ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Demonstrate the C code that generates the assembly output below:

```asm
<hash>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x348,%rsp
  mov    %r9d,0xc(%rsp)
  mov    %fs:0x28,%rax
  mov    %rax,0x338(%rsp)
  xor    %eax,%eax
  test   %rdi,%rdi
  je     L13
  mov    %rdi,%r12
  mov    %rsi,%r14
  mov    %rdx,%r13
  mov    %rcx,%rbp
  mov    %r8,%rbx
  test   %rdx,%rdx
  sete   %dl
  test   %rcx,%rcx
  setne  %al
  test   %al,%dl
  jne    L61
  cmp    $0xc7,%r8
  ja     L61
  lea    0x10(%rsp),%rdi
  mov    $0x64,%ecx
  mov    $0x0,%eax
  rep stos %rax,%es:(%rdi)
  cmp    %r8,%rbp
  jb     L3
  lea    0x0(,%r8,4),%rax
  mov    %rax,(%rsp)
  lea    0x10(%rsp),%r15
L95:
  mov    %rbx,%rdx
  mov    %r13,%rsi
  mov    %r15,%rdi
  call   <xorin@plt>
  mov    %r15,%rdi
  call   <keccakf@plt>
  add    (%rsp),%r13
  sub    %rbx,%rbp
  cmp    %rbp,%rbx
  jbe    L95
L3:
  mov    0xc(%rsp),%eax
  xor    %eax,0x10(%rsp,%rbp,4)
  xorl   $0x80,0xc(%rsp,%rbx,4)
  lea    0x10(%rsp),%r15
  mov    %rbp,%rdx
  mov    %r13,%rsi
  mov    %r15,%rdi
  call   <xorin@plt>
  mov    %r15,%rdi
  call   <keccakf@plt>
  cmp    %r14,%rbx
  ja     L58
  lea    0x0(,%rbx,4),%r13
  mov    %r15,%rbp
L86:
  mov    %rbx,%rdx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <setout@plt>
  mov    %rbp,%rdi
  call   <keccakf@plt>
  add    %r13,%r12
  sub    %rbx,%r14
  cmp    %r14,%rbx
  jbe    L86
L58:
  lea    0x10(%rsp),%rbx
  mov    %r14,%rdx
  mov    %r12,%rsi
  mov    %rbx,%rdi
  call   <setout@plt>
  mov    $0xc8,%ecx
  mov    $0x0,%edx
  mov    $0xc8,%esi
  mov    %rbx,%rdi
  call   <memset_s@plt>
  mov    $0x0,%eax
L63:
  mov    0x338(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L74
  add    $0x348,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L13:
  mov    $0xffffffff,%eax
  jmp    L63
L61:
  mov    $0xffffffff,%eax
  jmp    L63
L74:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int hash(uint8_t *out, size_t outlen, const uint8_t *in, size_t inlen, size_t rate, uint8_t delim) {
  if ((out == NULL) || ((in == NULL) && inlen != 0) || (rate >= 200)) {
    return -1;
  }
  uint8_t a[200] = {0};
  while (inlen >= rate) {
    xorin(a, in, rate);
    keccakf(a);
    in += rate;
    inlen -= rate;
  };
  a[inlen] ^= delim;
  a[rate - 1] ^= 0x80;
  xorin(a, in, inlen);
  keccakf(a);
  while (outlen >= rate) {
    setout(a, out, rate);
    keccakf(a);
    out += rate;
    outlen -= rate;
  };
  setout(a, out, outlen);
  memset_s(a, 200, 0, 200);
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<hash>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x348,%rsp
  mov    %r9d,0xc(%rsp)
  mov    %fs:0x28,%rax
  mov    %rax,0x338(%rsp)
  xor    %eax,%eax
  test   %rdi,%rdi
  je     L83
  test   %rdx,%rdx
  mov    %rdx,%r15
  mov    %rcx,%r14
  sete   %dl
  test   %rcx,%rcx
  setne  %al
  test   %al,%dl
  jne    L83
  mov    %r8,%rbx
  cmp    $0xc7,%r8
  ja     L83
  lea    0x10(%rsp),%rbp
  mov    %rdi,%r12
  mov    $0x64,%ecx
  xor    %eax,%eax
  mov    %rbp,%rdi
  mov    %rsi,%r13
  rep stos %rax,%es:(%rdi)
  cmp    %r8,%r14
  jb     L43
  lea    0x0(,%r8,4),%rax
  mov    %rax,(%rsp)
  nopl   0x0(%rax)
L38:
  mov    %r15,%rsi
  mov    %rbx,%rdx
  mov    %rbp,%rdi
  sub    %rbx,%r14
  call   <xorin@plt>
  mov    %rbp,%rdi
  call   <keccakf@plt>
  add    (%rsp),%r15
  cmp    %r14,%rbx
  jbe    L38
L43:
  mov    %rbp,%rdi
  mov    %r14,%rdx
  mov    %r15,%rsi
  mov    0xc(%rsp),%eax
  xor    %eax,0x10(%rsp,%r14,4)
  xorl   $0x80,0xc(%rsp,%rbx,4)
  call   <xorin@plt>
  mov    %rbp,%rdi
  call   <keccakf@plt>
  cmp    %r13,%rbx
  ja     L81
  lea    0x0(,%rbx,4),%r14
  cs nopw 0x0(%rax,%rax,1)
L41:
  mov    %r12,%rsi
  mov    %rbx,%rdx
  mov    %rbp,%rdi
  sub    %rbx,%r13
  call   <setout@plt>
  mov    %rbp,%rdi
  add    %r14,%r12
  call   <keccakf@plt>
  cmp    %r13,%rbx
  jbe    L41
L81:
  mov    %r13,%rdx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <setout@plt>
  mov    $0xc8,%ecx
  xor    %edx,%edx
  mov    %rbp,%rdi
  mov    $0xc8,%esi
  call   <memset_s@plt>
  xor    %eax,%eax
L89:
  mov    0x338(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L72
  add    $0x348,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L83:
  mov    $0xffffffff,%eax
  jmp    L89
L72:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int hash(uint8_t *out, size_t outlen, const uint8_t *in, size_t inlen, size_t rate, uint8_t delim) {
  if ((out == NULL) || ((in == NULL) && inlen != 0) || (rate >= 200)) {
    return -1;
  }
  uint8_t a[200] = {0};
  while (inlen >= rate) {
    xorin(a, in, rate);
    keccakf(a);
    in += rate;
    inlen -= rate;
  };
  a[inlen] ^= delim;
  a[rate - 1] ^= 0x80;
  xorin(a, in, inlen);
  keccakf(a);
  while (outlen >= rate) {
    setout(a, out, rate);
    keccakf(a);
    out += rate;
    outlen -= rate;
  };
  setout(a, out, outlen);
  memset_s(a, 200, 0, 200);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<hash>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x348,%rsp
  mov    %r9d,0xc(%rsp)
  mov    %fs:0x28,%rax
  mov    %rax,0x338(%rsp)
  xor    %eax,%eax
  test   %rdi,%rdi
  je     L71
  test   %rdx,%rdx
  mov    %rdx,%r15
  mov    %rcx,%r14
  sete   %dl
  test   %rcx,%rcx
  setne  %al
  test   %al,%dl
  jne    L71
  mov    %r8,%rbx
  cmp    $0xc7,%r8
  ja     L71
  lea    0x10(%rsp),%rbp
  mov    %rdi,%r12
  mov    $0x64,%ecx
  xor    %eax,%eax
  mov    %rbp,%rdi
  mov    %rsi,%r13
  rep stos %rax,%es:(%rdi)
  cmp    %r8,%r14
  jb     L60
  lea    0x0(,%r8,4),%rax
  mov    %rax,(%rsp)
  nopl   0x0(%rax)
L48:
  mov    %r15,%rsi
  mov    %rbx,%rdx
  mov    %rbp,%rdi
  sub    %rbx,%r14
  call   <xorin@plt>
  mov    %rbp,%rdi
  call   <keccakf@plt>
  add    (%rsp),%r15
  cmp    %r14,%rbx
  jbe    L48
L60:
  mov    %rbp,%rdi
  mov    %r14,%rdx
  mov    %r15,%rsi
  mov    0xc(%rsp),%eax
  xor    %eax,0x10(%rsp,%r14,4)
  xorl   $0x80,0xc(%rsp,%rbx,4)
  call   <xorin@plt>
  mov    %rbp,%rdi
  call   <keccakf@plt>
  cmp    %r13,%rbx
  ja     L51
  lea    0x0(,%rbx,4),%r14
  cs nopw 0x0(%rax,%rax,1)
L81:
  mov    %r12,%rsi
  mov    %rbx,%rdx
  mov    %rbp,%rdi
  sub    %rbx,%r13
  call   <setout@plt>
  mov    %rbp,%rdi
  add    %r14,%r12
  call   <keccakf@plt>
  cmp    %r13,%rbx
  jbe    L81
L51:
  mov    %r13,%rdx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <setout@plt>
  mov    $0xc8,%ecx
  xor    %edx,%edx
  mov    %rbp,%rdi
  mov    $0xc8,%esi
  call   <memset_s@plt>
  xor    %eax,%eax
L33:
  mov    0x338(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L83
  add    $0x348,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L71:
  mov    $0xffffffff,%eax
  jmp    L33
L83:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int hash(uint8_t *out, size_t outlen, const uint8_t *in, size_t inlen, size_t rate, uint8_t delim) {
  if ((out == NULL) || ((in == NULL) && inlen != 0) || (rate >= 200)) {
    return -1;
  }
  uint8_t a[200] = {0};
  while (inlen >= rate) {
    xorin(a, in, rate);
    keccakf(a);
    in += rate;
    inlen -= rate;
  };
  a[inlen] ^= delim;
  a[rate - 1] ^= 0x80;
  xorin(a, in, inlen);
  keccakf(a);
  while (outlen >= rate) {
    setout(a, out, rate);
    keccakf(a);
    out += rate;
    outlen -= rate;
  };
  setout(a, out, outlen);
  memset_s(a, 200, 0, 200);
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<test_nested>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    0x10(%rbp),%rax
  mov    %rax,-0x18(%rbp)
  mov    -0x24(%rbp),%eax
  mov    %eax,-0x20(%rbp)
  mov    -0x20(%rbp),%eax
  lea    0x2(%rax),%edx
  mov    -0x20(%rbp),%eax
  lea    0x1(%rax),%ecx
  mov    -0x20(%rbp),%eax
  lea    -0x20(%rbp),%rsi
  mov    %rsi,%r10
  mov    %ecx,%esi
  mov    %eax,%edi
  call   <foo.0>
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L2
  call   <__stack_chk_fail@plt>
L2:
  leave
  ret
```

