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

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %edi,-0x44(%rbp)
  mov    %rsi,-0x50(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x30(%rbp)
  movl   $0x1,-0x2c(%rbp)
  movl   $0x2,-0x28(%rbp)
  movl   $0x3,-0x24(%rbp)
  movl   $0x4,-0x20(%rbp)
  movl   $0x5,-0x1c(%rbp)
  movl   $0x6,-0x18(%rbp)
  movl   $0x7,-0x14(%rbp)
  movl   $0x8,-0x10(%rbp)
  movl   $0x9,-0xc(%rbp)
  mov    -0x30(%rbp),%eax
  mov    %eax,-0x34(%rbp)
  movl   $0x1,-0x38(%rbp)
  jmp    L29
L7:
  mov    -0x38(%rbp),%eax
  cltq
  mov    -0x30(%rbp,%rax,4),%eax
  cmp    %eax,-0x34(%rbp)
  jle    L13
  mov    -0x38(%rbp),%eax
  cltq
  mov    -0x30(%rbp,%rax,4),%eax
  mov    %eax,-0x34(%rbp)
L13:
  addl   $0x1,-0x38(%rbp)
L29:
  cmpl   $0x9,-0x38(%rbp)
  jle    L7
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L19
  call   <__stack_chk_fail@plt>
L19:
  leave
  ret
```

assistant: ```c
int main(int argc, const char *argv[]) {
  int array[10] = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9};
  int i;
  int min = array[0];
  for (i = 1; i < 10; ++i) {
    if (array[i] < min) {
      min = array[i];
    }
  }
  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
  mov    $0x0,%eax
  ret
```

assistant: ```c
int main(int argc, const char *argv[]) {
  int array[10] = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9};
  int i;
  int min = array[0];
  for (i = 1; i < 10; ++i) {
    if (array[i] < min) {
      min = array[i];
    }
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int main(int argc, const char *argv[]) {
  int array[10] = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9};
  int i;
  int min = array[0];
  for (i = 1; i < 10; ++i) {
    if (array[i] < min) {
      min = array[i];
    }
  }
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int main(int argc, const char *argv[]) {
  int array[10] = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9};
  int i;
  int min = array[0];
  for (i = 1; i < 10; ++i) {
    if (array[i] < min) {
      min = array[i];
    }
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<ResetIntHandler>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    D11(%rip),%rax
  mov    %rax,-0x10(%rbp)
  mov    D22(%rip),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L33
L36:
  mov    -0x10(%rbp),%rdx
  lea    0x8(%rdx),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x8(%rbp),%rax
  lea    0x8(%rax),%rcx
  mov    %rcx,-0x8(%rbp)
  mov    (%rdx),%rdx
  mov    %rdx,(%rax)
L33:
  mov    D32(%rip),%rax
  cmp    %rax,-0x8(%rbp)
  jb     L36
  mov    D14(%rip),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L3
L42:
  mov    -0x8(%rbp),%rax
  movq   $0x0,(%rax)
  addq   $0x8,-0x8(%rbp)
L3:
  mov    D27(%rip),%rax
  cmp    %rax,-0x8(%rbp)
  jb     L42
  mov    $0x0,%eax
  call   <main@plt>
L32:
  jmp    L32
```

assistant: ```c
void ResetIntHandler() {
  extern uint32_t _etext, _data, _edata, _bss, _ebss;
  uint32_t *src = &_etext, *dst = &_data;
  while (dst < &_edata)
    *(dst++) = *(src++);
  for (dst = &_bss; dst < &_ebss; dst++)
    *dst = 0;
  main();
  while (1)
    ;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<ResetIntHandler>:
  endbr64
  sub    $0x8,%rsp
  mov    D26(%rip),%rax
  cmp    %rax,D61(%rip)
  jae    L48
  lea    -0x1(%rax),%rcx
  mov    D61(%rip),%rdx
  sub    %rdx,%rcx
  shr    $0x3,%rcx
  mov    D59(%rip),%rax
  lea    0x8(%rax,%rcx,8),%rsi
L31:
  add    $0x8,%rax
  add    $0x8,%rdx
  mov    -0x8(%rax),%rcx
  mov    %rcx,-0x8(%rdx)
  cmp    %rsi,%rax
  jne    L31
L48:
  mov    D39(%rip),%rax
  cmp    %rax,D2(%rip)
  jae    L32
  lea    -0x1(%rax),%rdx
  mov    D2(%rip),%rax
  sub    %rax,%rdx
  shr    $0x3,%rdx
  lea    0x8(%rax,%rdx,8),%rdx
L39:
  movq   $0x0,(%rax)
  add    $0x8,%rax
  cmp    %rdx,%rax
  jne    L39
L32:
  mov    $0x0,%eax
  call   <main@plt>
L28:
  jmp    L28
```

