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

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rsi
  lea    D20(%rip),%rdi
  call   <scanf@plt>
  cmpl   $0x0,0xc(%rsp)
  jle    L9
  mov    $0x1,%ebx
  lea    0x14(%rsp),%r14
  lea    0x10(%rsp),%r13
  lea    D2(%rip),%r12
  lea    D8(%rip),%rbp
L31:
  mov    %r14,%rdx
  mov    %r13,%rsi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    0x10(%rsp),%eax
  mov    0x14(%rsp),%edi
  cmp    %edi,%eax
  jg     L49
  add    $0x1,%edi
  mov    $0x0,%ecx
L3:
  mov    %eax,%esi
  shr    $0x1f,%esi
  lea    (%rax,%rsi,1),%edx
  and    $0x1,%edx
  sub    %esi,%edx
  lea    (%rcx,%rax,1),%esi
  cmp    $0x1,%edx
  cmove  %esi,%ecx
  add    $0x1,%eax
  cmp    %eax,%edi
  jne    L3
L45:
  mov    %ecx,%edx
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  add    $0x1,%ebx
  cmp    %ebx,0xc(%rsp)
  jge    L31
L9:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L11
  mov    $0x0,%eax
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L49:
  mov    $0x0,%ecx
  jmp    L45
L11:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D20
type_guessed: string
name: parse_data
arguments: data_label: D2
type_guessed: string
name: parse_data
arguments: data_label: D8
type_guessed: string

tool: D20: %d

tool: D2: %d %d

tool: D8: Case %d: %d


assistant: ```c
int main() {
  int n, i, a, b, sum = 0, j;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d %d", &a, &b);
    sum = 0;
    for (j = a; j <= b; j++) {
      if (j % 2 == 1) {
        sum = sum + j;
      }
    }
    printf("Case %d: %d\n", i, sum);
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r14
  lea    D24(%rip),%rdi
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rsi
  call   <scanf@plt>
  mov    0xc(%rsp),%eax
  test   %eax,%eax
  jle    L33
  mov    $0x1,%r14d
  lea    0x14(%rsp),%r13
  lea    0x10(%rsp),%r12
  lea    D28(%rip),%rbp
  lea    D13(%rip),%rbx
  nopl   0x0(%rax,%rax,1)
L6:
  mov    %r12,%rsi
  mov    %r13,%rdx
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    0x10(%rsp),%eax
  mov    0x14(%rsp),%esi
  cmp    %esi,%eax
  jg     L36
  add    $0x1,%esi
  xor    %r8d,%r8d
  nopw   0x0(%rax,%rax,1)
L3:
  mov    %eax,%ecx
  shr    $0x1f,%ecx
  lea    (%rax,%rcx,1),%edx
  and    $0x1,%edx
  sub    %ecx,%edx
  lea    (%r8,%rax,1),%ecx
  cmp    $0x1,%edx
  cmove  %ecx,%r8d
  add    $0x1,%eax
  cmp    %eax,%esi
  jne    L3
L23:
  mov    %r14d,%esi
  mov    %r8d,%edx
  mov    %rbx,%rdi
  add    $0x1,%r14d
  call   <printf@plt>
  cmp    %r14d,0xc(%rsp)
  jge    L6
L33:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L29
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nop
L36:
  xor    %r8d,%r8d
  jmp    L23
L29:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D24
type_infered: string
name: disassemble_data
arguments: data_tag: D28
type_infered: string
name: disassemble_data
arguments: data_tag: D13
type_infered: string

tool: D24: %d

tool: D28: %d %d

tool: D13: Case %d: %d


assistant: ```c
int main() {
  int n, i, a, b, sum = 0, j;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d %d", &a, &b);
    sum = 0;
    for (j = a; j <= b; j++) {
      if (j % 2 == 1) {
        sum = sum + j;
      }
    }
    printf("Case %d: %d\n", i, sum);
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<mean>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  pxor   %xmm0,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L10
L2:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  movsd  -0x8(%rbp),%xmm1
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  addl   $0x1,-0xc(%rbp)
L10:
  mov    -0xc(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L2
  pxor   %xmm1,%xmm1
  cvtsi2sdl -0x1c(%rbp),%xmm1
  movsd  -0x8(%rbp),%xmm0
  divsd  %xmm1,%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  pop    %rbp
  ret
```

assistant: ```c
double mean(const double *x, int dim) {
  double mean = 0;
  for (int i = 0; i < dim; i++) {
    mean += x[i];
  }
  return mean / dim;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<mean>:
  endbr64
  test   %esi,%esi
  jle    L3
  mov    %rdi,%rax
  lea    -0x1(%rsi),%edx
  lea    0x8(%rdi,%rdx,8),%rdx
  pxor   %xmm0,%xmm0
L7:
  addsd  (%rax),%xmm0
  add    $0x8,%rax
  cmp    %rdx,%rax
  jne    L7
L16:
  pxor   %xmm1,%xmm1
  cvtsi2sd %esi,%xmm1
  divsd  %xmm1,%xmm0
  ret
L3:
  pxor   %xmm0,%xmm0
  jmp    L16
```

