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

```asm
<main>:
  endbr64
  push   %r12
  mov    $0x9,%edx
  xor    %esi,%esi
  push   %rbp
  lea    D16(%rip),%rbp
  push   %rbx
  sub    $0x30,%rsp
  movdqa D37(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  mov    D20(%rip),%rax
  lea    0x28(%rsp),%r12
  movaps %xmm0,(%rsp)
  movdqa D38(%rip),%xmm0
  mov    %rbx,%rdi
  mov    %rax,0x20(%rsp)
  movaps %xmm0,0x10(%rsp)
  call   <quickSort@plt>
  nop
L5:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %rbx,%r12
  jne    L5
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L17
  add    $0x30,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L17:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D16
data_type: string
name: extract_symbolic_data
arguments: data_label: D37
data_type: byte[16]
name: extract_symbolic_data
arguments: data_label: D20
data_type: byte[8]
name: extract_symbolic_data
arguments: data_label: D38
data_type: byte[16]

tool: D16: %d

tool: {"D37": [0x04,0x00,0x00,0x00,0x02,0x00,0x00,0x00,0x03,0x00,0x00,0x00,0x05,0x00,0x00,0x00]}

tool: {"D20": [0x09,0x00,0x00,0x00,0x01,0x00,0x00,0x00]}

tool: {"D38": [0x08,0x00,0x00,0x00,0x06,0x00,0x00,0x00,0x07,0x00,0x00,0x00,0x00,0x00,0x00,0x00]}

assistant: ```c
int main(int argc, char *argv[]) {
  int value[10] = {4, 2, 3, 5, 8, 6, 7, 0, 9, 1};
  quickSort(value, 0, 9);
  for (int i = 0; i < 10; i++) {
    printf("%d", value[i]);
  }
  printf("\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
<get_index>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    D6(%rip),%rax
  mov    (%rax),%ecx
  mov    -0x10(%rbp),%rdx
  mov    -0x4(%rbp),%eax
  mov    %ecx,%esi
  mov    %eax,%edi
  call   <ioctl@plt>
  test   %rax,%rax
  jns    L8
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <close@plt>
  mov    $0xffffffff,%eax
  jmp    L18
L8:
  mov    $0x0,%eax
L18:
  leave
  ret
```

assistant: ```c
int get_index(int sock, struct ifreq *ifr) {
  if (ioctl(sock, SIOCGIFINDEX, ifr) < 0) {
    close(sock);
    return -1;
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<get_index>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  mov    %rsi,%rdx
  mov    D7(%rip),%rax
  mov    (%rax),%esi
  call   <ioctl@plt>
  test   %rax,%rax
  js     L14
  mov    $0x0,%eax
L7:
  pop    %rbx
  ret
L14:
  mov    %ebx,%edi
  call   <close@plt>
  mov    $0xffffffff,%eax
  jmp    L7
```

assistant: ```c
int get_index(int sock, struct ifreq *ifr) {
  if (ioctl(sock, SIOCGIFINDEX, ifr) < 0) {
    close(sock);
    return -1;
  }
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<get_index>:
  endbr64
  mov    D6(%rip),%rax
  push   %rbp
  mov    %rsi,%rdx
  mov    %edi,%ebp
  mov    (%rax),%esi
  call   <ioctl@plt>
  test   %rax,%rax
  js     L6
  xor    %eax,%eax
  pop    %rbp
  ret
L6:
  mov    %ebp,%edi
  call   <close@plt>
  mov    $0xffffffff,%eax
  pop    %rbp
  ret
```

assistant: ```c
int get_index(int sock, struct ifreq *ifr) {
  if (ioctl(sock, SIOCGIFINDEX, ifr) < 0) {
    close(sock);
    return -1;
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<get_index>:
  endbr64
  mov    D2(%rip),%rax
  push   %rbp
  mov    %rsi,%rdx
  mov    %edi,%ebp
  mov    (%rax),%esi
  call   <ioctl@plt>
  test   %rax,%rax
  js     L7
  xor    %eax,%eax
  pop    %rbp
  ret
L7:
  mov    %ebp,%edi
  call   <close@plt>
  mov    $0xffffffff,%eax
  pop    %rbp
  ret
```

assistant: ```c
int get_index(int sock, struct ifreq *ifr) {
  if (ioctl(sock, SIOCGIFINDEX, ifr) < 0) {
    close(sock);
    return -1;
  }
