ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请从给定的汇编指令推导出C源代码：

```asm
<square_distance>:
  endbr64
  movss  (%rdi),%xmm0
  subss  (%rsi),%xmm0
  cvtss2sd %xmm0,%xmm0
  movss  0x4(%rdi),%xmm1
  subss  0x4(%rsi),%xmm1
  cvtss2sd %xmm1,%xmm1
  mulsd  %xmm0,%xmm0
  mulsd  %xmm1,%xmm1
  addsd  %xmm1,%xmm0
  cvtsd2ss %xmm0,%xmm0
  ret
```

assistant: ```c
float square_distance(float const a[2], float const b[2]) {
  double dx, dy;
  dx = a[0] - b[0];
  dy = a[1] - b[1];
  return dx * dx + dy * dy;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<square_distance>:
  endbr64
  movss  (%rdi),%xmm0
  movss  0x4(%rdi),%xmm1
  subss  (%rsi),%xmm0
  subss  0x4(%rsi),%xmm1
  cvtss2sd %xmm0,%xmm0
  cvtss2sd %xmm1,%xmm1
  mulsd  %xmm0,%xmm0
  mulsd  %xmm1,%xmm1
  addsd  %xmm1,%xmm0
  cvtsd2ss %xmm0,%xmm0
  ret
```

assistant: ```c
float square_distance(float const a[2], float const b[2]) {
  double dx, dy;
  dx = a[0] - b[0];
  dy = a[1] - b[1];
  return dx * dx + dy * dy;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<square_distance>:
  endbr64
  movss  (%rdi),%xmm0
  movss  0x4(%rdi),%xmm1
  subss  (%rsi),%xmm0
  subss  0x4(%rsi),%xmm1
  cvtss2sd %xmm0,%xmm0
  cvtss2sd %xmm1,%xmm1
  mulsd  %xmm0,%xmm0
  mulsd  %xmm1,%xmm1
  addsd  %xmm1,%xmm0
  cvtsd2ss %xmm0,%xmm0
  ret
```

assistant: ```c
float square_distance(float const a[2], float const b[2]) {
  double dx, dy;
  dx = a[0] - b[0];
  dy = a[1] - b[1];
  return dx * dx + dy * dy;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x48,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  mov    $0x0,%r15d
  mov    $0x1,%r12d
  lea    0x10(%rsp),%rax
  mov    %rax,0x8(%rsp)
  lea    D48(%rip),%r14
  lea    D17(%rip),%r13
  jmp    L12
L48:
  add    $0x1,%r12d
  add    $0x3,%r15
  cmp    $0x4,%r12d
  je     L46
L12:
  mov    0x8(%rsp),%rax
  lea    (%rax,%r15,4),%rbp
  mov    $0x0,%ebx
L9:
  add    $0x1,%ebx
  mov    %r12d,%edx
  mov    %ebx,%esi
  mov    %r14,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
  add    $0x4,%rbp
  cmp    $0x3,%ebx
  jne    L9
  jmp    L48
L46:
  mov    0x10(%rsp),%eax
  mov    %eax,%edx
  add    0x14(%rsp),%edx
  mov    0x18(%rsp),%esi
  mov    %esi,%ebp
  add    0x24(%rsp),%ebp
  add    0x30(%rsp),%ebp
  add    0x20(%rsp),%eax
  add    0x30(%rsp),%eax
  mov    %eax,%ebx
  add    %edx,%esi
  lea    D18(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %ebp,%esi
  lea    D28(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %ebx,%esi
  lea    D20(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L47
  mov    $0x0,%eax
  add    $0x48,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L47:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D48
type_infered: string
name: unpack_data
arguments: data_tag: D17
type_infered: string
name: unpack_data
arguments: data_tag: D18
type_infered: string
name: unpack_data
arguments: data_tag: D28
type_infered: string
name: unpack_data
arguments: data_tag: D20
type_infered: string

