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
<my_aes_encrypt>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  mov    %edx,%r13d
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  test   %rdi,%rdi
  sete   %al
  test   %rcx,%rcx
  sete   %dl
  or     %dl,%al
  jne    L43
  mov    %rsi,%rbp
  test   %rsi,%rsi
  je     L43
  lea    0x4(%rsp),%r14
  mov    $0x80,%esi
  mov    %rcx,%rbx
  mov    %r14,%rdx
  call   <AES_set_encrypt_key@plt>
  test   %rax,%rax
  js     L43
  test   %r13d,%r13d
  jle    L42
  mov    D4(%rip),%r15
  xor    %r12d,%r12d
  cs nopw 0x0(%rax,%rax,1)
L26:
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  mov    %r14,%rdx
  call   <AES_encrypt@plt>
  add    (%r15),%r12d
  add    (%r15),%rbp
  add    (%r15),%rbx
  cmp    %r12d,%r13d
  jg     L26
L42:
  mov    $0x1,%eax
  jmp    L28
  nopl   0x0(%rax,%rax,1)
L43:
  xor    %eax,%eax
L28:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L11
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L11:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int my_aes_encrypt(char *key, char *in, int inlen, char *out) {
  int i, en_len = 0;
  if (!in || !key || !out)
    return 0;
  AES_KEY aes;
  if (AES_set_encrypt_key((unsigned char *)key, 128, &aes) < 0) {
    return 0;
  }
  while (en_len < inlen) {
    AES_encrypt((unsigned char *)in, (unsigned char *)out, &aes);
    in += AES_BLOCK_SIZE;
    out += AES_BLOCK_SIZE;
    en_len += AES_BLOCK_SIZE;
  }
  return 1;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<my_aes_encrypt>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  mov    %edx,%r13d
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  test   %rdi,%rdi
  sete   %al
  test   %rcx,%rcx
  sete   %dl
  or     %dl,%al
  jne    L19
  mov    %rsi,%rbp
  test   %rsi,%rsi
  je     L19
  lea    0x4(%rsp),%r14
  mov    $0x80,%esi
  mov    %rcx,%rbx
  mov    %r14,%rdx
  call   <AES_set_encrypt_key@plt>
  test   %rax,%rax
  js     L19
  test   %r13d,%r13d
  jle    L33
  mov    D7(%rip),%r15
  xor    %r12d,%r12d
  cs nopw 0x0(%rax,%rax,1)
L7:
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  mov    %r14,%rdx
  call   <AES_encrypt@plt>
  add    (%r15),%r12d
  add    (%r15),%rbp
  add    (%r15),%rbx
  cmp    %r12d,%r13d
  jg     L7
L33:
  mov    $0x1,%eax
  jmp    L17
  nopl   0x0(%rax,%rax,1)
L19:
  xor    %eax,%eax
L17:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L26
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L26:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int my_aes_encrypt(char *key, char *in, int inlen, char *out) {
  int i, en_len = 0;
  if (!in || !key || !out)
    return 0;
  AES_KEY aes;
  if (AES_set_encrypt_key((unsigned char *)key, 128, &aes) < 0) {
    return 0;
  }
  while (en_len < inlen) {
    AES_encrypt((unsigned char *)in, (unsigned char *)out, &aes);
    in += AES_BLOCK_SIZE;
    out += AES_BLOCK_SIZE;
    en_len += AES_BLOCK_SIZE;
  }
  return 1;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %rsi,-0x30(%rbp)
  movl   $0x10000,-0x14(%rbp)
  movq   $0x0,-0x10(%rbp)
  cmpl   $0x2,-0x24(%rbp)
  jne    L9
  mov    -0x30(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  mov    $0xa,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <strtol@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %eax,-0x14(%rbp)
L9:
  mov    -0x14(%rbp),%eax
  mov    %eax,%edi
  call   <primelist@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x14(%rbp),%edx
  mov    -0x10(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <print_primes@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  movq   $0x0,-0x10(%rbp)
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  unsigned int max = (1 << 16);
  unsigned int *p = NULL;
  if (argc == 2) {
    long int tmp = strtol(argv[1], NULL, 10);
    max = (unsigned int)tmp;
  }
  p = primelist(max);
  print_primes(p, max);
  free(p);
  p = NULL;
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    $0x10000,%ebp
  cmp    $0x2,%edi
  je     L11
L4:
  mov    %ebp,%edi
  call   <primelist@plt>
  mov    %rax,%rbx
  mov    %ebp,%esi
  mov    %rax,%rdi
  call   <print_primes@plt>
  mov    %rbx,%rdi
  call   <free@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L11:
  mov    0x8(%rsi),%rdi
  mov    $0xa,%edx
  mov    $0x0,%esi
  call   <strtol@plt>
  mov    %eax,%ebp
  jmp    L4
```

assistant: ```c
int main(int argc, char *argv[]) {
  unsigned int max = (1 << 16);
  unsigned int *p = NULL;
  if (argc == 2) {
    long int tmp = strtol(argv[1], NULL, 10);
    max = (unsigned int)tmp;
  }
  p = primelist(max);
  print_primes(p, max);
  free(p);
  p = NULL;
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r12
  mov    $0x10000,%r12d
  push   %rbp
  sub    $0x8,%rsp
  cmp    $0x2,%edi
  je     L5
L7:
  mov    %r12d,%edi
  call   <primelist@plt>
  mov    %r12d,%esi
  mov    %rax,%rbp
  mov    %rax,%rdi
  call   <print_primes@plt>
  mov    %rbp,%rdi
  call   <free@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
L5:
  mov    0x8(%rsi),%rdi
  mov    $0xa,%edx
  xor    %esi,%esi
  call   <strtol@plt>
  mov    %eax,%r12d
  jmp    L7
```

