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

```asm
<atoi>:
  endbr64
  mov    $0x1,%eax
  cmpb   $0x2d,(%rdi)
  je     L31
L29:
  movzbl (%rdi),%edx
  mov    $0x0,%r8d
  test   %dl,%dl
  je     L34
L8:
  add    $0x1,%rdi
  movsbl %dl,%edx
  lea    -0x30(%rdx),%ecx
  sub    $0x37,%edx
  cmp    $0x9,%ecx
  cmovg  %edx,%ecx
  imul   %esi,%r8d
  add    %ecx,%r8d
  movzbl (%rdi),%edx
  test   %dl,%dl
  jne    L8
L34:
  imul   %r8d,%eax
  ret
L31:
  add    $0x1,%rdi
  mov    $0xffffffff,%eax
  jmp    L29
```

assistant: ```c
int atoi(char *s, int index) {
  int ret = 0;
  int sign = 1;
  if (*s == '-') {
    sign = -1;
    s++;
  }
  while (*s) {
    int r = *(s++);
    r -= '0';
    if (r > 9)
      r -= 7;
    ret = (ret * index) + r;
  }
  return sign * ret;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<atoi>:
  endbr64
  movsbl (%rdi),%edx
  mov    $0x1,%r8d
  cmp    $0x2d,%dl
  jne    L0
  movsbl 0x1(%rdi),%edx
  mov    $0xffffffff,%r8d
  add    $0x1,%rdi
L0:
  xor    %eax,%eax
  test   %dl,%dl
  je     L28
  cs nopw 0x0(%rax,%rax,1)
L2:
  lea    -0x30(%rdx),%ecx
  add    $0x1,%rdi
  sub    $0x37,%edx
  cmp    $0x9,%ecx
  cmovg  %edx,%ecx
  imul   %esi,%eax
  movsbl (%rdi),%edx
  add    %ecx,%eax
  test   %dl,%dl
  jne    L2
  imul   %r8d,%eax
  ret
  nopl   0x0(%rax)
L28:
  ret
```

assistant: ```c
int atoi(char *s, int index) {
  int ret = 0;
  int sign = 1;
  if (*s == '-') {
    sign = -1;
    s++;
  }
  while (*s) {
    int r = *(s++);
    r -= '0';
    if (r > 9)
      r -= 7;
    ret = (ret * index) + r;
  }
  return sign * ret;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<atoi>:
  endbr64
  movsbl (%rdi),%edx
  mov    $0x1,%r8d
  cmp    $0x2d,%dl
  jne    L3
  movsbl 0x1(%rdi),%edx
  mov    $0xffffffff,%r8d
  add    $0x1,%rdi
L3:
  xor    %eax,%eax
  test   %dl,%dl
  je     L15
  cs nopw 0x0(%rax,%rax,1)
L40:
  imul   %esi,%eax
  lea    -0x30(%rdx),%ecx
  add    $0x1,%rdi
  cmp    $0x9,%ecx
  jle    L44
  lea    -0x37(%rdx,%rax,1),%eax
  movsbl (%rdi),%edx
  test   %dl,%dl
  jne    L40
  imul   %r8d,%eax
  ret
  nop
L44:
  movsbl (%rdi),%edx
  add    %ecx,%eax
  test   %dl,%dl
  jne    L40
  imul   %r8d,%eax
  ret
  xchg   %ax,%ax
L15:
  ret
```

