ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 这段汇编代码用C语言表示是什么样的?

```asm
<evalChunk>:
  endbr64
  push   %rbx
  sub    $0xd0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xc8(%rsp)
  xor    %eax,%eax
  cmp    %edx,%esi
  jg     L19
  mov    %edx,%ebx
  mov    %rsp,%r8
  sub    %esi,%ebx
  movslq %esi,%rsi
  mov    %ebx,%edx
  add    %rdi,%rsi
  mov    %r8,%rdi
  add    $0x1,%rdx
  call   <memcpy@plt>
  lea    0x1(%rbx),%edx
  mov    %rax,%r8
L12:
  movslq %edx,%rdx
  mov    %r8,%rdi
  movb   $0x0,(%rsp,%rdx,1)
  call   <eval@plt>
  mov    0xc8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0xd0,%rsp
  pop    %rbx
  ret
  nop
L19:
  xor    %edx,%edx
  mov    %rsp,%r8
  jmp    L12
L0:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
double evalChunk(char str[200], int start, int end) {
  char exprstr[200];
  int i = start, j = 0;
  for (i = i; i <= end; i++) {
    exprstr[j] = str[i];
    j++;
  }
  exprstr[j] = '\0';
  return eval(exprstr);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<evalChunk>:
  endbr64
  push   %rbx
  sub    $0xd0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xc8(%rsp)
  xor    %eax,%eax
  cmp    %edx,%esi
  jg     L3
  mov    %edx,%ebx
  mov    %rsp,%r8
  sub    %esi,%ebx
  movslq %esi,%rsi
  mov    %ebx,%edx
  add    %rdi,%rsi
  mov    %r8,%rdi
  add    $0x1,%rdx
  call   <memcpy@plt>
  lea    0x1(%rbx),%edx
  mov    %rax,%r8
L12:
  movslq %edx,%rdx
  mov    %r8,%rdi
  movb   $0x0,(%rsp,%rdx,1)
  call   <eval@plt>
  mov    0xc8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L24
  add    $0xd0,%rsp
  pop    %rbx
  ret
  nop
L3:
  xor    %edx,%edx
  mov    %rsp,%r8
  jmp    L12
L24:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
double evalChunk(char str[200], int start, int end) {
  char exprstr[200];
  int i = start, j = 0;
  for (i = i; i <= end; i++) {
    exprstr[j] = str[i];
    j++;
  }
  exprstr[j] = '\0';
  return eval(exprstr);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<bwb_CSAVE>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <bwx_DEBUG@plt>
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <bwb_SAVE@plt>
  cltq
  leave
  ret
```

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

tool: D3: bwb_CSAVE

assistant: ```c
struct bwb_line *bwb_CSAVE(struct bwb_line *l) {
  bwx_DEBUG(__FUNCTION__);
  return bwb_SAVE(l);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<bwb_CSAVE>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  lea    D6(%rip),%rdi
  mov    $0x0,%eax
  call   <bwx_DEBUG@plt>
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <bwb_SAVE@plt>
  cltq
  pop    %rbx
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D6
type: string

tool: D6: bwb_CSAVE

assistant: ```c
struct bwb_line *bwb_CSAVE(struct bwb_line *l) {
  bwx_DEBUG(__FUNCTION__);
  return bwb_SAVE(l);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<bwb_CSAVE>:
  endbr64
  push   %rbp
  xor    %eax,%eax
  mov    %rdi,%rbp
  lea    D3(%rip),%rdi
  call   <bwx_DEBUG@plt>
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <bwb_SAVE@plt>
  pop    %rbp
  cltq
  ret
```

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

tool: D3: bwb_CSAVE

assistant: ```c
struct bwb_line *bwb_CSAVE(struct bwb_line *l) {
  bwx_DEBUG(__FUNCTION__);
  return bwb_SAVE(l);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<bwb_CSAVE>:
  endbr64
  push   %rbp
  xor    %eax,%eax
  mov    %rdi,%rbp
  lea    D5(%rip),%rdi
  call   <bwx_DEBUG@plt>
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <bwb_SAVE@plt>
  pop    %rbp
  cltq
  ret
```

tool_calls: name: unpack_data
arguments: tag: D5
type: string

tool: D5: bwb_CSAVE

assistant: ```c
struct bwb_line *bwb_CSAVE(struct bwb_line *l) {
  bwx_DEBUG(__FUNCTION__);
  return bwb_SAVE(l);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x0,-0x4(%rbp)
  sub    $0x10,%rsp
  movl   $0x2,-0x8(%rbp)
  movl   $0x4,-0xc(%rbp)
  mov    -0x8(%rbp),%eax
  mov    -0xc(%rbp),%edx
  add    %eax,%edx
  mov    %edx,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D0
datatype: string

