ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<unquant_intra_differences_first_frame>:
  endbr64
  lea    -0x2(%rsi),%eax
  xor    %r8d,%r8d
  lea    (%rax,%rax,2),%rax
  lea    0xc(%rdi,%rax,4),%rcx
L28:
  mov    (%rdi),%edx
  cmp    $0x1,%esi
  jle    L26
  mov    %rdi,%rax
  nopl   (%rax)
L13:
  add    0xc(%rax),%edx
  add    $0xc,%rax
  mov    %edx,(%rax)
  cmp    %rcx,%rax
  jne    L13
L26:
  add    $0x1,%r8d
  add    $0x4,%rdi
  add    $0x4,%rcx
  cmp    $0x3,%r8d
  jne    L28
  ret
```

assistant: ```c
void unquant_intra_differences_first_frame(int *quant, const int natoms) {
  int i, j;
  for (j = 0; j < 3; j++) {
    int q = quant[j];
    for (i = 1; i < natoms; i++) {
      q += quant[i * 3 + j];
      quant[i * 3 + j] = q;
    }
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<unquant_intra_differences_first_frame>:
  endbr64
  cmp    $0x1,%esi
  jle    L23
  lea    -0x2(%rsi),%eax
  xor    %esi,%esi
  lea    (%rax,%rax,2),%rax
  lea    0xc(%rdi,%rax,4),%rcx
L20:
  mov    (%rdi),%edx
  mov    %rdi,%rax
  nopl   0x0(%rax)
L5:
  add    0xc(%rax),%edx
  add    $0xc,%rax
  mov    %edx,(%rax)
  cmp    %rax,%rcx
  jne    L5
  add    $0x1,%esi
  add    $0x4,%rdi
  add    $0x4,%rcx
  cmp    $0x3,%esi
  jne    L20
L23:
  ret
```

assistant: ```c
void unquant_intra_differences_first_frame(int *quant, const int natoms) {
  int i, j;
  for (j = 0; j < 3; j++) {
    int q = quant[j];
    for (i = 1; i < natoms; i++) {
      q += quant[i * 3 + j];
      quant[i * 3 + j] = q;
    }
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<CloneString>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  jne    L2
  mov    $0x0,%eax
  jmp    L0
L2:
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  add    $0x1,%eax
  mov    %eax,%edi
  call   <Malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x18(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    -0x8(%rbp),%rax
L0:
  leave
  ret
```

assistant: ```c
char *CloneString(char *str) {
  char *clone;
  if (str == NULL)
    return NULL;
  strcpy((clone = (char *)Malloc((strlen(str) + 1) * sizeof(char))), str);
  return clone;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<CloneString>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  test   %rdi,%rdi
  je     L15
  call   <strlen@plt>
  lea    0x1(%rax),%edi
  call   <Malloc@plt>
  mov    %rax,%rbp
  mov    %rbx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
L3:
  mov    %rbp,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L15:
  mov    %rdi,%rbp
  jmp    L3
```

