ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What is the c source code of the assembly code below?

```asm
<crossVectors>:
  endbr64
  movss  0x4(%rdi),%xmm0
  mulss  0x8(%rsi),%xmm0
  movss  0x8(%rdi),%xmm1
  mulss  0x4(%rsi),%xmm1
  subss  %xmm1,%xmm0
  movss  %xmm0,(%rdx)
  movss  0x8(%rdi),%xmm0
  mulss  (%rsi),%xmm0
  movss  (%rdi),%xmm1
  mulss  0x8(%rsi),%xmm1
  subss  %xmm1,%xmm0
  movss  %xmm0,0x4(%rdx)
  movss  (%rdi),%xmm0
  mulss  0x4(%rsi),%xmm0
  movss  0x4(%rdi),%xmm1
  mulss  (%rsi),%xmm1
  subss  %xmm1,%xmm0
  movss  %xmm0,0x8(%rdx)
  ret
```

assistant: ```c
void crossVectors(const float *u, const float *v, float *res) {
  res[0] = u[1] * v[2] - u[2] * v[1];
  res[1] = u[2] * v[0] - u[0] * v[2];
  res[2] = u[0] * v[1] - u[1] * v[0];
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<crossVectors>:
  endbr64
  movss  0x4(%rdi),%xmm0
  movss  0x8(%rdi),%xmm1
  mulss  0x8(%rsi),%xmm0
  mulss  0x4(%rsi),%xmm1
  subss  %xmm1,%xmm0
  movss  %xmm0,(%rdx)
  movss  0x8(%rdi),%xmm0
  movss  (%rdi),%xmm1
  mulss  (%rsi),%xmm0
  mulss  0x8(%rsi),%xmm1
  subss  %xmm1,%xmm0
  movss  %xmm0,0x4(%rdx)
  movss  (%rdi),%xmm0
  movss  0x4(%rdi),%xmm1
  mulss  0x4(%rsi),%xmm0
  mulss  (%rsi),%xmm1
  subss  %xmm1,%xmm0
  movss  %xmm0,0x8(%rdx)
  ret
```

assistant: ```c
void crossVectors(const float *u, const float *v, float *res) {
  res[0] = u[1] * v[2] - u[2] * v[1];
  res[1] = u[2] * v[0] - u[0] * v[2];
  res[2] = u[0] * v[1] - u[1] * v[0];
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<crossVectors>:
  endbr64
  movss  0x4(%rdi),%xmm0
  movss  0x8(%rdi),%xmm1
  mulss  0x8(%rsi),%xmm0
  mulss  0x4(%rsi),%xmm1
  subss  %xmm1,%xmm0
  movss  %xmm0,(%rdx)
  movss  0x8(%rdi),%xmm0
  movss  (%rdi),%xmm1
  mulss  (%rsi),%xmm0
  mulss  0x8(%rsi),%xmm1
  subss  %xmm1,%xmm0
  movss  %xmm0,0x4(%rdx)
  movss  (%rdi),%xmm0
  movss  0x4(%rdi),%xmm1
  mulss  0x4(%rsi),%xmm0
  mulss  (%rsi),%xmm1
  subss  %xmm1,%xmm0
  movss  %xmm0,0x8(%rdx)
  ret
```

assistant: ```c
void crossVectors(const float *u, const float *v, float *res) {
  res[0] = u[1] * v[2] - u[2] * v[1];
  res[1] = u[2] * v[0] - u[0] * v[2];
  res[2] = u[0] * v[1] - u[1] * v[0];
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<rq_statistics_histogram>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    %rcx,-0x40(%rbp)
  mov    %r8,-0x48(%rbp)
  mov    %r9,-0x50(%rbp)
  mov    -0x40(%rbp),%rax
  sub    $0x1,%eax
  mov    %eax,-0xc(%rbp)
  cmpq   $0x0,-0x40(%rbp)
  jne    L89
  mov    $0x1,%eax
  jmp    L112
L89:
  movl   $0x0,-0x14(%rbp)
  jmp    L113
L101:
  mov    -0x14(%rbp),%eax
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  movsd  %xmm0,-0x8(%rbp)
  mov    -0x38(%rbp),%rax
  movsd  (%rax),%xmm0
  comisd -0x8(%rbp),%xmm0
  jbe    L102
  mov    -0x50(%rbp),%rax