assistant: ```c
void ResetIntHandler() {
  extern uint32_t _etext, _data, _edata, _bss, _ebss;
  uint32_t *src = &_etext, *dst = &_data;
  while (dst < &_edata)
    *(dst++) = *(src++);
  for (dst = &_bss; dst < &_ebss; dst++)
    *dst = 0;
  main();
  while (1)
    ;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<ResetIntHandler>:
  endbr64
  sub    $0x8,%rsp
  mov    D40(%rip),%rdi
  mov    D31(%rip),%rax
  cmp    %rax,%rdi
  jae    L28
  sub    $0x1,%rax
  mov    D32(%rip),%rsi
  sub    %rdi,%rax
  shr    $0x3,%rax
  lea    0x8(,%rax,8),%rdx
  call   <memcpy@plt>
L28:
  mov    D43(%rip),%rdi
  mov    D18(%rip),%rax
  cmp    %rax,%rdi
  jae    L4
  sub    $0x1,%rax
  xor    %esi,%esi
  sub    %rdi,%rax
  shr    $0x3,%rax
  lea    0x8(,%rax,8),%rdx
  call   <memset@plt>
L4:
  xor    %eax,%eax
  call   <main@plt>
L24:
  jmp    L24
```

assistant: ```c
void ResetIntHandler() {
  extern uint32_t _etext, _data, _edata, _bss, _ebss;
  uint32_t *src = &_etext, *dst = &_data;
  while (dst < &_edata)
    *(dst++) = *(src++);
  for (dst = &_bss; dst < &_ebss; dst++)
    *dst = 0;
  main();
  while (1)
    ;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ResetIntHandler>:
  endbr64
  sub    $0x8,%rsp
  mov    D40(%rip),%rdi
  mov    D42(%rip),%rax
  cmp    %rax,%rdi
  jae    L28
  sub    $0x1,%rax
  mov    D0(%rip),%rsi
  sub    %rdi,%rax
  shr    $0x3,%rax
  lea    0x8(,%rax,8),%rdx
  call   <memcpy@plt>
L28:
  mov    D20(%rip),%rdi
  mov    D6(%rip),%rax
  cmp    %rax,%rdi
  jae    L1
  sub    $0x1,%rax
  xor    %esi,%esi
  sub    %rdi,%rax
  shr    $0x3,%rax
  lea    0x8(,%rax,8),%rdx
  call   <memset@plt>
L1:
  xor    %eax,%eax
  call   <main@plt>
L22:
  jmp    L22
```

