ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <megaInit@plt>
  cmp    $0x2,%ebp
  je     L8
  lea    D19(%rip),%rdi
  mov    $0x1,%esi
  xor    %eax,%eax
  call   <printf@plt>
  or     $0xffffffff,%edi
  call   <exit@plt>
L8:
  mov    0x8(%rbx),%rdi
  mov    $0xa,%edx
  xor    %esi,%esi
  call   <strtoul@plt>
  lea    0x10(%rsp),%rsi
  lea    0x8(%rsp),%rdi
  mov    %rax,0x8(%rsp)
  call   <RandomFunc@plt>
  mov    0x10(%rsp),%rsi
  cmp    $0x1ea19042,%rsi
  je     L1
L11:
  xor    %eax,%eax
  lea    D6(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L34
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L1:
  lea    D21(%rip),%rdi
  call   <puts@plt>
  mov    0x10(%rsp),%rsi
  jmp    L11
L34:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D19
data_type: string
name: parse_data
arguments: data_tag: D6
data_type: string
name: parse_data
arguments: data_tag: D21
data_type: string

tool: D19: Call this program with %i arguments


tool: D6: %lu


tool: D21: You win!

assistant: ```c
int main(int argc, char *argv[]) {
  unsigned long input[1];
  unsigned long output[1];
  int randomFuns_i5;
  unsigned long randomFuns_value6;
  int randomFuns_main_i7;
  {
    megaInit();
    if (argc != 2) {
      printf("Call this program with %i arguments\n", 1);
      exit(-1);
    } else {
    }
    randomFuns_i5 = 0;
    while (randomFuns_i5 < 1) {
      randomFuns_value6 = strtoul(argv[randomFuns_i5 + 1], 0, 10);
      input[randomFuns_i5] = randomFuns_value6;
      randomFuns_i5++;
    }
    RandomFunc(input, output);
    if (output[0] == 513904706UL) {
      printf("You win!\n");
    } else {
    }
    randomFuns_main_i7 = 0;
    while (randomFuns_main_i7 < 1) {
      printf("%lu\n", output[randomFuns_main_i7]);
      randomFuns_main_i7++;
    }
  }
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <megaInit@plt>
  cmp    $0x2,%ebp
  je     L22
  lea    D9(%rip),%rdi
  mov    $0x1,%esi
  xor    %eax,%eax
  call   <printf@plt>
  or     $0xffffffff,%edi
  call   <exit@plt>
L22:
  mov    0x8(%rbx),%rdi
  mov    $0xa,%edx
  xor    %esi,%esi
  call   <strtoul@plt>
  lea    0x10(%rsp),%rsi
  lea    0x8(%rsp),%rdi
  mov    %rax,0x8(%rsp)
  call   <RandomFunc@plt>
  mov    0x10(%rsp),%rsi
  cmp    $0x1ea19042,%rsi
  je     L18
L36:
  xor    %eax,%eax
  lea    D8(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L34
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L18:
  lea    D29(%rip),%rdi
  call   <puts@plt>
  mov    0x10(%rsp),%rsi
  jmp    L36
L34:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D9
type_guessed: string
name: parse_data
arguments: data_label: D8
type_guessed: string
name: parse_data
arguments: data_label: D29
type_guessed: string

tool: D9: Call this program with %i arguments


tool: D8: %lu


tool: D29: You win!