assistant: ```c
int main(int argc, char *argv[]) {
  unsigned int max = (1 << 16);
  unsigned int *p = NULL;
  if (argc == 2) {
    long int tmp = strtol(argv[1], NULL, 10);
    max = (unsigned int)tmp;
  }
  p = primelist(max);
  print_primes(p, max);
  free(p);
  p = NULL;
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r12
  mov    $0x10000,%r12d
  push   %rbp
  sub    $0x8,%rsp
  cmp    $0x2,%edi
  je     L19
L5:
  mov    %r12d,%edi
  call   <primelist@plt>
  mov    %r12d,%esi
  mov    %rax,%rbp
  mov    %rax,%rdi
  call   <print_primes@plt>
  mov    %rbp,%rdi
  call   <free@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
L19:
  mov    0x8(%rsi),%rdi
  mov    $0xa,%edx
  xor    %esi,%esi
  call   <strtol@plt>
  mov    %eax,%r12d
  jmp    L5
```

assistant: ```c
int main(int argc, char *argv[]) {
  unsigned int max = (1 << 16);
  unsigned int *p = NULL;
  if (argc == 2) {
    long int tmp = strtol(argv[1], NULL, 10);
    max = (unsigned int)tmp;
  }
  p = primelist(max);
  print_primes(p, max);
  free(p);
  p = NULL;
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<bitstream_write_byte>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,%eax
  mov    %al,-0xc(%rbp)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rax
  movzbl -0xc(%rbp),%edx
  mov    %dl,(%rax)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rax
  lea    0x1(%rax),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void bitstream_write_byte(char **b, char value) {
  **b = value;
  *b += 1;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<bitstream_write_byte>:
  endbr64
  mov    (%rdi),%rax
  mov    %sil,(%rax)
  addq   $0x1,(%rdi)
  ret
```

assistant: ```c
void bitstream_write_byte(char **b, char value) {
  **b = value;
  *b += 1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<bitstream_write_byte>:
  endbr64
  mov    (%rdi),%rax
  mov    %sil,(%rax)
  addq   $0x1,(%rdi)
  ret
```

assistant: ```c
void bitstream_write_byte(char **b, char value) {
  **b = value;
  *b += 1;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<bitstream_write_byte>:
  endbr64
  mov    (%rdi),%rax
  mov    %sil,(%rax)
  addq   $0x1,(%rdi)
  ret
```