  mov    (%rax),%rax
  lea    0x1(%rax),%rdx
  mov    -0x50(%rbp),%rax
  mov    %rdx,(%rax)
  jmp    L67
L102:
  movl   $0x0,-0x10(%rbp)
  jmp    L125
L30:
  mov    -0x10(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jne    L76
  mov    -0x10(%rbp),%eax
  lea    0x0(,%rax,8),%rdx
  mov    -0x48(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  add    $0x1,%rdx
  mov    %rdx,(%rax)
  jmp    L75
L76:
  mov    -0x10(%rbp),%eax
  lea    0x0(,%rax,8),%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm1
  movsd  -0x8(%rbp),%xmm0
  comisd %xmm1,%xmm0
  jb     L75
  mov    -0x10(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,%eax
  lea    0x0(,%rax,8),%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  comisd -0x8(%rbp),%xmm0
  jbe    L75
  mov    -0x10(%rbp),%eax
  lea    0x0(,%rax,8),%rdx
  mov    -0x48(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  add    $0x1,%rdx
  mov    %rdx,(%rax)
  jmp    L67
L75:
  addl   $0x1,-0x10(%rbp)
L125:
  mov    -0x10(%rbp),%eax
  cmp    %rax,-0x40(%rbp)
  ja     L30
L67:
  addl   $0x1,-0x14(%rbp)
L113:
  mov    -0x14(%rbp),%eax
  cmp    %rax,-0x30(%rbp)
  ja     L101
  mov    $0x0,%eax
L112:
  pop    %rbp
  ret
```

assistant: ```c
short rq_statistics_histogram(double *inputs, unsigned long num_inputs, double *buckets, unsigned long num_buckets, unsigned long *counts, unsigned long *underflow) {
  unsigned int c;
  unsigned last_bucket_offset = num_buckets - 1;
  if (num_buckets == 0)
    return 1;
  for (c = 0; c < num_inputs; c++) {
    double val = inputs[c];
    unsigned int b;
    if (val < buckets[0])
      (*underflow)++;
    else {
      for (b = 0; b < num_buckets; b++) {
        if (b == last_bucket_offset)
          counts[b]++;
        else if (val >= buckets[b] && val < buckets[b + 1]) {
          counts[b]++;
          break;
        }
      }
    }
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<rq_statistics_histogram>:
  endbr64
  mov    $0x1,%eax
  test   %rcx,%rcx
  je     L83
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r11
  mov    %rsi,%rbx
  mov    %rdx,%rsi
  test   %rbx,%rbx
  je     L59
  lea    -0x1(%rcx),%edi
  mov    $0x0,%r10d
  mov    $0x0,%eax
  mov    $0x0,%r12d
  jmp    L100
L66:
  mov    $0x0,%eax
  mov    %r12,%rdx
  jmp    L50
L44:
  addq   $0x1,(%r8,%rdx,8)
L60:
  add    $0x1,%eax
  mov    %eax,%edx
  cmp    %rcx,%rdx
  jae    L35
L50:
  cmp    %eax,%edi
  je     L44
  lea    0x0(,%rdx,8),%rbp
  comisd (%rsi,%rdx,8),%xmm0
  jb     L60
  lea    0x1(%rax),%edx
  movsd  (%rsi,%rdx,8),%xmm1
  comisd %xmm0,%xmm1
  jbe    L60
  addq   $0x1,(%r8,%rbp,1)
L35:
  add    $0x1,%r10d
  mov    %r10d,%eax
  cmp    %rbx,%rax
  jae    L63
L100:
  movsd  (%r11,%rax,8),%xmm0
  movsd  (%rsi),%xmm1
  comisd %xmm0,%xmm1
  jbe    L66
  addq   $0x1,(%r9)
  jmp    L35
