ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 下面的汇编指令对应的C源代码是什么:

```asm
<splitargs>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%r12
  mov    %fs:0x28,%rax
  mov    %rax,-0x28(%rbp)
  xor    %eax,%eax
  mov    $0x0,%esi
  call   <splitargv@plt>
  mov    %eax,%ebx
  lea    0x1(%rax),%eax
  cltq
  lea    0xf(,%rax,8),%rax
  mov    %rax,%rcx
  and    $0xfffffffffffffff0,%rcx
  and    $0xfffffffffffff000,%rax
  mov    %rsp,%rdx
  sub    %rax,%rdx
L35:
  cmp    %rdx,%rsp
  je     L10
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L35
L10:
  mov    %rcx,%rax
  and    $0xfff,%eax
  sub    %rax,%rsp
  test   %rax,%rax
  je     L24
  orq    $0x0,-0x8(%rsp,%rax,1)
L24:
  mov    %rsp,%r13
  mov    %r13,%rsi
  mov    %r12,%rdi
  call   <splitargv@plt>
  mov    %r13,%rsi
  mov    %ebx,%edi
  call   <echoargs@plt>
  mov    -0x28(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L28
  lea    -0x18(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %rbp
  ret
L28:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void splitargs(char *args) {
  int newargc = splitargv(args, NULL);
  char *newargv[newargc + 1];
  splitargv(args, newargv);
  echoargs(newargc, newargv);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<splitargs>:
  endbr64
  push   %rbp
  xor    %esi,%esi
  mov    %rsp,%rbp
  push   %r14
  push   %r13
  mov    %rdi,%r13
  push   %r12
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x28(%rbp)
  xor    %eax,%eax
  call   <splitargv@plt>
  mov    %rsp,%rcx
  mov    %eax,%r12d
  lea    0x1(%rax),%eax
  cltq
  lea    0xf(,%rax,8),%rdx
  mov    %rdx,%rax
  and    $0xfffffffffffff000,%rdx
  sub    %rdx,%rcx
  and    $0xfffffffffffffff0,%rax
  cmp    %rcx,%rsp
  je     L2
L16:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rcx,%rsp
  jne    L16
L2:
  and    $0xfff,%eax
  sub    %rax,%rsp
  test   %rax,%rax
  jne    L19
L14:
  mov    %rsp,%r14
  mov    %r13,%rdi
  mov    %r14,%rsi
  call   <splitargv@plt>
  mov    %r14,%rsi
  mov    %r12d,%edi
  call   <echoargs@plt>
  mov    -0x28(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L44
  lea    -0x18(%rbp),%rsp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %rbp
  ret
  nopw   0x0(%rax,%rax,1)
L19:
  orq    $0x0,-0x8(%rsp,%rax,1)
  jmp    L14
L44:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void splitargs(char *args) {
  int newargc = splitargv(args, NULL);
  char *newargv[newargc + 1];
  splitargv(args, newargv);
  echoargs(newargc, newargv);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<splitargs>:
  endbr64
  push   %rbp
  xor    %esi,%esi
  mov    %rsp,%rbp
  push   %r14
  push   %r13
  mov    %rdi,%r13
  push   %r12
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x28(%rbp)
  xor    %eax,%eax
  call   <splitargv@plt>
  mov    %rsp,%rcx
  mov    %eax,%r12d
  lea    0x1(%rax),%eax
  cltq
  lea    0xf(,%rax,8),%rdx
  mov    %rdx,%rax
  and    $0xfffffffffffff000,%rdx
  sub    %rdx,%rcx
  and    $0xfffffffffffffff0,%rax
  cmp    %rcx,%rsp
  je     L24
L30:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rcx,%rsp
  jne    L30
L24:
  and    $0xfff,%eax
  sub    %rax,%rsp
  test   %rax,%rax
  jne    L47
L17:
  mov    %rsp,%r14
  mov    %r13,%rdi
  mov    %r14,%rsi
  call   <splitargv@plt>
  mov    %r14,%rsi
  mov    %r12d,%edi
  call   <echoargs@plt>
  mov    -0x28(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L36
  lea    -0x18(%rbp),%rsp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %rbp
  ret
  nopw   0x0(%rax,%rax,1)
L47:
  orq    $0x0,-0x8(%rsp,%rax,1)
  jmp    L17
L36:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void splitargs(char *args) {
  int newargc = splitargv(args, NULL);
  char *newargv[newargc + 1];
  splitargv(args, newargv);
  echoargs(newargc, newargv);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0xa,-0x4(%rbp)
  cmpl   $0x1,-0x14(%rbp)
  jle    L9
  mov    -0x20(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <atoi@plt>
  mov    %eax,-0x4(%rbp)
L9:
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <benchmark@plt>
  mov    D8(%rip),%rax
  mov    (%rax),%eax
  leave
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  int num_games_per_point = 10;
  if (argc > 1)
    num_games_per_point = atoi(argv[1]);
  benchmark(num_games_per_point);
  return EXIT_SUCCESS;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    %edi,%eax
  mov    $0xa,%edi
  cmp    $0x1,%eax
  jg     L6
L3:
  call   <benchmark@plt>
  mov    D6(%rip),%rax
  mov    (%rax),%eax
  add    $0x8,%rsp
  ret
L6:
  mov    0x8(%rsi),%rdi
  call   <atoi@plt>
  mov    %eax,%edi
  jmp    L3
```