assistant: ```c
double mean(const double *x, int dim) {
  double mean = 0;
  for (int i = 0; i < dim; i++) {
    mean += x[i];
  }
  return mean / dim;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<mean>:
  endbr64
  test   %esi,%esi
  jle    L1
  lea    -0x1(%rsi),%eax
  pxor   %xmm0,%xmm0
  lea    0x8(%rdi,%rax,8),%rax
  nopl   0x0(%rax)
L19:
  addsd  (%rdi),%xmm0
  add    $0x8,%rdi
  cmp    %rax,%rdi
  jne    L19
  pxor   %xmm1,%xmm1
  cvtsi2sd %esi,%xmm1
  divsd  %xmm1,%xmm0
  ret
  nopw   0x0(%rax,%rax,1)
L1:
  pxor   %xmm1,%xmm1
  pxor   %xmm0,%xmm0
  cvtsi2sd %esi,%xmm1
  divsd  %xmm1,%xmm0
  ret
```

assistant: ```c
double mean(const double *x, int dim) {
  double mean = 0;
  for (int i = 0; i < dim; i++) {
    mean += x[i];
  }
  return mean / dim;
}
```
user: Translate the below assembly code into C programming language:

```asm
<mean>:
  endbr64
  test   %esi,%esi
  jle    L13
  cmp    $0x1,%esi
  je     L38
  mov    %esi,%edx
  mov    %rdi,%rax
  pxor   %xmm0,%xmm0
  shr    %edx
  sub    $0x1,%edx
  shl    $0x4,%rdx
  lea    0x10(%rdi,%rdx,1),%rdx
  nopl   0x0(%rax)
L29:
  movsd  (%rax),%xmm1
  add    $0x10,%rax
  addsd  %xmm1,%xmm0
  movsd  -0x8(%rax),%xmm1
  addsd  %xmm1,%xmm0
  cmp    %rdx,%rax
  jne    L29
  mov    %esi,%eax
  and    $0xfffffffe,%eax
  test   $0x1,%sil
  je     L0
L25:
  cltq
  addsd  (%rdi,%rax,8),%xmm0
L0:
  pxor   %xmm1,%xmm1
  cvtsi2sd %esi,%xmm1
  divsd  %xmm1,%xmm0
  ret
  nopl   0x0(%rax)
L13:
  pxor   %xmm1,%xmm1
  pxor   %xmm0,%xmm0
  cvtsi2sd %esi,%xmm1
  divsd  %xmm1,%xmm0
  ret
L38:
  xor    %eax,%eax
  pxor   %xmm0,%xmm0
  jmp    L25
```

assistant: ```c
double mean(const double *x, int dim) {
  double mean = 0;
  for (int i = 0; i < dim; i++) {
    mean += x[i];
  }
  return mean / dim;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<countdownTicks>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L1
L19:
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <updateTickCount@plt>
  addl   $0x1,-0x4(%rbp)
L1:
  mov    -0x4(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L19
  nop
  nop
  leave
  ret
```

assistant: ```c
void countdownTicks(int tickNo, struct ThreadResources *thResources) {
  int i;
  for (i = 0; i < tickNo; i++) {
    updateTickCount(thResources);
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<countdownTicks>:
  endbr64
  test   %edi,%edi
  jle    L10
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%ebp
  mov    %rsi,%r12
  mov    $0x0,%ebx
L5:
  mov    %r12,%rdi
  call   <updateTickCount@plt>
  add    $0x1,%ebx
  cmp    %ebx,%ebp
  jne    L5
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L10:
  ret
```