assistant: ```c
char *CloneString(char *str) {
  char *clone;
  if (str == NULL)
    return NULL;
  strcpy((clone = (char *)Malloc((strlen(str) + 1) * sizeof(char))), str);
  return clone;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<CloneString>:
  endbr64
  push   %r12
  push   %rbp
  sub    $0x8,%rsp
  test   %rdi,%rdi
  je     L0
  mov    %rdi,%rbp
  call   <strlen@plt>
  lea    0x1(%rax),%edi
  call   <Malloc@plt>
  mov    %rbp,%rsi
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <strcpy@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L0:
  xor    %r12d,%r12d
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *CloneString(char *str) {
  char *clone;
  if (str == NULL)
    return NULL;
  strcpy((clone = (char *)Malloc((strlen(str) + 1) * sizeof(char))), str);
  return clone;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<CloneString>:
  endbr64
  push   %r12
  push   %rbp
  sub    $0x8,%rsp
  test   %rdi,%rdi
  je     L5
  mov    %rdi,%rbp
  call   <strlen@plt>
  lea    0x1(%rax),%edi
  call   <Malloc@plt>
  mov    %rbp,%rsi
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <strcpy@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L5:
  xor    %r12d,%r12d
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *CloneString(char *str) {
  char *clone;
  if (str == NULL)
    return NULL;
  strcpy((clone = (char *)Malloc((strlen(str) + 1) * sizeof(char))), str);
  return clone;
}
```
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    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x38(%rbp)
  movl   $0x0,-0x34(%rbp)
  lea    -0x3c(%rbp),%rax
  mov    %rax,%rsi
  lea    D15(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  lea    -0x30(%rbp),%rax
  lea    0x10(%rax),%rdi
  lea    -0x30(%rbp),%rax
  lea    0xc(%rax),%rsi
  lea    -0x30(%rbp),%rax
  lea    0x8(%rax),%rcx
  lea    -0x30(%rbp),%rax
  lea    0x4(%rax),%rdx
  lea    -0x30(%rbp),%rax
  mov    %rdi,%r9
  mov    %rsi,%r8
  mov    %rax,%rsi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  jmp    L29
L22:
  mov    -0x38(%rbp),%eax
  cltq
  mov    -0x30(%rbp,%rax,4),%eax
  add    %eax,-0x34(%rbp)
  addl   $0x1,-0x38(%rbp)
L29:
  mov    -0x3c(%rbp),%eax
  cmp    %eax,-0x38(%rbp)
  jl     L22
  mov    -0x34(%rbp),%eax
  mov    %eax,%esi
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L19
  call   <__stack_chk_fail@plt>
L19:
  leave
  ret
```

tool_calls: name: read_data
arguments: data_tag: D15
data_type: string
name: read_data
arguments: data_tag: D9
data_type: string
name: read_data
arguments: data_tag: D6
data_type: string

tool: D15: %d

tool: D9: %d %d %d %d %d

tool: D6: sum is %d 


assistant: ```c
int main() {
  int i, sum, n;
  int a[10];
  i = 0;
  sum = 0;
  scanf("%d", &n);
  scanf("%d %d %d %d %d", &a[0], &a[1], &a[2], &a[3], &a[4]);
  while (i < n) {
    sum = sum + a[i];
    i++;
  }
  printf("sum is %d \n", sum);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  sub    $0x48,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rsi
  lea    D24(%rip),%rdi
  call   <scanf@plt>
  lea    0x10(%rsp),%rsi
  lea    0x18(%rsp),%rcx
  lea    0x14(%rsp),%rdx
  lea    0x20(%rsp),%r9
  lea    0x1c(%rsp),%r8
  lea    D16(%rip),%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    0xc(%rsp),%edx
  test   %edx,%edx
  jle    L15
  lea    0x10(%rsp),%rax
  lea    -0x1(%rdx),%edx
  lea    0x14(%rsp,%rdx,4),%rdx
  mov    $0x0,%esi
L14:
  add    (%rax),%esi
  add    $0x4,%rax
  cmp    %rdx,%rax
  jne    L14
L28:
  lea    D10(%rip),%rdi
  call   <printf@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  mov    $0x0,%eax
  add    $0x48,%rsp
  ret
L15:
  mov    $0x0,%esi
  jmp    L28
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D24
type: string
name: disassemble_data
arguments: data_tag: D16
type: string
name: disassemble_data
arguments: data_tag: D10
type: string

