ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将以下的汇编代码转换成C语言代码:

```asm
<test11>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <DifferingObjectSize0@plt>
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <DifferingObjectSize1@plt>
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <DifferingObjectSize2@plt>
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <DifferingObjectSize3@plt>
  nop
  leave
  ret
```

assistant: ```c
void test11(void *const p __attribute__((pass_object_size(2)))) {
  DifferingObjectSize0(p);
  DifferingObjectSize1(p);
  DifferingObjectSize2(p);
  DifferingObjectSize3(p);
}
```
user: What is the c source code of the assembly code below?

```asm
<test11>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <DifferingObjectSize0@plt>
  mov    %rbx,%rdi
  call   <DifferingObjectSize1@plt>
  mov    %rbx,%rdi
  call   <DifferingObjectSize2@plt>
  mov    %rbx,%rdi
  call   <DifferingObjectSize3@plt>
  pop    %rbx
  ret
```

assistant: ```c
void test11(void *const p __attribute__((pass_object_size(2)))) {
  DifferingObjectSize0(p);
  DifferingObjectSize1(p);
  DifferingObjectSize2(p);
  DifferingObjectSize3(p);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<test11>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  call   <DifferingObjectSize0@plt>
  mov    %rbp,%rdi
  call   <DifferingObjectSize1@plt>
  mov    %rbp,%rdi
  call   <DifferingObjectSize2@plt>
  mov    %rbp,%rdi
  pop    %rbp
  jmp    10a0 <DifferingObjectSize3@plt>
```

assistant: ```c
void test11(void *const p __attribute__((pass_object_size(2)))) {
  DifferingObjectSize0(p);
  DifferingObjectSize1(p);
  DifferingObjectSize2(p);
  DifferingObjectSize3(p);
}
```
user: Translate the below assembly code into C programming language:

