ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请从给定的汇编指令推导出C源代码:

```asm
<ol>:
  endbr64
  mov    %edi,%ecx
  test   %edi,%edi
  je     L38
  movslq %edi,%rax
  mov    %edi,%esi
  imul   $0x66666667,%rax,%rdx
  sar    $0x1f,%esi
  sar    $0x22,%rdx
  sub    %esi,%edx
  lea    (%rdx,%rdx,4),%esi
  mov    %edi,%edx
  add    %esi,%esi
  sub    %esi,%edx
  mov    $0x12000,%esi
  add    $0x9,%edx
  bt     %rdx,%rsi
  jb     L22
  mov    $0x12000,%edi
  jmp    L7
  nopl   (%rax)
L41:
  movslq %ecx,%rax
  mov    %ecx,%esi
  imul   $0x66666667,%rax,%rdx
  sar    $0x1f,%esi
  sar    $0x22,%rdx
  sub    %esi,%edx
  lea    (%rdx,%rdx,4),%esi
  mov    %ecx,%edx
  add    %esi,%esi
  sub    %esi,%edx
  add    $0x9,%edx
  bt     %rdx,%rdi
  jb     L22
L7:
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%ecx
  sar    $0x22,%rax
  sub    %ecx,%eax
  mov    %eax,%ecx
  jne    L41
L38:
  xor    %eax,%eax
  ret
  xchg   %ax,%ax
L22:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int ol(int n) {
  while (n)
    if ((n % 10 == 4) || (n % 10 == 7))
      return 1;
    else
      n /= 10;
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<ol>:
  endbr64
  mov    %edi,%ecx
  test   %edi,%edi
  je     L45
  movslq %edi,%rax
  mov    %edi,%esi
  imul   $0x66666667,%rax,%rdx
  sar    $0x1f,%esi
  sar    $0x22,%rdx
  sub    %esi,%edx
  lea    (%rdx,%rdx,4),%esi
  mov    %edi,%edx
  add    %esi,%esi
  sub    %esi,%edx
  mov    $0x12000,%esi
  add    $0x9,%edx
  bt     %rdx,%rsi
  jb     L32
  mov    $0x12000,%edi
  jmp    L17
  nopl   (%rax)
L22:
  movslq %ecx,%rax
  mov    %ecx,%esi
  imul   $0x66666667,%rax,%rdx
  sar    $0x1f,%esi
  sar    $0x22,%rdx
  sub    %esi,%edx
  lea    (%rdx,%rdx,4),%esi
  mov    %ecx,%edx
  add    %esi,%esi
  sub    %esi,%edx
  add    $0x9,%edx
  bt     %rdx,%rdi
  jb     L32
L17:
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%ecx
  sar    $0x22,%rax
  sub    %ecx,%eax
  mov    %eax,%ecx
  jne    L22
L45:
  xor    %eax,%eax
  ret
  xchg   %ax,%ax
L32:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int ol(int n) {
  while (n)
    if ((n % 10 == 4) || (n % 10 == 7))
      return 1;
    else
      n /= 10;
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<allocate_buffer>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  cltq
  shl    $0x3,%eax
  mov    %eax,%edi
  call   <x_malloc@plt>
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L5
L13:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movq   $0x0,(%rax)
  addl   $0x1,-0xc(%rbp)
L5:
  mov    -0xc(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L13
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char **allocate_buffer(int lines_count) {
  char **lines = x_malloc(lines_count * sizeof(char *));
  for (int i = 0; i < lines_count; i++)
    lines[i] = NULL;
  return lines;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<allocate_buffer>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  lea    0x0(,%rdi,8),%edi
  call   <x_malloc@plt>
  test   %ebx,%ebx
  jle    L11
  mov    %rax,%rdx
  mov    %ebx,%ebx
  lea    (%rax,%rbx,8),%rcx
L1:
  movq   $0x0,(%rdx)
  add    $0x8,%rdx
  cmp    %rcx,%rdx
  jne    L1
L11:
  pop    %rbx
  ret
```

assistant: ```c
char **allocate_buffer(int lines_count) {