L63:
  mov    $0x0,%eax
L46:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L59:
  mov    $0x0,%eax
  jmp    L46
L83:
  ret
```

assistant: ```c
short rq_statistics_histogram(double *inputs, unsigned long num_inputs, double *buckets, unsigned long num_buckets, unsigned long *counts, unsigned long *underflow) {
  unsigned int c;
  unsigned last_bucket_offset = num_buckets - 1;
  if (num_buckets == 0)
    return 1;
  for (c = 0; c < num_inputs; c++) {
    double val = inputs[c];
    unsigned int b;
    if (val < buckets[0])
      (*underflow)++;
    else {
      for (b = 0; b < num_buckets; b++) {
        if (b == last_bucket_offset)
          counts[b]++;
        else if (val >= buckets[b] && val < buckets[b + 1]) {
          counts[b]++;
          break;
        }
      }
    }
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<rq_statistics_histogram>:
  endbr64
  test   %rcx,%rcx
  je     L47
  push   %r12
  mov    %rdx,%r10
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  mov    %rcx,%rdi
  push   %rbx
  test   %rsi,%rsi
  je     L75
  movsd  (%rdx),%xmm2
  lea    -0x1(%rcx),%r11d
  xor    %ebx,%ebx
  xor    %eax,%eax
  jmp    L118
L103:
  addq   $0x1,(%r9)
L6:
  lea    0x1(%rbx),%eax
  mov    %rax,%rbx
  cmp    %r12,%rax
  jae    L75
L118:
  movsd  0x0(%rbp,%rax,8),%xmm0
  comisd %xmm0,%xmm2
  ja     L103
  xor    %eax,%eax
  xor    %edx,%edx
  jmp    L38
  nopl   0x0(%rax)
L28:
  comisd (%r10,%rcx,1),%xmm0
  jb     L73
  movsd  (%r10,%rdx,8),%xmm1
  comisd %xmm0,%xmm1
  ja     L39
L73:
  cmp    %rdx,%rdi
  jbe    L6
L38:
  lea    0x0(,%rdx,8),%rcx
  mov    %eax,%esi
  lea    0x1(%rax),%edx
  mov    %rdx,%rax
  cmp    %esi,%r11d
  jne    L28
  addq   $0x1,(%r8,%rcx,1)
  jmp    L73
  nopl   (%rax)
L75:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L39:
  addq   $0x1,(%r8,%rcx,1)
  jmp    L6
  nopw   0x0(%rax,%rax,1)
L47:
  mov    $0x1,%eax
  ret
```

assistant: ```c
short rq_statistics_histogram(double *inputs, unsigned long num_inputs, double *buckets, unsigned long num_buckets, unsigned long *counts, unsigned long *underflow) {
  unsigned int c;
  unsigned last_bucket_offset = num_buckets - 1;
  if (num_buckets == 0)
    return 1;
  for (c = 0; c < num_inputs; c++) {
    double val = inputs[c];
    unsigned int b;
    if (val < buckets[0])
      (*underflow)++;
    else {
      for (b = 0; b < num_buckets; b++) {
        if (b == last_bucket_offset)
          counts[b]++;
        else if (val >= buckets[b] && val < buckets[b + 1]) {
          counts[b]++;
          break;
        }
      }
    }
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<rq_statistics_histogram>:
  endbr64
  test   %rcx,%rcx
  je     L44
  push   %r12
  mov    %rdx,%r10
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  mov    %rcx,%rdi
  push   %rbx
