ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0xf0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xe8(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%r12
  mov    %r12,%rsi
  lea    D18(%rip),%rbp
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    0xb(%rsp),%rsi
  lea    D2(%rip),%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  lea    0xc(%rsp),%rsi
  lea    D36(%rip),%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  lea    0x80(%rsp),%rbx
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    %r12,%rdi
  call   <length@plt>
  mov    %eax,%ebp
  mov    %rbx,%rdi
  call   <length@plt>
  mov    %eax,%ebx
  movsbl 0xb(%rsp),%edx
  mov    0xc(%rsp),%ecx
  mov    %ebp,%esi
  mov    %r12,%rdi
  call   <check_char@plt>
  mov    %eax,%r9d
  sub    %ebx,%ebp
  test   %ebp,%ebp
  jle    L103
  movzbl 0x80(%rsp),%edi
  mov    %r12,%rsi
  lea    -0x1(%rbp),%eax
  lea    0x11(%rsp,%rax,1),%r8
  mov    %r12,%rcx
  jmp    L47
L26:
  add    $0x1,%rcx
  cmp    %r8,%rcx
  je     L101
L47:
  cmp    %dil,(%rcx)
  jne    L26
  test   %ebx,%ebx
  jle    L26
  mov    $0x0,%eax
L60:
  mov    %eax,%edx
  add    $0x1,%eax
  cmp    %eax,%ebx
  jne    L60
  sub    %esi,%eax
  lea    -0x1(%rax,%rcx,1),%eax
  cltq
  movslq %edx,%rdx
  movzbl 0x80(%rsp,%rdx,1),%edx
  cmp    %dl,0x10(%rsp,%rax,1)
  jne    L26
  mov    $0x1,%eax
  jmp    L29
L101:
  mov    $0x0,%eax
L29:
  add    %r9d,%eax
  cmp    $0x1,%eax
  je     L61
  lea    D29(%rip),%rdi
  call   <printf@plt>
L96:
  mov    0xe8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L94
  mov    $0x0,%eax
  add    $0xf0,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L103:
  mov    $0x0,%eax
  jmp    L29
L61:
  lea    D11(%rip),%rdi
  call   <printf@plt>
  jmp    L96
L94:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D18
datatype: string
name: unpack_data
arguments: label: D2
datatype: string
name: unpack_data
arguments: label: D36
datatype: string
name: unpack_data
arguments: label: D29
datatype: string
name: unpack_data
arguments: label: D11
datatype: string

