ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请从给定的汇编指令推导出C源代码:

```asm
<profile_list_normalize>:
  endbr64
  movzbl (%rdi),%ecx
  mov    %rdi,%rax
  cmp    $0x2c,%cl
  jne    L78
  xor    %esi,%esi
  nopl   (%rax)
L4:
  mov    %rsi,%rdx
  add    $0x1,%rsi
  movzbl (%rax,%rsi,1),%ecx
  cmp    $0x2c,%cl
  je     L4
  add    $0x2,%rdx
L108:
  movsbl %cl,%edi
  test   %edi,%edi
  je     L103
  xor    %esi,%esi
  jmp    L51
  nopw   0x0(%rax,%rax,1)
L16:
  add    $0x1,%rsi
  add    $0x1,%rdx
  mov    %cl,-0x1(%rax,%rsi,1)
  movsbl (%rax,%r8,1),%edi
  mov    %edi,%ecx
  test   %edi,%edi
  je     L18
L51:
  mov    %rdx,%r8
  cmp    $0x2c,%edi
  jne    L16
  movzbl (%rax,%rdx,1),%edi
  cmp    $0x2c,%dil
  jne    L74
  nopl   0x0(%rax)
L92:
  add    $0x1,%rdx
  movzbl (%rax,%rdx,1),%edi
  cmp    $0x2c,%dil
  je     L92
  mov    %rdx,%r8
L74:
  test   %dil,%dil
  jne    L16
  cs nopw 0x0(%rax,%rax,1)
L18:
  add    %rax,%rsi
L19:
  movb   $0x0,(%rsi)
  ret
L78:
  mov    $0x1,%edx
  jmp    L108
L103:
  mov    %rax,%rsi
  jmp    L19
```

assistant: ```c
char *profile_list_normalize(char *list) {
  size_t i = 0;
  size_t j = 0;
  int c;
  while (list[i] == ',')
    ++i;
  while ((c = list[i++])) {
    if (c == ',') {
      while (list[i] == ',')
        ++i;
      if (list[i] == 0)
        break;
    }
    list[j++] = c;
  }
  list[j] = 0;
  return list;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<profile_list_normalize>:
  endbr64
  movzbl (%rdi),%ecx
  mov    %rdi,%rax
  cmp    $0x2c,%cl
  jne    L83
  xor    %esi,%esi
  nopl   (%rax)
L104:
  mov    %rsi,%rdx
  add    $0x1,%rsi
  movzbl (%rax,%rsi,1),%ecx
  cmp    $0x2c,%cl
  je     L104
  add    $0x2,%rdx
L15:
  movsbl %cl,%r8d
  test   %r8d,%r8d
  je     L6
  xor    %edi,%edi
  jmp    L12
  nopl   0x0(%rax)
L89:
  add    $0x1,%rdi
  add    $0x1,%rdx
  mov    %cl,-0x1(%rax,%rdi,1)
  movsbl (%rsi),%r8d
  mov    %r8d,%ecx
  test   %r8d,%r8d
  je     L85
L12:
  lea    (%rax,%rdx,1),%rsi
  cmp    $0x2c,%r8d
  jne    L89
  movzbl (%rsi),%r8d
  cmp    $0x2c,%r8b
  jne    L102
  nopl   0x0(%rax)
L69:
  add    $0x1,%rdx
  movzbl (%rax,%rdx,1),%r8d
  lea    (%rax,%rdx,1),%rsi
  cmp    $0x2c,%r8b
  je     L69
L102:
  test   %r8b,%r8b
  jne    L89
  nopl   0x0(%rax,%rax,1)
L85:
  add    %rax,%rdi
L101:
  movb   $0x0,(%rdi)
  ret
L83:
  mov    $0x1,%edx
  jmp    L15
L6:
  mov    %rax,%rdi
  jmp    L101
```

assistant: ```c
char *profile_list_normalize(char *list) {
  size_t i = 0;
  size_t j = 0;
  int c;
  while (list[i] == ',')
    ++i;
  while ((c = list[i++])) {
    if (c == ',') {
      while (list[i] == ',')
        ++i;
      if (list[i] == 0)
        break;
    }
    list[j++] = c;
  }
  list[j] = 0;
  return list;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<icmp_uint32_t>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x8(%rbp),%rax
  cmp    -0x10(%rbp),%rax
  jle    L23
  mov    $0x1,%eax
  jmp    L33
L23:
  mov    -0x8(%rbp),%rax
  cmp    -0x10(%rbp),%rax
  jge    L17
  mov    $0xffffffff,%eax
  jmp    L33
L17:
  mov    $0x0,%eax
L33:
  pop    %rbp
  ret
```