assistant: ```c
int atoi(char *s, int index) {
  int ret = 0;
  int sign = 1;
  if (*s == '-') {
    sign = -1;
    s++;
  }
  while (*s) {
    int r = *(s++);
    r -= '0';
    if (r > 9)
      r -= 7;
    ret = (ret * index) + r;
  }
  return sign * ret;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<getNext>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L7
L4:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  test   %eax,%eax
  je     L6
  addl   $0x1,-0x4(%rbp)
L7:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L4
  jmp    L29
L6:
  nop
L29:
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int getNext(bool *taken, int length) {
  int i = 0;
  while (i < length) {
    if (!taken[i])
      break;
    i++;
  }
  return i;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<getNext>:
  endbr64
  test   %esi,%esi
  jle    L28
  mov    $0x0,%eax
L15:
  cmpl   $0x0,(%rdi)
  je     L14
  add    $0x1,%eax
  add    $0x4,%rdi
  cmp    %eax,%esi
  jne    L15
  mov    %esi,%eax
  ret
L28:
  mov    $0x0,%eax
L14:
  ret
```

assistant: ```c
int getNext(bool *taken, int length) {
  int i = 0;
  while (i < length) {
    if (!taken[i])
      break;
    i++;
  }
  return i;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<getNext>:
  endbr64
  xor    %eax,%eax
  test   %esi,%esi
  jg     L13
  jmp    L33
  nopl   0x0(%rax)
L18:
  add    $0x1,%eax
  add    $0x4,%rdi
  cmp    %eax,%esi
  je     L35
L13:
  mov    (%rdi),%edx
  test   %edx,%edx
  jne    L18
L33:
  ret
  nopw   0x0(%rax,%rax,1)
L35:
  ret
```

assistant: ```c
int getNext(bool *taken, int length) {
  int i = 0;
  while (i < length) {
    if (!taken[i])
      break;
    i++;
  }
  return i;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<getNext>:
  endbr64
  xor    %eax,%eax
  test   %esi,%esi
  jg     L0
  jmp    L2
  nopl   0x0(%rax)
L10:
  add    $0x1,%eax
  add    $0x4,%rdi
  cmp    %eax,%esi
  je     L34
L0:
  mov    (%rdi),%edx
  test   %edx,%edx
  jne    L10
L2:
  ret
  nopw   0x0(%rax,%rax,1)
L34:
  ret
```

assistant: ```c
int getNext(bool *taken, int length) {
  int i = 0;
  while (i < length) {
    if (!taken[i])
      break;
    i++;
  }
  return i;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<OS_free>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L0
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
L0:
  nop
  leave
  ret
```

assistant: ```c
void OS_free(void *p_memory) {
  if (p_memory)
    free(p_memory);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<OS_free>:
  endbr64
  test   %rdi,%rdi
  je     L7
  sub    $0x8,%rsp
  call   <free@plt>
  add    $0x8,%rsp
  ret
L7:
  ret
```

assistant: ```c
void OS_free(void *p_memory) {
  if (p_memory)
    free(p_memory);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<OS_free>:
  endbr64
  test   %rdi,%rdi
  je     L5
  jmp    1050 <free@plt>
  xchg   %ax,%ax
L5:
  ret
```

assistant: ```c
void OS_free(void *p_memory) {
  if (p_memory)
    free(p_memory);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<OS_free>:
  endbr64
  test   %rdi,%rdi
  je     L0
  jmp    1050 <free@plt>
  xchg   %ax,%ax
L0:
  ret
```

assistant: ```c
void OS_free(void *p_memory) {
  if (p_memory)
    free(p_memory);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<rop3_242_0_1>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,%ecx
  mov    %edx,%eax
  mov    %ecx,%edx
  mov    %dl,-0x1c(%rbp)
  mov    %al,-0x20(%rbp)
  movzbl -0x1c(%rbp),%eax
  not    %eax
  mov    %al,-0x3(%rbp)
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  and    -0x3(%rbp),%al
  mov    %al,-0x2(%rbp)
  movzbl -0x20(%rbp),%eax
  or     -0x2(%rbp),%al
  mov    %al,-0x1(%rbp)
  movzbl -0x1(%rbp),%eax
  and    -0x1c(%rbp),%al
  mov    %eax,%ecx
  movzbl -0x20(%rbp),%eax
  not    %eax
  mov    %eax,%edx
  movzbl -0x1(%rbp),%eax
  and    %edx,%eax
  or     %eax,%ecx
  mov    %ecx,%edx
  movzbl -0x1c(%rbp),%eax
  not    %eax
  mov    %eax,%ecx
  movzbl -0x20(%rbp),%eax
  and    %eax,%ecx
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  and    %ecx,%eax
  or     %edx,%eax
  mov    %eax,%edx
  mov    -0x18(%rbp),%rax
  mov    %dl,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void rop3_242_0_1(unsigned char *D, unsigned char S, unsigned char T) {
  unsigned char stk1;
  unsigned char stk2;
  unsigned char stk3;
  stk3 = ~S;
  stk2 = *D & stk3;
  stk1 = T | stk2;
  *D = (stk1 & S) | (stk1 & (~T)) | (T & (~S) & *D);
}
```
user: Translate the below assembly code into C programming language:

```asm
<rop3_242_0_1>:
  endbr64
  mov    %rdi,%rax
  mov    %esi,%edi
  mov    %esi,%ecx
  not    %ecx
  mov    %ecx,%r8d
  and    (%rax),%r8b
  or     %edx,%r8d
  mov    %edx,%esi
  not    %esi
  or     %edi,%esi
  and    %r8d,%esi
  and    (%rax),%dl
  and    %edx,%ecx
  or     %ecx,%esi
  mov    %sil,(%rax)
  ret
```