assistant: ```c
int main(int argc, char **argv) {
  int num_games_per_point = 10;
  if (argc > 1)
    num_games_per_point = atoi(argv[1]);
  benchmark(num_games_per_point);
  return EXIT_SUCCESS;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  mov    %edi,%r8d
  sub    $0x8,%rsp
  mov    $0xa,%edi
  cmp    $0x1,%r8d
  jle    L5
  mov    0x8(%rsi),%rdi
  call   <atoi@plt>
  mov    %eax,%edi
L5:
  call   <benchmark@plt>
  mov    D2(%rip),%rax
  mov    (%rax),%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  int num_games_per_point = 10;
  if (argc > 1)
    num_games_per_point = atoi(argv[1]);
  benchmark(num_games_per_point);
  return EXIT_SUCCESS;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  mov    %edi,%r8d
  sub    $0x8,%rsp
  mov    $0xa,%edi
  cmp    $0x1,%r8d
  jle    L8
  mov    0x8(%rsi),%rdi
  call   <atoi@plt>
  mov    %eax,%edi
L8:
  call   <benchmark@plt>
  mov    D9(%rip),%rax
  mov    (%rax),%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  int num_games_per_point = 10;
  if (argc > 1)
    num_games_per_point = atoi(argv[1]);
  benchmark(num_games_per_point);
  return EXIT_SUCCESS;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<binary>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %edi,-0x34(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x20(%rbp),%rax
  mov    %rax,-0x28(%rbp)
  movl   $0x0,-0x2c(%rbp)
  cmpl   $0x0,-0x34(%rbp)
  jne    L31
  mov    -0x28(%rbp),%rax
  movb   $0x30,(%rax)
  addl   $0x1,-0x2c(%rbp)
  jmp    L31
L16:
  mov    -0x34(%rbp),%eax
  and    $0x1,%eax
  test   %eax,%eax
  je     L7
  mov    $0x31,%ecx
  jmp    L51
L7:
  mov    $0x30,%ecx
L51:
  mov    -0x2c(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x2c(%rbp)
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    %cl,(%rax)
  mov    -0x34(%rbp),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  mov    %eax,-0x34(%rbp)
L31:
  cmpl   $0x0,-0x34(%rbp)
  jg     L16
  mov    -0x2c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x28(%rbp),%rax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L49
  call   <__stack_chk_fail@plt>
L49:
  leave
  ret
```

assistant: ```c
char *binary(int arg) {
  char bin[17];
  char *ptr = bin;
  int i = 0;
  if (!arg) {
    *ptr = '0';
    ++i;
  }
  while (arg >= 1) {
    *(ptr + i++) = (arg % 2) ? '1' : '0';
    arg /= 2;
  }
  *(ptr + i) = '\0';
  return ptr;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<binary>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  test   %edi,%edi
  jle    L17
L3:
  mov    %edi,%eax
  mov    %edi,%edx
  shr    $0x1f,%edx
  add    %edx,%edi
  sar    %edi
  cmp    $0x1,%eax
  jg     L3
L17:
  mov    %rsp,%rax
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L18
  add    $0x28,%rsp
  ret
L18:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char *binary(int arg) {
  char bin[17];
  char *ptr = bin;
  int i = 0;
  if (!arg) {
    *ptr = '0';
    ++i;
  }
  while (arg >= 1) {
    *(ptr + i++) = (arg % 2) ? '1' : '0';
    arg /= 2;
  }
  *(ptr + i) = '\0';
  return ptr;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

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

assistant: ```c
char *binary(int arg) {
  char bin[17];
  char *ptr = bin;
  int i = 0;
  if (!arg) {
    *ptr = '0';
    ++i;
  }
  while (arg >= 1) {
    *(ptr + i++) = (arg % 2) ? '1' : '0';
    arg /= 2;
  }
  *(ptr + i) = '\0';
  return ptr;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

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

assistant: ```c
char *binary(int arg) {
  char bin[17];
  char *ptr = bin;
  int i = 0;
  if (!arg) {
    *ptr = '0';
    ++i;
  }
  while (arg >= 1) {
    *(ptr + i++) = (arg % 2) ? '1' : '0';
    arg /= 2;
  }
  *(ptr + i) = '\0';
  return ptr;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<FileString_StrCmpNoCase>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  jmp    L5
L78:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  mov    %al,-0x6(%rbp)
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%eax
  mov    %al,-0x5(%rbp)
  cmpb   $0x40,-0x6(%rbp)
  jle    L9
  cmpb   $0x5a,-0x6(%rbp)
  jg     L9
  movzbl -0x6(%rbp),%eax
  add    $0x20,%eax
  mov    %al,-0x6(%rbp)
L9:
  cmpb   $0x40,-0x5(%rbp)
  jle    L99
  cmpb   $0x5a,-0x5(%rbp)
  jg     L99
  movzbl -0x5(%rbp),%eax
  add    $0x20,%eax
  mov    %al,-0x5(%rbp)
L99:
  movsbl -0x6(%rbp),%eax
  movsbl -0x5(%rbp),%edx
  sub    %edx,%eax
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  je     L92
  mov    -0x4(%rbp),%eax
  jmp    L75
L92:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L61
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L61
  addq   $0x1,-0x18(%rbp)
  addq   $0x1,-0x20(%rbp)
L5:
  mov    -0x24(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0x24(%rbp)
  test   %eax,%eax
  jne    L78
L61:
  mov    $0x0,%eax
L75:
  pop    %rbp
  ret
```

