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

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D12(%rip),%rdi
  call   <printf@plt>
  lea    D5(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: label: D12
type: string
name: disassemble_data
arguments: label: D5
type: string

tool: D12: Hello World!


tool: D5: SAB Rube Goldberg Hackathon!!!

assistant: ```c
int main() {
  printf("Hello World!\n");
  printf("SAB Rube Goldberg Hackathon!!!");
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D4(%rip),%rdi
  call   <printf@plt>
  lea    D14(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D4
type: string
name: disassemble_data
arguments: data_label: D14
type: string

tool: D4: Hello World!


tool: D14: SAB Rube Goldberg Hackathon!!!

assistant: ```c
int main() {
  printf("Hello World!\n");
  printf("SAB Rube Goldberg Hackathon!!!");
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D16(%rip),%rdi
  call   <printf@plt>
  lea    D0(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D16
type_guessed: string
name: retrieve_data
arguments: tag: D0
type_guessed: string

tool: D16: Hello World!


tool: D0: SAB Rube Goldberg Hackathon!!!

assistant: ```c
int main() {
  printf("Hello World!\n");
  printf("SAB Rube Goldberg Hackathon!!!");
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<ntlibc_utoa>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x24(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %edx,-0x28(%rbp)
  mov    -0x30(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x30(%rbp),%rax
  mov    %rax,-0x8(%rbp)
L25:
  mov    -0x28(%rbp),%ecx
  mov    -0x24(%rbp),%eax
  mov    $0x0,%edx
  div    %ecx
  mov    %edx,%ecx
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  mov    %ecx,%ecx
  lea    D6(%rip),%rdx
  movzbl (%rcx,%rdx,1),%edx
  mov    %dl,(%rax)
  mov    -0x28(%rbp),%esi
  mov    -0x24(%rbp),%eax
  mov    $0x0,%edx
  div    %esi
  mov    %eax,-0x24(%rbp)
  cmpl   $0x0,-0x24(%rbp)
  jne    L25
  mov    -0x8(%rbp),%rax
  lea    -0x1(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  movb   $0x0,(%rax)
  jmp    L14
L3:
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  mov    %al,-0x11(%rbp)
  mov    -0x10(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x10(%rbp)
  mov    -0x8(%rbp),%rdx
  movzbl (%rdx),%edx
  mov    %dl,(%rax)
  mov    -0x8(%rbp),%rax
  lea    -0x1(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  movzbl -0x11(%rbp),%edx
  mov    %dl,(%rax)
L14:
  mov    -0x10(%rbp),%rax
  cmp    -0x8(%rbp),%rax
  jb     L3
  mov    -0x30(%rbp),%rax
  pop    %rbp
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D6
datatype: string

