ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 这段汇编代码用C语言表示是什么样的:

```asm
<exp_commands>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    $0x20,%ebx
  mov    D8(%rip),%rbp
L7:
  mov    0x0(%rbp),%edi
  call   <getc@plt>
  cmp    $0x21,%al
  sete   %dl
  cmp    %al,%bl
  setne  %al
  mov    $0x21,%ebx
  test   %al,%dl
  jne    L7
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void exp_commands() {
  char command = '!', mid = ' ';
  do {
    command = getc(stdin);
    if (command == mid)
      break;
    mid = command;
  } while (command == '!');
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<exp_commands>:
  endbr64
  push   %rbp
  push   %rbx
  mov    $0x20,%ebx
  sub    $0x8,%rsp
  mov    D8(%rip),%rbp
  cs nopw 0x0(%rax,%rax,1)
L9:
  mov    0x0(%rbp),%edi
  call   <getc@plt>
  cmp    $0x21,%al
  sete   %dl
  cmp    %al,%bl
  mov    $0x21,%ebx
  setne  %al
  test   %al,%dl
  jne    L9
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void exp_commands() {
  char command = '!', mid = ' ';
  do {
    command = getc(stdin);
    if (command == mid)
      break;
    mid = command;
  } while (command == '!');
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<exp_commands>:
  endbr64
  push   %rbp
  push   %rbx
  mov    $0x20,%ebx
  sub    $0x8,%rsp
  mov    D0(%rip),%rbp
  cs nopw 0x0(%rax,%rax,1)
L8:
  mov    0x0(%rbp),%edi
  call   <getc@plt>
  cmp    $0x21,%al
  sete   %dl
  cmp    %al,%bl
  mov    $0x21,%ebx
  setne  %al
  test   %al,%dl
  jne    L8
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void exp_commands() {
  char command = '!', mid = ' ';
  do {
    command = getc(stdin);
    if (command == mid)
      break;
    mid = command;
  } while (command == '!');
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<step>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  and    $0x1,%eax
  test   %eax,%eax
  jne    L0
  mov    -0x4(%rbp),%eax
  shr    %eax
  jmp    L6
L0:
  mov    -0x4(%rbp),%edx
  mov    %edx,%eax
  add    %eax,%eax
  add    %edx,%eax
  add    $0x1,%eax
L6:
  pop    %rbp
  ret
```

assistant: ```c
unsigned int step(unsigned int x) {
  if (x % 2 == 0) {
    return (x / 2);
  } else {
    return (3 * x + 1);
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<step>:
  endbr64
  test   $0x1,%dil
  jne    L7
  mov    %edi,%eax
  shr    %eax
  ret
L7:
  lea    0x1(%rdi,%rdi,2),%eax
  ret
```

assistant: ```c
unsigned int step(unsigned int x) {
  if (x % 2 == 0) {
    return (x / 2);
  } else {
    return (3 * x + 1);
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<step>:
  endbr64
  test   $0x1,%dil
  jne    L9
  mov    %edi,%eax
  shr    %eax
  ret
  nop
L9:
  lea    0x1(%rdi,%rdi,2),%eax
  ret
```

assistant: ```c
unsigned int step(unsigned int x) {
  if (x % 2 == 0) {
    return (x / 2);
  } else {
    return (3 * x + 1);
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<step>:
  endbr64
  test   $0x1,%dil
  jne    L1
  mov    %edi,%eax
  shr    %eax
  ret
  nop
L1:
  lea    0x1(%rdi,%rdi,2),%eax
  ret
```

assistant: ```c
unsigned int step(unsigned int x) {
  if (x % 2 == 0) {
    return (x / 2);
  } else {
    return (3 * x + 1);
  }
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<stoi>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  movl   $0xffffffff,-0x8(%rbp)
  nop
L42:
  addl   $0x1,-0x8(%rbp)
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L42
  movl   $0x0,-0x4(%rbp)
  jmp    L44
L18:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isdigit@plt>
  test   %rax,%rax
  jne    L43
  mov    $0xffffffff,%eax
  jmp    L35
L43:
  nop
  addl   $0x1,-0x4(%rbp)
L44:
  mov    -0x8(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0x4(%rbp)
  jl     L18
  mov    -0x18(%rbp),%rax
  mov    $0xa,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <strtol@plt>
L35:
  leave
  ret
```

assistant: ```c
int stoi(char string[]) {
  int length = -1;
  int index;
  while (string[++length] != '\0')
    ;
  for (index = 0; index < length - 1; index++)
    if ((isdigit(string[index])) != 0)
      continue;
    else
      return -1;
  return (int)strtol(string, (char **)NULL, 10);
}
```
user: Translate the below assembly code into C programming language:

