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

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  movaps D13(%rip),%xmm0
  mov    $0x6,%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    D8(%rip),%rax
  mov    %rsp,%rdi
  movaps %xmm0,(%rsp)
  mov    %rax,0x10(%rsp)
  call   <es9@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  xor    %eax,%eax
  add    $0x28,%rsp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D13
type_guessed: f32[4]
name: unpack_data
arguments: data_tag: D8
type_guessed: f32[2]

tool: {"D13": [35.400001525878906,46.70000076293945,77.55000305175781,11.100000381469727]}

tool: {"D8": [9.039999961853027,0.75]}

assistant: ```c
int main() {
  float array[] = {35.4, 46.7, 77.55, 11.1, 9.04, 0.75};
  es9(array, sizeof(array) / sizeof(array[0]));
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<prac3>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x0,-0x10(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L17
L8:
  movl   $0x0,-0x4(%rbp)
  mov    -0xc(%rbp),%eax
  mov    %eax,-0x8(%rbp)
  jmp    L40
L9:
  mov    -0x4(%rbp),%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  add    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  cmp    -0x10(%rbp),%eax
  jle    L55
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <is_prime@plt>
  test   %rax,%rax
  je     L55
  mov    -0x4(%rbp),%eax
  mov    %eax,-0x10(%rbp)
L55:
  addl   $0x1,-0x8(%rbp)
L40:
  mov    -0x8(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L9
  addl   $0x1,-0xc(%rbp)
L17:
  mov    -0xc(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L8
  mov    -0x10(%rbp),%eax
  leave
  ret
```

assistant: ```c
int prac3(const int n, const int arr[]) {
  int num = 0;
  int i, j;
  for (i = 0; i < n; ++i) {
    int t = 0;
    for (j = i; j < n; ++j) {
      t *= 10;
      t += arr[j];
      if (t > num && is_prime(t))
        num = t;
    }
  }
  return num;
}
```
user: What is the c source code of the assembly code below:

```asm
<prac3>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rsi,0x8(%rsp)
  test   %edi,%edi
  jle    L62
  mov    %edi,%r15d
  lea    -0x1(%rdi),%eax
  mov    %eax,0x14(%rsp)
  mov    %edi,%eax
  mov    %rax,(%rsp)
  mov    $0x0,%r14d
  mov    $0x0,%r12d
  lea    0x4(%rsi),%rax
  mov    %rax,0x18(%rsp)
  jmp    L23
L0:
  add    $0x4,%rbp
  cmp    %r13,%rbp
  je     L79
L63:
  lea    (%rbx,%rbx,4),%edx
  mov    0x0(%rbp),%eax
  lea    (%rax,%rdx,2),%ebx
  cmp    %r12d,%ebx
  jle    L0
  mov    %ebx,%edi
  call   <is_prime@plt>
  test   %rax,%rax
  cmovne %ebx,%r12d
  jmp    L0
L79:
  add    $0x1,%r14
  cmp    (%rsp),%r14
  je     L5
L23:
  cmp    %r14d,%r15d
  jle    L79
  mov    0x8(%rsp),%rax
  lea    (%rax,%r14,4),%rbp
  mov    0x14(%rsp),%eax
  sub    %r14d,%eax
  add    %r14,%rax
  mov    0x18(%rsp),%rcx
  lea    (%rcx,%rax,4),%r13
  mov    $0x0,%ebx
  jmp    L63
L62:
  mov    $0x0,%r12d
L5:
  mov    %r12d,%eax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
int prac3(const int n, const int arr[]) {
  int num = 0;
  int i, j;
  for (i = 0; i < n; ++i) {
    int t = 0;
    for (j = i; j < n; ++j) {
      t *= 10;
      t += arr[j];
      if (t > num && is_prime(t))
        num = t;
    }
  }
  return num;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<prac3>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  test   %edi,%edi
  jle    L48
  movslq %edi,%rax
  mov    %edi,%r15d
  mov    %rsi,%r13
  xor    %r12d,%r12d
  mov    %rax,0x8(%rsp)
  xor    %r14d,%r14d
  nopw   0x0(%rax,%rax,1)
L39:
  mov    %r12,%rbp
  xor    %ebx,%ebx
  jmp    L32
  nopw   0x0(%rax,%rax,1)
L23:
  add    $0x1,%rbp
  cmp    %ebp,%r15d
  jle    L9
L32:
  mov    0x0(%r13,%rbp,4),%eax
  lea    (%rbx,%rbx,4),%edx
  lea    (%rax,%rdx,2),%ebx
  cmp    %r14d,%ebx
  jle    L23
  mov    %ebx,%edi
  call   <is_prime@plt>
  test   %rax,%rax
  cmovne %ebx,%r14d
  add    $0x1,%rbp
  cmp    %ebp,%r15d
  jg     L32
L9:
  add    $0x1,%r12
  cmp    %r12,0x8(%rsp)
  jne    L39
L1:
  add    $0x18,%rsp
  mov    %r14d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L48:
  xor    %r14d,%r14d
  jmp    L1
```