assistant: ```c
int icmp_uint32_t(uint32_t a, uint32_t b) {
  if (a > b)
    return 1;
  else if (a < b)
    return -1;
  else
    return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<icmp_uint32_t>:
  endbr64
  mov    $0x1,%eax
  cmp    %rsi,%rdi
  jg     L8
  setl   %al
  movzbl %al,%eax
  neg    %eax
L8:
  ret
```

assistant: ```c
int icmp_uint32_t(uint32_t a, uint32_t b) {
  if (a > b)
    return 1;
  else if (a < b)
    return -1;
  else
    return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<icmp_uint32_t>:
  endbr64
  xor    %eax,%eax
  cmp    %rsi,%rdi
  mov    $0x1,%edx
  setl   %al
  neg    %eax
  cmp    %rsi,%rdi
  cmovg  %edx,%eax
  ret
```

assistant: ```c
int icmp_uint32_t(uint32_t a, uint32_t b) {
  if (a > b)
    return 1;
  else if (a < b)
    return -1;
  else
    return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<icmp_uint32_t>:
  endbr64
  xor    %eax,%eax
  cmp    %rsi,%rdi
  mov    $0x1,%edx
  setl   %al
  neg    %eax
  cmp    %rsi,%rdi
  cmovg  %edx,%eax
  ret
```

assistant: ```c
int icmp_uint32_t(uint32_t a, uint32_t b) {
  if (a > b)
    return 1;
  else if (a < b)
    return -1;
  else
    return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<convert_string>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  mov    -0x2c(%rbp),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0x14(%rbp)
  movl   $0x0,-0x10(%rbp)
  jmp    L10
L5:
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    -0x10(%rbp),%edx
  movslq %edx,%rcx
  mov    -0x8(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  addl   $0x2,-0x14(%rbp)
  addl   $0x1,-0x10(%rbp)
L10:
  mov    -0x14(%rbp),%eax
  cmp    -0x2c(%rbp),%eax
  jl     L5
  mov    -0xc(%rbp),%eax
  cltq
  lea    -0x1(%rax),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *convert_string(void *string, int len) {
  int i, k;
  int reallen = len / 2;
  char *cstring = malloc(reallen);
  for (i = 0, k = 0; i < len; i += 2, k++) {
    cstring[k] = ((char *)string)[i];
  }
  cstring[reallen - 1] = '\0';
  return cstring;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<convert_string>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %esi,%r12d
  mov    %esi,%ebp
  shr    $0x1f,%ebp
  add    %esi,%ebp
  sar    %ebp
  mov    %ebp,%edi
  call   <malloc@plt>
  test   %r12d,%r12d
  jle    L0
  sub    $0x1,%r12d
  shr    %r12d
  lea    0x1(%r12),%esi
  mov    $0x0,%edx
L18:
  movzbl (%rbx,%rdx,2),%ecx
  mov    %cl,(%rax,%rdx,1)
  add    $0x1,%rdx
  cmp    %rsi,%rdx
  jne    L18
L0:
  movslq %ebp,%rbp
  movb   $0x0,-0x1(%rax,%rbp,1)
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *convert_string(void *string, int len) {
  int i, k;
  int reallen = len / 2;
  char *cstring = malloc(reallen);
  for (i = 0, k = 0; i < len; i += 2, k++) {
    cstring[k] = ((char *)string)[i];
  }
  cstring[reallen - 1] = '\0';
  return cstring;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<convert_string>:
  endbr64
  push   %r12
  mov    %esi,%r12d
  shr    $0x1f,%r12d
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  add    %esi,%r12d
  mov    %rdi,%rbx
  sar    %r12d
  mov    %r12d,%edi
  call   <malloc@plt>
  test   %ebp,%ebp
  jle    L6
  sub    $0x1,%ebp
  xor    %edx,%edx
  shr    %ebp
  lea    0x1(%rbp),%esi
L11:
  movzbl (%rbx,%rdx,2),%ecx
  mov    %cl,(%rax,%rdx,1)
  add    $0x1,%rdx
  cmp    %rsi,%rdx
  jne    L11
L6:
  movslq %r12d,%r12
  movb   $0x0,-0x1(%rax,%r12,1)
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *convert_string(void *string, int len) {
  int i, k;
  int reallen = len / 2;
  char *cstring = malloc(reallen);
  for (i = 0, k = 0; i < len; i += 2, k++) {
    cstring[k] = ((char *)string)[i];
  }
  cstring[reallen - 1] = '\0';
  return cstring;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    $0x0,%eax
  call   <RunSub@plt>
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  RunSub();
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

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

assistant: ```c
int main(int argc, char **argv) {
  RunSub();
  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
  xor    %eax,%eax
  call   <RunSub@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  RunSub();
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <RunSub@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  RunSub();
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<callrts>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <ot@plt>
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <outstr@plt>
  mov    $0x0,%eax
  call   <nl@plt>
  nop
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D0
type_guessed: string

tool: D0: call	

assistant: ```c
void callrts(char *sname) {
  ot("call\t");
  outstr(sname);
  nl();
}
```
user: Translate the below assembly code into C programming language:

```asm
<callrts>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  lea    D2(%rip),%rdi
  call   <ot@plt>
  mov    %rbx,%rdi
  call   <outstr@plt>
  mov    $0x0,%eax
  call   <nl@plt>
  pop    %rbx
  ret
```