assistant: ```c
int test_nested(int i) {
  int __attribute__((__noinline__, __regparm__(3))) foo(int j, int k, int l) { return i + j + k + l; }
  return foo(i, i + 1, i + 2);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<test_nested>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x30(%rsp),%rax
  mov    %rax,0x8(%rsp)
  mov    %edi,(%rsp)
  lea    0x2(%rdi),%edx
  lea    0x1(%rdi),%esi
  mov    %rsp,%r10
  call   <foo.0>
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L5
  add    $0x28,%rsp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int test_nested(int i) {
  int __attribute__((__noinline__, __regparm__(3))) foo(int j, int k, int l) { return i + j + k + l; }
  return foo(i, i + 1, i + 2);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<test_nested>:
  endbr64
  sub    $0x28,%rsp
  lea    0x2(%rdi),%edx
  lea    0x1(%rdi),%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r10
  lea    0x30(%rsp),%rax
  mov    %edi,(%rsp)
  mov    %rax,0x8(%rsp)
  call   <foo.0>
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L9
  add    $0x28,%rsp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int test_nested(int i) {
  int __attribute__((__noinline__, __regparm__(3))) foo(int j, int k, int l) { return i + j + k + l; }
  return foo(i, i + 1, i + 2);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<test_nested>:
  endbr64
  sub    $0x28,%rsp
  lea    0x2(%rdi),%edx
  lea    0x1(%rdi),%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r10
  lea    0x30(%rsp),%rax
  mov    %edi,(%rsp)
  mov    %rax,0x8(%rsp)
  call   <foo.0>
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L9
  add    $0x28,%rsp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int test_nested(int i) {
  int __attribute__((__noinline__, __regparm__(3))) foo(int j, int k, int l) { return i + j + k + l; }
  return foo(i, i + 1, i + 2);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x78,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  lea    D48(%rip),%rdi
  call   <printf@plt>
  mov    %rsp,%rsi
  lea    D33(%rip),%rdi
  call   <scanf@plt>
  movsbl (%rsp),%edi
  call   <Roman_Value@plt>
  mov    $0x0,%r13d
  mov    $0x0,%ebx
  test   %eax,%eax
  jg     L60
  lea    D0(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L20
L47:
  movslq %ebx,%rax
  movsbl (%rsp,%rax,1),%edi
  call   <Roman_Value@plt>
  add    %eax,%r13d
  mov    %ebx,%ebp
L29:
  lea    0x1(%rbp),%ebx
L60:
  movslq %ebx,%rax
  movsbl (%rsp,%rax,1),%edi
  call   <Roman_Value@plt>
  test   %eax,%eax
  je     L6
  movslq %ebx,%rax
  movsbl (%rsp,%rax,1),%edi
  call   <Roman_Value@plt>
  mov    %eax,%r12d
  lea    0x1(%rbx),%ebp
  movslq %ebp,%rax
  movsbl (%rsp,%rax,1),%edi
  call   <Roman_Value@plt>
  cmp    %eax,%r12d
  jge    L47
  movslq %ebp,%rax
  movsbl (%rsp,%rax,1),%edi
  call   <Roman_Value@plt>
  mov    %eax,%r12d
  movslq %ebx,%rbx
  movsbl (%rsp,%rbx,1),%edi
  call   <Roman_Value@plt>
  sub    %eax,%r12d
  add    %r12d,%r13d
  jmp    L29
L6:
  cmp    $0x100,%r13d
  jg     L65
  mov    %r13d,%esi
  lea    D38(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L20:
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  mov    $0x0,%eax
  add    $0x78,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L65:
  lea    D0(%rip),%rdi
  call   <printf@plt>
  jmp    L20
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D48
datatype: string
name: extract_symbolic_data
arguments: data_label: D33
datatype: string
name: extract_symbolic_data
arguments: data_label: D0
datatype: string
name: extract_symbolic_data
arguments: data_label: D38
datatype: string

