ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What would the C source look like for the assembly code provided here?

```asm
<__hsail_cvt_zeroi_sat_s16_f32>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movss  %xmm0,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  movd   %eax,%xmm0
  call   <isnan@plt>
  test   %rax,%rax
  je     L15
  mov    $0x0,%eax
  jmp    L25
L15:
  mov    D18(%rip),%rax
  mov    (%rax),%rax
  pxor   %xmm1,%xmm1
  cvtsi2ss %rax,%xmm1
  movss  -0x4(%rbp),%xmm0
  comiss %xmm1,%xmm0
  jb     L56
  mov    D18(%rip),%rax
  mov    (%rax),%rax
  jmp    L25
L56:
  mov    D39(%rip),%rax
  mov    (%rax),%rax
  pxor   %xmm0,%xmm0
  cvtsi2ss %rax,%xmm0
  comiss -0x4(%rbp),%xmm0
  jb     L37
  mov    D39(%rip),%rax
  mov    (%rax),%rax
  jmp    L25
L37:
  movss  -0x4(%rbp),%xmm0
  cvttss2si %xmm0,%rax
L25:
  leave
  ret
```

assistant: ```c
int16_t __hsail_cvt_zeroi_sat_s16_f32(float a) {
  if (isnan(a))
    return 0;
  if (a >= (float)INT16_MAX)
    return INT16_MAX;
  if (a <= (float)INT16_MIN)
    return INT16_MIN;
  return (int16_t)a;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<__hsail_cvt_zeroi_sat_s16_f32>:
  endbr64
  sub    $0x18,%rsp
  movss  %xmm0,0xc(%rsp)
  call   <isnan@plt>
  mov    %rax,%rdx
  mov    $0x0,%eax
  test   %rdx,%rdx
  jne    L4
  mov    D9(%rip),%rax
  mov    (%rax),%rax
  pxor   %xmm0,%xmm0
  cvtsi2ss %rax,%xmm0
  movss  0xc(%rsp),%xmm1
  comiss %xmm0,%xmm1
  jae    L4
  mov    D2(%rip),%rax
  mov    (%rax),%rax
  pxor   %xmm0,%xmm0
  cvtsi2ss %rax,%xmm0
  comiss %xmm1,%xmm0
  jae    L4
  cvttss2si %xmm1,%rax
L4:
  add    $0x18,%rsp
  ret
```

assistant: ```c
int16_t __hsail_cvt_zeroi_sat_s16_f32(float a) {
  if (isnan(a))
    return 0;
  if (a >= (float)INT16_MAX)
    return INT16_MAX;
  if (a <= (float)INT16_MIN)
    return INT16_MIN;
  return (int16_t)a;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<__hsail_cvt_zeroi_sat_s16_f32>:
  endbr64
  sub    $0x18,%rsp
  movss  %xmm0,0xc(%rsp)
  call   <isnan@plt>
  mov    %rax,%r8
  xor    %eax,%eax
  test   %r8,%r8
  jne    L6
  mov    D19(%rip),%rax
  pxor   %xmm1,%xmm1
  movss  0xc(%rsp),%xmm0
  mov    (%rax),%rax
  cvtsi2ss %rax,%xmm1
  comiss %xmm1,%xmm0
  jae    L6
  mov    D2(%rip),%rax
  pxor   %xmm1,%xmm1
  mov    (%rax),%rax
  cvtsi2ss %rax,%xmm1
  comiss %xmm0,%xmm1
  jae    L6
  cvttss2si %xmm0,%rax
L6:
  add    $0x18,%rsp
  ret
```