assistant: ```c
int prac3(const int n, const int arr[]) {
  int num = 0;
  int i, j;
  for (i = 0; i < n; ++i) {
    int t = 0;
    for (j = i; j < n; ++j) {
      t *= 10;
      t += arr[j];
      if (t > num && is_prime(t))
        num = t;
    }
  }
  return num;
}
```
user: What is the c source code of the assembly code below:

```asm
<prac3>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  test   %edi,%edi
  jle    L51
  movslq %edi,%rax
  mov    %edi,%r15d
  mov    %rsi,%r13
  xor    %r12d,%r12d
  mov    %rax,0x8(%rsp)
  xor    %r14d,%r14d
  nopw   0x0(%rax,%rax,1)
L0:
  mov    %r12,%rbp
  xor    %ebx,%ebx
  jmp    L64
  nopw   0x0(%rax,%rax,1)
L65:
  add    $0x1,%rbp
  cmp    %ebp,%r15d
  jle    L50
L64:
  mov    0x0(%r13,%rbp,4),%eax
  lea    (%rbx,%rbx,4),%edx
  lea    (%rax,%rdx,2),%ebx
  cmp    %r14d,%ebx
  jle    L65
  mov    %ebx,%edi
  call   <is_prime@plt>
  test   %rax,%rax
  cmovne %ebx,%r14d
  add    $0x1,%rbp
  cmp    %ebp,%r15d
  jg     L64
L50:
  add    $0x1,%r12
  cmp    %r12,0x8(%rsp)
  jne    L0
L9:
  add    $0x18,%rsp
  mov    %r14d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L51:
  xor    %r14d,%r14d
  jmp    L9
```

assistant: ```c
int prac3(const int n, const int arr[]) {
  int num = 0;
  int i, j;
  for (i = 0; i < n; ++i) {
    int t = 0;
    for (j = i; j < n; ++j) {
      t *= 10;
      t += arr[j];
      if (t > num && is_prime(t))
        num = t;
    }
  }
  return num;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<psf_get_le64>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    -0x1c(%rbp),%eax
  cltq
  add    $0x7,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  shl    $0x18,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x1c(%rbp),%eax
  cltq
  add    $0x6,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  shl    $0x10,%eax
  add    %eax,-0x4(%rbp)
  mov    -0x1c(%rbp),%eax
  cltq
  add    $0x5,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  shl    $0x8,%eax
  add    %eax,-0x4(%rbp)
  mov    -0x1c(%rbp),%eax
  cltq
  add    $0x4,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  add    %eax,-0x4(%rbp)
  mov    $0x20,%eax
  mov    %eax,%ecx
  shll   %cl,-0x4(%rbp)
  mov    -0x1c(%rbp),%eax
  cltq
  add    $0x3,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  shl    $0x18,%eax
  add    %eax,-0x4(%rbp)
  mov    -0x1c(%rbp),%eax
  cltq
  add    $0x2,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  shl    $0x10,%eax
  add    %eax,-0x4(%rbp)
  mov    -0x1c(%rbp),%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  shl    $0x8,%eax
  add    %eax,-0x4(%rbp)
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  add    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int64_t psf_get_le64(uint8_t *ptr, int offset) {
  int64_t value;
  value = ((uint32_t)ptr[offset + 7]) << 24;
  value += ptr[offset + 6] << 16;
  value += ptr[offset + 5] << 8;
  value += ptr[offset + 4];
  value = ((uint64_t)value) << 32;
  value += ((uint32_t)ptr[offset + 3]) << 24;
  value += ptr[offset + 2] << 16;
  value += ptr[offset + 1] << 8;
  value += ptr[offset];
  return value;
}
```
user: Translate the below assembly code into C programming language:

```asm
<psf_get_le64>:
  endbr64
  movslq %esi,%rsi
  add    $0x3,%rsi
  lea    0x0(,%rsi,4),%rcx
  mov    -0x4(%rdi,%rcx,1),%edx
  shl    $0x10,%edx
  mov    (%rdi,%rsi,4),%eax
  shl    $0x18,%eax
  add    %eax,%edx
  mov    -0x8(%rdi,%rcx,1),%eax
  shl    $0x8,%eax
  add    %edx,%eax
  add    -0xc(%rdi,%rcx,1),%eax
  ret
```