tool: D24: %d

tool: D16: %d %d %d %d %d

tool: D10: sum is %d 


assistant: ```c
int main() {
  int i, sum, n;
  int a[10];
  i = 0;
  sum = 0;
  scanf("%d", &n);
  scanf("%d %d %d %d %d", &a[0], &a[1], &a[2], &a[3], &a[4]);
  while (i < n) {
    sum = sum + a[i];
    i++;
  }
  printf("sum is %d \n", sum);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D24(%rip),%rdi
  push   %rbx
  sub    $0x48,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rsi
  lea    0x10(%rsp),%rbx
  lea    0x14(%rsp),%rbp
  call   <scanf@plt>
  mov    %rbp,%rdx
  lea    0x18(%rsp),%rcx
  mov    %rbx,%rsi
  lea    0x20(%rsp),%r9
  lea    0x1c(%rsp),%r8
  xor    %eax,%eax
  lea    D16(%rip),%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%edx
  test   %edx,%edx
  jle    L7
  sub    $0x1,%edx
  mov    %rbx,%rax
  xor    %esi,%esi
  lea    0x0(%rbp,%rdx,4),%rdx
  nopl   (%rax)
L3:
  add    (%rax),%esi
  add    $0x4,%rax
  cmp    %rdx,%rax
  jne    L3
L30:
  lea    D20(%rip),%rdi
  call   <printf@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  add    $0x48,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L7:
  xor    %esi,%esi
  jmp    L30
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D24
type_infered: string
name: parse_data
arguments: data_label: D16
type_infered: string
name: parse_data
arguments: data_label: D20
type_infered: string

tool: D24: %d

tool: D16: %d %d %d %d %d

tool: D20: sum is %d 


assistant: ```c
int main() {
  int i, sum, n;
  int a[10];
  i = 0;
  sum = 0;
  scanf("%d", &n);
  scanf("%d %d %d %d %d", &a[0], &a[1], &a[2], &a[3], &a[4]);
  while (i < n) {
    sum = sum + a[i];
    i++;
  }
  printf("sum is %d \n", sum);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<main>:
  endbr64
  sub    $0x48,%rsp
  lea    D9(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rsi
  call   <scanf@plt>
  lea    0x10(%rsp),%rsi
  lea    0x18(%rsp),%rcx
  xor    %eax,%eax
  lea    0x14(%rsp),%rdx
  lea    0x20(%rsp),%r9
  lea    0x1c(%rsp),%r8
  lea    D29(%rip),%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%eax
  test   %eax,%eax
  jle    L70
  lea    -0x1(%rax),%edx
  cmp    $0x2,%edx
  jbe    L64
  mov    %eax,%edx
  movdqa 0x10(%rsp),%xmm0
  shr    $0x2,%edx
  cmp    $0x1,%edx
  je     L65
  paddd  0x20(%rsp),%xmm0
L65:
  movdqa %xmm0,%xmm1
  mov    %eax,%edx
  psrldq $0x8,%xmm1
  and    $0xfffffffc,%edx
  paddd  %xmm1,%xmm0
  movdqa %xmm0,%xmm1
  psrldq $0x4,%xmm1
  paddd  %xmm1,%xmm0
  movd   %xmm0,%esi
  test   $0x3,%al
  je     L36
L34:
  movslq %edx,%rcx
  add    0x10(%rsp,%rcx,4),%esi
  lea    0x1(%rdx),%ecx
  cmp    %eax,%ecx
  jge    L36
  movslq %ecx,%rcx
  add    $0x2,%edx
  add    0x10(%rsp,%rcx,4),%esi
  cmp    %edx,%eax
  jle    L36
  movslq %edx,%rdx
  add    0x10(%rsp,%rdx,4),%esi
L36:
  lea    D17(%rip),%rdi
  call   <printf@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L83
  xor    %eax,%eax
  add    $0x48,%rsp
  ret
L70:
  xor    %esi,%esi
  jmp    L36
L64:
  xor    %esi,%esi
  xor    %edx,%edx
  jmp    L34
L83:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D9
type_guessed: string
name: retrieve_data
arguments: tag: D29
type_guessed: string
name: retrieve_data
arguments: tag: D17
type_guessed: string

tool: D9: %d

tool: D29: %d %d %d %d %d

tool: D17: sum is %d 


assistant: ```c
int main() {
  int i, sum, n;
  int a[10];
  i = 0;
  sum = 0;
  scanf("%d", &n);
  scanf("%d %d %d %d %d", &a[0], &a[1], &a[2], &a[3], &a[4]);
  while (i < n) {
    sum = sum + a[i];
    i++;
  }
  printf("sum is %d \n", sum);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x1,-0xc(%rbp)
  lea    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x10(%rbp),%eax
  cmp    $0x270f,%eax
  jg     L44
  jmp    L2
L33:
  addl   $0x1,-0xc(%rbp)
  mov    -0x10(%rbp),%ecx
  mov    -0xc(%rbp),%eax
  cltd
  idiv   %ecx
  mov    %edx,%eax
  cmp    $0x2,%eax
  jne    L2
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L2:
  cmpl   $0x270f,-0xc(%rbp)
  jle    L33
L44:
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L42
  call   <__stack_chk_fail@plt>
L42:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D9
type_guessed: string
name: extract_symbolic_data
arguments: tag: D6
type_guessed: string

