ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 这段汇编代码用C语言表示是什么样的：

```asm
<jacexpfunc>:
  endbr64
  test   %ecx,%ecx
  jle    L5
  push   %r13
  mov    %rdi,%r13
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  mov    %ecx,%ebp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x18,%rsp
  nopl   0x0(%rax)
L14:
  pxor   %xmm2,%xmm2
  movsd  0x8(%r13),%xmm0
  add    $0x18,%rbx
  xorpd  D16(%rip),%xmm0
  cvtsi2sd %r12d,%xmm2
  add    $0x1,%r12d
  mulsd  %xmm2,%xmm0
  movsd  %xmm2,0x8(%rsp)
  call   <exp@plt>
  movsd  0x8(%rsp),%xmm2
  movsd  %xmm0,-0x18(%rbx)
  movsd  0x0(%r13),%xmm0
  xorpd  D16(%rip),%xmm0
  movsd  0x8(%r13),%xmm1
  xorpd  D16(%rip),%xmm1
  mulsd  %xmm2,%xmm0
  movsd  %xmm0,0x8(%rsp)
  movapd %xmm1,%xmm0
  mulsd  %xmm2,%xmm0
  call   <exp@plt>
  mulsd  0x8(%rsp),%xmm0
  mov    D37(%rip),%rax
  mov    %rax,-0x8(%rbx)
  movsd  %xmm0,-0x10(%rbx)
  cmp    %r12d,%ebp
  jne    L14
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  xchg   %ax,%ax
L5:
  ret
```

tool_calls: name: read_data
arguments: tag: D16
type_infered: byte[16]
name: read_data
arguments: tag: D37
type_infered: f64

tool: {"D16": [0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x80,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00]}

tool: D37: 1

assistant: ```c
void jacexpfunc(double *p, double *jac, int m, int n, void *data) {
  register int i, j;
  for (i = j = 0; i < n; ++i) {
    jac[j++] = exp(-p[1] * i);
    jac[j++] = -p[0] * i * exp(-p[1] * i);
    jac[j++] = 1.0;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<adts_hdr_up>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    -0x1c(%rbp),%eax
  add    $0x7,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  sar    $0xb,%eax
  and    $0x3,%eax
  mov    %eax,%ecx
  mov    -0x18(%rbp),%rax
  add    $0x5,%rax
  movzbl (%rax),%eax
  and    $0xfffffffc,%eax
  mov    %eax,%edx
  mov    -0x18(%rbp),%rax
  add    $0x3,%rax
  or     %ecx,%edx
  mov    %dl,(%rax)
  mov    -0x4(%rbp),%eax
  sar    $0x3,%eax
  mov    %eax,%edx
  mov    -0x18(%rbp),%rax
  add    $0x4,%rax
  mov    %dl,(%rax)
  mov    -0x4(%rbp),%eax
  shl    $0x5,%eax
  mov    %eax,%ecx
  mov    -0x18(%rbp),%rax
  add    $0x5,%rax
  movzbl (%rax),%eax
  and    $0x1f,%eax
  mov    %eax,%edx
  mov    -0x18(%rbp),%rax
  add    $0x5,%rax
  or     %ecx,%edx
  mov    %dl,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void adts_hdr_up(char *adts, int size) {
  int len = size + 7;
  adts[3] = ((len >> 11) & 0x03) | (adts[5] & 0xfc);
  ;
  adts[4] = (len >> 3) & 0xff;
  adts[5] = ((len & 7) << 5) | (adts[5] & 0x1f);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<adts_hdr_up>:
  endbr64
  add    $0x7,%esi
  movzbl 0x5(%rdi),%eax
  mov    %esi,%edx
  sar    $0xb,%edx
  and    $0x3,%edx
  mov    %eax,%ecx
  and    $0xfffffffc,%ecx
  or     %ecx,%edx
  mov    %dl,0x3(%rdi)
  mov    %esi,%edx
  sar    $0x3,%edx
  mov    %dl,0x4(%rdi)
  shl    $0x5,%esi
  and    $0x1f,%eax
  or     %esi,%eax
  mov    %al,0x5(%rdi)
  ret
```

assistant: ```c
void adts_hdr_up(char *adts, int size) {
  int len = size + 7;
  adts[3] = ((len >> 11) & 0x03) | (adts[5] & 0xfc);
  ;
  adts[4] = (len >> 3) & 0xff;
  adts[5] = ((len & 7) << 5) | (adts[5] & 0x1f);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<adts_hdr_up>:
  endbr64
  movzbl 0x5(%rdi),%eax
  add    $0x7,%esi
  mov    %esi,%edx
  mov    %eax,%ecx
  sar    $0xb,%edx
  and    $0x1f,%eax
  and    $0x3,%edx
  and    $0xfffffffc,%ecx
  or     %ecx,%edx
  mov    %dl,0x3(%rdi)
  mov    %esi,%edx
  shl    $0x5,%esi
  sar    $0x3,%edx
  or     %esi,%eax
  mov    %dl,0x4(%rdi)
  mov    %al,0x5(%rdi)
  ret
```

