ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What is the c source code of the assembly code below?

```asm
<select_sort_noninc>:
  endbr64
  test   %esi,%esi
  jle    L47
  xor    %r10d,%r10d
  push   %rbp
  lea    0x4(%rdi),%r11
  push   %rbx
  lea    0x1(%r10),%ebx
  mov    -0x4(%r11),%ebp
  cmp    %ebx,%esi
  je     L17
  nopl   (%rax)
L37:
  mov    %r11,%rax
  mov    %ebp,%r8d
  mov    %ebx,%edx
  jmp    L13
  nopw   0x0(%rax,%rax,1)
L31:
  movslq %r10d,%rcx
  add    $0x1,%edx
  add    $0x4,%rax
  lea    (%rdi,%rcx,4),%r9
  cmp    %edx,%esi
  je     L34
L13:
  mov    (%rax),%ecx
  mov    %rax,%r9
  cmp    %r8d,%ecx
  jle    L31
  mov    %edx,%r10d
  add    $0x1,%edx
  mov    %ecx,%r8d
  add    $0x4,%rax
  cmp    %edx,%esi
  jne    L13
L34:
  mov    %ebx,%r10d
  mov    %r8d,-0x4(%r11)
  add    $0x4,%r11
  lea    0x1(%r10),%ebx
  mov    %ebp,(%r9)
  mov    -0x4(%r11),%ebp
  cmp    %ebx,%esi
  jne    L37
L17:
  pop    %rbx
  pop    %rbp
  ret
L47:
  ret
```

assistant: ```c
void select_sort_noninc(int A[], int n) {
  int i, j, maxIdx, temp;
  for (i = 0; i < n; i++) {
    maxIdx = i;
    for (j = i + 1; j < n; j++) {
      if (A[maxIdx] < A[j])
        maxIdx = j;
    }
    temp = A[i];
    A[i] = A[maxIdx];
    A[maxIdx] = temp;
  }
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<select_sort_noninc>:
  endbr64
  test   %esi,%esi
  jle    L24
  push   %rbp
  mov    $0x1,%r11d
  mov    (%rdi),%ebp
  push   %rbx
  lea    0x4(%rdi),%rbx
  cmp    $0x1,%esi
  je     L10
  nopl   0x0(%rax,%rax,1)
L43:
  lea    -0x1(%r11),%r10d
  mov    %rbx,%rax
  mov    %ebp,%r8d
  mov    %r11d,%edx
  jmp    L49
  nop
L53:
  movslq %r10d,%rcx
  add    $0x1,%edx
  add    $0x4,%rax
  lea    (%rdi,%rcx,4),%r9
  cmp    %edx,%esi
  jle    L57
L49:
  mov    (%rax),%ecx
  mov    %rax,%r9
  cmp    %ecx,%r8d
  jge    L53
  mov    %edx,%r10d
  add    $0x1,%edx
  mov    %ecx,%r8d
  add    $0x4,%rax
  cmp    %edx,%esi
  jg     L49
L57:
  mov    %r8d,-0x4(%rbx)
  add    $0x1,%r11d
  add    $0x4,%rbx
  mov    %ebp,(%r9)
  mov    -0x4(%rbx),%ebp
  cmp    %r11d,%esi
  jne    L43
L10:
  pop    %rbx
  pop    %rbp
  ret
L24:
  ret
```

assistant: ```c
void select_sort_noninc(int A[], int n) {
  int i, j, maxIdx, temp;
  for (i = 0; i < n; i++) {
    maxIdx = i;
    for (j = i + 1; j < n; j++) {
      if (A[maxIdx] < A[j])
        maxIdx = j;
    }
    temp = A[i];
    A[i] = A[maxIdx];
    A[maxIdx] = temp;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x0,%eax
  call   <__VERIFIER_nondet_int@plt>
  mov    %eax,-0x8(%rbp)
  mov    $0x0,%eax
  call   <__VERIFIER_nondet_int@plt>
  mov    %eax,-0x4(%rbp)
  jmp    L18
L3:
  mov    -0x4(%rbp),%eax
  add    %eax,-0x8(%rbp)
  mov    -0x4(%rbp),%edx
  mov    $0x0,%eax
  sub    %edx,%eax
  add    %eax,%eax
  mov    %eax,-0x4(%rbp)
L18:
  cmpl   $0x0,-0x8(%rbp)
  jg     L3
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main() {
  int x, y;
  x = __VERIFIER_nondet_int();
  y = __VERIFIER_nondet_int();
  while (x > 0) {
    x = x + y;
    y = -2 * y;
