ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Translate the below assembly code into C programming language:

```asm
<poly1305aes_aix_authenticate>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %rdi,%rbp
  mov    %rsi,%rbx
  mov    %rcx,%r12
  mov    %r8d,%r13d
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r14
  mov    %r14,%rdi
  call   <aes_aix@plt>
  lea    0x10(%rbx),%rsi
  mov    %r13d,%r8d
  mov    %r12,%rcx
  mov    %r14,%rdx
  mov    %rbp,%rdi
  call   <poly1305_aix@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void poly1305aes_aix_authenticate(unsigned char out[16], const unsigned char kr[32], const unsigned char n[16], const unsigned char m[], unsigned int l) {
  unsigned char aeskn[16];
  aes_aix(aeskn, (kr + 0), n);
  poly1305_aix(out, (kr + 16), aeskn, m, l);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<poly1305aes_aix_authenticate>:
  endbr64
  push   %r14
  push   %r13
  mov    %r8d,%r13d
  push   %r12
  mov    %rcx,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r14
  mov    %r14,%rdi
  call   <aes_aix@plt>
  lea    0x10(%rbx),%rsi
  mov    %r13d,%r8d
  mov    %r12,%rcx
  mov    %r14,%rdx
  mov    %rbp,%rdi
  call   <poly1305_aix@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L6:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void poly1305aes_aix_authenticate(unsigned char out[16], const unsigned char kr[32], const unsigned char n[16], const unsigned char m[], unsigned int l) {
  unsigned char aeskn[16];
  aes_aix(aeskn, (kr + 0), n);
  poly1305_aix(out, (kr + 16), aeskn, m, l);
}
```
user: Translate the below assembly code into C programming language:

```asm
<poly1305aes_aix_authenticate>:
  endbr64
  push   %r14
  push   %r13
  mov    %r8d,%r13d
  push   %r12
  mov    %rcx,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r14
  mov    %r14,%rdi
  call   <aes_aix@plt>
  lea    0x10(%rbx),%rsi
  mov    %r13d,%r8d
  mov    %r12,%rcx
  mov    %r14,%rdx
  mov    %rbp,%rdi
  call   <poly1305_aix@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L0:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void poly1305aes_aix_authenticate(unsigned char out[16], const unsigned char kr[32], const unsigned char n[16], const unsigned char m[], unsigned int l) {
  unsigned char aeskn[16];
  aes_aix(aeskn, (kr + 0), n);
  poly1305_aix(out, (kr + 16), aeskn, m, l);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<rol>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%eax
  shr    $0x7,%al
  mov    %al,-0x2(%rbp)
  jmp    L9
L4:
  mov    -0x20(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  shr    $0x7,%al
  mov    %al,-0x1(%rbp)
  mov    -0x20(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x20(%rbp)
  movzbl (%rax),%eax
  movzbl %al,%eax
  add    %eax,%eax
  mov    %eax,%edx
  movzbl -0x1(%rbp),%eax
  mov    %edx,%ecx
  or     %eax,%ecx
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  mov    %ecx,%edx
  mov    %dl,(%rax)
L9:
  subl   $0x1,-0x24(%rbp)
  cmpl   $0x0,-0x24(%rbp)
  jne    L4
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  add    %eax,%eax
  mov    %eax,%edx
  movzbl -0x2(%rbp),%eax
  or     %edx,%eax
  mov    %eax,%edx
  mov    -0x18(%rbp),%rax
  mov    %dl,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void rol(unsigned char *out, const unsigned char *in, unsigned int s) {
  unsigned char firstbit = in[0] >> 7;
  while (--s > 0) {
    unsigned char bit = in[1] >> 7;
    *out++ = (*in++ << 1) | bit;
  }
  *out = (*in << 1) | firstbit;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<rol>:
  endbr64
  movzbl (%rsi),%r9d
  shr    $0x7,%r9b
  cmp    $0x1,%edx
  je     L11
  lea    -0x1(%rdx),%r8d
  mov    $0x0,%eax
L0:
  movzbl 0x1(%rsi,%rax,1),%edx
  shr    $0x7,%dl
  movzbl (%rsi,%rax,1),%ecx
  add    %ecx,%ecx
  or     %ecx,%edx
  mov    %dl,(%rdi,%rax,1)
  add    $0x1,%rax
  cmp    %r8,%rax
  jne    L0
  add    %r8,%rsi
  add    %r8,%rdi
L11:
  movzbl (%rsi),%eax
  add    %eax,%eax
  or     %r9d,%eax
  mov    %al,(%rdi)
  ret
```