assistant: ```c
void ResetIntHandler() {
  extern uint32_t _etext, _data, _edata, _bss, _ebss;
  uint32_t *src = &_etext, *dst = &_data;
  while (dst < &_edata)
    *(dst++) = *(src++);
  for (dst = &_bss; dst < &_ebss; dst++)
    *dst = 0;
  main();
  while (1)
    ;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<remove_leading_whitespace>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <skip_whitespace@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  cmp    -0x18(%rbp),%rax
  je     L8
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  lea    0x1(%rax),%rdx
  mov    -0x8(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memmove@plt>
L8:
  nop
  leave
  ret
```

assistant: ```c
void remove_leading_whitespace(char line[]) {
  const char *const non_whitespace = skip_whitespace(line);
  if (non_whitespace != line) {
    memmove(line, non_whitespace, strlen(non_whitespace) + 1);
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<remove_leading_whitespace>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  call   <skip_whitespace@plt>
  cmp    %rax,%rbx
  je     L2
  mov    %rax,%rbp
  mov    %rax,%rdi
  call   <strlen@plt>
  lea    0x1(%rax),%rdx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <memmove@plt>
L2:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void remove_leading_whitespace(char line[]) {
  const char *const non_whitespace = skip_whitespace(line);
  if (non_whitespace != line) {
    memmove(line, non_whitespace, strlen(non_whitespace) + 1);
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<remove_leading_whitespace>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <skip_whitespace@plt>
  cmp    %rax,%rbp
  je     L8
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <strlen@plt>
  add    $0x8,%rsp
  mov    %r12,%rsi
  mov    %rbp,%rdi
  lea    0x1(%rax),%rdx
  pop    %rbp
  pop    %r12
  jmp    1090 <memmove@plt>
  nopl   0x0(%rax)
L8:
  add    $0x8,%rsp
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void remove_leading_whitespace(char line[]) {
  const char *const non_whitespace = skip_whitespace(line);
  if (non_whitespace != line) {
    memmove(line, non_whitespace, strlen(non_whitespace) + 1);
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<remove_leading_whitespace>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <skip_whitespace@plt>
  cmp    %rax,%rbp
  je     L3
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <strlen@plt>
  add    $0x8,%rsp
  mov    %r12,%rsi
  mov    %rbp,%rdi
  lea    0x1(%rax),%rdx
  pop    %rbp
  pop    %r12
  jmp    1090 <memmove@plt>
  nopl   0x0(%rax)
L3:
  add    $0x8,%rsp
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void remove_leading_whitespace(char line[]) {
  const char *const non_whitespace = skip_whitespace(line);
  if (non_whitespace != line) {
    memmove(line, non_whitespace, strlen(non_whitespace) + 1);
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<std_printf>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xd0,%rsp
  mov    %rdi,-0xc8(%rbp)
  mov    %rsi,-0xa8(%rbp)
  mov    %rdx,-0xa0(%rbp)
  mov    %rcx,-0x98(%rbp)
  mov    %r8,-0x90(%rbp)
  mov    %r9,-0x88(%rbp)
  test   %al,%al
  je     L3
  movaps %xmm0,-0x80(%rbp)
  movaps %xmm1,-0x70(%rbp)
  movaps %xmm2,-0x60(%rbp)
  movaps %xmm3,-0x50(%rbp)
  movaps %xmm4,-0x40(%rbp)
  movaps %xmm5,-0x30(%rbp)
  movaps %xmm6,-0x20(%rbp)
  movaps %xmm7,-0x10(%rbp)
L3:
  mov    -0xc8(%rbp),%rdx
  mov    -0xb4(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <va_start@plt>
  mov    -0xb4(%rbp),%edx
  mov    -0xc8(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x0,%edi
  call   <print@plt>
  leave
  ret
```

assistant: ```c
int std_printf(const char *format, ...) {
  va_list args;
  va_start(args, format);
  return print(0, format, args);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<std_printf>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rdi,%rsi
  mov    $0x0,%ebp
  mov    %ebp,%edi
  call   <va_start@plt>
  mov    %ebp,%edx
  mov    %rbx,%rsi
  mov    $0x0,%edi
  call   <print@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int std_printf(const char *format, ...) {
  va_list args;
  va_start(args, format);
  return print(0, format, args);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<std_printf>:
  endbr64
  push   %r12
  xor    %r12d,%r12d
  mov    %rdi,%rsi
  push   %rbp
  mov    %rdi,%rbp
  mov    %r12d,%edi
  sub    $0x8,%rsp
  call   <va_start@plt>
  mov    %r12d,%edx
  mov    %rbp,%rsi
  xor    %edi,%edi
  call   <print@plt>
  add    $0x8,%rsp
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int std_printf(const char *format, ...) {
  va_list args;
  va_start(args, format);
  return print(0, format, args);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<std_printf>:
  endbr64
  push   %r12
  xor    %r12d,%r12d
  mov    %rdi,%rsi
  push   %rbp
  mov    %rdi,%rbp
  mov    %r12d,%edi
  sub    $0x8,%rsp
  call   <va_start@plt>
  mov    %r12d,%edx
  mov    %rbp,%rsi
  xor    %edi,%edi
  call   <print@plt>
  add    $0x8,%rsp
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int std_printf(const char *format, ...) {
  va_list args;
  va_start(args, format);
  return print(0, format, args);
}
```
user: What is the c source code of the assembly code below?

