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

```asm
<xchg>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  cmp    %rcx,%rdx
  jne    L22
L36:
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L21
  lea    -0x28(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
  nopl   0x0(%rax,%rax,1)
L22:
  lea    0xf(%rsi),%rax
  mov    %rsi,%r13
  mov    %rcx,%rbx
  mov    %rsp,%rsi
  mov    %rax,%rcx
  and    $0xfffffffffffff000,%rax
  mov    %rdi,%r14
  mov    %rsp,%r15
  sub    %rax,%rsi
  and    $0xfffffffffffffff0,%rcx
  cmp    %rsi,%rsp
  je     L47
L11:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rsi,%rsp
  jne    L11
L47:
  and    $0xfff,%ecx
  sub    %rcx,%rsp
  test   %rcx,%rcx
  je     L1
  orq    $0x0,-0x8(%rsp,%rcx,1)
L1:
  imul   %r13,%rdx
  mov    %rsp,%rdi
  imul   %r13,%rbx
  lea    (%r14,%rdx,1),%r12
  mov    %r13,%rdx
  mov    %r12,%rsi
  add    %rbx,%r14
  call   <memcpy@plt>
  mov    %r13,%rdx
  mov    %r14,%rsi
  mov    %r12,%rdi
  call   <memcpy@plt>
  mov    %rsp,%rsi
  mov    %r13,%rdx
  mov    %r14,%rdi
  call   <memcpy@plt>
  mov    %r15,%rsp
  jmp    L36
L21:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void xchg(char *base, size_t size, size_t a, size_t b) {
  if (a != b) {
    char buffer[size];
    memcpy(buffer, base + a * size, size);
    memcpy(base + a * size, base + b * size, size);
    memcpy(base + b * size, buffer, size);
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x0,-0x4(%rbp)
  jmp    L11
L9:
  mov    D19(%rip),%eax
  mov    %eax,%edx
  lea    D19(%rip),%rax
  mov    %rax,%rsi
  lea    D21(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  call   <function@plt>
  addl   $0x1,-0x4(%rbp)
L11:
  cmpl   $0x4,-0x4(%rbp)
  jle    L9
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: unpack_data
arguments: tag: D21
data_type: string

tool: D21: [in main] static_var @ %p = %d


assistant: ```c
int main() {
  int i;
  static int static_var = 1337;
  for (i = 0; i < 5; i++) {
    printf("[in main] static_var @ %p = %d\n", &static_var, static_var);
    function();
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    $0x5,%ebx
  lea    D0(%rip),%r12
  lea    D8(%rip),%rbp
L0:
  mov    D0(%rip),%edx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  call   <function@plt>
  sub    $0x1,%ebx
  jne    L0
  mov    $0x0,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D8
datatype: string

tool: D8: [in main] static_var @ %p = %d


assistant: ```c
int main() {
  int i;
  static int static_var = 1337;
  for (i = 0; i < 5; i++) {
    printf("[in main] static_var @ %p = %d\n", &static_var, static_var);
    function();
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r12
  lea    D28(%rip),%r12
  push   %rbp
  lea    D4(%rip),%rbp
  push   %rbx
  mov    $0x5,%ebx
L6:
  mov    D28(%rip),%edx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <printf@plt>
  xor    %eax,%eax
  call   <function@plt>
  sub    $0x1,%ebx
  jne    L6
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D4
type: string

tool: D4: [in main] static_var @ %p = %d


assistant: ```c
int main() {
  int i;
  static int static_var = 1337;
  for (i = 0; i < 5; i++) {
    printf("[in main] static_var @ %p = %d\n", &static_var, static_var);
    function();
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r12
  lea    D22(%rip),%r12
  push   %rbp
  lea    D16(%rip),%rbp
  push   %rbx
  mov    $0x5,%ebx
L9:
  mov    D22(%rip),%edx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <printf@plt>
  xor    %eax,%eax
  call   <function@plt>
  sub    $0x1,%ebx
  jne    L9
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: unpack_data
arguments: tag: D16
type_infered: string

tool: D16: [in main] static_var @ %p = %d


assistant: ```c
int main() {
  int i;
  static int static_var = 1337;
  for (i = 0; i < 5; i++) {
    printf("[in main] static_var @ %p = %d\n", &static_var, static_var);
    function();
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<isprime>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x28(%rbp)
  cmpq   $0x1f,-0x28(%rbp)
  ja     L142
  mov    -0x28(%rbp),%rax
  mov    %eax,%edx
  mov    $0xa08a28ac,%eax
  mov    %edx,%ecx
  shr    %cl,%rax
  and    $0x1,%eax
  jmp    L110
L142:
  mov    -0x28(%rbp),%rax
  and    $0x1,%eax
  test   %rax,%rax
  jne    L153
  mov    $0x0,%eax
  jmp    L110
L153:
  mov    -0x28(%rbp),%rcx
  movabs $0xaaaaaaaaaaaaaaab,%rdx
  mov    %rcx,%rax
  mul    %rdx
  mov    %rdx,%rax
  shr    %rax
  mov    %rax,%rdx
  add    %rdx,%rdx
  add    %rax,%rdx
  mov    %rcx,%rax
  sub    %rdx,%rax
  test   %rax,%rax
  jne    L157
  mov    $0x0,%eax
  jmp    L110
L157:
  mov    -0x28(%rbp),%rcx
  movabs $0xcccccccccccccccd,%rdx
  mov    %rcx,%rax
  mul    %rdx
  mov    %rdx,%rax
  shr    $0x2,%rax
  mov    %rax,%rdx
  shl    $0x2,%rdx
  add    %rax,%rdx
  mov    %rcx,%rax
  sub    %rdx,%rax
  test   %rax,%rax
  jne    L144
  mov    $0x0,%eax
  jmp    L110
L144:
  mov    -0x28(%rbp),%rcx
  movabs $0x2492492492492493,%rdx
  mov    %rcx,%rax
  mul    %rdx
  mov    %rcx,%rax
  sub    %rdx,%rax
  shr    %rax
  add    %rdx,%rax
  shr    $0x2,%rax
  mov    %rax,%rdx
  shl    $0x3,%rdx
  sub    %rax,%rdx
  mov    %rcx,%rax
  sub    %rdx,%rax
  test   %rax,%rax
  jne    L96
  mov    $0x0,%eax
  jmp    L110
L96:
  movq   $0xb,-0x18(%rbp)
L174:
  mov    -0x28(%rbp),%rax
  mov    $0x0,%edx
  divq   -0x18(%rbp)
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  imul   -0x18(%rbp),%rax
  mov    %rax,%rdx
  mov    -0x28(%rbp),%rax
  sub    %rdx,%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x10(%rbp),%rax
  cmp    -0x18(%rbp),%rax
  jae    L179
  mov    $0x1,%eax
  jmp    L110
L179:
  cmpq   $0x0,-0x8(%rbp)
  je     L163
  addq   $0x2,-0x18(%rbp)
  mov    -0x28(%rbp),%rax
  mov    $0x0,%edx
  divq   -0x18(%rbp)
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  imul   -0x18(%rbp),%rax
  mov    %rax,%rdx
  mov    -0x28(%rbp),%rax
  sub    %rdx,%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x10(%rbp),%rax
  cmp    -0x18(%rbp),%rax
  jae    L16
  mov    $0x1,%eax
  jmp    L110
L16:
  cmpq   $0x0,-0x8(%rbp)
  je     L44
  addq   $0x4,-0x18(%rbp)
  jmp    L174
L163:
  nop
  jmp    L150
L44:
  nop
L150:
  mov    $0x0,%eax
L110:
  pop    %rbp
  ret
```