tool: D48: Digite o %d numero da %d coluna

tool: D17: %d

tool: D18: a soma dos valores da primeira linha e %d


tool: D28: a soma dos valores da terceira coluna e %d


tool: D20: a soma dos valores da diagonal e%d


assistant: ```c
int main() {
  int matriz[3][3], i, j, soma1 = 0, soma2 = 0, soma3 = 0, v = 0;
  for (i = 0; i < 3; i++) {
    for (j = 0; j < 3; j++) {
      printf("Digite o %d numero da %d coluna", j + 1, i + 1);
      scanf("%d", &matriz[i][j]);
    }
  }
  for (i = 0; i < 3; i++) {
    soma1 = soma1 + matriz[0][i];
  }
  for (i = 0; i < 3; i++) {
    soma2 = soma2 + matriz[i][2];
  }
  for (i = 0; i < 3; i++) {
    soma3 = soma3 + matriz[i][v + i];
  }
  printf("a soma dos valores da primeira linha e %d\n", soma1);
  printf("a soma dos valores da terceira coluna e %d\n", soma2);
  printf("a soma dos valores da diagonal e%d\n", soma3);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r15
  xor    %r15d,%r15d
  push   %r14
  xor    %r14d,%r14d
  push   %r13
  lea    D47(%rip),%r13
  push   %r12
  lea    D9(%rip),%r12
  push   %rbp
  push   %rbx
  sub    $0x48,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rax
  mov    %rax,0x8(%rsp)
L22:
  mov    0x8(%rsp),%rax
  add    $0x1,%r15d
  xor    %ebx,%ebx
  lea    (%rax,%r14,4),%rbp
L23:
  add    $0x1,%ebx
  mov    %r15d,%edx
  mov    %r13,%rdi
  xor    %eax,%eax
  mov    %ebx,%esi
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %r12,%rdi
  add    $0x4,%rbp
  call   <scanf@plt>
  cmp    $0x3,%ebx
  jne    L23
  add    $0x3,%r14
  cmp    $0x3,%r15d
  jne    L22
  mov    0x10(%rsp),%eax
  mov    0x14(%rsp),%esi
  lea    D17(%rip),%rdi
  mov    0x18(%rsp),%edx
  mov    0x24(%rsp),%r12d
  add    %eax,%esi
  add    0x20(%rsp),%eax
  add    0x30(%rsp),%eax
  add    %edx,%r12d
  add    %edx,%esi
  add    0x30(%rsp),%r12d
  mov    %eax,%ebp
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r12d,%esi
  lea    D25(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  mov    %ebp,%esi
  lea    D18(%rip),%rdi
  call   <printf@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L17
  add    $0x48,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L17:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D47
datatype: string
name: unpack_data
arguments: tag: D9
datatype: string
name: unpack_data
arguments: tag: D17
datatype: string
name: unpack_data
arguments: tag: D25
datatype: string
name: unpack_data
arguments: tag: D18
datatype: string

tool: D47: Digite o %d numero da %d coluna

tool: D9: %d

tool: D17: a soma dos valores da primeira linha e %d


tool: D25: a soma dos valores da terceira coluna e %d


tool: D18: a soma dos valores da diagonal e%d


assistant: ```c
int main() {
  int matriz[3][3], i, j, soma1 = 0, soma2 = 0, soma3 = 0, v = 0;
  for (i = 0; i < 3; i++) {
    for (j = 0; j < 3; j++) {
      printf("Digite o %d numero da %d coluna", j + 1, i + 1);
      scanf("%d", &matriz[i][j]);
    }
  }
  for (i = 0; i < 3; i++) {
    soma1 = soma1 + matriz[0][i];
  }
  for (i = 0; i < 3; i++) {
    soma2 = soma2 + matriz[i][2];
  }
  for (i = 0; i < 3; i++) {
    soma3 = soma3 + matriz[i][v + i];
  }
  printf("a soma dos valores da primeira linha e %d\n", soma1);
  printf("a soma dos valores da terceira coluna e %d\n", soma2);
  printf("a soma dos valores da diagonal e%d\n", soma3);
  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   %r15
  xor    %r15d,%r15d
  push   %r14
  xor    %r14d,%r14d
  push   %r13
  lea    D13(%rip),%r13
  push   %r12
  lea    D20(%rip),%r12
  push   %rbp
  push   %rbx
  sub    $0x48,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rax
  mov    %rax,0x8(%rsp)
L7:
  mov    0x8(%rsp),%rax
  add    $0x1,%r15d
  xor    %ebx,%ebx
  lea    (%rax,%r14,4),%rbp
L17:
  add    $0x1,%ebx
  mov    %r15d,%edx
  mov    %r13,%rdi
  xor    %eax,%eax
  mov    %ebx,%esi
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %r12,%rdi
  add    $0x4,%rbp
  call   <scanf@plt>
  cmp    $0x3,%ebx
  jne    L17
  add    $0x3,%r14
  cmp    $0x3,%r15d
  jne    L7
  mov    0x10(%rsp),%eax
  mov    0x14(%rsp),%esi
  lea    D42(%rip),%rdi
  mov    0x18(%rsp),%edx
  mov    0x24(%rsp),%r12d
  add    %eax,%esi
  add    0x20(%rsp),%eax
  add    0x30(%rsp),%eax
  add    %edx,%r12d
  add    %edx,%esi
  add    0x30(%rsp),%r12d
  mov    %eax,%ebp
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r12d,%esi
  lea    D44(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  mov    %ebp,%esi
  lea    D34(%rip),%rdi
  call   <printf@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L18
  add    $0x48,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L18:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D13
data_type: string
name: parse_data
arguments: data_label: D20
data_type: string
name: parse_data
arguments: data_label: D42
data_type: string
name: parse_data
arguments: data_label: D44
data_type: string
name: parse_data
arguments: data_label: D34
data_type: string

