ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  movl   $0x9,(%rsp)
  movl   $0x8,0x4(%rsp)
  movl   $0x7,0x8(%rsp)
  movl   $0x6,0xc(%rsp)
  movl   $0x5,0x10(%rsp)
  movl   $0x4,0x14(%rsp)
  movl   $0x3,0x18(%rsp)
  movl   $0x2,0x1c(%rsp)
  movl   $0x1,0x20(%rsp)
  movl   $0x0,0x24(%rsp)
  mov    %rsp,%rbx
  lea    0x28(%rsp),%r12
  lea    D7(%rip),%rbp
L0:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  add    $0x4,%rbx
  cmp    %r12,%rbx
  jne    L0
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  mov    $0x0,%eax
  add    $0x30,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L1:
  call   <__stack_chk_fail@plt>
```

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

tool: D7: %d 

assistant: ```c
int main() {
  int i, a[10] = {9, 8, 7, 6, 5, 4, 3, 2, 1, 0};
  for (i = 0; i < 10; i++)
    printf("%d ", a[i]);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %r12
  mov    $0x9,%esi
  push   %rbp
  lea    D4(%rip),%rbp
  push   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rbx
  lea    0x28(%rsp),%r12
  movabs $0x600000007,%rax
  mov    %rax,0x8(%rsp)
  movabs $0x400000005,%rax
  mov    %rax,0x10(%rsp)
  movabs $0x200000003,%rax
  movl   $0x8,0x4(%rsp)
  mov    %rax,0x18(%rsp)
  movq   $0x1,0x20(%rsp)
  jmp    L9
  nopw   0x0(%rax,%rax,1)
L1:
  mov    (%rbx),%esi
  add    $0x4,%rbx
L9:
  mov    %rbp,%rdi
  call   <printf@plt>
  cmp    %r12,%rbx
  jne    L1
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L10
  add    $0x30,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L10:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D4
data_type: string

tool: D4: %d 

assistant: ```c
int main() {
  int i, a[10] = {9, 8, 7, 6, 5, 4, 3, 2, 1, 0};
  for (i = 0; i < 10; i++)
    printf("%d ", a[i]);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  lea    D10(%rip),%rbp
  push   %rbx
  sub    $0x30,%rsp
  movdqa D23(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    D27(%rip),%rax
  mov    %rsp,%rbx
  lea    0x28(%rsp),%r12
  movaps %xmm0,(%rsp)
  movdqa D31(%rip),%xmm0
  mov    %rax,0x20(%rsp)
  movaps %xmm0,0x10(%rsp)
L6:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %rbx,%r12
  jne    L6
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L11
  add    $0x30,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L11:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D10
type_guessed: string
name: retrieve_data
arguments: data_tag: D23
type_guessed: byte[16]
name: retrieve_data
arguments: data_tag: D27
type_guessed: byte[8]
name: retrieve_data
arguments: data_tag: D31
type_guessed: byte[16]

tool: D10: %d 

tool: {"D23": [0x09,0x00,0x00,0x00,0x08,0x00,0x00,0x00,0x07,0x00,0x00,0x00,0x06,0x00,0x00,0x00]}

tool: {"D27": [0x01,0x00,0x00,0x00,0x00,0x00,0x00,0x00]}

tool: {"D31": [0x05,0x00,0x00,0x00,0x04,0x00,0x00,0x00,0x03,0x00,0x00,0x00,0x02,0x00,0x00,0x00]}

assistant: ```c
int main() {
  int i, a[10] = {9, 8, 7, 6, 5, 4, 3, 2, 1, 0};
  for (i = 0; i < 10; i++)