assistant: ```c
int FileString_StrCmpNoCase(char *s1, char *s2, int n) {
  int diff;
  char a, b;
  while (n--) {
    a = *s1;
    b = *s2;
    if ((a >= 'A') && (a <= 'Z'))
      a += 32;
    if ((b >= 'A') && (b <= 'Z'))
      b += 32;
    diff = a - b;
    if (diff)
      return diff;
    if ((*s1 == 0) || (*s2 == 0))
      break;
    s1++;
    s2++;
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<FileString_StrCmpNoCase>:
  endbr64
  mov    %rdi,%r10
  mov    %rsi,%r11
  test   %edx,%edx
  je     L22
  mov    %edx,%edx
  mov    $0x0,%r8d
L20:
  movzbl (%r10,%r8,1),%edi
  movzbl (%r11,%r8,1),%esi
  lea    -0x41(%rdi),%ecx
  lea    0x20(%rdi),%eax
  cmp    $0x1a,%cl
  cmovae %edi,%eax
  lea    -0x41(%rsi),%r9d
  lea    0x20(%rsi),%ecx
  cmp    $0x1a,%r9b
  cmovae %esi,%ecx
  movsbl %al,%eax
  movsbl %cl,%ecx
  sub    %ecx,%eax
  jne    L30
  test   %sil,%sil
  je     L30
  test   %dil,%dil
  je     L30
  add    $0x1,%r8
  cmp    %rdx,%r8
  jne    L20
  ret
L22:
  mov    %edx,%eax
L30:
  ret
```

assistant: ```c
int FileString_StrCmpNoCase(char *s1, char *s2, int n) {
  int diff;
  char a, b;
  while (n--) {
    a = *s1;
    b = *s2;
    if ((a >= 'A') && (a <= 'Z'))
      a += 32;
    if ((b >= 'A') && (b <= 'Z'))
      b += 32;
    diff = a - b;
    if (diff)
      return diff;
    if ((*s1 == 0) || (*s2 == 0))
      break;
    s1++;
    s2++;
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<FileString_StrCmpNoCase>:
  endbr64
  mov    %rdi,%r10
  mov    %rsi,%r11
  test   %edx,%edx
  je     L58
  mov    %edx,%edx
  xor    %r8d,%r8d
  jmp    L45
  nopl   (%rax)
L22:
  test   %sil,%sil
  je     L52
  test   %dil,%dil
  je     L52
  add    $0x1,%r8
  cmp    %r8,%rdx
  je     L41
L45:
  movzbl (%r10,%r8,1),%edi
  movzbl (%r11,%r8,1),%esi
  lea    -0x41(%rdi),%ecx
  lea    0x20(%rdi),%eax
  cmp    $0x1a,%cl
  lea    -0x41(%rsi),%r9d
  lea    0x20(%rsi),%ecx
  cmovae %edi,%eax
  cmp    $0x1a,%r9b
  cmovae %esi,%ecx
  movsbl %al,%eax
  movsbl %cl,%ecx
  sub    %ecx,%eax
  je     L22
L52:
  ret
  nopw   0x0(%rax,%rax,1)
L41:
  ret
L58:
  xor    %eax,%eax
  ret
```