```asm
<stoi>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    $0xffffffffffffffff,%rax
L39:
  mov    %eax,%edx
  add    $0x1,%rax
  cmpb   $0x0,0x0(%rbp,%rax,1)
  jne    L39
  test   %edx,%edx
  jle    L37
  mov    %rbp,%rbx
  lea    -0x1(%rdx),%eax
  lea    0x1(%rbp,%rax,1),%r12
L43:
  movsbl (%rbx),%edi
  call   <isdigit@plt>
  test   %rax,%rax
  je     L14
  add    $0x1,%rbx
  cmp    %r12,%rbx
  jne    L43
L37:
  mov    $0xa,%edx
  mov    $0x0,%esi
  mov    %rbp,%rdi
  call   <strtol@plt>
L36:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L14:
  mov    $0xffffffff,%eax
  jmp    L36
```

assistant: ```c
int stoi(char string[]) {
  int length = -1;
  int index;
  while (string[++length] != '\0')
    ;
  for (index = 0; index < length - 1; index++)
    if ((isdigit(string[index])) != 0)
      continue;
    else
      return -1;
  return (int)strtol(string, (char **)NULL, 10);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<stoi>:
  endbr64
  push   %r12
  mov    $0xffffffffffffffff,%rax
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  nopw   0x0(%rax,%rax,1)
L28:
  mov    %eax,%edx
  add    $0x1,%rax
  cmpb   $0x0,0x0(%rbp,%rax,1)
  jne    L28
  test   %edx,%edx
  jle    L7
  lea    -0x1(%rdx),%eax
  mov    %rbp,%rbx
  lea    0x1(%rbp,%rax,1),%r12
  nopl   0x0(%rax)
L32:
  movsbl (%rbx),%edi
  call   <isdigit@plt>
  test   %rax,%rax
  je     L33
  add    $0x1,%rbx
  cmp    %r12,%rbx
  jne    L32
L7:
  mov    %rbp,%rdi
  mov    $0xa,%edx
  xor    %esi,%esi
  call   <strtol@plt>
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L33:
  pop    %rbx
  mov    $0xffffffff,%eax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int stoi(char string[]) {
  int length = -1;
  int index;
  while (string[++length] != '\0')
    ;
  for (index = 0; index < length - 1; index++)
    if ((isdigit(string[index])) != 0)
      continue;
    else
      return -1;
  return (int)strtol(string, (char **)NULL, 10);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<stoi>:
  endbr64
  push   %r12
  mov    $0xffffffffffffffff,%rax
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  nopw   0x0(%rax,%rax,1)
L21:
  mov    %eax,%edx
  add    $0x1,%rax
  cmpb   $0x0,0x0(%rbp,%rax,1)
  jne    L21
  test   %edx,%edx
  jle    L7
  lea    -0x1(%rdx),%eax
  mov    %rbp,%rbx
  lea    0x1(%rbp,%rax,1),%r12
  nopl   0x0(%rax)
L4:
  movsbl (%rbx),%edi
  call   <isdigit@plt>
  test   %rax,%rax
  je     L38
  add    $0x1,%rbx
  cmp    %r12,%rbx
  jne    L4
L7:
  mov    %rbp,%rdi
  mov    $0xa,%edx
  xor    %esi,%esi
  call   <strtol@plt>
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L38:
  pop    %rbx
  mov    $0xffffffff,%eax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int stoi(char string[]) {
  int length = -1;
  int index;
  while (string[++length] != '\0')
    ;
  for (index = 0; index < length - 1; index++)
    if ((isdigit(string[index])) != 0)
      continue;
    else
      return -1;
  return (int)strtol(string, (char **)NULL, 10);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<Introduction>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
  pop    %rbp
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D9
data_type: string

tool: D9: Welcome to The Land of CS 344 Adventure.



assistant: ```c
void Introduction() {
  printf("Welcome to The Land of CS 344 Adventure.\n\n");
  return;
}
```
user: 这段汇编代码用C语言表示是什么样的？