  char **lines = x_malloc(lines_count * sizeof(char *));
  for (int i = 0; i < lines_count; i++)
    lines[i] = NULL;
  return lines;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<allocate_buffer>:
  endbr64
  push   %rbp
  movslq %edi,%rbp
  push   %rbx
  lea    0x0(,%rbp,8),%edi
  sub    $0x8,%rsp
  call   <x_malloc@plt>
  mov    %rax,%r8
  test   %ebp,%ebp
  jle    L8
  lea    0x0(,%rbp,8),%rdx
  xor    %esi,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  mov    %rax,%r8
L8:
  add    $0x8,%rsp
  mov    %r8,%rax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
char **allocate_buffer(int lines_count) {
  char **lines = x_malloc(lines_count * sizeof(char *));
  for (int i = 0; i < lines_count; i++)
    lines[i] = NULL;
  return lines;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<allocate_buffer>:
  endbr64
  push   %rbp
  movslq %edi,%rbp
  push   %rbx
  lea    0x0(,%rbp,8),%edi
  sub    $0x8,%rsp
  call   <x_malloc@plt>
  mov    %rax,%r8
  test   %ebp,%ebp
  jle    L8
  lea    0x0(,%rbp,8),%rdx
  xor    %esi,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  mov    %rax,%r8
L8:
  add    $0x8,%rsp
  mov    %r8,%rax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
char **allocate_buffer(int lines_count) {
  char **lines = x_malloc(lines_count * sizeof(char *));
  for (int i = 0; i < lines_count; i++)
    lines[i] = NULL;
  return lines;
}
```
user: What is the c source code of the assembly code below?

```asm
<swap>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x20(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x1c(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  mov    -0x20(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  mov    %eax,(%rdx)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void swap(int a[], int m, int n) {
  int t;
  t = a[m];
  a[m] = a[n];
  a[n] = t;
  return;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<swap>:
  endbr64
  movslq %esi,%rsi
  lea    (%rdi,%rsi,4),%rcx
  mov    (%rcx),%esi
  movslq %edx,%rdx
  lea    (%rdi,%rdx,4),%rax
  mov    (%rax),%edx
  mov    %edx,(%rcx)
  mov    %esi,(%rax)
  ret
```

assistant: ```c
void swap(int a[], int m, int n) {
  int t;
  t = a[m];
  a[m] = a[n];
  a[n] = t;
  return;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<swap>:
  endbr64
  movslq %esi,%rsi
  movslq %edx,%rdx
  lea    (%rdi,%rsi,4),%rcx
  lea    (%rdi,%rdx,4),%rax
  mov    (%rcx),%esi
  mov    (%rax),%edx
  mov    %edx,(%rcx)
  mov    %esi,(%rax)
  ret
```

assistant: ```c
void swap(int a[], int m, int n) {
  int t;
  t = a[m];
  a[m] = a[n];
  a[n] = t;
  return;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<swap>:
  endbr64
  movslq %esi,%rsi
  movslq %edx,%rdx
  lea    (%rdi,%rsi,4),%rcx
  lea    (%rdi,%rdx,4),%rax
  mov    (%rcx),%esi
  mov    (%rax),%edx
  mov    %edx,(%rcx)
  mov    %esi,(%rax)
  ret
```

assistant: ```c
void swap(int a[], int m, int n) {
  int t;
  t = a[m];
  a[m] = a[n];
  a[n] = t;
  return;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<RandomFunc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x18(%rbp),%rax
  movzwl (%rax),%eax
  xor    $0x1cdb,%ax
  mov    %ax,-0xa(%rbp)
  movw   $0x0,-0xc(%rbp)
  jmp    L14
L40:
  movzwl -0xa(%rbp),%eax
  cmp    %ax,-0xc(%rbp)
  je     L3
  lea    -0xa(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  mov    %al,-0xd(%rbp)
  lea    -0xa(%rbp),%rdx
  lea    -0xa(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rdx),%edx
  mov    %dl,(%rax)
  lea    -0xa(%rbp),%rax
  movzbl -0xd(%rbp),%edx
  mov    %dl,(%rax)
  movzwl -0xc(%rbp),%eax
  cltq
  movzwl -0xa(%rbp,%rax,2),%ecx
  movzwl -0xa(%rbp),%edx
  movzwl -0xc(%rbp),%eax
  xor    %ecx,%edx
  cltq
  mov    %dx,-0xa(%rbp,%rax,2)
  jmp    L1
L3:
  movzwl -0xa(%rbp),%eax
  movzwl %ax,%edx
  movzwl -0xa(%rbp),%eax
  shr    %ax
  movzwl %ax,%eax
  and    $0x6,%eax
  or     $0x1,%eax
  mov    %eax,%ecx
  shl    %cl,%edx
  mov    %edx,%eax
  mov    %ax,-0xa(%rbp)
  movzwl -0xc(%rbp),%eax
  cltq
  movzwl -0xa(%rbp,%rax,2),%edx
  movzwl -0xa(%rbp),%eax
  or     %edx,%eax
  mov    %ax,-0xa(%rbp)
L1:
  addw   $0x2,-0xc(%rbp)
L14:
  cmpw   $0x0,-0xc(%rbp)
  je     L40
  movzwl -0xa(%rbp),%eax
  shr    $0x7,%ax
  mov    %eax,%edx
  mov    -0x20(%rbp),%rax
  mov    %dx,(%rax)
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L23
  call   <__stack_chk_fail@plt>
L23:
  leave
  ret
```

assistant: ```c
void RandomFunc(unsigned short input[1], unsigned short output[1]) {
  unsigned short state[1];
  unsigned short local1;
  char copy11;
  {
    state[0UL] = input[0UL] ^ (unsigned short)7387;
    local1 = 0UL;
    while (local1 < 1UL) {
      if (state[0UL] != local1) {
        copy11 = *((char *)(&state[0UL]) + 1);
        *((char *)(&state[0UL]) + 1) = *((char *)(&state[0UL]) + 0);
        *((char *)(&state[0UL]) + 0) = copy11;
        state[local1] ^= state[0UL];
      } else {
        state[0UL] <<= ((state[0UL] >> (unsigned short)1) & (unsigned short)7) | 1UL;
        state[0UL] = state[local1] | state[0UL];
      }
      local1 += 2UL;
    }
    output[0UL] = state[0UL] >> (unsigned short)7;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<RandomFunc>:
  endbr64
  movw   $0x0,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned short input[1], unsigned short output[1]) {
  unsigned short state[1];
  unsigned short local1;
  char copy11;
  {
    state[0UL] = input[0UL] ^ (unsigned short)7387;
    local1 = 0UL;
    while (local1 < 1UL) {
      if (state[0UL] != local1) {
        copy11 = *((char *)(&state[0UL]) + 1);
        *((char *)(&state[0UL]) + 1) = *((char *)(&state[0UL]) + 0);
        *((char *)(&state[0UL]) + 0) = copy11;
        state[local1] ^= state[0UL];
      } else {
        state[0UL] <<= ((state[0UL] >> (unsigned short)1) & (unsigned short)7) | 1UL;
        state[0UL] = state[local1] | state[0UL];
      }
      local1 += 2UL;
    }
    output[0UL] = state[0UL] >> (unsigned short)7;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<RandomFunc>:
  endbr64
  xor    %eax,%eax
  mov    %ax,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned short input[1], unsigned short output[1]) {
  unsigned short state[1];
  unsigned short local1;
  char copy11;
  {
    state[0UL] = input[0UL] ^ (unsigned short)7387;
    local1 = 0UL;
    while (local1 < 1UL) {
      if (state[0UL] != local1) {
        copy11 = *((char *)(&state[0UL]) + 1);
        *((char *)(&state[0UL]) + 1) = *((char *)(&state[0UL]) + 0);