  test   %rsi,%rsi
  je     L103
  movsd  (%rdx),%xmm2
  lea    -0x1(%rcx),%r11d
  xor    %ebx,%ebx
  xor    %eax,%eax
  jmp    L62
L102:
  addq   $0x1,(%r9)
L10:
  lea    0x1(%rbx),%eax
  mov    %rax,%rbx
  cmp    %r12,%rax
  jae    L103
L62:
  movsd  0x0(%rbp,%rax,8),%xmm0
  comisd %xmm0,%xmm2
  ja     L102
  xor    %eax,%eax
  xor    %edx,%edx
  jmp    L79
  nopl   0x0(%rax)
L16:
  comisd (%r10,%rcx,1),%xmm0
  jb     L28
  movsd  (%r10,%rdx,8),%xmm1
  comisd %xmm0,%xmm1
  ja     L85
L28:
  cmp    %rdx,%rdi
  jbe    L10
L79:
  lea    0x0(,%rdx,8),%rcx
  mov    %eax,%esi
  lea    0x1(%rax),%edx
  mov    %rdx,%rax
  cmp    %esi,%r11d
  jne    L16
  addq   $0x1,(%r8,%rcx,1)
  jmp    L28
  nopl   (%rax)
L103:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L85:
  addq   $0x1,(%r8,%rcx,1)
  jmp    L10
  nopw   0x0(%rax,%rax,1)
L44:
  mov    $0x1,%eax
  ret
```

assistant: ```c
short rq_statistics_histogram(double *inputs, unsigned long num_inputs, double *buckets, unsigned long num_buckets, unsigned long *counts, unsigned long *underflow) {
  unsigned int c;
  unsigned last_bucket_offset = num_buckets - 1;
  if (num_buckets == 0)
    return 1;
  for (c = 0; c < num_inputs; c++) {
    double val = inputs[c];
    unsigned int b;
    if (val < buckets[0])
      (*underflow)++;
    else {
      for (b = 0; b < num_buckets; b++) {
        if (b == last_bucket_offset)
          counts[b]++;
        else if (val >= buckets[b] && val < buckets[b + 1]) {
          counts[b]++;
          break;
        }
      }
    }
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<maxSlidingWindow>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,%rbp
  mov    %esi,%r12d
  mov    %edx,%ebx
  mov    %rcx,0x10(%rsp)
  mov    %esi,%eax
  sub    %edx,%eax
  add    $0x1,%eax
  mov    %eax,0x1c(%rsp)
  lea    0x0(,%rax,4),%edi
  call   <malloc@plt>
  mov    %rax,0x8(%rsp)
  test   %r12d,%r12d
  jne    L94
L52:
  mov    0x10(%rsp),%rax
  mov    %r12d,(%rax)
  mov    0x8(%rsp),%rax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L94:
  lea    0x0(,%rbx,4),%edi
  call   <malloc@plt>
  mov    %rax,%r8
  test   %r12d,%r12d
  jle    L120
  mov    %ebx,%r9d
  neg    %r9d
  sub    %ebx,%r12d
  mov    $0x0,%r11d
  mov    $0x0,%r14d
  mov    $0x0,%esi
  mov    $0xffffffff,%edi
  mov    $0x0,%ecx
  lea    -0x1(%rbx),%r15d
  jmp    L133
L96:
  mov    %edx,%ecx
  test   %esi,%esi
  je     L3
  mov    0x0(%rbp,%r11,4),%r10d
  mov    %edi,%edx
L22:
  movslq %edx,%rax
  movslq (%r8,%rax,4),%rax
  cmp    %r10d,0x0(%rbp,%rax,4)
  jge    L2
  lea    -0x1(%rdx,%rbx,1),%eax
  cltd
  idiv   %ebx
  sub    $0x1,%esi
  jne    L22
  mov    %edx,%edi
  jmp    L3
L2:
  mov    %edx,%edi
L3:
  lea    0x1(%rdi),%eax
  cltd
  idiv   %ebx
  mov    %edx,%edi
  movslq %edx,%rdx
  mov    %r13d,(%r8,%rdx,4)