tool: D6: 0123456789abcdefghijklmnopqrstuvwxyz

assistant: ```c
char *ntlibc_utoa(unsigned int value, char *s, int radix) {
  char *s1 = s;
  char *s2 = s;
  do {
    *s2++ = "0123456789abcdefghijklmnopqrstuvwxyz"[value % radix];
    value /= radix;
  } while (value > 0);
  *s2-- = '\0';
  while (s1 < s2) {
    char c = *s1;
    *s1++ = *s2;
    *s2-- = c;
  }
  return s;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<ntlibc_utoa>:
  endbr64
  mov    %rsi,%r9
  mov    %edx,%r8d
  lea    D1(%rip),%r10
L11:
  mov    %rsi,%rcx
  add    $0x1,%rsi
  mov    %edi,%eax
  mov    $0x0,%edx
  div    %r8d
  movzbl (%r10,%rdx,1),%edx
  mov    %dl,-0x1(%rsi)
  mov    %edi,%edx
  mov    %eax,%edi
  cmp    %edx,%r8d
  jbe    L11
  movb   $0x0,(%rsi)
  cmp    %r9,%rcx
  jbe    L26
  mov    %r9,%rax
L17:
  movzbl (%rax),%edx
  add    $0x1,%rax
  movzbl (%rcx),%esi
  mov    %sil,-0x1(%rax)
  sub    $0x1,%rcx
  mov    %dl,0x1(%rcx)
  cmp    %rcx,%rax
  jb     L17
L26:
  mov    %r9,%rax
  ret
```

tool_calls: name: unpack_data
arguments: tag: D1
type_infered: string

tool: D1: 0123456789abcdefghijklmnopqrstuvwxyz

assistant: ```c
char *ntlibc_utoa(unsigned int value, char *s, int radix) {
  char *s1 = s;
  char *s2 = s;
  do {
    *s2++ = "0123456789abcdefghijklmnopqrstuvwxyz"[value % radix];
    value /= radix;
  } while (value > 0);
  *s2-- = '\0';
  while (s1 < s2) {
    char c = *s1;
    *s1++ = *s2;
    *s2-- = c;
  }
  return s;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ntlibc_utoa>:
  endbr64
  push   %r12
  mov    %edx,%r8d
  mov    %rsi,%r12
  mov    %rsi,%r9
  lea    D0(%rip),%r11
  cs nopw 0x0(%rax,%rax,1)
L3:
  mov    %edi,%eax
  xor    %edx,%edx
  mov    %r9,%rcx
  mov    %edi,%r10d
  div    %r8d
  lea    0x1(%r9),%r9
  movzbl (%r11,%rdx,1),%edx
  mov    %eax,%edi
  mov    %dl,-0x1(%r9)
  cmp    %r8d,%r10d
  jae    L3
  movb   $0x0,(%r9)
  cmp    %r12,%rcx
  jbe    L20
  mov    %r12,%rax
  jmp    L13
  nop
L10:
  movzbl (%rcx),%edx
L13:
  movzbl (%rax),%edi
  sub    $0x1,%rcx
  add    $0x1,%rax
  mov    %dl,-0x1(%rax)
  mov    %dil,0x1(%rcx)
  cmp    %rcx,%rax
  jb     L10
L20:
  mov    %r12,%rax
  pop    %r12
  ret
```

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

tool: D0: 0123456789abcdefghijklmnopqrstuvwxyz

assistant: ```c
char *ntlibc_utoa(unsigned int value, char *s, int radix) {
  char *s1 = s;
  char *s2 = s;
  do {
    *s2++ = "0123456789abcdefghijklmnopqrstuvwxyz"[value % radix];
    value /= radix;
  } while (value > 0);
  *s2-- = '\0';
  while (s1 < s2) {
    char c = *s1;
    *s1++ = *s2;
    *s2-- = c;
  }
  return s;
}
```
user: Translate the below assembly code into C programming language:

```asm
<ntlibc_utoa>:
  endbr64
  mov    %rsi,%r9
  lea    D3(%rip),%r10
  mov    %edx,%esi
  mov    %r9,%r8
  nopl   0x0(%rax,%rax,1)
L21:
  mov    %edi,%eax
  xor    %edx,%edx
  mov    %r8,%rcx
  div    %esi
  lea    0x1(%r8),%r8
  movzbl (%r10,%rdx,1),%edx
  mov    %dl,-0x1(%r8)
  mov    %edi,%edx
  mov    %eax,%edi
  cmp    %esi,%edx
  jae    L21
  movb   $0x0,(%r8)
  mov    %r9,%rax
  cmp    %r9,%rcx
  jbe    L23
  nopw   0x0(%rax,%rax,1)
L12:
  movzbl (%rax),%edx
  movzbl (%rcx),%esi
  add    $0x1,%rax
  sub    $0x1,%rcx
  mov    %sil,-0x1(%rax)
  mov    %dl,0x1(%rcx)
  cmp    %rcx,%rax
  jb     L12
L23:
  mov    %r9,%rax
  ret
```

tool_calls: name: read_data
arguments: tag: D3
type_guessed: string

tool: D3: 0123456789abcdefghijklmnopqrstuvwxyz

assistant: ```c
char *ntlibc_utoa(unsigned int value, char *s, int radix) {
  char *s1 = s;
  char *s2 = s;
  do {
    *s2++ = "0123456789abcdefghijklmnopqrstuvwxyz"[value % radix];
    value /= radix;
  } while (value > 0);
  *s2-- = '\0';
  while (s1 < s2) {
    char c = *s1;
    *s1++ = *s2;
    *s2-- = c;
  }
  return s;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<f>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  add    $0x1,%eax
  test   %eax,%eax
  je     L45
  test   %eax,%eax
  js     L29
  sub    $0x1,%eax
  cmp    $0x1,%eax
  ja     L29
  jmp    L46
L45:
  mov    -0x14(%rbp),%eax
  mov    %eax,-0x4(%rbp)
L46:
  movl   $0x3,-0x4(%rbp)
  jmp    L32
L29:
  movl   $0x4,-0x4(%rbp)
L32:
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int f(int x) {
  int y;
  switch (x + 1) {
  case 0:
    y = x;
  case 1:
  case 2:
    y = 3;
    break;
  default:
    y = 4;
  }
  return y;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<f>:
  endbr64
  add    $0x1,%edi
  cmp    $0x2,%edi
  seta   %al
  movzbl %al,%eax
  add    $0x3,%eax
  ret
```