tool: D18: %s 

tool: D2: %c

tool: D36: %d

tool: D29: NO

tool: D11: YES

assistant: ```c
int main() {
  int n, i;
  char ch;
  char str[101];
  char sub[101];
  scanf("%s ", str);
  scanf("%c", &ch);
  scanf("%d", &n);
  scanf("%s ", sub);
  int l1 = length(str);
  int l2 = length(sub);
  int flag2 = 0;
  int flag1 = check_char(str, l1, ch, n);
  int j;
  for (i = 0; i < l1 - l2; i++) {
    if (str[i] == sub[0]) {
      for (j = 0; j < l2; j++) {
        if (str[i + j] == sub[j])
          flag2 = 1;
        else
          flag2 = 0;
      }
    }
    if (flag2 == 1)
      break;
  }
  if ((flag1 + flag2) == 1)
    printf("YES");
  else
    printf("NO");
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  lea    D31(%rip),%rbp
  push   %rbx
  mov    %rbp,%rdi
  sub    $0xf8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xe8(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rbx
  lea    0x80(%rsp),%r12
  mov    %rbx,%rsi
  call   <scanf@plt>
  lea    0xb(%rsp),%rsi
  lea    D37(%rip),%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  lea    0xc(%rsp),%rsi
  lea    D10(%rip),%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    %rbx,%rdi
  call   <length@plt>
  mov    %r12,%rdi
  mov    %eax,%ebp
  call   <length@plt>
  movsbl 0xb(%rsp),%edx
  mov    0xc(%rsp),%ecx
  mov    %ebp,%esi
  mov    %rbx,%rdi
  mov    %eax,%r13d
  call   <check_char@plt>
  mov    %ebp,%r8d
  sub    %r13d,%r8d
  test   %r8d,%r8d
  jle    L83
  movzbl 0x80(%rsp),%r11d
  mov    %rbx,%rcx
  xor    %edi,%edi
  movslq %r13d,%r10
  jmp    L41
  nopl   0x0(%rax)
L35:
  add    $0x1,%edi
  add    $0x1,%rcx
  cmp    %r8d,%edi
  je     L83
L41:
  movzbl (%rcx),%esi
  cmp    %r11b,%sil
  jne    L35
  test   %r13d,%r13d
  jle    L35
  mov    %r11d,%r9d
  xor    %edx,%edx
  jmp    L79
L67:
  movzbl (%rcx,%rdx,1),%esi
  movzbl (%r12,%rdx,1),%r9d
L79:
  add    $0x1,%rdx
  cmp    %rdx,%r10
  jne    L67
  cmp    %r9b,%sil
  jne    L35
  mov    $0x1,%edx
  jmp    L9
L83:
  xor    %edx,%edx
L9:
  add    %edx,%eax
  cmp    $0x1,%eax
  je     L25
  lea    D17(%rip),%rdi
  call   <printf@plt>
L11:
  mov    0xe8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L50
  add    $0xf8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L25:
  lea    D6(%rip),%rdi
  call   <printf@plt>
  jmp    L11
L50:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D31
type_infered: string
name: disassemble_data
arguments: data_label: D37
type_infered: string
name: disassemble_data
arguments: data_label: D10
type_infered: string
name: disassemble_data
arguments: data_label: D17
type_infered: string
name: disassemble_data
arguments: data_label: D6
type_infered: string