assistant: ```c
void adts_hdr_up(char *adts, int size) {
  int len = size + 7;
  adts[3] = ((len >> 11) & 0x03) | (adts[5] & 0xfc);
  ;
  adts[4] = (len >> 3) & 0xff;
  adts[5] = ((len & 7) << 5) | (adts[5] & 0x1f);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<adts_hdr_up>:
  endbr64
  movzbl 0x5(%rdi),%eax
  add    $0x7,%esi
  mov    %esi,%edx
  mov    %eax,%ecx
  sar    $0xb,%edx
  and    $0x1f,%eax
  and    $0x3,%edx
  and    $0xfffffffc,%ecx
  or     %ecx,%edx
  mov    %dl,0x3(%rdi)
  mov    %esi,%edx
  shl    $0x5,%esi
  sar    $0x3,%edx
  or     %esi,%eax
  mov    %dl,0x4(%rdi)
  mov    %al,0x5(%rdi)
  ret
```

assistant: ```c
void adts_hdr_up(char *adts, int size) {
  int len = size + 7;
  adts[3] = ((len >> 11) & 0x03) | (adts[5] & 0xfc);
  ;
  adts[4] = (len >> 3) & 0xff;
  adts[5] = ((len & 7) << 5) | (adts[5] & 0x1f);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<findMin>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x10(%rbp)
  mov    -0x1c(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0xc(%rbp)
  mov    D7(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,-0x8(%rbp)
  jmp    L4
L25:
  mov    -0xc(%rbp),%eax
  sub    -0x10(%rbp),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  mov    %eax,%edx
  mov    -0x10(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    -0x8(%rbp),%edx
  cmp    %eax,%edx
  cmovle %edx,%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%eax
  cmp    %eax,%edx
  jle    L36
  mov    -0x4(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0x10(%rbp)
  jmp    L4
L36:
  mov    -0x4(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0xc(%rbp)
L4:
  mov    -0x10(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jle    L25
  mov    -0x8(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int findMin(int *nums, int numsSize) {
  int lo = 0;
  int hi = numsSize - 1;
  int min = INT_MAX;
  while (lo <= hi) {
    int mid = lo + (hi - lo) / 2;
    min = min < nums[mid] ? min : nums[mid];
    if (nums[mid] > nums[hi]) {
      lo = mid + 1;
    } else {
      hi = mid - 1;
    }
  }
  return min;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<findMin>:
  endbr64
  mov    D4(%rip),%rax
  mov    (%rax),%r8d
  sub    $0x1,%esi
  js     L43
  mov    $0x0,%ecx
  jmp    L27
L31:
  lea    -0x1(%rax),%esi
L45:
  cmp    %esi,%ecx
  jg     L43
L27:
  mov    %esi,%edx
  sub    %ecx,%edx
  mov    %edx,%eax
  shr    $0x1f,%eax
  add    %edx,%eax
  sar    %eax
  add    %ecx,%eax
  movslq %eax,%rdx
  mov    (%rdi,%rdx,4),%edx
  cmp    %edx,%r8d
  cmovg  %edx,%r8d
  movslq %esi,%r9
  cmp    (%rdi,%r9,4),%edx
  jle    L31
  lea    0x1(%rax),%ecx
  jmp    L45
L43:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int findMin(int *nums, int numsSize) {
  int lo = 0;
  int hi = numsSize - 1;
  int min = INT_MAX;
  while (lo <= hi) {
    int mid = lo + (hi - lo) / 2;
    min = min < nums[mid] ? min : nums[mid];
    if (nums[mid] > nums[hi]) {
      lo = mid + 1;
    } else {
      hi = mid - 1;
    }
  }
  return min;
}
```
user: What is the c source code of the assembly code below?

```asm
<findMin>:
  endbr64
  mov    D8(%rip),%rax
  mov    (%rax),%r8d
  sub    $0x1,%esi
  js     L36
  xor    %ecx,%ecx
  jmp    L5
  nopw   0x0(%rax,%rax,1)
L47:
  lea    0x1(%rax),%ecx
  cmp    %ecx,%esi
  jl     L36
L5:
  mov    %esi,%eax
  movslq %esi,%r9
  sub    %ecx,%eax
  sar    %eax
  add    %ecx,%eax
  movslq %eax,%rdx
  mov    (%rdi,%rdx,4),%edx
  cmp    %edx,%r8d
  cmovg  %edx,%r8d
  cmp    (%rdi,%r9,4),%edx
  jg     L47
  lea    -0x1(%rax),%esi
  cmp    %ecx,%esi
  jge    L5
L36:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int findMin(int *nums, int numsSize) {
  int lo = 0;
  int hi = numsSize - 1;
  int min = INT_MAX;
  while (lo <= hi) {
    int mid = lo + (hi - lo) / 2;
    min = min < nums[mid] ? min : nums[mid];
    if (nums[mid] > nums[hi]) {
      lo = mid + 1;
    } else {
      hi = mid - 1;
    }
  }
  return min;
}
```
user: Translate the below assembly code into C programming language:

```asm
<findMin>:
  endbr64
  mov    D1(%rip),%rax
  mov    (%rax),%r8d
  sub    $0x1,%esi
  js     L31
  xor    %ecx,%ecx
  nopl   (%rax)
L10:
  movslq %esi,%r9
L34:
  mov    %esi,%eax
  sub    %ecx,%eax
  sar    %eax
  add    %ecx,%eax
  movslq %eax,%rdx
  mov    (%rdi,%rdx,4),%edx
  cmp    %edx,%r8d
  cmovg  %edx,%r8d
  cmp    %edx,(%rdi,%r9,4)
  jge    L25
  lea    0x1(%rax),%ecx
  cmp    %esi,%ecx
  jle    L34
L31:
  mov    %r8d,%eax
  ret
  nopl   0x0(%rax)
L25:
  lea    -0x1(%rax),%esi
  cmp    %ecx,%esi
  jge    L10
  mov    %r8d,%eax
  ret
```