assistant: ```c
int isprime(unsigned long int t) {
  unsigned long int q, r, d;
  if (t < 32)
    return (0xa08a28acUL >> t) & 1;
  if ((t & 1) == 0)
    return 0;
  if (t % 3 == 0)
    return 0;
  if (t % 5 == 0)
    return 0;
  if (t % 7 == 0)
    return 0;
  for (d = 11;;) {
    q = t / d;
    r = t - q * d;
    if (q < d)
      return 1;
    if (r == 0)
      break;
    d += 2;
    q = t / d;
    r = t - q * d;
    if (q < d)
      return 1;
    if (r == 0)
      break;
    d += 4;
  }
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<isprime>:
  endbr64
  mov    %rdi,%rcx
  cmp    $0x1f,%rdi
  jbe    L127
  mov    $0x0,%eax
  test   $0x1,%dil
  je     L68
  movabs $0xaaaaaaaaaaaaaaab,%rdx
  mov    %rdi,%rax
  mul    %rdx
  mov    %rdx,%rax
  shr    %rax
  and    $0xfffffffffffffffe,%rdx
  add    %rax,%rdx
  mov    $0x0,%eax
  cmp    %rdx,%rdi
  je     L68
  movabs $0xcccccccccccccccd,%rdx
  mov    %rdi,%rax
  mul    %rdx
  mov    %rdx,%rax
  shr    $0x2,%rax
  and    $0xfffffffffffffffc,%rdx
  add    %rax,%rdx
  mov    $0x0,%eax
  cmp    %rdx,%rdi
  je     L68
  movabs $0x2492492492492493,%rdx
  mov    %rdi,%rax
  mul    %rdx
  mov    %rdi,%rax
  sub    %rdx,%rax
  shr    %rax
  add    %rax,%rdx
  shr    $0x2,%rdx
  lea    0x0(,%rdx,8),%rsi
  sub    %rdx,%rsi
  mov    $0x0,%eax
  cmp    %rsi,%rdi
  je     L68
  movabs $0x2e8ba2e8ba2e8ba3,%rdx
  mov    %rdi,%rax
  mul    %rdx
  shr    %rdx
  lea    (%rdx,%rdx,4),%rax
  lea    (%rdx,%rax,2),%rax
  mov    %rdi,%rdx
  sub    %rax,%rdx
  cmp    $0x78,%rdi
  jbe    L86
  test   %rdx,%rdx
  je     L34
  movabs $0x4ec4ec4ec4ec4ec5,%rdx
  mov    %rdi,%rax
  mul    %rdx
  shr    $0x2,%rdx
  lea    (%rdx,%rdx,2),%rax
  lea    (%rdx,%rax,4),%rax
  mov    %rdi,%rdx
  sub    %rax,%rdx
  cmp    $0xa8,%rdi
  jbe    L107
  test   %rdx,%rdx
  je     L57
  mov    $0x13,%edi
  mov    $0xb,%esi
L11:
  add    $0x6,%rsi
  mov    %rcx,%rax
  mov    $0x0,%edx
  div    %rsi
  cmp    %rsi,%rax
  jb     L16
  test   %rdx,%rdx
  je     L125
  mov    %rcx,%rax
  mov    $0x0,%edx
  div    %rdi
  cmp    %rdi,%rax
  jb     L79
  add    $0x6,%rdi
  test   %rdx,%rdx
  jne    L11
  mov    $0x0,%eax
L68:
  ret
L127:
  mov    $0xa08a28ac,%eax
  shr    %cl,%rax
  and    $0x1,%eax
  ret
L16:
  mov    $0x1,%eax
  ret
L86:
  mov    $0x1,%eax
  ret
L34:
  mov    $0x0,%eax
  ret
L107:
  mov    $0x1,%eax
  ret
L57:
  mov    $0x0,%eax
  ret
L125:
  mov    $0x0,%eax
  ret
L79:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int isprime(unsigned long int t) {
  unsigned long int q, r, d;
  if (t < 32)
    return (0xa08a28acUL >> t) & 1;
  if ((t & 1) == 0)
    return 0;
  if (t % 3 == 0)
    return 0;
  if (t % 5 == 0)
    return 0;
  if (t % 7 == 0)
    return 0;
  for (d = 11;;) {
    q = t / d;
    r = t - q * d;
    if (q < d)
      return 1;
    if (r == 0)
      break;
    d += 2;
    q = t / d;
    r = t - q * d;
    if (q < d)
      return 1;
    if (r == 0)
      break;
    d += 4;
  }
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<isprime>:
  endbr64
  mov    %rdi,%rcx
  mov    $0xa08a28ac,%r8d
  shr    %cl,%r8
  and    $0x1,%r8d
  cmp    $0x1f,%rdi
  jbe    L100
  xor    %r8d,%r8d
  test   $0x1,%dil
  je     L100
  movabs $0xaaaaaaaaaaaaaaab,%rax
  movabs $0x5555555555555555,%rdx
  imul   %rdi,%rax
  cmp    %rdx,%rax