tool_calls: name: read_data
arguments: label: D2
type: string

tool: D2: call	

assistant: ```c
void callrts(char *sname) {
  ot("call\t");
  outstr(sname);
  nl();
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<callrts>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  lea    D3(%rip),%rdi
  call   <ot@plt>
  mov    %rbp,%rdi
  call   <outstr@plt>
  xor    %eax,%eax
  pop    %rbp
  jmp    1080 <nl@plt>
```

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

tool: D3: call	

assistant: ```c
void callrts(char *sname) {
  ot("call\t");
  outstr(sname);
  nl();
}
```
user: What is the c source code of the assembly code below:

```asm
<callrts>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  lea    D4(%rip),%rdi
  call   <ot@plt>
  mov    %rbp,%rdi
  call   <outstr@plt>
  xor    %eax,%eax
  pop    %rbp
  jmp    1080 <nl@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D4
datatype: string

tool: D4: call	

assistant: ```c
void callrts(char *sname) {
  ot("call\t");
  outstr(sname);
  nl();
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<minmax>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %rcx,-0x30(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L18
L2:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x30(%rbp),%rax
  mov    (%rax),%eax
  cmp    %eax,%edx
  jle    L19
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x30(%rbp),%rax
  mov    %edx,(%rax)
L19:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x28(%rbp),%rax
  mov    (%rax),%eax
  cmp    %eax,%edx
  jge    L39
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x28(%rbp),%rax
  mov    %edx,(%rax)
L39:
  addl   $0x1,-0x4(%rbp)
L18:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L2
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void minmax(int *arr, int size, int *min, int *max) {
  int i;
  for (i = 0; i < size; i++) {
    if (arr[i] > *max) {
      *max = arr[i];
    }
    if (arr[i] < *min) {
      *min = arr[i];
    }
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<minmax>:
  endbr64
  test   %esi,%esi
  jle    L38
  mov    %rdi,%rax
  lea    -0x1(%rsi),%esi
  lea    0x4(%rdi,%rsi,4),%r8
  jmp    L40
L4:
  add    $0x4,%rax
  cmp    %r8,%rax
  je     L38
L40:
  mov    %rax,%rdi
  mov    (%rax),%esi
  cmp    (%rcx),%esi
  jle    L7
  mov    %esi,(%rcx)
L7:
  mov    (%rdi),%esi
  cmp    (%rdx),%esi
  jge    L4
  mov    %esi,(%rdx)
  jmp    L4
L38:
  ret
```