assistant: ```c
void bitstream_write_byte(char **b, char value) {
  **b = value;
  *b += 1;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<instructions>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
  pop    %rbp
  ret
```

tool_calls: name: retrieve_data
arguments: label: D1
datatype: string

tool: D1: Enter your choice:
   1 to add an item to the queue
   2 to remove an item from the queue
   3 to end


assistant: ```c
void instructions(void) {
  printf("Enter your choice:\n"
         "   1 to add an item to the queue\n"
         "   2 to remove an item from the queue\n"
         "   3 to end\n");
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<instructions>:
  endbr64
  sub    $0x8,%rsp
  lea    D3(%rip),%rdi
  call   <printf@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D3
type_guessed: string

tool: D3: Enter your choice:
   1 to add an item to the queue
   2 to remove an item from the queue
   3 to end


assistant: ```c
void instructions(void) {
  printf("Enter your choice:\n"
         "   1 to add an item to the queue\n"
         "   2 to remove an item from the queue\n"
         "   3 to end\n");
}
```
user: Translate the below assembly code into C programming language:

```asm
<instructions>:
  endbr64
  lea    D1(%rip),%rdi
  jmp    1050 <printf@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D1
type_guessed: string

tool: D1: Enter your choice:
   1 to add an item to the queue
   2 to remove an item from the queue
   3 to end


assistant: ```c
void instructions(void) {
  printf("Enter your choice:\n"
         "   1 to add an item to the queue\n"
         "   2 to remove an item from the queue\n"
         "   3 to end\n");
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<instructions>:
  endbr64
  lea    D4(%rip),%rdi
  jmp    1050 <printf@plt>
```

tool_calls: name: read_data
arguments: data_tag: D4
type_infered: string

tool: D4: Enter your choice:
   1 to add an item to the queue
   2 to remove an item from the queue
   3 to end


assistant: ```c
void instructions(void) {
  printf("Enter your choice:\n"
         "   1 to add an item to the queue\n"
         "   2 to remove an item from the queue\n"
         "   3 to end\n");
}
```
user: What is the c source code of the assembly code below:

```asm
<square>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <topDynArr@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <popDynArr@plt>
  mov    -0x8(%rbp),%rax
  mov    $0x2,%edi
  movq   %rax,%xmm0
  call   <pow@plt>
  mov    %eax,%edx
  mov    -0x18(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <pushDynArr@plt>
  nop
  leave
  ret
```

assistant: ```c
void square(struct DynArr *stack) {
  double i = topDynArr(stack);
  popDynArr(stack);
  pushDynArr(stack, pow(i, 2));
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<square>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  call   <topDynArr@plt>
  movq   %xmm0,%rbp
  mov    %rbx,%rdi
  call   <popDynArr@plt>
  mov    $0x2,%edi
  movq   %rbp,%xmm0
  call   <pow@plt>
  mov    %eax,%esi
  mov    %rbx,%rdi
  call   <pushDynArr@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void square(struct DynArr *stack) {
  double i = topDynArr(stack);
  popDynArr(stack);
  pushDynArr(stack, pow(i, 2));
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<square>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x10,%rsp
  call   <topDynArr@plt>
  mov    %rbp,%rdi
  movsd  %xmm0,0x8(%rsp)
  call   <popDynArr@plt>
  movsd  0x8(%rsp),%xmm0
  mov    $0x2,%edi
  call   <pow@plt>
  add    $0x10,%rsp
  mov    %rbp,%rdi
  mov    %eax,%esi
  pop    %rbp
  jmp    1080 <pushDynArr@plt>
```

assistant: ```c
void square(struct DynArr *stack) {
  double i = topDynArr(stack);
  popDynArr(stack);
  pushDynArr(stack, pow(i, 2));
}
```
user: What is the c source code of the assembly code below?