  jbe    L100
  movabs $0xcccccccccccccccd,%rax
  movabs $0x3333333333333333,%rdx
  imul   %rdi,%rax
  cmp    %rdx,%rax
  jbe    L100
  movabs $0x6db6db6db6db6db7,%rax
  movabs $0x2492492492492492,%rdx
  imul   %rdi,%rax
  cmp    %rdx,%rax
  jbe    L100
  movabs $0x2e8ba2e8ba2e8ba3,%rdx
  mov    %rdi,%rax
  mul    %rdx
  shr    %rdx
  lea    (%rdx,%rdx,4),%rax
  lea    (%rdx,%rax,2),%rax
  mov    %rdi,%rdx
  sub    %rax,%rdx
  cmp    $0x78,%rdi
  jbe    L37
  test   %rdx,%rdx
  je     L100
  movabs $0x4ec4ec4ec4ec4ec5,%rdx
  mov    %rdi,%rax
  mul    %rdx
  shr    $0x2,%rdx
  lea    (%rdx,%rdx,2),%rax
  lea    (%rdx,%rax,4),%rax
  mov    %rdi,%rdx
  sub    %rax,%rdx
  cmp    $0xa8,%rdi
  jbe    L37
  test   %rdx,%rdx
  je     L100
  mov    $0x13,%edi
  mov    $0xb,%esi
  jmp    L126
  cs nopw 0x0(%rax,%rax,1)
L65:
  test   %rdx,%rdx
  je     L114
  mov    %rcx,%rax
  xor    %edx,%edx
  div    %rdi
  cmp    %rdi,%rax
  jb     L37
  add    $0x6,%rdi
  test   %rdx,%rdx
  je     L114
L126:
  add    $0x6,%rsi
  mov    %rcx,%rax
  xor    %edx,%edx
  div    %rsi
  cmp    %rsi,%rax
  jae    L65
L37:
  mov    $0x1,%r8d
L100:
  mov    %r8d,%eax
  ret
  cs nopw 0x0(%rax,%rax,1)
L114:
  xor    %r8d,%r8d
  jmp    L100
```

assistant: ```c
int isprime(unsigned long int t) {
  unsigned long int q, r, d;
  if (t < 32)
    return (0xa08a28acUL >> t) & 1;
  if ((t & 1) == 0)
    return 0;
  if (t % 3 == 0)
    return 0;
  if (t % 5 == 0)
    return 0;
  if (t % 7 == 0)
    return 0;
  for (d = 11;;) {
    q = t / d;
    r = t - q * d;
    if (q < d)
      return 1;
    if (r == 0)
      break;
    d += 2;
    q = t / d;
    r = t - q * d;
    if (q < d)
      return 1;
    if (r == 0)
      break;
    d += 4;
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<isprime>:
  endbr64
  mov    %rdi,%rcx
  mov    $0xa08a28ac,%r8d
  shr    %cl,%r8
  and    $0x1,%r8d
  cmp    $0x1f,%rdi
  jbe    L117
  xor    %r8d,%r8d
  test   $0x1,%dil
  je     L117
  movabs $0xaaaaaaaaaaaaaaab,%rax
  movabs $0x5555555555555555,%rdx
  imul   %rdi,%rax
  cmp    %rdx,%rax
  jbe    L117
  movabs $0xcccccccccccccccd,%rax
  movabs $0x3333333333333333,%rdx
  imul   %rdi,%rax
  cmp    %rdx,%rax
  jbe    L117
  movabs $0x6db6db6db6db6db7,%rax
  movabs $0x2492492492492492,%rdx
  imul   %rdi,%rax
  cmp    %rdx,%rax
  jbe    L117
  movabs $0x2e8ba2e8ba2e8ba3,%rdx
  mov    %rdi,%rax
  mul    %rdx
  shr    %rdx
  lea    (%rdx,%rdx,4),%rax
  lea    (%rdx,%rax,2),%rax
  mov    %rdi,%rdx
  sub    %rax,%rdx
  cmp    $0x78,%rdi
  jbe    L99
  test   %rdx,%rdx
  je     L117
  movabs $0x4ec4ec4ec4ec4ec5,%rdx
  mov    %rdi,%rax
  mul    %rdx
  shr    $0x2,%rdx
  lea    (%rdx,%rdx,2),%rax
  lea    (%rdx,%rax,4),%rax
  mov    %rdi,%rdx
  sub    %rax,%rdx
  cmp    $0xa8,%rdi
  jbe    L99
  test   %rdx,%rdx
  je     L117
  mov    $0x13,%edi
  mov    $0xb,%esi
  jmp    L20
  cs nopw 0x0(%rax,%rax,1)
L84:
  test   %rdx,%rdx
  je     L16
  mov    %rcx,%rax
  xor    %edx,%edx
  div    %rdi
  cmp    %rdi,%rax
  jb     L99
  add    $0x6,%rdi
  test   %rdx,%rdx
  je     L16
L20:
  add    $0x6,%rsi
  mov    %rcx,%rax
  xor    %edx,%edx
  div    %rsi
  cmp    %rsi,%rax
  jae    L84
L99:
  mov    $0x1,%r8d
L117:
  mov    %r8d,%eax
  ret
  cs nopw 0x0(%rax,%rax,1)
L16:
  xor    %r8d,%r8d
  jmp    L117
```

