ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<time_func>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x186a0,-0xc(%rbp)
  jmp    L22
L13:
  mov    -0xc(%rbp),%eax
  lea    0x3(%rax),%edx
  test   %eax,%eax
  cmovs  %edx,%eax
  sar    $0x2,%eax
  mov    %eax,-0x10(%rbp)
  jmp    L41
L40:
  mov    $0x0,%eax
  call   <benchmark_func@plt>
  mov    $0x0,%eax
  call   <benchmark_func@plt>
  mov    $0x0,%eax
  call   <benchmark_func@plt>
  mov    $0x0,%eax
  call   <benchmark_func@plt>
L41:
  mov    -0x10(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0x10(%rbp)
  test   %eax,%eax
  jne    L40
L22:
  mov    $0x0,%eax
  call   <calls_per_sec@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x8(%rbp)
  pxor   %xmm0,%xmm0
  ucomisd -0x8(%rbp),%xmm0
  jp     L30
  pxor   %xmm0,%xmm0
  ucomisd -0x8(%rbp),%xmm0
  je     L13
L30:
  pxor   %xmm0,%xmm0
  cvtsi2sdl -0xc(%rbp),%xmm0
  mulsd  -0x8(%rbp),%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  leave
  ret
```

assistant: ```c
double time_func(void) {
  int const iter = 100000;
  double rate;
  while (!(rate = calls_per_sec())) {
    int n;
    for (n = iter / 4; n--;) {
      benchmark_func();
      benchmark_func();
      benchmark_func();
      benchmark_func();
    }
  }
  return rate * iter;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<time_func>:
  endbr64
  push   %rbx
L34:
  mov    $0x0,%eax
  call   <calls_per_sec@plt>
  pxor   %xmm1,%xmm1
  ucomisd %xmm1,%xmm0
  jp     L20
  jne    L20
  mov    $0x61a8,%ebx
L11:
  mov    $0x0,%eax
  call   <benchmark_func@plt>
  mov    $0x0,%eax
  call   <benchmark_func@plt>
  mov    $0x0,%eax
  call   <benchmark_func@plt>
  mov    $0x0,%eax
  call   <benchmark_func@plt>
  sub    $0x1,%ebx
  jne    L11
  jmp    L34
L20:
  mulsd  D2(%rip),%xmm0
  pop    %rbx
  ret
```

tool_calls: name: disassemble_data
arguments: label: D2
type_infered: byte[8]

tool: {"D2": [0x00,0x00,0x00,0x00,0x00,0x6a,0xf8,0x40]}

assistant: ```c
double time_func(void) {
  int const iter = 100000;
  double rate;
  while (!(rate = calls_per_sec())) {
    int n;
    for (n = iter / 4; n--;) {
      benchmark_func();
      benchmark_func();
      benchmark_func();
      benchmark_func();
    }
  }
  return rate * iter;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<time_func>:
  endbr64
  push   %rbx
  nopl   (%rax)
L30:
  xor    %eax,%eax
  call   <calls_per_sec@plt>
  pxor   %xmm1,%xmm1
  ucomisd %xmm1,%xmm0
  jp     L8
  jne    L8
  mov    $0x61a8,%ebx
L35:
  xor    %eax,%eax
  call   <benchmark_func@plt>
  xor    %eax,%eax
  call   <benchmark_func@plt>
  xor    %eax,%eax
  call   <benchmark_func@plt>
  xor    %eax,%eax
  call   <benchmark_func@plt>
  sub    $0x1,%ebx
  jne    L35
  jmp    L30
L8:
  mulsd  D2(%rip),%xmm0
  pop    %rbx
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D2
type_infered: byte[8]

