ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Translate the below assembly code into C programming language:

```asm
<byte2wav>:
  endbr64
  push   %r13
  xor    %r13d,%r13d
  push   %r12
  mov    %edi,%r12d
  mov    %rsi,%rdi
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  call   <generate_0@plt>
  jmp    L27
  nopl   0x0(%rax,%rax,1)
L5:
  mov    %rbp,%rdi
  add    $0x1,%ebx
  call   <generate_0@plt>
  cmp    $0x8,%ebx
  je     L14
L27:
  mov    %ebx,%esi
  mov    $0x2,%edi
  call   <pow@plt>
  mov    %r12d,%edx
  and    %eax,%edx
  cmp    %eax,%edx
  jne    L5
  mov    %rbp,%rdi
  add    $0x1,%ebx
  add    $0x1,%r13d
  call   <generate_1@plt>
  cmp    $0x8,%ebx
  jne    L27
L14:
  and    $0x1,%r13d
  mov    %rbp,%rdi
  jne    L19
  call   <generate_1@plt>
L8:
  mov    %rbp,%rdi
  call   <generate_1@plt>
  mov    %rbp,%rdi
  call   <generate_1@plt>
  add    $0x8,%rsp
  mov    %rbp,%rdi
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    1090 <generate_1@plt>
  nopl   0x0(%rax)
L19:
  call   <generate_0@plt>
  jmp    L8
```