assistant: ```c
int FileString_StrCmpNoCase(char *s1, char *s2, int n) {
  int diff;
  char a, b;
  while (n--) {
    a = *s1;
    b = *s2;
    if ((a >= 'A') && (a <= 'Z'))
      a += 32;
    if ((b >= 'A') && (b <= 'Z'))
      b += 32;
    diff = a - b;
    if (diff)
      return diff;
    if ((*s1 == 0) || (*s2 == 0))
      break;
    s1++;
    s2++;
  }
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<FileString_StrCmpNoCase>:
  endbr64
  mov    %rdi,%r10
  mov    %rsi,%r11
  test   %edx,%edx
  je     L57
  mov    %edx,%edx
  xor    %r8d,%r8d
  jmp    L6
  nopl   (%rax)
L32:
  test   %sil,%sil
  je     L36
  test   %dil,%dil
  je     L36
  add    $0x1,%r8
  cmp    %r8,%rdx
  je     L10
L6:
  movzbl (%r10,%r8,1),%edi
  movzbl (%r11,%r8,1),%esi
  lea    -0x41(%rdi),%ecx
  lea    0x20(%rdi),%eax
  cmp    $0x1a,%cl
  lea    -0x41(%rsi),%r9d
  lea    0x20(%rsi),%ecx
  cmovae %edi,%eax
  cmp    $0x1a,%r9b
  cmovae %esi,%ecx
  movsbl %al,%eax
  movsbl %cl,%ecx
  sub    %ecx,%eax
  je     L32
L36:
  ret
  nopw   0x0(%rax,%rax,1)
L10:
  ret
L57:
  xor    %eax,%eax
  ret
```

assistant: ```c
int FileString_StrCmpNoCase(char *s1, char *s2, int n) {
  int diff;
  char a, b;
  while (n--) {
    a = *s1;
    b = *s2;
    if ((a >= 'A') && (a <= 'Z'))
      a += 32;
    if ((b >= 'A') && (b <= 'Z'))
      b += 32;
    diff = a - b;
    if (diff)
      return diff;
    if ((*s1 == 0) || (*s2 == 0))
      break;
    s1++;
    s2++;
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x1,-0x14(%rbp)
  movl   $0x1,-0x10(%rbp)
  movl   $0x2,-0xc(%rbp)
  movl   $0x3,-0x1c(%rbp)
  mov    -0x1c(%rbp),%edx
  lea    -0x14(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <removeDuplicates@plt>
  mov    %eax,-0x18(%rbp)
  mov    -0x18(%rbp),%eax
  mov    %eax,%esi
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x0,-0x20(%rbp)
  jmp    L12
L18:
  mov    -0x20(%rbp),%eax
  cltq
  mov    -0x14(%rbp,%rax,4),%eax
  mov    %eax,%esi
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x20(%rbp)
L12:
  mov    -0x20(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jl     L18
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L1
  call   <__stack_chk_fail@plt>
L1:
  leave
  ret
```

tool_calls: name: read_data
arguments: label: D10
type_infered: string
name: read_data
arguments: label: D0
type_infered: string

tool: D10: Size: %d
Nums: 

tool: D0: %d	

assistant: ```c
int main() {
  int nums[] = {1, 1, 2};
  int size = sizeof(nums) / sizeof(nums[0]);
  int ss = removeDuplicates(nums, size);
  printf("Size: %d\nNums: ", ss);
  for (int i = 0; i < ss; i++) {
    printf("%d\t", nums[i]);
  }
  printf("\n");
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movl   $0x1,0xc(%rsp)
  movl   $0x1,0x10(%rsp)
  movl   $0x2,0x14(%rsp)
  lea    0xc(%rsp),%rdi
  mov    $0x3,%esi
  call   <removeDuplicates@plt>
  mov    %eax,%ebp
  mov    %eax,%esi
  lea    D1(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  test   %ebp,%ebp
  jle    L22
  mov    $0x0,%ebx
  lea    D3(%rip),%r12
L12:
  mov    0xc(%rsp,%rbx,4),%esi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x1,%rbx
  cmp    %ebx,%ebp
  jg     L12
L22:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L17
  mov    $0x0,%eax
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L17:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D1
data_type: string
name: unpack_data
arguments: tag: D3
data_type: string

tool: D1: Size: %d
Nums: 