assistant: ```c
int findMin(int *nums, int numsSize) {
  int lo = 0;
  int hi = numsSize - 1;
  int min = INT_MAX;
  while (lo <= hi) {
    int mid = lo + (hi - lo) / 2;
    min = min < nums[mid] ? min : nums[mid];
    if (nums[mid] > nums[hi]) {
      lo = mid + 1;
    } else {
      hi = mid - 1;
    }
  }
  return min;
}
```
user: What is the c source code of the assembly code below?

```asm
<numcmp>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x8(%rbp),%rax
  cmp    -0x10(%rbp),%rax
  jge    L23
  mov    $0xffffffff,%eax
  jmp    L21
L23:
  mov    -0x8(%rbp),%rax
  cmp    -0x10(%rbp),%rax
  jle    L1
  mov    $0x1,%eax
  jmp    L21
L1:
  mov    $0x0,%eax
L21:
  pop    %rbp
  ret
```

assistant: ```c
int numcmp(const long long a, const long long b) {
  if (a < b)
    return -1;
  if (a > b)
    return 1;
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<numcmp>:
  endbr64
  mov    $0xffffffff,%eax
  cmp    %rsi,%rdi
  jl     L8
  setg   %al
  movzbl %al,%eax
L8:
  ret
```

assistant: ```c
int numcmp(const long long a, const long long b) {
  if (a < b)
    return -1;
  if (a > b)
    return 1;
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<numcmp>:
  endbr64
  xor    %eax,%eax
  cmp    %rsi,%rdi
  mov    $0xffffffff,%edx
  setg   %al
  cmovl  %edx,%eax
  ret
```

assistant: ```c
int numcmp(const long long a, const long long b) {
  if (a < b)
    return -1;
  if (a > b)
    return 1;
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<numcmp>:
  endbr64
  xor    %eax,%eax
  cmp    %rsi,%rdi
  mov    $0xffffffff,%edx
  setg   %al
  cmovl  %edx,%eax
  ret
```

