ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Demonstrate the C code that generates the assembly output below:

```asm
<part2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %esi,-0x28(%rbp)
  mov    -0x24(%rbp),%eax
  mov    %eax,-0x20(%rbp)
  mov    -0x20(%rbp),%eax
  mov    %eax,-0x1c(%rbp)
  mov    -0x28(%rbp),%eax
  mov    %eax,-0x18(%rbp)
  mov    -0x18(%rbp),%eax
  mov    %eax,-0x14(%rbp)
  addl   $0x1,-0x1c(%rbp)
  mov    -0x1c(%rbp),%eax
  lea    (%rax,%rax,1),%ecx
  addl   $0x1,-0x14(%rbp)
  mov    -0x14(%rbp),%edx
  mov    %edx,%eax
  add    %eax,%eax
  add    %edx,%eax
  lea    (%rcx,%rax,1),%edx
  addl   $0x1,-0x1c(%rbp)
  addl   $0x1,-0x14(%rbp)
  mov    -0x1c(%rbp),%eax
  imul   -0x14(%rbp),%eax
  shl    $0x2,%eax
  lea    (%rdx,%rax,1),%ecx
  addl   $0x1,-0x1c(%rbp)
  addl   $0x1,-0x14(%rbp)
  mov    -0x1c(%rbp),%eax
  imul   -0x14(%rbp),%eax
  mov    %eax,%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  neg    %eax
  lea    (%rcx,%rax,1),%ebx
  addl   $0x1,-0x18(%rbp)
  addl   $0x1,-0x20(%rbp)
  mov    -0x18(%rbp),%edx
  mov    -0x20(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <function_1@plt>
  mov    %ebx,%edx
  mov    %eax,%esi
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D25(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  nop
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

tool_calls: name: parse_data
arguments: tag: D1
data_type: string
name: parse_data
arguments: tag: D25
data_type: string
name: parse_data
arguments: tag: D10
data_type: string
name: parse_data
arguments: tag: D2
data_type: string

tool: D1: function_1(x, y) = %d
macro_2(x, y) = %d



tool: D25: In VS : the result of function_1(x, y) = -2 and macro_2(x, y) = -18 

tool: D10: In GCC: the result of function_1(x, y) = -2 and macro_2(x, y) = -84 


tool: D2: Explanation: One explination for the variation could be attributed to how the macro equations expand (order of evaluation & value production). The order of evaluation is completely compiler dependent so in this case the way GCC goes through the macro could be different from VS. Adding auto keywords such as ++ could be buggy when moving from compiler to compiler. 


assistant: ```c
void part2(int x, int y) {
  int i, j, s, t;
  s = i = x;
  t = j = y;
  printf("function_1(x, y) = %d\nmacro_2(x, y) = %d\n\n", function_1(++i, ++j), (2 * ++s + 3 * ++t + 4 * ++s * ++t - ++s * ++t * 5));
  printf("In VS : the result of function_1(x, y) = -2 and macro_2(x, y) = -18 \n");
  printf("In GCC: the result of function_1(x, y) = -2 and macro_2(x, y) = -84 \n\n");
  printf("Explanation: One explination for the variation could be attributed to how the macro equations expand (order of evaluation & value production). The order of evaluation is completely compiler dependent so in this case the way GCC goes through the macro could be different from VS. Adding auto keywords such as ++ could be buggy when moving from compiler to compiler. \n\n");
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<part2>:
  endbr64
  push   %rbx
  lea    0x1(%rdi),%ecx
  lea    0x1(%rsi),%eax
  lea    0x2(%rdi),%edx
  lea    0x2(%rsi),%r8d
  imul   %r8d,%edx
  lea    (%rax,%rax,2),%r8d
  lea    (%r8,%rcx,2),%r8d
  lea    (%r8,%rdx,4),%ebx
  add    $0x3,%edi
  add    $0x3,%esi
  imul   %esi,%edi
  lea    (%rdi,%rdi,4),%edx
  sub    %edx,%ebx
  mov    %eax,%esi
  mov    %ecx,%edi
  call   <function_1@plt>
  mov    %eax,%esi
  mov    %ebx,%edx
  lea    D21(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D14(%rip),%rdi
  call   <puts@plt>
  lea    D38(%rip),%rdi
  call   <puts@plt>
  lea    D34(%rip),%rdi
  call   <puts@plt>
  pop    %rbx
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D21
datatype: string
name: disassemble_data
arguments: tag: D14
datatype: string
name: disassemble_data
arguments: tag: D38
datatype: string
name: disassemble_data
arguments: tag: D34
datatype: string

tool: D21: function_1(x, y) = %d
macro_2(x, y) = %d



tool: D14: In VS : the result of function_1(x, y) = -2 and macro_2(x, y) = -18 

tool: D38: In GCC: the result of function_1(x, y) = -2 and macro_2(x, y) = -84 


tool: D34: Explanation: One explination for the variation could be attributed to how the macro equations expand (order of evaluation & value production). The order of evaluation is completely compiler dependent so in this case the way GCC goes through the macro could be different from VS. Adding auto keywords such as ++ could be buggy when moving from compiler to compiler. 


assistant: ```c
void part2(int x, int y) {
  int i, j, s, t;
  s = i = x;
  t = j = y;
  printf("function_1(x, y) = %d\nmacro_2(x, y) = %d\n\n", function_1(++i, ++j), (2 * ++s + 3 * ++t + 4 * ++s * ++t - ++s * ++t * 5));
  printf("In VS : the result of function_1(x, y) = -2 and macro_2(x, y) = -18 \n");
  printf("In GCC: the result of function_1(x, y) = -2 and macro_2(x, y) = -84 \n\n");
  printf("Explanation: One explination for the variation could be attributed to how the macro equations expand (order of evaluation & value production). The order of evaluation is completely compiler dependent so in this case the way GCC goes through the macro could be different from VS. Adding auto keywords such as ++ could be buggy when moving from compiler to compiler. \n\n");
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<part2>:
  endbr64
  lea    0x1(%rdi),%r9d
  lea    0x1(%rsi),%r8d
  push   %r12
  lea    0x2(%rsi),%edx
  lea    0x2(%rdi),%eax
  add    $0x3,%esi
  add    $0x3,%edi
  imul   %edx,%eax
  lea    (%r8,%r8,2),%edx
  imul   %esi,%edi
  lea    (%rdx,%r9,2),%edx
  mov    %r8d,%esi
  lea    (%rdx,%rax,4),%r12d
  lea    (%rdi,%rdi,4),%eax
  mov    %r9d,%edi
  sub    %eax,%r12d
  call   <function_1@plt>
  mov    %r12d,%edx
  lea    D21(%rip),%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  lea    D27(%rip),%rdi
  call   <puts@plt>
  lea    D16(%rip),%rdi
  call   <puts@plt>
  lea    D11(%rip),%rdi
  pop    %r12
  jmp    1070 <puts@plt>
```