assistant: ```c
int main(int argc, char *argv[]) {
  unsigned long input[1];
  unsigned long output[1];
  int randomFuns_i5;
  unsigned long randomFuns_value6;
  int randomFuns_main_i7;
  {
    megaInit();
    if (argc != 2) {
      printf("Call this program with %i arguments\n", 1);
      exit(-1);
    } else {
    }
    randomFuns_i5 = 0;
    while (randomFuns_i5 < 1) {
      randomFuns_value6 = strtoul(argv[randomFuns_i5 + 1], 0, 10);
      input[randomFuns_i5] = randomFuns_value6;
      randomFuns_i5++;
    }
    RandomFunc(input, output);
    if (output[0] == 513904706UL) {
      printf("You win!\n");
    } else {
    }
    randomFuns_main_i7 = 0;
    while (randomFuns_main_i7 < 1) {
      printf("%lu\n", output[randomFuns_main_i7]);
      randomFuns_main_i7++;
    }
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<sumVectors>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x3,-0x1c(%rbp)
  movl   $0x3,-0x18(%rbp)
  pxor   %xmm0,%xmm0
  movss  %xmm0,-0x14(%rbp)
  pxor   %xmm0,%xmm0
  movss  %xmm0,-0x10(%rbp)
  pxor   %xmm0,%xmm0
  movss  %xmm0,-0xc(%rbp)
  mov    -0x1c(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  je     L17
  mov    $0x0,%eax
  jmp    L46
L17:
  movl   $0x0,-0x20(%rbp)
  jmp    L49
L4:
  mov    -0x20(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movss  (%rax),%xmm1
  mov    -0x20(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  movss  (%rax),%xmm0
  addss  %xmm1,%xmm0
  mov    -0x20(%rbp),%eax
  cltq
  movss  %xmm0,-0x14(%rbp,%rax,4)
  addl   $0x1,-0x20(%rbp)
L49:
  mov    -0x20(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L4
  mov    $0x0,%eax
L46:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L25
  call   <__stack_chk_fail@plt>
L25:
  leave
  ret
```

assistant: ```c
float *sumVectors(const float *v1, const float *v2) {
  int flen = (sizeof(v1) / sizeof((v1)[0])) + 1;
  int slen = (sizeof(v2) / sizeof((v2)[0])) + 1;
  float result[] = {0, 0, 0};
  if (flen != slen) {
    return result;
  }
  int i;
  for (i = 0; i < flen; i++) {
    result[i] = v1[i] + v2[i];
  }
  return result;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

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

assistant: ```c
float *sumVectors(const float *v1, const float *v2) {
  int flen = (sizeof(v1) / sizeof((v1)[0])) + 1;
  int slen = (sizeof(v2) / sizeof((v2)[0])) + 1;
  float result[] = {0, 0, 0};
  if (flen != slen) {
    return result;
  }
  int i;
  for (i = 0; i < flen; i++) {
    result[i] = v1[i] + v2[i];
  }
  return result;
}
```
user: 请将以下的汇编代码转换成C语言代码:

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

assistant: ```c
float *sumVectors(const float *v1, const float *v2) {
  int flen = (sizeof(v1) / sizeof((v1)[0])) + 1;
  int slen = (sizeof(v2) / sizeof((v2)[0])) + 1;
  float result[] = {0, 0, 0};
  if (flen != slen) {
    return result;
  }
  int i;
  for (i = 0; i < flen; i++) {
    result[i] = v1[i] + v2[i];
  }
  return result;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