tool: D3: %d	

assistant: ```c
int main() {
  int nums[] = {1, 1, 2};
  int size = sizeof(nums) / sizeof(nums[0]);
  int ss = removeDuplicates(nums, size);
  printf("Size: %d\nNums: ", ss);
  for (int i = 0; i < ss; i++) {
    printf("%d\t", nums[i]);
  }
  printf("\n");
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r13
  mov    $0x3,%esi
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  movabs $0x100000001,%rax
  lea    0xc(%rsp),%r12
  movl   $0x2,0x14(%rsp)
  mov    %r12,%rdi
  mov    %rax,0xc(%rsp)
  call   <removeDuplicates@plt>
  lea    D11(%rip),%rdi
  mov    %eax,%ebp
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  test   %ebp,%ebp
  jle    L27
  xor    %ebx,%ebx
  lea    D15(%rip),%r13
L3:
  mov    (%r12,%rbx,4),%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    %ebx,%ebp
  jg     L3
L27:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L15
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L15:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D11
type_infered: string
name: extract_symbolic_data
arguments: label: D15
type_infered: string

tool: D11: Size: %d
Nums: 

tool: D15: %d	

assistant: ```c
int main() {
  int nums[] = {1, 1, 2};
  int size = sizeof(nums) / sizeof(nums[0]);
  int ss = removeDuplicates(nums, size);
  printf("Size: %d\nNums: ", ss);
  for (int i = 0; i < ss; i++) {
    printf("%d\t", nums[i]);
  }
  printf("\n");
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %r13
  mov    $0x3,%esi
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x8(%rsp),%r12
  mov    D5(%rip),%rax
  movl   $0x2,0x10(%rsp)
  mov    %r12,%rdi
  mov    %rax,0x8(%rsp)
  call   <removeDuplicates@plt>
  lea    D24(%rip),%rdi
  mov    %eax,%ebp
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  test   %ebp,%ebp
  jle    L22
  xor    %ebx,%ebx
  lea    D3(%rip),%r13
L10:
  mov    (%r12,%rbx,4),%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    %ebx,%ebp
  jg     L10
L22:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L25
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L25:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D5
type: byte[8]
name: disassemble_data
arguments: label: D24
type: string
name: disassemble_data
arguments: label: D3
type: string

tool: {"D5": [0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00]}

tool: D24: Size: %d
Nums: 

tool: D3: %d	

assistant: ```c
int main() {
  int nums[] = {1, 1, 2};
  int size = sizeof(nums) / sizeof(nums[0]);
  int ss = removeDuplicates(nums, size);
  printf("Size: %d\nNums: ", ss);
  for (int i = 0; i < ss; i++) {
    printf("%d\t", nums[i]);
  }
  printf("\n");
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<strsize>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,%eax
  mov    %al,-0x1c(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L14
L17:
  addl   $0x1,-0x4(%rbp)
L14:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    %al,-0x1c(%rbp)
  jne    L17
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int strsize(char *str, char delim) {
  int i = 0;
  while (*(str + i) != delim)
    i++;
  return i;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<strsize>:
  endbr64
  mov    %esi,%ecx
  cmp    (%rdi),%sil
  je     L18
  mov    $0x1,%eax
L1:
  mov    %eax,%edx
  add    $0x1,%rax
  cmp    %cl,-0x1(%rdi,%rax,1)
  jne    L1
L9:
  mov    %edx,%eax
  ret
L18:
  mov    $0x0,%edx
  jmp    L9
```

assistant: ```c
int strsize(char *str, char delim) {
  int i = 0;
  while (*(str + i) != delim)
    i++;
  return i;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<strsize>:
  endbr64
  mov    %esi,%edx
  cmp    (%rdi),%sil
  je     L8
  mov    $0x1,%eax
L10:
  mov    %eax,%r8d
  add    $0x1,%rax
  cmp    %dl,-0x1(%rdi,%rax,1)
  jne    L10
  mov    %r8d,%eax
  ret
  nopl   0x0(%rax)
L8:
  xor    %r8d,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
int strsize(char *str, char delim) {
  int i = 0;
  while (*(str + i) != delim)
    i++;
  return i;
}
```
user: What is the c source code of the assembly code below?

```asm
<strsize>:
  endbr64
  mov    %esi,%edx
  cmp    (%rdi),%sil
  je     L0
  mov    $0x1,%eax
L16:
  mov    %eax,%r8d
  add    $0x1,%rax
  cmp    %dl,-0x1(%rdi,%rax,1)
  jne    L16
  mov    %r8d,%eax
  ret
  nopl   0x0(%rax)
L0:
  xor    %r8d,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
int strsize(char *str, char delim) {
  int i = 0;
  while (*(str + i) != delim)
    i++;
  return i;
}
```