tool: D13: Digite o %d numero da %d coluna

tool: D20: %d

tool: D42: a soma dos valores da primeira linha e %d


tool: D44: a soma dos valores da terceira coluna e %d


tool: D34: a soma dos valores da diagonal e%d


assistant: ```c
int main() {
  int matriz[3][3], i, j, soma1 = 0, soma2 = 0, soma3 = 0, v = 0;
  for (i = 0; i < 3; i++) {
    for (j = 0; j < 3; j++) {
      printf("Digite o %d numero da %d coluna", j + 1, i + 1);
      scanf("%d", &matriz[i][j]);
    }
  }
  for (i = 0; i < 3; i++) {
    soma1 = soma1 + matriz[0][i];
  }
  for (i = 0; i < 3; i++) {
    soma2 = soma2 + matriz[i][2];
  }
  for (i = 0; i < 3; i++) {
    soma3 = soma3 + matriz[i][v + i];
  }
  printf("a soma dos valores da primeira linha e %d\n", soma1);
  printf("a soma dos valores da terceira coluna e %d\n", soma2);
  printf("a soma dos valores da diagonal e%d\n", soma3);
  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
<stinger_init>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D17(%rip),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x24(%rbp),%eax
  mov    $0x5,%edx
  lea    D16(%rip),%rcx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <write@plt>
  cmp    $0x5,%eax
  je     L47
  mov    $0xffffffff,%eax
  jmp    L12
L47:
  movl   $0x0,-0x14(%rbp)
  jmp    L60
L18:
  lea    -0x15(%rbp),%rcx
  mov    -0x24(%rbp),%eax
  mov    $0xc8,%edx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <readchar@plt>
  test   %rax,%rax
  jne    L64
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%edx
  movzbl -0x15(%rbp),%eax
  cmp    %al,%dl
  je     L9
L64:
  mov    $0xffffffff,%eax
  jmp    L12
L9:
  addl   $0x1,-0x14(%rbp)
L60:
  cmpl   $0xf,-0x14(%rbp)
  jle    L18
  mov    $0x0,%eax
L12:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L31
  call   <__stack_chk_fail@plt>
L31:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: label: D17
type_guessed: string
name: retrieve_data
arguments: label: D16
type_guessed: string