assistant: ```c
void minmax(int *arr, int size, int *min, int *max) {
  int i;
  for (i = 0; i < size; i++) {
    if (arr[i] > *max) {
      *max = arr[i];
    }
    if (arr[i] < *min) {
      *min = arr[i];
    }
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<minmax>:
  endbr64
  test   %esi,%esi
  jle    L14
  lea    -0x1(%rsi),%eax
  lea    0x4(%rdi,%rax,4),%rsi
L36:
  mov    (%rdi),%eax
  cmp    (%rcx),%eax
  jle    L39
  mov    %eax,(%rcx)
L39:
  cmp    (%rdx),%eax
  jge    L25
  mov    %eax,(%rdx)
L25:
  add    $0x4,%rdi
  cmp    %rsi,%rdi
  jne    L36
L14:
  ret
```

assistant: ```c
void minmax(int *arr, int size, int *min, int *max) {
  int i;
  for (i = 0; i < size; i++) {
    if (arr[i] > *max) {
      *max = arr[i];
    }
    if (arr[i] < *min) {
      *min = arr[i];
    }
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<minmax>:
  endbr64
  test   %esi,%esi
  jle    L28
  lea    -0x1(%rsi),%eax
  lea    0x4(%rdi,%rax,4),%rsi
L10:
  mov    (%rdi),%eax
  cmp    (%rcx),%eax
  jle    L1
  mov    %eax,(%rcx)
L1:
  cmp    (%rdx),%eax
  jge    L26
  mov    %eax,(%rdx)
L26:
  add    $0x4,%rdi
  cmp    %rsi,%rdi
  jne    L10
L28:
  ret
```

assistant: ```c
void minmax(int *arr, int size, int *min, int *max) {
  int i;
  for (i = 0; i < size; i++) {
    if (arr[i] > *max) {
      *max = arr[i];
    }
    if (arr[i] < *min) {
      *min = arr[i];
    }
  }
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<glaguerre>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x28(%rbp)
  movsd  %xmm0,-0x30(%rbp)
  movsd  %xmm1,-0x38(%rbp)
  cmpq   $0x0,-0x28(%rbp)
  jne    L58
  movsd  D5(%rip),%xmm0
  jmp    L43
L58:
  cmpq   $0x1,-0x28(%rbp)
  jne    L17
  movsd  -0x30(%rbp),%xmm1
  movsd  D5(%rip),%xmm0
  addsd  %xmm1,%xmm0
  subsd  -0x38(%rbp),%xmm0
  jmp    L43
L17:
  movsd  -0x30(%rbp),%xmm1
  movsd  D5(%rip),%xmm0
  addsd  %xmm1,%xmm0
  subsd  -0x38(%rbp),%xmm0
  movsd  %xmm0,-0x18(%rbp)
  movsd  D5(%rip),%xmm0
  movsd  %xmm0,-0x10(%rbp)
  movq   $0x2,-0x20(%rbp)
  jmp    L46
L34:
  mov    -0x20(%rbp),%rax
  sub    $0x1,%rax
  add    %rax,%rax
  pxor   %xmm0,%xmm0
  cvtsi2sd %rax,%xmm0
  movapd %xmm0,%xmm1
  addsd  -0x30(%rbp),%xmm1
  movsd  D5(%rip),%xmm0
  addsd  %xmm1,%xmm0
  subsd  -0x38(%rbp),%xmm0
  mulsd  -0x18(%rbp),%xmm0
  mov    -0x20(%rbp),%rax
  sub    $0x1,%rax
  pxor   %xmm1,%xmm1
  cvtsi2sd %rax,%xmm1
  addsd  -0x30(%rbp),%xmm1
  mulsd  -0x10(%rbp),%xmm1
  subsd  %xmm1,%xmm0
  pxor   %xmm1,%xmm1
  cvtsi2sdq -0x20(%rbp),%xmm1
  divsd  %xmm1,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movsd  -0x18(%rbp),%xmm0
  movsd  %xmm0,-0x10(%rbp)
  movsd  -0x8(%rbp),%xmm0
  movsd  %xmm0,-0x18(%rbp)
  addq   $0x1,-0x20(%rbp)
L46:
  mov    -0x20(%rbp),%rax
  cmp    -0x28(%rbp),%rax
  jle    L34
  movsd  -0x18(%rbp),%xmm0
L43:
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D5
type_infered: f64

