);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<WriteSword>:
  endbr64
  sub    $0x18,%rsp
  mov    %esi,%eax
  mov    %fs:0x28,%rdx
  mov    %rdx,0x8(%rsp)
  xor    %edx,%edx
  mov    %sil,0x6(%rsp)
  movzbl %ah,%eax
  mov    %al,0x7(%rsp)
  lea    0x6(%rsp),%rsi
  mov    $0x2,%edx
  call   <write@plt>
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L3
  add    $0x18,%rsp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int WriteSword(int hFilename, unsigned dValue) {
  unsigned char cBuffer[2];
  cBuffer[0] = dValue & 0xFF;
  cBuffer[1] = (dValue >> 8) & 0xFF;
  return write(hFilename, cBuffer, 2);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<WriteSword>:
  endbr64
  sub    $0x18,%rsp
  mov    $0x2,%edx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %si,0x6(%rsp)
  lea    0x6(%rsp),%rsi
  call   <write@plt>
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L5
  add    $0x18,%rsp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int WriteSword(int hFilename, unsigned dValue) {
  unsigned char cBuffer[2];
  cBuffer[0] = dValue & 0xFF;
  cBuffer[1] = (dValue >> 8) & 0xFF;
  return write(hFilename, cBuffer, 2);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<WriteSword>:
  endbr64
  sub    $0x18,%rsp
  mov    $0x2,%edx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %si,0x6(%rsp)
  lea    0x6(%rsp),%rsi
  call   <write@plt>
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L9
  add    $0x18,%rsp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int WriteSword(int hFilename, unsigned dValue) {
  unsigned char cBuffer[2];
  cBuffer[0] = dValue & 0xFF;
  cBuffer[1] = (dValue >> 8) & 0xFF;
  return write(hFilename, cBuffer, 2);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<h264_calloc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    -0x14(%rbp),%eax
  imul   -0x18(%rbp),%eax
  mov    %eax,%edi
  mov    $0x0,%eax
  call   <h264_malloc@plt>
  cltq
  mov    %rax,-0x8(%rbp)
  mov    -0x14(%rbp),%eax
  imul   -0x18(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
void *h264_calloc(int n, int i_size) {
  unsigned char *buf;
  buf = (unsigned char *)h264_malloc(n * i_size);
  memset(buf, 0, n * i_size);
  return buf;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<h264_calloc>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebx
  imul   %esi,%ebx
  mov    %ebx,%edi
  mov    $0x0,%eax
  call   <h264_malloc@plt>
  movslq %eax,%rbp
  movslq %ebx,%rdx
  mov    $0x0,%esi
  mov    %rbp,%rdi
  call   <memset@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void *h264_calloc(int n, int i_size) {
  unsigned char *buf;
  buf = (unsigned char *)h264_malloc(n * i_size);
  memset(buf, 0, n * i_size);
  return buf;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<h264_calloc>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  xor    %eax,%eax
  imul   %esi,%ebx
  mov    %ebx,%edi
  call   <h264_malloc@plt>
  movslq %ebx,%rdx
  xor    %esi,%esi
  movslq %eax,%r8
  mov    %r8,%rdi
  call   <memset@plt>
  pop    %rbx
  ret
```

assistant: ```c
void *h264_calloc(int n, int i_size) {
  unsigned char *buf;
  buf = (unsigned char *)h264_malloc(n * i_size);
  memset(buf, 0, n * i_size);
  return buf;
}
```
user: Translate the below assembly code into C programming language:

```asm
<h264_calloc>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  xor    %eax,%eax
  imul   %esi,%ebx
  mov    %ebx,%edi
  call   <h264_malloc@plt>
  movslq %ebx,%rdx
  xor    %esi,%esi
  movslq %eax,%r8
  mov    %r8,%rdi
  call   <memset@plt>
  pop    %rbx
  ret
```

assistant: ```c
void *h264_calloc(int n, int i_size) {