tool: D0: Assembly added value of z is %d


assistant: ```c
int main() {
  int z = 0;
  __asm__("subq $16, %rsp\n\t"
          "movl $2, -8(%rbp)\n\t"
          "movl $4, -12(%rbp)\n\t"
          "movl -8(%rbp), %eax\n\t"
          "movl -12(%rbp), %edx\n\t"
          "addl %eax, %edx\n\t"
          "movl %edx, -4(%rbp)\n\t");
  printf("Assembly added value of z is %d\n", z);
  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
  sub    $0x10,%rsp
  movl   $0x2,-0x8(%rbp)
  movl   $0x4,-0xc(%rbp)
  mov    -0x8(%rbp),%eax
  mov    -0xc(%rbp),%edx
  add    %eax,%edx
  mov    %edx,-0x4(%rbp)
  mov    $0x0,%esi
  lea    D9(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D9
type_guessed: string

tool: D9: Assembly added value of z is %d


assistant: ```c
int main() {
  int z = 0;
  __asm__("subq $16, %rsp\n\t"
          "movl $2, -8(%rbp)\n\t"
          "movl $4, -12(%rbp)\n\t"
          "movl -8(%rbp), %eax\n\t"
          "movl -12(%rbp), %edx\n\t"
          "addl %eax, %edx\n\t"
          "movl %edx, -4(%rbp)\n\t");
  printf("Assembly added value of z is %d\n", z);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  sub    $0x10,%rsp
  movl   $0x2,-0x8(%rbp)
  movl   $0x4,-0xc(%rbp)
  mov    -0x8(%rbp),%eax
  mov    -0xc(%rbp),%edx
  add    %eax,%edx
  mov    %edx,-0x4(%rbp)
  xor    %esi,%esi
  lea    D1(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D1
type_infered: string

tool: D1: Assembly added value of z is %d


assistant: ```c
int main() {
  int z = 0;
  __asm__("subq $16, %rsp\n\t"
          "movl $2, -8(%rbp)\n\t"
          "movl $4, -12(%rbp)\n\t"
          "movl -8(%rbp), %eax\n\t"
          "movl -12(%rbp), %edx\n\t"
          "addl %eax, %edx\n\t"
          "movl %edx, -4(%rbp)\n\t");
  printf("Assembly added value of z is %d\n", z);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  sub    $0x10,%rsp
  movl   $0x2,-0x8(%rbp)
  movl   $0x4,-0xc(%rbp)
  mov    -0x8(%rbp),%eax
  mov    -0xc(%rbp),%edx
  add    %eax,%edx
  mov    %edx,-0x4(%rbp)
  xor    %esi,%esi
  lea    D9(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D9
data_type: string

