ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<test3>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x2,%esi
  xor    %edi,%edi
  call   <InsertInFront@plt>
  mov    $0x4,%esi
  mov    %rax,%rdi
  call   <InsertInFront@plt>
  mov    $0x6,%esi
  mov    %rax,%rdi
  call   <InsertInFront@plt>
  mov    %rax,%rdi
  call   <PrintList@plt>
  mov    $0x2,%esi
  xor    %edi,%edi
  call   <InsertAtEnd@plt>
  mov    $0x4,%esi
  mov    %rax,%rdi
  call   <InsertAtEnd@plt>
  mov    $0x6,%esi
  mov    %rax,%rdi
  call   <InsertAtEnd@plt>
  mov    %rax,%rdi
  call   <PrintList@plt>
  mov    $0x5,%edx
  mov    $0x2,%esi
  xor    %edi,%edi
  call   <InsertAtN@plt>
  xor    %edx,%edx
  mov    $0x3,%esi
  mov    %rax,%rdi
  call   <InsertAtN@plt>
  mov    $0x2,%edx
  mov    $0x4,%esi
  mov    %rax,%rdi
  call   <InsertAtN@plt>
  mov    $0x1,%edx
  mov    $0x5,%esi
  mov    %rax,%rdi
  call   <InsertAtN@plt>
  mov    $0x1,%edx
  mov    $0x6,%esi
  mov    %rax,%rdi
  call   <InsertAtN@plt>
  mov    $0x3,%edx
  mov    $0x7,%esi
  mov    %rax,%rdi
  call   <InsertAtN@plt>
  xor    %edx,%edx
  mov    $0x8,%esi
  mov    %rax,%rdi
  call   <InsertAtN@plt>
  add    $0x8,%rsp
  mov    %rax,%rdi
  jmp    10a0 <PrintList@plt>
```

assistant: ```c
void test3() {
  struct node *head = NULL;
  head = InsertInFront(head, 2);
  head = InsertInFront(head, 4);
  head = InsertInFront(head, 6);
  PrintList(head);
  head = NULL;
  head = InsertAtEnd(head, 2);
  head = InsertAtEnd(head, 4);
  head = InsertAtEnd(head, 6);
  PrintList(head);
  head = NULL;
  head = InsertAtN(head, 2, 5);
  head = InsertAtN(head, 3, 0);
  head = InsertAtN(head, 4, 2);
  head = InsertAtN(head, 5, 1);
  head = InsertAtN(head, 6, 1);
  head = InsertAtN(head, 7, 3);
  head = InsertAtN(head, 8, 0);
  PrintList(head);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<rjhash>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  shl    $0xc,%eax
  mov    %eax,%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  add    $0x7ed55d16,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  shr    $0x13,%eax
  xor    -0x4(%rbp),%eax
  xor    $0xc761c23c,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  shl    $0x5,%eax
  mov    %eax,%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  add    $0x165667b1,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  lea    -0x2c5d9b94(%rax),%edx
  mov    -0x4(%rbp),%eax
  shl    $0x9,%eax
  xor    %edx,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  lea    0x0(,%rax,8),%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  sub    $0x28fb93b,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  shr    $0x10,%eax
  xor    -0x4(%rbp),%eax
  xor    $0xb55a4f09,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned int rjhash(unsigned int a) {
  a = (a + 0x7ed55d16) + (a << 12);
  a = (a ^ 0xc761c23c) ^ (a >> 19);
  a = (a + 0x165667b1) + (a << 5);
  a = (a + 0xd3a2646c) ^ (a << 9);
  a = (a + 0xfd7046c5) + (a << 3);
  a = (a ^ 0xb55a4f09) ^ (a >> 16);
  return a;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<rjhash>:
  endbr64
  mov    %edi,%eax
  shl    $0xc,%eax
  lea    0x7ed55d16(%rax,%rdi,1),%eax
  mov    %eax,%edx
  shr    $0x13,%edx
  xor    %eax,%edx
  xor    $0xc761c23c,%edx
  mov    %edx,%eax
  shl    $0x5,%eax
  add    %edx,%eax
  lea    0x165667b1(%rax),%edx
  shl    $0x9,%edx
  sub    $0x160733e3,%eax
  xor    %edx,%eax
  lea    -0x28fb93b(%rax,%rax,8),%edx
  mov    %edx,%eax
  shr    $0x10,%eax
  xor    %edx,%eax
  xor    $0xb55a4f09,%eax
  ret
```