  unsigned char *buf;
  buf = (unsigned char *)h264_malloc(n * i_size);
  memset(buf, 0, n * i_size);
  return buf;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<dc_armor_radix64>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  movl   $0x0,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L14
L11:
  mov    -0x28(%rbp),%rax
  mov    -0xc(%rbp),%edx
  sub    %edx,%eax
  mov    %eax,-0x4(%rbp)
  cmpl   $0x3,-0x4(%rbp)
  jle    L2
  movl   $0x3,-0x4(%rbp)
L2:
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  lea    (%rdx,%rax,1),%rsi
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0x4(%rbp),%eax
  mov    %eax,%edx
  mov    %rcx,%rdi
  call   <dc_armor_chunk24@plt>
  addl   $0x3,-0xc(%rbp)
  addl   $0x4,-0x8(%rbp)
L14:
  mov    -0xc(%rbp),%eax
  cltq
  cmp    %rax,-0x28(%rbp)
  ja     L11
  nop
  nop
  leave
  ret
```

assistant: ```c
void dc_armor_radix64(char *octets, char *index, size_t len) {
  int i, k;
  int chunk;
  for (k = 0, i = 0; i < len; i += 3, k += 4) {
    chunk = len - i;
    if (chunk > 3) {
      chunk = 3;
    }
    dc_armor_chunk24(octets + i, index + k, chunk);
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<dc_armor_radix64>:
  endbr64
  test   %rdx,%rdx
  je     L8
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r13
  mov    %rdx,%r12
  mov    %rsi,%rbp
  mov    $0x0,%ebx
L5:
  mov    %r12d,%edx
  sub    %ebx,%edx
  mov    $0x3,%eax
  cmp    %eax,%edx
  cmovg  %eax,%edx
  lea    0x0(%r13,%rbx,1),%rdi
  mov    %rbp,%rsi
  call   <dc_armor_chunk24@plt>
  add    $0x3,%rbx
  add    $0x4,%rbp
  cmp    %rbx,%r12
  ja     L5
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L8:
  ret
```

assistant: ```c
void dc_armor_radix64(char *octets, char *index, size_t len) {
  int i, k;
  int chunk;
  for (k = 0, i = 0; i < len; i += 3, k += 4) {
    chunk = len - i;
    if (chunk > 3) {
      chunk = 3;
    }
    dc_armor_chunk24(octets + i, index + k, chunk);
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<dc_armor_radix64>:
  endbr64
  test   %rdx,%rdx
  je     L8
  push   %r14
  mov    %edx,%r14d
  push   %r13
  mov    %rdi,%r13
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  nop
L12:
  mov    %r14d,%edx
  mov    $0x3,%eax
  lea    0x0(%r13,%rbx,1),%rdi
  mov    %rbp,%rsi
  sub    %ebx,%edx
  cmp    %eax,%edx
  cmovg  %eax,%edx
  add    $0x3,%rbx
  add    $0x4,%rbp
  call   <dc_armor_chunk24@plt>
  cmp    %rbx,%r12
  ja     L12
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopw   0x0(%rax,%rax,1)
L8:
  ret
```

assistant: ```c
void dc_armor_radix64(char *octets, char *index, size_t len) {
  int i, k;
  int chunk;
  for (k = 0, i = 0; i < len; i += 3, k += 4) {
    chunk = len - i;
    if (chunk > 3) {
      chunk = 3;
    }
    dc_armor_chunk24(octets + i, index + k, chunk);
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<dc_armor_radix64>:
  endbr64
  test   %rdx,%rdx
  je     L15
  push   %r14
  mov    %edx,%r14d
  push   %r13
  mov    %rdi,%r13
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  nop
L9:
  mov    %r14d,%edx
  mov    $0x3,%eax
  lea    0x0(%r13,%rbx,1),%rdi
  mov    %rbp,%rsi
  sub    %ebx,%edx
  cmp    %eax,%edx
  cmovg  %eax,%edx
  add    $0x3,%rbx
  add    $0x4,%rbp
  call   <dc_armor_chunk24@plt>
  cmp    %rbx,%r12
  ja     L9
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopw   0x0(%rax,%rax,1)
L15:
  ret
```

assistant: ```c
void dc_armor_radix64(char *octets, char *index, size_t len) {
  int i, k;
  int chunk;
  for (k = 0, i = 0; i < len; i += 3, k += 4) {
    chunk = len - i;
    if (chunk > 3) {
      chunk = 3;
    }