tool: {"D2": [0x00,0x00,0x00,0x00,0x00,0x6a,0xf8,0x40]}

assistant: ```c
double time_func(void) {
  int const iter = 100000;
  double rate;
  while (!(rate = calls_per_sec())) {
    int n;
    for (n = iter / 4; n--;) {
      benchmark_func();
      benchmark_func();
      benchmark_func();
      benchmark_func();
    }
  }
  return rate * iter;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<time_func>:
  endbr64
  push   %rbx
  nopl   (%rax)
L22:
  xor    %eax,%eax
  call   <calls_per_sec@plt>
  pxor   %xmm1,%xmm1
  ucomisd %xmm1,%xmm0
  jp     L16
  jne    L16
  mov    $0x61a8,%ebx
L38:
  xor    %eax,%eax
  call   <benchmark_func@plt>
  xor    %eax,%eax
  call   <benchmark_func@plt>
  xor    %eax,%eax
  call   <benchmark_func@plt>
  xor    %eax,%eax
  call   <benchmark_func@plt>
  sub    $0x1,%ebx
  jne    L38
  jmp    L22
L16:
  mulsd  D1(%rip),%xmm0
  pop    %rbx
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D1
datatype: byte[8]

tool: {"D1": [0x00,0x00,0x00,0x00,0x00,0x6a,0xf8,0x40]}

assistant: ```c
double time_func(void) {
  int const iter = 100000;
  double rate;
  while (!(rate = calls_per_sec())) {
    int n;
    for (n = iter / 4; n--;) {
      benchmark_func();
      benchmark_func();
      benchmark_func();
      benchmark_func();
    }
  }
  return rate * iter;
}
```
user: What is the c source code of the assembly code below?

```asm
<psvec4i_multiply_mat4>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    -0x18(%rbp),%rdx
  mov    -0x10(%rbp),%rcx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <vec4i_multiply_mat4@plt>
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
__attribute__((always_)) struct vec4i *psvec4i_multiply_mat4(struct vec4i *result, struct vec4i *a, struct mat4 *m) {
  vec4i_multiply_mat4((int32_t *)result, (int32_t *)a, (float *)m);
  return result;
}
```
user: What is the c source code of the assembly code below?

```asm
<psvec4i_multiply_mat4>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <vec4i_multiply_mat4@plt>
  mov    %rbx,%rax
  pop    %rbx
  ret
```

assistant: ```c
__attribute__((always_)) struct vec4i *psvec4i_multiply_mat4(struct vec4i *result, struct vec4i *a, struct mat4 *m) {
  vec4i_multiply_mat4((int32_t *)result, (int32_t *)a, (float *)m);
  return result;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<psvec4i_multiply_mat4>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  call   <vec4i_multiply_mat4@plt>
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
__attribute__((always_)) struct vec4i *psvec4i_multiply_mat4(struct vec4i *result, struct vec4i *a, struct mat4 *m) {
  vec4i_multiply_mat4((int32_t *)result, (int32_t *)a, (float *)m);
  return result;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<psvec4i_multiply_mat4>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  call   <vec4i_multiply_mat4@plt>
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
__attribute__((always_)) struct vec4i *psvec4i_multiply_mat4(struct vec4i *result, struct vec4i *a, struct mat4 *m) {
  vec4i_multiply_mat4((int32_t *)result, (int32_t *)a, (float *)m);
  return result;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x38(%rbp)
  jmp    L0
L49:
  lea    D52(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x30(%rbp),%rdx
  mov    -0x38(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rdx,%rax
  mov    %rax,%rsi
  lea    D15(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  addl   $0x1,-0x38(%rbp)
L0:
  cmpl   $0x9,-0x38(%rbp)
  jle    L49
  movl   $0xa,-0x34(%rbp)
  lea    D51(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x3c(%rbp),%rax
  mov    %rax,%rsi
  lea    D15(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x3c(%rbp),%eax
  mov    %eax,-0x38(%rbp)
  jmp    L39
L64:
  mov    -0x38(%rbp),%eax
  add    $0x1,%eax
  cltq
  mov    -0x30(%rbp,%rax,4),%edx
  mov    -0x38(%rbp),%eax
  cltq
  mov    %edx,-0x30(%rbp,%rax,4)
  addl   $0x1,-0x38(%rbp)
L39:
  mov    -0x38(%rbp),%eax
  cmp    -0x34(%rbp),%eax
  jl     L64
  subl   $0x1,-0x34(%rbp)
  movl   $0x0,-0x38(%rbp)
  jmp    L48
L58:
  mov    -0x38(%rbp),%eax
  cltq
  mov    -0x30(%rbp,%rax,4),%eax
  mov    %eax,%esi
  lea    D36(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x38(%rbp)
L48:
  mov    -0x38(%rbp),%eax
  cmp    -0x34(%rbp),%eax
  jl     L58
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  call   <system@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L27
  call   <__stack_chk_fail@plt>
L27:
  leave
  ret
```

tool_calls: name: read_data
arguments: tag: D52
type_guessed: string
name: read_data
arguments: tag: D15
type_guessed: string
name: read_data
arguments: tag: D51
type_guessed: string
name: read_data
arguments: tag: D36
type_guessed: string
name: read_data
arguments: tag: D10
type_guessed: string

tool: D52: 
Inserisci numero:

tool: D15: %d

tool: D51: 
Inserisci posizione dell'elemento da eliminare:

tool: D36: 