tool: D48: Enter Roman number:

tool: D33: %s

tool: D0: Invalid number.

tool: D38: %d

assistant: ```c
int main() {
  char string[100];
  int counter = 0, number = 0;
  printf("Enter Roman number:");
  scanf("%s", string);
  if (Roman_Value(string[counter]) <= 0) {
    printf("Invalid number.");
    return 0;
  }
  while (Roman_Value(string[counter]) != 0) {
    if (Roman_Value(string[counter]) < Roman_Value(string[counter + 1])) {
      number += Roman_Value(string[counter + 1]) - Roman_Value(string[counter]);
      counter++;
    } else
      number = number + Roman_Value(string[counter]);
    counter++;
  }
  if (number > 256) {
    printf("Invalid number.");
    return 0;
  }
  printf("%d", number);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r15
  lea    D3(%rip),%rdi
  push   %r14
  push   %r13
  xor    %r13d,%r13d
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  push   %rbx
  sub    $0x78,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rsp,%rsi
  lea    D26(%rip),%rdi
  call   <scanf@plt>
  movsbl (%rsp),%edi
  call   <Roman_Value@plt>
  test   %eax,%eax
  jle    L23
  nopl   0x0(%rax)
L39:
  movslq %r12d,%rbx
  movsbl (%rsp,%rbx,1),%edi
  call   <Roman_Value@plt>
  test   %eax,%eax
  je     L77
L50:
  movsbl (%rsp,%rbx,1),%edi
  lea    0x1(%r12),%r14d
  movslq %r14d,%r15
  call   <Roman_Value@plt>
  movsbl (%rsp,%r15,1),%edi
  mov    %eax,%ebp
  call   <Roman_Value@plt>
  cmp    %eax,%ebp
  jge    L42
  movsbl (%rsp,%r15,1),%edi
  add    $0x2,%r12d
  call   <Roman_Value@plt>
  movsbl (%rsp,%rbx,1),%edi
  movslq %r12d,%rbx
  mov    %eax,%ebp
  call   <Roman_Value@plt>
  movsbl (%rsp,%rbx,1),%edi
  sub    %eax,%ebp
  add    %ebp,%r13d
  call   <Roman_Value@plt>
  test   %eax,%eax
  jne    L50
L77:
  cmp    $0x100,%r13d
  jg     L23
  mov    %r13d,%esi
  lea    D35(%rip),%rdi
  call   <printf@plt>
L53:
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L59
  add    $0x78,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L42:
  movsbl (%rsp,%rbx,1),%edi
  mov    %r14d,%r12d
  call   <Roman_Value@plt>
  add    %eax,%r13d
  jmp    L39
L23:
  lea    D22(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L53
L59:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D3
data_type: string
name: retrieve_data
arguments: data_label: D26
data_type: string
name: retrieve_data
arguments: data_label: D35
data_type: string
name: retrieve_data
arguments: data_label: D22
data_type: string