```asm
<test11>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  call   <DifferingObjectSize0@plt>
  mov    %rbp,%rdi
  call   <DifferingObjectSize1@plt>
  mov    %rbp,%rdi
  call   <DifferingObjectSize2@plt>
  mov    %rbp,%rdi
  pop    %rbp
  jmp    10a0 <DifferingObjectSize3@plt>
```

assistant: ```c
void test11(void *const p __attribute__((pass_object_size(2)))) {
  DifferingObjectSize0(p);
  DifferingObjectSize1(p);
  DifferingObjectSize2(p);
  DifferingObjectSize3(p);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x10(%rbp)
  movl   $0x0,-0xc(%rbp)
  lea    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  jmp    L20
L17:
  mov    -0x18(%rbp),%eax
  mov    %eax,-0x14(%rbp)
  movl   $0x0,-0x10(%rbp)
  jmp    L58
L48:
  mov    -0x14(%rbp),%ecx
  movslq %ecx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %ecx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  sub    %eax,%ecx
  mov    %ecx,%edx
  cmp    %edx,-0xc(%rbp)
  jne    L26
  addl   $0x1,-0x10(%rbp)
L26:
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x14(%rbp)
L58:
  cmpl   $0x0,-0x14(%rbp)
  jg     L48
  cmpl   $0x1,-0x10(%rbp)
  jle    L54
  mov    -0x10(%rbp),%edx
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L54:
  addl   $0x1,-0xc(%rbp)
L20:
  cmpl   $0x9,-0xc(%rbp)
  jle    L17
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L28
  call   <__stack_chk_fail@plt>
L28:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D7
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D6
type_infered: string

tool: D7: %d

tool: D6: %d is met exactly %d times


assistant: ```c
int main() {
  int num, tmp, cnt = 0, i = 0;
  scanf("%d", &num);
  while (i < 10) {
    tmp = num;
    cnt = 0;
    while (tmp > 0) {
      if (i == tmp % 10)
        cnt++;
      tmp /= 10;
    }
    if (cnt >= 2) {
      printf("%d is met exactly %d times\n", i, cnt);
    }
    i++;
  }
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  lea    D3(%rip),%rdi
  call   <scanf@plt>
  mov    $0x0,%ebx
  lea    D7(%rip),%rbp
  jmp    L71
L50:
  mov    %eax,%edx
L1:
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  mov    %edx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  lea    (%rax,%rax,4),%eax
  add    %eax,%eax
  mov    %edx,%esi
  sub    %eax,%esi
  cmp    %ebx,%esi
  sete   %al
  movzbl %al,%eax
  add    %eax,%ecx
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  mov    %edx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  cmp    $0x9,%edx
  jg     L50
  cmp    $0x1,%ecx
  jg     L46
L66:
  add    $0x1,%ebx
  cmp    $0xa,%ebx
  je     L14
L71:
  mov    0x4(%rsp),%edx
  test   %edx,%edx
  jle    L66
  mov    $0x0,%ecx
  jmp    L1
L46:
  mov    %ecx,%edx
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  jmp    L66
L14:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L61
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L61:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D3
type: string
name: disassemble_data
arguments: label: D7
type: string

tool: D3: %d

tool: D7: %d is met exactly %d times


assistant: ```c
int main() {
  int num, tmp, cnt = 0, i = 0;
  scanf("%d", &num);
  while (i < 10) {
    tmp = num;
    cnt = 0;
    while (tmp > 0) {
      if (i == tmp % 10)
        cnt++;
      tmp /= 10;
    }
    if (cnt >= 2) {
      printf("%d is met exactly %d times\n", i, cnt);
    }
    i++;
  }
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %r12
  lea    D8(%rip),%rdi
  xor    %r12d,%r12d
  push   %rbp
  mov    $0xcccccccd,%ebp
  push   %rbx
  lea    D16(%rip),%rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  nopl   0x0(%rax)
L58:
  mov    0x4(%rsp),%edx
  test   %edx,%edx
  jle    L40
  xor    %r8d,%r8d
  jmp    L6
  nopl   (%rax)
L46:
  mov    %eax,%edx
L6:
  mov    %edx,%eax
  mov    %edx,%esi
  imul   %rbp,%rax
  shr    $0x23,%rax
  lea    (%rax,%rax,4),%ecx
  add    %ecx,%ecx
  sub    %ecx,%esi
  xor    %ecx,%ecx
  cmp    %r12d,%esi
  sete   %cl
  add    %ecx,%r8d
  cmp    $0x9,%edx
  jg     L46
  cmp    $0x1,%r8d
  jle    L40
  mov    %r8d,%edx
  mov    %r12d,%esi
  mov    %rbx,%rdi
  call   <printf@plt>
L40:
  add    $0x1,%r12d
  cmp    $0xa,%r12d
  jne    L58
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L29
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L29:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D8
type: string
name: unpack_data
arguments: data_label: D16
type: string

tool: D8: %d

tool: D16: %d is met exactly %d times


assistant: ```c
int main() {
  int num, tmp, cnt = 0, i = 0;
  scanf("%d", &num);
  while (i < 10) {
    tmp = num;
    cnt = 0;
    while (tmp > 0) {
      if (i == tmp % 10)
        cnt++;
      tmp /= 10;
    }
    if (cnt >= 2) {
      printf("%d is met exactly %d times\n", i, cnt);
    }
    i++;
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r12
  lea    D17(%rip),%rdi
  xor    %r12d,%r12d
  push   %rbp
  mov    $0xcccccccd,%ebp
  push   %rbx
  lea    D15(%rip),%rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  nopl   0x0(%rax)
L12:
  mov    0x4(%rsp),%edx
  test   %edx,%edx
  jle    L55
  xor    %r8d,%r8d
  jmp    L44
  nopl   (%rax)
L0:
  mov    %eax,%edx
L44:
  mov    %edx,%eax
  mov    %edx,%esi
  imul   %rbp,%rax
  shr    $0x23,%rax
  lea    (%rax,%rax,4),%ecx
  add    %ecx,%ecx
  sub    %ecx,%esi
  xor    %ecx,%ecx
  cmp    %esi,%r12d
  sete   %cl
  add    %ecx,%r8d
  cmp    $0x9,%edx
  jg     L0
  cmp    $0x1,%r8d
  jle    L29
  mov    %r8d,%edx
  mov    %r12d,%esi
  mov    %rbx,%rdi
  call   <printf@plt>
L29:
  add    $0x1,%r12d
  cmp    $0xa,%r12d
  jne    L12
  nopl   0x0(%rax,%rax,1)
L55:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D17
type: string
name: unpack_data
arguments: tag: D15
type: string

tool: D17: %d

tool: D15: %d is met exactly %d times


assistant: ```c
int main() {
  int num, tmp, cnt = 0, i = 0;
  scanf("%d", &num);
  while (i < 10) {
    tmp = num;
    cnt = 0;
    while (tmp > 0) {
      if (i == tmp % 10)
        cnt++;
      tmp /= 10;
    }
    if (cnt >= 2) {
      printf("%d is met exactly %d times\n", i, cnt);
    }
    i++;
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<ecc_gen>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %rcx,-0x30(%rbp)
  mov    %r8,-0x38(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0x0,-0x10(%rbp)
  jmp    L14
L1:
  lea    -0x10(%rbp),%rcx
  mov    -0x30(%rbp),%rdx
  mov    -0x28(%rbp),%rsi
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <ecc_gen_stripe@plt>
  mov    -0x28(%rbp),%rax
  add    %rax,-0x18(%rbp)
  mov    -0x10(%rbp),%rax
  add    %rax,-0x30(%rbp)
  subq   $0x1,-0x20(%rbp)
L14:
  cmpq   $0x0,-0x20(%rbp)
  jne    L1
  cmpq   $0x0,-0x38(%rbp)
  je     L3
  mov    -0x10(%rbp),%rdx
  mov    -0x38(%rbp),%rax
  mov    %rdx,(%rax)
L3:
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L27
  call   <__stack_chk_fail@plt>
L27:
  leave
  ret
```