    printf("%d ", a[i]);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<aml_array_dbl_4d_variable_2nd_and_4th>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x58,%rsp
  mov    %edi,%ebx
  mov    %rsi,%rbp
  mov    %edx,%r14d
  mov    %edx,0x24(%rsp)
  mov    %rcx,%r13
  mov    %rcx,0x28(%rsp)
  movq   %xmm0,%r15
  lea    0x0(,%rdi,8),%edi
  call   <malloc@plt>
  mov    %rax,0x40(%rsp)
  test   %ebx,%ebx
  jle    L56
  lea    0x0(,%r14,8),%ecx
  mov    %ecx,0x20(%rsp)
  mov    %rbp,0x30(%rsp)
  mov    %rax,0x38(%rsp)
  lea    -0x1(%rbx),%eax
  lea    0x4(%rbp,%rax,4),%rax
  mov    %rax,0x48(%rsp)
  lea    -0x1(%r14),%eax
  lea    0x4(%r13,%rax,4),%rax
  mov    %rax,0x8(%rsp)
L10:
  mov    0x30(%rsp),%rbx
  mov    %rbx,0x18(%rsp)
  mov    (%rbx),%eax
  mov    %eax,0x10(%rsp)
  lea    0x0(,%rax,8),%edi
  call   <malloc@plt>
  mov    0x38(%rsp),%rsi
  mov    %rsi,%rbp
  mov    %rax,(%rsi)
  cmpl   $0x0,(%rbx)
  jle    L40
  movq   $0x0,0x10(%rsp)
L31:
  mov    0x20(%rsp),%edi
  call   <malloc@plt>
  mov    0x10(%rsp),%rcx
  lea    0x0(,%rcx,8),%r12
  mov    0x0(%rbp),%rdx
  mov    %rax,(%rdx,%rcx,8)
  cmpl   $0x0,0x24(%rsp)
  jle    L51
  mov    0x28(%rsp),%r14
  mov    $0x0,%ebx
L29:
  mov    %r14,%r13
  mov    (%r14),%eax
  lea    0x0(,%rax,8),%edi
  call   <malloc@plt>
  mov    %rax,%rdx
  mov    0x0(%rbp),%rax
  mov    (%rax,%r12,1),%rax
  mov    %rdx,(%rax,%rbx,1)
  cmpl   $0x0,(%r14)
  jle    L19
  mov    $0x0,%eax
  movq   %r15,%xmm0
L52:
  mov    0x0(%rbp),%rdx
  mov    (%rdx,%r12,1),%rdx
  mov    (%rdx,%rbx,1),%rdx
  movsd  %xmm0,(%rdx,%rax,8)
  add    $0x1,%rax
  cmp    %eax,0x0(%r13)
  jg     L52
L19:
  add    $0x4,%r14
  add    $0x8,%rbx
  cmp    0x8(%rsp),%r14
  jne    L29
L51:
  addq   $0x1,0x10(%rsp)
  mov    0x10(%rsp),%rax
  mov    0x18(%rsp),%rsi
  cmp    %eax,(%rsi)
  jg     L31
L40:
  addq   $0x4,0x30(%rsp)
  mov    0x30(%rsp),%rax
  addq   $0x8,0x38(%rsp)
  cmp    0x48(%rsp),%rax
  jne    L10
L56:
  mov    0x40(%rsp),%rax
  add    $0x58,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
double ****aml_array_dbl_4d_variable_2nd_and_4th(int s1, int *s2, int s3, int *s4, double val) {
  int i, j, k, l;
  double ****arr = (double ****)malloc(s1 * sizeof(double ***));
  for (i = 0; i < s1; i++) {
    arr[i] = (double ***)malloc(s2[i] * sizeof(double **));
    for (j = 0; j < s2[i]; j++) {
      arr[i][j] = (double **)malloc(s3 * sizeof(double *));
      for (k = 0; k < s3; k++) {
        arr[i][j][k] = (double *)malloc(s4[k] * sizeof(double));
        for (l = 0; l < s4[k]; l++)
          arr[i][j][k][l] = val;
      }
    }
  }
  return arr;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<aml_array_dbl_4d_variable_2nd_and_4th>:
  endbr64
  push   %r15
  mov    %rcx,%r15
  push   %r14
  push   %r13
  mov    %rsi,%r13
  push   %r12
  push   %rbp
  mov    %edx,%ebp
  push   %rbx
  mov    %edi,%ebx
  shl    $0x3,%edi
  sub    $0x38,%rsp
  mov    %edx,0x14(%rsp)
  movsd  %xmm0,(%rsp)
  call   <malloc@plt>
  mov    %rax,0x20(%rsp)
  test   %ebx,%ebx
  jle    L50
  movslq %ebp,%rbp