tool_calls: name: parse_data
arguments: data_label: D21
type: string
name: parse_data
arguments: data_label: D27
type: string
name: parse_data
arguments: data_label: D16
type: string
name: parse_data
arguments: data_label: D11
type: string

tool: D21: function_1(x, y) = %d
macro_2(x, y) = %d



tool: D27: In VS : the result of function_1(x, y) = -2 and macro_2(x, y) = -18 

tool: D16: In GCC: the result of function_1(x, y) = -2 and macro_2(x, y) = -84 


tool: D11: Explanation: One explination for the variation could be attributed to how the macro equations expand (order of evaluation & value production). The order of evaluation is completely compiler dependent so in this case the way GCC goes through the macro could be different from VS. Adding auto keywords such as ++ could be buggy when moving from compiler to compiler. 


assistant: ```c
void part2(int x, int y) {
  int i, j, s, t;
  s = i = x;
  t = j = y;
  printf("function_1(x, y) = %d\nmacro_2(x, y) = %d\n\n", function_1(++i, ++j), (2 * ++s + 3 * ++t + 4 * ++s * ++t - ++s * ++t * 5));
  printf("In VS : the result of function_1(x, y) = -2 and macro_2(x, y) = -18 \n");
  printf("In GCC: the result of function_1(x, y) = -2 and macro_2(x, y) = -84 \n\n");
  printf("Explanation: One explination for the variation could be attributed to how the macro equations expand (order of evaluation & value production). The order of evaluation is completely compiler dependent so in this case the way GCC goes through the macro could be different from VS. Adding auto keywords such as ++ could be buggy when moving from compiler to compiler. \n\n");
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<part2>:
  endbr64
  lea    0x1(%rdi),%r9d
  lea    0x1(%rsi),%r8d
  push   %r12
  lea    0x2(%rsi),%edx
  lea    0x2(%rdi),%eax
  add    $0x3,%esi
  add    $0x3,%edi
  imul   %edx,%eax
  lea    (%r8,%r8,2),%edx
  imul   %esi,%edi
  lea    (%rdx,%r9,2),%edx
  mov    %r8d,%esi
  lea    (%rdx,%rax,4),%r12d
  lea    (%rdi,%rdi,4),%eax
  mov    %r9d,%edi
  sub    %eax,%r12d
  call   <function_1@plt>
  mov    %r12d,%edx
  lea    D29(%rip),%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  lea    D39(%rip),%rdi
  call   <puts@plt>
  lea    D3(%rip),%rdi
  call   <puts@plt>
  lea    D8(%rip),%rdi
  pop    %r12
  jmp    1070 <puts@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D29
type_guessed: string
name: parse_data
arguments: data_tag: D39
type_guessed: string
name: parse_data
arguments: data_tag: D3
type_guessed: string
name: parse_data
arguments: data_tag: D8
type_guessed: string