tool: D31: %s 

tool: D37: %c

tool: D10: %d

tool: D17: NO

tool: D6: YES

assistant: ```c
int main() {
  int n, i;
  char ch;
  char str[101];
  char sub[101];
  scanf("%s ", str);
  scanf("%c", &ch);
  scanf("%d", &n);
  scanf("%s ", sub);
  int l1 = length(str);
  int l2 = length(sub);
  int flag2 = 0;
  int flag1 = check_char(str, l1, ch, n);
  int j;
  for (i = 0; i < l1 - l2; i++) {
    if (str[i] == sub[0]) {
      for (j = 0; j < l2; j++) {
        if (str[i + j] == sub[j])
          flag2 = 1;
        else
          flag2 = 0;
      }
    }
    if (flag2 == 1)
      break;
  }
  if ((flag1 + flag2) == 1)
    printf("YES");
  else
    printf("NO");
  return 0;
}
```
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   %r12
  push   %rbp
  lea    D6(%rip),%rbp
  push   %rbx
  mov    %rbp,%rdi
  sub    $0xf0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xe8(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rbx
  lea    0x80(%rsp),%r12
  mov    %rbx,%rsi
  call   <scanf@plt>
  lea    0xb(%rsp),%rsi
  lea    D16(%rip),%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  lea    0xc(%rsp),%rsi
  lea    D49(%rip),%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    %rbx,%rdi
  call   <length@plt>
  mov    %r12,%rdi
  mov    %eax,%ebp
  call   <length@plt>
  movsbl 0xb(%rsp),%edx
  mov    0xc(%rsp),%ecx
  mov    %ebp,%esi
  mov    %eax,%r12d
  mov    %rbx,%rdi
  call   <check_char@plt>
  sub    %r12d,%ebp
  mov    %eax,%r9d
  test   %ebp,%ebp
  jle    L32
  mov    %r12d,%r8d
  lea    -0x1(%rbp),%eax
  lea    -0x1(%r12),%edi
  movzbl 0x80(%rsp),%edx
  sub    %ebx,%r8d
  lea    0x11(%rsp,%rax,1),%rcx
  movslq %edi,%rdi
  mov    %rbx,%rax
  sub    $0x1,%r8d
  jmp    L52
  nopl   (%rax)
L66:
  add    $0x1,%rax
  cmp    %rcx,%rax
  je     L32
L52:
  cmp    %dl,(%rax)
  jne    L66
  test   %r12d,%r12d
  jle    L66
  lea    (%r8,%rax,1),%esi
  movzbl 0x80(%rsp,%rdi,1),%ebx
  movslq %esi,%rsi
  cmp    %bl,0x10(%rsp,%rsi,1)
  jne    L66
  mov    $0x1,%eax
  jmp    L19
L32:
  xor    %eax,%eax
L19:
  add    %r9d,%eax
  cmp    $0x1,%eax
  je     L17
  lea    D9(%rip),%rdi
  call   <printf@plt>
L73:
  mov    0xe8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L51
  add    $0xf0,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L17:
  lea    D32(%rip),%rdi
  call   <printf@plt>
  jmp    L73
L51:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D6
type_infered: string
name: parse_data
arguments: tag: D16
type_infered: string
name: parse_data
arguments: tag: D49
type_infered: string
name: parse_data
arguments: tag: D9
type_infered: string
name: parse_data
arguments: tag: D32
type_infered: string

tool: D6: %s 

tool: D16: %c

tool: D49: %d

tool: D9: NO

tool: D32: YES

assistant: ```c
int main() {
  int n, i;
  char ch;
  char str[101];
  char sub[101];
  scanf("%s ", str);
  scanf("%c", &ch);
  scanf("%d", &n);
  scanf("%s ", sub);
  int l1 = length(str);
  int l2 = length(sub);
  int flag2 = 0;
  int flag1 = check_char(str, l1, ch, n);
  int j;
  for (i = 0; i < l1 - l2; i++) {
    if (str[i] == sub[0]) {
      for (j = 0; j < l2; j++) {
        if (str[i + j] == sub[j])
          flag2 = 1;
        else
          flag2 = 0;
      }
    }
    if (flag2 == 1)
      break;
  }
  if ((flag1 + flag2) == 1)
    printf("YES");
  else
    printf("NO");
  return 0;
}
```
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    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    $0x0,%eax
  call   <int_shifts_are_arithemetic@plt>
  mov    %eax,%edi
  call   <assert@plt>
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  assert(int_shifts_are_arithemetic());
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <int_shifts_are_arithemetic@plt>
  mov    %eax,%edi
  call   <assert@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  assert(int_shifts_are_arithemetic());
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

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

assistant: ```c
int main(int argc, char *argv[]) {
  assert(int_shifts_are_arithemetic());
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

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

assistant: ```c
int main(int argc, char *argv[]) {
  assert(int_shifts_are_arithemetic());
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<sarprintf_ret>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x48,%rsp
  mov    %rdi,-0xb8(%rbp)
  mov    %rsi,-0xa8(%rbp)
  mov    %rdx,-0xa0(%rbp)
  mov    %rcx,-0x98(%rbp)
  mov    %r8,-0x90(%rbp)
  mov    %r9,-0x88(%rbp)
  test   %al,%al
  je     L3
  movaps %xmm0,-0x80(%rbp)
  movaps %xmm1,-0x70(%rbp)
  movaps %xmm2,-0x60(%rbp)
  movaps %xmm3,-0x50(%rbp)
  movaps %xmm4,-0x40(%rbp)
  movaps %xmm5,-0x30(%rbp)
  movaps %xmm6,-0x20(%rbp)
  movaps %xmm7,-0x10(%rbp)
L3:
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int sarprintf_ret(const char *format, ...) {
  format = format;
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

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

assistant: ```c
int sarprintf_ret(const char *format, ...) {
  format = format;
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

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

assistant: ```c
int sarprintf_ret(const char *format, ...) {
  format = format;
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