assistant: ```c
void rop3_242_0_1(unsigned char *D, unsigned char S, unsigned char T) {
  unsigned char stk1;
  unsigned char stk2;
  unsigned char stk3;
  stk3 = ~S;
  stk2 = *D & stk3;
  stk1 = T | stk2;
  *D = (stk1 & S) | (stk1 & (~T)) | (T & (~S) & *D);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<rop3_242_0_1>:
  endbr64
  mov    %rdi,%rax
  mov    %esi,%ecx
  mov    %esi,%edi
  mov    %edx,%esi
  movzbl (%rax),%r8d
  not    %ecx
  not    %esi
  or     %edi,%esi
  and    %ecx,%r8d
  or     %edx,%r8d
  and    (%rax),%dl
  and    %r8d,%esi
  and    %edx,%ecx
  or     %ecx,%esi
  mov    %sil,(%rax)
  ret
```

assistant: ```c
void rop3_242_0_1(unsigned char *D, unsigned char S, unsigned char T) {
  unsigned char stk1;
  unsigned char stk2;
  unsigned char stk3;
  stk3 = ~S;
  stk2 = *D & stk3;
  stk1 = T | stk2;
  *D = (stk1 & S) | (stk1 & (~T)) | (T & (~S) & *D);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<rop3_242_0_1>:
  endbr64
  mov    %rdi,%rax
  mov    %esi,%ecx
  mov    %esi,%edi
  mov    %edx,%esi
  movzbl (%rax),%r8d
  not    %ecx
  not    %esi
  or     %edi,%esi
  and    %ecx,%r8d
  or     %edx,%r8d
  and    (%rax),%dl
  and    %r8d,%esi
  and    %edx,%ecx
  or     %ecx,%esi
  mov    %sil,(%rax)
  ret
```