tool: D29: function_1(x, y) = %d
macro_2(x, y) = %d



tool: D39: In VS : the result of function_1(x, y) = -2 and macro_2(x, y) = -18 

tool: D3: In GCC: the result of function_1(x, y) = -2 and macro_2(x, y) = -84 


tool: D8: Explanation: One explination for the variation could be attributed to how the macro equations expand (order of evaluation & value production). The order of evaluation is completely compiler dependent so in this case the way GCC goes through the macro could be different from VS. Adding auto keywords such as ++ could be buggy when moving from compiler to compiler. 


assistant: ```c
void part2(int x, int y) {
  int i, j, s, t;
  s = i = x;
  t = j = y;
  printf("function_1(x, y) = %d\nmacro_2(x, y) = %d\n\n", function_1(++i, ++j), (2 * ++s + 3 * ++t + 4 * ++s * ++t - ++s * ++t * 5));
  printf("In VS : the result of function_1(x, y) = -2 and macro_2(x, y) = -18 \n");
  printf("In GCC: the result of function_1(x, y) = -2 and macro_2(x, y) = -84 \n\n");
  printf("Explanation: One explination for the variation could be attributed to how the macro equations expand (order of evaluation & value production). The order of evaluation is completely compiler dependent so in this case the way GCC goes through the macro could be different from VS. Adding auto keywords such as ++ could be buggy when moving from compiler to compiler. \n\n");
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<stdlib_strided_dsnannsumors>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    %rcx,-0x40(%rbp)
  pxor   %xmm0,%xmm0
  movsd  %xmm0,-0x18(%rbp)
  mov    -0x40(%rbp),%rax
  movq   $0x0,(%rax)
  cmpq   $0x0,-0x28(%rbp)
  jg     L101
  movsd  -0x18(%rbp),%xmm0
  jmp    L102
L101:
  cmpq   $0x1,-0x28(%rbp)
  je     L99
  cmpq   $0x0,-0x38(%rbp)
  jne    L18
L99:
  mov    -0x30(%rbp),%rax
  mov    (%rax),%eax
  movd   %eax,%xmm0
  call   <stdlib_base_is_nanf@plt>
  test   %rax,%rax
  je     L24
  movsd  -0x18(%rbp),%xmm0
  jmp    L102
L24:
  mov    -0x40(%rbp),%rax
  mov    (%rax),%rax
  lea    0x1(%rax),%rdx
  mov    -0x40(%rbp),%rax
  mov    %rdx,(%rax)
  mov    -0x30(%rbp),%rax
  movss  (%rax),%xmm0
  cvtss2sd %xmm0,%xmm0
  jmp    L102
L18:
  cmpq   $0x0,-0x38(%rbp)
  jns    L100
  mov    $0x1,%eax
  sub    -0x28(%rbp),%rax
  mov    -0x38(%rbp),%rdx
  imul   %rdx,%rax
  mov    %rax,-0x10(%rbp)
  jmp    L85
L100:
  movq   $0x0,-0x10(%rbp)
L85:
  movq   $0x0,-0x8(%rbp)
  jmp    L90
L104:
  mov    -0x10(%rbp),%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  movd   %eax,%xmm0
  call   <stdlib_base_is_nanf@plt>
  test   %rax,%rax
  jne    L38
  mov    -0x10(%rbp),%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  movss  (%rax),%xmm0
  cvtss2sd %xmm0,%xmm0
  movsd  -0x18(%rbp),%xmm1
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x18(%rbp)
  mov    -0x40(%rbp),%rax
  mov    (%rax),%rax
  lea    0x1(%rax),%rdx
  mov    -0x40(%rbp),%rax
  mov    %rdx,(%rax)
L38:
  mov    -0x38(%rbp),%rax
  add    %rax,-0x10(%rbp)
  addq   $0x1,-0x8(%rbp)
L90:
  mov    -0x8(%rbp),%rax
  cmp    -0x28(%rbp),%rax
  jl     L104
  movsd  -0x18(%rbp),%xmm0
L102:
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  leave
  ret
```

assistant: ```c
double stdlib_strided_dsnannsumors(const int64_t N, const float *X, const int64_t stride, int64_t *n) {
  double sum;
  int64_t ix;
  int64_t i;
  sum = 0.0;