tool: D9: Assembly added value of z is %d


assistant: ```c
int main() {
  int z = 0;
  __asm__("subq $16, %rsp\n\t"
          "movl $2, -8(%rbp)\n\t"
          "movl $4, -12(%rbp)\n\t"
          "movl -8(%rbp), %eax\n\t"
          "movl -12(%rbp), %edx\n\t"
          "addl %eax, %edx\n\t"
          "movl %edx, -4(%rbp)\n\t");
  printf("Assembly added value of z is %d\n", z);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<Decrypt>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x19,-0x4(%rbp)
  jmp    L3
L32:
  mov    -0x4(%rbp),%eax
  shl    $0x3,%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rax,%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <Iround_function@plt>
  subl   $0x1,-0x4(%rbp)
L3:
  cmpl   $0x0,-0x4(%rbp)
  jne    L32
  movl   $0x0,-0x4(%rbp)
  jmp    L25
L0:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,%edi
  call   <READ_ROUND_KEY_BYTE@plt>
  mov    %eax,%edx
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%ecx
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rsi
  mov    -0x18(%rbp),%rax
  add    %rsi,%rax
  xor    %ecx,%edx
  mov    %edx,(%rax)
  addl   $0x1,-0x4(%rbp)
L25:
  cmpl   $0x7,-0x4(%rbp)
  jle    L0
  nop
  nop
  leave
  ret
```

assistant: ```c
void Decrypt(uint8_t *block, uint8_t *roundKeys) {
  uint8_t i;
  for (i = 25; i != 0; i--)
    Iround_function((uint16_t *)block, (uint16_t *)(roundKeys + 8 * i));
  for (i = 0; i < 8; i++)
    block[i] ^= READ_ROUND_KEY_BYTE(roundKeys[i]);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<Decrypt>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    %rsi,%r12
  lea    0x320(%rsi),%rbx
L6:
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <Iround_function@plt>
  sub    $0x20,%rbx
  cmp    %r12,%rbx
  jne    L6
  mov    $0x0,%ebx
L7:
  mov    (%r12,%rbx,1),%edi
  call   <READ_ROUND_KEY_BYTE@plt>
  xor    %eax,0x0(%rbp,%rbx,1)
  add    $0x4,%rbx
  cmp    $0x20,%rbx
  jne    L7
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void Decrypt(uint8_t *block, uint8_t *roundKeys) {
  uint8_t i;
  for (i = 25; i != 0; i--)
    Iround_function((uint16_t *)block, (uint16_t *)(roundKeys + 8 * i));
  for (i = 0; i < 8; i++)
    block[i] ^= READ_ROUND_KEY_BYTE(roundKeys[i]);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<Decrypt>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  lea    0x320(%rsi),%rbx
  nopl   (%rax)
L15:
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  sub    $0x20,%rbx
  call   <Iround_function@plt>
  cmp    %rbx,%r12
  jne    L15
  xor    %ebx,%ebx
  xchg   %ax,%ax
L10:
  mov    (%r12,%rbx,1),%edi
  call   <READ_ROUND_KEY_BYTE@plt>
  xor    %eax,0x0(%rbp,%rbx,1)
  add    $0x4,%rbx
  cmp    $0x20,%rbx
  jne    L10
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void Decrypt(uint8_t *block, uint8_t *roundKeys) {
  uint8_t i;
  for (i = 25; i != 0; i--)
    Iround_function((uint16_t *)block, (uint16_t *)(roundKeys + 8 * i));
  for (i = 0; i < 8; i++)
    block[i] ^= READ_ROUND_KEY_BYTE(roundKeys[i]);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<Decrypt>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  lea    0x320(%rsi),%rbx
  nopl   (%rax)
L9:
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  sub    $0x20,%rbx
  call   <Iround_function@plt>
  cmp    %rbx,%r12
  jne    L9
  xor    %ebx,%ebx
  xchg   %ax,%ax
L13:
  mov    (%r12,%rbx,1),%edi
  call   <READ_ROUND_KEY_BYTE@plt>
  xor    %eax,0x0(%rbp,%rbx,1)
  add    $0x4,%rbx
  cmp    $0x20,%rbx
  jne    L13
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void Decrypt(uint8_t *block, uint8_t *roundKeys) {
  uint8_t i;
  for (i = 25; i != 0; i--)
    Iround_function((uint16_t *)block, (uint16_t *)(roundKeys + 8 * i));
  for (i = 0; i < 8; i++)
    block[i] ^= READ_ROUND_KEY_BYTE(roundKeys[i]);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<CalculateCounts>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    $0x50,%edi
  mov    $0x0,%eax
  call   <mymalloc@plt>
  cltq
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L0