assistant: ```c
int f(int x) {
  int y;
  switch (x + 1) {
  case 0:
    y = x;
  case 1:
  case 2:
    y = 3;
    break;
  default:
    y = 4;
  }
  return y;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<f>:
  endbr64
  add    $0x1,%edi
  xor    %eax,%eax
  cmp    $0x2,%edi
  seta   %al
  add    $0x3,%eax
  ret
```

assistant: ```c
int f(int x) {
  int y;
  switch (x + 1) {
  case 0:
    y = x;
  case 1:
  case 2:
    y = 3;
    break;
  default:
    y = 4;
  }
  return y;
}
```
user: Translate the below assembly code into C programming language:

```asm
<f>:
  endbr64
  add    $0x1,%edi
  xor    %eax,%eax
  cmp    $0x2,%edi
  seta   %al
  add    $0x3,%eax
  ret
```

assistant: ```c
int f(int x) {
  int y;
  switch (x + 1) {
  case 0:
    y = x;
  case 1:
  case 2:
    y = 3;
    break;
  default:
    y = 4;
  }
  return y;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<input>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  movl   $0x0,-0x14(%rbp)
L15:
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  lea    (%rdx,%rax,1),%rbx
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %al,(%rbx)
  mov    -0x14(%rbp),%eax
  cltq
  cmp    %rax,-0x30(%rbp)
  ja     L35
  addq   $0x14,-0x30(%rbp)
  mov    -0x30(%rbp),%rdx
  mov    -0x28(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <realloc@plt>
  mov    %rax,-0x28(%rbp)
L35:
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0xa,%al
  jne    L38
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  jmp    L33
L38:
  addl   $0x1,-0x14(%rbp)
  jmp    L15
L33:
  mov    -0x28(%rbp),%rax
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
char *input(char *dest, size_t size) {
  int i = 0;
  while (1) {
    dest[i] = getchar();
    if (i >= size) {
      size += 20;
      dest = realloc(dest, size);
    }
    if (dest[i] == '\n') {
      dest[i] = '\0';
      break;
    }
    i++;
  }
  return dest;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<input>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    %rsi,%r12
  mov    $0x0,%ebx
  jmp    L39
L33:
  lea    0x0(%rbp,%rbx,1),%rax
  add    $0x1,%rbx
  cmpb   $0xa,(%rax)
  je     L11
L39:
  lea    0x0(%rbp,%rbx,1),%r13
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %al,0x0(%r13)
  cmp    %rbx,%r12
  ja     L33
  add    $0x14,%r12
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <realloc@plt>
  mov    %rax,%rbp
  jmp    L33
L11:
  movb   $0x0,(%rax)
  mov    %rbp,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *input(char *dest, size_t size) {
  int i = 0;
  while (1) {
    dest[i] = getchar();
    if (i >= size) {
      size += 20;
      dest = realloc(dest, size);
    }
    if (dest[i] == '\n') {
      dest[i] = '\0';
      break;
    }
    i++;
  }
  return dest;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<input>:
  endbr64
  push   %r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  jmp    L3
  nopl   0x0(%rax,%rax,1)
L24:
  add    $0x1,%rbx
  cmp    $0xa,%al
  je     L7
L3:
  xor    %eax,%eax
  lea    0x0(%rbp,%rbx,1),%r13
  call   <getchar@plt>
  mov    %al,0x0(%r13)
  cmp    %r12,%rbx
  jb     L24
  add    $0x14,%r12
  mov    %rbp,%rdi
  mov    %r12,%rsi
  call   <realloc@plt>
  lea    (%rax,%rbx,1),%r13
  mov    %rax,%rbp
  add    $0x1,%rbx
  movzbl 0x0(%r13),%eax
  cmp    $0xa,%al
  jne    L3
L7:
  movb   $0x0,0x0(%r13)
  add    $0x8,%rsp