assistant: ```c
void byte2wav(int data, struct wav_data *wave_data) {
  int i;
  int counter = 0;
  generate_0(wave_data);
  for (i = 0; i < 8; i++) {
    int bit = (int)(pow(2, i));
    if ((data & bit) == bit) {
      counter++;
      generate_1(wave_data);
    } else
      generate_0(wave_data);
  }
  if ((counter % 2) == 0)
    generate_1(wave_data);
  else
    generate_0(wave_data);
  generate_1(wave_data);
  generate_1(wave_data);
  generate_1(wave_data);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<byte2wav>:
  endbr64
  push   %r13
  xor    %r13d,%r13d
  push   %r12
  mov    %edi,%r12d
  mov    %rsi,%rdi
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  call   <generate_0@plt>
  jmp    L31
  nopl   0x0(%rax,%rax,1)
L30:
  mov    %rbp,%rdi
  add    $0x1,%ebx
  call   <generate_0@plt>
  cmp    $0x8,%ebx
  je     L56
L31:
  mov    %ebx,%esi
  mov    $0x2,%edi
  call   <pow@plt>
  mov    %r12d,%edx
  and    %eax,%edx
  cmp    %eax,%edx
  jne    L30
  mov    %rbp,%rdi
  add    $0x1,%ebx
  add    $0x1,%r13d
  call   <generate_1@plt>
  cmp    $0x8,%ebx
  jne    L31
L56:
  and    $0x1,%r13d
  mov    %rbp,%rdi
  jne    L6
  call   <generate_1@plt>
L46:
  mov    %rbp,%rdi
  call   <generate_1@plt>
  mov    %rbp,%rdi
  call   <generate_1@plt>
  add    $0x8,%rsp
  mov    %rbp,%rdi
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    1090 <generate_1@plt>
  nopl   0x0(%rax)
L6:
  call   <generate_0@plt>
  jmp    L46
```

assistant: ```c
void byte2wav(int data, struct wav_data *wave_data) {
  int i;
  int counter = 0;
  generate_0(wave_data);
  for (i = 0; i < 8; i++) {
    int bit = (int)(pow(2, i));
    if ((data & bit) == bit) {
      counter++;
      generate_1(wave_data);
    } else
      generate_0(wave_data);
  }
  if ((counter % 2) == 0)
    generate_1(wave_data);
  else
    generate_0(wave_data);
  generate_1(wave_data);
  generate_1(wave_data);
  generate_1(wave_data);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D31(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x2010(%rbp),%rax
  mov    %rax,%rsi
  lea    D38(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    -0x2010(%rbp),%rax
  lea    D21(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <fopen@plt>
  mov    %rax,-0x2018(%rbp)
  cmpq   $0x0,-0x2018(%rbp)
  je     L35
  jmp    L23
L11:
  mov    -0x2018(%rbp),%rdx
  lea    -0x1010(%rbp),%rax
  mov    %rdx,%rcx
  mov    $0x1000,%edx
  mov    $0x1,%esi
  mov    %rax,%rdi
  call   <fread@plt>
  mov    %eax,-0x201c(%rbp)
  mov    -0x201c(%rbp),%eax
  cltq
  movb   $0x0,-0x1010(%rbp,%rax,1)
  lea    -0x1010(%rbp),%rax
  mov    %rax,%rsi
  lea    D38(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L23:
  mov    -0x2018(%rbp),%rax
  mov    %rax,%rdi
  call   <feof@plt>
  test   %eax,%eax
  je     L11
L35:
  mov    -0x2018(%rbp),%rax
  mov    %rax,%rdi
  call   <fclose@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L3
  call   <__stack_chk_fail@plt>
L3:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D31
type: string
name: disassemble_data
arguments: data_tag: D38
type: string
name: disassemble_data
arguments: data_tag: D21
type: string

tool: D31: Please enter path to file: 

tool: D38: %s

tool: D21: r

assistant: ```c
int main() {
  char path[4096];
  printf("Please enter path to file: ");
  scanf("%s", path);
  FILE *file = fopen(path, "r");
  char buffer[4096 + 1];
  if (file) {
    while (!feof(file)) {
      int readBytes = fread(buffer, 1, 4096, file);
      buffer[readBytes] = '\0';
      printf("%s", buffer);
    }
  }
  fclose(file);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x2008(%rsp)
  xor    %eax,%eax
  lea    D11(%rip),%rdi
  call   <printf@plt>
  mov    %rsp,%rbx
  mov    %rbx,%rsi
  lea    D34(%rip),%rdi
  call   <scanf@plt>
  lea    D14(%rip),%rsi
  mov    %rbx,%rdi
  call   <fopen@plt>
  mov    %rax,%rbx
  lea    D34(%rip),%r12
  test   %rax,%rax
  jne    L23
  jmp    L3
L32:
  lea    0x1000(%rsp),%rbp
  mov    %rbx,%rcx
  mov    $0x1000,%edx
  mov    $0x1,%esi
  mov    %rbp,%rdi
  call   <fread@plt>
  cltq
  movb   $0x0,0x1000(%rsp,%rax,1)
  mov    %rbp,%rsi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L23:
  mov    %rbx,%rdi
  call   <feof@plt>
  test   %eax,%eax
  je     L32
L3:
  mov    %rbx,%rdi
  call   <fclose@plt>
  mov    0x2008(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  mov    $0x0,%eax
  add    $0x2010,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D11
data_type: string
name: unpack_data
arguments: tag: D34
data_type: string
name: unpack_data
arguments: tag: D14
data_type: string

tool: D11: Please enter path to file: 

tool: D34: %s

tool: D14: r

assistant: ```c
int main() {
  char path[4096];
  printf("Please enter path to file: ");
  scanf("%s", path);
  FILE *file = fopen(path, "r");
  char buffer[4096 + 1];
  if (file) {
    while (!feof(file)) {
      int readBytes = fread(buffer, 1, 4096, file);
      buffer[readBytes] = '\0';
      printf("%s", buffer);
    }
  }
  fclose(file);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x10,%rsp
  lea    D26(%rip),%rdi
  lea    D21(%rip),%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x2008(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  lea    0x1000(%rsp),%r12
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    %rbp,%rdi
  lea    D10(%rip),%rsi
  call   <fopen@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  jne    L22
  jmp    L33
  nopl   0x0(%rax)
L29:
  mov    $0x1,%esi
  mov    %r12,%rdi
  mov    %rbp,%rcx
  mov    $0x1000,%edx
  call   <fread@plt>
  mov    %r12,%rsi
  mov    %rbx,%rdi
  cltq
  movb   $0x0,0x1000(%rsp,%rax,1)
  xor    %eax,%eax
  call   <printf@plt>
L22:
  mov    %rbp,%rdi
  call   <feof@plt>
  test   %eax,%eax
  je     L29
L33:
  mov    %rbp,%rdi
  call   <fclose@plt>
  mov    0x2008(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L19
  add    $0x2010,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L19:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D26
datatype: string
name: extract_symbolic_data
arguments: label: D21
datatype: string
name: extract_symbolic_data
arguments: label: D10
datatype: string

tool: D26: Please enter path to file: 

tool: D21: %s

tool: D10: r

assistant: ```c
int main() {
  char path[4096];
  printf("Please enter path to file: ");
  scanf("%s", path);
  FILE *file = fopen(path, "r");
  char buffer[4096 + 1];
  if (file) {
    while (!feof(file)) {
      int readBytes = fread(buffer, 1, 4096, file);
      buffer[readBytes] = '\0';
      printf("%s", buffer);
    }
  }
  fclose(file);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x10,%rsp
  lea    D19(%rip),%rdi
  lea    D18(%rip),%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x2008(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  lea    0x1000(%rsp),%r12
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    %rbp,%rdi
  lea    D4(%rip),%rsi
  call   <fopen@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  jne    L26
  jmp    L1
  nopl   0x0(%rax)
L39:
  mov    $0x1,%esi
  mov    %r12,%rdi
  mov    %rbp,%rcx
  mov    $0x1000,%edx
  call   <fread@plt>
  mov    %r12,%rsi
  mov    %rbx,%rdi
  cltq
  movb   $0x0,0x1000(%rsp,%rax,1)
  xor    %eax,%eax
  call   <printf@plt>
L26:
  mov    %rbp,%rdi
  call   <feof@plt>
  test   %eax,%eax
  je     L39
L1:
  mov    %rbp,%rdi
  call   <fclose@plt>
  mov    0x2008(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L10
  add    $0x2010,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L10:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D19
type_infered: string
name: extract_symbolic_data
arguments: label: D18
type_infered: string
name: extract_symbolic_data
arguments: label: D4
type_infered: string

tool: D19: Please enter path to file: 

tool: D18: %s

tool: D4: r

assistant: ```c
int main() {
  char path[4096];
  printf("Please enter path to file: ");
  scanf("%s", path);
  FILE *file = fopen(path, "r");
  char buffer[4096 + 1];
  if (file) {
    while (!feof(file)) {
      int readBytes = fread(buffer, 1, 4096, file);
      buffer[readBytes] = '\0';
      printf("%s", buffer);
    }
  }
  fclose(file);
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<string_count_chr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L11
L9:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  cmp    %eax,-0x1c(%rbp)
  jne    L28
  addl   $0x1,-0x4(%rbp)
L28:
  addq   $0x1,-0x18(%rbp)
L11:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L9
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int string_count_chr(const char *string, int c) {
  int count;
  for (count = 0; *string; string++)
    if (*string == c)
      count++;
  return count;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<string_count_chr>:
  endbr64
  movzbl (%rdi),%eax
  test   %al,%al
  je     L9
  mov    $0x0,%edx
L13:
  movsbl %al,%eax
  cmp    %esi,%eax
  sete   %al
  movzbl %al,%eax
  add    %eax,%edx
  add    $0x1,%rdi
  movzbl (%rdi),%eax
  test   %al,%al
  jne    L13
L28:
  mov    %edx,%eax
  ret
L9:
  mov    $0x0,%edx
  jmp    L28
```