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<write_int32>:
  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),%ecx
  lea    -0xc(%rbp),%rax
  mov    $0x0,%edx
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <buf_int32@plt>
  lea    -0xc(%rbp),%rcx
  mov    -0x14(%rbp),%eax
  mov    $0x4,%edx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <write@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L8
  call   <__stack_chk_fail@plt>
L8:
  leave
  ret
```

assistant: ```c
void write_int32(int fd, u_int32_t n) {
  char buf[4];
  buf_int32(buf, n, 0);
  write(fd, buf, 4);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<write_int32>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %edi,%ebx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rbp
  mov    $0x0,%edx
  mov    %rbp,%rdi
  call   <buf_int32@plt>
  mov    $0x4,%edx
  mov    %rbp,%rsi
  mov    %ebx,%edi
  call   <write@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L6:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void write_int32(int fd, u_int32_t n) {
  char buf[4];
  buf_int32(buf, n, 0);
  write(fd, buf, 4);
}
```
user: Translate the below assembly code into C programming language:

```asm
<write_int32>:
  endbr64
  push   %r12
  xor    %edx,%edx
  push   %rbp
  mov    %edi,%ebp
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%r12
  mov    %r12,%rdi
  call   <buf_int32@plt>
  mov    $0x4,%edx
  mov    %r12,%rsi
  mov    %ebp,%edi
  call   <write@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x18,%rsp
  pop    %rbp
  pop    %r12
  ret
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void write_int32(int fd, u_int32_t n) {
  char buf[4];
  buf_int32(buf, n, 0);
  write(fd, buf, 4);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<write_int32>:
  endbr64
  push   %r12
  xor    %edx,%edx
  push   %rbp
  mov    %edi,%ebp
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%r12
  mov    %r12,%rdi
  call   <buf_int32@plt>
  mov    $0x4,%edx
  mov    %r12,%rsi
  mov    %ebp,%edi
  call   <write@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x18,%rsp
  pop    %rbp
  pop    %r12
  ret
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void write_int32(int fd, u_int32_t n) {
  char buf[4];
  buf_int32(buf, n, 0);
  write(fd, buf, 4);
}
```
user: Translate the below assembly code into C programming language:

```asm
<c25519_prepare>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%eax
  and    $0xf8,%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x8(%rbp),%rax
  add    $0x7c,%rax
  mov    (%rax),%edx
  mov    -0x8(%rbp),%rax
  add    $0x7c,%rax
  and    $0x7f,%edx
  mov    %edx,(%rax)
  mov    -0x8(%rbp),%rax
  add    $0x7c,%rax
  mov    (%rax),%edx
  mov    -0x8(%rbp),%rax
  add    $0x7c,%rax
  or     $0x40,%edx
  mov    %edx,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void c25519_prepare(uint8_t *key) {
  key[0] &= 0xf8;
  key[31] &= 0x7f;
  key[31] |= 0x40;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<c25519_prepare>:
  endbr64
  andl   $0xf8,(%rdi)
  mov    0x7c(%rdi),%eax
  and    $0x7f,%eax
  or     $0x40,%eax
  mov    %eax,0x7c(%rdi)
  ret
```

assistant: ```c
void c25519_prepare(uint8_t *key) {
  key[0] &= 0xf8;
  key[31] &= 0x7f;
  key[31] |= 0x40;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<c25519_prepare>:
  endbr64
  mov    0x7c(%rdi),%eax
  andl   $0xf8,(%rdi)
  and    $0x7f,%eax
  or     $0x40,%eax
  mov    %eax,0x7c(%rdi)
  ret
```

assistant: ```c
void c25519_prepare(uint8_t *key) {
  key[0] &= 0xf8;
  key[31] &= 0x7f;