assistant: ```c
void countdownTicks(int tickNo, struct ThreadResources *thResources) {
  int i;
  for (i = 0; i < tickNo; i++) {
    updateTickCount(thResources);
  }
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<countdownTicks>:
  endbr64
  test   %edi,%edi
  jle    L4
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  xor    %ebx,%ebx
  nopl   0x0(%rax,%rax,1)
L6:
  mov    %r12,%rdi
  add    $0x1,%ebx
  call   <updateTickCount@plt>
  cmp    %ebx,%ebp
  jne    L6
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L4:
  ret
```

assistant: ```c
void countdownTicks(int tickNo, struct ThreadResources *thResources) {
  int i;
  for (i = 0; i < tickNo; i++) {
    updateTickCount(thResources);
  }
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<countdownTicks>:
  endbr64
  test   %edi,%edi
  jle    L14
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  xor    %ebx,%ebx
  nopl   0x0(%rax,%rax,1)
L7:
  mov    %r12,%rdi
  add    $0x1,%ebx
  call   <updateTickCount@plt>
  cmp    %ebx,%ebp
  jne    L7
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L14:
  ret
```

assistant: ```c
void countdownTicks(int tickNo, struct ThreadResources *thResources) {
  int i;
  for (i = 0; i < tickNo; i++) {
    updateTickCount(thResources);
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<MergeSort>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  mov    %ecx,-0x28(%rbp)
  mov    -0x24(%rbp),%eax
  cmp    -0x28(%rbp),%eax
  jge    L9
  mov    -0x24(%rbp),%edx
  mov    -0x28(%rbp),%eax
  add    %edx,%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%ecx
  mov    -0x24(%rbp),%edx
  mov    -0x20(%rbp),%rsi
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <MergeSort>
  mov    -0x4(%rbp),%eax
  lea    0x1(%rax),%edi
  mov    -0x28(%rbp),%edx
  mov    -0x20(%rbp),%rsi
  mov    -0x18(%rbp),%rax
  mov    %edx,%ecx
  mov    %edi,%edx
  mov    %rax,%rdi
  call   <MergeSort>
  mov    -0x28(%rbp),%edi
  mov    -0x4(%rbp),%ecx
  mov    -0x24(%rbp),%edx
  mov    -0x20(%rbp),%rsi
  mov    -0x18(%rbp),%rax
  mov    %edi,%r8d
  mov    %rax,%rdi
  call   <Merge@plt>
L9:
  nop
  leave
  ret
```

assistant: ```c
void MergeSort(int array[], int tmp[], int start, int end) {
  int middle;
  if (start < end) {
    middle = (start + end) / 2;
    MergeSort(array, tmp, start, middle);
    MergeSort(array, tmp, middle + 1, end);
    Merge(array, tmp, start, middle, end);
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<MergeSort>:
  endbr64
  cmp    %ecx,%edx
  jl     L9
  ret
L9:
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    %rsi,%r13
  mov    %edx,%ebx
  mov    %ecx,%ebp
  lea    (%rdx,%rcx,1),%eax
  mov    %eax,%r14d
  shr    $0x1f,%r14d
  add    %eax,%r14d
  sar    %r14d
  mov    %r14d,%ecx
  call   <MergeSort>
  lea    0x1(%r14),%edx
  mov    %ebp,%ecx
  mov    %r13,%rsi
  mov    %r12,%rdi
  call   <MergeSort>
  mov    %ebp,%r8d
  mov    %r14d,%ecx
  mov    %ebx,%edx
  mov    %r13,%rsi
  mov    %r12,%rdi
  call   <Merge@plt>
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
void MergeSort(int array[], int tmp[], int start, int end) {
  int middle;
  if (start < end) {
    middle = (start + end) / 2;
    MergeSort(array, tmp, start, middle);
    MergeSort(array, tmp, middle + 1, end);
    Merge(array, tmp, start, middle, end);
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<MergeSort>:
  endbr64
  cmp    %ecx,%edx
  jl     L1
  ret
  nopl   0x0(%rax)
L1:
  lea    (%rdx,%rcx,1),%eax
  push   %r14
  mov    %rdi,%r14
  push   %r13
  mov    %eax,%r13d
  shr    $0x1f,%r13d
  push   %r12
  mov    %edx,%r12d
  push   %rbp
  add    %eax,%r13d
  mov    %rsi,%rbp
  push   %rbx
  sar    %r13d
  mov    %ecx,%ebx
  mov    %r13d,%ecx
  call   <MergeSort>
  lea    0x1(%r13),%edx
  mov    %ebx,%ecx
  mov    %rbp,%rsi
  mov    %r14,%rdi
  call   <MergeSort>
  mov    %ebx,%r8d
  mov    %r13d,%ecx
  pop    %rbx
  mov    %r12d,%edx
  mov    %rbp,%rsi
  mov    %r14,%rdi
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  jmp    1050 <Merge@plt>
```