%d

tool: D10: PAUSE

assistant: ```c
int main() {
  int v[10];
  int cont;
  int pos;
  int tot;
  cont = 0;
  while (cont <= 9) {
    printf("\nInserisci numero:");
    scanf("%d", &v[cont]);
    cont++;
  }
  tot = 10;
  printf("\nInserisci posizione dell'elemento da eliminare:");
  scanf("%d", &pos);
  for (cont = pos; cont < tot; cont++) {
    v[cont] = v[cont + 1];
  }
  tot--;
  for (cont = 0; cont < tot; cont++) {
    printf("\n\n%d", v[cont]);
  }
  system("PAUSE");
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rbp
  lea    0x38(%rsp),%r14
  mov    %rbp,%rbx
  lea    D25(%rip),%r13
  lea    D18(%rip),%r12
L46:
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %rbx,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  add    $0x4,%rbx
  cmp    %r14,%rbx
  jne    L46
  lea    D43(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0xc(%rsp),%rsi
  lea    D18(%rip),%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%ecx
  cmp    $0x9,%ecx
  jg     L37
  movslq %ecx,%rsi
  lea    0x10(%rsp,%rsi,4),%rax
  mov    $0x9,%edx
  sub    %ecx,%edx
  add    %rsi,%rdx
  lea    0x4(%rbp,%rdx,4),%rcx
L12:
  mov    0x4(%rax),%edx
  mov    %edx,(%rax)
  add    $0x4,%rax
  cmp    %rcx,%rax
  jne    L12
L37:
  lea    0x24(%rbp),%r12
  lea    D58(%rip),%rbx
L35:
  mov    0x0(%rbp),%esi
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x4,%rbp
  cmp    %r12,%rbp
  jne    L35
  lea    D2(%rip),%rdi
  call   <system@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L24
  mov    $0x0,%eax
  add    $0x40,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L24:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D25
data_type: string
name: unpack_data
arguments: tag: D18
data_type: string
name: unpack_data
arguments: tag: D43
data_type: string
name: unpack_data
arguments: tag: D58
data_type: string
name: unpack_data
arguments: tag: D2
data_type: string

tool: D25: 
Inserisci numero:

tool: D18: %d

tool: D43: 
Inserisci posizione dell'elemento da eliminare:

tool: D58: 

%d

tool: D2: PAUSE

assistant: ```c
int main() {
  int v[10];
  int cont;
  int pos;
  int tot;
  cont = 0;
  while (cont <= 9) {
    printf("\nInserisci numero:");
    scanf("%d", &v[cont]);
    cont++;
  }
  tot = 10;
  printf("\nInserisci posizione dell'elemento da eliminare:");
  scanf("%d", &pos);
  for (cont = pos; cont < tot; cont++) {
    v[cont] = v[cont + 1];
  }
  tot--;
  for (cont = 0; cont < tot; cont++) {
    printf("\n\n%d", v[cont]);
  }
  system("PAUSE");
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  lea    D48(%rip),%r14
  push   %r13
  lea    D35(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x48,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%r12
  lea    0x38(%rsp),%r15
  mov    %r12,%rbp
  mov    %r12,%rbx
L24:
  mov    %r14,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbx,%rsi
  mov    %r13,%rdi
  add    $0x4,%rbx
  call   <scanf@plt>
  cmp    %r15,%rbx
  jne    L24
  lea    D40(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0xc(%rsp),%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%eax
  cmp    $0x9,%eax
  jg     L28
  movslq %eax,%rdi
  mov    $0xa,%edx
  shl    $0x2,%rdi
  sub    %eax,%edx
  lea    0x4(%r12,%rdi,1),%rsi
  shl    $0x2,%rdx
  add    %r12,%rdi
  call   <memmove@plt>
L28:
  lea    0x24(%r12),%rbx
  lea    D41(%rip),%r12
  nopl   (%rax)
L36:
  mov    0x0(%rbp),%esi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x4,%rbp
  call   <printf@plt>
  cmp    %rbx,%rbp
  jne    L36
  lea    D33(%rip),%rdi
  call   <system@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L14
  add    $0x48,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L14:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D48
type_infered: string
name: disassemble_data
arguments: label: D35
type_infered: string
name: disassemble_data
arguments: label: D40
type_infered: string
name: disassemble_data
arguments: label: D41
type_infered: string
name: disassemble_data
arguments: label: D33
type_infered: string