assistant: ```c
int16_t __hsail_cvt_zeroi_sat_s16_f32(float a) {
  if (isnan(a))
    return 0;
  if (a >= (float)INT16_MAX)
    return INT16_MAX;
  if (a <= (float)INT16_MIN)
    return INT16_MIN;
  return (int16_t)a;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<__hsail_cvt_zeroi_sat_s16_f32>:
  endbr64
  sub    $0x18,%rsp
  movss  %xmm0,0xc(%rsp)
  call   <isnan@plt>
  mov    %rax,%r8
  xor    %eax,%eax
  test   %r8,%r8
  jne    L29
  mov    D10(%rip),%rax
  pxor   %xmm1,%xmm1
  movss  0xc(%rsp),%xmm0
  mov    (%rax),%rax
  cvtsi2ss %rax,%xmm1
  comiss %xmm1,%xmm0
  jae    L29
  mov    D9(%rip),%rax
  pxor   %xmm1,%xmm1
  mov    (%rax),%rax
  cvtsi2ss %rax,%xmm1
  comiss %xmm0,%xmm1
  jae    L29
  cvttss2si %xmm0,%rax
L29:
  add    $0x18,%rsp
  ret
```

assistant: ```c
int16_t __hsail_cvt_zeroi_sat_s16_f32(float a) {
  if (isnan(a))
    return 0;
  if (a >= (float)INT16_MAX)
    return INT16_MAX;
  if (a <= (float)INT16_MIN)
    return INT16_MIN;
  return (int16_t)a;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
L0:
  lea    D14(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x1c(%rbp),%rax
  mov    %rax,%rsi
  lea    D57(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    -0x1c(%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
  test   %edx,%edx
  jne    L0
  mov    -0x1c(%rbp),%eax
  lea    -0xc(%rbp),%rdi
  lea    -0x10(%rbp),%rcx
  lea    -0x14(%rbp),%rdx
  lea    -0x18(%rbp),%rsi
  mov    %rdi,%r8
  mov    %eax,%edi
  call   <dispense_cash@plt>
  lea    D36(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    -0x18(%rbp),%eax
  mov    %eax,%esi
  lea    D55(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x14(%rbp),%eax
  mov    %eax,%esi
  lea    D17(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x10(%rbp),%eax
  mov    %eax,%esi
  lea    D13(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D24(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L17
  call   <__stack_chk_fail@plt>
L17:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D14
type: string
name: disassemble_data
arguments: data_tag: D57
type: string
name: disassemble_data
arguments: data_tag: D36
type: string
name: disassemble_data
arguments: data_tag: D55
type: string
name: disassemble_data
arguments: data_tag: D17
type: string
name: disassemble_data
arguments: data_tag: D13
type: string
name: disassemble_data
arguments: data_tag: D24
type: string

tool: D14: Please enter amount divisible by 10: 

tool: D57: %d

tool: D36: ATM machine should dispense following amount of notes:

tool: D55: 100: %d


tool: D17: 50:  %d


tool: D13: 20:  %d


tool: D24: 10:  %d


assistant: ```c
int main(void) {
  int amount, notes100, notes50, notes20, notes10;
  do {
    printf("Please enter amount divisible by 10: ");
    scanf("%d", &amount);
    printf("\n");
  } while (amount % 10 != 0);
  dispense_cash(amount, &notes100, &notes50, &notes20, &notes10);
  printf("ATM machine should dispense following amount of notes:\n");
  printf("100: %d\n", notes100);
  printf("50:  %d\n", notes50);
  printf("20:  %d\n", notes20);
  printf("10:  %d\n", notes10);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    D6(%rip),%r12
  lea    0x4(%rsp),%rbp
  lea    D9(%rip),%rbx
L15:
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x4(%rsp),%edi
  movslq %edi,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  mov    %edi,%edx
  sar    $0x1f,%edx
  sub    %edx,%eax
  lea    (%rax,%rax,4),%eax
  add    %eax,%eax
  cmp    %eax,%edi
  jne    L15
  lea    0x10(%rsp),%rcx
  lea    0xc(%rsp),%rdx
  lea    0x8(%rsp),%rsi
  lea    0x14(%rsp),%r8
  call   <dispense_cash@plt>
  lea    D28(%rip),%rdi
  call   <puts@plt>
  mov    0x8(%rsp),%esi
  lea    D54(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0xc(%rsp),%esi
  lea    D10(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x10(%rsp),%esi
  lea    D5(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x14(%rsp),%esi
  lea    D64(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  mov    $0x0,%eax
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D6
type_guessed: string
name: unpack_data
arguments: data_label: D9
type_guessed: string
name: unpack_data
arguments: data_label: D28
type_guessed: string
name: unpack_data
arguments: data_label: D54
type_guessed: string
name: unpack_data
arguments: data_label: D10
type_guessed: string
name: unpack_data
arguments: data_label: D5
type_guessed: string
name: unpack_data
arguments: data_label: D64
type_guessed: string

tool: D6: Please enter amount divisible by 10: 

tool: D9: %d

tool: D28: ATM machine should dispense following amount of notes:

tool: D54: 100: %d


tool: D10: 50:  %d


tool: D5: 20:  %d


tool: D64: 10:  %d


assistant: ```c
int main(void) {
  int amount, notes100, notes50, notes20, notes10;
  do {
    printf("Please enter amount divisible by 10: ");
    scanf("%d", &amount);
    printf("\n");
  } while (amount % 10 != 0);
  dispense_cash(amount, &notes100, &notes50, &notes20, &notes10);
  printf("ATM machine should dispense following amount of notes:\n");
  printf("100: %d\n", notes100);
  printf("50:  %d\n", notes50);
  printf("20:  %d\n", notes20);
  printf("10:  %d\n", notes10);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r12
  lea    D36(%rip),%r12
  push   %rbp
  push   %rbx
  lea    D13(%rip),%rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rbp
  nop
L16:
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x4(%rsp),%edi
  imul   $0xcccccccd,%edi,%eax
  add    $0x19999998,%eax
  ror    %eax
  cmp    $0x19999998,%eax
  ja     L16
  lea    0x10(%rsp),%rcx
  lea    0xc(%rsp),%rdx
  lea    0x14(%rsp),%r8
  lea    0x8(%rsp),%rsi
  call   <dispense_cash@plt>
  lea    D2(%rip),%rdi
  call   <puts@plt>
  mov    0x8(%rsp),%esi