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

assistant: ```c
float *sumVectors(const float *v1, const float *v2) {
  int flen = (sizeof(v1) / sizeof((v1)[0])) + 1;
  int slen = (sizeof(v2) / sizeof((v2)[0])) + 1;
  float result[] = {0, 0, 0};
  if (flen != slen) {
    return result;
  }
  int i;
  for (i = 0; i < flen; i++) {
    result[i] = v1[i] + v2[i];
  }
  return result;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<itoa>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x18(%rbp)
  mov    -0x14(%rbp),%eax
  cltd
  idivl  -0x18(%rbp)
  mov    %eax,-0x8(%rbp)
  mov    -0x14(%rbp),%eax
  cltd
  idivl  -0x18(%rbp)
  mov    %edx,-0x4(%rbp)
  cmpl   $0x0,-0x8(%rbp)
  je     L1
  mov    -0x18(%rbp),%edx
  mov    -0x20(%rbp),%rcx
  mov    -0x8(%rbp),%eax
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <itoa>
  mov    %rax,-0x20(%rbp)
L1:
  cmpl   $0x9,-0x4(%rbp)
  jg     L11
  mov    -0x4(%rbp),%eax
  add    $0x30,%eax
  mov    %eax,%ecx
  jmp    L19
L11:
  mov    -0x4(%rbp),%eax
  add    $0x37,%eax
  mov    %eax,%ecx
L19:
  mov    -0x20(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x20(%rbp)
  mov    %cl,(%rax)
  mov    -0x20(%rbp),%rax
  movb   $0x0,(%rax)
  mov    -0x20(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *itoa(int d, char *buf, int base) {
  int a = d / base;
  int b = d % base;
  if (a)
    buf = itoa(a, buf, base);
  *buf++ = b < 10 ? b + '0' : b + '0' + 7;
  *buf = '\0';
  return buf;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<itoa>:
  endbr64
  push   %rbx
  mov    %edx,%ecx
  mov    %edi,%eax
  cltd
  idiv   %ecx
  mov    %edx,%ebx
  test   %eax,%eax
  jne    L3
L4:
  lea    0x30(%rbx),%edx
  lea    0x37(%rbx),%eax
  cmp    $0x9,%ebx
  cmovle %edx,%eax
  mov    %al,(%rsi)
  movb   $0x0,0x1(%rsi)
  lea    0x1(%rsi),%rax
  pop    %rbx
  ret
L3:
  mov    %eax,%edi
  mov    %ecx,%edx
  call   <itoa>
  mov    %rax,%rsi
  jmp    L4
```

assistant: ```c
char *itoa(int d, char *buf, int base) {
  int a = d / base;
  int b = d % base;
  if (a)
    buf = itoa(a, buf, base);
  *buf++ = b < 10 ? b + '0' : b + '0' + 7;
  *buf = '\0';
  return buf;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<itoa>:
  endbr64
  mov    %edi,%eax
  mov    %edx,%r8d
  push   %r12
  cltd
  push   %rbp
  idiv   %r8d
  push   %rbx
  mov    %edx,%ebx
  test   %eax,%eax
  jne    L28
L7:
  lea    0x30(%rbx),%edx
  lea    0x37(%rbx),%eax
  cmp    $0x9,%ebx
  movb   $0x0,0x1(%rsi)
  cmovle %edx,%eax
  mov    %al,(%rsi)
  lea    0x1(%rsi),%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L28:
  cltd
  idiv   %r8d
  mov    %edx,%ebp
  test   %eax,%eax
  jne    L40
L21:
  cmp    $0x9,%ebp
  lea    0x30(%rbp),%edx
  lea    0x37(%rbp),%eax
  movb   $0x0,0x1(%rsi)
  cmovle %edx,%eax
  add    $0x1,%rsi
  mov    %al,-0x1(%rsi)
  jmp    L7
  nopl   0x0(%rax,%rax,1)
L40:
  cltd
  idiv   %r8d
  mov    %edx,%r12d
  test   %eax,%eax
  jne    L36
L38:
  cmp    $0x9,%r12d
  lea    0x30(%r12),%edx
  lea    0x37(%r12),%eax
  movb   $0x0,0x1(%rsi)
  cmovle %edx,%eax
  add    $0x1,%rsi
  mov    %al,-0x1(%rsi)
  jmp    L21
  nopl   0x0(%rax)
L36:
  mov    %r8d,%edx
  mov    %eax,%edi
  call   <itoa>
  mov    %rax,%rsi
  jmp    L38
```