assistant: ```c
int64_t psf_get_le64(uint8_t *ptr, int offset) {
  int64_t value;
  value = ((uint32_t)ptr[offset + 7]) << 24;
  value += ptr[offset + 6] << 16;
  value += ptr[offset + 5] << 8;
  value += ptr[offset + 4];
  value = ((uint64_t)value) << 32;
  value += ((uint32_t)ptr[offset + 3]) << 24;
  value += ptr[offset + 2] << 16;
  value += ptr[offset + 1] << 8;
  value += ptr[offset];
  return value;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<psf_get_le64>:
  endbr64
  movslq %esi,%rsi
  add    $0x3,%rsi
  lea    0x0(,%rsi,4),%rcx
  mov    (%rdi,%rsi,4),%eax
  mov    -0x4(%rdi,%rcx,1),%edx
  shl    $0x18,%eax
  shl    $0x10,%edx
  add    %eax,%edx
  mov    -0x8(%rdi,%rcx,1),%eax
  shl    $0x8,%eax
  add    %edx,%eax
  add    -0xc(%rdi,%rcx,1),%eax
  ret
```

assistant: ```c
int64_t psf_get_le64(uint8_t *ptr, int offset) {
  int64_t value;
  value = ((uint32_t)ptr[offset + 7]) << 24;
  value += ptr[offset + 6] << 16;
  value += ptr[offset + 5] << 8;
  value += ptr[offset + 4];
  value = ((uint64_t)value) << 32;
  value += ((uint32_t)ptr[offset + 3]) << 24;
  value += ptr[offset + 2] << 16;
  value += ptr[offset + 1] << 8;
  value += ptr[offset];
  return value;
}
```
user: Translate the below assembly code into C programming language:

```asm
<psf_get_le64>:
  endbr64
  movslq %esi,%rsi
  add    $0x3,%rsi
  lea    0x0(,%rsi,4),%rcx
  mov    (%rdi,%rsi,4),%eax
  mov    -0x4(%rdi,%rcx,1),%edx
  shl    $0x18,%eax
  shl    $0x10,%edx
  add    %eax,%edx
  mov    -0x8(%rdi,%rcx,1),%eax
  shl    $0x8,%eax
  add    %edx,%eax
  add    -0xc(%rdi,%rcx,1),%eax
  ret
```

assistant: ```c
int64_t psf_get_le64(uint8_t *ptr, int offset) {
  int64_t value;
  value = ((uint32_t)ptr[offset + 7]) << 24;
  value += ptr[offset + 6] << 16;
  value += ptr[offset + 5] << 8;
  value += ptr[offset + 4];
  value = ((uint64_t)value) << 32;
  value += ((uint32_t)ptr[offset + 3]) << 24;
  value += ptr[offset + 2] << 16;
  value += ptr[offset + 1] << 8;
  value += ptr[offset];
  return value;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<end_radial_dist>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <end@plt>
  mov    $0x0,%eax
  call   <end@plt>
  mov    $0x0,%eax
  call   <end@plt>
  nop
  pop    %rbp
  ret
```

assistant: ```c
void end_radial_dist(void) {
  end();
  end();
  end();
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<end_radial_dist>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <end@plt>
  mov    $0x0,%eax
  call   <end@plt>
  mov    $0x0,%eax
  call   <end@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void end_radial_dist(void) {
  end();
  end();
  end();
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<end_radial_dist>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <end@plt>
  xor    %eax,%eax
  call   <end@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1050 <end@plt>
```