  }
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  push   %rbx
  mov    $0x0,%eax
  call   <__VERIFIER_nondet_int@plt>
  mov    %eax,%ebx
  mov    $0x0,%eax
  call   <__VERIFIER_nondet_int@plt>
  test   %ebx,%ebx
  jle    L8
L2:
  add    %eax,%ebx
  neg    %eax
  add    %eax,%eax
  test   %ebx,%ebx
  jg     L2
L8:
  mov    $0x0,%eax
  pop    %rbx
  ret
```

assistant: ```c
int main() {
  int x, y;
  x = __VERIFIER_nondet_int();
  y = __VERIFIER_nondet_int();
  while (x > 0) {
    x = x + y;
    y = -2 * y;
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  push   %rbx
  xor    %eax,%eax
  call   <__VERIFIER_nondet_int@plt>
  mov    %eax,%ebx
  xor    %eax,%eax
  call   <__VERIFIER_nondet_int@plt>
  test   %ebx,%ebx
  jle    L2
  nopl   0x0(%rax)
L5:
  add    %eax,%ebx
  neg    %eax
  add    %eax,%eax
  test   %ebx,%ebx
  jg     L5
L2:
  xor    %eax,%eax
  pop    %rbx
  ret
```

assistant: ```c
int main() {
  int x, y;
  x = __VERIFIER_nondet_int();
  y = __VERIFIER_nondet_int();
  while (x > 0) {
    x = x + y;
    y = -2 * y;
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  push   %rbx
  xor    %eax,%eax
  call   <__VERIFIER_nondet_int@plt>
  mov    %eax,%ebx
  xor    %eax,%eax
  call   <__VERIFIER_nondet_int@plt>
  test   %ebx,%ebx
  jle    L0
  nopl   0x0(%rax)
L14:
  add    %eax,%ebx
  neg    %eax
  add    %eax,%eax
  test   %ebx,%ebx
  jg     L14
L0:
  xor    %eax,%eax
  pop    %rbx
  ret
```

assistant: ```c
int main() {
  int x, y;
  x = __VERIFIER_nondet_int();
  y = __VERIFIER_nondet_int();
  while (x > 0) {
    x = x + y;
    y = -2 * y;
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %edi,-0x34(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D8(%rip),%rax
  mov    %rax,-0x18(%rbp)
  movq   $0x0,-0x10(%rbp)
  movq   $0x0,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strdup@plt>
  mov    %rax,-0x10(%rbp)
  lea    -0x28(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  lea    D18(%rip),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strtok_r@plt>
  mov    %rax,-0x20(%rbp)
  cmpq   $0x0,-0x20(%rbp)
  jne    L36
  lea    D36(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0xffffffff,%eax
  jmp    L30
L36:
  mov    -0x20(%rbp),%rax
  mov    %rax,%rsi
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x28(%rbp),%rax
  mov    %rax,%rdx
  lea    D18(%rip),%rax
  mov    %rax,%rsi
  mov    $0x0,%edi
  call   <strtok_r@plt>
  mov    %rax,-0x20(%rbp)
  cmpq   $0x0,-0x20(%rbp)
  jne    L36
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    $0x0,%eax
L30:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L0
  call   <__stack_chk_fail@plt>
L0:
  leave
  ret
```

tool_calls: name: read_data
arguments: data_tag: D8
type_infered: string
name: read_data
arguments: data_tag: D18
type_infered: string
name: read_data
arguments: data_tag: D36
type_infered: string
name: read_data
arguments: data_tag: D12
type_infered: string

tool: D8: /obix/test/testdevice/bool/////

tool: D18: /

tool: D36: %s: no token in provided string.