assistant: ```c
void MergeSort(int array[], int tmp[], int start, int end) {
  int middle;
  if (start < end) {
    middle = (start + end) / 2;
    MergeSort(array, tmp, start, middle);
    MergeSort(array, tmp, middle + 1, end);
    Merge(array, tmp, start, middle, end);
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<open_udp_socket>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x1,-0x1c(%rbp)
  mov    D20(%rip),%rax
  mov    (%rax),%edx
  mov    D11(%rip),%rax
  mov    (%rax),%ecx
  mov    D60(%rip),%rax
  mov    (%rax),%eax
  mov    %ecx,%esi
  mov    %eax,%edi
  call   <socket@plt>
  mov    %eax,-0x18(%rbp)
  cmpl   $0xffffffff,-0x18(%rbp)
  jne    L38
  mov    $0xffffffff,%eax
  jmp    L21
L38:
  mov    D67(%rip),%rax
  mov    (%rax),%edx
  mov    D47(%rip),%rax
  mov    (%rax),%esi
  lea    -0x1c(%rbp),%rcx
  mov    -0x18(%rbp),%eax
  mov    $0x4,%r8d
  mov    %eax,%edi
  call   <setsockopt@plt>
  test   %rax,%rax
  je     L68
  mov    $0xffffffff,%eax
  jmp    L21
L68:
  lea    -0x14(%rbp),%rax
  mov    $0xc,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  mov    D60(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x24(%rbp),%eax
  mov    %eax,%edi
  call   <htons@plt>
  mov    %eax,-0x10(%rbp)
  mov    D63(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,%edi
  call   <htonl@plt>
  mov    %eax,-0x14(%rbp)
  lea    -0x14(%rbp),%rcx
  mov    -0x18(%rbp),%eax
  mov    $0xc,%edx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <bind@plt>
  test   %rax,%rax
  je     L26
  mov    $0xffffffff,%eax
  jmp    L21
L26:
  mov    -0x18(%rbp),%eax
L21:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L31
  call   <__stack_chk_fail@plt>
L31:
  leave
  ret
```

assistant: ```c
int open_udp_socket(int port) {
  struct sockaddr_in si_me;
  int s;
  int reuse = 1;
  if ((s = socket(AF_INET, SOCK_DGRAM, IPPROTO_UDP)) == -1)
    return (-1);
  if (setsockopt(s, SOL_SOCKET, SO_REUSEADDR, (const char *)&reuse, sizeof(reuse)) != 0)
    return (-1);
  memset((char *)&si_me, 0, sizeof(si_me));
  si_me.sin_family = AF_INET;
  si_me.sin_port = htons(port);
  si_me.sin_addr.s_addr = htonl(INADDR_ANY);
  if (bind(s, (struct sockaddr *)&si_me, sizeof(si_me)) != 0)
    return (-1);
  return s;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<open_udp_socket>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %edi,%r12d
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movl   $0x1,0x8(%rsp)
  mov    D11(%rip),%rax
  mov    (%rax),%edx
  mov    D4(%rip),%rax
  mov    (%rax),%esi
  mov    D31(%rip),%rax
  mov    (%rax),%edi
  call   <socket@plt>
  mov    %eax,%ebx
  cmp    $0xffffffff,%eax
  je     L12
  lea    0x8(%rsp),%rcx
  mov    $0x4,%r8d
  mov    D16(%rip),%rax
  mov    (%rax),%edx
  mov    D62(%rip),%rax
  mov    (%rax),%esi
  mov    %ebx,%edi
  call   <setsockopt@plt>
  test   %rax,%rax
  jne    L0
  lea    0xc(%rsp),%rbp
  mov    $0xc,%edx
  mov    $0x0,%esi
  mov    %rbp,%rdi
  call   <memset@plt>
  mov    D31(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,0x14(%rsp)
  mov    %r12d,%edi
  call   <htons@plt>
  mov    %eax,0x10(%rsp)
  mov    D63(%rip),%rax
  mov    (%rax),%edi
  call   <htonl@plt>
  mov    %eax,0xc(%rsp)
  mov    $0xc,%edx
  mov    %rbp,%rsi
  mov    %ebx,%edi
  call   <bind@plt>
  test   %rax,%rax
  mov    $0xffffffff,%eax
  cmovne %eax,%ebx
L12:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L14
  mov    %ebx,%eax
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L0:
  mov    $0xffffffff,%ebx
  jmp    L12
L14:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int open_udp_socket(int port) {
  struct sockaddr_in si_me;
  int s;
  int reuse = 1;
  if ((s = socket(AF_INET, SOCK_DGRAM, IPPROTO_UDP)) == -1)
    return (-1);
  if (setsockopt(s, SOL_SOCKET, SO_REUSEADDR, (const char *)&reuse, sizeof(reuse)) != 0)
    return (-1);
  memset((char *)&si_me, 0, sizeof(si_me));
  si_me.sin_family = AF_INET;
  si_me.sin_port = htons(port);
  si_me.sin_addr.s_addr = htonl(INADDR_ANY);
  if (bind(s, (struct sockaddr *)&si_me, sizeof(si_me)) != 0)
    return (-1);