tool: D9: %d

tool: D6: %d


assistant: ```c
int main() {
  int N, I;
  I = 1;
  scanf("%d", &N);
  if (N < 10000) {
    while (I < 10000) {
      I++;
      if (I % N == 2)
        printf("%d\n", I);
    }
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  lea    D13(%rip),%rdi
  call   <scanf@plt>
  cmpl   $0x270f,0x4(%rsp)
  jg     L26
  mov    $0x1,%ebx
  lea    D2(%rip),%rbp
  jmp    L8
L13:
  cmp    $0x2710,%ebx
  je     L26
L8:
  add    $0x1,%ebx
  mov    %ebx,%eax
  cltd
  idivl  0x4(%rsp)
  cmp    $0x2,%edx
  jne    L13
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  jmp    L13
L26:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L19
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L19:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D13
data_type: string
name: read_data
arguments: label: D2
data_type: string

tool: D13: %d

tool: D2: %d


assistant: ```c
int main() {
  int N, I;
  I = 1;
  scanf("%d", &N);
  if (N < 10000) {
    while (I < 10000) {
      I++;
      if (I % N == 2)
        printf("%d\n", I);
    }
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r12
  lea    D19(%rip),%rdi
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  cmpl   $0x270f,0x4(%rsp)
  jg     L19
  mov    $0x1,%r12d
  lea    D14(%rip),%rbx
  jmp    L44
  nopl   (%rax)
L7:
  cmp    $0x2710,%r12d
  je     L19
L44:
  add    $0x1,%r12d
  mov    %r12d,%eax
  cltd
  idivl  0x4(%rsp)
  cmp    $0x2,%edx
  jne    L7
  mov    %r12d,%esi
  mov    %rbx,%rdi
  call   <printf@plt>
  cmp    $0x2710,%r12d
  jne    L44
  cs nopw 0x0(%rax,%rax,1)
L19:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %r12
  ret
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D19
data_type: string
name: read_data
arguments: tag: D14
data_type: string

tool: D19: %d

tool: D14: %d


assistant: ```c
int main() {
  int N, I;
  I = 1;
  scanf("%d", &N);
  if (N < 10000) {
    while (I < 10000) {
      I++;
      if (I % N == 2)
        printf("%d\n", I);
    }
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r12
  lea    D18(%rip),%rdi
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  cmpl   $0x270f,0x4(%rsp)
  jg     L47
  mov    $0x1,%r12d
  lea    D15(%rip),%rbx
  jmp    L56
  nopl   (%rax)
L37:
  cmp    $0x2710,%r12d
  je     L47
L56:
  add    $0x1,%r12d
  mov    %r12d,%eax
  cltd
  idivl  0x4(%rsp)
  cmp    $0x2,%edx
  jne    L37
  mov    %r12d,%esi
  mov    %rbx,%rdi
  call   <printf@plt>
  cmp    $0x2710,%r12d
  jne    L56
  cs nopw 0x0(%rax,%rax,1)
L47:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L21
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %r12
  ret
L21:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D18
data_type: string
name: retrieve_data
arguments: tag: D15
data_type: string