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

tool_calls: name: disassemble_data
arguments: label: D5
data_type: string

tool: D5: Welcome to The Land of CS 344 Adventure.



assistant: ```c
void Introduction() {
  printf("Welcome to The Land of CS 344 Adventure.\n\n");
  return;
}
```
user: What is the c source code of the assembly code below:

```asm
<Introduction>:
  endbr64
  lea    D7(%rip),%rdi
  jmp    1050 <printf@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D7
type_guessed: string

tool: D7: Welcome to The Land of CS 344 Adventure.



assistant: ```c
void Introduction() {
  printf("Welcome to The Land of CS 344 Adventure.\n\n");
  return;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<Introduction>:
  endbr64
  lea    D0(%rip),%rdi
  jmp    1050 <printf@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D0
datatype: string

tool: D0: Welcome to The Land of CS 344 Adventure.



assistant: ```c
void Introduction() {
  printf("Welcome to The Land of CS 344 Adventure.\n\n");
  return;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <VG_print_translation_stats@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main(void) {
  VG_print_translation_stats();
  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   <VG_print_translation_stats@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  VG_print_translation_stats();
  return 0;
}
```
user: What is the c source code of the assembly code below:

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

assistant: ```c
int main(void) {
  VG_print_translation_stats();
  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
  xor    %eax,%eax
  call   <VG_print_translation_stats@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  VG_print_translation_stats();
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<roundToInt16>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  add    $0x8000,%eax
  sar    $0x10,%eax
  mov    %eax,-0x4(%rbp)
  cmpl   $0xffff8001,-0x4(%rbp)
  jge    L4
  mov    $0x8000,%eax
  jmp    L33
L4:
  cmpl   $0x7fff,-0x4(%rbp)
  jle    L9
  mov    $0x7fff,%eax
  jmp    L33
L9:
  mov    -0x4(%rbp),%eax
L33:
  pop    %rbp
  ret
```

assistant: ```c
uint16_t roundToInt16(int64_t f) {
  int r = (f + (1 << 15)) >> 16;
  if (r < -0x7FFF)
    return 0x8000;
  else if (r > 0x7FFF)
    return 0x7FFF;
  else
    return r;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<roundToInt16>:
  endbr64
  add    $0x8000,%edi
  sar    $0x10,%edi
  mov    $0x7fff,%eax
  cmp    %eax,%edi
  cmovle %edi,%eax
  cmp    $0xffff8000,%edi
  mov    $0x8000,%edx
  cmovle %edx,%eax
  ret
```

assistant: ```c
uint16_t roundToInt16(int64_t f) {
  int r = (f + (1 << 15)) >> 16;
  if (r < -0x7FFF)
    return 0x8000;
  else if (r > 0x7FFF)
    return 0x7FFF;
  else
    return r;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<roundToInt16>:
  endbr64
  lea    0x8000(%rdi),%eax
  mov    $0x8000,%edx
  sar    $0x10,%eax
  cmp    $0xffff8000,%eax
  cmove  %edx,%eax
  ret
```

assistant: ```c
uint16_t roundToInt16(int64_t f) {
  int r = (f + (1 << 15)) >> 16;
  if (r < -0x7FFF)
    return 0x8000;
  else if (r > 0x7FFF)
    return 0x7FFF;
  else
    return r;
}
```
user: Translate the below assembly code into C programming language:

```asm
<roundToInt16>:
  endbr64
  lea    0x8000(%rdi),%eax
  mov    $0x8000,%edx
  sar    $0x10,%eax
  cmp    $0xffff8000,%eax
  cmove  %edx,%eax
  ret
```

assistant: ```c
uint16_t roundToInt16(int64_t f) {
  int r = (f + (1 << 15)) >> 16;
  if (r < -0x7FFF)
    return 0x8000;
  else if (r > 0x7FFF)
    return 0x7FFF;
  else
    return r;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<bspline>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %esi,%r12d
  mov    %edx,%ebp
  mov    %rcx,%rbx
  movsd  %xmm0,(%rsp)
  lea    (%rsi,%rdx,1),%r13d
  mov    %edi,0x8(%rsp)
  cmp    %edi,%r13d
  setb   %dil
  movzbl %dil,%edi
  mov    $0x0,%eax
  call   <assert@plt>
  mov    %r12d,%eax
  lea    (%rbx,%rax,8),%r14
  lea    0x1(%r12),%eax
  mov    %eax,0x14(%rsp)
  mov    %eax,%eax
  lea    (%rbx,%rax,8),%r15
  movsd  (%r15),%xmm0
  comisd (%r14),%xmm0
  setae  %dil
  movzbl %dil,%edi
  mov    $0x0,%eax
  call   <assert@plt>
  movsd  (%rsp),%xmm3
  mov    $0x0,%edi
  comisd (%rbx),%xmm3
  jb     L21
  mov    0x8(%rsp),%eax
  movsd  (%rbx,%rax,8),%xmm0
  comisd %xmm3,%xmm0
  setae  %dil
  movzbl %dil,%edi
L21:
  mov    $0x0,%eax
  call   <assert@plt>
  test   %ebp,%ebp
  jne    L39
  movsd  (%rsp),%xmm4
  pxor   %xmm0,%xmm0
  comisd (%r14),%xmm4
  jb     L4
  movsd  (%r15),%xmm1
  movsd  D4(%rip),%xmm0
  comisd %xmm4,%xmm1
  ja     L4
  pxor   %xmm0,%xmm0
  jmp    L4
L39:
  lea    0x1(%r13),%eax
  lea    (%rbx,%rax,8),%rax
  mov    %rax,0x18(%rsp)
  movsd  (%rax),%xmm0
  comisd (%r14),%xmm0
  setae  %dil
  movzbl %dil,%edi
  mov    $0x0,%eax
  call   <assert@plt>
  movsd  (%r14),%xmm0
  mov    %r13d,%r13d
  movsd  (%rbx,%r13,8),%xmm1
  movsd  (%rsp),%xmm2
  subsd  %xmm0,%xmm2
  subsd  %xmm0,%xmm1
  movapd %xmm2,%xmm0
  mov    $0x2,%eax
  call   <frac@plt>
  mov    %eax,%r13d
  mov    0x18(%rsp),%rax
  movsd  (%rax),%xmm1
  movapd %xmm1,%xmm0
  subsd  (%rsp),%xmm0
  subsd  (%r15),%xmm1
  mov    $0x2,%eax
  call   <frac@plt>
  mov    %eax,%r14d
  sub    $0x1,%ebp
  movsd  (%rsp),%xmm0
  mov    %rbx,%rcx
  mov    %ebp,%edx
  mov    %r12d,%esi
  mov    0x8(%rsp),%r15d
  mov    %r15d,%edi
  call   <bspline>
  pxor   %xmm1,%xmm1
  cvtsi2sd %r13d,%xmm1
  mulsd  %xmm0,%xmm1
  movsd  %xmm1,0x8(%rsp)
  movsd  (%rsp),%xmm0
  mov    %rbx,%rcx
  mov    %ebp,%edx
  mov    0x14(%rsp),%esi
  mov    %r15d,%edi
  call   <bspline>
  movapd %xmm0,%xmm1
  pxor   %xmm0,%xmm0
  cvtsi2sd %r14d,%xmm0
  mulsd  %xmm1,%xmm0
  addsd  0x8(%rsp),%xmm0
L4:
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D4
type_guessed: f64

tool: D4: 1

assistant: ```c
double bspline(unsigned int n, unsigned int i, unsigned int p, const double tau[n + 1], double u) {
  assert(i + p < n);
  assert(tau[i] <= tau[i + 1]);
  assert((tau[0] <= u) && (u <= tau[n]));
  if (0 == p)
    return ((tau[i] <= u) && (u < tau[i + 1])) ? 1. : 0.;
  assert(tau[i] <= tau[i + p + 1]);
  double a = frac(u - tau[i], tau[i + p] - tau[i]);
  double b = frac(tau[i + p + 1] - u, tau[i + p + 1] - tau[i + 1]);
  return a * bspline(n, i, p - 1, tau, u) + b * bspline(n, i + 1, p - 1, tau, u);
}
```
user: Translate the below assembly code into C programming language:

```asm