assistant: ```c
void rol(unsigned char *out, const unsigned char *in, unsigned int s) {
  unsigned char firstbit = in[0] >> 7;
  while (--s > 0) {
    unsigned char bit = in[1] >> 7;
    *out++ = (*in++ << 1) | bit;
  }
  *out = (*in << 1) | firstbit;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<rol>:
  endbr64
  movzbl (%rsi),%eax
  mov    %eax,%r9d
  shr    $0x7,%r9b
  cmp    $0x1,%edx
  je     L28
  lea    -0x2(%rdx),%r8d
  xor    %edx,%edx
  lea    0x1(%r8),%r10
  jmp    L7
  nop
L15:
  movzbl 0x1(%rsi,%rdx,1),%eax
  mov    %rcx,%rdx
L7:
  movzbl 0x1(%rsi,%rdx,1),%ecx
  add    %eax,%eax
  shr    $0x7,%cl
  or     %ecx,%eax
  lea    0x1(%rdx),%rcx
  mov    %al,(%rdi,%rdx,1)
  cmp    %rdx,%r8
  jne    L15
  movzbl 0x1(%rsi,%r8,1),%eax
  add    %r10,%rdi
L28:
  add    %eax,%eax
  or     %r9d,%eax
  mov    %al,(%rdi)
  ret
```

assistant: ```c
void rol(unsigned char *out, const unsigned char *in, unsigned int s) {
  unsigned char firstbit = in[0] >> 7;
  while (--s > 0) {
    unsigned char bit = in[1] >> 7;
    *out++ = (*in++ << 1) | bit;
  }
  *out = (*in << 1) | firstbit;
}
```
user: Translate the below assembly code into C programming language:

```asm
<filterString2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x8(%rbp)
  movl   $0x0,-0x10(%rbp)
  movl   $0x0,-0xc(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L3
L37:
  cmpl   $0x40,-0x4(%rbp)
  jle    L13
  cmpl   $0x5a,-0x4(%rbp)
  jle    L29
L13:
  cmpl   $0x60,-0x4(%rbp)
  jle    L28
  cmpl   $0x7a,-0x4(%rbp)
  jle    L29
L28:
  addl   $0x1,-0x10(%rbp)
  jmp    L3
L29:
  mov    -0x10(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x10(%rbp)
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0xc(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0xc(%rbp)
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  movzbl (%rcx),%eax
  mov    %al,(%rdx)
L3:
  mov    -0x10(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jne    L37
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x18(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *filterString2(char *input) {
  int length = strlen(input);
  int i = 0;
  int j = 0;
  int ch = 0;
  while (ch = input[i]) {
    if (!((ch >= 'A' && ch <= 'Z') || (ch >= 'a' && ch <= 'z'))) {
      i++;
    } else {
      input[j++] = input[i++];
    }
  }
  input[j] = '\0';
  return input;
}
```
user: What is the c source code of the assembly code below:

```asm
<filterString2>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  movzbl (%rbx),%edx
  test   %dl,%dl
  je     L46
  mov    $0x1,%ecx
  mov    $0x0,%esi
  jmp    L27
L12:
  mov    %ecx,%eax
  movslq %esi,%rdi
  mov    %dl,(%rbx,%rdi,1)
  lea    0x1(%rsi),%esi
L5:
  cltq
  movzbl (%rbx,%rax,1),%edx
  add    $0x1,%ecx
  test   %dl,%dl
  je     L23
L27:
  mov    %edx,%eax
  and    $0xffffffdf,%eax
  sub    $0x41,%eax
  cmp    $0x19,%al
  jbe    L12
  mov    %ecx,%eax
  jmp    L5
L46:
  mov    $0x0,%esi
L23:
  movslq %esi,%rsi
  movb   $0x0,(%rbx,%rsi,1)
  mov    %rbx,%rax
  pop    %rbx
  ret
```