  mov    %rax,%r14
  lea    -0x1(%rbx),%eax
  mov    %r13,0x8(%rsp)
  lea    0x0(,%rbp,8),%esi
  lea    0x4(%r13,%rax,4),%rax
  movsd  (%rsp),%xmm0
  mov    %esi,0x10(%rsp)
  mov    %rax,0x18(%rsp)
L1:
  mov    0x8(%rsp),%rbx
  movsd  %xmm0,0x28(%rsp)
  mov    (%rbx),%eax
  lea    0x0(,%rax,8),%edi
  mov    %eax,(%rsp)
  call   <malloc@plt>
  mov    (%rbx),%edx
  movsd  0x28(%rsp),%xmm0
  mov    %rax,(%r14)
  test   %edx,%edx
  jle    L79
  xor    %r12d,%r12d
  nopl   0x0(%rax)
L45:
  mov    0x10(%rsp),%edi
  movsd  %xmm0,(%rsp)
  lea    0x0(,%r12,8),%rbx
  call   <malloc@plt>
  mov    (%r14),%rdx
  movsd  (%rsp),%xmm0
  mov    %rax,(%rdx,%r12,8)
  mov    0x14(%rsp),%eax
  test   %eax,%eax
  jle    L3
  xor    %r13d,%r13d
  nopl   (%rax)
L46:
  mov    (%r15,%r13,4),%eax
  movsd  %xmm0,(%rsp)
  lea    0x0(,%rax,8),%edi
  call   <malloc@plt>
  mov    (%r14),%rdx
  movsd  (%rsp),%xmm0
  mov    (%rdx,%rbx,1),%rdx
  mov    %rax,(%rdx,%r13,8)
  mov    (%r15,%r13,4),%edx
  test   %edx,%edx
  jle    L37
  sub    $0x1,%edx
  lea    0x8(%rax,%rdx,8),%rdx
  nopl   (%rax)
L12:
  movsd  %xmm0,(%rax)
  add    $0x8,%rax
  cmp    %rdx,%rax
  jne    L12
L37:
  add    $0x1,%r13
  cmp    %r13,%rbp
  jne    L46
L3:
  mov    0x8(%rsp),%rax
  add    $0x1,%r12
  cmp    %r12d,(%rax)
  jg     L45
L79:
  addq   $0x4,0x8(%rsp)
  add    $0x8,%r14
  mov    0x8(%rsp),%rax
  cmp    0x18(%rsp),%rax
  jne    L1
L50:
  mov    0x20(%rsp),%rax
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
double ****aml_array_dbl_4d_variable_2nd_and_4th(int s1, int *s2, int s3, int *s4, double val) {
  int i, j, k, l;
  double ****arr = (double ****)malloc(s1 * sizeof(double ***));
  for (i = 0; i < s1; i++) {
    arr[i] = (double ***)malloc(s2[i] * sizeof(double **));
    for (j = 0; j < s2[i]; j++) {
      arr[i][j] = (double **)malloc(s3 * sizeof(double *));
      for (k = 0; k < s3; k++) {
        arr[i][j][k] = (double *)malloc(s4[k] * sizeof(double));
        for (l = 0; l < s4[k]; l++)
          arr[i][j][k][l] = val;
      }
    }
  }
  return arr;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<aml_array_dbl_4d_variable_2nd_and_4th>:
  endbr64
  push   %r15
  mov    %rcx,%r15
  push   %r14
  push   %r13
  mov    %rsi,%r13
  push   %r12
  push   %rbp
  mov    %edx,%ebp
  push   %rbx
  mov    %edi,%ebx
  shl    $0x3,%edi
  sub    $0x48,%rsp
  mov    %edx,0x2c(%rsp)
  movsd  %xmm0,0x8(%rsp)
  call   <malloc@plt>
  mov    %rax,0x38(%rsp)
  test   %ebx,%ebx
  jle    L89
  mov    %rax,%r14
  movslq %ebp,%rbp
  lea    -0x1(%rbx),%eax
  mov    %r13,0x20(%rsp)
  lea    0x4(%r13,%rax,4),%rax
  lea    0x0(,%rbp,8),%esi
  movsd  0x8(%rsp),%xmm0
  mov    %r15,%r13
  mov    %esi,0x28(%rsp)
  mov    %r14,%r15
  mov    %rax,0x30(%rsp)
L17:
  mov    0x20(%rsp),%rbx
  movsd  %xmm0,0x10(%rsp)
  mov    (%rbx),%eax
  lea    0x0(,%rax,8),%edi
  mov    %eax,0x8(%rsp)
  call   <malloc@plt>
  mov    (%rbx),%edx
  movsd  0x10(%rsp),%xmm0
  mov    %rax,(%r15)
  test   %edx,%edx
  jle    L25
  movapd %xmm0,%xmm1
  mov    %r13,%rax