tool: D5: 1

assistant: ```c
double glaguerre(long m, double b, double x) {
  long i;
  double gln, glnm1, glnp1;
  if (m == 0)
    return 1.0;
  else {
    if (m == 1)
      return 1.0 + b - x;
    else {
      gln = 1.0 + b - x;
      glnm1 = 1.0;
      for (i = 2; i <= m; i++) {
        glnp1 = ((2 * (i - 1) + b + 1.0 - x) * gln - (i - 1 + b) * glnm1) / i;
        glnm1 = gln;
        gln = glnp1;
      }
      return gln;
    }
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<glaguerre>:
  endbr64
  mov    %rdi,%rcx
  movapd %xmm0,%xmm4
  test   %rdi,%rdi
  je     L11
  cmp    $0x1,%rdi
  je     L8
  movapd %xmm0,%xmm3
  addsd  D12(%rip),%xmm3
  subsd  %xmm1,%xmm3
  jle    L17
  mov    $0x1,%eax
  movsd  D12(%rip),%xmm5
  movapd %xmm5,%xmm6
  jmp    L12
L8:
  addsd  D12(%rip),%xmm4
  movapd %xmm4,%xmm0
  subsd  %xmm1,%xmm0
  ret
L25:
  movapd %xmm0,%xmm3
L12:
  mov    %rax,%rdx
  add    $0x1,%rax
  lea    (%rdx,%rdx,1),%rsi
  pxor   %xmm0,%xmm0
  cvtsi2sd %rsi,%xmm0
  addsd  %xmm4,%xmm0
  addsd  %xmm6,%xmm0
  subsd  %xmm1,%xmm0
  mulsd  %xmm3,%xmm0
  pxor   %xmm2,%xmm2
  cvtsi2sd %rdx,%xmm2
  addsd  %xmm4,%xmm2
  mulsd  %xmm5,%xmm2
  subsd  %xmm2,%xmm0
  pxor   %xmm2,%xmm2
  cvtsi2sd %rax,%xmm2
  divsd  %xmm2,%xmm0
  movapd %xmm3,%xmm5
  cmp    %rcx,%rax
  jne    L25
  ret
L11:
  movsd  D12(%rip),%xmm0
  ret
L17:
  movapd %xmm3,%xmm0
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D12
type_infered: f64

tool: D12: 1

assistant: ```c
double glaguerre(long m, double b, double x) {
  long i;
  double gln, glnm1, glnp1;
  if (m == 0)
    return 1.0;
  else {
    if (m == 1)
      return 1.0 + b - x;
    else {
      gln = 1.0 + b - x;
      glnm1 = 1.0;
      for (i = 2; i <= m; i++) {
        glnp1 = ((2 * (i - 1) + b + 1.0 - x) * gln - (i - 1 + b) * glnm1) / i;
        glnm1 = gln;
        gln = glnp1;
      }
      return gln;
    }
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<glaguerre>:
  endbr64
  movapd %xmm0,%xmm4
  test   %rdi,%rdi
  je     L11
  movsd  D17(%rip),%xmm6
  movapd %xmm0,%xmm3
  addsd  %xmm6,%xmm3
  subsd  %xmm1,%xmm3
  cmp    $0x1,%rdi
  jle    L23
  mov    $0x1,%eax
  movapd %xmm6,%xmm5
  jmp    L16
  cs nopw 0x0(%rax,%rax,1)
L8:
  movapd %xmm0,%xmm3
L16:
  mov    %rax,%rdx
  pxor   %xmm0,%xmm0
  pxor   %xmm2,%xmm2
  add    $0x1,%rax
  cvtsi2sd %rdx,%xmm2
  lea    (%rdx,%rdx,1),%rcx
  cvtsi2sd %rcx,%xmm0
  addsd  %xmm4,%xmm2
  addsd  %xmm4,%xmm0
  mulsd  %xmm5,%xmm2
  movapd %xmm3,%xmm5
  addsd  %xmm6,%xmm0
  subsd  %xmm1,%xmm0
  mulsd  %xmm3,%xmm0
  subsd  %xmm2,%xmm0
  pxor   %xmm2,%xmm2
  cvtsi2sd %rax,%xmm2
  divsd  %xmm2,%xmm0
  cmp    %rax,%rdi
  jne    L8
  ret
  nopl   0x0(%rax)
L11:
  movsd  D17(%rip),%xmm0
  ret
  nopl   0x0(%rax)
L23:
  movapd %xmm3,%xmm0
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D17
type: f64