assistant: ```c
void ecc_gen(const void *data, size_t count, size_t size, void *ecc_buf, size_t *ecc_sizep) {
  size_t ecc_sz;
  ecc_sz = 0;
  while (count > 0) {
    (void)ecc_gen_stripe(data, size, ecc_buf, &ecc_sz);
    data = (void *)((char *)data + size);
    ecc_buf = (void *)((char *)ecc_buf + ecc_sz);
    --count;
  }
  if (ecc_sizep != NULL) {
    *ecc_sizep = ecc_sz;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<ecc_gen>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %r8,%r15
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movq   $0x0,(%rsp)
  test   %rsi,%rsi
  je     L39
  mov    %rdi,%rbp
  mov    %rsi,%rbx
  mov    %rdx,%r13
  mov    %rcx,%r12
  mov    %rsp,%r14
L10:
  mov    %r14,%rcx
  mov    %r12,%rdx
  mov    %r13,%rsi
  mov    %rbp,%rdi
  call   <ecc_gen_stripe@plt>
  add    %r13,%rbp
  add    (%rsp),%r12
  sub    $0x1,%rbx
  jne    L10
L39:
  test   %r15,%r15
  je     L30
  mov    (%rsp),%rax
  mov    %rax,(%r15)
L30:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L21
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L21:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void ecc_gen(const void *data, size_t count, size_t size, void *ecc_buf, size_t *ecc_sizep) {
  size_t ecc_sz;
  ecc_sz = 0;
  while (count > 0) {
    (void)ecc_gen_stripe(data, size, ecc_buf, &ecc_sz);
    data = (void *)((char *)data + size);
    ecc_buf = (void *)((char *)ecc_buf + ecc_sz);
    --count;
  }
  if (ecc_sizep != NULL) {
    *ecc_sizep = ecc_sz;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<ecc_gen>:
  endbr64
  push   %r15
  mov    %r8,%r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movq   $0x0,(%rsp)
  test   %rsi,%rsi
  je     L15
  mov    %rdi,%rbp
  mov    %rsi,%rbx
  mov    %rdx,%r13
  mov    %rcx,%r12
  mov    %rsp,%r14
  nopl   0x0(%rax)
L20:
  mov    %r12,%rdx
  mov    %rbp,%rdi
  mov    %r14,%rcx
  mov    %r13,%rsi
  call   <ecc_gen_stripe@plt>
  add    %r13,%rbp
  add    (%rsp),%r12
  sub    $0x1,%rbx
  jne    L20
L15:
  test   %r15,%r15
  je     L17
  mov    (%rsp),%rax
  mov    %rax,(%r15)
L17:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void ecc_gen(const void *data, size_t count, size_t size, void *ecc_buf, size_t *ecc_sizep) {
  size_t ecc_sz;
  ecc_sz = 0;
  while (count > 0) {
    (void)ecc_gen_stripe(data, size, ecc_buf, &ecc_sz);
    data = (void *)((char *)data + size);
    ecc_buf = (void *)((char *)ecc_buf + ecc_sz);
    --count;
  }
  if (ecc_sizep != NULL) {
    *ecc_sizep = ecc_sz;
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<ecc_gen>:
  endbr64
  push   %r15
  mov    %r8,%r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movq   $0x0,(%rsp)
  test   %rsi,%rsi
  je     L18
  mov    %rdi,%rbp
  mov    %rsi,%rbx
  mov    %rdx,%r13
  mov    %rcx,%r12
  mov    %rsp,%r14
  nopl   0x0(%rax)
L42:
  mov    %r12,%rdx
  mov    %rbp,%rdi
  mov    %r14,%rcx
  mov    %r13,%rsi
  call   <ecc_gen_stripe@plt>
  mov    (%rsp),%rax
  add    %r13,%rbp
  add    %rax,%r12
  sub    $0x1,%rbx
  jne    L42
L29:
  test   %r15,%r15
  je     L23
  mov    %rax,(%r15)
L23:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L45
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L18:
  xor    %eax,%eax
  jmp    L29
L45:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void ecc_gen(const void *data, size_t count, size_t size, void *ecc_buf, size_t *ecc_sizep) {
  size_t ecc_sz;
  ecc_sz = 0;
  while (count > 0) {
    (void)ecc_gen_stripe(data, size, ecc_buf, &ecc_sz);
    data = (void *)((char *)data + size);
    ecc_buf = (void *)((char *)ecc_buf + ecc_sz);
    --count;
  }
  if (ecc_sizep != NULL) {
    *ecc_sizep = ecc_sz;
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<rec2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jg     L5
  mov    $0x0,%eax
  jmp    L7
L5:
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <rec1@plt>
  sub    $0x1,%eax
  mov    %eax,%edi
  call   <rec2>
  sub    $0x1,%eax
L7:
  leave
  ret
```

assistant: ```c
int rec2(int j) {
  if (j <= 0)
    return 0;
  return rec2(rec1(j) - 1) - 1;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<rec2>:
  endbr64
  mov    $0x0,%eax
  test   %edi,%edi
  jle    L1
  sub    $0x8,%rsp
  call   <rec1@plt>
  lea    -0x1(%rax),%edi
  call   <rec2>
  sub    $0x1,%eax
  add    $0x8,%rsp
  ret
L1:
  ret
```