```asm
<square>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x10,%rsp
  call   <topDynArr@plt>
  mov    %rbp,%rdi
  movsd  %xmm0,0x8(%rsp)
  call   <popDynArr@plt>
  movsd  0x8(%rsp),%xmm0
  mov    $0x2,%edi
  call   <pow@plt>
  add    $0x10,%rsp
  mov    %rbp,%rdi
  mov    %eax,%esi
  pop    %rbp
  jmp    1080 <pushDynArr@plt>
```

assistant: ```c
void square(struct DynArr *stack) {
  double i = topDynArr(stack);
  popDynArr(stack);
  pushDynArr(stack, pow(i, 2));
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<copy>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  add    $0x1,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0xc(%rbp),%edx
  mov    -0x18(%rbp),%rcx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *copy(char *str) {
  int size = strlen(str) + 1;
  char *result = malloc(size);
  memcpy(result, str, size);
  return result;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<copy>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  call   <strlen@plt>
  lea    0x1(%rax),%r12d
  mov    %r12d,%edi
  call   <malloc@plt>
  mov    %rax,%rbx
  mov    %r12d,%edx
  mov    %rbp,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    %rbx,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *copy(char *str) {
  int size = strlen(str) + 1;
  char *result = malloc(size);
  memcpy(result, str, size);
  return result;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<copy>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  call   <strlen@plt>
  lea    0x1(%rax),%r13d
  mov    %r13d,%edi
  call   <malloc@plt>
  mov    %r13d,%edx
  mov    %rbp,%rsi
  mov    %rax,%r12
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *copy(char *str) {
  int size = strlen(str) + 1;
  char *result = malloc(size);
  memcpy(result, str, size);
  return result;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<copy>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  call   <strlen@plt>
  lea    0x1(%rax),%r13d
  mov    %r13d,%edi
  call   <malloc@plt>
  mov    %r13d,%edx
  mov    %rbp,%rsi
  mov    %rax,%r12
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *copy(char *str) {
  int size = strlen(str) + 1;
  char *result = malloc(size);
  memcpy(result, str, size);
  return 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
<isLegit>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x410,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  jmp    L0
L35:
  lea    -0x410(%rbp),%rax
  lea    D1(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strstr@plt>
  test   %rax,%rax
  je     L0
  mov    $0x0,%eax
  jmp    L49
L0:
  lea    -0x410(%rbp),%rax
  mov    $0x400,%edx
  mov    %rax,%rsi
  mov    $0x0,%edi
  call   <read@plt>
  test   %rax,%rax
  jg     L35
  mov    $0x1,%eax
L49:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L27
  call   <__stack_chk_fail@plt>
L27:
  leave
  ret
```

tool_calls: name: read_data
arguments: label: D1
data_type: string

tool: D1: SPAM

assistant: ```c
int isLegit() {
  char buf[1024];
  while (read(0, buf, 1024) > 0) {
    if (strstr(buf, "SPAM") != NULL) {
      return 0;
    }
  }
  return 1;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<isLegit>:
  endbr64
  push   %rbx
  sub    $0x410,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x408(%rsp)
  xor    %eax,%eax
  lea    D3(%rip),%rbx
L1:
  mov    %rsp,%rsi
  mov    $0x400,%edx
  mov    $0x0,%edi
  call   <read@plt>
  test   %rax,%rax
  jle    L2
  mov    %rsp,%rdi
  mov    %rbx,%rsi
  call   <strstr@plt>
  test   %rax,%rax
  je     L1
  mov    $0x0,%eax
  jmp    L20
L2:
  mov    $0x1,%eax
L20:
  mov    0x408(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L0
  add    $0x410,%rsp
  pop    %rbx
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D3
datatype: string

tool: D3: SPAM

assistant: ```c
int isLegit() {
  char buf[1024];
  while (read(0, buf, 1024) > 0) {
    if (strstr(buf, "SPAM") != NULL) {
      return 0;
    }
  }
  return 1;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<isLegit>:
  endbr64
  push   %rbp
  lea    D3(%rip),%rbp
  push   %rbx
  sub    $0x418,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x408(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  jmp    L42
  nopl   0x0(%rax)
L16:
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <strstr@plt>
  test   %rax,%rax
  jne    L34
L42:
  xor    %edi,%edi
  mov    $0x400,%edx
  mov    %rbx,%rsi
  call   <read@plt>
  test   %rax,%rax
  jg     L16
  mov    $0x1,%eax
L45:
  mov    0x408(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L17
  add    $0x418,%rsp
  pop    %rbx
  pop    %rbp
  ret
  cs nopw 0x0(%rax,%rax,1)
L34:
  xor    %eax,%eax
  jmp    L45
L17:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D3
data_type: string