tool: D18: %d

tool: D15: %d


assistant: ```c
int main() {
  int N, I;
  I = 1;
  scanf("%d", &N);
  if (N < 10000) {
    while (I < 10000) {
      I++;
      if (I % N == 2)
        printf("%d\n", I);
    }
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<TimerIntClear>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x8(%rbp),%rax
  add    $0x24,%rax
  mov    %rax,%rdx
  mov    -0x10(%rbp),%rax
  mov    %rax,(%rdx)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void TimerIntClear(unsigned long ulBase, unsigned long ulIntFlags) {
  ;
  (*((volatile unsigned long *)(ulBase + 0x00000024))) = ulIntFlags;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<TimerIntClear>:
  endbr64
  mov    %rsi,0x24(%rdi)
  ret
```

assistant: ```c
void TimerIntClear(unsigned long ulBase, unsigned long ulIntFlags) {
  ;
  (*((volatile unsigned long *)(ulBase + 0x00000024))) = ulIntFlags;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<TimerIntClear>:
  endbr64
  mov    %rsi,0x24(%rdi)
  ret
```

assistant: ```c
void TimerIntClear(unsigned long ulBase, unsigned long ulIntFlags) {
  ;
  (*((volatile unsigned long *)(ulBase + 0x00000024))) = ulIntFlags;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  jmp    L11
L1:
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <rot13@plt>
  mov    %eax,%edi
  call   <putchar@plt>
L11:
  mov    D9(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,%edi
  call   <fgetc@plt>
  mov    %eax,-0x4(%rbp)
  mov    D8(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,-0x4(%rbp)
  jne    L1
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main(void) {
  int c;
  while ((c = fgetc(stdin)) != EOF) {
    putchar(rot13(c));
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbx
  mov    D13(%rip),%rbx
  jmp    L1
L14:
  call   <rot13@plt>
  mov    %eax,%edi
  call   <putchar@plt>
L1:
  mov    (%rbx),%edi
  call   <fgetc@plt>
  mov    %eax,%edi
  mov    D4(%rip),%rax
  cmp    %edi,(%rax)
  jne    L14
  mov    $0x0,%eax
  pop    %rbx
  ret
```

assistant: ```c
int main(void) {
  int c;
  while ((c = fgetc(stdin)) != EOF) {
    putchar(rot13(c));
  }
  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   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    D13(%rip),%rbp
  mov    D8(%rip),%rbx
  jmp    L0
  nopw   0x0(%rax,%rax,1)
L10:
  call   <rot13@plt>
  mov    %eax,%edi
  call   <putchar@plt>
L0:
  mov    0x0(%rbp),%edi
  call   <fgetc@plt>
  mov    %eax,%edi
  cmp    %eax,(%rbx)
  jne    L10
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int main(void) {
  int c;
  while ((c = fgetc(stdin)) != EOF) {
    putchar(rot13(c));
  }
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    D5(%rip),%rbp
  mov    D6(%rip),%rbx
  jmp    L12
  nopw   0x0(%rax,%rax,1)
L13:
  call   <rot13@plt>
  mov    %eax,%edi
  call   <putchar@plt>
L12:
  mov    0x0(%rbp),%edi
  call   <fgetc@plt>
  mov    %eax,%edi
  cmp    %eax,(%rbx)
  jne    L13
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int main(void) {
  int c;
  while ((c = fgetc(stdin)) != EOF) {
    putchar(rot13(c));
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed 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,-0x38(%rbp)
  movl   $0x0,-0x34(%rbp)
  jmp    L20
L27:
  mov    -0x38(%rbp),%eax
  cltq
  mov    -0x38(%rbp),%edx
  mov    %edx,-0x30(%rbp,%rax,4)
  addl   $0x1,-0x38(%rbp)
  addl   $0x2,-0x34(%rbp)
L20:
  cmpl   $0x9,-0x38(%rbp)
  jle    L27
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L23
  call   <__stack_chk_fail@plt>
L23:
  leave
  ret
```