        *((char *)(&state[0UL]) + 0) = copy11;
        state[local1] ^= state[0UL];
      } else {
        state[0UL] <<= ((state[0UL] >> (unsigned short)1) & (unsigned short)7) | 1UL;
        state[0UL] = state[local1] | state[0UL];
      }
      local1 += 2UL;
    }
    output[0UL] = state[0UL] >> (unsigned short)7;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<RandomFunc>:
  endbr64
  xor    %eax,%eax
  mov    %ax,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned short input[1], unsigned short output[1]) {
  unsigned short state[1];
  unsigned short local1;
  char copy11;
  {
    state[0UL] = input[0UL] ^ (unsigned short)7387;
    local1 = 0UL;
    while (local1 < 1UL) {
      if (state[0UL] != local1) {
        copy11 = *((char *)(&state[0UL]) + 1);
        *((char *)(&state[0UL]) + 1) = *((char *)(&state[0UL]) + 0);
        *((char *)(&state[0UL]) + 0) = copy11;
        state[local1] ^= state[0UL];
      } else {
        state[0UL] <<= ((state[0UL] >> (unsigned short)1) & (unsigned short)7) | 1UL;
        state[0UL] = state[local1] | state[0UL];
      }
      local1 += 2UL;
    }
    output[0UL] = state[0UL] >> (unsigned short)7;
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<fps_intersperse>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    %ecx,%eax
  mov    %al,-0x1c(%rbp)
  jmp    L17
L12:
  mov    -0x10(%rbp),%rdx
  lea    0x1(%rdx),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rcx
  mov    %rcx,-0x8(%rbp)
  movzbl (%rdx),%edx
  mov    %dl,(%rax)
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  movzbl -0x1c(%rbp),%edx
  mov    %dl,(%rax)
  subq   $0x1,-0x18(%rbp)
L17:
  cmpq   $0x1,-0x18(%rbp)
  ja     L12
  cmpq   $0x1,-0x18(%rbp)
  jne    L6
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%edx
  mov    -0x8(%rbp),%rax
  mov    %dl,(%rax)
L6:
  nop
  pop    %rbp
  ret
```

assistant: ```c
void fps_intersperse(unsigned char *q, unsigned char *p, unsigned long n, unsigned char c) {
  while (n > 1) {
    *q++ = *p++;
    *q++ = c;
    n--;
  }
  if (n == 1)
    *q = *p;
}
```
user: What is the c source code of the assembly code below?

```asm
<fps_intersperse>:
  endbr64
  cmp    $0x1,%rdx
  jbe    L15
  sub    $0x1,%rdx
  lea    (%rsi,%rdx,1),%r9
  mov    %rdi,%rax
L36:
  add    $0x1,%rsi
  movzbl -0x1(%rsi),%r8d
  mov    %r8b,(%rax)
  add    $0x2,%rax
  mov    %cl,-0x1(%rax)
  cmp    %r9,%rsi
  jne    L36
  lea    (%rdi,%rdx,2),%rdi
L14:
  movzbl (%r9),%eax
  mov    %al,(%rdi)
  jmp    L40
L15:
  je     L26
L40:
  ret
L26:
  mov    %rsi,%r9
  jmp    L14
```

assistant: ```c
void fps_intersperse(unsigned char *q, unsigned char *p, unsigned long n, unsigned char c) {
  while (n > 1) {
    *q++ = *p++;
    *q++ = c;
    n--;
  }
  if (n == 1)
    *q = *p;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<fps_intersperse>:
  endbr64
  cmp    $0x1,%rdx
  jbe    L5
  sub    $0x1,%rdx
  mov    %rdi,%rax
  lea    (%rsi,%rdx,1),%r9
  nopl   (%rax)
L2:
  movzbl (%rsi),%r8d
  add    $0x1,%rsi
  add    $0x2,%rax
  mov    %r8b,-0x2(%rax)
  mov    %cl,-0x1(%rax)
  cmp    %rsi,%r9
  jne    L2
  lea    (%rdi,%rdx,2),%rdi
L25:
  movzbl (%r9),%eax
  mov    %al,(%rdi)
L1:
  ret
  nopl   0x0(%rax,%rax,1)
L5:
  jne    L1
  mov    %rsi,%r9