    dc_armor_chunk24(octets + i, index + k, chunk);
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<calc_6j_long_double>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  mov    %edx,-0x10(%rbp)
  mov    %ecx,-0x14(%rbp)
  mov    %r8d,-0x18(%rbp)
  mov    %r9d,-0x1c(%rbp)
  mov    -0x1c(%rbp),%edi
  mov    -0x18(%rbp),%ecx
  mov    -0x14(%rbp),%edx
  mov    -0x10(%rbp),%esi
  mov    -0xc(%rbp),%eax
  mov    0x10(%rbp),%r9d
  mov    %edi,%r8d
  mov    %eax,%edi
  call   <trivial_zero_6j@plt>
  test   %rax,%rax
  je     L13
  mov    -0x8(%rbp),%rax
  fldz
  fstpt  (%rax)
  jmp    L15
L13:
  mov    -0x1c(%rbp),%r9d
  mov    -0x18(%rbp),%r8d
  mov    -0x14(%rbp),%ecx
  mov    -0x10(%rbp),%edx
  mov    -0xc(%rbp),%esi
  mov    0x18(%rbp),%rax
  sub    $0x8,%rsp
  mov    0x10(%rbp),%edi
  push   %rdi
  mov    %rax,%rdi
  call   <calcsum_6j@plt>
  add    $0x10,%rsp
  mov    0x18(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <eval_calcsum_info_long_double@plt>
L15:
  leave
  ret
```

assistant: ```c
void calc_6j_long_double(long double *result, int two_j1, int two_j2, int two_j3, int two_j4, int two_j5, int two_j6, struct wigxjpf_temp *temp) {
  if (trivial_zero_6j(two_j1, two_j2, two_j3, two_j4, two_j5, two_j6)) {
    *result = 0;
    return;
  }
  calcsum_6j(temp, two_j1, two_j2, two_j3, two_j4, two_j5, two_j6);
  eval_calcsum_info_long_double(result, temp);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<calc_6j_long_double>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r15
  mov    %esi,%ebx
  mov    %edx,%ebp
  mov    %ecx,%r12d
  mov    %r8d,%r13d
  mov    %r9d,%r14d
  mov    0x40(%rsp),%r9d
  mov    %r14d,%r8d
  mov    %r13d,%ecx
  mov    %r12d,%edx
  mov    %ebp,%esi
  mov    %ebx,%edi
  call   <trivial_zero_6j@plt>
  test   %rax,%rax
  je     L18
  fldz
  fstpt  (%r15)
L8:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L18:
  sub    $0x8,%rsp
  mov    0x48(%rsp),%eax
  push   %rax
  mov    %r14d,%r9d
  mov    %r13d,%r8d
  mov    %r12d,%ecx
  mov    %ebp,%edx
  mov    %ebx,%esi
  mov    0x58(%rsp),%rdi
  call   <calcsum_6j@plt>
  mov    0x58(%rsp),%rsi
  mov    %r15,%rdi
  call   <eval_calcsum_info_long_double@plt>
  add    $0x10,%rsp
  jmp    L8
```

assistant: ```c
void calc_6j_long_double(long double *result, int two_j1, int two_j2, int two_j3, int two_j4, int two_j5, int two_j6, struct wigxjpf_temp *temp) {
  if (trivial_zero_6j(two_j1, two_j2, two_j3, two_j4, two_j5, two_j6)) {
    *result = 0;
    return;
  }
  calcsum_6j(temp, two_j1, two_j2, two_j3, two_j4, two_j5, two_j6);
  eval_calcsum_info_long_double(result, temp);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<calc_6j_long_double>:
  endbr64
  push   %r15
  push   %r14
  mov    %rdi,%r14
  push   %r13
  mov    %ecx,%r13d
  push   %r12
  mov    %edx,%r12d
  mov    %r13d,%edx
  push   %rbp
  mov    %esi,%ebp
  mov    %r12d,%esi
  push   %rbx
  mov    %r8d,%ebx
  mov    %r9d,%r8d
  mov    %ebp,%edi
  mov    %ebx,%ecx
  sub    $0x18,%rsp
  mov    0x50(%rsp),%r9d
  mov    0x58(%rsp),%r15
  mov    %r8d,0x8(%rsp)
  mov    %r9d,0xc(%rsp)
  call   <trivial_zero_6j@plt>
  mov    0x8(%rsp),%r8d
  mov    0xc(%rsp),%r9d
  test   %rax,%rax
  je     L1
  fldz
  fstpt  (%r14)
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L1:
  sub    $0x8,%rsp
  mov    %ebp,%esi
  mov    %r15,%rdi
  mov    %r13d,%ecx
  push   %r9
  mov    %r12d,%edx
  mov    %r8d,%r9d
  mov    %ebx,%r8d
  call   <calcsum_6j@plt>
  add    $0x28,%rsp
  mov    %r15,%rsi
  mov    %r14,%rdi
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  jmp    1080 <eval_calcsum_info_long_double@plt>
```

assistant: ```c