assistant: ```c
void rop3_242_0_1(unsigned char *D, unsigned char S, unsigned char T) {
  unsigned char stk1;
  unsigned char stk2;
  unsigned char stk3;
  stk3 = ~S;
  stk2 = *D & stk3;
  stk1 = T | stk2;
  *D = (stk1 & S) | (stk1 & (~T)) | (T & (~S) & *D);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<f>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L38
L19:
  movl   $0x1,-0x8(%rbp)
  jmp    L0
L4:
  mov    -0x8(%rbp),%eax
  cltq
  shl    $0x3,%rax
  lea    -0x8(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x4(%rbp),%eax
  cltq
  shl    $0x2,%rax
  sub    $0x4,%rax
  add    %rdx,%rax
  movss  (%rax),%xmm1
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x4(%rbp),%eax
  cltq
  shl    $0x2,%rax
  sub    $0x4,%rax
  add    %rdx,%rax
  movss  (%rax),%xmm0
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x4(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rdx,%rax
  addss  %xmm1,%xmm0
  movss  %xmm0,(%rax)
  addl   $0x1,-0x8(%rbp)
L0:
  mov    -0x1c(%rbp),%eax
  sub    -0x4(%rbp),%eax
  cmp    %eax,-0x8(%rbp)
  jl     L4
  addl   $0x1,-0x4(%rbp)
L38:
  mov    -0x1c(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0x4(%rbp)
  jl     L19
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void f(float **a, int n) {
  int t4;
  int t2;
  int i;
  int j;
  for (t2 = 0; t2 <= n - 2; t2 += 1)
    for (t4 = 1; t4 <= n - t2 - 1; t4 += 1)
      a[t4][t2] = (a[t4 - 1][t2 - 1] + a[t4][t2 - 1]);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<f>:
  endbr64
  cmp    $0x1,%esi
  jle    L58
  mov    $0xfffffffffffffffc,%rcx
  lea    0x8(%rdi),%r11
  jmp    L52
L23:
  lea    0x4(%rcx),%r10
  mov    %rdi,%rax
  lea    -0x2(%rsi),%edx
  lea    (%r11,%rdx,8),%r9
L35:
  mov    0x8(%rax),%rdx
  mov    (%rax),%r8
  movss  (%r8,%rcx,1),%xmm0
  addss  (%rdx,%rcx,1),%xmm0
  movss  %xmm0,(%rdx,%r10,1)
  add    $0x8,%rax
  cmp    %r9,%rax
  jne    L35
L8:
  sub    $0x1,%esi
  add    $0x4,%rcx
  cmp    $0x1,%esi
  je     L58
L52:
  cmp    $0x1,%esi
  jg     L23
  jmp    L8
L58:
  ret
```

assistant: ```c
void f(float **a, int n) {
  int t4;
  int t2;
  int i;
  int j;
  for (t2 = 0; t2 <= n - 2; t2 += 1)
    for (t4 = 1; t4 <= n - t2 - 1; t4 += 1)
      a[t4][t2] = (a[t4 - 1][t2 - 1] + a[t4][t2 - 1]);
}
```
user: What is the c source code of the assembly code below:

```asm
<f>:
  endbr64
  cmp    $0x1,%esi
  jle    L30
  mov    $0xfffffffffffffffc,%rcx
  lea    0x8(%rdi),%r11
  nopl   0x0(%rax)
L12:
  lea    0x4(%rcx),%r9
  cmp    $0x1,%esi
  jle    L14
  lea    -0x2(%rsi),%edx
  mov    %rdi,%rax
  lea    (%r11,%rdx,8),%r10
  nopl   0x0(%rax,%rax,1)
L39:
  mov    0x8(%rax),%rdx
  mov    (%rax),%r8
  add    $0x8,%rax
  movss  (%r8,%rcx,1),%xmm0
  addss  (%rdx,%rcx,1),%xmm0
  movss  %xmm0,(%rdx,%r9,1)
  cmp    %rax,%r10
  jne    L39
L14:
  sub    $0x1,%esi
  mov    %r9,%rcx
  cmp    $0x1,%esi
  jne    L12
L30:
  ret
```

assistant: ```c
void f(float **a, int n) {
  int t4;
  int t2;
  int i;
  int j;
  for (t2 = 0; t2 <= n - 2; t2 += 1)
    for (t4 = 1; t4 <= n - t2 - 1; t4 += 1)
      a[t4][t2] = (a[t4 - 1][t2 - 1] + a[t4][t2 - 1]);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<f>:
  endbr64
  cmp    $0x1,%esi
  jle    L11
  mov    $0xfffffffffffffffc,%rcx
  lea    0x8(%rdi),%r11
  nopl   0x0(%rax)
L30:
  lea    0x4(%rcx),%r9
  cmp    $0x1,%esi
  jle    L36
  lea    -0x2(%rsi),%edx
  mov    %rdi,%rax
  lea    (%r11,%rdx,8),%r10
  nopl   0x0(%rax,%rax,1)
L24:
  mov    0x8(%rax),%rdx
  mov    (%rax),%r8
  add    $0x8,%rax
  movss  (%r8,%rcx,1),%xmm0
  addss  (%rdx,%rcx,1),%xmm0
  movss  %xmm0,(%rdx,%r9,1)
  cmp    %r10,%rax
  jne    L24
L36:
  sub    $0x1,%esi
  mov    %r9,%rcx
  cmp    $0x1,%esi
  jne    L30
L11:
  ret
```

assistant: ```c
void f(float **a, int n) {
  int t4;
  int t2;
  int i;
  int j;
  for (t2 = 0; t2 <= n - 2; t2 += 1)
    for (t4 = 1; t4 <= n - t2 - 1; t4 += 1)
      a[t4][t2] = (a[t4 - 1][t2 - 1] + a[t4][t2 - 1]);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<moment3>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  pxor   %xmm0,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L5
L13:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm1
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  mulsd  %xmm0,%xmm1
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  mulsd  %xmm1,%xmm0
  movsd  -0x8(%rbp),%xmm1
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  addl   $0x1,-0xc(%rbp)
L5:
  mov    -0xc(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L13
  pxor   %xmm1,%xmm1
  cvtsi2sdl -0x14(%rbp),%xmm1
  movsd  -0x8(%rbp),%xmm0
  divsd  %xmm1,%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  pop    %rbp
  ret
```