  xor    %r12d,%r12d
  mov    %r15,%r13
  unpcklpd %xmm1,%xmm1
  mov    %rax,%r15
  movaps %xmm1,0x10(%rsp)
  nopl   0x0(%rax,%rax,1)
L62:
  mov    0x28(%rsp),%edi
  movsd  %xmm0,0x8(%rsp)
  lea    0x0(,%r12,8),%rbx
  call   <malloc@plt>
  mov    0x0(%r13),%rdx
  movsd  0x8(%rsp),%xmm0
  mov    %rax,(%rdx,%r12,8)
  mov    0x2c(%rsp),%eax
  test   %eax,%eax
  jle    L44
  xor    %r14d,%r14d
  nopl   0x0(%rax)
L94:
  mov    (%r15,%r14,4),%eax
  movsd  %xmm0,0x8(%rsp)
  lea    0x0(,%rax,8),%edi
  call   <malloc@plt>
  mov    (%r15,%r14,4),%esi
  movsd  0x8(%rsp),%xmm0
  mov    %rax,%rdx
  mov    0x0(%r13),%rax
  test   %esi,%esi
  mov    (%rax,%rbx,1),%rax
  mov    %rdx,(%rax,%r14,8)
  jle    L88
  cmp    $0x1,%esi
  je     L106
  mov    %esi,%ecx
  movapd 0x10(%rsp),%xmm1
  mov    %rdx,%rax
  shr    %ecx
  sub    $0x1,%ecx
  shl    $0x4,%rcx
  lea    0x10(%rdx,%rcx,1),%rcx
  nopl   (%rax)
L109:
  movups %xmm1,(%rax)
  add    $0x10,%rax
  cmp    %rcx,%rax
  jne    L109
  mov    %esi,%eax
  and    $0xfffffffe,%eax
  and    $0x1,%esi
  je     L88
L50:
  cltq
  movsd  %xmm0,(%rdx,%rax,8)
L88:
  add    $0x1,%r14
  cmp    %r14,%rbp
  jne    L94
L44:
  mov    0x20(%rsp),%rax
  add    $0x1,%r12
  cmp    %r12d,(%rax)
  jg     L62
  mov    %r15,%rax
  mov    %r13,%r15
  mov    %rax,%r13
L25:
  addq   $0x4,0x20(%rsp)
  add    $0x8,%r15
  mov    0x20(%rsp),%rax
  cmp    0x30(%rsp),%rax
  jne    L17
L89:
  mov    0x38(%rsp),%rax
  add    $0x48,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L106:
  xor    %eax,%eax
  jmp    L50
```

assistant: ```c
double ****aml_array_dbl_4d_variable_2nd_and_4th(int s1, int *s2, int s3, int *s4, double val) {
  int i, j, k, l;
  double ****arr = (double ****)malloc(s1 * sizeof(double ***));
  for (i = 0; i < s1; i++) {
    arr[i] = (double ***)malloc(s2[i] * sizeof(double **));
    for (j = 0; j < s2[i]; j++) {
      arr[i][j] = (double **)malloc(s3 * sizeof(double *));
      for (k = 0; k < s3; k++) {
        arr[i][j][k] = (double *)malloc(s4[k] * sizeof(double));
        for (l = 0; l < s4[k]; l++)
          arr[i][j][k][l] = val;
      }
    }
  }
  return arr;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x10(%rbp),%rdx
  mov    -0x4(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <init_test@plt>
  mov    $0x0,%eax
  call   <test_simple_jump@plt>
  mov    $0x0,%eax
  call   <test_multiple_jumps_to_same_label@plt>
  mov    $0x0,%eax
  leave
  ret
```

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

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  call   <init_test@plt>
  mov    $0x0,%eax
  call   <test_simple_jump@plt>
  mov    $0x0,%eax
  call   <test_multiple_jumps_to_same_label@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  init_test(argc, argv);
  test_simple_jump();
  test_multiple_jumps_to_same_label();
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  call   <init_test@plt>
  xor    %eax,%eax
  call   <test_simple_jump@plt>
  xor    %eax,%eax