  jmp    L25
```

assistant: ```c
void fps_intersperse(unsigned char *q, unsigned char *p, unsigned long n, unsigned char c) {
  while (n > 1) {
    *q++ = *p++;
    *q++ = c;
    n--;
  }
  if (n == 1)
    *q = *p;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<fps_intersperse>:
  endbr64
  cmp    $0x1,%rdx
  jbe    L40
  mov    %ecx,%r8d
  mov    %ecx,%r10d
  lea    -0x1(%rdx),%rcx
  push   %rbx
  lea    (%rsi,%rcx,1),%r9
  lea    (%rdi,%rcx,2),%r11
  cmp    %r9,%rdi
  setae  %al
  cmp    %r11,%rsi
  setae  %bl
  or     %bl,%al
  je     L214
  sub    $0x2,%rdx
  cmp    $0xe,%rdx
  jbe    L214
  movd   %r8d,%xmm1
  mov    %rcx,%rdx
  xor    %eax,%eax
  punpcklbw %xmm1,%xmm1
  and    $0xfffffffffffffff0,%rdx
  punpcklwd %xmm1,%xmm1
  pshufd $0x0,%xmm1,%xmm1
  xchg   %ax,%ax
L130:
  movdqu (%rsi,%rax,1),%xmm0
  movdqa %xmm0,%xmm2
  punpckhbw %xmm1,%xmm0
  punpcklbw %xmm1,%xmm2
  movups %xmm0,0x10(%rdi,%rax,2)
  movups %xmm2,(%rdi,%rax,2)
  add    $0x10,%rax
  cmp    %rdx,%rax
  jne    L130
  mov    %rcx,%r10
  and    $0xfffffffffffffff0,%r10
  lea    (%rdi,%r10,2),%rax
  add    %r10,%rsi
  cmp    %r10,%rcx
  je     L191
  movzbl (%rsi),%edx
  sub    %r10,%rcx
  mov    %r8b,0x1(%rax)
  mov    %dl,(%rax)
  mov    %rcx,%rdx
  cmp    $0x1,%rcx
  je     L191
  movzbl 0x1(%rsi),%ecx
  mov    %r8b,0x3(%rax)
  mov    %cl,0x2(%rax)
  cmp    $0x2,%rdx
  je     L191
  movzbl 0x2(%rsi),%ecx
  mov    %r8b,0x5(%rax)
  mov    %cl,0x4(%rax)
  cmp    $0x3,%rdx
  je     L191
  movzbl 0x3(%rsi),%ecx
  mov    %r8b,0x7(%rax)
  mov    %cl,0x6(%rax)
  cmp    $0x4,%rdx
  je     L191
  movzbl 0x4(%rsi),%ecx
  mov    %r8b,0x9(%rax)
  mov    %cl,0x8(%rax)
  cmp    $0x5,%rdx
  je     L191
  movzbl 0x5(%rsi),%ecx
  mov    %r8b,0xb(%rax)
  mov    %cl,0xa(%rax)
  cmp    $0x6,%rdx
  je     L191
  movzbl 0x6(%rsi),%ecx
  mov    %r8b,0xd(%rax)
  mov    %cl,0xc(%rax)
  cmp    $0x7,%rdx
  je     L191
  movzbl 0x7(%rsi),%ecx
  mov    %r8b,0xf(%rax)
  mov    %cl,0xe(%rax)
  cmp    $0x8,%rdx
  je     L191
  movzbl 0x8(%rsi),%ecx
  mov    %r8b,0x11(%rax)
  mov    %cl,0x10(%rax)
  cmp    $0x9,%rdx
  je     L191
  movzbl 0x9(%rsi),%ecx
  mov    %r8b,0x13(%rax)
  mov    %cl,0x12(%rax)
  cmp    $0xa,%rdx
  je     L191
  movzbl 0xa(%rsi),%ecx
  mov    %r8b,0x15(%rax)
  mov    %cl,0x14(%rax)
  cmp    $0xb,%rdx
  je     L191
  movzbl 0xb(%rsi),%ecx
  mov    %r8b,0x17(%rax)
  mov    %cl,0x16(%rax)
  cmp    $0xc,%rdx
  je     L191
  movzbl 0xc(%rsi),%ecx
  mov    %r8b,0x19(%rax)
  mov    %cl,0x18(%rax)
  cmp    $0xd,%rdx
  je     L191
  movzbl 0xd(%rsi),%ecx
  mov    %r8b,0x1b(%rax)
  mov    %cl,0x1a(%rax)
  cmp    $0xe,%rdx
  je     L191
  movzbl 0xe(%rsi),%edx
  mov    %r8b,0x1d(%rax)
  mov    %dl,0x1c(%rax)
L191:
  movzbl (%r9),%eax
  mov    %al,(%r11)
  pop    %rbx
  ret
  nopl   0x0(%rax)
L214:
  movzbl (%rsi),%eax
  add    $0x1,%rsi
  add    $0x2,%rdi
  mov    %al,-0x2(%rdi)
  mov    %r10b,-0x1(%rdi)
  cmp    %rsi,%r9
  jne    L214
  jmp    L191
  nopl   0x0(%rax)
L40:
  je     L113
  ret
L113:
  movzbl (%rsi),%eax
  mov    %al,(%rdi)
  ret
```

assistant: ```c
void fps_intersperse(unsigned char *q, unsigned char *p, unsigned long n, unsigned char c) {
  while (n > 1) {
    *q++ = *p++;
    *q++ = c;
    n--;
  }
  if (n == 1)