assistant: ```c
double moment3(int n, double *v) {
  double suma = 0.0;
  int i;
  for (i = 0; i < n; ++i)
    suma += v[i] * v[i] * v[i];
  return suma / (double)n;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<moment3>:
  endbr64
  test   %edi,%edi
  jle    L1
  mov    %rsi,%rax
  lea    -0x1(%rdi),%edx
  lea    0x8(%rsi,%rdx,8),%rdx
  pxor   %xmm0,%xmm0
L20:
  movsd  (%rax),%xmm2
  movapd %xmm2,%xmm1
  mulsd  %xmm2,%xmm1
  mulsd  %xmm2,%xmm1
  addsd  %xmm1,%xmm0
  add    $0x8,%rax
  cmp    %rdx,%rax
  jne    L20
L24:
  pxor   %xmm1,%xmm1
  cvtsi2sd %edi,%xmm1
  divsd  %xmm1,%xmm0
  ret
L1:
  pxor   %xmm0,%xmm0
  jmp    L24
```

assistant: ```c
double moment3(int n, double *v) {
  double suma = 0.0;
  int i;
  for (i = 0; i < n; ++i)
    suma += v[i] * v[i] * v[i];
  return suma / (double)n;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<moment3>:
  endbr64
  test   %edi,%edi
  jle    L14
  lea    -0x1(%rdi),%eax
  pxor   %xmm0,%xmm0
  lea    0x8(%rsi,%rax,8),%rax
  nopl   0x0(%rax)
L19:
  movsd  (%rsi),%xmm2
  add    $0x8,%rsi
  movapd %xmm2,%xmm1
  mulsd  %xmm2,%xmm1
  mulsd  %xmm2,%xmm1
  addsd  %xmm1,%xmm0
  cmp    %rax,%rsi
  jne    L19
  pxor   %xmm1,%xmm1
  cvtsi2sd %edi,%xmm1
  divsd  %xmm1,%xmm0
  ret
  nopw   0x0(%rax,%rax,1)
L14:
  pxor   %xmm1,%xmm1
  pxor   %xmm0,%xmm0
  cvtsi2sd %edi,%xmm1
  divsd  %xmm1,%xmm0
  ret
```

assistant: ```c
double moment3(int n, double *v) {
  double suma = 0.0;
  int i;
  for (i = 0; i < n; ++i)
    suma += v[i] * v[i] * v[i];
  return suma / (double)n;
}
```
user: Translate the below assembly code into C programming language:

```asm
<moment3>:
  endbr64
  test   %edi,%edi
  jle    L12
  cmp    $0x1,%edi
  je     L16
  mov    %edi,%edx
  mov    %rsi,%rax
  pxor   %xmm0,%xmm0
  shr    %edx
  sub    $0x1,%edx
  shl    $0x4,%rdx
  lea    0x10(%rsi,%rdx,1),%rdx
  nopl   0x0(%rax)
L46:
  movupd (%rax),%xmm2
  add    $0x10,%rax
  movapd %xmm2,%xmm1
  mulpd  %xmm2,%xmm1
  mulpd  %xmm2,%xmm1
  addsd  %xmm1,%xmm0
  unpckhpd %xmm1,%xmm1
  addsd  %xmm1,%xmm0
  cmp    %rax,%rdx
  jne    L46
  mov    %edi,%eax
  and    $0xfffffffe,%eax
  test   $0x1,%dil
  je     L28
L30:
  cltq
  movsd  (%rsi,%rax,8),%xmm2
  movapd %xmm2,%xmm1
  mulsd  %xmm2,%xmm1
  mulsd  %xmm2,%xmm1
  addsd  %xmm1,%xmm0
L28:
  pxor   %xmm1,%xmm1
  cvtsi2sd %edi,%xmm1
  divsd  %xmm1,%xmm0
  ret
  nopl   0x0(%rax)
L12:
  pxor   %xmm1,%xmm1
  pxor   %xmm0,%xmm0
  cvtsi2sd %edi,%xmm1
  divsd  %xmm1,%xmm0
  ret
L16:
  xor    %eax,%eax
  pxor   %xmm0,%xmm0
  jmp    L30
```

assistant: ```c
double moment3(int n, double *v) {
  double suma = 0.0;
  int i;
  for (i = 0; i < n; ++i)
    suma += v[i] * v[i] * v[i];
  return suma / (double)n;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<memparse>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x18(%rbp),%rcx
  mov    -0x28(%rbp),%rax
  mov    $0x0,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <simple_strtoull@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  sub    $0x47,%eax
  cmp    $0x26,%eax
  seta   %dl
  test   %dl,%dl
  jne    L44
  mov    $0x1,%edx
  mov    %eax,%ecx
  shl    %cl,%rdx
  mov    %rdx,%rax
  movabs $0x1000000010,%rdx
  and    %rax,%rdx
  test   %rdx,%rdx
  setne  %dl
  test   %dl,%dl
  jne    L54
  movabs $0x4000000040,%rdx
  and    %rax,%rdx
  test   %rdx,%rdx
  setne  %dl
  test   %dl,%dl
  jne    L58
  movabs $0x100000001,%rdx
  and    %rdx,%rax
  test   %rax,%rax
  setne  %al
  test   %al,%al
  je     L44
  shlq   $0xa,-0x10(%rbp)
L58:
  shlq   $0xa,-0x10(%rbp)
L54:
  shlq   $0xa,-0x10(%rbp)
  mov    -0x18(%rbp),%rax
  add    $0x1,%rax
  mov    %rax,-0x18(%rbp)
L44:
  nop
  cmpq   $0x0,-0x30(%rbp)
  je     L19
  mov    -0x18(%rbp),%rdx
  mov    -0x30(%rbp),%rax
  mov    %rdx,(%rax)
L19:
  mov    -0x10(%rbp),%rax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L36
  call   <__stack_chk_fail@plt>
L36:
  leave
  ret
```

assistant: ```c
unsigned long long memparse(const char *ptr, char **retptr) {
  char *endptr;
  unsigned long long ret = simple_strtoull(ptr, &endptr, 0);
  switch (*endptr) {
  case 'G':
  case 'g':
    ret <<= 10;
  case 'M':
  case 'm':
    ret <<= 10;
  case 'K':
  case 'k':
    ret <<= 10;
    endptr++;
  default:
    break;
  }
  if (retptr)
    *retptr = endptr;
  return ret;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<memparse>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %rsi,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  mov    $0x0,%edx
  call   <simple_strtoull@plt>
  mov    (%rsp),%rsi
  movzbl (%rsi),%edi
  lea    -0x47(%rdi),%ecx
  cmp    $0x26,%cl
  ja     L55
  mov    $0x1,%edx
  shl    %cl,%rdx
  movabs $0x1000000010,%rcx
  test   %rcx,%rdx
  jne    L28
  movabs $0x4000000040,%rcx
  test   %rcx,%rdx
  jne    L24
  movabs $0x100000001,%rcx
  test   %rcx,%rdx
  je     L55
  shl    $0xa,%rax
L24:
  shl    $0xa,%rax
L28:
  shl    $0xa,%rax
  add    $0x1,%rsi
  mov    %rsi,(%rsp)
L55:
  test   %rbx,%rbx
  je     L30
  mov    (%rsp),%rdx
  mov    %rdx,(%rbx)
L30:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L40
  add    $0x10,%rsp
  pop    %rbx
  ret
L40:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
unsigned long long memparse(const char *ptr, char **retptr) {
  char *endptr;
  unsigned long long ret = simple_strtoull(ptr, &endptr, 0);
  switch (*endptr) {
  case 'G':
  case 'g':
    ret <<= 10;
  case 'M':
  case 'm':
    ret <<= 10;
  case 'K':
  case 'k':
    ret <<= 10;
    endptr++;
  default:
    break;
  }
  if (retptr)
    *retptr = endptr;
  return ret;
}
```
user: What is the c source code of the assembly code below?