  add    $0x1,%esi
  cmp    %r11d,%r15d
  jg     L69
  movslq %ecx,%rax
  movslq (%r8,%rax,4),%rax
  mov    0x0(%rbp,%rax,4),%edx
  movslq %r14d,%rax
  mov    0x8(%rsp),%r10
  mov    %edx,(%r10,%rax,4)
  lea    0x1(%r14),%r14d
L69:
  add    $0x1,%r11
  add    $0x1,%r9d
  cmp    %r12d,%r9d
  je     L120
L133:
  mov    %r11d,%r13d
  mov    %ecx,%edx
  test   %esi,%esi
  je     L131
L59:
  movslq %edx,%rax
  cmp    %r9d,(%r8,%rax,4)
  jg     L96
  lea    0x1(%rdx),%eax
  cltd
  idiv   %ebx
  sub    $0x1,%esi
  jne    L59
  mov    %edx,%ecx
L131:
  mov    $0x0,%esi
  jmp    L3
L120:
  mov    %r8,%rdi
  call   <free@plt>
  mov    0x1c(%rsp),%r12d
  jmp    L52
```

assistant: ```c
int *maxSlidingWindow(int *nums, int numsSize, int k, int *returnSize) {
  int resSize = numsSize - k + 1;
  int *res = (int *)malloc(resSize * sizeof(int));
  if (numsSize == 0) {
    *returnSize = 0;
    return res;
  }
  int *can = (int *)malloc(k * sizeof(int));
  int first = 0;
  int last = -1;
  int size = 0;
  int r = 0;
  for (int i = 0; i < numsSize; i++) {
    while (size != 0 && can[first] < i - k + 1) {
      first = (first + 1) % k;
      size--;
    }
    while (size != 0 && nums[can[last]] < nums[i]) {
      last = (last - 1 + k) % k;
      size--;
    }
    last = (last + 1) % k;
    can[last] = i;
    size++;
    if (i >= k - 1) {
      res[r++] = nums[can[first]];
    }
  }
  free(can);
  *returnSize = resSize;
  return res;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<maxSlidingWindow>:
  endbr64
  push   %r15
  push   %r14
  mov    %esi,%r14d
  push   %r13
  sub    %edx,%r14d
  push   %r12
  lea    0x1(%r14),%eax
  mov    %esi,%r12d
  push   %rbp
  mov    %rdi,%rbp
  lea    0x0(,%rax,4),%edi
  push   %rbx
  mov    %edx,%ebx
  sub    $0x18,%rsp
  mov    %rcx,(%rsp)
  mov    %eax,0xc(%rsp)
  call   <malloc@plt>
  mov    %rax,%r13
  test   %r12d,%r12d
  jne    L13
L33:
  mov    (%rsp),%rax
  mov    %r12d,(%rax)
  add    $0x18,%rsp
  mov    %r13,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L13:
  lea    0x0(,%rbx,4),%edi
  call   <malloc@plt>
  mov    %rax,%r8
  test   %r12d,%r12d
  jle    L122
  mov    $0x1,%r9d
  lea    -0x1(%rbx),%r15d
  xor    %r11d,%r11d
  xor    %r12d,%r12d
  sub    %ebx,%r9d
  xor    %ecx,%ecx
  mov    $0xffffffff,%edi
  xor    %esi,%esi
  nopl   0x0(%rax)
L116:
  lea    0x1(%rdi),%eax
  add    $0x1,%esi
  cltd
  idiv   %ebx
  movslq %edx,%rax
  mov    %r11d,(%r8,%rax,4)
  mov    %rax,%rdi
  cmp    %r11d,%r15d
  jg     L95
  movslq %ecx,%rax
  movslq (%r8,%rax,4),%rax
  mov    0x0(%rbp,%rax,4),%edx
  movslq %r12d,%rax
  add    $0x1,%r12d
  mov    %edx,0x0(%r13,%rax,4)
L95:
  cmp    %r9d,%r14d
  je     L122
  mov    %ecx,%edx
  jmp    L109
L45:
  lea    0x1(%rdx),%eax
  cltd
  idiv   %ebx
  sub    $0x1,%esi
  je     L127