tool: D48: 
Inserisci numero:

tool: D35: %d

tool: D40: 
Inserisci posizione dell'elemento da eliminare:

tool: D41: 

%d

tool: D33: PAUSE

assistant: ```c
int main() {
  int v[10];
  int cont;
  int pos;
  int tot;
  cont = 0;
  while (cont <= 9) {
    printf("\nInserisci numero:");
    scanf("%d", &v[cont]);
    cont++;
  }
  tot = 10;
  printf("\nInserisci posizione dell'elemento da eliminare:");
  scanf("%d", &pos);
  for (cont = pos; cont < tot; cont++) {
    v[cont] = v[cont + 1];
  }
  tot--;
  for (cont = 0; cont < tot; cont++) {
    printf("\n\n%d", v[cont]);
  }
  system("PAUSE");
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  lea    D25(%rip),%r14
  push   %r13
  lea    D5(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x48,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%r12
  lea    0x38(%rsp),%r15
  mov    %r12,%rbp
  mov    %r12,%rbx
L1:
  mov    %r14,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbx,%rsi
  mov    %r13,%rdi
  add    $0x4,%rbx
  call   <scanf@plt>
  cmp    %r15,%rbx
  jne    L1
  lea    D29(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0xc(%rsp),%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%eax
  cmp    $0x9,%eax
  jg     L8
  movslq %eax,%rdi
  mov    $0xa,%edx
  shl    $0x2,%rdi
  sub    %eax,%edx
  lea    0x4(%r12,%rdi,1),%rsi
  shl    $0x2,%rdx
  add    %r12,%rdi
  call   <memmove@plt>
L8:
  lea    0x24(%r12),%rbx
  lea    D31(%rip),%r12
  nopl   (%rax)
L39:
  mov    0x0(%rbp),%esi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x4,%rbp
  call   <printf@plt>
  cmp    %rbx,%rbp
  jne    L39
  lea    D46(%rip),%rdi
  call   <system@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L35
  add    $0x48,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L35:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D25
type: string
name: extract_symbolic_data
arguments: data_tag: D5
type: string
name: extract_symbolic_data
arguments: data_tag: D29
type: string
name: extract_symbolic_data
arguments: data_tag: D31
type: string
name: extract_symbolic_data
arguments: data_tag: D46
type: string

tool: D25: 
Inserisci numero:

tool: D5: %d

tool: D29: 
Inserisci posizione dell'elemento da eliminare:

tool: D31: 

%d

tool: D46: PAUSE

assistant: ```c
int main() {
  int v[10];
  int cont;
  int pos;
  int tot;
  cont = 0;
  while (cont <= 9) {
    printf("\nInserisci numero:");
    scanf("%d", &v[cont]);
    cont++;
  }
  tot = 10;
  printf("\nInserisci posizione dell'elemento da eliminare:");
  scanf("%d", &pos);
  for (cont = pos; cont < tot; cont++) {
    v[cont] = v[cont + 1];
  }
  tot--;
  for (cont = 0; cont < tot; cont++) {
    printf("\n\n%d", v[cont]);
  }
  system("PAUSE");
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<ta_asprintf>:
  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     L6
  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)
L6:
  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   <ta_vasprintf@plt>
  mov    %rax,-0xb8(%rbp)
  mov    -0xbc(%rbp),%eax
  mov    %eax,%edi
  call   <va_end@plt>
  mov    -0xb8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *ta_asprintf(void *ta_parent, const char *fmt, ...) {
  char *res;
  va_list ap;
  va_start(ap, fmt);
  res = ta_vasprintf(ta_parent, fmt, ap);
  va_end(ap);
  return res;
}
```
user: What is the c source code of the assembly code below?