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

assistant: ```c
int sarprintf_ret(const char *format, ...) {
  format = format;
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<RandomFunc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  sub    $0x3f60a4d,%eax
  mov    %eax,-0xc(%rbp)
  movl   $0x0,-0x10(%rbp)
  nop
  mov    -0xc(%rbp),%eax
  lea    0x282cec05(%rax),%edx
  mov    -0x20(%rbp),%rax
  mov    %edx,(%rax)
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L3
  call   <__stack_chk_fail@plt>
L3:
  leave
  ret
```

assistant: ```c
void RandomFunc(unsigned int input[1], unsigned int output[1]) {
  unsigned int state[1];
  unsigned int local2;
  unsigned int local1;
  unsigned short copy12;
  {
    state[0UL] = (input[0UL] + 914778474UL) - 981234615U;
    local1 = 0UL;
    while (local1 < 0U) {
      local2 = 0UL;
      while (local2 < 0U) {
        if (state[0UL] < local2 + local1) {
          state[local2] = state[0UL] + state[0UL];
        } else {
          copy12 = *((unsigned short *)(&state[local2]) + 1);
          *((unsigned short *)(&state[local2]) + 1) = *((unsigned short *)(&state[local2]) + 0);
          *((unsigned short *)(&state[local2]) + 0) = copy12;
          state[local2] = state[0UL] + state[0UL];
        }
        local2++;
      }
      local1++;
    }
    output[0UL] = state[0UL] + 674032645UL;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<RandomFunc>:
  endbr64
  mov    (%rdi),%eax
  add    $0x2436e1b8,%eax
  mov    %eax,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned int input[1], unsigned int output[1]) {
  unsigned int state[1];
  unsigned int local2;
  unsigned int local1;
  unsigned short copy12;
  {
    state[0UL] = (input[0UL] + 914778474UL) - 981234615U;
    local1 = 0UL;
    while (local1 < 0U) {
      local2 = 0UL;
      while (local2 < 0U) {
        if (state[0UL] < local2 + local1) {
          state[local2] = state[0UL] + state[0UL];
        } else {
          copy12 = *((unsigned short *)(&state[local2]) + 1);
          *((unsigned short *)(&state[local2]) + 1) = *((unsigned short *)(&state[local2]) + 0);
          *((unsigned short *)(&state[local2]) + 0) = copy12;
          state[local2] = state[0UL] + state[0UL];
        }
        local2++;
      }
      local1++;
    }
    output[0UL] = state[0UL] + 674032645UL;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<RandomFunc>:
  endbr64
  mov    (%rdi),%eax
  add    $0x2436e1b8,%eax
  mov    %eax,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned int input[1], unsigned int output[1]) {
  unsigned int state[1];
  unsigned int local2;
  unsigned int local1;
  unsigned short copy12;
  {
    state[0UL] = (input[0UL] + 914778474UL) - 981234615U;
    local1 = 0UL;
    while (local1 < 0U) {
      local2 = 0UL;
      while (local2 < 0U) {
        if (state[0UL] < local2 + local1) {
          state[local2] = state[0UL] + state[0UL];
        } else {
          copy12 = *((unsigned short *)(&state[local2]) + 1);
          *((unsigned short *)(&state[local2]) + 1) = *((unsigned short *)(&state[local2]) + 0);
          *((unsigned short *)(&state[local2]) + 0) = copy12;
          state[local2] = state[0UL] + state[0UL];
        }
        local2++;
      }
      local1++;
    }
    output[0UL] = state[0UL] + 674032645UL;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<RandomFunc>:
  endbr64
  mov    (%rdi),%eax
  add    $0x2436e1b8,%eax
  mov    %eax,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned int input[1], unsigned int output[1]) {
  unsigned int state[1];
  unsigned int local2;
  unsigned int local1;
  unsigned short copy12;
  {
    state[0UL] = (input[0UL] + 914778474UL) - 981234615U;
    local1 = 0UL;