assistant: ```c
int isprime(unsigned long int t) {
  unsigned long int q, r, d;
  if (t < 32)
    return (0xa08a28acUL >> t) & 1;
  if ((t & 1) == 0)
    return 0;
  if (t % 3 == 0)
    return 0;
  if (t % 5 == 0)
    return 0;
  if (t % 7 == 0)
    return 0;
  for (d = 11;;) {
    q = t / d;
    r = t - q * d;
    if (q < d)
      return 1;
    if (r == 0)
      break;
    d += 2;
    q = t / d;
    r = t - q * d;
    if (q < d)
      return 1;
    if (r == 0)
      break;
    d += 4;
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<vocal>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,%eax
  mov    %al,-0x4(%rbp)
  cmpb   $0x61,-0x4(%rbp)
  je     L6
  cmpb   $0x41,-0x4(%rbp)
  je     L6
  cmpb   $0x65,-0x4(%rbp)
  je     L6
  cmpb   $0x45,-0x4(%rbp)
  je     L6
  cmpb   $0x69,-0x4(%rbp)
  je     L6
  cmpb   $0x49,-0x4(%rbp)
  je     L6
  cmpb   $0x6f,-0x4(%rbp)
  je     L6
  cmpb   $0x4f,-0x4(%rbp)
  je     L6
  cmpb   $0x75,-0x4(%rbp)
  je     L6
  cmpb   $0x55,-0x4(%rbp)
  jne    L102
L6:
  mov    $0x1,%eax
  jmp    L96
L102:
  mov    $0x0,%eax
L96:
  pop    %rbp
  ret
```

assistant: ```c
char vocal(char c) {
  if (c == 'a' || c == 'A' || c == 'e' || c == 'E' || c == 'i' || c == 'I' || c == 'o' || c == 'O' || c == 'u' || c == 'U')
    return 1;
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<vocal>:
  endbr64
  lea    -0x41(%rdi),%ecx
  cmp    $0x34,%cl
  ja     L2
  movabs $0x10411100104111,%rax
  shr    %cl,%rax
  and    $0x1,%eax
  ret
L2:
  mov    $0x0,%eax
  ret
```