  call   <test_multiple_jumps_to_same_label@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  init_test(argc, argv);
  test_simple_jump();
  test_multiple_jumps_to_same_label();
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  call   <init_test@plt>
  xor    %eax,%eax
  call   <test_simple_jump@plt>
  xor    %eax,%eax
  call   <test_multiple_jumps_to_same_label@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  init_test(argc, argv);
  test_simple_jump();
  test_multiple_jumps_to_same_label();
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<string_ncopy_do>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    %rcx,-0x20(%rbp)
  mov    -0x10(%rbp),%rax
  sub    $0x1,%rax
  cmp    %rax,-0x20(%rbp)
  jbe    L8
  mov    -0x10(%rbp),%rax
  sub    $0x1,%rax
  mov    %rax,-0x20(%rbp)
L8:
  mov    -0x20(%rbp),%rdx
  mov    -0x18(%rbp),%rcx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strncpy@plt>
  mov    -0x8(%rbp),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  nop
  leave
  ret
```

assistant: ```c
void string_ncopy_do(char *dst, size_t dstlen, const char *src, size_t srclen) {
  if (srclen > dstlen - 1)
    srclen = dstlen - 1;
  strncpy(dst, src, srclen);
  dst[srclen] = 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<string_ncopy_do>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    %rdx,%rax
  lea    -0x1(%rsi),%rbx
  cmp    %rcx,%rbx
  cmova  %rcx,%rbx
  mov    %rbx,%rdx
  mov    %rax,%rsi
  call   <strncpy@plt>
  movb   $0x0,0x0(%rbp,%rbx,1)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void string_ncopy_do(char *dst, size_t dstlen, const char *src, size_t srclen) {
  if (srclen > dstlen - 1)
    srclen = dstlen - 1;
  strncpy(dst, src, srclen);
  dst[srclen] = 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
<string_ncopy_do>:
  endbr64
  push   %rbp
  mov    %rdx,%r8
  mov    %rdi,%rbp
  push   %rbx
  lea    -0x1(%rsi),%rbx
  mov    %r8,%rsi
  sub    $0x8,%rsp
  cmp    %rcx,%rbx
  cmova  %rcx,%rbx
  mov    %rbx,%rdx
  call   <strncpy@plt>
  movb   $0x0,0x0(%rbp,%rbx,1)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void string_ncopy_do(char *dst, size_t dstlen, const char *src, size_t srclen) {
  if (srclen > dstlen - 1)
    srclen = dstlen - 1;
  strncpy(dst, src, srclen);
  dst[srclen] = 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<clear_keyboard_buffers>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
L14:
  mov    $0x64,%edi
  call   <inportb@plt>
  mov    %al,-0x1(%rbp)
  movzbl -0x1(%rbp),%eax
  and    $0x1,%eax
  test   %eax,%eax
  je     L9
  mov    $0x60,%edi
  call   <inportb@plt>
L9:
  movzbl -0x1(%rbp),%eax
  and    $0x2,%eax
  test   %eax,%eax
  jne    L14
  nop
  nop
  leave
  ret
```

assistant: ```c
void clear_keyboard_buffers() {
  unsigned char temp;
  do {
    temp = inportb(0x64);
    if (((temp) & (1 << (0))) != 0)
      inportb(0x60);
  } while (((temp) & (1 << (1))) != 0);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<clear_keyboard_buffers>:
  endbr64
  push   %rbx
  jmp    L28
L30:
  test   $0x2,%bl