L37:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movl   $0x0,(%rax)
  addl   $0x1,-0xc(%rbp)
L0:
  cmpl   $0x13,-0xc(%rbp)
  jle    L37
  movl   $0x1,-0xc(%rbp)
  jmp    L35
L15:
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  lea    0x0(,%rax,4),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  add    $0x1,%edx
  mov    %edx,(%rax)
  addl   $0x1,-0xc(%rbp)
L35:
  mov    -0xc(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jle    L15
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
int *CalculateCounts(sequence, lsequence)
unsigned char *sequence;
int lsequence;
{
  int i;
  int *counts = mymalloc((long)sizeof(int) * 20);
  for (i = 0; i < 20; i++)
    counts[i] = 0;
  for (i = 1; i <= lsequence; i++)
    counts[sequence[i]]++;
  return counts;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<CalculateCounts>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    %esi,%ebx
  mov    $0x50,%edi
  mov    $0x0,%eax
  call   <mymalloc@plt>
  cltq
  mov    %rax,%rdx
  lea    0x50(%rax),%rcx
L1:
  movl   $0x0,(%rdx)
  add    $0x4,%rdx
  cmp    %rcx,%rdx
  jne    L1
  test   %ebx,%ebx
  jle    L6
  lea    0x1(%rbp),%rdx
  lea    -0x1(%rbx),%ecx
  lea    0x2(%rbp,%rcx,1),%rsi
L25:
  movzbl (%rdx),%ecx
  addl   $0x1,(%rax,%rcx,4)
  add    $0x1,%rdx
  cmp    %rsi,%rdx
  jne    L25
L6:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int *CalculateCounts(sequence, lsequence)
unsigned char *sequence;
int lsequence;
{
  int i;
  int *counts = mymalloc((long)sizeof(int) * 20);
  for (i = 0; i < 20; i++)
    counts[i] = 0;
  for (i = 1; i <= lsequence; i++)
    counts[sequence[i]]++;
  return counts;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<CalculateCounts>:
  endbr64
  push   %rbp
  xor    %eax,%eax
  mov    %rdi,%rbp
  mov    $0x50,%edi
  push   %rbx
  mov    %esi,%ebx
  sub    $0x8,%rsp
  call   <mymalloc@plt>
  pxor   %xmm0,%xmm0
  cltq
  movups %xmm0,(%rax)
  movups %xmm0,0x10(%rax)
  movups %xmm0,0x20(%rax)
  movups %xmm0,0x30(%rax)
  movups %xmm0,0x40(%rax)
  test   %ebx,%ebx
  jle    L3
  lea    -0x1(%rbx),%ecx
  lea    0x1(%rbp),%rdx
  lea    0x2(%rbp,%rcx,1),%rsi
  nopl   0x0(%rax)
L11:
  movzbl (%rdx),%ecx
  add    $0x1,%rdx
  addl   $0x1,(%rax,%rcx,4)
  cmp    %rsi,%rdx
  jne    L11
L3:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int *CalculateCounts(sequence, lsequence)
unsigned char *sequence;
int lsequence;
{
  int i;
  int *counts = mymalloc((long)sizeof(int) * 20);
  for (i = 0; i < 20; i++)
    counts[i] = 0;
  for (i = 1; i <= lsequence; i++)
    counts[sequence[i]]++;
  return counts;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<CalculateCounts>:
  endbr64
  push   %rbp
  xor    %eax,%eax
  mov    %rdi,%rbp
  mov    $0x50,%edi
  push   %rbx
  mov    %esi,%ebx
  sub    $0x8,%rsp
  call   <mymalloc@plt>
  pxor   %xmm0,%xmm0
  cltq
  movups %xmm0,(%rax)
  movups %xmm0,0x10(%rax)
  movups %xmm0,0x20(%rax)
  movups %xmm0,0x30(%rax)
  movups %xmm0,0x40(%rax)
  test   %ebx,%ebx
  jle    L6
  lea    -0x1(%rbx),%ecx
  lea    0x1(%rbp),%rdx
  lea    0x2(%rbp,%rcx,1),%rsi
  nopl   0x0(%rax)
L19:
  movzbl (%rdx),%ecx
  add    $0x1,%rdx
  addl   $0x1,(%rax,%rcx,4)
  cmp    %rsi,%rdx
  jne    L19
L6:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int *CalculateCounts(sequence, lsequence)
unsigned char *sequence;
int lsequence;
{
  int i;
  int *counts = mymalloc((long)sizeof(int) * 20);
  for (i = 0; i < 20; i++)
    counts[i] = 0;
  for (i = 1; i <= lsequence; i++)
    counts[sequence[i]]++;
  return counts;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<HexCharToNibble>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,%eax
  mov    %al,-0x4(%rbp)
  cmpb   $0x2f,-0x4(%rbp)
  jle    L51
  cmpb   $0x39,-0x4(%rbp)
  jg     L51
  movsbl -0x4(%rbp),%eax
  sub    $0x30,%eax
  jmp    L49
L51:
  cmpb   $0x40,-0x4(%rbp)
  jle    L30
  cmpb   $0x46,-0x4(%rbp)
  jg     L30
  movsbl -0x4(%rbp),%eax
  sub    $0x37,%eax
  jmp    L49
L30:
  mov    $0xffffffff,%eax
L49:
  pop    %rbp
  ret
```