  mov    %rbp,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *input(char *dest, size_t size) {
  int i = 0;
  while (1) {
    dest[i] = getchar();
    if (i >= size) {
      size += 20;
      dest = realloc(dest, size);
    }
    if (dest[i] == '\n') {
      dest[i] = '\0';
      break;
    }
    i++;
  }
  return dest;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<input>:
  endbr64
  push   %r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  cs nopw 0x0(%rax,%rax,1)
L39:
  xor    %eax,%eax
  lea    0x0(%rbp,%rbx,1),%r13
  call   <getchar@plt>
  mov    %al,0x0(%r13)
  cmp    %r12,%rbx
  jae    L5
  add    $0x1,%rbx
  cmp    $0xa,%al
  jne    L39
L11:
  movb   $0x0,0x0(%r13)
  add    $0x8,%rsp
  mov    %rbp,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L5:
  add    $0x14,%r12
  mov    %rbp,%rdi
  mov    %r12,%rsi
  call   <realloc@plt>
  lea    (%rax,%rbx,1),%r13
  add    $0x1,%rbx
  mov    %rax,%rbp
  cmpb   $0xa,0x0(%r13)
  jne    L39
  jmp    L11
```

assistant: ```c
char *input(char *dest, size_t size) {
  int i = 0;
  while (1) {
    dest[i] = getchar();
    if (i >= size) {
      size += 20;
      dest = realloc(dest, size);
    }
    if (dest[i] == '\n') {
      dest[i] = '\0';
      break;
    }
    i++;
  }
  return dest;
}
```
user: What is the c source code of the assembly code below:

```asm
<shift>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  mov    %edx,-0x10(%rbp)
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x10(%rbp),%eax
  add    %edx,%eax
  lea    0xa(%rax),%ecx
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x8(%rbp),%rax
  lea    (%rdx,%rax,1),%rsi
  movslq %ecx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %ecx,%edi
  sar    $0x1f,%edi
  sub    %edi,%eax
  mov    %eax,%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  sub    %eax,%ecx
  mov    %ecx,%edx
  mov    %edx,(%rsi)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void shift(int x[], int d, int steps) {
  x[d] = (x[d] + steps + 10) % 10;
  return;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<shift>:
  endbr64
  movslq %esi,%rsi
  lea    (%rdi,%rsi,4),%rcx
  add    (%rcx),%edx
  add    $0xa,%edx
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  mov    %edx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  lea    (%rax,%rax,4),%eax
  add    %eax,%eax
  sub    %eax,%edx
  mov    %edx,(%rcx)
  ret
```

assistant: ```c
void shift(int x[], int d, int steps) {
  x[d] = (x[d] + steps + 10) % 10;
  return;
}
```
user: Translate the below assembly code into C programming language:

```asm
<shift>:
  endbr64
  movslq %esi,%rsi
  lea    (%rdi,%rsi,4),%rcx
  add    (%rcx),%edx
  add    $0xa,%edx
  movslq %edx,%rax
  mov    %edx,%esi
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%esi
  sar    $0x22,%rax
  sub    %esi,%eax
  lea    (%rax,%rax,4),%eax
  add    %eax,%eax
  sub    %eax,%edx
  mov    %edx,(%rcx)
  ret
```

assistant: ```c
void shift(int x[], int d, int steps) {
  x[d] = (x[d] + steps + 10) % 10;
  return;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<shift>:
  endbr64
  movslq %esi,%rsi
  lea    (%rdi,%rsi,4),%rcx
  add    (%rcx),%edx
  add    $0xa,%edx
  movslq %edx,%rax
  mov    %edx,%esi
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%esi
  sar    $0x22,%rax
  sub    %esi,%eax
  lea    (%rax,%rax,4),%eax
  add    %eax,%eax
  sub    %eax,%edx
  mov    %edx,(%rcx)
  ret
```