  je     L24
L28:
  mov    $0x64,%edi
  call   <inportb@plt>
  mov    %eax,%ebx
  test   $0x1,%al
  je     L30
  mov    $0x60,%edi
  call   <inportb@plt>
  jmp    L30
L24:
  pop    %rbx
  ret
```

assistant: ```c
void clear_keyboard_buffers() {
  unsigned char temp;
  do {
    temp = inportb(0x64);
    if (((temp) & (1 << (0))) != 0)
      inportb(0x60);
  } while (((temp) & (1 << (1))) != 0);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<clear_keyboard_buffers>:
  endbr64
  push   %rbx
  jmp    L28
  nopw   0x0(%rax,%rax,1)
L5:
  and    $0x2,%ebx
  je     L11
L28:
  mov    $0x64,%edi
  call   <inportb@plt>
  mov    %eax,%ebx
  test   $0x1,%al
  je     L5
  mov    $0x60,%edi
  call   <inportb@plt>
  and    $0x2,%ebx
  jne    L28
L11:
  pop    %rbx
  ret
```

assistant: ```c
void clear_keyboard_buffers() {
  unsigned char temp;
  do {
    temp = inportb(0x64);
    if (((temp) & (1 << (0))) != 0)
      inportb(0x60);
  } while (((temp) & (1 << (1))) != 0);
}
```
user: What is the c source code of the assembly code below:

```asm
<clear_keyboard_buffers>:
  endbr64
  push   %rbx
  jmp    L24
  nopw   0x0(%rax,%rax,1)
L19:
  and    $0x2,%ebx
  je     L16
L24:
  mov    $0x64,%edi
  call   <inportb@plt>
  mov    %eax,%ebx
  test   $0x1,%al
  je     L19
  mov    $0x60,%edi
  call   <inportb@plt>
  and    $0x2,%ebx
  jne    L24
L16:
  pop    %rbx
  ret
```

assistant: ```c
void clear_keyboard_buffers() {
  unsigned char temp;
  do {
    temp = inportb(0x64);
    if (((temp) & (1 << (0))) != 0)
      inportb(0x60);
  } while (((temp) & (1 << (1))) != 0);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<edflib_sprint_int_number_nonlocalized>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  mov    %edx,-0x30(%rbp)
  mov    %ecx,-0x34(%rbp)
  movl   $0x0,-0x14(%rbp)
  movl   $0x0,-0xc(%rbp)
  movl   $0x3b9aca00,-0x8(%rbp)
  cmpl   $0x0,-0x30(%rbp)
  jns    L79
  movl   $0x0,-0x30(%rbp)
L79:
  cmpl   $0x9,-0x30(%rbp)
  jle    L58
  movl   $0x1,-0x14(%rbp)
L58:
  cmpl   $0x0,-0x2c(%rbp)
  jns    L29
  mov    -0xc(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0xc(%rbp)
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movb   $0x2d,(%rax)
  negl   -0x2c(%rbp)
  jmp    L45
L29:
  cmpl   $0x0,-0x34(%rbp)
  je     L45
  mov    -0xc(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0xc(%rbp)
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movb   $0x2b,(%rax)
L45:
  movl   $0xa,-0x10(%rbp)
  jmp    L38
L5:
  mov    -0x30(%rbp),%eax
  cmp    -0x10(%rbp),%eax
  jne    L11
  movl   $0x1,-0x14(%rbp)
L11:
  mov    -0x2c(%rbp),%eax
  cltd
  idivl  -0x8(%rbp)
  mov    %eax,-0x4(%rbp)
  mov    -0x2c(%rbp),%eax
  cltd
  idivl  -0x8(%rbp)
  mov    %edx,-0x2c(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jne    L89
  cmpl   $0x0,-0x14(%rbp)
  je     L97
L89:
  mov    -0x4(%rbp),%eax
  lea    0x30(%rax),%ecx
  mov    -0xc(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0xc(%rbp)
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    %ecx,%edx