  lea    D49(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0xc(%rsp),%esi
  lea    D22(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x10(%rsp),%esi
  lea    D59(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x14(%rsp),%esi
  xor    %eax,%eax
  lea    D66(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L10
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L10:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D36
datatype: string
name: disassemble_data
arguments: label: D13
datatype: string
name: disassemble_data
arguments: label: D2
datatype: string
name: disassemble_data
arguments: label: D49
datatype: string
name: disassemble_data
arguments: label: D22
datatype: string
name: disassemble_data
arguments: label: D59
datatype: string
name: disassemble_data
arguments: label: D66
datatype: string

tool: D36: Please enter amount divisible by 10: 

tool: D13: %d

tool: D2: ATM machine should dispense following amount of notes:

tool: D49: 100: %d


tool: D22: 50:  %d


tool: D59: 20:  %d


tool: D66: 10:  %d


assistant: ```c
int main(void) {
  int amount, notes100, notes50, notes20, notes10;
  do {
    printf("Please enter amount divisible by 10: ");
    scanf("%d", &amount);
    printf("\n");
  } while (amount % 10 != 0);
  dispense_cash(amount, &notes100, &notes50, &notes20, &notes10);
  printf("ATM machine should dispense following amount of notes:\n");
  printf("100: %d\n", notes100);
  printf("50:  %d\n", notes50);
  printf("20:  %d\n", notes20);
  printf("10:  %d\n", notes10);
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<main>:
  endbr64
  push   %r12
  lea    D7(%rip),%r12
  push   %rbp
  push   %rbx
  lea    D24(%rip),%rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rbp
  nop
L1:
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x4(%rsp),%edi
  imul   $0xcccccccd,%edi,%eax
  add    $0x19999998,%eax
  ror    %eax
  cmp    $0x19999998,%eax
  ja     L1
  lea    0x10(%rsp),%rcx
  lea    0xc(%rsp),%rdx
  lea    0x14(%rsp),%r8
  lea    0x8(%rsp),%rsi
  call   <dispense_cash@plt>
  lea    D11(%rip),%rdi
  call   <puts@plt>
  mov    0x8(%rsp),%esi
  lea    D50(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0xc(%rsp),%esi
  lea    D1(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x10(%rsp),%esi
  lea    D45(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x14(%rsp),%esi
  xor    %eax,%eax
  lea    D6(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L11
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L11:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D7
type: string
name: extract_symbolic_data
arguments: data_label: D24
type: string
name: extract_symbolic_data
arguments: data_label: D11
type: string
name: extract_symbolic_data
arguments: data_label: D50
type: string
name: extract_symbolic_data
arguments: data_label: D1
type: string
name: extract_symbolic_data
arguments: data_label: D45
type: string
name: extract_symbolic_data
arguments: data_label: D6
type: string

tool: D7: Please enter amount divisible by 10: 

tool: D24: %d

tool: D11: ATM machine should dispense following amount of notes:

tool: D50: 100: %d


tool: D1: 50:  %d


tool: D45: 20:  %d


tool: D6: 10:  %d


assistant: ```c
int main(void) {
  int amount, notes100, notes50, notes20, notes10;
  do {
    printf("Please enter amount divisible by 10: ");
    scanf("%d", &amount);
    printf("\n");
  } while (amount % 10 != 0);
  dispense_cash(amount, &notes100, &notes50, &notes20, &notes10);
  printf("ATM machine should dispense following amount of notes:\n");
  printf("100: %d\n", notes100);
  printf("50:  %d\n", notes50);
  printf("20:  %d\n", notes20);
  printf("10:  %d\n", notes10);
  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
<qre_il_pc_pack_item_destroy>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int qre_il_pc_pack_item_destroy(struct qre_il_pc_pack_item *qre_il_pc_pack_item) {
  free(qre_il_pc_pack_item);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<qre_il_pc_pack_item_destroy>:
  endbr64
  sub    $0x8,%rsp
  call   <free@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int qre_il_pc_pack_item_destroy(struct qre_il_pc_pack_item *qre_il_pc_pack_item) {
  free(qre_il_pc_pack_item);
  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
<qre_il_pc_pack_item_destroy>:
  endbr64
  sub    $0x8,%rsp
  call   <free@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int qre_il_pc_pack_item_destroy(struct qre_il_pc_pack_item *qre_il_pc_pack_item) {
  free(qre_il_pc_pack_item);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<qre_il_pc_pack_item_destroy>:
  endbr64
  sub    $0x8,%rsp
  call   <free@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int qre_il_pc_pack_item_destroy(struct qre_il_pc_pack_item *qre_il_pc_pack_item) {
  free(qre_il_pc_pack_item);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<timediff>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  mov    $0x0,%edx
  movd   %xmm0,%esi
  mov    %rax,%rdi
  movabs $0xffffffff00000000,%rcx
  and    %rdi,%rcx
  or     %rsi,%rcx
  mov    %rcx,%rax
  movq   %xmm1,%rdx
  mov    %rax,-0x20(%rbp)
  mov    %rdx,-0x18(%rbp)
  movaps %xmm2,%xmm1
  movapd %xmm3,%xmm0
  mov    $0x0,%eax
  mov    $0x0,%edx
  movd   %xmm1,%esi
  mov    %rax,%rdi
  movabs $0xffffffff00000000,%rcx
  and    %rdi,%rcx
  or     %rsi,%rcx
  mov    %rcx,%rax
  movq   %xmm0,%rdx
  mov    %rax,-0x30(%rbp)
  mov    %rdx,-0x28(%rbp)
  movss  -0x20(%rbp),%xmm0
  movss  -0x30(%rbp),%xmm1
  subss  %xmm1,%xmm0
  movss  %xmm0,-0x4(%rbp)
  pxor   %xmm1,%xmm1
  cvtss2sd -0x4(%rbp),%xmm1
  movsd  -0x18(%rbp),%xmm0
  movsd  -0x28(%rbp),%xmm2
  subsd  %xmm2,%xmm0
  movsd  D6(%rip),%xmm2
  divsd  %xmm2,%xmm0
  addsd  %xmm1,%xmm0
  cvtsd2ss %xmm0,%xmm0
  movss  %xmm0,-0x4(%rbp)
  movss  -0x4(%rbp),%xmm0
  pop    %rbp
  ret
```