assistant: ```c
unsigned int rjhash(unsigned int a) {
  a = (a + 0x7ed55d16) + (a << 12);
  a = (a ^ 0xc761c23c) ^ (a >> 19);
  a = (a + 0x165667b1) + (a << 5);
  a = (a + 0xd3a2646c) ^ (a << 9);
  a = (a + 0xfd7046c5) + (a << 3);
  a = (a ^ 0xb55a4f09) ^ (a >> 16);
  return a;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<rjhash>:
  endbr64
  mov    %edi,%eax
  shl    $0xc,%eax
  lea    0x7ed55d16(%rax,%rdi,1),%eax
  mov    %eax,%edx
  shr    $0x13,%edx
  xor    %eax,%edx
  xor    $0xc761c23c,%edx
  mov    %edx,%eax
  shl    $0x5,%eax
  add    %edx,%eax
  lea    0x165667b1(%rax),%edx
  sub    $0x160733e3,%eax
  shl    $0x9,%edx
  xor    %edx,%eax
  lea    -0x28fb93b(%rax,%rax,8),%edx
  mov    %edx,%eax
  shr    $0x10,%eax
  xor    %edx,%eax
  xor    $0xb55a4f09,%eax
  ret
```

assistant: ```c
unsigned int rjhash(unsigned int a) {
  a = (a + 0x7ed55d16) + (a << 12);
  a = (a ^ 0xc761c23c) ^ (a >> 19);
  a = (a + 0x165667b1) + (a << 5);
  a = (a + 0xd3a2646c) ^ (a << 9);
  a = (a + 0xfd7046c5) + (a << 3);
  a = (a ^ 0xb55a4f09) ^ (a >> 16);
  return a;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<rjhash>:
  endbr64
  mov    %edi,%eax
  shl    $0xc,%eax
  lea    0x7ed55d16(%rax,%rdi,1),%eax
  mov    %eax,%edx
  shr    $0x13,%edx
  xor    %eax,%edx
  xor    $0xc761c23c,%edx
  mov    %edx,%eax
  shl    $0x5,%eax
  add    %edx,%eax
  lea    0x165667b1(%rax),%edx
  sub    $0x160733e3,%eax
  shl    $0x9,%edx
  xor    %edx,%eax
  lea    -0x28fb93b(%rax,%rax,8),%edx
  mov    %edx,%eax
  shr    $0x10,%eax
  xor    %edx,%eax
  xor    $0xb55a4f09,%eax
  ret
```

assistant: ```c
unsigned int rjhash(unsigned int a) {
  a = (a + 0x7ed55d16) + (a << 12);
  a = (a ^ 0xc761c23c) ^ (a >> 19);
  a = (a + 0x165667b1) + (a << 5);
  a = (a + 0xd3a2646c) ^ (a << 9);
  a = (a + 0xfd7046c5) + (a << 3);
  a = (a ^ 0xb55a4f09) ^ (a >> 16);
  return a;
}
```
user: What is the c source code of the assembly code below:

```asm
<test_simdlen>:
  endbr64
  ret
```

assistant: ```c
void test_simdlen() {
  int i;
#pragma omp for simd simdlen
  for (i = 0; i < 16; ++i)
    ;
#pragma omp for simd simdlen(
  for (i = 0; i < 16; ++i)
    ;
#pragma omp for simd simdlen()
  for (i = 0; i < 16; ++i)
    ;
#pragma omp for simd simdlen(,
  for (i = 0; i < 16; ++i)
    ;
#pragma omp for simd simdlen(, )
  for (i = 0; i < 16; ++i)
    ;
#pragma omp for simd simdlen 4)
  for (i = 0; i < 16; ++i)
    ;
#pragma omp for simd simdlen(4
  for (i = 0; i < 16; ++i)
    ;
#pragma omp for simd simdlen(4,
  for (i = 0; i < 16; ++i)
    ;
#pragma omp for simd simdlen(4, )
  for (i = 0; i < 16; ++i)
    ;
#pragma omp for simd simdlen(4)
  for (i = 0; i < 16; ++i)
    ;
#pragma omp for simd simdlen(4 4)
  for (i = 0; i < 16; ++i)
    ;
#pragma omp for simd simdlen(4, , 4)
  for (i = 0; i < 16; ++i)
    ;
#pragma omp for simd simdlen(4)
  for (i = 0; i < 16; ++i)
    ;
#pragma omp for simd simdlen(4, 8)
  for (i = 0; i < 16; ++i)
    ;
#pragma omp for simd simdlen(2.5)
  for (i = 0; i < 16; ++i)
    ;
#pragma omp for simd simdlen(foo())
  for (i = 0; i < 16; ++i)
    ;
#pragma omp for simd simdlen(-5)
  for (i = 0; i < 16; ++i)
    ;
#pragma omp for simd simdlen(0)
  for (i = 0; i < 16; ++i)
    ;
#pragma omp for simd simdlen(5 - 5)
  for (i = 0; i < 16; ++i)
    ;
}
```
user: Translate the below assembly code into C programming language:

```asm
<test_simdlen>:
  endbr64
  ret
```