L109:
  movslq %edx,%rax
  cmp    %r9d,(%r8,%rax,4)
  jle    L45
  mov    %edx,%ecx
  mov    0x4(%rbp,%r11,4),%r10d
  mov    %edi,%edx
  jmp    L97
  nop
L102:
  lea    -0x1(%rdx,%rbx,1),%eax
  cltd
  idiv   %ebx
  sub    $0x1,%esi
  je     L106
L97:
  movslq %edx,%rax
  movslq (%r8,%rax,4),%rax
  cmp    %r10d,0x0(%rbp,%rax,4)
  jl     L102
L106:
  mov    %edx,%edi
L60:
  add    $0x1,%r11
  add    $0x1,%r9d
  jmp    L116
  cs nopw 0x0(%rax,%rax,1)
L127:
  mov    %edx,%ecx
  jmp    L60
  nopl   0x0(%rax)
L122:
  mov    %r8,%rdi
  call   <free@plt>
  mov    0xc(%rsp),%r12d
  jmp    L33
```

assistant: ```c
int *maxSlidingWindow(int *nums, int numsSize, int k, int *returnSize) {
  int resSize = numsSize - k + 1;
  int *res = (int *)malloc(resSize * sizeof(int));
  if (numsSize == 0) {
    *returnSize = 0;
    return res;
  }
  int *can = (int *)malloc(k * sizeof(int));
  int first = 0;
  int last = -1;
  int size = 0;
  int r = 0;
  for (int i = 0; i < numsSize; i++) {
    while (size != 0 && can[first] < i - k + 1) {
      first = (first + 1) % k;
      size--;
    }
    while (size != 0 && nums[can[last]] < nums[i]) {
      last = (last - 1 + k) % k;
      size--;
    }
    last = (last + 1) % k;
    can[last] = i;
    size++;
    if (i >= k - 1) {
      res[r++] = nums[can[first]];
    }
  }
  free(can);
  *returnSize = resSize;
  return res;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<maxSlidingWindow>:
  endbr64
  push   %r15
  push   %r14
  mov    %esi,%r14d
  push   %r13
  sub    %edx,%r14d
  push   %r12
  lea    0x1(%r14),%eax
  mov    %esi,%r12d
  push   %rbp
  mov    %rdi,%rbp
  lea    0x0(,%rax,4),%edi
  push   %rbx
  mov    %edx,%ebx
  sub    $0x18,%rsp
  mov    %rcx,(%rsp)
  mov    %eax,0xc(%rsp)
  call   <malloc@plt>
  mov    %rax,%r13
  test   %r12d,%r12d
  jne    L12
L30:
  mov    (%rsp),%rax
  mov    %r12d,(%rax)
  add    $0x18,%rsp
  mov    %r13,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L12:
  lea    0x0(,%rbx,4),%edi
  call   <malloc@plt>
  mov    %rax,%r8
  test   %r12d,%r12d
  jle    L40
  mov    $0x1,%r9d
  lea    -0x1(%rbx),%r15d
  xor    %r11d,%r11d
  xor    %r12d,%r12d
  sub    %ebx,%r9d
  xor    %ecx,%ecx
  mov    $0xffffffff,%edi
  xor    %esi,%esi
  nopl   0x0(%rax)
L4:
  lea    0x1(%rdi),%eax
  add    $0x1,%esi
  cltd
  idiv   %ebx
  movslq %edx,%rax
  mov    %r11d,(%r8,%rax,4)
  mov    %rax,%rdi
  cmp    %r11d,%r15d
  jg     L106
  movslq %ecx,%rax
  movslq (%r8,%rax,4),%rax
  mov    0x0(%rbp,%rax,4),%edx
  movslq %r12d,%rax
  add    $0x1,%r12d
  mov    %edx,0x0(%r13,%rax,4)
L106:
  cmp    %r9d,%r14d
  je     L40
  mov    %ecx,%edx
  jmp    L5
L98:
  lea    0x1(%rdx),%eax
  cltd
  idiv   %ebx
  sub    $0x1,%esi
  je     L48
L5:
  movslq %edx,%rax
  cmp    %r9d,(%r8,%rax,4)
  jle    L98
  mov    %edx,%ecx