assistant: ```c
int rec2(int j) {
  if (j <= 0)
    return 0;
  return rec2(rec1(j) - 1) - 1;
}
```
user: What is the c source code of the assembly code below:

```asm
<rec2>:
  endbr64
  test   %edi,%edi
  jle    L4
  push   %rbx
  mov    $0xffffffff,%ebx
  xchg   %ax,%ax
L2:
  call   <rec1@plt>
  lea    -0x1(%rax),%edi
  mov    %ebx,%eax
  sub    $0x1,%ebx
  test   %edi,%edi
  jg     L2
  pop    %rbx
  ret
  nopl   0x0(%rax,%rax,1)
L4:
  xor    %eax,%eax
  ret
```

assistant: ```c
int rec2(int j) {
  if (j <= 0)
    return 0;
  return rec2(rec1(j) - 1) - 1;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<rec2>:
  endbr64
  test   %edi,%edi
  jle    L16
  push   %rbx
  mov    $0xffffffff,%ebx
  xchg   %ax,%ax
L3:
  call   <rec1@plt>
  lea    -0x1(%rax),%edi
  mov    %ebx,%eax
  sub    $0x1,%ebx
  test   %edi,%edi
  jg     L3
  pop    %rbx
  ret
  nopl   0x0(%rax,%rax,1)
L16:
  xor    %eax,%eax
  ret
```