tool: D17: 
0600520058C272

tool: D16:  E5E5

assistant: ```c
int stinger_init(int fd, long *id, long *extra) {
  int i;
  unsigned char c;
  unsigned char *response = "\r\n0600520058C272";
  if (write(fd, " E5E5", 5) != 5)
    return -1;
  for (i = 0; i < 16; i++)
    if (readchar(fd, &c, 200) || (c != response[i]))
      return -1;
  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
<stinger_init>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %edi,%ebp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    $0x5,%edx
  lea    D19(%rip),%rsi
  call   <write@plt>
  cmp    $0x5,%eax
  jne    L57
  lea    D4(%rip),%rbx
  lea    0x10(%rbx),%r13
  lea    0x7(%rsp),%r12
L8:
  mov    $0xc8,%edx
  mov    %r12,%rsi
  mov    %ebp,%edi
  call   <readchar@plt>
  test   %rax,%rax
  jne    L30
  movzbl 0x7(%rsp),%eax
  cmp    %al,(%rbx)
  jne    L2
  add    $0x1,%rbx
  cmp    %r13,%rbx
  jne    L8
  mov    $0x0,%eax
L1:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L9
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L57:
  mov    $0xffffffff,%eax
  jmp    L1
L30:
  mov    $0xffffffff,%eax
  jmp    L1
L2:
  mov    $0xffffffff,%eax
  jmp    L1
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D19
datatype: string
name: extract_symbolic_data
arguments: tag: D4
datatype: string

tool: D19:  E5E5

tool: D4: 
0600520058C272

assistant: ```c
int stinger_init(int fd, long *id, long *extra) {
  int i;
  unsigned char c;
  unsigned char *response = "\r\n0600520058C272";
  if (write(fd, " E5E5", 5) != 5)
    return -1;
  for (i = 0; i < 16; i++)
    if (readchar(fd, &c, 200) || (c != response[i]))
      return -1;
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<stinger_init>:
  endbr64
  push   %r13
  mov    $0x5,%edx
  lea    D8(%rip),%rsi
  push   %r12
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <write@plt>
  cmp    $0x5,%eax
  jne    L17
  lea    D16(%rip),%rbx
  lea    0x7(%rsp),%r12
  lea    0x10(%rbx),%r13
  cs nopw 0x0(%rax,%rax,1)
L58:
  mov    $0xc8,%edx
  mov    %r12,%rsi
  mov    %ebp,%edi
  call   <readchar@plt>
  test   %rax,%rax
  jne    L17
  movzbl 0x7(%rsp),%eax
  cmp    %al,(%rbx)
  jne    L17
  add    $0x1,%rbx
  cmp    %r13,%rbx
  jne    L58
  xor    %eax,%eax
  jmp    L6
  nopw   0x0(%rax,%rax,1)
L17:
  mov    $0xffffffff,%eax
L6:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L29
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L29:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D8
type_infered: string
name: unpack_data
arguments: label: D16
type_infered: string

tool: D8:  E5E5

tool: D16: 
0600520058C272

assistant: ```c
int stinger_init(int fd, long *id, long *extra) {
  int i;
  unsigned char c;
  unsigned char *response = "\r\n0600520058C272";
  if (write(fd, " E5E5", 5) != 5)
    return -1;
  for (i = 0; i < 16; i++)
    if (readchar(fd, &c, 200) || (c != response[i]))
      return -1;
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<stinger_init>:
  endbr64
  push   %r13
  mov    $0x5,%edx
  lea    D11(%rip),%rsi
  push   %r12
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <write@plt>
  cmp    $0x5,%eax
  jne    L54
  lea    D4(%rip),%rbx
  lea    0x7(%rsp),%r12
  lea    0x10(%rbx),%r13
  cs nopw 0x0(%rax,%rax,1)
L32:
  mov    $0xc8,%edx
  mov    %r12,%rsi
  mov    %ebp,%edi
  call   <readchar@plt>
  test   %rax,%rax
  jne    L54
  movzbl 0x7(%rsp),%eax
  cmp    %al,(%rbx)
  jne    L54
  add    $0x1,%rbx
  cmp    %r13,%rbx
  jne    L32
  xor    %eax,%eax
  jmp    L47
  nopw   0x0(%rax,%rax,1)
L54:
  mov    $0xffffffff,%eax
L47:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L34
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L34:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D11
type: string
name: read_data
arguments: data_tag: D4
type: string