  mov    0x4(%rbp,%r11,4),%r10d
  mov    %edi,%edx
  jmp    L36
  nop
L70:
  lea    -0x1(%rdx,%rbx,1),%eax
  cltd
  idiv   %ebx
  sub    $0x1,%esi
  je     L113
L36:
  movslq %edx,%rax
  movslq (%r8,%rax,4),%rax
  cmp    %r10d,0x0(%rbp,%rax,4)
  jl     L70
L113:
  mov    %edx,%edi
L27:
  add    $0x1,%r11
  add    $0x1,%r9d
  jmp    L4
  cs nopw 0x0(%rax,%rax,1)
L48:
  mov    %edx,%ecx
  jmp    L27
  nopl   0x0(%rax)
L40:
  mov    %r8,%rdi
  call   <free@plt>
  mov    0xc(%rsp),%r12d
  jmp    L30
```

assistant: ```c
int *maxSlidingWindow(int *nums, int numsSize, int k, int *returnSize) {
  int resSize = numsSize - k + 1;
  int *res = (int *)malloc(resSize * sizeof(int));
  if (numsSize == 0) {
    *returnSize = 0;
    return res;
  }
  int *can = (int *)malloc(k * sizeof(int));
  int first = 0;
  int last = -1;
  int size = 0;
  int r = 0;
  for (int i = 0; i < numsSize; i++) {
    while (size != 0 && can[first] < i - k + 1) {
      first = (first + 1) % k;
      size--;
    }
    while (size != 0 && nums[can[last]] < nums[i]) {
      last = (last - 1 + k) % k;
      size--;
    }
    last = (last + 1) % k;
    can[last] = i;
    size++;
    if (i >= k - 1) {
      res[r++] = nums[can[first]];
    }
  }
  free(can);
  *returnSize = resSize;
  return res;
}
```
user: Translate the below assembly code into C programming language:

```asm
<do_recvfile>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    -0x4000(%rsp),%r11
L17:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L17
  sub    $0x30,%rsp
  mov    %edi,-0x4024(%rbp)
  mov    %esi,-0x4028(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
L49:
  lea    -0x4010(%rbp),%rcx
  mov    -0x4024(%rbp),%eax
  mov    $0x4000,%edx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <do_rnd_read@plt>
  mov    %rax,-0x4018(%rbp)
  cmpq   $0x0,-0x4018(%rbp)
  jle    L19
  mov    -0x4018(%rbp),%rdx
  lea    -0x4010(%rbp),%rcx
  mov    -0x4028(%rbp),%eax
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <write@plt>
  cmp    %rax,-0x4018(%rbp)
  je     L15
  jmp    L4
L19:
  cmpq   $0x0,-0x4018(%rbp)
  jns    L15
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <perror@plt>
L15:
  cmpq   $0x0,-0x4018(%rbp)
  jg     L49
L4:
  mov    -0x4018(%rbp),%rax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L16
  call   <__stack_chk_fail@plt>
L16:
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D5
type_guessed: string

tool: D5: read

assistant: ```c
int do_recvfile(int infd, int outfd) {
  ssize_t r;
  do {
    char buf[16384];
    r = do_rnd_read(infd, buf, sizeof(buf));
    if (r > 0) {
      if (write(outfd, buf, r) != r)
        break;
    } else if (r < 0) {
      perror("read");
    }
  } while (r > 0);
  return (int)r;
}
```
user: What is the c source code of the assembly code below?