  *n = 0;
  if (N <= 0) {
    return sum;
  }
  if (N == 1 || stride == 0) {
    if (stdlib_base_is_nanf(X[0])) {
      return sum;
    }
    *n += 1;
    return X[0];
  }
  if (stride < 0) {
    ix = (1 - N) * stride;
  } else {
    ix = 0;
  }
  for (i = 0; i < N; i++) {
    if (!stdlib_base_is_nanf(X[ix])) {
      sum += (double)X[ix];
      *n += 1;
    }
    ix += stride;
  }
  return sum;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<stdlib_strided_dsnannsumors>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  movq   $0x0,(%rcx)
  test   %rdi,%rdi
  jle    L15
  mov    %rdi,%r13
  mov    %rsi,%rbx
  mov    %rcx,%r14
  cmp    $0x1,%rdi
  je     L6
  test   %rdx,%rdx
  je     L6
  mov    $0x0,%eax
  js     L92
L23:
  lea    0x0(,%rdx,4),%r12
  lea    (%rbx,%rax,4),%rbx
  mov    $0x0,%ebp
  mov    D4(%rip),%r15
  jmp    L17
L6:
  movss  (%rbx),%xmm0
  call   <stdlib_base_is_nanf@plt>
  mov    D4(%rip),%r15
  test   %rax,%rax
  jne    L78
  addq   $0x1,(%r14)
  pxor   %xmm2,%xmm2
  cvtss2sd (%rbx),%xmm2
  movq   %xmm2,%r15
  jmp    L78
L92:
  mov    $0x1,%eax
  sub    %rdi,%rax
  imul   %rdx,%rax
  jmp    L23
L95:
  add    $0x1,%rbp
  add    %r12,%rbx
  cmp    %rbp,%r13
  je     L78
L17:
  movss  (%rbx),%xmm0
  call   <stdlib_base_is_nanf@plt>
  test   %rax,%rax
  jne    L95
  pxor   %xmm0,%xmm0
  cvtss2sd (%rbx),%xmm0
  movq   %r15,%xmm1
  addsd  %xmm0,%xmm1
  movq   %xmm1,%r15
  addq   $0x1,(%r14)
  jmp    L95
L15:
  mov    D4(%rip),%r15
L78:
  movq   %r15,%xmm0
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D4
type: f64

tool: D4: 0

assistant: ```c
double stdlib_strided_dsnannsumors(const int64_t N, const float *X, const int64_t stride, int64_t *n) {
  double sum;
  int64_t ix;
  int64_t i;
  sum = 0.0;
  *n = 0;
  if (N <= 0) {
    return sum;
  }
  if (N == 1 || stride == 0) {
    if (stdlib_base_is_nanf(X[0])) {
      return sum;
    }
    *n += 1;
    return X[0];
  }
  if (stride < 0) {
    ix = (1 - N) * stride;
  } else {
    ix = 0;
  }
  for (i = 0; i < N; i++) {
    if (!stdlib_base_is_nanf(X[ix])) {
      sum += (double)X[ix];
      *n += 1;
    }
    ix += stride;
  }
  return sum;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<stdlib_strided_dsnannsumors>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  movq   $0x0,(%rcx)
  test   %rdi,%rdi
  jle    L47
  mov    %rdi,%r13
  mov    %rsi,%rbx
  mov    %rcx,%r14
  cmp    $0x1,%rdi
  je     L22
  test   %rdx,%rdx
  je     L22
  mov    $0x0,%eax
  jns    L76
  mov    $0x1,%eax
  sub    %rdi,%rax
  imul   %rdx,%rax
L76:
  pxor   %xmm2,%xmm2
  lea    0x0(,%rdx,4),%r12
  lea    (%rbx,%rax,4),%rbx
  xor    %ebp,%ebp
  movsd  %xmm2,0x8(%rsp)
  nop
L7:
  movss  (%rbx),%xmm0
  call   <stdlib_base_is_nanf@plt>
  test   %rax,%rax
  jne    L38
  pxor   %xmm0,%xmm0
  cvtss2sd (%rbx),%xmm0
  addq   $0x1,(%r14)
  addsd  0x8(%rsp),%xmm0
  movsd  %xmm0,0x8(%rsp)
L38:
  add    $0x1,%rbp
  add    %r12,%rbx
  cmp    %rbp,%r13
  jne    L7
L39:
  movsd  0x8(%rsp),%xmm0
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   (%rax)
L22:
  movss  (%rbx),%xmm0
  call   <stdlib_base_is_nanf@plt>
  test   %rax,%rax
  je     L8
L47:
  pxor   %xmm3,%xmm3
  movsd  %xmm3,0x8(%rsp)
  jmp    L39
  nopw   0x0(%rax,%rax,1)
L8:
  pxor   %xmm4,%xmm4
  addq   $0x1,(%r14)
  cvtss2sd (%rbx),%xmm4
  movsd  %xmm4,0x8(%rsp)
  jmp    L39
```