    *q = *p;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<backup_senser_cmd_ID1>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,%eax
  mov    %rsi,-0x20(%rbp)
  mov    %al,-0x14(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0x1,-0x10(%rbp)
  movsbl -0x14(%rbp),%ecx
  lea    -0x10(%rbp),%rdx
  mov    -0x20(%rbp),%rax
  mov    %ecx,%r9d
  mov    $0x1,%r8d
  mov    %rdx,%rcx
  mov    %rax,%rdx
  mov    $0x0,%esi
  mov    $0xf,%edi
  call   <backup_senser_sync_msg@plt>
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L1
  call   <__stack_chk_fail@plt>
L1:
  leave
  ret
```

assistant: ```c
int backup_senser_cmd_ID1(char set_value, char *get_value) {
  size_t len = sizeof(char);
  return backup_senser_sync_msg(15, 0, get_value, &len, 1, set_value);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<backup_senser_cmd_ID1>:
  endbr64
  sub    $0x18,%rsp
  mov    %rsi,%rdx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movq   $0x1,(%rsp)
  mov    %rsp,%rcx
  movsbl %dil,%r9d
  mov    $0x1,%r8d
  mov    $0x0,%esi
  mov    $0xf,%edi
  call   <backup_senser_sync_msg@plt>
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L3
  add    $0x18,%rsp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int backup_senser_cmd_ID1(char set_value, char *get_value) {
  size_t len = sizeof(char);
  return backup_senser_sync_msg(15, 0, get_value, &len, 1, set_value);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<backup_senser_cmd_ID1>:
  endbr64
  sub    $0x18,%rsp
  mov    %rsi,%rdx
  movsbl %dil,%r9d
  xor    %esi,%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rcx
  mov    $0x1,%r8d
  mov    $0xf,%edi
  movq   $0x1,(%rsp)
  call   <backup_senser_sync_msg@plt>
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L8
  add    $0x18,%rsp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int backup_senser_cmd_ID1(char set_value, char *get_value) {
  size_t len = sizeof(char);
  return backup_senser_sync_msg(15, 0, get_value, &len, 1, set_value);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<backup_senser_cmd_ID1>:
  endbr64
  sub    $0x18,%rsp
  mov    %rsi,%rdx
  movsbl %dil,%r9d
  xor    %esi,%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rcx
  mov    $0x1,%r8d
  mov    $0xf,%edi
  movq   $0x1,(%rsp)
  call   <backup_senser_sync_msg@plt>
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L3
  add    $0x18,%rsp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int backup_senser_cmd_ID1(char set_value, char *get_value) {
  size_t len = sizeof(char);
  return backup_senser_sync_msg(15, 0, get_value, &len, 1, set_value);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<bingham_F_1d>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  movsd  %xmm0,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  movq   %rax,%xmm0
  call   <fabs@plt>
  mov    %eax,%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  cmp    $0xa,%eax
  jle    L17
  mov    -0x18(%rbp),%rax
  movq   %rax,%xmm0
  call   <fabs@plt>
  mov    %eax,%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  jmp    L12
L17:
  mov    $0xa,%eax
L12:
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%edx
  mov    -0x18(%rbp),%rax
  mov    %edx,%esi
  movq   %rax,%xmm0