assistant: ```c
char *itoa(int d, char *buf, int base) {
  int a = d / base;
  int b = d % base;
  if (a)
    buf = itoa(a, buf, base);
  *buf++ = b < 10 ? b + '0' : b + '0' + 7;
  *buf = '\0';
  return buf;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<itoa>:
  endbr64
  mov    %edi,%eax
  mov    %edx,%r8d
  push   %r12
  cltd
  push   %rbp
  idiv   %r8d
  push   %rbx
  mov    %edx,%ebx
  test   %eax,%eax
  jne    L58
L21:
  lea    0x30(%rbx),%edx
  lea    0x37(%rbx),%eax
  cmp    $0x9,%ebx
  movb   $0x0,0x1(%rsi)
  cmovle %edx,%eax
  mov    %al,(%rsi)
  lea    0x1(%rsi),%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L58:
  cltd
  idiv   %r8d
  mov    %edx,%ebp
  test   %eax,%eax
  jne    L18
L54:
  cmp    $0x9,%ebp
  lea    0x30(%rbp),%edx
  lea    0x37(%rbp),%eax
  movb   $0x0,0x1(%rsi)
  cmovle %edx,%eax
  add    $0x1,%rsi
  mov    %al,-0x1(%rsi)
  jmp    L21
  nopl   0x0(%rax,%rax,1)
L18:
  cltd
  idiv   %r8d
  mov    %edx,%r12d
  test   %eax,%eax
  jne    L56
L20:
  cmp    $0x9,%r12d
  lea    0x30(%r12),%edx
  lea    0x37(%r12),%eax
  movb   $0x0,0x1(%rsi)
  cmovle %edx,%eax
  add    $0x1,%rsi
  mov    %al,-0x1(%rsi)
  jmp    L54
  nopl   0x0(%rax)
L56:
  mov    %r8d,%edx
  mov    %eax,%edi
  call   <itoa>
  mov    %rax,%rsi
  jmp    L20
```

assistant: ```c
char *itoa(int d, char *buf, int base) {
  int a = d / base;
  int b = d % base;
  if (a)
    buf = itoa(a, buf, base);
  *buf++ = b < 10 ? b + '0' : b + '0' + 7;
  *buf = '\0';
  return buf;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<abs>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jns    L8
  mov    -0x8(%rbp),%rax
  neg    %rax
  jmp    L3
L8:
  mov    -0x8(%rbp),%rax
L3:
  pop    %rbp
  ret
```

assistant: ```c
long abs(long number) {
  if (number < 0)
    return number * (-1);
  return number;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<abs>:
  endbr64
  mov    %rdi,%rax
  neg    %rax
  cmovs  %rdi,%rax
  ret
```

assistant: ```c
long abs(long number) {
  if (number < 0)
    return number * (-1);
  return number;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<abs>:
  endbr64
  mov    %rdi,%rax
  neg    %rax
  cmovs  %rdi,%rax
  ret
```

assistant: ```c
long abs(long number) {
  if (number < 0)
    return number * (-1);
  return number;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<abs>:
  endbr64
  mov    %rdi,%rax
  neg    %rax
  cmovs  %rdi,%rax
  ret
```

assistant: ```c
long abs(long number) {
  if (number < 0)
    return number * (-1);
  return number;
}
```
user: What is the c source code of the assembly code below:

```asm
<append_end_katcl>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    $0x2,%esi
  mov    %rax,%rdi
  call   <before_append_katcl@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L16
  mov    $0xffffffff,%eax
  jmp    L13
L16:
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <add_end_parse_katcl@plt>
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%edx
  mov    -0x18(%rbp),%rax
  mov    $0x2,%esi
  mov    %rax,%rdi
  call   <after_append_katcl@plt>
L13:
  leave
  ret
```