assistant: ```c
int numcmp(const long long a, const long long b) {
  if (a < b)
    return -1;
  if (a > b)
    return 1;
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<ft_putchar_fd>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,%eax
  mov    %esi,-0x8(%rbp)
  mov    %al,-0x4(%rbp)
  lea    -0x4(%rbp),%rcx
  mov    -0x8(%rbp),%eax
  mov    $0x1,%edx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <write@plt>
  nop
  leave
  ret
```

assistant: ```c
void ft_putchar_fd(char c, int fd) {
  write(fd, &c, 1);
  return;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<ft_putchar_fd>:
  endbr64
  sub    $0x18,%rsp
  mov    %edi,%eax
  mov    %esi,%edi
  mov    %al,0xc(%rsp)
  lea    0xc(%rsp),%rsi
  mov    $0x1,%edx
  call   <write@plt>
  add    $0x18,%rsp
  ret
```

assistant: ```c
void ft_putchar_fd(char c, int fd) {
  write(fd, &c, 1);
  return;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<ft_putchar_fd>:
  endbr64
  sub    $0x18,%rsp
  mov    %edi,%r8d
  mov    $0x1,%edx
  mov    %esi,%edi
  lea    0xc(%rsp),%rsi
  mov    %r8b,0xc(%rsp)
  call   <write@plt>
  add    $0x18,%rsp
  ret
```

assistant: ```c
void ft_putchar_fd(char c, int fd) {
  write(fd, &c, 1);
  return;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ft_putchar_fd>:
  endbr64
  sub    $0x18,%rsp
  mov    %edi,%r8d
  mov    $0x1,%edx
  mov    %esi,%edi
  lea    0xc(%rsp),%rsi
  mov    %r8b,0xc(%rsp)
  call   <write@plt>
  add    $0x18,%rsp
  ret
```

assistant: ```c
void ft_putchar_fd(char c, int fd) {
  write(fd, &c, 1);
  return;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<bubble_sort>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L16
L41:
  movl   $0x0,-0x8(%rbp)
  jmp    L19
L24:
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x8(%rbp),%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,4),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%eax
  cmp    %eax,%edx
  jle    L8
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x8(%rbp),%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x8(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  mov    -0x8(%rbp),%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  mov    %eax,(%rdx)
L8:
  addl   $0x1,-0x8(%rbp)
L19:
  mov    -0x1c(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0x8(%rbp)
  jl     L24
  addl   $0x1,-0xc(%rbp)
L16:
  mov    -0xc(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L41
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void bubble_sort(int *collection, int length) {
  int temp;
  for (int _ = 0; _ < length; _++) {
    for (int index = 0; index < length - 1; index++) {
      if (*(collection + index) > *(collection + index + 1)) {
        temp = *(collection + index);
        *(collection + index) = *(collection + index + 1);
        *(collection + index + 1) = temp;
      }
    }
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<bubble_sort>:
  endbr64
  test   %esi,%esi
  jle    L22
  lea    -0x2(%rsi),%eax
  lea    0x4(%rdi,%rax,4),%r8
  mov    $0x0,%r9d
  jmp    L9
L33:
  add    $0x4,%rax
  cmp    %r8,%rax
  je     L64
L37:
  mov    (%rax),%edx
  mov    0x4(%rax),%ecx
  cmp    %ecx,%edx
  jle    L33
  mov    %ecx,(%rax)
  mov    %edx,0x4(%rax)
  jmp    L33
L64:
  add    $0x1,%r9d
  cmp    %r9d,%esi
  je     L22
L9:
  mov    %rdi,%rax
  cmp    $0x1,%esi
  jg     L37
  jmp    L64
L22:
  ret
```

assistant: ```c
void bubble_sort(int *collection, int length) {
  int temp;
  for (int _ = 0; _ < length; _++) {
    for (int index = 0; index < length - 1; index++) {
      if (*(collection + index) > *(collection + index + 1)) {
        temp = *(collection + index);
        *(collection + index) = *(collection + index + 1);
        *(collection + index + 1) = temp;
      }
    }
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<bubble_sort>:
  endbr64
  test   %esi,%esi
  jle    L15
  lea    -0x2(%rsi),%eax
  xor    %r9d,%r9d
  lea    0x4(%rdi,%rax,4),%r8
  nopl   0x0(%rax,%rax,1)
L5:
  mov    %rdi,%rax
  cmp    $0x1,%esi
  je     L19
L47:
  mov    (%rax),%edx
  mov    0x4(%rax),%ecx
  cmp    %ecx,%edx
  jle    L23
  mov    %ecx,(%rax)
  mov    %edx,0x4(%rax)
L23:
  add    $0x4,%rax
  cmp    %r8,%rax
  jne    L47
L19:
  add    $0x1,%r9d
  cmp    %r9d,%esi
  jne    L5
L15:
  ret
```