assistant: ```c
void end_radial_dist(void) {
  end();
  end();
  end();
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<end_radial_dist>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <end@plt>
  xor    %eax,%eax
  call   <end@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1050 <end@plt>
```

assistant: ```c
void end_radial_dist(void) {
  end();
  end();
  end();
}
```
user: Translate the below assembly code into C programming language:

```asm
<clear_buffers>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <ch_getflags@plt>
  and    $0x1,%eax
  test   %eax,%eax
  je     L2
  mov    $0x0,%eax
  call   <ch_flush@plt>
  mov    $0x0,%eax
  call   <clr_linenum@plt>
  mov    $0x0,%eax
  call   <clr_hilite@plt>
  jmp    L19
L2:
  nop
L19:
  pop    %rbp
  ret
```

assistant: ```c
void clear_buffers(void) {
  if (!(ch_getflags() & 001))
    return;
  ch_flush();
  clr_linenum();
  clr_hilite();
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<clear_buffers>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <ch_getflags@plt>
  test   $0x1,%al
  jne    L4
L14:
  add    $0x8,%rsp
  ret
L4:
  mov    $0x0,%eax
  call   <ch_flush@plt>
  mov    $0x0,%eax
  call   <clr_linenum@plt>
  mov    $0x0,%eax
  call   <clr_hilite@plt>
  jmp    L14
```

assistant: ```c
void clear_buffers(void) {
  if (!(ch_getflags() & 001))
    return;
  ch_flush();
  clr_linenum();
  clr_hilite();
}
```
user: Translate the below assembly code into C programming language:

```asm
<clear_buffers>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <ch_getflags@plt>
  test   $0x1,%al
  jne    L6
  add    $0x8,%rsp
  ret
  nopl   0x0(%rax,%rax,1)
L6:
  xor    %eax,%eax
  call   <ch_flush@plt>
  xor    %eax,%eax
  call   <clr_linenum@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    10b0 <clr_hilite@plt>
```

assistant: ```c
void clear_buffers(void) {
  if (!(ch_getflags() & 001))
    return;
  ch_flush();
  clr_linenum();
  clr_hilite();
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<clear_buffers>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <ch_getflags@plt>
  test   $0x1,%al
  jne    L4
  add    $0x8,%rsp
  ret
  nopl   0x0(%rax,%rax,1)
L4:
  xor    %eax,%eax
  call   <ch_flush@plt>
  xor    %eax,%eax
  call   <clr_linenum@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    10b0 <clr_hilite@plt>
```

assistant: ```c
void clear_buffers(void) {
  if (!(ch_getflags() & 001))
    return;
  ch_flush();
  clr_linenum();
  clr_hilite();
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<makeSTUtil>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r12
  push   %rbx
  sub    $0x30,%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    L19
  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    L11
L19:
  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    $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   <makeSTUtil>
  mov    %eax,%r12d
  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   <makeSTUtil>
  mov    -0x3c(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x38(%rbp),%rdx
  lea    (%rcx,%rdx,1),%rbx
  mov    %r12d,%esi
  mov    %eax,%edi
  call   <minVal@plt>
  mov    %eax,(%rbx)
  mov    -0x3c(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
L11:
  add    $0x30,%rsp
  pop    %rbx
  pop    %r12
  pop    %rbp
  ret
```