assistant: ```c
int main(int argc, const char *argv[]) {
  int buffer[10];
  int i = 0;
  int a = 0;
  while (i < 10) {
    buffer[i] = i;
    ++i;
    a = a + 2;
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  mov    $0x0,%eax
  ret
```

assistant: ```c
int main(int argc, const char *argv[]) {
  int buffer[10];
  int i = 0;
  int a = 0;
  while (i < 10) {
    buffer[i] = i;
    ++i;
    a = a + 2;
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

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

assistant: ```c
int main(int argc, const char *argv[]) {
  int buffer[10];
  int i = 0;
  int a = 0;
  while (i < 10) {
    buffer[i] = i;
    ++i;
    a = a + 2;
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

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

assistant: ```c
int main(int argc, const char *argv[]) {
  int buffer[10];
  int i = 0;
  int a = 0;
  while (i < 10) {
    buffer[i] = i;
    ++i;
    a = a + 2;
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<gen_FPE>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    D5(%rip),%rax
  mov    (%rax),%ebx
  mov    $0x0,%eax
  call   <getpid@plt>
  mov    %ebx,%esi
  mov    %eax,%edi
  call   <kill@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int gen_FPE(void) {
  kill(getpid(), SIGFPE);
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<gen_FPE>:
  endbr64
  push   %rbx
  mov    D7(%rip),%rax
  mov    (%rax),%ebx
  mov    $0x0,%eax
  call   <getpid@plt>
  mov    %eax,%edi
  mov    %ebx,%esi
  call   <kill@plt>
  mov    $0x0,%eax
  pop    %rbx
  ret
```

assistant: ```c
int gen_FPE(void) {
  kill(getpid(), SIGFPE);
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<gen_FPE>:
  endbr64
  push   %rbp
  mov    D8(%rip),%rax
  mov    (%rax),%ebp
  xor    %eax,%eax
  call   <getpid@plt>
  mov    %eax,%edi
  mov    %ebp,%esi
  call   <kill@plt>
  xor    %eax,%eax
  pop    %rbp
  ret
```

assistant: ```c
int gen_FPE(void) {
  kill(getpid(), SIGFPE);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<gen_FPE>:
  endbr64
  push   %rbp
  mov    D9(%rip),%rax
  mov    (%rax),%ebp
  xor    %eax,%eax
  call   <getpid@plt>
  mov    %eax,%edi
  mov    %ebp,%esi
  call   <kill@plt>
  xor    %eax,%eax
  pop    %rbp
  ret
```

assistant: ```c
int gen_FPE(void) {
  kill(getpid(), SIGFPE);
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<_nand>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L17
L46:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  pxor   %xmm1,%xmm1
  ucomisd %xmm1,%xmm0
  jp     L40
  pxor   %xmm1,%xmm1
  ucomisd %xmm1,%xmm0
  jne    L40
  movsd  D8(%rip),%xmm0
  jmp    L18
L40:
  addl   $0x1,-0x4(%rbp)
L17:
  mov    -0x4(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L46
  pxor   %xmm0,%xmm0
L18:
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  pop    %rbp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D8
type: byte[8]

tool: {"D8": [0x00,0x00,0x00,0x00,0x00,0x00,0xf0,0x3f]}

assistant: ```c
double _nand(const int numInputs, const double *inputs, const double *connectionWeights) {
  int i;
  for (i = 0; i < numInputs; i++) {
    if (inputs[i] == 0) {
      return 1;
    }
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<_nand>:
  endbr64
  test   %edi,%edi
  jle    L3
  mov    %rsi,%rax
  lea    -0x1(%rdi),%edx
  lea    0x8(%rsi,%rdx,8),%rdx
  pxor   %xmm0,%xmm0
  movapd %xmm0,%xmm1
L15:
  ucomisd (%rax),%xmm0
  jp     L13
  ucomisd (%rax),%xmm1
  je     L36
L13:
  add    $0x8,%rax
  cmp    %rdx,%rax
  jne    L15
  pxor   %xmm0,%xmm0
  ret
L3:
  pxor   %xmm0,%xmm0
  ret
L36:
  movsd  D0(%rip),%xmm0
  ret
```

tool_calls: name: retrieve_data
arguments: label: D0
type: byte[8]

tool: {"D0": [0x00,0x00,0x00,0x00,0x00,0x00,0xf0,0x3f]}

assistant: ```c
double _nand(const int numInputs, const double *inputs, const double *connectionWeights) {
  int i;
  for (i = 0; i < numInputs; i++) {
    if (inputs[i] == 0) {
      return 1;
    }
  }
  return 0;
}
```