assistant: ```c
int string_count_chr(const char *string, int c) {
  int count;
  for (count = 0; *string; string++)
    if (*string == c)
      count++;
  return count;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<string_count_chr>:
  endbr64
  movsbl (%rdi),%eax
  xor    %r8d,%r8d
  test   %al,%al
  je     L11
  xchg   %ax,%ax
L0:
  cmp    %esi,%eax
  sete   %al
  add    $0x1,%rdi
  movzbl %al,%eax
  add    %eax,%r8d
  movsbl (%rdi),%eax
  test   %al,%al
  jne    L0
L11:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int string_count_chr(const char *string, int c) {
  int count;
  for (count = 0; *string; string++)
    if (*string == c)
      count++;
  return count;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<string_count_chr>:
  endbr64
  movsbl (%rdi),%eax
  xor    %r8d,%r8d
  test   %al,%al
  je     L5
  xchg   %ax,%ax
L6:
  cmp    %eax,%esi
  sete   %al
  add    $0x1,%rdi
  movzbl %al,%eax
  add    %eax,%r8d
  movsbl (%rdi),%eax
  test   %al,%al
  jne    L6
L5:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int string_count_chr(const char *string, int c) {
  int count;
  for (count = 0; *string; string++)
    if (*string == c)
      count++;
  return count;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<eh_palindromo>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x4(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L8
L9:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x1c(%rbp),%eax
  sub    $0x1,%eax
  sub    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%eax
  cmp    %eax,%edx
  je     L22
  mov    $0x0,%eax
  jmp    L0
L22:
  addl   $0x1,-0x4(%rbp)
L8:
  mov    -0x1c(%rbp),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  cmp    %eax,-0x4(%rbp)
  jl     L9
  mov    $0x1,%eax
L0:
  pop    %rbp
  ret
```

assistant: ```c
bool eh_palindromo(int *vetor, int tamanho) {
  int i = 0;
  for (i = 0; i < tamanho / 2; i++) {
    if (vetor[i] != vetor[tamanho - 1 - i]) {
      return false;
    }
  }
  return true;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<eh_palindromo>:
  endbr64
  mov    %esi,%ecx
  shr    $0x1f,%ecx
  add    %esi,%ecx
  sar    %ecx
  cmp    $0x1,%esi
  jle    L6
  movslq %esi,%rsi
  lea    (%rdi,%rsi,4),%rdx
  mov    $0x0,%eax
L25:
  mov    -0x4(%rdx),%esi
  cmp    %esi,(%rdi,%rax,4)
  jne    L12
  add    $0x1,%rax
  sub    $0x4,%rdx
  cmp    %eax,%ecx
  jg     L25
  mov    $0x1,%eax
  ret
L6:
  mov    $0x1,%eax
  ret
L12:
  mov    $0x0,%eax
  ret
```