  mov    %dl,(%rax)
  movl   $0x1,-0x14(%rbp)
L97:
  mov    -0x8(%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,-0x8(%rbp)
  subl   $0x1,-0x10(%rbp)
L38:
  cmpl   $0x0,-0x10(%rbp)
  jne    L5
  cmpl   $0x0,-0x14(%rbp)
  jne    L59
  mov    -0xc(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0xc(%rbp)
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movb   $0x30,(%rax)
L59:
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0xc(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int edflib_sprint_int_number_nonlocalized(char *str, int q, int minimum, int sign) {
  int flag = 0, z, i, j = 0, base = 1000000000;
  if (minimum < 0) {
    minimum = 0;
  }
  if (minimum > 9) {
    flag = 1;
  }
  if (q < 0) {
    str[j++] = '-';
    q = -q;
  } else {
    if (sign) {
      str[j++] = '+';
    }
  }
  for (i = 10; i; i--) {
    if (minimum == i) {
      flag = 1;
    }
    z = q / base;
    q %= base;
    if (z || flag) {
      str[j++] = '0' + z;
      flag = 1;
    }
    base /= 10;
  }
  if (!flag) {
    str[j++] = '0';
  }
  str[j] = 0;
  return (j);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<edflib_sprint_int_number_nonlocalized>:
  endbr64
  push   %rbx
  mov    %rdi,%r10
  mov    %esi,%edi
  test   %edx,%edx
  js     L61
  mov    %edx,%r11d
  cmp    $0x9,%edx
  setg   %r9b
  movzbl %r9b,%r9d
L86:
  test   %edi,%edi
  js     L85
  mov    %ecx,%ebx
  test   %ecx,%ecx
  je     L44
  movb   $0x2b,(%r10)
  mov    $0x1,%ebx
  jmp    L44
L61:
  mov    $0x0,%r11d
  mov    $0x0,%r9d
  jmp    L86
L85:
  movb   $0x2d,(%r10)
  neg    %edi
  mov    $0x1,%ebx
L44:
  mov    $0x3b9aca00,%esi
  mov    $0xa,%r8d
  jmp    L93
L60:
  test   %r9d,%r9d
  jne    L15
  movslq %ebx,%rax
  movb   $0x30,(%r10,%rax,1)
  lea    0x1(%rbx),%ebx
L15:
  movslq %ebx,%rax
  movb   $0x0,(%r10,%rax,1)
  mov    %ebx,%eax
  pop    %rbx
  ret
L39:
  mov    %edi,%eax
  cltd
  idiv   %esi
  mov    %edx,%edi
  mov    %eax,%ecx
L84:
  movslq %ebx,%rax
  add    $0x30,%ecx
  mov    %cl,(%r10,%rax,1)
  lea    0x1(%rbx),%ebx
  mov    $0x1,%r9d
L10:
  movslq %esi,%rcx
  imul   $0x66666667,%rcx,%rcx
  sar    $0x22,%rcx
  mov    %esi,%eax
  sar    $0x1f,%eax
  mov    %ecx,%esi
  sub    %eax,%esi
  sub    $0x1,%r8d
  je     L60
L93:
  cmp    %r8d,%r11d
  je     L39
  mov    %edi,%eax
  cltd
  idiv   %esi
  mov    %eax,%ecx
  mov    %edx,%edi
  or     %eax,%r9d
  je     L10
  jmp    L84
```

assistant: ```c
int edflib_sprint_int_number_nonlocalized(char *str, int q, int minimum, int sign) {
  int flag = 0, z, i, j = 0, base = 1000000000;
  if (minimum < 0) {
    minimum = 0;
  }
  if (minimum > 9) {
    flag = 1;
  }
  if (q < 0) {
    str[j++] = '-';
    q = -q;
  } else {
    if (sign) {
      str[j++] = '+';
    }
  }
  for (i = 10; i; i--) {
    if (minimum == i) {
      flag = 1;
    }
    z = q / base;
    q %= base;
    if (z || flag) {
      str[j++] = '0' + z;
      flag = 1;
    }
    base /= 10;
  }
  if (!flag) {
    str[j++] = '0';
  }
  str[j] = 0;
  return (j);
}
```