  return s;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<open_udp_socket>:
  endbr64
  push   %r13
  mov    %edi,%r13d
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    D6(%rip),%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    D23(%rip),%rax
  movl   $0x1,0x8(%rsp)
  mov    (%rbx),%edi
  mov    (%rax),%edx
  mov    D8(%rip),%rax
  mov    (%rax),%esi
  call   <socket@plt>
  cmp    $0xffffffff,%eax
  je     L42
  mov    %eax,%r12d
  mov    D15(%rip),%rax
  lea    0x8(%rsp),%rcx
  mov    $0x4,%r8d
  mov    %r12d,%edi
  mov    (%rax),%edx
  mov    D12(%rip),%rax
  mov    (%rax),%esi
  call   <setsockopt@plt>
  test   %rax,%rax
  jne    L42
  lea    0xc(%rsp),%rbp
  xor    %esi,%esi
  mov    $0xc,%edx
  mov    %rbp,%rdi
  call   <memset@plt>
  mov    (%rbx),%eax
  mov    %r13d,%edi
  mov    %eax,0x14(%rsp)
  call   <htons@plt>
  mov    %eax,0x10(%rsp)
  mov    D50(%rip),%rax
  mov    (%rax),%edi
  call   <htonl@plt>
  mov    $0xc,%edx
  mov    %rbp,%rsi
  mov    %r12d,%edi
  mov    %eax,0xc(%rsp)
  call   <bind@plt>
  test   %rax,%rax
  jne    L42
L10:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L46
  add    $0x28,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L42:
  mov    $0xffffffff,%r12d
  jmp    L10
L46:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int open_udp_socket(int port) {
  struct sockaddr_in si_me;
  int s;
  int reuse = 1;
  if ((s = socket(AF_INET, SOCK_DGRAM, IPPROTO_UDP)) == -1)
    return (-1);
  if (setsockopt(s, SOL_SOCKET, SO_REUSEADDR, (const char *)&reuse, sizeof(reuse)) != 0)
    return (-1);
  memset((char *)&si_me, 0, sizeof(si_me));
  si_me.sin_family = AF_INET;
  si_me.sin_port = htons(port);
  si_me.sin_addr.s_addr = htonl(INADDR_ANY);
  if (bind(s, (struct sockaddr *)&si_me, sizeof(si_me)) != 0)
    return (-1);
  return s;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<open_udp_socket>:
  endbr64
  push   %r13
  mov    %edi,%r13d
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    D45(%rip),%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    D0(%rip),%rax
  movl   $0x1,0x8(%rsp)
  mov    (%rbx),%edi
  mov    (%rax),%edx
  mov    D5(%rip),%rax
  mov    (%rax),%esi
  call   <socket@plt>
  cmp    $0xffffffff,%eax
  je     L31
  mov    %eax,%r12d
  mov    D14(%rip),%rax
  lea    0x8(%rsp),%rcx
  mov    $0x4,%r8d
  mov    %r12d,%edi
  mov    (%rax),%edx
  mov    D39(%rip),%rax
  mov    (%rax),%esi
  call   <setsockopt@plt>
  test   %rax,%rax
  jne    L31
  lea    0xc(%rsp),%rbp
  xor    %esi,%esi
  mov    $0xc,%edx
  mov    %rbp,%rdi
  call   <memset@plt>
  mov    (%rbx),%eax
  mov    %r13d,%edi
  mov    %eax,0x14(%rsp)
  call   <htons@plt>
  mov    %eax,0x10(%rsp)
  mov    D25(%rip),%rax
  mov    (%rax),%edi
  call   <htonl@plt>
  mov    $0xc,%edx
  mov    %rbp,%rsi
  mov    %r12d,%edi
  mov    %eax,0xc(%rsp)
  call   <bind@plt>
  test   %rax,%rax
  jne    L31
L0:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L12
  add    $0x28,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L31:
  mov    $0xffffffff,%r12d
  jmp    L0
L12:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int open_udp_socket(int port) {
  struct sockaddr_in si_me;
  int s;
  int reuse = 1;
  if ((s = socket(AF_INET, SOCK_DGRAM, IPPROTO_UDP)) == -1)
    return (-1);
  if (setsockopt(s, SOL_SOCKET, SO_REUSEADDR, (const char *)&reuse, sizeof(reuse)) != 0)
    return (-1);
  memset((char *)&si_me, 0, sizeof(si_me));
  si_me.sin_family = AF_INET;
  si_me.sin_port = htons(port);
  si_me.sin_addr.s_addr = htonl(INADDR_ANY);
  if (bind(s, (struct sockaddr *)&si_me, sizeof(si_me)) != 0)
    return (-1);
  return s;
}
```
user: Translate the below assembly code into C programming language:

```asm
<split_get_size>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L26
  mov    $0x0,%eax
  jmp    L27
L7:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  cmp    %eax,-0x1c(%rbp)
  jne    L21
  addl   $0x1,-0x8(%rbp)
L21:
  addl   $0x1,-0x4(%rbp)
L26:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L7
  mov    -0x8(%rbp),%eax
  add    $0x1,%eax
L27:
  pop    %rbp
  ret
```