  key[31] |= 0x40;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<c25519_prepare>:
  endbr64
  mov    0x7c(%rdi),%eax
  andl   $0xf8,(%rdi)
  and    $0x7f,%eax
  or     $0x40,%eax
  mov    %eax,0x7c(%rdi)
  ret
```

assistant: ```c
void c25519_prepare(uint8_t *key) {
  key[0] &= 0xf8;
  key[31] &= 0x7f;
  key[31] |= 0x40;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<super_factorial>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  movl   $0x1,-0x8(%rbp)
  movl   $0x1,-0x4(%rbp)
  jmp    L8
L16:
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <factorial@plt>
  mov    -0x8(%rbp),%edx
  imul   %edx,%eax
  mov    %eax,-0x8(%rbp)
  addl   $0x1,-0x4(%rbp)
L8:
  mov    -0x4(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jle    L16
  mov    -0x8(%rbp),%eax
  leave
  ret
```

assistant: ```c
uint64_t super_factorial(int n) {
  uint64_t result = 1;
  int i;
  for (i = 1; i <= n; i++) {
    result *= factorial(i);
  }
  return result;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<super_factorial>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  test   %edi,%edi
  jle    L1
  mov    %edi,%r12d
  mov    $0x1,%ebx
  mov    $0x1,%ebp
L29:
  mov    %ebx,%edi
  call   <factorial@plt>
  imul   %eax,%ebp
  add    $0x1,%ebx
  cmp    %ebx,%r12d
  jge    L29
L12:
  mov    %ebp,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L1:
  mov    $0x1,%ebp
  jmp    L12
```

assistant: ```c
uint64_t super_factorial(int n) {
  uint64_t result = 1;
  int i;
  for (i = 1; i <= n; i++) {
    result *= factorial(i);
  }
  return result;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<super_factorial>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  test   %edi,%edi
  jle    L16
  mov    %edi,%ebp
  mov    $0x1,%ebx
  mov    $0x1,%r12d
  nopl   0x0(%rax)
L13:
  mov    %ebx,%edi
  add    $0x1,%ebx
  call   <factorial@plt>
  imul   %eax,%r12d
  cmp    %ebx,%ebp
  jge    L13
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L16:
  mov    $0x1,%r12d
  pop    %rbx
  pop    %rbp
  mov    %r12d,%eax
  pop    %r12
  ret
```

assistant: ```c
uint64_t super_factorial(int n) {
  uint64_t result = 1;
  int i;
  for (i = 1; i <= n; i++) {
    result *= factorial(i);
  }
  return result;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<super_factorial>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  test   %edi,%edi
  jle    L9
  mov    %edi,%ebp
  mov    $0x1,%ebx
  mov    $0x1,%r12d
  nopl   0x0(%rax)
L17:
  mov    %ebx,%edi
  add    $0x1,%ebx
  call   <factorial@plt>
  imul   %eax,%r12d
  cmp    %ebx,%ebp
  jge    L17
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L9:
  mov    $0x1,%r12d
  pop    %rbx
  pop    %rbp
  mov    %r12d,%eax
  pop    %r12
  ret
```

assistant: ```c
uint64_t super_factorial(int n) {
  uint64_t result = 1;
  int i;
  for (i = 1; i <= n; i++) {
    result *= factorial(i);
  }
  return result;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<hybridFindMaxSubarray>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %rdi,%rbp
  mov    %rdx,%r12
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    (%rdx),%eax
  mov    %eax,0x8(%rsp)
  mov    (%rsi),%edx
  mov    %edx,0xc(%rsp)
  mov    %eax,%ecx
  sub    %edx,%ecx
  cmp    $0xa,%ecx
  jle    L25
  mov    %rsi,%r13
  lea    (%rax,%rdx,1),%ecx
  mov    %ecx,%r14d
  shr    $0x1f,%r14d
  add    %ecx,%r14d
  sar    %r14d
  mov    %edx,0x10(%rsp)
  mov    %r14d,0x14(%rsp)
  lea    0x1(%r14),%edx
  mov    %edx,0x18(%rsp)
  mov    %eax,0x1c(%rsp)
  lea    0x14(%rsp),%rdx