```asm
<GetMaxDouble>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movabs $0xfffffffffffff,%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  movabs $0x7fe0000000000000,%rdx
  or     %rdx,%rax
  mov    %rax,-0x10(%rbp)
  lea    -0x10(%rbp),%rax
  movsd  (%rax),%xmm0
  movq   %xmm0,%rax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L6
  call   <__stack_chk_fail@plt>
L6:
  movq   %rax,%xmm0
  leave
  ret
```

assistant: ```c
double GetMaxDouble(void) {
  unsigned long long bits = ((unsigned long long)-1) >> 12;
  bits = bits | ((((unsigned long long)-1) >> 1) ^ (((unsigned long long)-1) >> 11));
  return *(double *)&bits;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<GetMaxDouble>:
  endbr64
  movsd  D4(%rip),%xmm0
  ret
```

tool_calls: name: read_data
arguments: label: D4
type_guessed: byte[8]

tool: {"D4": [0xff,0xff,0xff,0xff,0xff,0xff,0xef,0x7f]}

assistant: ```c
double GetMaxDouble(void) {
  unsigned long long bits = ((unsigned long long)-1) >> 12;
  bits = bits | ((((unsigned long long)-1) >> 1) ^ (((unsigned long long)-1) >> 11));
  return *(double *)&bits;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<GetMaxDouble>:
  endbr64
  movsd  D5(%rip),%xmm0
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D5
type: byte[8]

tool: {"D5": [0xff,0xff,0xff,0xff,0xff,0xff,0xef,0x7f]}

assistant: ```c
double GetMaxDouble(void) {
  unsigned long long bits = ((unsigned long long)-1) >> 12;
  bits = bits | ((((unsigned long long)-1) >> 1) ^ (((unsigned long long)-1) >> 11));
  return *(double *)&bits;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<GetMaxDouble>:
  endbr64
  movsd  D7(%rip),%xmm0
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D7
datatype: byte[8]

tool: {"D7": [0xff,0xff,0xff,0xff,0xff,0xff,0xef,0x7f]}

assistant: ```c
double GetMaxDouble(void) {
  unsigned long long bits = ((unsigned long long)-1) >> 12;
  bits = bits | ((((unsigned long long)-1) >> 1) ^ (((unsigned long long)-1) >> 11));
  return *(double *)&bits;
}
```
user: Translate the below assembly code into C programming language:

```asm
<_mosquitto_calloc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x20(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <calloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
void *_mosquitto_calloc(size_t nmemb, size_t size) {
  void *mem = calloc(nmemb, size);
  return mem;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<_mosquitto_calloc>:
  endbr64
  sub    $0x8,%rsp
  call   <calloc@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void *_mosquitto_calloc(size_t nmemb, size_t size) {
  void *mem = calloc(nmemb, size);
  return mem;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<_mosquitto_calloc>:
  endbr64
  jmp    1050 <calloc@plt>
```

assistant: ```c
void *_mosquitto_calloc(size_t nmemb, size_t size) {
  void *mem = calloc(nmemb, size);
  return mem;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<_mosquitto_calloc>:
  endbr64
  jmp    1050 <calloc@plt>
```