assistant: ```c
bool eh_palindromo(int *vetor, int tamanho) {
  int i = 0;
  for (i = 0; i < tamanho / 2; i++) {
    if (vetor[i] != vetor[tamanho - 1 - i]) {
      return false;
    }
  }
  return true;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<eh_palindromo>:
  endbr64
  mov    %esi,%ecx
  shr    $0x1f,%ecx
  add    %esi,%ecx
  sar    %ecx
  cmp    $0x1,%esi
  jle    L36
  sub    $0x1,%esi
  xor    %eax,%eax
  movslq %esi,%rsi
  lea    (%rdi,%rsi,4),%rdx
  jmp    L35
L38:
  add    $0x1,%rax
  sub    $0x4,%rdx
  cmp    %eax,%ecx
  jle    L36
L35:
  mov    (%rdx),%esi
  cmp    %esi,(%rdi,%rax,4)
  je     L38
  xor    %eax,%eax
  ret
  cs nopw 0x0(%rax,%rax,1)
L36:
  mov    $0x1,%eax
  ret
```

assistant: ```c
bool eh_palindromo(int *vetor, int tamanho) {
  int i = 0;
  for (i = 0; i < tamanho / 2; i++) {
    if (vetor[i] != vetor[tamanho - 1 - i]) {
      return false;
    }
  }
  return true;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<eh_palindromo>:
  endbr64
  mov    %esi,%ecx
  shr    $0x1f,%ecx
  add    %esi,%ecx
  sar    %ecx
  cmp    $0x1,%esi
  jle    L14
  sub    $0x1,%esi
  xor    %eax,%eax
  movslq %esi,%rsi
  lea    (%rdi,%rsi,4),%rdx
  jmp    L20
L18:
  add    $0x1,%rax
  sub    $0x4,%rdx
  cmp    %eax,%ecx
  jle    L14
L20:
  mov    (%rdx),%esi
  cmp    %esi,(%rdi,%rax,4)
  je     L18
  xor    %eax,%eax
  ret
  cs nopw 0x0(%rax,%rax,1)
L14:
  mov    $0x1,%eax
  ret
```