assistant: ```c
void shift(int x[], int d, int steps) {
  x[d] = (x[d] + steps + 10) % 10;
  return;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x10(%rbp),%rdi
  lea    -0x14(%rbp),%rsi
  lea    -0x18(%rbp),%rcx
  lea    -0x1c(%rbp),%rdx
  lea    -0x20(%rbp),%rax
  mov    %rdi,%r9
  mov    %rsi,%r8
  mov    %rax,%rsi
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  movss  -0x20(%rbp),%xmm0
  movss  -0x14(%rbp),%xmm2
  movaps %xmm0,%xmm1
  subss  %xmm2,%xmm1
  movss  -0x20(%rbp),%xmm0
  movss  -0x14(%rbp),%xmm2
  subss  %xmm2,%xmm0
  mulss  %xmm0,%xmm1
  movss  -0x1c(%rbp),%xmm0
  movss  -0x10(%rbp),%xmm3
  movaps %xmm0,%xmm2
  subss  %xmm3,%xmm2
  movss  -0x1c(%rbp),%xmm0
  movss  -0x10(%rbp),%xmm3
  subss  %xmm3,%xmm0
  mulss  %xmm2,%xmm0
  addss  %xmm0,%xmm1
  movd   %xmm1,%eax
  movd   %eax,%xmm0
  call   <sqrt@plt>
  movd   %xmm0,%eax
  mov    %eax,-0xc(%rbp)
  movss  -0x18(%rbp),%xmm0
  comiss -0xc(%rbp),%xmm0
  jbe    L43
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L7
L43:
  movss  -0x18(%rbp),%xmm0
  ucomiss -0xc(%rbp),%xmm0
  jp     L26
  ucomiss -0xc(%rbp),%xmm0
  jne    L26
  lea    D25(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L7
L26:
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L7:
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L46
  call   <__stack_chk_fail@plt>
L46:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D10
datatype: string
name: extract_symbolic_data
arguments: data_label: D12
datatype: string
name: extract_symbolic_data
arguments: data_label: D25
datatype: string
name: extract_symbolic_data
arguments: data_label: D8
datatype: string

tool: D10: %f %f %f %f %f

tool: D12: Point is inside the Circle.

tool: D25: Point is on the Circle.

tool: D8: Point is outside the Circle.

assistant: ```c
int main() {
  float x, y, r, x1, y1, s;
  scanf("%f %f %f %f %f", &x, &y, &r, &x1, &y1);
  s = sqrt((x - x1) * (x - x1) + (y - y1) * (y - y1));
  if (r > s) {
    printf("Point is inside the Circle.");
  } else if (r == s) {
    printf("Point is on the Circle.");
  } else {
    printf("Point is outside the Circle.");
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rcx
  lea    0x8(%rsp),%rdx
  lea    0x4(%rsp),%rsi
  lea    0x14(%rsp),%r9
  lea    0x10(%rsp),%r8
  lea    D16(%rip),%rdi
  call   <scanf@plt>
  movss  0x4(%rsp),%xmm0
  subss  0x10(%rsp),%xmm0
  movss  0x8(%rsp),%xmm1
  subss  0x14(%rsp),%xmm1
  mulss  %xmm0,%xmm0
  mulss  %xmm1,%xmm1
  addss  %xmm1,%xmm0
  call   <sqrt@plt>
  movss  0xc(%rsp),%xmm1
  comiss %xmm0,%xmm1
  ja     L32
  ucomiss %xmm0,%xmm1
  jp     L54
  jne    L54
  lea    D10(%rip),%rdi
  call   <printf@plt>
L38:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  mov    $0x0,%eax
  add    $0x28,%rsp
  ret
L32:
  lea    D8(%rip),%rdi
  call   <printf@plt>
  jmp    L38
L54:
  lea    D21(%rip),%rdi
  call   <printf@plt>
  jmp    L38
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D16
type_guessed: string
name: retrieve_data
arguments: data_label: D10
type_guessed: string
name: retrieve_data
arguments: data_label: D8
type_guessed: string
name: retrieve_data
arguments: data_label: D21
type_guessed: string

tool: D16: %f %f %f %f %f

tool: D10: Point is on the Circle.

tool: D8: Point is inside the Circle.

tool: D21: Point is outside the Circle.