tool: D12: Token: %s


assistant: ```c
int main(int argc, char **argv) {
  char *test = "/obix/test/testdevice/bool/////";
  char *src = NULL;
  char *token = NULL;
  char *reentrant_ptr;
  src = strdup(test);
  token = strtok_r(src, "/", &reentrant_ptr);
  if (token == NULL) {
    printf("%s: no token in provided string.\n");
    return -1;
  }
  do {
    printf("Token: %s\n", token);
  } while ((token = strtok_r(NULL, "/", &reentrant_ptr)) != NULL);
  free(src);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D15(%rip),%rdi
  call   <strdup@plt>
  mov    %rax,%r12
  mov    %rsp,%rdx
  lea    D26(%rip),%rsi
  mov    %rax,%rdi
  call   <strtok_r@plt>
  test   %rax,%rax
  je     L2
  mov    %rax,%rsi
  lea    D27(%rip),%r13
  mov    %rsp,%rbp
  lea    D26(%rip),%rbx
L11:
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %rbp,%rdx
  mov    %rbx,%rsi
  mov    $0x0,%edi
  call   <strtok_r@plt>
  mov    %rax,%rsi
  test   %rax,%rax
  jne    L11
  mov    %r12,%rdi
  call   <free@plt>
  mov    $0x0,%eax
L3:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L20
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L2:
  lea    D46(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0xffffffff,%eax
  jmp    L3
L20:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D15
datatype: string
name: extract_symbolic_data
arguments: data_label: D26
datatype: string
name: extract_symbolic_data
arguments: data_label: D27
datatype: string
name: extract_symbolic_data
arguments: data_label: D46
datatype: string

tool: D15: /obix/test/testdevice/bool/////

tool: D26: /

tool: D27: Token: %s


tool: D46: %s: no token in provided string.


assistant: ```c
int main(int argc, char **argv) {
  char *test = "/obix/test/testdevice/bool/////";
  char *src = NULL;
  char *token = NULL;
  char *reentrant_ptr;
  src = strdup(test);
  token = strtok_r(src, "/", &reentrant_ptr);
  if (token == NULL) {
    printf("%s: no token in provided string.\n");
    return -1;
  }
  do {
    printf("Token: %s\n", token);
  } while ((token = strtok_r(NULL, "/", &reentrant_ptr)) != NULL);
  free(src);
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %r13
  lea    D21(%rip),%rdi
  push   %r12
  push   %rbp
  lea    D26(%rip),%rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  call   <strdup@plt>
  mov    %rbx,%rdx
  mov    %rbp,%rsi
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <strtok_r@plt>
  test   %rax,%rax
  je     L17
  mov    %rax,%rsi
  lea    D0(%rip),%r13
  nopl   0x0(%rax)
L32:
  mov    %r13,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdx
  xor    %edi,%edi
  call   <strtok_r@plt>
  mov    %rax,%rsi
  test   %rax,%rax
  jne    L32
  mov    %r12,%rdi
  call   <free@plt>
  xor    %eax,%eax
L13:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L24
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L17:
  lea    D16(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  or     $0xffffffff,%eax
  jmp    L13
L24:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D21
type_guessed: string
name: retrieve_data
arguments: data_label: D26
type_guessed: string
name: retrieve_data
arguments: data_label: D0
type_guessed: string
name: retrieve_data
arguments: data_label: D16
type_guessed: string

tool: D21: /obix/test/testdevice/bool/////

tool: D26: /

tool: D0: Token: %s


tool: D16: %s: no token in provided string.


assistant: ```c
int main(int argc, char **argv) {
  char *test = "/obix/test/testdevice/bool/////";
  char *src = NULL;
  char *token = NULL;
  char *reentrant_ptr;
  src = strdup(test);
  token = strtok_r(src, "/", &reentrant_ptr);
  if (token == NULL) {
    printf("%s: no token in provided string.\n");
    return -1;
  }
  do {
    printf("Token: %s\n", token);
  } while ((token = strtok_r(NULL, "/", &reentrant_ptr)) != NULL);
  free(src);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r13
  lea    D8(%rip),%rdi
  push   %r12
  push   %rbp
  lea    D22(%rip),%rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  call   <strdup@plt>
  mov    %rbx,%rdx
  mov    %rbp,%rsi
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <strtok_r@plt>
  test   %rax,%rax
  je     L28
  mov    %rax,%rsi
  lea    D35(%rip),%r13
  nopl   0x0(%rax)
L32:
  mov    %r13,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdx
  xor    %edi,%edi
  call   <strtok_r@plt>
  mov    %rax,%rsi
  test   %rax,%rax
  jne    L32
  mov    %r12,%rdi
  call   <free@plt>
  xor    %eax,%eax
L1:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L12
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L28:
  lea    D2(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  or     $0xffffffff,%eax
  jmp    L1
L12:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D8
data_type: string
name: retrieve_data
arguments: data_tag: D22
data_type: string
name: retrieve_data
arguments: data_tag: D35
data_type: string
name: retrieve_data
arguments: data_tag: D2
data_type: string

tool: D8: /obix/test/testdevice/bool/////

tool: D22: /

tool: D35: Token: %s


tool: D2: %s: no token in provided string.