assistant: ```c
bool eh_palindromo(int *vetor, int tamanho) {
  int i = 0;
  for (i = 0; i < tamanho / 2; i++) {
    if (vetor[i] != vetor[tamanho - 1 - i]) {
      return false;
    }
  }
  return true;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x1,%edi
  call   <foo@plt>
  mov    %eax,%edx
  mov    D8(%rip),%rax
  mov    %edx,%esi
  movq   %rax,%xmm0
  lea    D32(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    D8(%rip),%rax
  movq   %rax,%xmm0
  call   <wiz@plt>
  mov    %eax,%edx
  mov    D8(%rip),%rax
  mov    %edx,%esi
  movq   %rax,%xmm0
  lea    D33(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

tool_calls: name: read_data
arguments: tag: D8
type: f64
name: read_data
arguments: tag: D32
type: string
name: read_data
arguments: tag: D33
type: string

tool: D8: 1

tool: D32: foo(%d): %d


tool: D33: Success! wiz(%f): %d


assistant: ```c
int main() {
  printf("foo(%d): %d\n", 1, foo(1));
  printf("Success! wiz(%f): %d\n", 1.0, wiz(1.0));
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x1,%edi
  call   <foo@plt>
  mov    %eax,%esi
  mov    D41(%rip),%rax
  movq   %rax,%xmm0
  lea    D8(%rip),%rdi
  call   <printf@plt>
  mov    D41(%rip),%rax
  movq   %rax,%xmm0
  call   <wiz@plt>
  mov    %eax,%esi
  mov    D41(%rip),%rax
  movq   %rax,%xmm0
  lea    D13(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: data_label: D41
type: f64
name: read_data
arguments: data_label: D8
type: string
name: read_data
arguments: data_label: D13
type: string

tool: D41: 1

tool: D8: foo(%d): %d


tool: D13: Success! wiz(%f): %d


assistant: ```c
int main() {
  printf("foo(%d): %d\n", 1, foo(1));
  printf("Success! wiz(%f): %d\n", 1.0, wiz(1.0));
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x1,%edi
  call   <foo@plt>
  movsd  D28(%rip),%xmm0
  lea    D22(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    D28(%rip),%rax
  movq   %rax,%xmm0
  call   <wiz@plt>
  lea    D47(%rip),%rdi
  mov    %eax,%esi
  mov    D28(%rip),%rax
  movq   %rax,%xmm0
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D28
type_infered: f64
name: unpack_data
arguments: data_label: D22
type_infered: string
name: unpack_data
arguments: data_label: D47
type_infered: string

tool: D28: 1

tool: D22: foo(%d): %d


tool: D47: Success! wiz(%f): %d


assistant: ```c
int main() {
  printf("foo(%d): %d\n", 1, foo(1));
  printf("Success! wiz(%f): %d\n", 1.0, wiz(1.0));
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x1,%edi
  call   <foo@plt>
  movsd  D49(%rip),%xmm0
  lea    D44(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    D49(%rip),%rax
  movq   %rax,%xmm0
  call   <wiz@plt>
  lea    D31(%rip),%rdi
  mov    %eax,%esi
  mov    D49(%rip),%rax
  movq   %rax,%xmm0
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D49
datatype: f64
name: disassemble_data
arguments: data_tag: D44
datatype: string
name: disassemble_data
arguments: data_tag: D31
datatype: string

tool: D49: 1

tool: D44: foo(%d): %d


tool: D31: Success! wiz(%f): %d


assistant: ```c
int main() {
  printf("foo(%d): %d\n", 1, foo(1));
  printf("Success! wiz(%f): %d\n", 1.0, wiz(1.0));
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<XFree>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L1
L16:
  jmp    L16
L1:
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  nop
  leave
  ret
```

assistant: ```c
void XFree(void *ptr) {
  if (!ptr)
    while (1)
      ;
  free(ptr);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<XFree>:
  endbr64
  test   %rdi,%rdi
  je     L13
  sub    $0x8,%rsp
  call   <free@plt>
  add    $0x8,%rsp
  ret
L13:
  jmp    L13
```