assistant: ```c
char vocal(char c) {
  if (c == 'a' || c == 'A' || c == 'e' || c == 'E' || c == 'i' || c == 'I' || c == 'o' || c == 'O' || c == 'u' || c == 'U')
    return 1;
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<vocal>:
  endbr64
  lea    -0x41(%rdi),%ecx
  cmp    $0x34,%cl
  ja     L4
  movabs $0x10411100104111,%rax
  shr    %cl,%rax
  and    $0x1,%eax
  ret
  nopl   (%rax)
L4:
  xor    %eax,%eax
  ret
```

assistant: ```c
char vocal(char c) {
  if (c == 'a' || c == 'A' || c == 'e' || c == 'E' || c == 'i' || c == 'I' || c == 'o' || c == 'O' || c == 'u' || c == 'U')
    return 1;
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<vocal>:
  endbr64
  lea    -0x41(%rdi),%ecx
  cmp    $0x34,%cl
  ja     L5
  movabs $0x10411100104111,%rax
  shr    %cl,%rax
  and    $0x1,%eax
  ret
  nopl   (%rax)
L5:
  xor    %eax,%eax
  ret
```

assistant: ```c
char vocal(char c) {
  if (c == 'a' || c == 'A' || c == 'e' || c == 'E' || c == 'i' || c == 'I' || c == 'o' || c == 'O' || c == 'u' || c == 'U')
    return 1;
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<select_one>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x0,%eax
  call   <__VERIFIER_nondet_int@plt>
  mov    %eax,-0x8(%rbp)
  mov    -0x8(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int select_one(void) {
  int retValue_acc;
  int choice = __VERIFIER_nondet_int();
  {
    retValue_acc = choice;
    return (retValue_acc);
    return (retValue_acc);
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