assistant: ```c
int split_get_size(char *buffer, int delimeter) {
  int c = 0, i = 0;
  if (buffer[0] == 0)
    return 0;
  while (buffer[i] != 0) {
    if (buffer[i] == delimeter)
      c++;
    i++;
  }
  return c + 1;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<split_get_size>:
  endbr64
  movzbl (%rdi),%edx
  mov    $0x0,%eax
  test   %dl,%dl
  je     L16
  add    $0x1,%rdi
L10:
  movsbl %dl,%edx
  cmp    %esi,%edx
  sete   %dl
  movzbl %dl,%edx
  add    %edx,%eax
  add    $0x1,%rdi
  movzbl -0x1(%rdi),%edx
  test   %dl,%dl
  jne    L10
  add    $0x1,%eax
L16:
  ret
```

assistant: ```c
int split_get_size(char *buffer, int delimeter) {
  int c = 0, i = 0;
  if (buffer[0] == 0)
    return 0;
  while (buffer[i] != 0) {
    if (buffer[i] == delimeter)
      c++;
    i++;
  }
  return c + 1;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<split_get_size>:
  endbr64
  movsbl (%rdi),%edx
  xor    %eax,%eax
  test   %dl,%dl
  je     L14
  add    $0x1,%rdi
  nopl   0x0(%rax)
L15:
  cmp    %esi,%edx
  sete   %dl
  add    $0x1,%rdi
  movzbl %dl,%edx
  add    %edx,%eax
  movsbl -0x1(%rdi),%edx
  test   %dl,%dl
  jne    L15
  add    $0x1,%eax
L14:
  ret
```

assistant: ```c
int split_get_size(char *buffer, int delimeter) {
  int c = 0, i = 0;
  if (buffer[0] == 0)
    return 0;
  while (buffer[i] != 0) {
    if (buffer[i] == delimeter)
      c++;
    i++;
  }
  return c + 1;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<split_get_size>:
  endbr64
  movsbl (%rdi),%edx
  xor    %eax,%eax
  test   %dl,%dl
  je     L2
  add    $0x1,%rdi
  nopl   0x0(%rax)
L10:
  cmp    %edx,%esi
  sete   %dl
  add    $0x1,%rdi
  movzbl %dl,%edx
  add    %edx,%eax
  movsbl -0x1(%rdi),%edx
  test   %dl,%dl
  jne    L10
  add    $0x1,%eax
L2:
  ret
```

assistant: ```c
int split_get_size(char *buffer, int delimeter) {
  int c = 0, i = 0;
  if (buffer[0] == 0)
    return 0;
  while (buffer[i] != 0) {
    if (buffer[i] == delimeter)
      c++;
    i++;
  }
  return c + 1;
}
```