<bspline>:
  endbr64
  push   %r15
  lea    (%rsi,%rdx,1),%eax
  push   %r14
  push   %r13
  mov    %edi,%r13d
  push   %r12
  mov    %rcx,%r12
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %edx,%ebx
  sub    $0x28,%rsp
  cmp    %edi,%eax
  setb   %dil
  mov    %eax,0x8(%rsp)
  xor    %eax,%eax
  movzbl %dil,%edi
  movsd  %xmm0,(%rsp)
  call   <assert@plt>
  mov    %ebp,%eax
  xor    %edi,%edi
  lea    (%r12,%rax,8),%r14
  lea    0x1(%rbp),%eax
  lea    (%r12,%rax,8),%r15
  mov    %eax,0x14(%rsp)
  movsd  (%r15),%xmm0
  comisd (%r14),%xmm0
  setae  %dil
  xor    %eax,%eax
  call   <assert@plt>
  movsd  (%rsp),%xmm3
  xor    %edi,%edi
  comisd (%r12),%xmm3
  jb     L24
  mov    %r13d,%eax
  xor    %edi,%edi
  movsd  (%r12,%rax,8),%xmm0
  comisd %xmm3,%xmm0
  setae  %dil
L24:
  xor    %eax,%eax
  call   <assert@plt>
  test   %ebx,%ebx
  jne    L33
  movsd  (%rsp),%xmm4
  comisd (%r14),%xmm4
  jae    L7
L9:
  add    $0x28,%rsp
  pxor   %xmm0,%xmm0
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   (%rax)
L7:
  movsd  (%r15),%xmm1
  movsd  D1(%rip),%xmm0
  comisd %xmm4,%xmm1
  jbe    L9
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L33:
  mov    0x8(%rsp),%eax
  xor    %edi,%edi
  add    $0x1,%eax
  lea    (%r12,%rax,8),%rdx
  movsd  (%rdx),%xmm0
  comisd (%r14),%xmm0
  mov    %rdx,0x18(%rsp)
  setae  %dil
  xor    %eax,%eax
  call   <assert@plt>
  mov    0x8(%rsp),%eax
  movsd  (%r14),%xmm0
  lea    -0x1(%rbx),%r14d
  movsd  (%rsp),%xmm2
  movsd  (%r12,%rax,8),%xmm1
  mov    $0x2,%eax
  subsd  %xmm0,%xmm2
  subsd  %xmm0,%xmm1
  movapd %xmm2,%xmm0
  call   <frac@plt>
  mov    0x18(%rsp),%rdx
  mov    %eax,0x8(%rsp)
  mov    $0x2,%eax
  movsd  (%rdx),%xmm1
  movapd %xmm1,%xmm0
  subsd  (%r15),%xmm1
  subsd  (%rsp),%xmm0
  call   <frac@plt>
  movsd  (%rsp),%xmm0
  mov    %r12,%rcx
  mov    %r14d,%edx
  mov    %ebp,%esi
  mov    %r13d,%edi
  mov    %eax,%r15d
  call   <bspline>
  mov    0x8(%rsp),%r8d
  mov    %r12,%rcx
  mov    %r14d,%edx
  pxor   %xmm1,%xmm1
  mov    0x14(%rsp),%esi
  mov    %r13d,%edi
  cvtsi2sd %r8d,%xmm1
  mulsd  %xmm0,%xmm1
  movsd  (%rsp),%xmm0
  movsd  %xmm1,0x8(%rsp)
  call   <bspline>
  movapd %xmm0,%xmm1
  pxor   %xmm0,%xmm0
  cvtsi2sd %r15d,%xmm0
  mulsd  %xmm1,%xmm0
  addsd  0x8(%rsp),%xmm0
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: unpack_data
arguments: label: D1
data_type: f64

tool: D1: 1

assistant: ```c
double bspline(unsigned int n, unsigned int i, unsigned int p, const double tau[n + 1], double u) {
  assert(i + p < n);
  assert(tau[i] <= tau[i + 1]);
  assert((tau[0] <= u) && (u <= tau[n]));
  if (0 == p)
    return ((tau[i] <= u) && (u < tau[i + 1])) ? 1. : 0.;
  assert(tau[i] <= tau[i + p + 1]);
  double a = frac(u - tau[i], tau[i + p] - tau[i]);
  double b = frac(tau[i + p + 1] - u, tau[i + p + 1] - tau[i + 1]);
  return a * bspline(n, i, p - 1, tau, u) + b * bspline(n, i + 1, p - 1, tau, u);
}
```
user: Translate the below assembly code into C programming language:

```asm
<x3main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <x3sync@plt>
  mov    $0x0,%eax
  call   <RunApplicationEventLoop@plt>
  nop
  pop    %rbp
  ret
```