assistant: ```c
int append_end_katcl(struct katcl_line *l) {
  struct katcl_parse *p;
  int result;
  p = before_append_katcl(l, 0x02);
  if (p == NULL) {
    return -1;
  }
  result = add_end_parse_katcl(p);
  return after_append_katcl(l, 0x02, result);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<append_end_katcl>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  mov    $0x2,%esi
  call   <before_append_katcl@plt>
  test   %rax,%rax
  je     L18
  mov    %rax,%rdi
  call   <add_end_parse_katcl@plt>
  mov    %eax,%edx
  mov    $0x2,%esi
  mov    %rbx,%rdi
  call   <after_append_katcl@plt>
L7:
  pop    %rbx
  ret
L18:
  mov    $0xffffffff,%eax
  jmp    L7
```

assistant: ```c
int append_end_katcl(struct katcl_line *l) {
  struct katcl_parse *p;
  int result;
  p = before_append_katcl(l, 0x02);
  if (p == NULL) {
    return -1;
  }
  result = add_end_parse_katcl(p);
  return after_append_katcl(l, 0x02, result);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<append_end_katcl>:
  endbr64
  push   %rbp
  mov    $0x2,%esi
  mov    %rdi,%rbp
  call   <before_append_katcl@plt>
  test   %rax,%rax
  je     L7
  mov    %rax,%rdi
  call   <add_end_parse_katcl@plt>
  mov    %rbp,%rdi
  mov    $0x2,%esi
  pop    %rbp
  mov    %eax,%edx
  jmp    1080 <after_append_katcl@plt>
L7:
  mov    $0xffffffff,%eax
  pop    %rbp
  ret
```

assistant: ```c
int append_end_katcl(struct katcl_line *l) {
  struct katcl_parse *p;
  int result;
  p = before_append_katcl(l, 0x02);
  if (p == NULL) {
    return -1;
  }
  result = add_end_parse_katcl(p);
  return after_append_katcl(l, 0x02, result);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<append_end_katcl>:
  endbr64
  push   %rbp
  mov    $0x2,%esi
  mov    %rdi,%rbp
  call   <before_append_katcl@plt>
  test   %rax,%rax
  je     L0
  mov    %rax,%rdi
  call   <add_end_parse_katcl@plt>
  mov    %rbp,%rdi
  mov    $0x2,%esi
  pop    %rbp
  mov    %eax,%edx
  jmp    1080 <after_append_katcl@plt>
L0:
  mov    $0xffffffff,%eax
  pop    %rbp
  ret
```

assistant: ```c
int append_end_katcl(struct katcl_line *l) {
  struct katcl_parse *p;
  int result;
  p = before_append_katcl(l, 0x02);
  if (p == NULL) {
    return -1;
  }
  result = add_end_parse_katcl(p);
  return after_append_katcl(l, 0x02, result);
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  call   <doUpper@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

tool_calls: name: disassemble_data
arguments: label: D9
type_infered: string

tool: D9: Enter A String: 

assistant: ```c
int main() {
  printf("Enter A String: ");
  doUpper();
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D9(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  call   <doUpper@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D9
type_infered: string

tool: D9: Enter A String: 

assistant: ```c
int main() {
  printf("Enter A String: ");
  doUpper();
  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
  sub    $0x8,%rsp
  lea    D2(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  call   <doUpper@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D2
datatype: string

tool: D2: Enter A String: 

assistant: ```c
int main() {
  printf("Enter A String: ");
  doUpper();
  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
  lea    D1(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  call   <doUpper@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: tag: D1
type: string

tool: D1: Enter A String: 

assistant: ```c
int main() {
  printf("Enter A String: ");
  doUpper();
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<AESReset>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x44036000,-0x4(%rbp)
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <ASSERT@plt>
  mov    -0x4(%rbp),%eax
  add    $0x84,%eax
  cltq
  mov    (%rax),%eax
  mov    -0x4(%rbp),%edx
  add    $0x84,%edx
  movslq %edx,%rdx
  or     $0x2,%eax
  mov    %eax,(%rdx)
  nop
L7:
  mov    -0x4(%rbp),%eax
  add    $0x88,%eax
  cltq
  mov    (%rax),%eax
  and    $0x1,%eax
  test   %eax,%eax
  je     L7
  nop
  nop
  leave
  ret
```