void calc_6j_long_double(long double *result, int two_j1, int two_j2, int two_j3, int two_j4, int two_j5, int two_j6, struct wigxjpf_temp *temp) {
  if (trivial_zero_6j(two_j1, two_j2, two_j3, two_j4, two_j5, two_j6)) {
    *result = 0;
    return;
  }
  calcsum_6j(temp, two_j1, two_j2, two_j3, two_j4, two_j5, two_j6);
  eval_calcsum_info_long_double(result, temp);
}
```
user: Translate the below assembly code into C programming language:

```asm
<calc_6j_long_double>:
  endbr64
  push   %r15
  push   %r14
  mov    %rdi,%r14
  push   %r13
  mov    %ecx,%r13d
  push   %r12
  mov    %edx,%r12d
  mov    %r13d,%edx
  push   %rbp
  mov    %esi,%ebp
  mov    %r12d,%esi
  push   %rbx
  mov    %r8d,%ebx
  mov    %r9d,%r8d
  mov    %ebp,%edi
  mov    %ebx,%ecx
  sub    $0x18,%rsp
  mov    0x50(%rsp),%r9d
  mov    0x58(%rsp),%r15
  mov    %r8d,0x8(%rsp)
  mov    %r9d,0xc(%rsp)
  call   <trivial_zero_6j@plt>
  mov    0x8(%rsp),%r8d
  mov    0xc(%rsp),%r9d
  test   %rax,%rax
  je     L9
  fldz
  fstpt  (%r14)
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L9:
  sub    $0x8,%rsp
  mov    %ebp,%esi
  mov    %r15,%rdi
  mov    %r13d,%ecx
  push   %r9
  mov    %r12d,%edx
  mov    %r8d,%r9d
  mov    %ebx,%r8d
  call   <calcsum_6j@plt>
  add    $0x28,%rsp
  mov    %r15,%rsi
  mov    %r14,%rdi
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  jmp    1080 <eval_calcsum_info_long_double@plt>
```

assistant: ```c
void calc_6j_long_double(long double *result, int two_j1, int two_j2, int two_j3, int two_j4, int two_j5, int two_j6, struct wigxjpf_temp *temp) {
  if (trivial_zero_6j(two_j1, two_j2, two_j3, two_j4, two_j5, two_j6)) {
    *result = 0;
    return;
  }
  calcsum_6j(temp, two_j1, two_j2, two_j3, two_j4, two_j5, two_j6);
  eval_calcsum_info_long_double(result, temp);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<SetTerminal>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    D28(%rip),%rax
  mov    (%rax),%eax
  lea    -0x20(%rbp),%rdx
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <tcgetattr@plt>
  mov    -0x20(%rbp),%edx
  mov    D68(%rip),%rax
  mov    (%rax),%ecx
  mov    D6(%rip),%rax
  mov    (%rax),%eax
  or     %eax,%ecx
  mov    D63(%rip),%rax
  mov    (%rax),%eax
  or     %ecx,%eax
  not    %eax
  and    %edx,%eax
  mov    %eax,-0x20(%rbp)
  mov    -0x20(%rbp),%edx
  mov    D68(%rip),%rax
  mov    (%rax),%ecx
  mov    D6(%rip),%rax
  mov    (%rax),%eax
  or     %ecx,%eax
  not    %eax
  and    %edx,%eax
  mov    %eax,-0x20(%rbp)
  mov    -0x10(%rbp),%edx
  mov    D64(%rip),%rax
  mov    (%rax),%eax
  not    %eax
  and    %edx,%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x18(%rbp),%rdx
  mov    D44(%rip),%rax
  mov    (%rax),%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  movl   $0x0,(%rax)
  mov    -0x18(%rbp),%rdx
  mov    D109(%rip),%rax
  mov    (%rax),%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  movl   $0x1,(%rax)
  mov    D92(%rip),%rax
  mov    (%rax),%edx
  mov    -0x24(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <tcflush@plt>
  mov    D75(%rip),%rax
  mov    (%rax),%ecx
  lea    -0x20(%rbp),%rdx
  mov    -0x24(%rbp),%eax
  mov    %ecx,%esi
  mov    %eax,%edi
  call   <tcsetattr@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L3
  call   <__stack_chk_fail@plt>
L3:
  leave
  ret
```

assistant: ```c
void SetTerminal(int fd) {
  struct termios t;
  tcgetattr(STDIN_FILENO, &t);
  t.c_lflag &= ~(ECHO | ICANON | ISIG);
  t.c_lflag &= ~(ECHO | ICANON);
  t.c_iflag &= ~ICRNL;
  t.c_cc[VTIME] = 0;
  t.c_cc[VMIN] = 1;
  tcflush(fd, TCIFLUSH);
  tcsetattr(fd, TCSANOW, &t);
}
```