assistant: ```c
void x3main(void) {
  x3sync();
  RunApplicationEventLoop();
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<x3main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <x3sync@plt>
  mov    $0x0,%eax
  call   <RunApplicationEventLoop@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void x3main(void) {
  x3sync();
  RunApplicationEventLoop();
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<x3main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <x3sync@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1070 <RunApplicationEventLoop@plt>
```

assistant: ```c
void x3main(void) {
  x3sync();
  RunApplicationEventLoop();
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<x3main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <x3sync@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1070 <RunApplicationEventLoop@plt>
```

assistant: ```c
void x3main(void) {
  x3sync();
  RunApplicationEventLoop();
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  addq   $0x1,-0x8(%rbp)
  addq   $0x1,-0x8(%rbp)
  addq   $0x1,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rsi
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
  leave
  ret
```

tool_calls: name: read_data
arguments: data_label: D0
data_type: string

tool: D0: %d

assistant: ```c
void main() {
  void *pvoid;
  pvoid++;
  pvoid += 1;
  (char *)pvoid++;
  printf("%d", pvoid);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x3,%esi
  lea    D2(%rip),%rdi
  call   <printf@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D2
type: string

tool: D2: %d

assistant: ```c
void main() {
  void *pvoid;
  pvoid++;
  pvoid += 1;
  (char *)pvoid++;
  printf("%d", pvoid);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  mov    $0x3,%esi
  lea    D0(%rip),%rdi
  jmp    1050 <printf@plt>
```

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

tool: D0: %d

assistant: ```c
void main() {
  void *pvoid;
  pvoid++;
  pvoid += 1;
  (char *)pvoid++;
  printf("%d", pvoid);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  mov    $0x3,%esi
  lea    D6(%rip),%rdi
  jmp    1050 <printf@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D6
data_type: string

tool: D6: %d

assistant: ```c
void main() {
  void *pvoid;
  pvoid++;
  pvoid += 1;
  (char *)pvoid++;
  printf("%d", pvoid);
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x5,-0x8(%rbp)
  cmpl   $0x2,-0x14(%rbp)
  jne    L7
  mov    -0x20(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <atoi@plt>
  mov    %eax,-0x8(%rbp)
L7:
  mov    $0xffffffff,%edi
  call   <suspend_block@plt>
  mov    %eax,-0x4(%rbp)
  lea    D23(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x8(%rbp),%eax
  mov    %eax,%edi
  call   <sleep@plt>
  lea    D26(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <suspend_allow@plt>
  mov    $0x1,%edi
  call   <sleep@plt>
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <suspend_block@plt>
  mov    %eax,-0x4(%rbp)
  lea    D32(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x8(%rbp),%eax
  mov    %eax,%edi
  call   <sleep@plt>
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <suspend_close@plt>
  lea    D28(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D23
datatype: string
name: disassemble_data
arguments: data_label: D26
datatype: string
name: disassemble_data
arguments: data_label: D32
datatype: string
name: disassemble_data
arguments: data_label: D28
datatype: string

tool: D23: Have the block - waiting


tool: D26: Releasing...


tool: D32: Blocked again


tool: D28: OK all done


assistant: ```c
main(int argc, char *argv[]) {
  int sleep_time = 5;
  int handle;
  if (argc == 2)
    sleep_time = atoi(argv[1]);
  handle = suspend_block(-1);
  printf("Have the block - waiting\n");
  sleep(sleep_time);
  printf("Releasing...\n");
  suspend_allow(handle);
  sleep(1);
  handle = suspend_block(handle);
  printf("Blocked again\n");
  sleep(sleep_time);
  suspend_close(handle);
  printf("OK all done\n");
  exit(1);
}
```