assistant: ```c
int makeSTUtil(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] = minVal(makeSTUtil(arr, ss, mid, st, si * 2 + 1), makeSTUtil(arr, mid + 1, se, st, si * 2 + 2));
  return st[si];
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<makeSTUtil>:
  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     L16
  mov    %edx,%r12d
  mov    %edx,%esi
  mov    %ebp,%edi
  call   <getMid@plt>
  lea    0x2(%rbx,%rbx,1),%r15d
  mov    %eax,0xc(%rsp)
  lea    0x1(%rax),%esi
  mov    %r15d,%r8d
  mov    %r13,%rcx
  mov    %r12d,%edx
  mov    %r14,%rdi
  call   <makeSTUtil>
  mov    %eax,%r12d
  lea    -0x1(%r15),%r8d
  mov    %r13,%rcx
  mov    0xc(%rsp),%edx
  mov    %ebp,%esi
  mov    %r14,%rdi
  call   <makeSTUtil>
  mov    %eax,%edi
  movslq %ebx,%rbx
  lea    0x0(%r13,%rbx,4),%rbx
  mov    %r12d,%esi
  call   <minVal@plt>
  mov    %eax,(%rbx)
L17:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L16:
  movslq %esi,%rbp
  mov    (%rdi,%rbp,4),%eax
  movslq %r8d,%rbx
  mov    %eax,(%rcx,%rbx,4)
  jmp    L17
```

assistant: ```c
int makeSTUtil(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] = minVal(makeSTUtil(arr, ss, mid, st, si * 2 + 1), makeSTUtil(arr, mid + 1, se, st, si * 2 + 2));
  return st[si];
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<makeSTUtil>:
  endbr64
  push   %r15
  movslq %r8d,%rax
  push   %r14
  lea    (%rcx,%rax,4),%r15
  mov    %rdi,%r14
  push   %r13
  push   %r12
  push   %rbp
  movslq %esi,%rbp
  push   %rbx
  sub    $0x18,%rsp
  cmp    %edx,%ebp
  je     L6
  mov    %edx,%esi
  mov    %rax,%rbx
  mov    %ebp,%edi
  mov    %edx,%r13d
  mov    %rcx,%r12
  lea    0x2(%rbx,%rbx,1),%ebx
  call   <getMid@plt>
  mov    %r13d,%edx
  mov    %ebx,%r8d
  mov    %r12,%rcx
  lea    0x1(%rax),%esi
  mov    %r14,%rdi
  mov    %eax,0xc(%rsp)
  call   <makeSTUtil>
  mov    0xc(%rsp),%edx
  mov    %ebp,%esi
  mov    %r14,%rdi
  lea    -0x1(%rbx),%r8d
  mov    %r12,%rcx
  mov    %eax,%r13d
  call   <makeSTUtil>
  mov    %r13d,%esi
  mov    %eax,%edi
  call   <minVal@plt>
L14:
  mov    %eax,(%r15)
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   (%rax)
L6:
  mov    (%rdi,%rbp,4),%eax
  jmp    L14
```

assistant: ```c
int makeSTUtil(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] = minVal(makeSTUtil(arr, ss, mid, st, si * 2 + 1), makeSTUtil(arr, mid + 1, se, st, si * 2 + 2));
  return st[si];
}
```
user: What is the c source code of the assembly code below?