assistant: ```c
int main(int argc, char **argv) {
  char *test = "/obix/test/testdevice/bool/////";
  char *src = NULL;
  char *token = NULL;
  char *reentrant_ptr;
  src = strdup(test);
  token = strtok_r(src, "/", &reentrant_ptr);
  if (token == NULL) {
    printf("%s: no token in provided string.\n");
    return -1;
  }
  do {
    printf("Token: %s\n", token);
  } while ((token = strtok_r(NULL, "/", &reentrant_ptr)) != NULL);
  free(src);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<GArray_growing_policy>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movq   $0x1,-0x8(%rbp)
  cmpq   $0x0,-0x20(%rbp)
  setne  %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  jmp    L40
L23:
  shlq   -0x8(%rbp)
L40:
  mov    -0x8(%rbp),%rax
  cmp    -0x18(%rbp),%rax
  jae    L36
  cmpq   $0x0,-0x8(%rbp)
  jne    L23
L36:
  cmpq   $0x0,-0x8(%rbp)
  je     L44
  mov    -0x8(%rbp),%rax
  jmp    L18
L44:
  mov    -0x18(%rbp),%rax
L18:
  leave
  ret
```

assistant: ```c
size_t GArray_growing_policy(size_t capacity, size_t elem_size) {
  size_t new_capacity = 1;
  assert(elem_size > 0);
  while (new_capacity < capacity && new_capacity > 0)
    new_capacity <<= 1;
  return new_capacity ? new_capacity : capacity;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<GArray_growing_policy>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  test   %rsi,%rsi
  setne  %dil
  movzbl %dil,%edi
  call   <assert@plt>
  cmp    $0x1,%rbx
  jbe    L17
  mov    $0x1,%eax
L12:
  add    %rax,%rax
  cmp    %rax,%rbx
  jbe    L24
  test   %rax,%rax
  jne    L12
L24:
  test   %rax,%rax
  cmovne %rax,%rbx
L20:
  mov    %rbx,%rax
  pop    %rbx
  ret
L17:
  mov    $0x1,%ebx
  jmp    L20
```

assistant: ```c
size_t GArray_growing_policy(size_t capacity, size_t elem_size) {
  size_t new_capacity = 1;
  assert(elem_size > 0);
  while (new_capacity < capacity && new_capacity > 0)
    new_capacity <<= 1;
  return new_capacity ? new_capacity : capacity;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<GArray_growing_policy>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  xor    %edi,%edi
  test   %rsi,%rsi
  setne  %dil
  call   <assert@plt>
  mov    $0x1,%eax
  cmp    $0x1,%rbx
  jbe    L17
  nopl   0x0(%rax)
L19:
  add    %rax,%rax
  cmp    %rax,%rbx
  jbe    L2
  test   %rax,%rax
  jne    L19
L2:
  test   %rax,%rax
  cmove  %rbx,%rax
L17:
  pop    %rbx
  ret
```

assistant: ```c
size_t GArray_growing_policy(size_t capacity, size_t elem_size) {
  size_t new_capacity = 1;
  assert(elem_size > 0);
  while (new_capacity < capacity && new_capacity > 0)
    new_capacity <<= 1;
  return new_capacity ? new_capacity : capacity;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<GArray_growing_policy>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  xor    %edi,%edi
  test   %rsi,%rsi
  setne  %dil
  call   <assert@plt>
  mov    $0x1,%eax
  cmp    $0x1,%rbx
  jbe    L27
  nopl   0x0(%rax)
L1:
  add    %rax,%rax
  cmp    %rax,%rbx
  jbe    L7
  test   %rax,%rax
  jne    L1
L7:
  test   %rax,%rax
  cmove  %rbx,%rax
L27:
  pop    %rbx
  ret
```

assistant: ```c
size_t GArray_growing_policy(size_t capacity, size_t elem_size) {
  size_t new_capacity = 1;
  assert(elem_size > 0);
  while (new_capacity < capacity && new_capacity > 0)
    new_capacity <<= 1;
  return new_capacity ? new_capacity : capacity;
}
```
user: What is the c source code of the assembly code below?