tool: D11:  E5E5

tool: D4: 
0600520058C272

assistant: ```c
int stinger_init(int fd, long *id, long *extra) {
  int i;
  unsigned char c;
  unsigned char *response = "\r\n0600520058C272";
  if (write(fd, " E5E5", 5) != 5)
    return -1;
  for (i = 0; i < 16; i++)
    if (readchar(fd, &c, 200) || (c != response[i]))
      return -1;
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<a_sprintf>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xd0,%rsp
  mov    %rdi,-0xc8(%rbp)
  mov    %rsi,-0xd0(%rbp)
  mov    %rdx,-0xa0(%rbp)
  mov    %rcx,-0x98(%rbp)
  mov    %r8,-0x90(%rbp)
  mov    %r9,-0x88(%rbp)
  test   %al,%al
  je     L2
  movaps %xmm0,-0x80(%rbp)
  movaps %xmm1,-0x70(%rbp)
  movaps %xmm2,-0x60(%rbp)
  movaps %xmm3,-0x50(%rbp)
  movaps %xmm4,-0x40(%rbp)
  movaps %xmm5,-0x30(%rbp)
  movaps %xmm6,-0x20(%rbp)
  movaps %xmm7,-0x10(%rbp)
L2:
  mov    -0xd0(%rbp),%rdx
  mov    -0xbc(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <va_start@plt>
  mov    -0xbc(%rbp),%edx
  mov    -0xd0(%rbp),%rcx
  mov    -0xc8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <a_sprintfv@plt>
  mov    %rax,-0xb8(%rbp)
  mov    -0xbc(%rbp),%eax
  mov    %eax,%edi
  call   <va_end@plt>
  mov    -0xb8(%rbp),%rax
  leave
  ret
```

assistant: ```c
int a_sprintf(char **buf, const char *format, ...) {
  size_t ret;
  va_list lst;
  va_start(lst, format);
  ret = a_sprintfv(buf, format, lst);
  va_end(lst);
  return ret;
}
```
user: What is the c source code of the assembly code below?

```asm
<a_sprintf>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    %rsi,%rbx
  mov    $0x0,%ebp
  mov    %ebp,%edi
  call   <va_start@plt>
  mov    %ebp,%edx
  mov    %rbx,%rsi
  mov    %r12,%rdi
  call   <a_sprintfv@plt>
  mov    %rax,%rbx
  mov    %ebp,%edi
  call   <va_end@plt>
  mov    %ebx,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int a_sprintf(char **buf, const char *format, ...) {
  size_t ret;
  va_list lst;
  va_start(lst, format);
  ret = a_sprintfv(buf, format, lst);
  va_end(lst);
  return ret;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<a_sprintf>:
  endbr64
  push   %r13
  mov    %rdi,%r13
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  mov    %r12d,%edi
  mov    %rsi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <va_start@plt>
  mov    %r12d,%edx
  mov    %rbp,%rsi
  mov    %r13,%rdi
  call   <a_sprintfv@plt>
  mov    %r12d,%edi
  mov    %rax,%rbx
  call   <va_end@plt>
  add    $0x8,%rsp
  mov    %ebx,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int a_sprintf(char **buf, const char *format, ...) {
  size_t ret;
  va_list lst;
  va_start(lst, format);
  ret = a_sprintfv(buf, format, lst);
  va_end(lst);
  return ret;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<a_sprintf>:
  endbr64
  push   %r13
  mov    %rdi,%r13
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  mov    %r12d,%edi
  mov    %rsi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <va_start@plt>
  mov    %r12d,%edx
  mov    %rbp,%rsi
  mov    %r13,%rdi
  call   <a_sprintfv@plt>
  mov    %r12d,%edi
  mov    %rax,%rbx
  call   <va_end@plt>
  add    $0x8,%rsp
  mov    %ebx,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int a_sprintf(char **buf, const char *format, ...) {
  size_t ret;
  va_list lst;
  va_start(lst, format);
  ret = a_sprintfv(buf, format, lst);
  va_end(lst);
  return ret;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<sort>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L2
L5:
  movl   $0x0,-0x8(%rbp)
  jmp    L25
L43:
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x8(%rbp),%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,4),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%eax
  cmp    %eax,%edx
  jle    L33
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x8(%rbp),%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x8(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  mov    -0x8(%rbp),%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  mov    %eax,(%rdx)
L33:
  addl   $0x1,-0x8(%rbp)
L25:
  mov    -0x1c(%rbp),%eax
  sub    $0x1,%eax
  sub    -0xc(%rbp),%eax
  cmp    %eax,-0x8(%rbp)
  jl     L43
  addl   $0x1,-0xc(%rbp)
L2:
  mov    -0x1c(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0xc(%rbp)
  jl     L5
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void sort(int numbers[4], int totalNumbers) {
  int i, j, temporaryNumber;
  for (i = 0; i < totalNumbers - 1; i++) {
    for (j = 0; j < (totalNumbers - 1 - i); j++) {
      if (numbers[j] > numbers[j + 1]) {
        temporaryNumber = numbers[j];
        numbers[j] = numbers[j + 1];
        numbers[j + 1] = temporaryNumber;
      }
    }
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<sort>:
  endbr64
  lea    -0x1(%rsi),%r8d
  test   %r8d,%r8d
  jle    L57
  lea    0x4(%rdi),%r9
  jmp    L45
L36:
  add    $0x4,%rax
  cmp    %rsi,%rax
  je     L41
L28:
  mov    (%rax),%edx
  mov    0x4(%rax),%ecx
  cmp    %ecx,%edx
  jle    L36
  mov    %ecx,(%rax)
  mov    %edx,0x4(%rax)
  jmp    L36
L41:
  sub    $0x1,%r8d
  je     L57
L45:
  test   %r8d,%r8d
  jle    L41
  mov    %rdi,%rax
  lea    -0x1(%r8),%edx
  lea    (%r9,%rdx,4),%rsi
  jmp    L28
L57:
  ret
```