assistant: ```c
void bubble_sort(int *collection, int length) {
  int temp;
  for (int _ = 0; _ < length; _++) {
    for (int index = 0; index < length - 1; index++) {
      if (*(collection + index) > *(collection + index + 1)) {
        temp = *(collection + index);
        *(collection + index) = *(collection + index + 1);
        *(collection + index + 1) = temp;
      }
    }
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<bubble_sort>:
  endbr64
  cmp    $0x1,%esi
  jle    L10
  lea    -0x2(%rsi),%eax
  xor    %r9d,%r9d
  lea    0x4(%rdi,%rax,4),%r8
  nopl   0x0(%rax)
L2:
  mov    %rdi,%rdx
  nopl   0x0(%rax,%rax,1)
L30:
  movq   (%rdx),%xmm0
  pshufd $0xe5,%xmm0,%xmm2
  movd   %xmm0,%ecx
  movd   %xmm2,%eax
  pshufd $0xe1,%xmm0,%xmm1
  cmp    %eax,%ecx
  jle    L34
  movq   %xmm1,(%rdx)
L34:
  add    $0x4,%rdx
  cmp    %r8,%rdx
  jne    L30
  add    $0x1,%r9d
  cmp    %r9d,%esi
  jne    L2
L10:
  ret
```

assistant: ```c
void bubble_sort(int *collection, int length) {
  int temp;
  for (int _ = 0; _ < length; _++) {
    for (int index = 0; index < length - 1; index++) {
      if (*(collection + index) > *(collection + index + 1)) {
        temp = *(collection + index);
        *(collection + index) = *(collection + index + 1);
        *(collection + index + 1) = temp;
      }
    }
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<newton>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  cmpl   $0x0,-0x18(%rbp)
  je     L9
  mov    -0x18(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jne    L22
L9:
  mov    $0x1,%eax
  jmp    L27
L22:
  mov    -0x18(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    -0x14(%rbp),%eax
  sub    $0x1,%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <newton>
  mov    %eax,%ebx
  mov    -0x14(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    -0x18(%rbp),%eax
  mov    %eax,%esi
  mov    %edx,%edi
  call   <newton>
  add    %ebx,%eax
L27:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int newton(int n, int k) {
  if (k == 0 || k == n)
    return 1;
  else
    return newton(n - 1, k - 1) + newton(n - 1, k);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<newton>:
  endbr64
  test   %esi,%esi
  je     L16
  push   %r12
  push   %rbp
  push   %rbx
  mov    %esi,%ebx
  mov    $0x1,%eax
  cmp    %edi,%esi
  jne    L18
L29:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L18:
  lea    -0x1(%rdi),%ebp
  lea    -0x1(%rsi),%esi
  mov    %ebp,%edi
  call   <newton>
  mov    %eax,%r12d
  mov    %ebx,%esi
  mov    %ebp,%edi
  call   <newton>
  add    %r12d,%eax
  jmp    L29
L16:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int newton(int n, int k) {
  if (k == 0 || k == n)
    return 1;
  else
    return newton(n - 1, k - 1) + newton(n - 1, k);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<strtok>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x10(%rbp),%rcx
  mov    -0x8(%rbp),%rax
  lea    D3(%rip),%rdx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strtok_r@plt>
  leave
  ret
```