tool: D3: Enter Roman number:

tool: D26: %s

tool: D35: %d

tool: D22: Invalid number.

assistant: ```c
int main() {
  char string[100];
  int counter = 0, number = 0;
  printf("Enter Roman number:");
  scanf("%s", string);
  if (Roman_Value(string[counter]) <= 0) {
    printf("Invalid number.");
    return 0;
  }
  while (Roman_Value(string[counter]) != 0) {
    if (Roman_Value(string[counter]) < Roman_Value(string[counter + 1])) {
      number += Roman_Value(string[counter + 1]) - Roman_Value(string[counter]);
      counter++;
    } else
      number = number + Roman_Value(string[counter]);
    counter++;
  }
  if (number > 256) {
    printf("Invalid number.");
    return 0;
  }
  printf("%d", number);
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  push   %r15
  lea    D26(%rip),%rdi
  push   %r14
  push   %r13
  xor    %r13d,%r13d
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  push   %rbx
  sub    $0x78,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rsp,%rsi
  lea    D4(%rip),%rdi
  call   <scanf@plt>
  movsbl (%rsp),%edi
  call   <Roman_Value@plt>
  test   %eax,%eax
  jle    L18
  nopl   0x0(%rax)
L55:
  movslq %r12d,%rbx
  movsbl (%rsp,%rbx,1),%edi
  call   <Roman_Value@plt>
  test   %eax,%eax
  je     L21
L71:
  movsbl (%rsp,%rbx,1),%edi
  lea    0x1(%r12),%r14d
  movslq %r14d,%r15
  call   <Roman_Value@plt>
  movsbl (%rsp,%r15,1),%edi
  mov    %eax,%ebp
  call   <Roman_Value@plt>
  cmp    %eax,%ebp
  jge    L0
  movsbl (%rsp,%r15,1),%edi
  add    $0x2,%r12d
  call   <Roman_Value@plt>
  movsbl (%rsp,%rbx,1),%edi
  movslq %r12d,%rbx
  mov    %eax,%ebp
  call   <Roman_Value@plt>
  movsbl (%rsp,%rbx,1),%edi
  sub    %eax,%ebp
  add    %ebp,%r13d
  call   <Roman_Value@plt>
  test   %eax,%eax
  jne    L71
L21:
  cmp    $0x100,%r13d
  jg     L18
  mov    %r13d,%esi
  lea    D20(%rip),%rdi
  call   <printf@plt>
L66:
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L10
  add    $0x78,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L0:
  movsbl (%rsp,%rbx,1),%edi
  mov    %r14d,%r12d
  call   <Roman_Value@plt>
  add    %eax,%r13d
  jmp    L55
L18:
  lea    D13(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L66
L10:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D26
data_type: string
name: retrieve_data
arguments: tag: D4
data_type: string
name: retrieve_data
arguments: tag: D20
data_type: string
name: retrieve_data
arguments: tag: D13
data_type: string

tool: D26: Enter Roman number:

tool: D4: %s

tool: D20: %d

tool: D13: Invalid number.