  mov    $0x1,%edi
  call   <compute_1F1_1d@plt>
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  leave
  ret
```

assistant: ```c
double bingham_F_1d(double z) {
  int iter = (((int)fabs(z) * 10) > (10) ? ((int)fabs(z) * 10) : (10));
  return compute_1F1_1d(1, z, iter);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<bingham_F_1d>:
  endbr64
  push   %rbx
  movq   %xmm0,%rbx
  call   <fabs@plt>
  lea    (%rax,%rax,4),%eax
  add    %eax,%eax
  mov    $0xa,%esi
  cmp    $0xa,%eax
  jg     L1
L2:
  movq   %rbx,%xmm0
  mov    $0x1,%edi
  call   <compute_1F1_1d@plt>
  pop    %rbx
  ret
L1:
  movq   %rbx,%xmm0
  call   <fabs@plt>
  lea    (%rax,%rax,4),%esi
  add    %esi,%esi
  jmp    L2
```

assistant: ```c
double bingham_F_1d(double z) {
  int iter = (((int)fabs(z) * 10) > (10) ? ((int)fabs(z) * 10) : (10));
  return compute_1F1_1d(1, z, iter);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<bingham_F_1d>:
  endbr64
  sub    $0x18,%rsp
  movsd  %xmm0,0x8(%rsp)
  call   <fabs@plt>
  movsd  0x8(%rsp),%xmm0
  mov    $0xa,%esi
  lea    (%rax,%rax,4),%eax
  add    %eax,%eax
  cmp    $0xa,%eax
  jg     L8
  mov    $0x1,%edi
  add    $0x18,%rsp
  jmp    1070 <compute_1F1_1d@plt>
  cs nopw 0x0(%rax,%rax,1)
L8:
  call   <fabs@plt>
  movsd  0x8(%rsp),%xmm0
  mov    $0x1,%edi
  add    $0x18,%rsp
  lea    (%rax,%rax,4),%esi
  add    %esi,%esi
  jmp    1070 <compute_1F1_1d@plt>
```

assistant: ```c
double bingham_F_1d(double z) {
  int iter = (((int)fabs(z) * 10) > (10) ? ((int)fabs(z) * 10) : (10));
  return compute_1F1_1d(1, z, iter);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<bingham_F_1d>:
  endbr64
  sub    $0x18,%rsp
  movsd  %xmm0,0x8(%rsp)
  call   <fabs@plt>
  movsd  0x8(%rsp),%xmm0
  mov    $0xa,%esi
  lea    (%rax,%rax,4),%eax
  add    %eax,%eax
  cmp    $0xa,%eax
  jg     L9
  mov    $0x1,%edi
  add    $0x18,%rsp
  jmp    1070 <compute_1F1_1d@plt>
  cs nopw 0x0(%rax,%rax,1)
L9:
  call   <fabs@plt>
  movsd  0x8(%rsp),%xmm0
  mov    $0x1,%edi
  add    $0x18,%rsp
  lea    (%rax,%rax,4),%esi
  add    %esi,%esi
  jmp    1070 <compute_1F1_1d@plt>
```

assistant: ```c
double bingham_F_1d(double z) {
  int iter = (((int)fabs(z) * 10) > (10) ? ((int)fabs(z) * 10) : (10));
  return compute_1F1_1d(1, z, iter);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<memset>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L17
L18:
  mov    -0xc(%rbp),%edx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    -0x1c(%rbp),%edx
  mov    %dl,(%rax)
  addl   $0x1,-0xc(%rbp)
L17:
  mov    -0xc(%rbp),%eax
  cmp    -0x20(%rbp),%eax
  jb     L18
  mov    -0x18(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
void *memset(void *ptr, int c, unsigned int len) {
  unsigned int i;
  char *p;
  p = ptr;
  for (i = 0; i < len; i++)
    p[i] = c;
  return ptr;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<memset>:
  endbr64
  mov    %rdi,%rax
  test   %edx,%edx
  je     L6
  mov    %rdi,%rcx
  mov    %edx,%edx
  add    %rdi,%rdx
L13:
  mov    %sil,(%rcx)
  add    $0x1,%rcx
  cmp    %rdx,%rcx
  jne    L13
L6:
  ret
```

assistant: ```c
void *memset(void *ptr, int c, unsigned int len) {
  unsigned int i;
  char *p;
  p = ptr;
  for (i = 0; i < len; i++)
    p[i] = c;
  return ptr;
}
```