```asm
<get_dht>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    $0x0,%eax
  call   <get2@plt>
  sub    $0x2,%eax
  mov    %eax,-0x4(%rbp)
  jmp    L6
L10:
  mov    $0x0,%eax
  call   <get1dht@plt>
  mov    -0x4(%rbp),%edx
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x4(%rbp)
L6:
  cmpl   $0x0,-0x4(%rbp)
  jg     L10
  nop
  nop
  leave
  ret
```

assistant: ```c
void get_dht(int kind) {
  int length;
  length = get2() - 2;
  while (length > 0)
    length -= get1dht();
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<get_dht>:
  endbr64
  push   %rbx
  mov    $0x0,%eax
  call   <get2@plt>
  lea    -0x2(%rax),%ebx
  test   %ebx,%ebx
  jle    L19
L4:
  mov    $0x0,%eax
  call   <get1dht@plt>
  mov    %rax,%rdx
  mov    %ebx,%eax
  sub    %edx,%eax
  mov    %eax,%ebx
  test   %eax,%eax
  jg     L4
L19:
  pop    %rbx
  ret
```

assistant: ```c
void get_dht(int kind) {
  int length;
  length = get2() - 2;
  while (length > 0)
    length -= get1dht();
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<get_dht>:
  endbr64
  push   %rbx
  xor    %eax,%eax
  call   <get2@plt>
  lea    -0x2(%rax),%ebx
  test   %ebx,%ebx
  jle    L13
  nopl   0x0(%rax,%rax,1)
L3:
  xor    %eax,%eax
  call   <get1dht@plt>
  sub    %eax,%ebx
  test   %ebx,%ebx
  jg     L3
L13:
  pop    %rbx
  ret
```

assistant: ```c
void get_dht(int kind) {
  int length;
  length = get2() - 2;
  while (length > 0)
    length -= get1dht();
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<get_dht>:
  endbr64
  push   %rbx
  xor    %eax,%eax
  call   <get2@plt>
  lea    -0x2(%rax),%ebx
  test   %ebx,%ebx
  jle    L8
  nopl   0x0(%rax,%rax,1)
L17:
  xor    %eax,%eax
  call   <get1dht@plt>
  sub    %eax,%ebx
  test   %ebx,%ebx
  jg     L17
L8:
  pop    %rbx
  ret
```

assistant: ```c
void get_dht(int kind) {
  int length;
  length = get2() - 2;
  while (length > 0)
    length -= get1dht();
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<is_palindrome>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    -0x24(%rbp),%eax
  mov    %eax,%edi
  call   <log10@plt>
  add    $0x1,%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x10(%rbp),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  mov    %eax,-0xc(%rbp)
  movl   $0x0,-0x14(%rbp)
  jmp    L20
L16:
  mov    -0x10(%rbp),%eax
  sub    -0x14(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,%esi
  mov    $0xa,%edi
  call   <pow@plt>
  mov    %eax,%esi
  mov    -0x24(%rbp),%eax
  cltd
  idiv   %esi
  mov    %eax,%edx
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x8(%rbp),%ecx
  mov    %ecx,%eax
  shl    $0x2,%eax
  add    %ecx,%eax
  add    %eax,%eax
  sub    %eax,%edx
  mov    %edx,-0x8(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,%esi
  mov    $0xa,%edi
  call   <pow@plt>
  mov    %eax,%esi
  mov    -0x24(%rbp),%eax
  cltd
  idiv   %esi
  mov    %eax,%edx
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%ecx
  mov    %ecx,%eax
  shl    $0x2,%eax
  add    %ecx,%eax
  add    %eax,%eax
  sub    %eax,%edx
  mov    %edx,-0x4(%rbp)
  mov    -0x8(%rbp),%eax
  cmp    -0x4(%rbp),%eax
  je     L6
  mov    $0x0,%eax
  jmp    L12
L6:
  addl   $0x1,-0x14(%rbp)
L20:
  mov    -0x14(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jl     L16
  mov    $0x1,%eax
L12:
  leave
  ret
```

assistant: ```c
int is_palindrome(const int val) {
  int digits = log10(val) + 1;