tool_calls: name: read_data
arguments: tag: D6
data_type: f64

tool: D6: 1000000000

assistant: ```c
float timediff(struct timespec t2, struct timespec t1) {
  float diff = t2.tv_sec - t1.tv_sec;
  diff += (t2.tv_nsec - t1.tv_nsec) / 1000000000.0;
  return diff;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<timediff>:
  endbr64
  subsd  %xmm3,%xmm1
  divsd  D0(%rip),%xmm1
  subss  %xmm2,%xmm0
  cvtss2sd %xmm0,%xmm0
  addsd  %xmm0,%xmm1
  pxor   %xmm0,%xmm0
  cvtsd2ss %xmm1,%xmm0
  ret
```

tool_calls: name: parse_data
arguments: tag: D0
type_guessed: f64

tool: D0: 1000000000

assistant: ```c
float timediff(struct timespec t2, struct timespec t1) {
  float diff = t2.tv_sec - t1.tv_sec;
  diff += (t2.tv_nsec - t1.tv_nsec) / 1000000000.0;
  return diff;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<timediff>:
  endbr64
  subss  %xmm2,%xmm0
  subsd  %xmm3,%xmm1
  divsd  D8(%rip),%xmm1
  cvtss2sd %xmm0,%xmm0
  addsd  %xmm1,%xmm0
  cvtsd2ss %xmm0,%xmm0
  ret
```