assistant: ```c
int rec2(int j) {
  if (j <= 0)
    return 0;
  return rec2(rec1(j) - 1) - 1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<postorderTraversal>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    $0xfa0,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x20(%rbp),%rax
  movl   $0x0,(%rax)
  mov    -0x20(%rbp),%rdx
  mov    -0x8(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <function@plt>
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
int *postorderTraversal(struct TreeNode *root, int *returnSize) {
  int *result = (int *)malloc(sizeof(int) * 1000);
  *returnSize = 0;
  function(root, result, returnSize);
  return result;
}
```
user: What is the c source code of the assembly code below?

```asm
<postorderTraversal>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    %rsi,%rbp
  mov    $0xfa0,%edi
  call   <malloc@plt>
  mov    %rax,%rbx
  movl   $0x0,0x0(%rbp)
  mov    %rbp,%rdx
  mov    %rax,%rsi
  mov    %r12,%rdi
  call   <function@plt>
  mov    %rbx,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int *postorderTraversal(struct TreeNode *root, int *returnSize) {
  int *result = (int *)malloc(sizeof(int) * 1000);
  *returnSize = 0;
  function(root, result, returnSize);
  return result;
}
```
user: What is the c source code of the assembly code below:

```asm
<postorderTraversal>:
  endbr64
  push   %r13
  mov    %rsi,%r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  mov    $0xfa0,%edi
  call   <malloc@plt>
  movl   $0x0,0x0(%r13)
  mov    %r13,%rdx
  mov    %rbp,%rdi
  mov    %rax,%r12
  mov    %rax,%rsi
  call   <function@plt>
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int *postorderTraversal(struct TreeNode *root, int *returnSize) {
  int *result = (int *)malloc(sizeof(int) * 1000);
  *returnSize = 0;
  function(root, result, returnSize);
  return result;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<postorderTraversal>:
  endbr64
  push   %r13
  mov    %rsi,%r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  mov    $0xfa0,%edi
  call   <malloc@plt>
  movl   $0x0,0x0(%r13)
  mov    %r13,%rdx
  mov    %rbp,%rdi
  mov    %rax,%r12
  mov    %rax,%rsi
  call   <function@plt>
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int *postorderTraversal(struct TreeNode *root, int *returnSize) {
  int *result = (int *)malloc(sizeof(int) * 1000);
  *returnSize = 0;
  function(root, result, returnSize);
  return result;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<_dl_get_last_path_component>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r12
  push   %rbx
  mov    %rdi,-0x28(%rbp)
  mov    -0x28(%rbp),%rbx
  mov    $0x0,%r12d
  jmp    L12
L79:
  add    $0x1,%rbx
L12:
  movzbl (%rbx),%eax
  test   %al,%al
  jne    L79
  lea    -0x1(%rbx),%rax
  mov    %rax,-0x18(%rbp)
  jmp    L39
L31:
  mov    -0x18(%rbp),%rax
  lea    -0x1(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  movb   $0x0,(%rax)
L39:
  mov    -0x18(%rbp),%rax
  cmp    -0x28(%rbp),%rax
  je     L69
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x2f,%al
  je     L31
L69:
  mov    -0x28(%rbp),%rbx
  jmp    L43
L90:
  movzbl (%rbx),%eax
  cmp    $0x2f,%al
  jne    L86
  mov    %rbx,%r12
L86:
  add    $0x1,%rbx
L43:
  movzbl (%rbx),%eax
  test   %al,%al
  jne    L90
  mov    %r12,-0x18(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  je     L71
  mov    -0x18(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L9
L71:
  mov    -0x28(%rbp),%rax
  jmp    L60
L9:
  mov    -0x18(%rbp),%rax
  add    $0x1,%rax
L60:
  pop    %rbx
  pop    %r12
  pop    %rbp
  ret
```

assistant: ```c
char *_dl_get_last_path_component(char *path) {
  char *s;
  register char *ptr = path;
  register char *prev = 0;
  while (*ptr)
    ptr++;
  s = ptr - 1;
  while (s != path && *s == '/') {
    *s-- = '\0';
  }
  ptr = path;
  while (*ptr != '\0') {
    if (*ptr == '/')
      prev = ptr;
    ptr++;
  }
  s = prev;
  if (s == ((void *)0) || s[1] == '\0')
    return path;
  else
    return s + 1;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<_dl_get_last_path_component>:
  endbr64
  mov    %rdi,%rax
  cmpb   $0x0,(%rdi)
  je     L2
  mov    %rdi,%rdx
L97:
  add    $0x1,%rdx
  cmpb   $0x0,(%rdx)
  jne    L97
L63:
  sub    $0x1,%rdx
  cmp    %rdx,%rax
  je     L39
L82:
  cmpb   $0x2f,(%rdx)
  jne    L49
  sub    $0x1,%rdx
  movb   $0x0,0x1(%rdx)
  cmp    %rdx,%rax
  jne    L82
  jmp    L39
L2:
  mov    %rdi,%rdx
  jmp    L63
L15:
  ret
L49:
  mov    %rax,%rdx
L39:
  movzbl (%rdx),%ecx
  mov    $0x0,%esi
  test   %cl,%cl
  je     L15
L58:
  cmp    $0x2f,%cl
  cmove  %rdx,%rsi
  add    $0x1,%rdx
  movzbl (%rdx),%ecx
  test   %cl,%cl
  jne    L58
  test   %rsi,%rsi
  je     L40
  cmpb   $0x0,0x1(%rsi)
  je     L40
  lea    0x1(%rsi),%rax
L40:
  ret
```