assistant: ```c
double stdlib_strided_dsnannsumors(const int64_t N, const float *X, const int64_t stride, int64_t *n) {
  double sum;
  int64_t ix;
  int64_t i;
  sum = 0.0;
  *n = 0;
  if (N <= 0) {
    return sum;
  }
  if (N == 1 || stride == 0) {
    if (stdlib_base_is_nanf(X[0])) {
      return sum;
    }
    *n += 1;
    return X[0];
  }
  if (stride < 0) {
    ix = (1 - N) * stride;
  } else {
    ix = 0;
  }
  for (i = 0; i < N; i++) {
    if (!stdlib_base_is_nanf(X[ix])) {
      sum += (double)X[ix];
      *n += 1;
    }
    ix += stride;
  }
  return sum;
}
```
user: Translate the below assembly code into C programming language:

```asm
<stdlib_strided_dsnannsumors>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  movq   $0x0,(%rcx)
  test   %rdi,%rdi
  jle    L52
  mov    %rdi,%r13
  mov    %rsi,%rbx
  mov    %rcx,%r14
  cmp    $0x1,%rdi
  je     L5
  test   %rdx,%rdx
  je     L5
  mov    $0x0,%eax
  jns    L80
  mov    $0x1,%eax
  sub    %rdi,%rax
  imul   %rdx,%rax
L80:
  pxor   %xmm2,%xmm2
  lea    0x0(,%rdx,4),%r12
  lea    (%rbx,%rax,4),%rbx
  xor    %ebp,%ebp
  movsd  %xmm2,0x8(%rsp)
  nop
L16:
  movss  (%rbx),%xmm0
  call   <stdlib_base_is_nanf@plt>
  test   %rax,%rax
  jne    L70
  pxor   %xmm0,%xmm0
  cvtss2sd (%rbx),%xmm0
  addq   $0x1,(%r14)
  addsd  0x8(%rsp),%xmm0
  movsd  %xmm0,0x8(%rsp)
L70:
  add    $0x1,%rbp
  add    %r12,%rbx
  cmp    %rbp,%r13
  jne    L16
L21:
  movsd  0x8(%rsp),%xmm0
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   (%rax)
L5:
  movss  (%rbx),%xmm0
  call   <stdlib_base_is_nanf@plt>
  test   %rax,%rax
  je     L25
L52:
  pxor   %xmm3,%xmm3
  movsd  %xmm3,0x8(%rsp)
  jmp    L21
  nopw   0x0(%rax,%rax,1)
L25:
  pxor   %xmm4,%xmm4
  addq   $0x1,(%r14)
  cvtss2sd (%rbx),%xmm4
  movsd  %xmm4,0x8(%rsp)
  jmp    L21
```

assistant: ```c
double stdlib_strided_dsnannsumors(const int64_t N, const float *X, const int64_t stride, int64_t *n) {
  double sum;
  int64_t ix;
  int64_t i;
  sum = 0.0;
  *n = 0;
  if (N <= 0) {
    return sum;
  }
  if (N == 1 || stride == 0) {
    if (stdlib_base_is_nanf(X[0])) {
      return sum;
    }
    *n += 1;
    return X[0];
  }
  if (stride < 0) {
    ix = (1 - N) * stride;
  } else {
    ix = 0;
  }
  for (i = 0; i < N; i++) {
    if (!stdlib_base_is_nanf(X[ix])) {
      sum += (double)X[ix];
      *n += 1;
    }
    ix += stride;
  }
  return sum;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<create_address>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,%edi
  call   <inet_ntoa@plt>
  mov    %eax,%edi
  call   <strdup@plt>
  mov    %rax,-0x10(%rbp)
  cmpq   $0x0,-0x10(%rbp)
  jne    L18
  mov    $0x0,%eax
  jmp    L11
L18:
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <node_create@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L28
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
L28:
  mov    -0x8(%rbp),%rax
L11:
  leave
  ret
```