assistant: ```c
void AESReset(uint32_t ui32Base) {
  ASSERT(ui32Base == 0x44036000);
  (*((volatile uint32_t *)(ui32Base + 0x00000084))) |= 0x00000002;
  while (((*((volatile uint32_t *)(ui32Base + 0x00000088))) & 0x00000001) == 0) {
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<AESReset>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  cmp    $0x44036000,%edi
  sete   %dil
  movzbl %dil,%edi
  call   <ASSERT@plt>
  lea    0x84(%rbx),%eax
  cltq
  mov    (%rax),%edx
  or     $0x2,%edx
  mov    %edx,(%rax)
  lea    0x88(%rbx),%edx
  movslq %edx,%rdx
L5:
  mov    (%rdx),%eax
  test   $0x1,%al
  je     L5
  pop    %rbx
  ret
```

assistant: ```c
void AESReset(uint32_t ui32Base) {
  ASSERT(ui32Base == 0x44036000);
  (*((volatile uint32_t *)(ui32Base + 0x00000084))) |= 0x00000002;
  while (((*((volatile uint32_t *)(ui32Base + 0x00000088))) & 0x00000001) == 0) {
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<AESReset>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  xor    %edi,%edi
  cmp    $0x44036000,%ebx
  sete   %dil
  call   <ASSERT@plt>
  lea    0x84(%rbx),%eax
  cltq
  mov    (%rax),%edx
  or     $0x2,%edx
  mov    %edx,(%rax)
  lea    0x88(%rbx),%edx
  movslq %edx,%rdx
L7:
  mov    (%rdx),%eax
  test   $0x1,%al
  je     L7
  pop    %rbx
  ret
```

assistant: ```c
void AESReset(uint32_t ui32Base) {
  ASSERT(ui32Base == 0x44036000);
  (*((volatile uint32_t *)(ui32Base + 0x00000084))) |= 0x00000002;
  while (((*((volatile uint32_t *)(ui32Base + 0x00000088))) & 0x00000001) == 0) {
  }
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<AESReset>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  xor    %edi,%edi
  cmp    $0x44036000,%ebx
  sete   %dil
  call   <ASSERT@plt>
  lea    0x84(%rbx),%eax
  cltq
  mov    (%rax),%edx
  or     $0x2,%edx
  mov    %edx,(%rax)
  lea    0x88(%rbx),%edx
  movslq %edx,%rdx
L8:
  mov    (%rdx),%eax
  test   $0x1,%al
  je     L8
  pop    %rbx
  ret
```

assistant: ```c
void AESReset(uint32_t ui32Base) {
  ASSERT(ui32Base == 0x44036000);
  (*((volatile uint32_t *)(ui32Base + 0x00000084))) |= 0x00000002;
  while (((*((volatile uint32_t *)(ui32Base + 0x00000088))) & 0x00000001) == 0) {
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<fn4>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    D0(%rip),%eax
  pop    %rbp
  ret
```

assistant: ```c
int fn4(void) {
  static int d = 6;
  return d;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<fn4>:
  endbr64
  mov    $0x6,%eax
  ret
```

assistant: ```c
int fn4(void) {
  static int d = 6;
  return d;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<fn4>:
  endbr64
  mov    $0x6,%eax
  ret
```