    while (local1 < 0U) {
      local2 = 0UL;
      while (local2 < 0U) {
        if (state[0UL] < local2 + local1) {
          state[local2] = state[0UL] + state[0UL];
        } else {
          copy12 = *((unsigned short *)(&state[local2]) + 1);
          *((unsigned short *)(&state[local2]) + 1) = *((unsigned short *)(&state[local2]) + 0);
          *((unsigned short *)(&state[local2]) + 0) = copy12;
          state[local2] = state[0UL] + state[0UL];
        }
        local2++;
      }
      local1++;
    }
    output[0UL] = state[0UL] + 674032645UL;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<putint_uart>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xa0,%rsp
  mov    %edi,-0x94(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x94(%rbp),%edx
  lea    -0x90(%rbp),%rax
  lea    D7(%rip),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <sprintf@plt>
  lea    -0x90(%rbp),%rax
  mov    %rax,%rdi
  call   <putst_uart@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L6
  call   <__stack_chk_fail@plt>
L6:
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D7
type_guessed: string

tool: D7: %ld

assistant: ```c
void putint_uart(int32_t v) {
  uint8_t buf[32];
  sprintf((char *)buf, "%ld", v);
  putst_uart(buf);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<putint_uart>:
  endbr64
  push   %rbx
  sub    $0x90,%rsp
  mov    %edi,%edx
  mov    %fs:0x28,%rax
  mov    %rax,0x88(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  lea    D2(%rip),%rsi
  mov    %rbx,%rdi
  call   <sprintf@plt>
  mov    %rbx,%rdi
  call   <putst_uart@plt>
  mov    0x88(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x90,%rsp
  pop    %rbx
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D2
data_type: string

tool: D2: %ld

assistant: ```c
void putint_uart(int32_t v) {
  uint8_t buf[32];
  sprintf((char *)buf, "%ld", v);
  putst_uart(buf);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<putint_uart>:
  endbr64
  push   %rbp
  mov    %edi,%edx
  lea    D7(%rip),%rsi
  sub    $0x90,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x88(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%rdi
  call   <sprintf@plt>
  mov    %rbp,%rdi
  call   <putst_uart@plt>
  mov    0x88(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0x90,%rsp
  pop    %rbp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D7
type_guessed: string

tool: D7: %ld

assistant: ```c
void putint_uart(int32_t v) {
  uint8_t buf[32];
  sprintf((char *)buf, "%ld", v);
  putst_uart(buf);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<putint_uart>:
  endbr64
  push   %rbp
  mov    %edi,%edx
  lea    D1(%rip),%rsi
  sub    $0x90,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x88(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%rdi
  call   <sprintf@plt>
  mov    %rbp,%rdi
  call   <putst_uart@plt>
  mov    0x88(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x90,%rsp
  pop    %rbp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

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

tool: D1: %ld

assistant: ```c
void putint_uart(int32_t v) {
  uint8_t buf[32];
  sprintf((char *)buf, "%ld", v);
  putst_uart(buf);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<QEIIntEnable>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  cmpl   $0x4002c000,-0x4(%rbp)
  je     L26
  cmpl   $0x4002d000,-0x4(%rbp)
  jne    L21
L26:
  mov    $0x1,%eax
  jmp    L25
L21:
  mov    $0x0,%eax
L25:
  mov    %eax,%edi
  call   <ASSERT@plt>
  mov    -0x4(%rbp),%eax
  add    $0x20,%eax
  cltq
  mov    (%rax),%eax
  mov    -0x4(%rbp),%edx
  add    $0x20,%edx
  movslq %edx,%rdx
  or     -0x8(%rbp),%eax
  mov    %eax,(%rdx)
  nop
  leave
  ret
```