assistant: ```c
void XFree(void *ptr) {
  if (!ptr)
    while (1)
      ;
  free(ptr);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<XFree>:
  endbr64
  test   %rdi,%rdi
  je     L18
  jmp    1050 <free@plt>
L18:
  jmp    L18
```

assistant: ```c
void XFree(void *ptr) {
  if (!ptr)
    while (1)
      ;
  free(ptr);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<XFree>:
  endbr64
  test   %rdi,%rdi
  je     L0
  jmp    1050 <free@plt>
L0:
  jmp    L0
```

assistant: ```c
void XFree(void *ptr) {
  if (!ptr)
    while (1)
      ;
  free(ptr);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<helval>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  movsd  (%rdi),%xmm1
  pxor   %xmm0,%xmm0
  comisd %xmm1,%xmm0
  ja     L63
  comisd D73(%rip),%xmm1
  jbe    L0
  movsd  0x8(%rdi),%xmm0
  divsd  %xmm1,%xmm0
  call   <atan@plt>
  divsd  D47(%rip),%xmm0
  jmp    L55
L63:
  movsd  0x8(%rdi),%xmm0
  divsd  %xmm1,%xmm0
  call   <atan@plt>
  divsd  D47(%rip),%xmm0
  addsd  D89(%rip),%xmm0
L55:
  movsd  D20(%rip),%xmm2
  mulsd  %xmm2,%xmm0
  movsd  0x10(%rbx),%xmm1
  subsd  %xmm0,%xmm1
  mulsd  %xmm2,%xmm1
  movsd  %xmm1,0x0(%rbp)
  movsd  (%rbx),%xmm0
  movsd  0x8(%rbx),%xmm1
  mulsd  %xmm0,%xmm0
  mulsd  %xmm1,%xmm1
  addsd  %xmm1,%xmm0
  pxor   %xmm1,%xmm1
  ucomisd %xmm0,%xmm1
  ja     L47
  sqrtsd %xmm0,%xmm0
L67:
  subsd  D68(%rip),%xmm0
  mulsd  D20(%rip),%xmm0
  movsd  %xmm0,0x8(%rbp)
  movsd  0x10(%rbx),%xmm0
  movsd  %xmm0,0x10(%rbp)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L0:
  movsd  0x8(%rdi),%xmm1
  movsd  D26(%rip),%xmm0
  comisd D73(%rip),%xmm1
  jae    L55
  movsd  D53(%rip),%xmm0
  jmp    L55
L47:
  call   <sqrt@plt>
  jmp    L67
```

tool_calls: name: parse_data
arguments: tag: D73
type_infered: f64[2]
name: parse_data
arguments: tag: D47
type_infered: byte[8]
name: parse_data
arguments: tag: D89
type_infered: f64
name: parse_data
arguments: tag: D20
type_infered: f64
name: parse_data
arguments: tag: D68
type_infered: f64
name: parse_data
arguments: tag: D26
type_infered: f64
name: parse_data
arguments: tag: D53
type_infered: byte[8]

tool: {"D73": [0.0,6.283185307179586]}

tool: {"D47": [0x18,0x2d,0x44,0x54,0xfb,0x21,0x19,0x40]}

tool: D89: 0.5

tool: D20: 10

tool: D68: 1

tool: D26: 0.25

tool: {"D53": [0x00,0x00,0x00,0x00,0x00,0x00,0xd0,0xbf]}

assistant: ```c
void helval(double *p, double *x, int m, int n, void *data) {
  double theta;
  if (p[0] < 0.0)
    theta = atan(p[1] / p[0]) / (2.0 * 3.14159265358979323846) + 0.5;
  else if (0.0 < p[0])
    theta = atan(p[1] / p[0]) / (2.0 * 3.14159265358979323846);
  else
    theta = (p[1] >= 0) ? 0.25 : -0.25;
  x[0] = 10.0 * (p[2] - 10.0 * theta);
  x[1] = 10.0 * (sqrt(p[0] * p[0] + p[1] * p[1]) - 1.0);
  x[2] = p[2];
}
```