assistant: ```c
int HexCharToNibble(char hex_char) {
  if (hex_char >= '0' && hex_char <= '9') {
    return hex_char - '0';
  } else if (hex_char >= 'A' && hex_char <= 'F') {
    return hex_char + 10 - 'A';
  } else {
    return -1;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<HexCharToNibble>:
  endbr64
  lea    -0x30(%rdi),%eax
  cmp    $0x9,%al
  jbe    L8
  lea    -0x41(%rdi),%eax
  cmp    $0x5,%al
  ja     L10
  movsbl %dil,%edi
  lea    -0x37(%rdi),%eax
  ret
L8:
  movsbl %dil,%edi
  lea    -0x30(%rdi),%eax
  ret
L10:
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int HexCharToNibble(char hex_char) {
  if (hex_char >= '0' && hex_char <= '9') {
    return hex_char - '0';
  } else if (hex_char >= 'A' && hex_char <= 'F') {
    return hex_char + 10 - 'A';
  } else {
    return -1;
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<HexCharToNibble>:
  endbr64
  lea    -0x30(%rdi),%eax
  cmp    $0x9,%al
  jbe    L4
  lea    -0x41(%rdi),%eax
  cmp    $0x5,%al
  ja     L9
  sub    $0x37,%edi
  movsbl %dil,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L4:
  movsbl %al,%eax
  ret
  nopl   0x0(%rax)
L9:
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int HexCharToNibble(char hex_char) {
  if (hex_char >= '0' && hex_char <= '9') {
    return hex_char - '0';
  } else if (hex_char >= 'A' && hex_char <= 'F') {
    return hex_char + 10 - 'A';
  } else {
    return -1;
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<HexCharToNibble>:
  endbr64
  lea    -0x30(%rdi),%eax
  cmp    $0x9,%al
  jbe    L14
  lea    -0x41(%rdi),%eax
  cmp    $0x5,%al
  ja     L15
  sub    $0x37,%edi
  movsbl %dil,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L14:
  movsbl %al,%eax
  ret
  nopl   0x0(%rax)
L15:
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int HexCharToNibble(char hex_char) {
  if (hex_char >= '0' && hex_char <= '9') {
    return hex_char - '0';
  } else if (hex_char >= 'A' && hex_char <= 'F') {
    return hex_char + 10 - 'A';
  } else {
    return -1;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<is_tree_valid>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    %ecx,-0x3c(%rbp)
  mov    %r8d,-0x40(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L5
L70:
  mov    -0xc(%rbp),%eax
  lea    0x0(,%rax,4),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x14(%rbp)
  jmp    L68
L66:
  mov    -0x14(%rbp),%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x10(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jne    L68
  mov    $0xffffffff,%eax
  jmp    L76
L68:
  mov    D3(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,-0x14(%rbp)
  jl     L66
  mov    -0xc(%rbp),%eax
  lea    0x0(,%rax,4),%rdx
  mov    -0x38(%rbp),%rax
  add    %rax,%rdx
  mov    -0x10(%rbp),%eax
  mov    %eax,(%rdx)
  addl   $0x1,-0xc(%rbp)
L5:
  mov    -0xc(%rbp),%eax
  cmp    -0x40(%rbp),%eax
  jb     L70
  movl   $0x0,-0x8(%rbp)
  movl   $0x1,-0x4(%rbp)
  jmp    L41
L9:
  mov    -0x38(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x4(%rbp),%eax
  lea    0x0(,%rax,4),%rcx
  mov    -0x38(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%eax
  cmp    %eax,%edx
  setne  %al
  movzbl %al,%eax
  add    %eax,-0x8(%rbp)
  addl   $0x1,-0x4(%rbp)
L41:
  mov    -0x4(%rbp),%eax
  cmp    -0x40(%rbp),%eax
  jb     L9
  cmpl   $0x0,-0x8(%rbp)
  sete   %al
  movzbl %al,%eax
L76:
  pop    %rbp
  ret
```

assistant: ```c
int is_tree_valid(unsigned int *predecessor, unsigned int *terminal, unsigned int *source, const unsigned int n_nodes, const unsigned int n_term) {
  int next;
  int current;
  for (unsigned int i = 0; i < n_term; i++) {
    current = terminal[i];
    next = predecessor[current];
    while (INT_MAX > next) {
      current = next;
      next = predecessor[current];
      if (current == next)
        return -1;
    }
    source[i] = current;
  }
  unsigned int sum = 0;
  for (unsigned int i = 1; i < n_term; i++) {
    sum += (source[0] != source[i]);
  }
  return (0 == sum);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<is_tree_valid>:
  endbr64
  test   %r8d,%r8d
  je     L80
  push   %rbp
  push   %rbx
  mov    %rdi,%rcx
  mov    %rsi,%r11
  mov    %rdx,%rbx
  mov    %r8d,%ebp
  mov    $0x0,%r9d
  mov    D6(%rip),%r10
L36:
  mov    (%r11,%r9,4),%esi
  movslq %esi,%rax
  mov    (%rcx,%rax,4),%eax
  mov    (%r10),%edi
L46:
  cmp    %eax,%edi
  jle    L57
  movslq %eax,%rdx
  mov    (%rcx,%rdx,4),%edx
  mov    %eax,%esi
  cmp    %edx,%eax
  je     L21
  mov    %edx,%eax
  jmp    L46
L57:
  mov    %esi,(%rbx,%r9,4)
  add    $0x1,%r9
  cmp    %rbp,%r9
  jne    L36
  cmp    $0x1,%r8d
  jbe    L16
  mov    (%rbx),%esi
  lea    0x4(%rbx),%rax
  lea    -0x2(%r8),%edx
  lea    0x8(%rbx,%rdx,4),%rcx
  mov    $0x0,%r8d
L2:
  cmp    %esi,(%rax)
  setne  %dl
  movzbl %dl,%edx
  add    %edx,%r8d
  add    $0x4,%rax
  cmp    %rcx,%rax
  jne    L2
L9:
  test   %r8d,%r8d
  sete   %al
  movzbl %al,%eax
  jmp    L28
L16:
  mov    $0x0,%r8d
  jmp    L9
L21:
  mov    $0xffffffff,%eax
L28:
  pop    %rbx
  pop    %rbp
  ret
L80:
  test   %r8d,%r8d
  sete   %al
  movzbl %al,%eax
  ret
```