```asm
<makeSTUtil>:
  endbr64
  push   %r15
  movslq %r8d,%rax
  push   %r14
  lea    (%rcx,%rax,4),%r15
  mov    %rdi,%r14
  push   %r13
  push   %r12
  push   %rbp
  movslq %esi,%rbp
  push   %rbx
  sub    $0x18,%rsp
  cmp    %edx,%ebp
  je     L4
  mov    %edx,%esi
  mov    %rax,%rbx
  mov    %ebp,%edi
  mov    %edx,%r13d
  mov    %rcx,%r12
  lea    0x2(%rbx,%rbx,1),%ebx
  call   <getMid@plt>
  mov    %r13d,%edx
  mov    %ebx,%r8d
  mov    %r12,%rcx
  lea    0x1(%rax),%esi
  mov    %r14,%rdi
  mov    %eax,0xc(%rsp)
  call   <makeSTUtil>
  mov    0xc(%rsp),%edx
  mov    %ebp,%esi
  mov    %r14,%rdi
  lea    -0x1(%rbx),%r8d
  mov    %r12,%rcx
  mov    %eax,%r13d
  call   <makeSTUtil>
  mov    %r13d,%esi
  mov    %eax,%edi
  call   <minVal@plt>
L18:
  mov    %eax,(%r15)
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   (%rax)
L4:
  mov    (%rdi,%rbp,4),%eax
  jmp    L18
```

assistant: ```c
int makeSTUtil(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] = minVal(makeSTUtil(arr, ss, mid, st, si * 2 + 1), makeSTUtil(arr, mid + 1, se, st, si * 2 + 2));
  return st[si];
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x1,%esi
  mov    $0x1,%edi
  mov    $0x0,%eax
  call   <param_lister1_aux@plt>
  movsd  D10(%rip),%xmm0
  mov    D13(%rip),%rax
  movapd %xmm0,%xmm1
  mov    $0x3,%r9d
  mov    $0x3,%r8d
  mov    $0x1,%ecx
  movq   %rax,%xmm0
  mov    $0x3,%edx
  mov    $0x1,%esi
  mov    $0x1,%edi
  mov    $0x2,%eax
  call   <param_lister4_aux@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

tool_calls: name: read_data
arguments: label: D10
type: f64
name: read_data
arguments: label: D13
type: f64

tool: D10: 6

tool: D13: 2

assistant: ```c
int main() {
  param_lister1_aux(__builtin_types_compatible_p(typeof(1), int) * 1 + __builtin_types_compatible_p(typeof(1), long long) * 2 + __builtin_types_compatible_p(typeof(1), double) * 3 + __builtin_types_compatible_p(typeof(1), long double) * 4, 1);
  param_lister4_aux(__builtin_types_compatible_p(typeof(1), int) * 1 + __builtin_types_compatible_p(typeof(1), long long) * 2 + __builtin_types_compatible_p(typeof(1), double) * 3 + __builtin_types_compatible_p(typeof(1), long double) * 4, 1, __builtin_types_compatible_p(typeof(2.0), int) * 1 + __builtin_types_compatible_p(typeof(2.0), long long) * 2 + __builtin_types_compatible_p(typeof(2.0), double) * 3 + __builtin_types_compatible_p(typeof(2.0), long double) * 4, 2.0, __builtin_types_compatible_p(typeof(3), int) * 1 + __builtin_types_compatible_p(typeof(3), long long) * 2 + __builtin_types_compatible_p(typeof(3), double) * 3 + __builtin_types_compatible_p(typeof(3), long double) * 4, 3, __builtin_types_compatible_p(typeof(6.0), int) * 1 + __builtin_types_compatible_p(typeof(6.0), long long) * 2 + __builtin_types_compatible_p(typeof(6.0), double) * 3 + __builtin_types_compatible_p(typeof(6.0), long double) * 4, 6.0);
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x1,%esi
  mov    $0x1,%edi
  mov    $0x0,%eax
  call   <param_lister1_aux@plt>
  movsd  D3(%rip),%xmm1
  mov    $0x3,%r9d
  mov    $0x3,%r8d
  mov    $0x1,%ecx
  movsd  D15(%rip),%xmm0
  mov    $0x3,%edx
  mov    $0x1,%esi
  mov    $0x1,%edi
  mov    $0x2,%eax
  call   <param_lister4_aux@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D3
type_infered: f64
name: retrieve_data
arguments: data_tag: D15
type_infered: f64

tool: D3: 6

tool: D15: 2

assistant: ```c
int main() {
  param_lister1_aux(__builtin_types_compatible_p(typeof(1), int) * 1 + __builtin_types_compatible_p(typeof(1), long long) * 2 + __builtin_types_compatible_p(typeof(1), double) * 3 + __builtin_types_compatible_p(typeof(1), long double) * 4, 1);
  param_lister4_aux(__builtin_types_compatible_p(typeof(1), int) * 1 + __builtin_types_compatible_p(typeof(1), long long) * 2 + __builtin_types_compatible_p(typeof(1), double) * 3 + __builtin_types_compatible_p(typeof(1), long double) * 4, 1, __builtin_types_compatible_p(typeof(2.0), int) * 1 + __builtin_types_compatible_p(typeof(2.0), long long) * 2 + __builtin_types_compatible_p(typeof(2.0), double) * 3 + __builtin_types_compatible_p(typeof(2.0), long double) * 4, 2.0, __builtin_types_compatible_p(typeof(3), int) * 1 + __builtin_types_compatible_p(typeof(3), long long) * 2 + __builtin_types_compatible_p(typeof(3), double) * 3 + __builtin_types_compatible_p(typeof(3), long double) * 4, 3, __builtin_types_compatible_p(typeof(6.0), int) * 1 + __builtin_types_compatible_p(typeof(6.0), long long) * 2 + __builtin_types_compatible_p(typeof(6.0), double) * 3 + __builtin_types_compatible_p(typeof(6.0), long double) * 4, 6.0);
  return 0;
}
```