assistant: ```c
char *filterString2(char *input) {
  int length = strlen(input);
  int i = 0;
  int j = 0;
  int ch = 0;
  while (ch = input[i]) {
    if (!((ch >= 'A' && ch <= 'Z') || (ch >= 'a' && ch <= 'z'))) {
      i++;
    } else {
      input[j++] = input[i++];
    }
  }
  input[j] = '\0';
  return input;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<filterString2>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  call   <strlen@plt>
  movzbl (%r12),%edx
  test   %dl,%dl
  je     L21
  lea    0x1(%r12),%rcx
  xor    %esi,%esi
  xchg   %ax,%ax
L26:
  mov    %edx,%eax
  and    $0xffffffdf,%eax
  sub    $0x41,%eax
  cmp    $0x19,%al
  ja     L1
  movslq %esi,%rax
  add    $0x1,%esi
  mov    %dl,(%r12,%rax,1)
L1:
  movzbl (%rcx),%edx
  add    $0x1,%rcx
  test   %dl,%dl
  jne    L26
  movslq %esi,%rsi
  mov    %r12,%rax
  add    %r12,%rsi
  movb   $0x0,(%rsi)
  pop    %r12
  ret
L21:
  mov    %r12,%rsi
  mov    %r12,%rax
  movb   $0x0,(%rsi)
  pop    %r12
  ret
```

assistant: ```c
char *filterString2(char *input) {
  int length = strlen(input);
  int i = 0;
  int j = 0;
  int ch = 0;
  while (ch = input[i]) {
    if (!((ch >= 'A' && ch <= 'Z') || (ch >= 'a' && ch <= 'z'))) {
      i++;
    } else {
      input[j++] = input[i++];
    }
  }
  input[j] = '\0';
  return input;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<filterString2>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  call   <strlen@plt>
  movzbl (%r12),%edx
  test   %dl,%dl
  je     L15
  lea    0x1(%r12),%rsi
  xor    %ecx,%ecx
  xchg   %ax,%ax
L23:
  mov    %edx,%eax
  and    $0xffffffdf,%eax
  sub    $0x41,%eax
  cmp    $0x19,%al
  ja     L16
  add    $0x1,%rsi
  mov    %dl,(%r12,%rcx,1)
  lea    0x1(%rcx),%eax
  movzbl -0x1(%rsi),%edx
  movslq %eax,%rcx
  test   %dl,%dl
  jne    L23
L8:
  add    %r12,%rcx
  mov    %r12,%rax
  movb   $0x0,(%rcx)
  pop    %r12
  ret
  xchg   %ax,%ax
L16:
  movzbl (%rsi),%edx
  add    $0x1,%rsi
  test   %dl,%dl
  jne    L23
  jmp    L8
  nopl   (%rax)
L15:
  mov    %r12,%rcx
  mov    %r12,%rax
  movb   $0x0,(%rcx)
  pop    %r12
  ret
```