assistant: ```c
void sort(int numbers[4], int totalNumbers) {
  int i, j, temporaryNumber;
  for (i = 0; i < totalNumbers - 1; i++) {
    for (j = 0; j < (totalNumbers - 1 - i); j++) {
      if (numbers[j] > numbers[j + 1]) {
        temporaryNumber = numbers[j];
        numbers[j] = numbers[j + 1];
        numbers[j + 1] = temporaryNumber;
      }
    }
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<sort>:
  endbr64
  lea    -0x1(%rsi),%edx
  test   %edx,%edx
  jle    L25
  lea    0x4(%rdi),%r9
  nop
L1:
  sub    $0x1,%edx
  mov    %rdi,%rax
  mov    %rdx,%r8
  lea    (%r9,%rdx,4),%rsi
  nopl   (%rax)
L9:
  mov    (%rax),%edx
  mov    0x4(%rax),%ecx
  cmp    %ecx,%edx
  jle    L22
  mov    %ecx,(%rax)
  mov    %edx,0x4(%rax)
L22:
  add    $0x4,%rax
  cmp    %rsi,%rax
  jne    L9
  mov    %r8d,%edx
  test   %r8d,%r8d
  jne    L1
L25:
  ret
```

assistant: ```c
void sort(int numbers[4], int totalNumbers) {
  int i, j, temporaryNumber;
  for (i = 0; i < totalNumbers - 1; i++) {
    for (j = 0; j < (totalNumbers - 1 - i); j++) {
      if (numbers[j] > numbers[j + 1]) {
        temporaryNumber = numbers[j];
        numbers[j] = numbers[j + 1];
        numbers[j + 1] = temporaryNumber;
      }
    }
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<sort>:
  endbr64
  lea    -0x1(%rsi),%eax
  test   %eax,%eax
  jle    L25
  lea    0x4(%rdi),%r9
  nop
L30:
  sub    $0x1,%eax
  mov    %rdi,%rdx
  mov    %rax,%r8
  lea    (%r9,%rax,4),%rsi
  nopl   (%rax)
L21:
  movq   (%rdx),%xmm0
  pshufd $0xe5,%xmm0,%xmm2
  movd   %xmm0,%ecx
  movd   %xmm2,%eax
  pshufd $0xe1,%xmm0,%xmm1
  cmp    %eax,%ecx
  jle    L16
  movq   %xmm1,(%rdx)
L16:
  add    $0x4,%rdx
  cmp    %rsi,%rdx
  jne    L21
  mov    %r8d,%eax
  test   %r8d,%r8d
  jne    L30
L25:
  ret
```

assistant: ```c
void sort(int numbers[4], int totalNumbers) {
  int i, j, temporaryNumber;
  for (i = 0; i < totalNumbers - 1; i++) {
    for (j = 0; j < (totalNumbers - 1 - i); j++) {
      if (numbers[j] > numbers[j + 1]) {
        temporaryNumber = numbers[j];
        numbers[j] = numbers[j + 1];
        numbers[j + 1] = temporaryNumber;
      }
    }
  }
}
```