assistant: ```c
void QEIIntEnable(uint32_t ui32Base, uint32_t ui32IntFlags) {
  ASSERT((ui32Base == 0x4002C000) || (ui32Base == 0x4002D000));
  (*((volatile uint32_t *)(ui32Base + 0x00000020))) |= ui32IntFlags;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<QEIIntEnable>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebx
  mov    %esi,%ebp
  mov    %edi,%eax
  and    $0xef,%ah
  cmp    $0x4002c000,%eax
  sete   %dil
  movzbl %dil,%edi
  call   <ASSERT@plt>
  add    $0x20,%ebx
  movslq %ebx,%rbx
  mov    (%rbx),%eax
  or     %ebp,%eax
  mov    %eax,(%rbx)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void QEIIntEnable(uint32_t ui32Base, uint32_t ui32IntFlags) {
  ASSERT((ui32Base == 0x4002C000) || (ui32Base == 0x4002D000));
  (*((volatile uint32_t *)(ui32Base + 0x00000020))) |= ui32IntFlags;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<QEIIntEnable>:
  endbr64
  push   %rbp
  mov    %edi,%eax
  mov    %esi,%ebp
  push   %rbx
  and    $0xef,%ah
  mov    %edi,%ebx
  xor    %edi,%edi
  sub    $0x8,%rsp
  cmp    $0x4002c000,%eax
  sete   %dil
  add    $0x20,%ebx
  call   <ASSERT@plt>
  movslq %ebx,%rbx
  mov    (%rbx),%eax
  or     %ebp,%eax
  mov    %eax,(%rbx)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void QEIIntEnable(uint32_t ui32Base, uint32_t ui32IntFlags) {
  ASSERT((ui32Base == 0x4002C000) || (ui32Base == 0x4002D000));
  (*((volatile uint32_t *)(ui32Base + 0x00000020))) |= ui32IntFlags;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<QEIIntEnable>:
  endbr64
  push   %rbp
  mov    %edi,%eax
  mov    %esi,%ebp
  push   %rbx
  and    $0xef,%ah
  mov    %edi,%ebx
  xor    %edi,%edi
  sub    $0x8,%rsp
  cmp    $0x4002c000,%eax
  sete   %dil
  add    $0x20,%ebx
  call   <ASSERT@plt>
  movslq %ebx,%rbx
  mov    (%rbx),%eax
  or     %ebp,%eax
  mov    %eax,(%rbx)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void QEIIntEnable(uint32_t ui32Base, uint32_t ui32IntFlags) {
  ASSERT((ui32Base == 0x4002C000) || (ui32Base == 0x4002D000));
  (*((volatile uint32_t *)(ui32Base + 0x00000020))) |= ui32IntFlags;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<strspt>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  mov    $0x1,%edi
  call   <malloc@plt>
  mov    %rax,-0x10(%rbp)
  mov    $0x1,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0x20(%rbp)
  movl   $0x0,-0x1c(%rbp)
  movl   $0x0,-0x18(%rbp)
  jmp    L38
L1:
  mov    -0x14(%rbp),%eax
  cmp    -0x2c(%rbp),%eax
  jne    L30
  mov    -0x1c(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x1c(%rbp)
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x10(%rbp),%rax
  add    %rax,%rdx
  mov    -0x8(%rbp),%rax
  mov    %rax,(%rdx)
  mov    -0x20(%rbp),%edx
  mov    -0x8(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  movl   $0x0,-0x20(%rbp)
  jmp    L15
L30:
  mov    -0x20(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x20(%rbp)
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    -0x14(%rbp),%edx
  mov    %dl,(%rax)
L15:
  addl   $0x1,-0x18(%rbp)
L38:
  mov    -0x18(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,-0x14(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  jne    L1
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x10(%rbp),%rax
  add    %rax,%rdx
  mov    -0x8(%rbp),%rax
  mov    %rax,(%rdx)
  mov    -0x10(%rbp),%rax
  leave
  ret
```

assistant: ```c
char **strspt(char *s, int c) {
  char **list = malloc(sizeof(char));
  char *buffer = malloc(sizeof(char));
  int buffer_i = 0, list_i = 0;
  int i, read;
  for (i = 0; (read = *(s + i)) != '\0'; i++) {
    if (read == c) {
      list[list_i++] = buffer;
      memset(buffer, 0, buffer_i * sizeof(char));
      buffer_i = 0;
    } else
      *(buffer + buffer_i++) = read;
  }
  list[list_i] = buffer;
  return list;
}
```