assistant: ```c
char *filterString2(char *input) {
  int length = strlen(input);
  int i = 0;
  int j = 0;
  int ch = 0;
  while (ch = input[i]) {
    if (!((ch >= 'A' && ch <= 'Z') || (ch >= 'a' && ch <= 'z'))) {
      i++;
    } else {
      input[j++] = input[i++];
    }
  }
  input[j] = '\0';
  return input;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<args_str_array>:
  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
  movw   $0x20,-0xa(%rbp)
  movl   $0x0,-0x1c(%rbp)
  lea    -0xa(%rbp),%rdx
  mov    -0x28(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strtok@plt>
  mov    %rax,-0x18(%rbp)
  jmp    L29
L7:
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    -0x18(%rbp),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  addl   $0x1,-0x1c(%rbp)
  lea    -0xa(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x0,%edi
  call   <strtok@plt>
  mov    %rax,-0x18(%rbp)
L29:
  cmpq   $0x0,-0x18(%rbp)
  jne    L7
  mov    -0x1c(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L4
  call   <__stack_chk_fail@plt>
L4:
  leave
  ret
```

assistant: ```c
int args_str_array(char *str, char **args) {
  const char s[2] = " ";
  char *token;
  int index = 0;
  int ii;
  token = strtok(str, s);
  while (token != NULL) {
    strcpy(args[index], token);
    index++;
    token = strtok(NULL, s);
  }
  return index;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<args_str_array>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rsi,%rbp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movw   $0x20,0x6(%rsp)
  lea    0x6(%rsp),%rsi
  call   <strtok@plt>
  test   %rax,%rax
  je     L28
  mov    %rax,%rsi
  mov    $0x1,%ebx
  lea    0x6(%rsp),%r13
L38:
  mov    -0x8(%rbp,%rbx,8),%rdi
  call   <strcpy@plt>
  mov    %ebx,%r12d
  mov    %r13,%rsi
  mov    $0x0,%edi
  call   <strtok@plt>
  mov    %rax,%rsi
  add    $0x1,%rbx
  test   %rax,%rax
  jne    L38
L24:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L29
  mov    %r12d,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L28:
  mov    $0x0,%r12d
  jmp    L24
L29:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int args_str_array(char *str, char **args) {
  const char s[2] = " ";
  char *token;
  int index = 0;
  int ii;
  token = strtok(str, s);
  while (token != NULL) {
    strcpy(args[index], token);
    index++;
    token = strtok(NULL, s);
  }
  return index;
}
```
user: Translate the below assembly code into C programming language:

```asm
<args_str_array>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x6(%rsp),%r12
  mov    $0x20,%eax
  mov    %r12,%rsi
  mov    %ax,0x6(%rsp)
  call   <strtok@plt>
  test   %rax,%rax
  je     L9
  mov    %rax,%rsi
  mov    $0x1,%ebx
  nopl   (%rax)
L13:
  mov    -0x8(%rbp,%rbx,8),%rdi
  mov    %ebx,%r13d
  add    $0x1,%rbx
  call   <strcpy@plt>
  mov    %r12,%rsi
  xor    %edi,%edi
  call   <strtok@plt>
  mov    %rax,%rsi
  test   %rax,%rax
  jne    L13
L30:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L11
  add    $0x18,%rsp
  mov    %r13d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L9:
  xor    %r13d,%r13d
  jmp    L30
L11:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int args_str_array(char *str, char **args) {
  const char s[2] = " ";
  char *token;
  int index = 0;
  int ii;
  token = strtok(str, s);
  while (token != NULL) {
    strcpy(args[index], token);
    index++;
    token = strtok(NULL, s);
  }
  return index;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<args_str_array>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x6(%rsp),%r12
  mov    $0x20,%eax
  mov    %r12,%rsi
  mov    %ax,0x6(%rsp)
  call   <strtok@plt>
  test   %rax,%rax
  je     L24
  mov    %rax,%rsi
  mov    $0x1,%ebx
  nopl   (%rax)
L17:
  mov    -0x8(%rbp,%rbx,8),%rdi
  mov    %ebx,%r13d
  add    $0x1,%rbx
  call   <strcpy@plt>
  mov    %r12,%rsi
  xor    %edi,%edi
  call   <strtok@plt>
  mov    %rax,%rsi
  test   %rax,%rax
  jne    L17
L33:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0x18,%rsp
  mov    %r13d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L24:
  xor    %r13d,%r13d
  jmp    L33
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int args_str_array(char *str, char **args) {
  const char s[2] = " ";
  char *token;
  int index = 0;
  int ii;
  token = strtok(str, s);
  while (token != NULL) {
    strcpy(args[index], token);
    index++;
    token = strtok(NULL, s);
  }
  return index;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<error>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <perror@plt>
  mov    $0x2,%edi
  call   <exit@plt>
  nop
  leave
  ret
```