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

assistant: ```c
int select_one(void) {
  int retValue_acc;
  int choice = __VERIFIER_nondet_int();
  {
    retValue_acc = choice;
    return (retValue_acc);
    return (retValue_acc);
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<select_one>:
  endbr64
  xor    %eax,%eax
  jmp    1050 <__VERIFIER_nondet_int@plt>
```

assistant: ```c
int select_one(void) {
  int retValue_acc;
  int choice = __VERIFIER_nondet_int();
  {
    retValue_acc = choice;
    return (retValue_acc);
    return (retValue_acc);
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<select_one>:
  endbr64
  xor    %eax,%eax
  jmp    1050 <__VERIFIER_nondet_int@plt>
```

assistant: ```c
int select_one(void) {
  int retValue_acc;
  int choice = __VERIFIER_nondet_int();
  {
    retValue_acc = choice;
    return (retValue_acc);
    return (retValue_acc);
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <start@plt>
  mov    $0x320,%edi
  call   <multiply_matrix@plt>
  mov    $0x0,%eax
  call   <stop@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main() {
  start();
  multiply_matrix(800);
  stop();
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <start@plt>
  mov    $0x320,%edi
  call   <multiply_matrix@plt>
  mov    $0x0,%eax
  call   <stop@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  start();
  multiply_matrix(800);
  stop();
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <start@plt>
  mov    $0x320,%edi
  call   <multiply_matrix@plt>
  xor    %eax,%eax
  call   <stop@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  start();
  multiply_matrix(800);
  stop();
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <start@plt>
  mov    $0x320,%edi
  call   <multiply_matrix@plt>
  xor    %eax,%eax
  call   <stop@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  start();
  multiply_matrix(800);
  stop();
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<LPeek>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movzbl %al,%edx
  mov    -0x8(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  shl    $0x8,%eax
  add    %eax,%edx
  mov    -0x8(%rbp),%rax
  add    $0x2,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  shl    $0x10,%eax
  add    %eax,%edx
  mov    -0x8(%rbp),%rax
  add    $0x3,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  shl    $0x18,%eax
  add    %edx,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int LPeek(const void *a) {
  int r;
  unsigned char *c = (unsigned char *)a;
  r = c[0] + (c[1] << 8) + (c[2] << 16) + ((int)(signed char)c[3] << 24);
  return r;
}
```
user: What is the c source code of the assembly code below?