tool: D3: SPAM

assistant: ```c
int isLegit() {
  char buf[1024];
  while (read(0, buf, 1024) > 0) {
    if (strstr(buf, "SPAM") != NULL) {
      return 0;
    }
  }
  return 1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<isLegit>:
  endbr64
  push   %rbp
  lea    D6(%rip),%rbp
  push   %rbx
  sub    $0x418,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x408(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  jmp    L13
  nopl   0x0(%rax)
L14:
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <strstr@plt>
  test   %rax,%rax
  jne    L15
L13:
  xor    %edi,%edi
  mov    $0x400,%edx
  mov    %rbx,%rsi
  call   <read@plt>
  test   %rax,%rax
  jg     L14
  mov    $0x1,%eax
L25:
  mov    0x408(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L45
  add    $0x418,%rsp
  pop    %rbx
  pop    %rbp
  ret
  cs nopw 0x0(%rax,%rax,1)
L15:
  xor    %eax,%eax
  jmp    L25
L45:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D6
data_type: string

tool: D6: SPAM

assistant: ```c
int isLegit() {
  char buf[1024];
  while (read(0, buf, 1024) > 0) {
    if (strstr(buf, "SPAM") != NULL) {
      return 0;
    }
  }
  return 1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<fchksum>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xb0,%rsp
  mov    %edi,-0xa4(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movw   $0x0,-0x9a(%rbp)
  movl   $0x0,-0x98(%rbp)
  mov    D2(%rip),%rax
  mov    (%rax),%edx
  mov    -0xa4(%rbp),%eax
  mov    $0x0,%esi
  mov    %eax,%edi
  call   <lseek@plt>
  jmp    L10
L12:
  movzwl -0x9a(%rbp),%edx
  mov    -0x94(%rbp),%ecx
  lea    -0x90(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <ipchksum@plt>
  mov    %ax,-0x9a(%rbp)
L10:
  lea    -0x90(%rbp),%rcx
  mov    -0xa4(%rbp),%eax
  mov    $0x80,%edx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <read@plt>
  mov    %eax,-0x94(%rbp)
  cmpl   $0x0,-0x94(%rbp)
  jg     L12
  movzwl -0x9a(%rbp),%eax
  not    %eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L8
  call   <__stack_chk_fail@plt>
L8:
  leave
  ret
```

assistant: ```c
unsigned short fchksum(int fd) {
  unsigned char buf[128];
  int byteRead;
  unsigned short resid = 0;
  int i = 0;
  lseek(fd, 0, SEEK_SET);
  while ((byteRead = read(fd, buf, sizeof(buf))) > 0) {
    resid = ipchksum(buf, byteRead, resid);
  }
  return ~resid;
}
```
user: Translate the below assembly code into C programming language:

```asm
<fchksum>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x90,%rsp
  mov    %edi,%r12d
  mov    %fs:0x28,%rax
  mov    %rax,0x88(%rsp)
  xor    %eax,%eax
  mov    D1(%rip),%rax
  mov    (%rax),%edx
  mov    $0x0,%esi
  call   <lseek@plt>
  mov    $0x0,%ebx
  mov    %rsp,%rbp
  jmp    L8
L6:
  movzwl %bx,%edx
  mov    %rbp,%rdi
  call   <ipchksum@plt>
  mov    %eax,%ebx
L8:
  mov    $0x80,%edx
  mov    %rbp,%rsi
  mov    %r12d,%edi
  call   <read@plt>
  mov    %eax,%esi
  test   %eax,%eax
  jg     L6
  mov    %ebx,%eax
  not    %eax
  mov    0x88(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L23
  add    $0x90,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L23:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
unsigned short fchksum(int fd) {
  unsigned char buf[128];
  int byteRead;
  unsigned short resid = 0;
  int i = 0;
  lseek(fd, 0, SEEK_SET);