assistant: ```c
void *_mosquitto_calloc(size_t nmemb, size_t size) {
  void *mem = calloc(nmemb, size);
  return mem;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x1,-0x10(%rbp)
  pxor   %xmm0,%xmm0
  movss  %xmm0,-0xc(%rbp)
L9:
  lea    -0x14(%rbp),%rax
  mov    %rax,%rsi
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  movss  -0x14(%rbp),%xmm0
  movss  -0xc(%rbp),%xmm1
  addss  %xmm1,%xmm0
  movss  %xmm0,-0xc(%rbp)
  addl   $0x1,-0x10(%rbp)
  cmpl   $0xc,-0x10(%rbp)
  jle    L9
  movss  -0xc(%rbp),%xmm0
  movss  D17(%rip),%xmm1
  divss  %xmm1,%xmm0
  movd   %xmm0,%eax
  movd   %eax,%xmm0
  lea    D11(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L6
  call   <__stack_chk_fail@plt>
L6:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D4
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D17
type_infered: byte[4]
name: extract_symbolic_data
arguments: data_tag: D11
type_infered: string

tool: D4: %f

tool: {"D17": [0x00,0x00,0x40,0x41]}

tool: D11: $%.2f

assistant: ```c
int main(void) {
  int i = 1;
  float a, sum = 0;
  do {
    scanf("%f", &a);
    sum += a;
    i++;
  } while (i <= 12);
  printf("$%.2f", sum / 12);
  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   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    $0xc,%ebx
  mov    D30(%rip),%ebp
  lea    D28(%rip),%r12
L13:
  lea    0x4(%rsp),%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  movd   %ebp,%xmm1
  addss  0x4(%rsp),%xmm1
  movd   %xmm1,%ebp
  sub    $0x1,%ebx
  jne    L13
  movaps %xmm1,%xmm0
  divss  D4(%rip),%xmm0
  lea    D38(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  mov    $0x0,%eax
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D30
type_infered: f32
name: disassemble_data
arguments: data_label: D28
type_infered: string
name: disassemble_data
arguments: data_label: D4
type_infered: byte[4]
name: disassemble_data
arguments: data_label: D38
type_infered: string

tool: D30: 0

tool: D28: %f

tool: {"D4": [0x00,0x00,0x40,0x41]}

tool: D38: $%.2f

assistant: ```c
int main(void) {
  int i = 1;
  float a, sum = 0;
  do {
    scanf("%f", &a);
    sum += a;
    i++;
  } while (i <= 12);
  printf("$%.2f", sum / 12);
  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
  push   %r12
  pxor   %xmm0,%xmm0
  push   %rbp
  lea    D10(%rip),%rbp
  push   %rbx
  mov    $0xc,%ebx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x14(%rsp),%r12
  nopl   0x0(%rax)
L5:
  mov    %r12,%rsi
  mov    %rbp,%rdi
  movss  %xmm0,0xc(%rsp)
  call   <scanf@plt>
  movss  0xc(%rsp),%xmm0
  addss  0x14(%rsp),%xmm0
  sub    $0x1,%ebx
  jne    L5
  lea    D22(%rip),%rdi
  divss  D27(%rip),%xmm0
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D10
datatype: string
name: extract_symbolic_data
arguments: tag: D22
datatype: string
name: extract_symbolic_data
arguments: tag: D27
datatype: byte[4]

tool: D10: %f

tool: D22: $%.2f

tool: {"D27": [0x00,0x00,0x40,0x41]}

assistant: ```c
int main(void) {
  int i = 1;
  float a, sum = 0;
  do {
    scanf("%f", &a);
    sum += a;
    i++;
  } while (i <= 12);
  printf("$%.2f", sum / 12);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r12
  pxor   %xmm0,%xmm0
  push   %rbp
  lea    D13(%rip),%rbp
  push   %rbx
  mov    $0xc,%ebx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x14(%rsp),%r12
  nopl   0x0(%rax)
L10:
  mov    %r12,%rsi
  mov    %rbp,%rdi
  movss  %xmm0,0xc(%rsp)
  call   <scanf@plt>
  movss  0xc(%rsp),%xmm0
  addss  0x14(%rsp),%xmm0
  sub    $0x1,%ebx
  jne    L10
  lea    D16(%rip),%rdi
  divss  D7(%rip),%xmm0
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L15
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L15:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D13
type: string
name: extract_symbolic_data
arguments: data_label: D16
type: string
name: extract_symbolic_data
arguments: data_label: D7
type: byte[4]