```asm
<memparse>:
  endbr64
  push   %rbx
  xor    %edx,%edx
  mov    %rsi,%rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  call   <simple_strtoull@plt>
  mov    (%rsp),%rsi
  movzbl (%rsi),%edi
  lea    -0x47(%rdi),%ecx
  cmp    $0x26,%cl
  ja     L3
  mov    $0x1,%edx
  shl    %cl,%rdx
  movabs $0x1000000010,%rcx
  test   %rcx,%rdx
  jne    L68
  movabs $0x4000000040,%rcx
  test   %rcx,%rdx
  jne    L61
  movabs $0x100000001,%rcx
  test   %rcx,%rdx
  jne    L36
L3:
  test   %rbx,%rbx
  je     L21
  mov    (%rsp),%rdx
  mov    %rdx,(%rbx)
L21:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L12
  add    $0x10,%rsp
  pop    %rbx
  ret
  nopl   0x0(%rax)
L36:
  shl    $0xa,%rax
L61:
  shl    $0xa,%rax
L68:
  add    $0x1,%rsi
  shl    $0xa,%rax
  mov    %rsi,(%rsp)
  jmp    L3
L12:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
unsigned long long memparse(const char *ptr, char **retptr) {
  char *endptr;
  unsigned long long ret = simple_strtoull(ptr, &endptr, 0);
  switch (*endptr) {
  case 'G':
  case 'g':
    ret <<= 10;
  case 'M':
  case 'm':
    ret <<= 10;
  case 'K':
  case 'k':
    ret <<= 10;
    endptr++;
  default:
    break;
  }
  if (retptr)
    *retptr = endptr;
  return ret;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<memparse>:
  endbr64
  push   %rbx
  xor    %edx,%edx
  mov    %rsi,%rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  call   <simple_strtoull@plt>
  mov    (%rsp),%rsi
  movzbl (%rsi),%edi
  lea    -0x47(%rdi),%ecx
  cmp    $0x26,%cl
  ja     L56
  mov    $0x1,%edx
  shl    %cl,%rdx
  movabs $0x1000000010,%rcx
  test   %rcx,%rdx
  jne    L16
  movabs $0x4000000040,%rcx
  test   %rcx,%rdx
  jne    L57
  movabs $0x100000001,%rcx
  test   %rcx,%rdx
  jne    L43
L56:
  test   %rbx,%rbx
  je     L2
  mov    %rsi,(%rbx)
L2:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L8
  add    $0x10,%rsp
  pop    %rbx
  ret
  nopl   0x0(%rax,%rax,1)
L43:
  shl    $0xa,%rax
L57:
  shl    $0xa,%rax
L16:
  shl    $0xa,%rax
  add    $0x1,%rsi
  jmp    L56
L8:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
unsigned long long memparse(const char *ptr, char **retptr) {
  char *endptr;
  unsigned long long ret = simple_strtoull(ptr, &endptr, 0);
  switch (*endptr) {
  case 'G':
  case 'g':
    ret <<= 10;
  case 'M':
  case 'm':
    ret <<= 10;
  case 'K':
  case 'k':
    ret <<= 10;
    endptr++;
  default:
    break;
  }
  if (retptr)
    *retptr = endptr;
  return ret;
}
```