  lea    0x10(%rsp),%rsi
  call   <hybridFindMaxSubarray>
  mov    %eax,%ebx
  lea    0x1c(%rsp),%rdx
  lea    0x18(%rsp),%rsi
  mov    %rbp,%rdi
  call   <hybridFindMaxSubarray>
  mov    %eax,%r15d
  lea    0x24(%rsp),%r9
  lea    0x20(%rsp),%r8
  mov    (%r12),%ecx
  mov    %r14d,%edx
  mov    0x0(%r13),%esi
  mov    %rbp,%rdi
  call   <findMaxCrossingSubarray@plt>
  cmp    %eax,%r15d
  mov    %eax,%edx
  cmovge %r15d,%edx
  cmp    %edx,%ebx
  jge    L8
  cmp    %eax,%ebx
  cmovl  %eax,%ebx
  cmp    %ebx,%r15d
  jl     L48
  mov    0x18(%rsp),%eax
  mov    %eax,0x0(%r13)
  mov    0x1c(%rsp),%eax
  mov    %eax,(%r12)
  mov    %r15d,%ebx
  jmp    L63
L25:
  lea    0x8(%rsp),%rdx
  lea    0xc(%rsp),%rsi
  call   <bruteFindMaxSubarray@plt>
  mov    %eax,%ebx
L63:
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L24
  mov    %ebx,%eax
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L8:
  mov    0x10(%rsp),%eax
  mov    %eax,0x0(%r13)
  mov    0x14(%rsp),%eax
  mov    %eax,(%r12)
  jmp    L63
L48:
  mov    0x20(%rsp),%edx
  mov    %edx,0x0(%r13)
  mov    0x24(%rsp),%edx
  mov    %edx,(%r12)
  mov    %eax,%ebx
  jmp    L63
L24:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int hybridFindMaxSubarray(int *arr, int *low, int *high) {
  int deref_high = *high, deref_low = *low;
  if (*high - *low + 1 <= 11)
    return bruteFindMaxSubarray(arr, &deref_low, &deref_high);
  int mid = (*low + *high) / 2;
  int leftLow = *low, leftHigh = mid, leftSum;
  int rightLow = mid + 1, rightHigh = *high, rightSum;
  int crossLow, crossHigh, crossSum;
  leftSum = hybridFindMaxSubarray(arr, &leftLow, &leftHigh);
  rightSum = hybridFindMaxSubarray(arr, &rightLow, &rightHigh);
  crossSum = findMaxCrossingSubarray(arr, *low, mid, *high, &crossLow, &crossHigh);
  if (leftSum >= rightSum && leftSum >= crossSum) {
    *low = leftLow;
    *high = leftHigh;
    return leftSum;
  } else if (rightSum >= leftSum && rightSum >= crossSum) {
    *low = rightLow;
    *high = rightHigh;
    return rightSum;
  } else {
    *low = crossLow;
    *high = crossHigh;
    return crossSum;
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<hybridFindMaxSubarray>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rdx,%rbx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  mov    (%rdx),%eax
  mov    (%rsi),%edx
  mov    %edx,0xc(%rsp)
  mov    %eax,%ecx
  mov    %eax,0x8(%rsp)
  sub    %edx,%ecx
  cmp    $0xa,%ecx
  jle    L6
  lea    (%rax,%rdx,1),%ecx
  mov    %edx,0x10(%rsp)
  mov    %rsi,%r13
  lea    0x10(%rsp),%rsi
  mov    %ecx,%r14d
  mov    %eax,0x1c(%rsp)
  shr    $0x1f,%r14d
  add    %ecx,%r14d
  sar    %r14d
  lea    0x1(%r14),%edx
  mov    %r14d,0x14(%rsp)
  mov    %edx,0x18(%rsp)
  lea    0x14(%rsp),%rdx
  call   <hybridFindMaxSubarray>
  lea    0x1c(%rsp),%rdx
  lea    0x18(%rsp),%rsi
  mov    %rbp,%rdi
  mov    %eax,%r12d
  call   <hybridFindMaxSubarray>
  mov    (%rbx),%ecx
  mov    0x0(%r13),%esi
  mov    %r14d,%edx
  lea    0x24(%rsp),%r9
  lea    0x20(%rsp),%r8
  mov    %rbp,%rdi
  mov    %eax,%r15d
  call   <findMaxCrossingSubarray@plt>
  cmp    %eax,%r15d
  mov    %eax,%edx
  cmovge %r15d,%edx
  cmp    %edx,%r12d
  jge    L36
  cmp    %eax,%r12d
  cmovl  %eax,%r12d
  cmp    %r12d,%r15d
  jl     L17
  mov    0x18(%rsp),%eax
  mov    %r15d,%r12d
  mov    %eax,0x0(%r13)
  mov    0x1c(%rsp),%eax
  mov    %eax,(%rbx)
L13:
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L26
  add    $0x38,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  cs nopw 0x0(%rax,%rax,1)
L36:
  mov    0x10(%rsp),%eax