assistant: ```c
int is_tree_valid(unsigned int *predecessor, unsigned int *terminal, unsigned int *source, const unsigned int n_nodes, const unsigned int n_term) {
  int next;
  int current;
  for (unsigned int i = 0; i < n_term; i++) {
    current = terminal[i];
    next = predecessor[current];
    while (INT_MAX > next) {
      current = next;
      next = predecessor[current];
      if (current == next)
        return -1;
    }
    source[i] = current;
  }
  unsigned int sum = 0;
  for (unsigned int i = 1; i < n_term; i++) {
    sum += (source[0] != source[i]);
  }
  return (0 == sum);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<is_tree_valid>:
  endbr64
  test   %r8d,%r8d
  je     L15
  push   %rbp
  mov    %rsi,%r10
  mov    %rdx,%r11
  mov    %r8d,%ebp
  push   %rbx
  mov    D4(%rip),%rbx
  xor    %r9d,%r9d
L51:
  movslq (%r10,%r9,4),%rax
  mov    (%rbx),%esi
  mov    %rax,%rcx
  mov    (%rdi,%rax,4),%eax
  jmp    L64
L81:
  movslq %eax,%rdx
  mov    %eax,%ecx
  mov    (%rdi,%rdx,4),%edx
  cmp    %edx,%eax
  je     L59
  mov    %edx,%eax
L64:
  cmp    %eax,%esi
  jg     L81
  mov    %ecx,(%r11,%r9,4)
  add    $0x1,%r9
  cmp    %r9,%rbp
  jne    L51
  cmp    $0x1,%r8d
  je     L85
  lea    -0x2(%r8),%edx
  mov    (%r11),%edi
  lea    0x4(%r11),%rax
  lea    0x8(%r11,%rdx,4),%rsi
  xor    %edx,%edx
  nopw   0x0(%rax,%rax,1)
L82:
  xor    %ecx,%ecx
  cmp    %edi,(%rax)
  setne  %cl
  add    $0x4,%rax
  add    %ecx,%edx
  cmp    %rax,%rsi
  jne    L82
  xor    %eax,%eax
  test   %edx,%edx
  sete   %al
  jmp    L34
  nopl   0x0(%rax,%rax,1)
L59:
  mov    $0xffffffff,%eax
L34:
  pop    %rbx
  pop    %rbp
  ret
L85:
  mov    $0x1,%eax
  jmp    L34
L15:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int is_tree_valid(unsigned int *predecessor, unsigned int *terminal, unsigned int *source, const unsigned int n_nodes, const unsigned int n_term) {
  int next;
  int current;
  for (unsigned int i = 0; i < n_term; i++) {
    current = terminal[i];
    next = predecessor[current];
    while (INT_MAX > next) {
      current = next;
      next = predecessor[current];
      if (current == next)
        return -1;
    }
    source[i] = current;
  }
  unsigned int sum = 0;
  for (unsigned int i = 1; i < n_term; i++) {
    sum += (source[0] != source[i]);
  }
  return (0 == sum);
}
```