assistant: ```c
error(char *s) {
  perror(s);
  exit(2);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<error>:
  endbr64
  sub    $0x8,%rsp
  call   <perror@plt>
  mov    $0x2,%edi
  call   <exit@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
error(char *s) {
  perror(s);
  exit(2);
}
```
user: What is the c source code of the assembly code below?

```asm
<error>:
  endbr64
  sub    $0x8,%rsp
  call   <perror@plt>
  mov    $0x2,%edi
  add    $0x8,%rsp
  jmp    1070 <exit@plt>
```

assistant: ```c
error(char *s) {
  perror(s);
  exit(2);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<error>:
  endbr64
  sub    $0x8,%rsp
  call   <perror@plt>
  mov    $0x2,%edi
  add    $0x8,%rsp
  jmp    1070 <exit@plt>
```

assistant: ```c
error(char *s) {
  perror(s);
  exit(2);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    -0xc(%rbp),%edx
  mov    -0x8(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0xc(%rbp),%edx
  mov    -0x8(%rbp),%eax
  add    %edx,%eax
  cmp    %eax,-0x4(%rbp)
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main() {
  int resp, a, b;
  resp = a + b;
  assert(resp == a + b);
}
```
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
  mov    $0x1,%edi
  call   <assert@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  int resp, a, b;
  resp = a + b;
  assert(resp == a + b);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x1,%edi
  call   <assert@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  int resp, a, b;
  resp = a + b;
  assert(resp == a + b);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x1,%edi
  call   <assert@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  int resp, a, b;
  resp = a + b;
  assert(resp == a + b);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<strncpy_good1>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x90,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x90(%rbp),%rcx
  lea    -0x60(%rbp),%rax
  mov    $0x28,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strncpy@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L0
  call   <__stack_chk_fail@plt>
L0:
  leave
  ret
```

assistant: ```c
void strncpy_good1() {
  int arr1[10];
  int arr2[20];
  strncpy(arr2, arr1, 40);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<strncpy_good1>:
  endbr64
  sub    $0x98,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x88(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  lea    0x30(%rsp),%rdi
  mov    $0x28,%edx
  call   <strncpy@plt>
  mov    0x88(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0x98,%rsp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void strncpy_good1() {
  int arr1[10];
  int arr2[20];
  strncpy(arr2, arr1, 40);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<strncpy_good1>:
  endbr64
  sub    $0x98,%rsp
  mov    $0x28,%edx
  mov    %fs:0x28,%rax
  mov    %rax,0x88(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  lea    0x30(%rsp),%rdi
  call   <strncpy@plt>
  mov    0x88(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x98,%rsp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void strncpy_good1() {
  int arr1[10];
  int arr2[20];
  strncpy(arr2, arr1, 40);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<strncpy_good1>:
  endbr64
  sub    $0x98,%rsp
  mov    $0x28,%edx
  mov    %fs:0x28,%rax
  mov    %rax,0x88(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  lea    0x30(%rsp),%rdi
  call   <strncpy@plt>
  mov    0x88(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  add    $0x98,%rsp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void strncpy_good1() {
  int arr1[10];
  int arr2[20];
  strncpy(arr2, arr1, 40);
}
```
user: What is the c source code of the assembly code below:

```asm
<mymalloc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  jne    L0
  movl   $0x1,-0x14(%rbp)
L0:
  mov    D4(%rip),%rax
  test   %rax,%rax
  jne    L12
  mov    $0x40000,%edi
  call   <malloc@plt>
  mov    %rax,D32(%rip)
  mov    D32(%rip),%rax
  test   %rax,%rax
  jne    L12
  lea    D32(%rip),%rax
  mov    %rax,%rsi
  mov    $0x1,%edi
  call   <logx@plt>
  lea    D32(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x3,%edi
  call   <exit@plt>
L12:
  mov    -0x14(%rbp),%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L14
  lea    D22(%rip),%rax
  mov    %rax,%rsi
  mov    $0x1,%edi
  call   <logx@plt>
  mov    D32(%rip),%rax
  test   %rax,%rax
  je     L41
  mov    D32(%rip),%rax
  mov    %rax,%rdi
  call   <free@plt>
L41:
  mov    -0x14(%rbp),%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L14
  lea    D32(%rip),%rax
  mov    %rax,%rsi
  mov    $0x1,%edi
  call   <logx@plt>
  lea    D32(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x3,%edi
  call   <exit@plt>
L14:
  mov    -0x8(%rbp),%rax
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D32
datatype: string
name: retrieve_data
arguments: data_tag: D22
datatype: string

tool: D32: Out of memory!


tool: D22: Running low on memory!


assistant: ```c
void *mymalloc(int size) {
  static void *safe;
  void *p;
  if (!size)
    size = 1;
  if (!safe) {
    safe = malloc(256 * 1024);
    if (!safe) {
      logx(1, "Out of memory!\n");
      printf("Out of memory!\n");
      exit(3);
    }
  }
  p = malloc(size);
  if (!p) {
    logx(1, "Running low on memory!\n");
    if (safe) {
      free(safe);
    }
    p = malloc(size);
    if (!p) {
      logx(1, "Out of memory!\n");
      printf("Out of memory!\n");
      exit(3);
    }
  }
  return (p);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<mymalloc>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebx
  test   %edi,%edi
  mov    $0x1,%eax
  cmove  %eax,%ebx
  cmpq   $0x0,D51(%rip)
  je     L53
L35:
  mov    %ebx,%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  je     L49
L36:
  mov    %rbp,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L53:
  mov    $0x40000,%edi
  call   <malloc@plt>
  mov    %rax,D41(%rip)
  test   %rax,%rax
  jne    L35
  lea    D41(%rip),%rbp
  mov    %rbp,%rsi
  mov    $0x1,%edi
  call   <logx@plt>
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    $0x3,%edi
  call   <exit@plt>
  jmp    L35
L49:
  lea    D31(%rip),%rsi
  mov    $0x1,%edi
  call   <logx@plt>
  mov    D41(%rip),%rdi
  test   %rdi,%rdi
  je     L16
  call   <free@plt>
L16:
  mov    %ebx,%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  jne    L36
  lea    D41(%rip),%rbx
  mov    %rbx,%rsi
  mov    $0x1,%edi
  call   <logx@plt>
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    $0x3,%edi
  call   <exit@plt>
  jmp    L36
```

tool_calls: name: parse_data
arguments: data_tag: D41
type_guessed: string
name: parse_data
arguments: data_tag: D31
type_guessed: string

tool: D41: Out of memory!


tool: D31: Running low on memory!


assistant: ```c
void *mymalloc(int size) {
  static void *safe;
  void *p;
  if (!size)
    size = 1;
  if (!safe) {
    safe = malloc(256 * 1024);
    if (!safe) {
      logx(1, "Out of memory!\n");
      printf("Out of memory!\n");
      exit(3);
    }
  }
  p = malloc(size);
  if (!p) {
    logx(1, "Running low on memory!\n");
    if (safe) {
      free(safe);
    }
    p = malloc(size);
    if (!p) {
      logx(1, "Out of memory!\n");
      printf("Out of memory!\n");
      exit(3);
    }
  }
  return (p);
}
```