  mov    %eax,0x0(%r13)
  mov    0x14(%rsp),%eax
  mov    %eax,(%rbx)
  jmp    L13
L6:
  lea    0x8(%rsp),%rdx
  lea    0xc(%rsp),%rsi
  call   <bruteFindMaxSubarray@plt>
  mov    %eax,%r12d
  jmp    L13
  nopl   0x0(%rax)
L17:
  mov    0x20(%rsp),%edx
  mov    %eax,%r12d
  mov    %edx,0x0(%r13)
  mov    0x24(%rsp),%edx
  mov    %edx,(%rbx)
  jmp    L13
L26:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int hybridFindMaxSubarray(int *arr, int *low, int *high) {
  int deref_high = *high, deref_low = *low;
  if (*high - *low + 1 <= 11)
    return bruteFindMaxSubarray(arr, &deref_low, &deref_high);
  int mid = (*low + *high) / 2;
  int leftLow = *low, leftHigh = mid, leftSum;
  int rightLow = mid + 1, rightHigh = *high, rightSum;
  int crossLow, crossHigh, crossSum;
  leftSum = hybridFindMaxSubarray(arr, &leftLow, &leftHigh);
  rightSum = hybridFindMaxSubarray(arr, &rightLow, &rightHigh);
  crossSum = findMaxCrossingSubarray(arr, *low, mid, *high, &crossLow, &crossHigh);
  if (leftSum >= rightSum && leftSum >= crossSum) {
    *low = leftLow;
    *high = leftHigh;
    return leftSum;
  } else if (rightSum >= leftSum && rightSum >= crossSum) {
    *low = rightLow;
    *high = rightHigh;
    return rightSum;
  } else {
    *low = crossLow;
    *high = crossHigh;
    return crossSum;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<_ftok>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x10(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <stat@plt>
  test   %rax,%rax
  jns    L5
  mov    $0xffffffffffffffff,%rax
  jmp    L26
L5:
  mov    -0x10(%rbp),%eax
  shl    $0x10,%eax
  mov    %eax,%edx
  mov    -0xc(%rbp),%eax
  xor    %edx,%eax
  movslq %eax,%rdx
  mov    -0x1c(%rbp),%eax
  cltq
  add    %rdx,%rax
L26:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L27
  call   <__stack_chk_fail@plt>
L27:
  leave
  ret
```

assistant: ```c
key_t _ftok(const char *path, int id) {
  struct stat st;
  if (stat(path, &st) < 0)
    return (key_t)-1;
  return (key_t)(((st.st_dev & 0xffff) << 16) ^ st.st_ino) + id;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<_ftok>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %esi,%ebx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  call   <stat@plt>
  test   %rax,%rax
  js     L29
  mov    (%rsp),%eax
  shl    $0x10,%eax
  xor    0x4(%rsp),%eax
  cltq
  movslq %ebx,%rbx
  add    %rbx,%rax
L26:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L20
  add    $0x10,%rsp
  pop    %rbx
  ret
L29:
  mov    $0xffffffffffffffff,%rax
  jmp    L26
L20:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
key_t _ftok(const char *path, int id) {
  struct stat st;
  if (stat(path, &st) < 0)
    return (key_t)-1;
  return (key_t)(((st.st_dev & 0xffff) << 16) ^ st.st_ino) + id;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<_ftok>:
  endbr64
  push   %rbx
  movslq %esi,%rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  call   <stat@plt>
  test   %rax,%rax
  js     L27
  mov    (%rsp),%eax
  shl    $0x10,%eax
  xor    0x4(%rsp),%eax
  cltq
  add    %rbx,%rax
L2:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L4
  add    $0x10,%rsp
  pop    %rbx
  ret
L27:
  mov    $0xffffffffffffffff,%rax
  jmp    L2
L4:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
key_t _ftok(const char *path, int id) {
  struct stat st;
  if (stat(path, &st) < 0)
    return (key_t)-1;
  return (key_t)(((st.st_dev & 0xffff) << 16) ^ st.st_ino) + id;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<_ftok>:
  endbr64
  push   %rbx