tool_calls: name: retrieve_data
arguments: label: D8
datatype: f64

tool: D8: 1000000000

assistant: ```c
float timediff(struct timespec t2, struct timespec t1) {
  float diff = t2.tv_sec - t1.tv_sec;
  diff += (t2.tv_nsec - t1.tv_nsec) / 1000000000.0;
  return diff;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<timediff>:
  endbr64
  subss  %xmm2,%xmm0
  subsd  %xmm3,%xmm1
  divsd  D6(%rip),%xmm1
  cvtss2sd %xmm0,%xmm0
  addsd  %xmm1,%xmm0
  cvtsd2ss %xmm0,%xmm0
  ret
```

tool_calls: name: disassemble_data
arguments: label: D6
type_guessed: f64

tool: D6: 1000000000

assistant: ```c
float timediff(struct timespec t2, struct timespec t1) {
  float diff = t2.tv_sec - t1.tv_sec;
  diff += (t2.tv_nsec - t1.tv_nsec) / 1000000000.0;
  return diff;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<strncmp>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  jmp    L9
L26:
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  movzbl (%rax),%eax
  mov    %al,-0x2(%rbp)
  mov    -0x20(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x20(%rbp)
  movzbl (%rax),%eax
  mov    %al,-0x1(%rbp)
  movzbl -0x2(%rbp),%eax
  cmp    -0x1(%rbp),%al
  je     L42
  movsbl -0x2(%rbp),%eax
  movsbl -0x1(%rbp),%edx
  sub    %edx,%eax
  jmp    L49
L42:
  cmpb   $0x0,-0x2(%rbp)
  jne    L9
  mov    $0x0,%eax
  jmp    L49
L9:
  mov    -0x28(%rbp),%rax
  lea    -0x1(%rax),%rdx
  mov    %rdx,-0x28(%rbp)
  test   %rax,%rax
  jne    L26
  mov    $0x0,%eax
L49:
  pop    %rbp
  ret
```

assistant: ```c
int strncmp(const char *a, const char *b, size_t n) {
  while (n-- > 0) {
    char c1, c2;
    if ((c1 = *a++) != (c2 = *b++))
      return c1 - c2;
    if (c1 == 0)
      return 0;
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<strncmp>:
  endbr64
  mov    $0x0,%eax
L13:
  cmp    %rax,%rdx
  je     L10
  movzbl (%rdi,%rax,1),%ecx
  movzbl (%rsi,%rax,1),%r8d
  cmp    %r8b,%cl
  jne    L19
  add    $0x1,%rax
  test   %cl,%cl
  jne    L13
  mov    $0x0,%eax
  ret
L19:
  movsbl %cl,%eax
  movsbl %r8b,%r8d
  sub    %r8d,%eax
  ret
L10:
  mov    $0x0,%eax
  ret
```

assistant: ```c
int strncmp(const char *a, const char *b, size_t n) {
  while (n-- > 0) {
    char c1, c2;
    if ((c1 = *a++) != (c2 = *b++))
      return c1 - c2;
    if (c1 == 0)
      return 0;
  }
  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
<strncmp>:
  endbr64
  xor    %eax,%eax
  jmp    L33
  nopl   0x0(%rax,%rax,1)
L27:
  movzbl (%rdi,%rax,1),%ecx
  movsbl (%rsi,%rax,1),%r8d
  cmp    %r8b,%cl
  jne    L21
  add    $0x1,%rax
  test   %cl,%cl
  je     L30
L33:
  cmp    %rax,%rdx
  jne    L27
L30:
  xor    %eax,%eax
  ret
  xchg   %ax,%ax
L21:
  movsbl %cl,%eax
  sub    %r8d,%eax
  ret
```