  while ((byteRead = read(fd, buf, sizeof(buf))) > 0) {
    resid = ipchksum(buf, byteRead, resid);
  }
  return ~resid;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<fchksum>:
  endbr64
  push   %r12
  xor    %esi,%esi
  mov    %edi,%r12d
  push   %rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x90,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x88(%rsp)
  mov    D7(%rip),%rax
  mov    %rsp,%rbp
  mov    (%rax),%edx
  call   <lseek@plt>
  jmp    L11
  nopw   0x0(%rax,%rax,1)
L28:
  movzwl %bx,%edx
  mov    %rbp,%rdi
  call   <ipchksum@plt>
  mov    %eax,%ebx
L11:
  mov    %rbp,%rsi
  mov    $0x80,%edx
  mov    %r12d,%edi
  call   <read@plt>
  mov    %eax,%esi
  test   %eax,%eax
  jg     L28
  mov    %ebx,%eax
  not    %eax
  mov    0x88(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L15
  add    $0x90,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L15:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
unsigned short fchksum(int fd) {
  unsigned char buf[128];
  int byteRead;
  unsigned short resid = 0;
  int i = 0;
  lseek(fd, 0, SEEK_SET);
  while ((byteRead = read(fd, buf, sizeof(buf))) > 0) {
    resid = ipchksum(buf, byteRead, resid);
  }
  return ~resid;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<fchksum>:
  endbr64
  push   %r12
  xor    %esi,%esi
  mov    %edi,%r12d
  push   %rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x90,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x88(%rsp)
  mov    D3(%rip),%rax
  mov    %rsp,%rbp
  mov    (%rax),%edx
  call   <lseek@plt>
  jmp    L10
  nopw   0x0(%rax,%rax,1)
L21:
  movzwl %bx,%edx
  mov    %rbp,%rdi
  call   <ipchksum@plt>
  mov    %eax,%ebx
L10:
  mov    %rbp,%rsi
  mov    $0x80,%edx
  mov    %r12d,%edi
  call   <read@plt>
  mov    %eax,%esi
  test   %eax,%eax
  jg     L21
  mov    %ebx,%eax
  not    %eax
  mov    0x88(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L3
  add    $0x90,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
unsigned short fchksum(int fd) {
  unsigned char buf[128];
  int byteRead;
  unsigned short resid = 0;
  int i = 0;
  lseek(fd, 0, SEEK_SET);
  while ((byteRead = read(fd, buf, sizeof(buf))) > 0) {
    resid = ipchksum(buf, byteRead, resid);
  }
  return ~resid;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<colour_to_green>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  movl   $0x0,-0x4(%rbp)
  cmpl   $0x1,-0x14(%rbp)
  jg     L8
  cmpl   $0x0,-0x14(%rbp)
  jns    L18
  jmp    L37
L8:
  mov    -0x14(%rbp),%eax
  sub    $0x5,%eax
  cmp    $0x1,%eax
  ja     L37
L18:
  movl   $0xff,-0x4(%rbp)
  nop
L37:
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int colour_to_green(int colour) {
  int return_val = 0;
  switch (colour) {
  case 0:
  case 1:
  case 5:
  case 6:
    return_val = 255;
    break;
  }
  return return_val;
}
```
user: What is the c source code of the assembly code below?

```asm
<colour_to_green>:
  endbr64
  mov    %edi,%eax
  cmp    $0x1,%edi
  jle    L4
  sub    $0x5,%eax
  cmp    $0x2,%eax
  sbb    %eax,%eax
  movzbl %al,%eax
  ret
L4:
  sar    $0x1f,%eax
  not    %eax
  movzbl %al,%eax
  ret
```

assistant: ```c
int colour_to_green(int colour) {
  int return_val = 0;
  switch (colour) {
  case 0:
  case 1:
  case 5:
  case 6:
    return_val = 255;
    break;
  }
  return return_val;
}
```