  int mid = digits / 2;
  int iter;
  int left, right;
  for (iter = 0; iter < mid; ++iter) {
    left = (val / ((int)pow(10, digits - iter - 1))) % 10;
    right = (val / ((int)pow(10, iter))) % 10;
    if (left != right)
      return 0;
  }
  return 1;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<is_palindrome>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%ebp
  call   <log10@plt>
  mov    %eax,%r13d
  lea    0x1(%rax),%eax
  mov    %eax,%r12d
  shr    $0x1f,%r12d
  add    %eax,%r12d
  sar    %r12d
  cmp    $0x1,%eax
  jle    L34
  mov    $0x0,%ebx
L10:
  mov    %r13d,%esi
  sub    %ebx,%esi
  mov    $0xa,%edi
  call   <pow@plt>
  mov    %rax,%r14
  mov    %ebx,%esi
  mov    $0xa,%edi
  call   <pow@plt>
  mov    %rax,%rsi
  mov    %ebp,%eax
  cltd
  idiv   %r14d
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  sar    $0x22,%rdx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  lea    (%rdx,%rdx,4),%edx
  add    %edx,%edx
  sub    %edx,%eax
  mov    %eax,%ecx
  mov    %ebp,%eax
  cltd
  idiv   %esi
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  sar    $0x22,%rdx
  mov    %eax,%esi
  sar    $0x1f,%esi
  sub    %esi,%edx
  lea    (%rdx,%rdx,4),%edx
  add    %edx,%edx
  sub    %edx,%eax
  cmp    %eax,%ecx
  jne    L40
  add    $0x1,%ebx
  cmp    %ebx,%r12d
  jg     L10
  mov    $0x1,%eax
  jmp    L29
L34:
  mov    $0x1,%eax
  jmp    L29
L40:
  mov    $0x0,%eax
L29:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int is_palindrome(const int val) {
  int digits = log10(val) + 1;
  int mid = digits / 2;
  int iter;
  int left, right;
  for (iter = 0; iter < mid; ++iter) {
    left = (val / ((int)pow(10, digits - iter - 1))) % 10;
    right = (val / ((int)pow(10, iter))) % 10;
    if (left != right)
      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
<is_palindrome>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  mov    %edi,%r12d
  push   %rbp
  push   %rbx
  call   <log10@plt>
  mov    %eax,%ebp
  add    $0x1,%eax
  mov    %eax,%ebx
  shr    $0x1f,%ebx
  add    %eax,%ebx
  sar    %ebx
  cmp    $0x1,%eax
  jle    L11
  xor    %r13d,%r13d
  jmp    L31
L0:
  add    $0x1,%r13d
  cmp    %r13d,%ebx
  jle    L11
L31:
  mov    %ebp,%esi
  mov    $0xa,%edi
  sub    %r13d,%esi
  call   <pow@plt>
  mov    %r13d,%esi
  mov    $0xa,%edi
  mov    %rax,%r14
  call   <pow@plt>
  mov    %rax,%r8
  mov    %r12d,%eax
  cltd
  idiv   %r14d
  movslq %eax,%rdx
  mov    %eax,%ecx
  imul   $0x66666667,%rdx,%rdx
  sar    $0x1f,%ecx
  sar    $0x22,%rdx
  sub    %ecx,%edx
  lea    (%rdx,%rdx,4),%edx
  add    %edx,%edx
  sub    %edx,%eax
  mov    %eax,%ecx
  mov    %r12d,%eax
  cltd
  idiv   %r8d
  movslq %eax,%rdx
  mov    %eax,%esi
  imul   $0x66666667,%rdx,%rdx
  sar    $0x1f,%esi
  sar    $0x22,%rdx
  sub    %esi,%edx
  lea    (%rdx,%rdx,4),%edx
  add    %edx,%edx
  sub    %edx,%eax
  cmp    %eax,%ecx
  je     L0
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopw   0x0(%rax,%rax,1)
L11:
  pop    %rbx
  mov    $0x1,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int is_palindrome(const int val) {
  int digits = log10(val) + 1;
  int mid = digits / 2;
  int iter;
  int left, right;
  for (iter = 0; iter < mid; ++iter) {
    left = (val / ((int)pow(10, digits - iter - 1))) % 10;
    right = (val / ((int)pow(10, iter))) % 10;
    if (left != right)
      return 0;
  }
  return 1;
}
```