assistant: ```c
int main() {
  char string[100];
  int counter = 0, number = 0;
  printf("Enter Roman number:");
  scanf("%s", string);
  if (Roman_Value(string[counter]) <= 0) {
    printf("Invalid number.");
    return 0;
  }
  while (Roman_Value(string[counter]) != 0) {
    if (Roman_Value(string[counter]) < Roman_Value(string[counter + 1])) {
      number += Roman_Value(string[counter + 1]) - Roman_Value(string[counter]);
      counter++;
    } else
      number = number + Roman_Value(string[counter]);
    counter++;
  }
  if (number > 256) {
    printf("Invalid number.");
    return 0;
  }
  printf("%d", number);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<vecscale_float>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  movss  %xmm0,-0x24(%rbp)
  mov    %rcx,-0x30(%rbp)
  mov    -0x1c(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rax,-0x18(%rbp)
  mov    -0x1c(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rax,-0x30(%rbp)
  mov    -0x20(%rbp),%eax
  sub    -0x1c(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0x4(%rbp)
  jmp    L15
L7:
  mov    -0x30(%rbp),%rax
  lea    0x4(%rax),%rdx
  mov    %rdx,-0x30(%rbp)
  movss  (%rax),%xmm0
  mov    -0x18(%rbp),%rax
  lea    0x4(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  mulss  -0x24(%rbp),%xmm0
  movss  %xmm0,(%rax)
  subl   $0x1,-0x4(%rbp)
L15:
  cmpl   $0x0,-0x4(%rbp)
  jne    L7
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void vecscale_float(float *vec1, int beg, int end, float alpha, float *vec2) {
  int i;
  vec1 += beg;
  vec2 += beg;
  for (i = end - beg + 1; i; i--) {
    (*vec1++) = alpha * (*vec2++);
  }
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<vecscale_float>:
  endbr64
  movslq %esi,%rax
  shl    $0x2,%rax
  add    %rax,%rdi
  add    %rax,%rcx
  sub    %esi,%edx
  cmp    $0xffffffff,%edx
  je     L9
  lea    0x1(%rdx),%edx
  shl    $0x2,%rdx
  mov    $0x0,%eax
L12:
  movaps %xmm0,%xmm1
  mulss  (%rcx,%rax,1),%xmm1
  movss  %xmm1,(%rdi,%rax,1)
  add    $0x4,%rax
  cmp    %rdx,%rax
  jne    L12
L9:
  ret
```

assistant: ```c
void vecscale_float(float *vec1, int beg, int end, float alpha, float *vec2) {
  int i;
  vec1 += beg;
  vec2 += beg;
  for (i = end - beg + 1; i; i--) {
    (*vec1++) = alpha * (*vec2++);
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<vecscale_float>:
  endbr64
  movslq %esi,%rax
  sub    %esi,%edx
  shl    $0x2,%rax
  add    %rax,%rdi
  add    %rax,%rcx
  cmp    $0xffffffff,%edx
  je     L14
  add    $0x1,%edx
  xor    %eax,%eax
  shl    $0x2,%rdx
  nopl   0x0(%rax)
L11:
  movss  (%rcx,%rax,1),%xmm1
  mulss  %xmm0,%xmm1
  movss  %xmm1,(%rdi,%rax,1)
  add    $0x4,%rax
  cmp    %rdx,%rax
  jne    L11
L14:
  ret
```