```asm
<ta_asprintf>:
  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   <ta_vasprintf@plt>
  mov    %rax,%rbx
  mov    %ebp,%edi
  call   <va_end@plt>
  mov    %rbx,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *ta_asprintf(void *ta_parent, const char *fmt, ...) {
  char *res;
  va_list ap;
  va_start(ap, fmt);
  res = ta_vasprintf(ta_parent, fmt, ap);
  va_end(ap);
  return res;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<ta_asprintf>:
  endbr64
  push   %r13
  xor    %r13d,%r13d
  push   %r12
  mov    %rdi,%r12
  mov    %r13d,%edi
  push   %rbp
  mov    %rsi,%rbp
  call   <va_start@plt>
  mov    %r13d,%edx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <ta_vasprintf@plt>
  mov    %r13d,%edi
  mov    %rax,%r12
  call   <va_end@plt>
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *ta_asprintf(void *ta_parent, const char *fmt, ...) {
  char *res;
  va_list ap;
  va_start(ap, fmt);
  res = ta_vasprintf(ta_parent, fmt, ap);
  va_end(ap);
  return res;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<ta_asprintf>:
  endbr64
  push   %r13
  xor    %r13d,%r13d
  push   %r12
  mov    %rdi,%r12
  mov    %r13d,%edi
  push   %rbp
  mov    %rsi,%rbp
  call   <va_start@plt>
  mov    %r13d,%edx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <ta_vasprintf@plt>
  mov    %r13d,%edi
  mov    %rax,%r12
  call   <va_end@plt>
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *ta_asprintf(void *ta_parent, const char *fmt, ...) {
  char *res;
  va_list ap;
  va_start(ap, fmt);
  res = ta_vasprintf(ta_parent, fmt, ap);
  va_end(ap);
  return res;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<hashing3>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x10(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x4(%rbp)
  movl   $0x3e8,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L2
L5:
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  imul   -0x8(%rbp),%eax
  add    %eax,-0x10(%rbp)
  cmpl   $0x1,-0x8(%rbp)
  je     L12
  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)
L12:
  addl   $0x1,-0xc(%rbp)
L2:
  mov    -0xc(%rbp),%eax
  cmp    -0x4(%rbp),%eax
  jl     L5
  mov    -0x10(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x57619f1,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x6,%edx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  imul   $0xbb8,%edx,%ecx
  sub    %ecx,%eax
  mov    %eax,%edx
  mov    %edx,%eax
  leave
  ret
```

assistant: ```c
int hashing3(char str[]) {
  int nr = 0, i, len = strlen(str), x = 1000;
  for (i = 0; i < len; i++) {
    nr = nr + (str[i] * x);
    if (x != 1)
      x = x / 10;
  }
  return nr % 3000;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<hashing3>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  test   %eax,%eax
  jle    L20
  mov    %rbx,%rdx
  lea    -0x1(%rax),%eax
  lea    0x1(%rbx,%rax,1),%r8
  mov    $0x3e8,%eax
  mov    $0x0,%esi
  jmp    L40
L25:
  add    $0x1,%rdx
  cmp    %r8,%rdx
  je     L30
L40:
  movsbl (%rdx),%ecx
  imul   %eax,%ecx
  add    %ecx,%esi
  cmp    $0x1,%eax
  je     L25
  movslq %eax,%rcx
  imul   $0x66666667,%rcx,%rcx
  sar    $0x22,%rcx
  sar    $0x1f,%eax
  mov    %eax,%edi
  mov    %ecx,%eax
  sub    %edi,%eax
  jmp    L25
L20:
  mov    $0x0,%esi
L30:
  movslq %esi,%rax
  imul   $0x57619f1,%rax,%rax
  sar    $0x26,%rax
  mov    %esi,%edx
  sar    $0x1f,%edx
  sub    %edx,%eax
  imul   $0xbb8,%eax,%edx
  mov    %esi,%eax
  sub    %edx,%eax
  pop    %rbx
  ret
```

assistant: ```c
int hashing3(char str[]) {
  int nr = 0, i, len = strlen(str), x = 1000;
  for (i = 0; i < len; i++) {
    nr = nr + (str[i] * x);
    if (x != 1)
      x = x / 10;
  }
  return nr % 3000;
}
```
user: Translate the below assembly code into C programming language:

```asm
<hashing3>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  test   %eax,%eax
  jle    L15
  sub    $0x1,%eax
  mov    %rbx,%rdi
  xor    %ecx,%ecx
  lea    0x1(%rbx,%rax,1),%r8
  mov    $0x3e8,%eax
  nopl   0x0(%rax,%rax,1)
L6:
  movsbl (%rdi),%edx
  imul   %eax,%edx
  add    %edx,%ecx
  cmp    $0x1,%eax
  je     L29
  movslq %eax,%rdx
  sar    $0x1f,%eax
  imul   $0x66666667,%rdx,%rdx
  mov    %eax,%esi
  sar    $0x22,%rdx
  mov    %edx,%eax
  sub    %esi,%eax
L29:
  add    $0x1,%rdi
  cmp    %rdi,%r8
  jne    L6
  movslq %ecx,%rax
  mov    %ecx,%edx
  pop    %rbx
  imul   $0x57619f1,%rax,%rax
  sar    $0x1f,%edx
  sar    $0x26,%rax
  sub    %edx,%eax
  imul   $0xbb8,%eax,%edx
  mov    %ecx,%eax
  sub    %edx,%eax
  ret
  cs nopw 0x0(%rax,%rax,1)
L15:
  xor    %eax,%eax
  pop    %rbx
  ret
```

assistant: ```c
int hashing3(char str[]) {
  int nr = 0, i, len = strlen(str), x = 1000;
  for (i = 0; i < len; i++) {
    nr = nr + (str[i] * x);
    if (x != 1)
      x = x / 10;
  }
  return nr % 3000;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<hashing3>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  test   %eax,%eax
  jle    L25
  sub    $0x1,%eax
  mov    %rbx,%rdi
  xor    %ecx,%ecx
  lea    0x1(%rbx,%rax,1),%r8
  mov    $0x3e8,%eax
  nopl   0x0(%rax,%rax,1)
L16:
  movsbl (%rdi),%edx
  imul   %eax,%edx
  add    %edx,%ecx
  cmp    $0x1,%eax
  je     L10
  movslq %eax,%rdx
  sar    $0x1f,%eax
  imul   $0x66666667,%rdx,%rdx
  mov    %eax,%esi
  sar    $0x22,%rdx
  mov    %edx,%eax
  sub    %esi,%eax
L10:
  add    $0x1,%rdi
  cmp    %rdi,%r8
  jne    L16
  movslq %ecx,%rax
  mov    %ecx,%edx
  pop    %rbx
  imul   $0x57619f1,%rax,%rax
  sar    $0x1f,%edx
  sar    $0x26,%rax
  sub    %edx,%eax
  imul   $0xbb8,%eax,%edx
  mov    %ecx,%eax
  sub    %edx,%eax
  ret
  cs nopw 0x0(%rax,%rax,1)
L25:
  xor    %eax,%eax
  pop    %rbx
  ret
```

assistant: ```c
int hashing3(char str[]) {
  int nr = 0, i, len = strlen(str), x = 1000;
  for (i = 0; i < len; i++) {
    nr = nr + (str[i] * x);
    if (x != 1)
      x = x / 10;
  }
  return nr % 3000;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<init_buffer>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    $0x1,%edi
  call   <srand@plt>
  movq   $0x0,-0x8(%rbp)
  jmp    L16
L13:
  mov    $0x0,%eax
  call   <rand@plt>
  mov    %rax,%rdx
  mov    -0x18(%rbp),%rcx
  mov    -0x8(%rbp),%rax
  add    %rcx,%rax
  mov    %dl,(%rax)
  addq   $0x1,-0x8(%rbp)
L16:
  mov    -0x8(%rbp),%rax
  cmp    -0x20(%rbp),%rax
  jb     L13
  nop
  nop
  leave
  ret
```

assistant: ```c
void init_buffer(char *bufp, size_t bufsz) {
  size_t i;
  srand(1);
  for (i = 0; i < bufsz; i++) {
    bufp[i] = (char)rand();
  }
}
```