assistant: ```c
struct node *create_address(const struct in_addr address) {
  char *const ip = strdup(inet_ntoa(address));
  if (ip == NULL) {
    return NULL;
  }
  struct node *const node = node_create(ip);
  if (node == NULL) {
    free(ip);
  }
  return node;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<create_address>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <inet_ntoa@plt>
  mov    %eax,%edi
  call   <strdup@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L29
  mov    %rax,%rdi
  call   <node_create@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  je     L15
L32:
  mov    %rbp,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L15:
  mov    %rbx,%rdi
  call   <free@plt>
  jmp    L32
L29:
  mov    %rax,%rbp
  jmp    L32
```

assistant: ```c
struct node *create_address(const struct in_addr address) {
  char *const ip = strdup(inet_ntoa(address));
  if (ip == NULL) {
    return NULL;
  }
  struct node *const node = node_create(ip);
  if (node == NULL) {
    free(ip);
  }
  return node;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<create_address>:
  endbr64
  push   %r12
  push   %rbp
  sub    $0x8,%rsp
  call   <inet_ntoa@plt>
  mov    %eax,%edi
  call   <strdup@plt>
  test   %rax,%rax
  je     L2
  mov    %rax,%rdi
  mov    %rax,%rbp
  call   <node_create@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L10
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L2:
  xor    %r12d,%r12d
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
  xchg   %ax,%ax
L10:
  mov    %rbp,%rdi
  call   <free@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
struct node *create_address(const struct in_addr address) {
  char *const ip = strdup(inet_ntoa(address));
  if (ip == NULL) {
    return NULL;
  }
  struct node *const node = node_create(ip);
  if (node == NULL) {
    free(ip);
  }
  return node;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<create_address>:
  endbr64
  push   %r12
  push   %rbp
  sub    $0x8,%rsp
  call   <inet_ntoa@plt>
  mov    %eax,%edi
  call   <strdup@plt>
  test   %rax,%rax
  je     L12
  mov    %rax,%rdi
  mov    %rax,%rbp
  call   <node_create@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L2
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L12:
  xor    %r12d,%r12d
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
  xchg   %ax,%ax
L2:
  mov    %rbp,%rdi
  call   <free@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
struct node *create_address(const struct in_addr address) {
  char *const ip = strdup(inet_ntoa(address));
  if (ip == NULL) {
    return NULL;
  }
  struct node *const node = node_create(ip);
  if (node == NULL) {
    free(ip);
  }
  return node;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<freeMemoryBlock>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <free@plt>
  nop
  leave
  ret
```

assistant: ```c
void freeMemoryBlock(void **memory_pointer, int memory_block_index) {
  free(memory_pointer[memory_block_index]);
  return;
}
```
user: Translate the below assembly code into C programming language:

```asm
<freeMemoryBlock>:
  endbr64
  sub    $0x8,%rsp
  movslq %esi,%rsi
  mov    (%rdi,%rsi,8),%rdi
  call   <free@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void freeMemoryBlock(void **memory_pointer, int memory_block_index) {
  free(memory_pointer[memory_block_index]);
  return;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<freeMemoryBlock>:
  endbr64
  movslq %esi,%rsi
  mov    (%rdi,%rsi,8),%rdi
  jmp    1050 <free@plt>
```

assistant: ```c
void freeMemoryBlock(void **memory_pointer, int memory_block_index) {
  free(memory_pointer[memory_block_index]);
  return;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<freeMemoryBlock>:
  endbr64
  movslq %esi,%rsi
  mov    (%rdi,%rsi,8),%rdi
  jmp    1050 <free@plt>
```

assistant: ```c
void freeMemoryBlock(void **memory_pointer, int memory_block_index) {
  free(memory_pointer[memory_block_index]);
  return;
}
```