tool: D13: %f

tool: D16: $%.2f

tool: {"D7": [0x00,0x00,0x40,0x41]}

assistant: ```c
int main(void) {
  int i = 1;
  float a, sum = 0;
  do {
    scanf("%f", &a);
    sum += a;
    i++;
  } while (i <= 12);
  printf("$%.2f", sum / 12);
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<count_words>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movq   $0x0,-0x8(%rbp)
  cmpq   $0x0,-0x20(%rbp)
  jne    L14
  mov    $0x0,%eax
  jmp    L22
L14:
  mov    -0x20(%rbp),%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  and    $0x2,%eax
  test   %rax,%rax
  je     L2
  addq   $0x1,-0x8(%rbp)
L2:
  mov    -0x20(%rbp),%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  shr    $0xa,%rax
  and    $0x3fffff,%eax
  mov    %rax,%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <count_words>
  add    %rax,-0x8(%rbp)
  mov    -0x20(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x20(%rbp)
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  and    $0x1,%eax
  test   %rax,%rax
  je     L14
  mov    -0x8(%rbp),%rax
L22:
  leave
  ret
```

assistant: ```c
uint32_t count_words(const uint32_t *transitions, uint32_t pos) {
  uint32_t count = 0;
  if (!pos)
    return 0;
  do {
    if (((transitions[pos]) & 0x2))
      count++;
    count += count_words(transitions, (uint32_t)(((transitions[pos]) >> 10) & ((1 << 22) - 1)));
  } while (!((transitions[pos++]) & 0x1));
  return count;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<count_words>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rsi,%rbx
  test   %rsi,%rsi
  je     L8
  mov    %rdi,%r13
  lea    (%rdi,%rsi,8),%r12
  mov    $0x0,%ebx
L0:
  mov    (%r12),%rbp
  mov    %rbp,%rax
  and    $0x2,%eax
  cmp    $0x1,%rax
  sbb    $0xffffffffffffffff,%rbx
  mov    %rbp,%rsi
  shr    $0xa,%rsi
  and    $0x3fffff,%esi
  mov    %r13,%rdi
  call   <count_words>
  add    %rax,%rbx
  add    $0x8,%r12
  test   $0x1,%bpl
  je     L0
L8:
  mov    %rbx,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
uint32_t count_words(const uint32_t *transitions, uint32_t pos) {
  uint32_t count = 0;
  if (!pos)
    return 0;
  do {
    if (((transitions[pos]) & 0x2))
      count++;
    count += count_words(transitions, (uint32_t)(((transitions[pos]) >> 10) & ((1 << 22) - 1)));
  } while (!((transitions[pos++]) & 0x1));
  return count;
}
```
user: What is the c source code of the assembly code below:

```asm
<count_words>:
  endbr64
  push   %r13
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  test   %rsi,%rsi
  je     L4
  mov    %rdi,%r13
  lea    (%rdi,%rsi,8),%rbp
L1:
  mov    0x0(%rbp),%rbx
  mov    %r13,%rdi
  mov    %rbx,%rax
  mov    %rbx,%rsi
  and    $0x2,%eax
  cmp    $0x1,%rax
  sbb    $0xffffffffffffffff,%r12
  shr    $0xa,%rsi
  add    $0x8,%rbp
  and    $0x3fffff,%esi
  call   <count_words>
  add    %rax,%r12
  and    $0x1,%ebx
  je     L1
L4:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
uint32_t count_words(const uint32_t *transitions, uint32_t pos) {
  uint32_t count = 0;
  if (!pos)
    return 0;
  do {
    if (((transitions[pos]) & 0x2))
      count++;
    count += count_words(transitions, (uint32_t)(((transitions[pos]) >> 10) & ((1 << 22) - 1)));
  } while (!((transitions[pos++]) & 0x1));
  return count;
}
```