assistant: ```c
int main() {
  float x, y, r, x1, y1, s;
  scanf("%f %f %f %f %f", &x, &y, &r, &x1, &y1);
  s = sqrt((x - x1) * (x - x1) + (y - y1) * (y - y1));
  if (r > s) {
    printf("Point is inside the Circle.");
  } else if (r == s) {
    printf("Point is on the Circle.");
  } else {
    printf("Point is outside the Circle.");
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  lea    D0(%rip),%rdi
  lea    0xc(%rsp),%rcx
  lea    0x8(%rsp),%rdx
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  lea    0x14(%rsp),%r9
  lea    0x10(%rsp),%r8
  call   <scanf@plt>
  movss  0x4(%rsp),%xmm0
  movss  0x8(%rsp),%xmm1
  subss  0x10(%rsp),%xmm0
  subss  0x14(%rsp),%xmm1
  mulss  %xmm0,%xmm0
  mulss  %xmm1,%xmm1
  addss  %xmm1,%xmm0
  call   <sqrt@plt>
  movss  0xc(%rsp),%xmm1
  comiss %xmm0,%xmm1
  ja     L14
  ucomiss %xmm0,%xmm1
  jp     L0
  jne    L0
  lea    D29(%rip),%rdi
  call   <printf@plt>
L8:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L27
  xor    %eax,%eax
  add    $0x28,%rsp
  ret
L0:
  lea    D4(%rip),%rdi
  call   <printf@plt>
  jmp    L8
L14:
  lea    D11(%rip),%rdi
  call   <printf@plt>
  jmp    L8
L27:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D0
type_guessed: string
name: unpack_data
arguments: data_tag: D29
type_guessed: string
name: unpack_data
arguments: data_tag: D4
type_guessed: string
name: unpack_data
arguments: data_tag: D11
type_guessed: string

tool: D0: %f %f %f %f %f

tool: D29: Point is on the Circle.

tool: D4: Point is outside the Circle.

tool: D11: Point is inside the Circle.

assistant: ```c
int main() {
  float x, y, r, x1, y1, s;
  scanf("%f %f %f %f %f", &x, &y, &r, &x1, &y1);
  s = sqrt((x - x1) * (x - x1) + (y - y1) * (y - y1));
  if (r > s) {
    printf("Point is inside the Circle.");
  } else if (r == s) {
    printf("Point is on the Circle.");
  } else {
    printf("Point is outside the Circle.");
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  lea    D19(%rip),%rdi
  lea    0xc(%rsp),%rcx
  lea    0x8(%rsp),%rdx
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  lea    0x14(%rsp),%r9
  lea    0x10(%rsp),%r8
  call   <scanf@plt>
  movss  0x4(%rsp),%xmm0
  movss  0x8(%rsp),%xmm1
  subss  0x10(%rsp),%xmm0
  subss  0x14(%rsp),%xmm1
  mulss  %xmm0,%xmm0
  mulss  %xmm1,%xmm1
  addss  %xmm1,%xmm0
  call   <sqrt@plt>
  movss  0xc(%rsp),%xmm1
  comiss %xmm0,%xmm1
  ja     L30
  ucomiss %xmm0,%xmm1
  jp     L49
  jne    L49
  lea    D22(%rip),%rdi
  call   <printf@plt>
L45:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L34
  xor    %eax,%eax
  add    $0x28,%rsp
  ret
L49:
  lea    D15(%rip),%rdi
  call   <printf@plt>
  jmp    L45
L30:
  lea    D10(%rip),%rdi
  call   <printf@plt>
  jmp    L45
L34:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D19
datatype: string
name: read_data
arguments: label: D22
datatype: string
name: read_data
arguments: label: D15
datatype: string
name: read_data
arguments: label: D10
datatype: string

tool: D19: %f %f %f %f %f

tool: D22: Point is on the Circle.

tool: D15: Point is outside the Circle.

tool: D10: Point is inside the Circle.

assistant: ```c
int main() {
  float x, y, r, x1, y1, s;
  scanf("%f %f %f %f %f", &x, &y, &r, &x1, &y1);
  s = sqrt((x - x1) * (x - x1) + (y - y1) * (y - y1));
  if (r > s) {
    printf("Point is inside the Circle.");
  } else if (r == s) {
    printf("Point is on the Circle.");
  } else {
    printf("Point is outside the Circle.");
  }
  return 0;
}
```