assistant: ```c
void vecscale_float(float *vec1, int beg, int end, float alpha, float *vec2) {
  int i;
  vec1 += beg;
  vec2 += beg;
  for (i = end - beg + 1; i; i--) {
    (*vec1++) = alpha * (*vec2++);
  }
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<vecscale_float>:
  endbr64
  mov    %rdi,%r10
  movslq %esi,%r11
  mov    %edx,%edi
  mov    %rcx,%r8
  lea    0x0(,%r11,4),%rax
  sub    %esi,%edi
  lea    (%r10,%rax,1),%rdx
  add    %rax,%rcx
  cmp    $0xffffffff,%edi
  je     L70
  lea    0x4(%r8,%rax,1),%rsi
  mov    %rdx,%rax
  sub    %rsi,%rax
  cmp    $0x8,%rax
  jbe    L93
  test   %edi,%edi
  je     L93
  push   %rbx
  lea    0x1(%rdi),%r9d
  cmp    $0x2,%edi
  jbe    L84
  mov    %r9d,%esi
  movaps %xmm0,%xmm2
  xor    %eax,%eax
  shr    $0x2,%esi
  shufps $0x0,%xmm2,%xmm2
  shl    $0x4,%rsi
  nopw   0x0(%rax,%rax,1)
L30:
  movups (%rcx,%rax,1),%xmm1
  mulps  %xmm2,%xmm1
  movups %xmm1,(%rdx,%rax,1)
  add    $0x10,%rax
  cmp    %rax,%rsi
  jne    L30
  mov    %r9d,%esi
  and    $0xfffffffc,%esi
  mov    %esi,%eax
  lea    0x0(,%rax,4),%rbx
  add    %rbx,%rdx
  add    %rbx,%rcx
  cmp    %esi,%r9d
  je     L15
  sub    %esi,%r9d
  cmp    %esi,%edi
  je     L19
L36:
  add    %r11,%rax
  movaps %xmm0,%xmm2
  mov    %r9d,%esi
  movq   (%r8,%rax,4),%xmm1
  shufps $0xe0,%xmm2,%xmm2
  and    $0xfffffffe,%esi
  mulps  %xmm2,%xmm1
  movlps %xmm1,(%r10,%rax,4)
  mov    %esi,%eax
  shl    $0x2,%rax
  add    %rax,%rdx
  add    %rax,%rcx
  cmp    %esi,%r9d
  je     L15
L19:
  mulss  (%rcx),%xmm0
  movss  %xmm0,(%rdx)
L15:
  pop    %rbx
  ret
  nop
L93:
  lea    0x1(%rdi),%esi
  xor    %eax,%eax
  shl    $0x2,%rsi
  nopl   0x0(%rax)
L91:
  movss  (%rcx,%rax,1),%xmm1
  mulss  %xmm0,%xmm1
  movss  %xmm1,(%rdx,%rax,1)
  add    $0x4,%rax
  cmp    %rsi,%rax
  jne    L91
  ret
  nopl   0x0(%rax,%rax,1)
L70:
  ret
L84:
  xor    %eax,%eax
  jmp    L36
```

assistant: ```c
void vecscale_float(float *vec1, int beg, int end, float alpha, float *vec2) {
  int i;
  vec1 += beg;
  vec2 += beg;
  for (i = end - beg + 1; i; i--) {
    (*vec1++) = alpha * (*vec2++);
  }
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<qui>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  movsd  %xmm0,-0x18(%rbp)
  movsd  -0x18(%rbp),%xmm1
  movsd  D2(%rip),%xmm0
  addsd  %xmm0,%xmm1
  movq   %xmm1,%rax
  movq   %rax,%xmm0
  call   <floor@plt>
  pxor   %xmm1,%xmm1
  cvtsi2sd %rax,%xmm1
  movsd  -0x18(%rbp),%xmm0
  subsd  %xmm1,%xmm0
  movsd  %xmm0,-0x18(%rbp)
  movsd  -0x18(%rbp),%xmm0
  mulsd  %xmm0,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movsd  -0x8(%rbp),%xmm1
  movsd  D16(%rip),%xmm0
  mulsd  %xmm0,%xmm1
  movsd  D34(%rip),%xmm0
  subsd  %xmm1,%xmm0
  movapd %xmm0,%xmm1
  mulsd  -0x8(%rbp),%xmm1
  movsd  D26(%rip),%xmm0
  subsd  %xmm1,%xmm0
  mulsd  -0x18(%rbp),%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D2
type_guessed: f64
name: retrieve_data
arguments: data_label: D16
type_guessed: f64
name: retrieve_data
arguments: data_label: D34
type_guessed: f64
name: retrieve_data
arguments: data_label: D26
type_guessed: f64

tool: D2: 0.5

tool: D16: 40.886098457876805

tool: D34: 34.07174871489734

tool: D26: 5.962556025107034

assistant: ```c
double qui(double phase) {
  phase -= floor(phase + 0.5);
  double x2 = phase * phase;
  return phase * (5.96255602510703402 - x2 * (34.0717487148973373 - 40.8860984578768047 * x2));
}
```