```asm
<LPeek>:
  endbr64
  movzbl 0x1(%rdi),%eax
  shl    $0x8,%eax
  movzbl (%rdi),%edx
  add    %edx,%eax
  movzbl 0x2(%rdi),%edx
  shl    $0x10,%edx
  add    %edx,%eax
  movsbl 0x3(%rdi),%edx
  shl    $0x18,%edx
  add    %edx,%eax
  ret
```

assistant: ```c
int LPeek(const void *a) {
  int r;
  unsigned char *c = (unsigned char *)a;
  r = c[0] + (c[1] << 8) + (c[2] << 16) + ((int)(signed char)c[3] << 24);
  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
<LPeek>:
  endbr64
  movzbl 0x1(%rdi),%eax
  movzbl (%rdi),%edx
  shl    $0x8,%eax
  add    %edx,%eax
  movzbl 0x2(%rdi),%edx
  shl    $0x10,%edx
  add    %edx,%eax
  movsbl 0x3(%rdi),%edx
  shl    $0x18,%edx
  add    %edx,%eax
  ret
```

assistant: ```c
int LPeek(const void *a) {
  int r;
  unsigned char *c = (unsigned char *)a;
  r = c[0] + (c[1] << 8) + (c[2] << 16) + ((int)(signed char)c[3] << 24);
  return r;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  lea    D15(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    -0x8(%rbp),%eax
  mov    %eax,%esi
  lea    D21(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    D26(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D21(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    D14(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    -0x4(%rbp),%edx
  mov    -0x8(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <table@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_label: D15
data_type: string
name: parse_data
arguments: data_label: D21
data_type: string
name: parse_data
arguments: data_label: D26
data_type: string
name: parse_data
arguments: data_label: D14
data_type: string

tool: D15: Number of rows :

tool: D21: %d

tool: D26: Number of cols :

tool: D14: Here is your table! 


assistant: ```c
int main(void) {
  int rows;
  int cols;
  puts("Number of rows :");
  scanf("%d", rows);
  puts("Number of cols :");
  scanf("%d", cols);
  puts("Here is your table! \n");
  table(rows, cols);
  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
  push   %rbp
  push   %rbx
  lea    D29(%rip),%rdi
  call   <puts@plt>
  mov    $0x0,%ebx
  mov    %ebx,%esi
  lea    D31(%rip),%r12
  mov    %r12,%rdi
  call   <scanf@plt>
  lea    D34(%rip),%rdi
  call   <puts@plt>
  mov    $0x0,%ebp
  mov    %ebp,%esi
  mov    %r12,%rdi
  call   <scanf@plt>
  lea    D15(%rip),%rdi
  call   <puts@plt>
  mov    %ebp,%esi
  mov    %ebx,%edi
  call   <table@plt>
  mov    $0x0,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: read_data
arguments: data_tag: D29
type: string
name: read_data
arguments: data_tag: D31
type: string
name: read_data
arguments: data_tag: D34
type: string
name: read_data
arguments: data_tag: D15
type: string

tool: D29: Number of rows :

tool: D31: %d

tool: D34: Number of cols :

tool: D15: Here is your table! 


assistant: ```c
int main(void) {
  int rows;
  int cols;
  puts("Number of rows :");
  scanf("%d", rows);
  puts("Number of cols :");
  scanf("%d", cols);
  puts("Here is your table! \n");
  table(rows, cols);
  return 0;
}
```