assistant: ```c
int strncmp(const char *a, const char *b, size_t n) {
  while (n-- > 0) {
    char c1, c2;
    if ((c1 = *a++) != (c2 = *b++))
      return c1 - c2;
    if (c1 == 0)
      return 0;
  }
  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
<strncmp>:
  endbr64
  xor    %eax,%eax
  jmp    L37
  nopl   0x0(%rax,%rax,1)
L33:
  movzbl (%rdi,%rax,1),%ecx
  movsbl (%rsi,%rax,1),%r8d
  cmp    %r8b,%cl
  jne    L27
  add    $0x1,%rax
  test   %cl,%cl
  je     L17
L37:
  cmp    %rdx,%rax
  jne    L33
L17:
  xor    %eax,%eax
  ret
  xchg   %ax,%ax
L27:
  movsbl %cl,%eax
  sub    %r8d,%eax
  ret
```

assistant: ```c
int strncmp(const char *a, const char *b, size_t n) {
  while (n-- > 0) {
    char c1, c2;
    if ((c1 = *a++) != (c2 = *b++))
      return c1 - c2;
    if (c1 == 0)
      return 0;
  }
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<test2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  movq   $0x2328,-0x30(%rbp)
  movq   $0x3e8,-0x28(%rbp)
  mov    -0x30(%rbp),%rax
  mov    %rax,%rdi
  call   <heap_create@plt>
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <ant_alloc@plt>
  mov    %rax,-0x20(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <ant_alloc@plt>
  mov    %rax,-0x18(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <ant_alloc@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <ant_alloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    $0x0,%eax
  call   <heap_dump@plt>
  mov    -0x28(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <ant_free@plt>
  mov    -0x28(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <ant_free@plt>
  mov    $0x0,%eax
  call   <heap_dump@plt>
  mov    -0x28(%rbp),%rdx
  mov    -0x20(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <ant_free@plt>
  mov    -0x28(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <ant_free@plt>
  mov    $0x0,%eax
  call   <heap_dump@plt>
  mov    $0x0,%eax
  call   <heap_destroy@plt>
  nop
  leave
  ret
```

assistant: ```c
void test2() {
  size_t heap_sz = 9000;
  size_t request_sz = 1000;
  heap_create(heap_sz);
  void *p1 = ant_alloc(request_sz);
  void *p2 = ant_alloc(request_sz);
  void *p3 = ant_alloc(request_sz);
  void *p4 = ant_alloc(request_sz);
  heap_dump();
  ant_free(p2, request_sz);
  ant_free(p4, request_sz);
  heap_dump();
  ant_free(p1, request_sz);
  ant_free(p3, request_sz);
  heap_dump();
  heap_destroy();
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<test2>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    $0x2328,%edi
  call   <heap_create@plt>
  mov    $0x3e8,%edi
  call   <ant_alloc@plt>
  mov    %rax,%rbp
  mov    $0x3e8,%edi
  call   <ant_alloc@plt>
  mov    %rax,%r13
  mov    $0x3e8,%edi
  call   <ant_alloc@plt>
  mov    %rax,%rbx
  mov    $0x3e8,%edi
  call   <ant_alloc@plt>
  mov    %rax,%r12
  mov    $0x0,%eax
  call   <heap_dump@plt>
  mov    $0x3e8,%esi
  mov    %r13,%rdi
  call   <ant_free@plt>
  mov    $0x3e8,%esi
  mov    %r12,%rdi
  call   <ant_free@plt>
  mov    $0x0,%eax
  call   <heap_dump@plt>
  mov    $0x3e8,%esi
  mov    %rbp,%rdi
  call   <ant_free@plt>
  mov    $0x3e8,%esi
  mov    %rbx,%rdi
  call   <ant_free@plt>
  mov    $0x0,%eax
  call   <heap_dump@plt>
  mov    $0x0,%eax
  call   <heap_destroy@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
void test2() {
  size_t heap_sz = 9000;
  size_t request_sz = 1000;
  heap_create(heap_sz);
  void *p1 = ant_alloc(request_sz);
  void *p2 = ant_alloc(request_sz);
  void *p3 = ant_alloc(request_sz);
  void *p4 = ant_alloc(request_sz);
  heap_dump();
  ant_free(p2, request_sz);
  ant_free(p4, request_sz);
  heap_dump();
  ant_free(p1, request_sz);
  ant_free(p3, request_sz);
  heap_dump();
  heap_destroy();
}
```