  movslq %esi,%rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  call   <stat@plt>
  test   %rax,%rax
  js     L0
  mov    (%rsp),%eax
  shl    $0x10,%eax
  xor    0x4(%rsp),%eax
  cltq
  add    %rbx,%rax
L25:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L12
  add    $0x10,%rsp
  pop    %rbx
  ret
L0:
  mov    $0xffffffffffffffff,%rax
  jmp    L25
L12:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
key_t _ftok(const char *path, int id) {
  struct stat st;
  if (stat(path, &st) < 0)
    return (key_t)-1;
  return (key_t)(((st.st_dev & 0xffff) << 16) ^ st.st_ino) + id;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<constructSTUtil>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  mov    %edx,-0x30(%rbp)
  mov    %rcx,-0x38(%rbp)
  mov    %r8d,-0x3c(%rbp)
  mov    -0x2c(%rbp),%eax
  cmp    -0x30(%rbp),%eax
  jne    L10
  mov    -0x2c(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    -0x3c(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x38(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  mov    -0x2c(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  jmp    L12
L10:
  mov    -0x30(%rbp),%edx
  mov    -0x2c(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <getMid@plt>
  mov    %eax,-0x14(%rbp)
  mov    -0x3c(%rbp),%eax
  add    %eax,%eax
  lea    0x1(%rax),%edi
  mov    -0x38(%rbp),%rcx
  mov    -0x14(%rbp),%edx
  mov    -0x2c(%rbp),%esi
  mov    -0x28(%rbp),%rax
  mov    %edi,%r8d
  mov    %rax,%rdi
  call   <constructSTUtil>
  mov    %eax,%ebx
  mov    -0x3c(%rbp),%eax
  add    $0x1,%eax
  lea    (%rax,%rax,1),%edi
  mov    -0x14(%rbp),%eax
  lea    0x1(%rax),%esi
  mov    -0x38(%rbp),%rcx
  mov    -0x30(%rbp),%edx
  mov    -0x28(%rbp),%rax
  mov    %edi,%r8d
  mov    %rax,%rdi
  call   <constructSTUtil>
  mov    %eax,%edx
  mov    -0x3c(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x38(%rbp),%rax
  add    %rcx,%rax
  add    %ebx,%edx
  mov    %edx,(%rax)
  mov    -0x3c(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
L12:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int constructSTUtil(int arr[], int ss, int se, int *st, int si) {
  if (ss == se) {
    st[si] = arr[ss];
    return arr[ss];
  }
  int mid = getMid(ss, se);
  st[si] = constructSTUtil(arr, ss, mid, st, si * 2 + 1) + constructSTUtil(arr, mid + 1, se, st, si * 2 + 2);
  return st[si];
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<constructSTUtil>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%r14
  mov    %esi,%ebp
  mov    %rcx,%r13
  mov    %r8d,%ebx
  cmp    %edx,%esi
  je     L6
  mov    %edx,%r12d
  mov    %edx,%esi
  mov    %ebp,%edi
  call   <getMid@plt>
  lea    (%rbx,%rbx,1),%r15d
  lea    0x1(%r15),%r8d
  mov    %r13,%rcx
  mov    %eax,0xc(%rsp)
  mov    %eax,%edx
  mov    %ebp,%esi
  mov    %r14,%rdi
  call   <constructSTUtil>
  mov    %eax,%ebp
  mov    0xc(%rsp),%esi
  add    $0x1,%esi
  lea    0x2(%r15),%r8d
  mov    %r13,%rcx
  mov    %r12d,%edx
  mov    %r14,%rdi
  call   <constructSTUtil>
  add    %ebp,%eax
  movslq %ebx,%rbx
  mov    %eax,0x0(%r13,%rbx,4)
L0:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L6:
  movslq %esi,%rbp
  mov    (%rdi,%rbp,4),%eax
  movslq %r8d,%rbx
  mov    %eax,(%rcx,%rbx,4)
  jmp    L0
```

assistant: ```c
int constructSTUtil(int arr[], int ss, int se, int *st, int si) {
  if (ss == se) {
    st[si] = arr[ss];
    return arr[ss];
  }
  int mid = getMid(ss, se);
  st[si] = constructSTUtil(arr, ss, mid, st, si * 2 + 1) + constructSTUtil(arr, mid + 1, se, st, si * 2 + 2);
  return st[si];
}
```