assistant: ```c
char *_dl_get_last_path_component(char *path) {
  char *s;
  register char *ptr = path;
  register char *prev = 0;
  while (*ptr)
    ptr++;
  s = ptr - 1;
  while (s != path && *s == '/') {
    *s-- = '\0';
  }
  ptr = path;
  while (*ptr != '\0') {
    if (*ptr == '/')
      prev = ptr;
    ptr++;
  }
  s = prev;
  if (s == ((void *)0) || s[1] == '\0')
    return path;
  else
    return s + 1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<_dl_get_last_path_component>:
  endbr64
  cmpb   $0x0,(%rdi)
  je     L25
  mov    %rdi,%rdx
  nopl   0x0(%rax)
L6:
  mov    %rdx,%rax
  add    $0x1,%rdx
  cmpb   $0x0,(%rdx)
  jne    L6
L87:
  movzbl -0x1(%rdx),%edx
  cmp    %rdi,%rax
  jne    L51
  jmp    L119
  nopw   0x0(%rax,%rax,1)
L103:
  sub    $0x1,%rax
  movb   $0x0,0x1(%rax)
  cmp    %rax,%rdi
  je     L109
L51:
  cmpb   $0x2f,(%rax)
  je     L103
  mov    %rdi,%rax
L109:
  movzbl (%rax),%edx
  test   %dl,%dl
  je     L1
L42:
  xor    %ecx,%ecx
  xchg   %ax,%ax
L4:
  cmp    $0x2f,%dl
  cmove  %rax,%rcx
  movzbl 0x1(%rax),%edx
  add    $0x1,%rax
  test   %dl,%dl
  jne    L4
  test   %rcx,%rcx
  je     L1
  cmpb   $0x0,0x1(%rcx)
  jne    L27
L1:
  mov    %rdi,%rax
  ret
  nopw   0x0(%rax,%rax,1)
L27:
  lea    0x1(%rcx),%rax
  ret
  nopl   (%rax)
L25:
  lea    -0x1(%rdi),%rax
  mov    %rdi,%rdx
  jmp    L87
L119:
  test   %dl,%dl
  jne    L42
  ret
```

assistant: ```c
char *_dl_get_last_path_component(char *path) {
  char *s;
  register char *ptr = path;
  register char *prev = 0;
  while (*ptr)
    ptr++;
  s = ptr - 1;
  while (s != path && *s == '/') {
    *s-- = '\0';
  }
  ptr = path;
  while (*ptr != '\0') {
    if (*ptr == '/')
      prev = ptr;
    ptr++;
  }
  s = prev;
  if (s == ((void *)0) || s[1] == '\0')
    return path;
  else
    return s + 1;
}
```