```asm
<do_recvfile>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0x4000(%rsp),%r11
L59:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L59
  sub    $0x18,%rsp
  mov    %edi,%r12d
  mov    %esi,%r13d
  mov    %fs:0x28,%rax
  mov    %rax,0x4008(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  jmp    L48
L3:
  mov    %rax,%rdx
  mov    %rbp,%rsi
  mov    %r13d,%edi
  call   <write@plt>
  cmp    %rbx,%rax
  jne    L40
L48:
  mov    $0x4000,%edx
  mov    %rbp,%rsi
  mov    %r12d,%edi
  call   <do_rnd_read@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  jg     L3
  js     L8
L40:
  mov    %ebx,%eax
  mov    0x4008(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L57
  add    $0x4018,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L8:
  lea    D3(%rip),%rdi
  call   <perror@plt>
  jmp    L40
L57:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D3
data_type: string

tool: D3: read

assistant: ```c
int do_recvfile(int infd, int outfd) {
  ssize_t r;
  do {
    char buf[16384];
    r = do_rnd_read(infd, buf, sizeof(buf));
    if (r > 0) {
      if (write(outfd, buf, r) != r)
        break;
    } else if (r < 0) {
      perror("read");
    }
  } while (r > 0);
  return (int)r;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<do_recvfile>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0x4000(%rsp),%r11
L69:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L69
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x4008(%rsp)
  xor    %eax,%eax
  mov    %edi,%r12d
  mov    %esi,%r13d
  mov    %rsp,%rbp
  nopl   0x0(%rax,%rax,1)
L19:
  mov    $0x4000,%edx
  mov    %rbp,%rsi
  mov    %r12d,%edi
  call   <do_rnd_read@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  jg     L13
  jne    L54
L29:
  mov    %ebx,%eax
  mov    0x4008(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L14
  add    $0x4018,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   (%rax)
L13:
  mov    %rax,%rdx
  mov    %rbp,%rsi
  mov    %r13d,%edi
  call   <write@plt>
  cmp    %rbx,%rax
  je     L19
  jmp    L29
  nopl   (%rax)
L54:
  lea    D4(%rip),%rdi
  call   <perror@plt>
  jmp    L29
L14:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D4
type: string

tool: D4: read

assistant: ```c
int do_recvfile(int infd, int outfd) {
  ssize_t r;
  do {
    char buf[16384];
    r = do_rnd_read(infd, buf, sizeof(buf));
    if (r > 0) {
      if (write(outfd, buf, r) != r)
        break;
    } else if (r < 0) {
      perror("read");
    }
  } while (r > 0);
  return (int)r;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<do_recvfile>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0x4000(%rsp),%r11
L49:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L49
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x4008(%rsp)
  xor    %eax,%eax
  mov    %edi,%r12d
  mov    %esi,%r13d
  mov    %rsp,%rbp
  nopl   0x0(%rax,%rax,1)
L32:
  mov    $0x4000,%edx
  mov    %rbp,%rsi
  mov    %r12d,%edi
  call   <do_rnd_read@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  jg     L42
  jne    L23
L31:
  mov    %ebx,%eax
  mov    0x4008(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L28
  add    $0x4018,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   (%rax)
L42:
  mov    %rax,%rdx
  mov    %rbp,%rsi
  mov    %r13d,%edi
  call   <write@plt>
  cmp    %rbx,%rax
  je     L32
  jmp    L31
  nopl   (%rax)
L23:
  lea    D0(%rip),%rdi
  call   <perror@plt>
  jmp    L31
L28:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D0
type: string

tool: D0: read

assistant: ```c
int do_recvfile(int infd, int outfd) {
  ssize_t r;
  do {
    char buf[16384];
    r = do_rnd_read(infd, buf, sizeof(buf));
    if (r > 0) {
      if (write(outfd, buf, r) != r)
        break;
    } else if (r < 0) {
      perror("read");
    }
  } while (r > 0);
  return (int)r;
}
```