assistant: ```c
int fn4(void) {
  static int d = 6;
  return d;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<fn4>:
  endbr64
  mov    $0x6,%eax
  ret
```

assistant: ```c
int fn4(void) {
  static int d = 6;
  return d;
}
```
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
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xfc0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0xffffffff,-0xfb4(%rbp)
  lea    -0xfb0(%rbp),%rdx
  mov    $0x0,%eax
  mov    $0x1f4,%ecx
  mov    %rdx,%rdi
  rep stos %rax,%es:(%rdi)
  movl   $0x2,-0xfb0(%rbp)
  movl   $0x2,-0xfac(%rbp)
  movl   $0x2,-0xfa8(%rbp)
  movl   $0xffffffff,-0xfb4(%rbp)
  jmp    L17
L75:
  mov    -0xfb4(%rbp),%eax
  mov    %eax,%edi
  call   <input@plt>
L17:
  cmpl   $0x0,-0xfb4(%rbp)
  jle    L75
  cmpl   $0x3e7,-0xfb4(%rbp)
  jg     L75
  movl   $0x2,-0xfc0(%rbp)
  jmp    L30
L18:
  mov    -0xfc0(%rbp),%eax
  cltq
  movl   $0x0,-0xfb0(%rbp,%rax,4)
  addl   $0x1,-0xfc0(%rbp)
L30:
  mov    -0xfc0(%rbp),%eax
  cmp    -0xfb4(%rbp),%eax
  jle    L18
  movl   $0x1,-0xfb0(%rbp)
  movl   $0x1,-0xfac(%rbp)
  movl   $0x2,-0xfbc(%rbp)
  jmp    L104
L31:
  mov    -0xfbc(%rbp),%eax
  cltq
  mov    -0xfb0(%rbp,%rax,4),%eax
  test   %eax,%eax
  jne    L76
  mov    -0xfbc(%rbp),%eax
  mov    %eax,-0xfc0(%rbp)
  mov    -0xfc0(%rbp),%eax
  add    %eax,%eax
  mov    %eax,-0xfb8(%rbp)
  jmp    L9
L132:
  mov    -0xfb8(%rbp),%eax
  cltq
  movl   $0x1,-0xfb0(%rbp,%rax,4)
  mov    -0xfc0(%rbp),%eax
  add    %eax,-0xfb8(%rbp)
L9:
  mov    -0xfb4(%rbp),%eax
  add    $0x1,%eax
  cmp    %eax,-0xfb8(%rbp)
  jle    L132
L76:
  addl   $0x1,-0xfbc(%rbp)
L104:
  mov    -0xfb4(%rbp),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  add    $0x1,%eax
  cmp    %eax,-0xfbc(%rbp)
  jle    L31
  movl   $0x0,-0xfc0(%rbp)
  jmp    L88
L52:
  mov    -0xfc0(%rbp),%eax
  cltq
  mov    -0xfb0(%rbp,%rax,4),%eax
  test   %eax,%eax
  jne    L102
  mov    -0xfc0(%rbp),%eax
  mov    %eax,%edi
  call   <output@plt>
L102:
  addl   $0x1,-0xfc0(%rbp)
L88:
  mov    -0xfb4(%rbp),%eax
  cmp    -0xfc0(%rbp),%eax
  jge    L52
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L10
  call   <__stack_chk_fail@plt>
L10:
  leave
  ret
```

assistant: ```c
main() {
  int i = -1;
  int prime[1000] = {2, 2, 2};
  int c1, c2, c3;
  for (i = -1; i < 1 || i > 999; 1)
    input(i);
  for (c1 = 2; c1 <= i; c1++) {
    prime[c1] = 0;
  }
  prime[0] = 1;
  prime[1] = 1;
  for (c2 = 2; c2 <= i / 2 + 1; c2++) {
    if (prime[c2] == 0) {
      c1 = c2;
      for (c3 = 2 * c1; c3 <= i + 1; c3 = c3 + c1) {
        prime[c3] = 1;
      }
    }
  }
  for (c1 = 0; c1 < i + 1; c1++) {
    if (prime[c1] == 0)
      output(c1);
  }
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
L4:
  mov    $0xffffffff,%edi
  call   <input@plt>
  jmp    L4
```

assistant: ```c
main() {
  int i = -1;
  int prime[1000] = {2, 2, 2};
  int c1, c2, c3;
  for (i = -1; i < 1 || i > 999; 1)
    input(i);
  for (c1 = 2; c1 <= i; c1++) {
    prime[c1] = 0;
  }
  prime[0] = 1;
  prime[1] = 1;
  for (c2 = 2; c2 <= i / 2 + 1; c2++) {
    if (prime[c2] == 0) {
      c1 = c2;
      for (c3 = 2 * c1; c3 <= i + 1; c3 = c3 + c1) {
        prime[c3] = 1;
      }
    }
  }
  for (c1 = 0; c1 < i + 1; c1++) {
    if (prime[c1] == 0)
      output(c1);
  }
}
```