tool: D17: 1

assistant: ```c
double glaguerre(long m, double b, double x) {
  long i;
  double gln, glnm1, glnp1;
  if (m == 0)
    return 1.0;
  else {
    if (m == 1)
      return 1.0 + b - x;
    else {
      gln = 1.0 + b - x;
      glnm1 = 1.0;
      for (i = 2; i <= m; i++) {
        glnp1 = ((2 * (i - 1) + b + 1.0 - x) * gln - (i - 1 + b) * glnm1) / i;
        glnm1 = gln;
        gln = glnp1;
      }
      return gln;
    }
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<glaguerre>:
  endbr64
  movapd %xmm0,%xmm4
  test   %rdi,%rdi
  je     L9
  movsd  D19(%rip),%xmm6
  movapd %xmm0,%xmm3
  addsd  %xmm6,%xmm3
  subsd  %xmm1,%xmm3
  cmp    $0x1,%rdi
  jle    L32
  mov    $0x1,%eax
  movapd %xmm6,%xmm5
  jmp    L24
  cs nopw 0x0(%rax,%rax,1)
L33:
  movapd %xmm0,%xmm3
L24:
  mov    %rax,%rdx
  pxor   %xmm0,%xmm0
  pxor   %xmm2,%xmm2
  add    $0x1,%rax
  cvtsi2sd %rdx,%xmm2
  lea    (%rdx,%rdx,1),%rcx
  cvtsi2sd %rcx,%xmm0
  addsd  %xmm4,%xmm2
  addsd  %xmm4,%xmm0
  mulsd  %xmm5,%xmm2
  movapd %xmm3,%xmm5
  addsd  %xmm6,%xmm0
  subsd  %xmm1,%xmm0
  mulsd  %xmm3,%xmm0
  subsd  %xmm2,%xmm0
  pxor   %xmm2,%xmm2
  cvtsi2sd %rax,%xmm2
  divsd  %xmm2,%xmm0
  cmp    %rax,%rdi
  jne    L33
  ret
  nopl   0x0(%rax)
L9:
  movsd  D19(%rip),%xmm0
  ret
  nopl   0x0(%rax)
L32:
  movapd %xmm3,%xmm0
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D19
type: f64