assistant: ```c
char *strtok(char *string, const char *control) {
  static char *nexttoken = 0;
  return strtok_r(string, control, &nexttoken);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<strtok>:
  endbr64
  sub    $0x8,%rsp
  lea    D6(%rip),%rdx
  call   <strtok_r@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
char *strtok(char *string, const char *control) {
  static char *nexttoken = 0;
  return strtok_r(string, control, &nexttoken);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<strtok>:
  endbr64
  lea    D6(%rip),%rdx
  jmp    1050 <strtok_r@plt>
```

assistant: ```c
char *strtok(char *string, const char *control) {
  static char *nexttoken = 0;
  return strtok_r(string, control, &nexttoken);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<strtok>:
  endbr64
  lea    D9(%rip),%rdx
  jmp    1050 <strtok_r@plt>
```

assistant: ```c
char *strtok(char *string, const char *control) {
  static char *nexttoken = 0;
  return strtok_r(string, control, &nexttoken);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x10(%rbp)
  movl   $0x0,-0xc(%rbp)
  lea    -0x14(%rbp),%rax
  mov    %rax,%rsi
  lea    D33(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  jmp    L10
L20:
  mov    -0x14(%rbp),%ecx
  movslq %ecx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %ecx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  sub    %eax,%ecx
  mov    %ecx,%edx
  add    %edx,-0x10(%rbp)
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x14(%rbp)
  addl   $0x1,-0xc(%rbp)
L10:
  mov    -0x14(%rbp),%eax
  test   %eax,%eax
  jg     L20
  pxor   %xmm0,%xmm0
  cvtsi2ssl -0x10(%rbp),%xmm0
  pxor   %xmm2,%xmm2
  cvtsi2ssl -0xc(%rbp),%xmm2
  movaps %xmm0,%xmm1
  divss  %xmm2,%xmm1
  movss  D16(%rip),%xmm0
  comiss %xmm1,%xmm0
  jbe    L41
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L13
L41:
  lea    D25(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L13:
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L15
  call   <__stack_chk_fail@plt>
L15:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: label: D33
type_infered: string
name: disassemble_data
arguments: label: D16
type_infered: byte[4]
name: disassemble_data
arguments: label: D0
type_infered: string
name: disassemble_data
arguments: label: D25
type_infered: string

tool: D33: %d

tool: {"D16": [0x00,0x00,0xe0,0x40]}

tool: D0: light


tool: D25: heavy


assistant: ```c
int main() {
  int number, sum = 0, total = 0;
  scanf("%d", &number);
  for (; number > 0; number /= 10, total += 1) {
    sum += number % 10;
  }
  if ((float)sum / total < 7)
    printf("light\n");
  else
    printf("heavy\n");
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  lea    D17(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%edx
  test   %edx,%edx
  jle    L18
  mov    $0x0,%esi
  mov    $0x0,%edi
L50:
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  lea    (%rax,%rax,4),%ecx
  add    %ecx,%ecx
  mov    %edx,%r8d
  sub    %ecx,%r8d
  add    %r8d,%edi
  mov    %edx,%ecx
  mov    %eax,%edx
  add    $0x1,%esi
  cmp    $0x9,%ecx
  jg     L50
  mov    %eax,0x4(%rsp)
L28:
  pxor   %xmm0,%xmm0
  cvtsi2ss %edi,%xmm0
  pxor   %xmm1,%xmm1
  cvtsi2ss %esi,%xmm1
  divss  %xmm1,%xmm0
  movss  D14(%rip),%xmm1
  comiss %xmm0,%xmm1
  jbe    L9
  lea    D30(%rip),%rdi
  call   <printf@plt>
L3:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  mov    $0x0,%eax
  add    $0x18,%rsp
  ret
L18:
  mov    $0x0,%esi
  mov    $0x0,%edi
  jmp    L28
L9:
  lea    D38(%rip),%rdi
  call   <printf@plt>
  jmp    L3
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D17
type: string
name: disassemble_data
arguments: label: D14
type: byte[4]
name: disassemble_data
arguments: label: D30
type: string
name: disassemble_data
arguments: label: D38
type: string

tool: D17: %d

tool: {"D14": [0x00,0x00,0xe0,0x40]}

tool: D30: light


tool: D38: heavy


assistant: ```c
int main() {
  int number, sum = 0, total = 0;
  scanf("%d", &number);
  for (; number > 0; number /= 10, total += 1) {
    sum += number % 10;
  }
  if ((float)sum / total < 7)
    printf("light\n");
  else
    printf("heavy\n");
}
```