tool: D19: 1

assistant: ```c
double glaguerre(long m, double b, double x) {
  long i;
  double gln, glnm1, glnp1;
  if (m == 0)
    return 1.0;
  else {
    if (m == 1)
      return 1.0 + b - x;
    else {
      gln = 1.0 + b - x;
      glnm1 = 1.0;
      for (i = 2; i <= m; i++) {
        glnp1 = ((2 * (i - 1) + b + 1.0 - x) * gln - (i - 1 + b) * glnm1) / i;
        glnm1 = gln;
        gln = glnp1;
      }
      return gln;
    }
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<RandomFunc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  sub    $0x6a,%eax
  mov    %al,-0x9(%rbp)
  movzbl -0x9(%rbp),%eax
  shr    $0x4,%al
  movzbl %al,%eax
  and    $0x1,%eax
  test   %eax,%eax
  je     L46
  movzbl -0x9(%rbp),%eax
  shr    $0x3,%al
  movzbl %al,%eax
  and    $0x1,%eax
  test   %eax,%eax
  je     L57
  movzbl -0x9(%rbp),%eax
  add    %eax,%eax
  mov    %al,-0x9(%rbp)
  movzbl -0x9(%rbp),%eax
  add    %eax,%eax
  mov    %al,-0x9(%rbp)
  jmp    L5
L57:
  movzbl -0x9(%rbp),%eax
  add    %eax,%eax
  mov    %al,-0x9(%rbp)
  movzbl -0x9(%rbp),%eax
  add    %eax,%eax
  mov    %al,-0x9(%rbp)
  jmp    L5
L46:
  movzbl -0x9(%rbp),%eax
  shr    %al
  movzbl %al,%eax
  and    $0x1,%eax
  test   %eax,%eax
  je     L53
  movzbl -0x9(%rbp),%eax
  add    %eax,%eax
  mov    %al,-0x9(%rbp)
  movzbl -0x9(%rbp),%eax
  add    %eax,%eax
  mov    %al,-0x9(%rbp)
  jmp    L5
L53:
  movzbl -0x9(%rbp),%eax
  add    %eax,%eax
  mov    %al,-0x9(%rbp)
  movzbl -0x9(%rbp),%eax
  add    %eax,%eax
  mov    %al,-0x9(%rbp)
L5:
  movzbl -0x9(%rbp),%eax
  lea    0x4(%rax),%edx
  mov    -0x20(%rbp),%rax
  mov    %dl,(%rax)
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L10
  call   <__stack_chk_fail@plt>
L10:
  leave
  ret
```

assistant: ```c
void RandomFunc(unsigned char input[1], unsigned char output[1]) {
  unsigned char state[1];
  {
    state[0UL] = (input[0UL] + 51238316UL) + (unsigned char)234;
    if ((state[0UL] >> (unsigned char)4) & (unsigned char)1) {
      if ((state[0UL] >> (unsigned char)3) & (unsigned char)1) {
        state[0UL] += state[0UL];
        state[0UL] += state[0UL];
      } else {
        state[0UL] += state[0UL];
        state[0UL] += state[0UL];
      }
    } else if ((state[0UL] >> (unsigned char)1) & (unsigned char)1) {
      state[0UL] += state[0UL];
      state[0UL] += state[0UL];
    } else {
      state[0UL] += state[0UL];
      state[0UL] += state[0UL];
    }
    output[0UL] = state[0UL] + 792485636UL;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<RandomFunc>:
  endbr64
  movzbl (%rdi),%eax
  sub    $0x6a,%eax
  shl    $0x2,%eax
  add    $0x4,%eax
  mov    %al,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned char input[1], unsigned char output[1]) {
  unsigned char state[1];
  {
    state[0UL] = (input[0UL] + 51238316UL) + (unsigned char)234;
    if ((state[0UL] >> (unsigned char)4) & (unsigned char)1) {
      if ((state[0UL] >> (unsigned char)3) & (unsigned char)1) {
        state[0UL] += state[0UL];
        state[0UL] += state[0UL];
      } else {
        state[0UL] += state[0UL];
        state[0UL] += state[0UL];
      }
    } else if ((state[0UL] >> (unsigned char)1) & (unsigned char)1) {
      state[0UL] += state[0UL];
      state[0UL] += state[0UL];
    } else {
      state[0UL] += state[0UL];
      state[0UL] += state[0UL];
    }
    output[0UL] = state[0UL] + 792485636UL;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<RandomFunc>:
  endbr64
  movzbl (%rdi),%eax
  lea    0x5c(,%rax,4),%eax
  mov    %al,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned char input[1], unsigned char output[1]) {
  unsigned char state[1];
  {
    state[0UL] = (input[0UL] + 51238316UL) + (unsigned char)234;
    if ((state[0UL] >> (unsigned char)4) & (unsigned char)1) {
      if ((state[0UL] >> (unsigned char)3) & (unsigned char)1) {
        state[0UL] += state[0UL];
        state[0UL] += state[0UL];
      } else {
        state[0UL] += state[0UL];
        state[0UL] += state[0UL];
      }
    } else if ((state[0UL] >> (unsigned char)1) & (unsigned char)1) {
      state[0UL] += state[0UL];
      state[0UL] += state[0UL];
    } else {
      state[0UL] += state[0UL];
      state[0UL] += state[0UL];
    }
    output[0UL] = state[0UL] + 792485636UL;
  }
}
```