assistant: ```c
void test_simdlen() {
  int i;
#pragma omp for simd simdlen
  for (i = 0; i < 16; ++i)
    ;
#pragma omp for simd simdlen(
  for (i = 0; i < 16; ++i)
    ;
#pragma omp for simd simdlen()
  for (i = 0; i < 16; ++i)
    ;
#pragma omp for simd simdlen(,
  for (i = 0; i < 16; ++i)
    ;
#pragma omp for simd simdlen(, )
  for (i = 0; i < 16; ++i)
    ;
#pragma omp for simd simdlen 4)
  for (i = 0; i < 16; ++i)
    ;
#pragma omp for simd simdlen(4
  for (i = 0; i < 16; ++i)
    ;
#pragma omp for simd simdlen(4,
  for (i = 0; i < 16; ++i)
    ;
#pragma omp for simd simdlen(4, )
  for (i = 0; i < 16; ++i)
    ;
#pragma omp for simd simdlen(4)
  for (i = 0; i < 16; ++i)
    ;
#pragma omp for simd simdlen(4 4)
  for (i = 0; i < 16; ++i)
    ;
#pragma omp for simd simdlen(4, , 4)
  for (i = 0; i < 16; ++i)
    ;
#pragma omp for simd simdlen(4)
  for (i = 0; i < 16; ++i)
    ;
#pragma omp for simd simdlen(4, 8)
  for (i = 0; i < 16; ++i)
    ;
#pragma omp for simd simdlen(2.5)
  for (i = 0; i < 16; ++i)
    ;
#pragma omp for simd simdlen(foo())
  for (i = 0; i < 16; ++i)
    ;
#pragma omp for simd simdlen(-5)
  for (i = 0; i < 16; ++i)
    ;
#pragma omp for simd simdlen(0)
  for (i = 0; i < 16; ++i)
    ;
#pragma omp for simd simdlen(5 - 5)
  for (i = 0; i < 16; ++i)
    ;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<test_simdlen>:
  endbr64
  ret
```

assistant: ```c
void test_simdlen() {
  int i;
#pragma omp for simd simdlen
  for (i = 0; i < 16; ++i)
    ;
#pragma omp for simd simdlen(
  for (i = 0; i < 16; ++i)
    ;
#pragma omp for simd simdlen()
  for (i = 0; i < 16; ++i)
    ;
#pragma omp for simd simdlen(,
  for (i = 0; i < 16; ++i)
    ;
#pragma omp for simd simdlen(, )
  for (i = 0; i < 16; ++i)
    ;
#pragma omp for simd simdlen 4)
  for (i = 0; i < 16; ++i)
    ;
#pragma omp for simd simdlen(4
  for (i = 0; i < 16; ++i)
    ;
#pragma omp for simd simdlen(4,
  for (i = 0; i < 16; ++i)
    ;
#pragma omp for simd simdlen(4, )
  for (i = 0; i < 16; ++i)
    ;
#pragma omp for simd simdlen(4)
  for (i = 0; i < 16; ++i)
    ;
#pragma omp for simd simdlen(4 4)
  for (i = 0; i < 16; ++i)
    ;
#pragma omp for simd simdlen(4, , 4)
  for (i = 0; i < 16; ++i)
    ;
#pragma omp for simd simdlen(4)
  for (i = 0; i < 16; ++i)
    ;
#pragma omp for simd simdlen(4, 8)
  for (i = 0; i < 16; ++i)
    ;
#pragma omp for simd simdlen(2.5)
  for (i = 0; i < 16; ++i)
    ;
#pragma omp for simd simdlen(foo())
  for (i = 0; i < 16; ++i)
    ;
#pragma omp for simd simdlen(-5)
  for (i = 0; i < 16; ++i)
    ;
#pragma omp for simd simdlen(0)
  for (i = 0; i < 16; ++i)
    ;
#pragma omp for simd simdlen(5 - 5)
  for (i = 0; i < 16; ++i)
    ;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<turn_on_buzzer>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x7f0080a4,%eax
  mov    (%rax),%eax
  mov    $0x7f0080a4,%edx
  or     $0x80,%ah
  mov    %eax,(%rdx)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void turn_on_buzzer(void) {
  (*(volatile unsigned int *)(0x7F0080A4)) |= (1 << 15);
  return;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<turn_on_buzzer>:
  endbr64
  mov    0x7f0080a4,%eax
  or     $0x80,%ah
  mov    %eax,0x7f0080a4
  ret
```

assistant: ```c
void turn_on_buzzer(void) {
  (*(volatile unsigned int *)(0x7F0080A4)) |= (1 << 15);
  return;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<turn_on_buzzer>:
  endbr64
  mov    0x7f0080a4,%eax
  or     $0x80,%ah
  mov    %eax,0x7f0080a4
  ret
```

assistant: ```c
void turn_on_buzzer(void) {
  (*(volatile unsigned int *)(0x7F0080A4)) |= (1 << 15);
  return;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<BringBlockBack>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  mov    %ecx,-0x28(%rbp)
  mov    %r8d,-0x2c(%rbp)
  mov    %r9d,-0x30(%rbp)
  mov    -0x2c(%rbp),%eax
  imul   -0x28(%rbp),%eax
  mov    %eax,-0x8(%rbp)
  jmp    L6
L4:
  mov    -0x30(%rbp),%eax
  imul   -0x28(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  jmp    L10
L24:
  mov    -0x2c(%rbp),%eax
  imul   -0x28(%rbp),%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%eax
  sub    %edx,%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x30(%rbp),%eax
  imul   -0x28(%rbp),%eax
  mov    %eax,%ecx
  mov    -0x4(%rbp),%eax
  sub    %ecx,%eax
  cltq
  shl    $0x2,%rax
  add    %rdx,%rax
  mov    -0x8(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,8),%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rdx),%rcx
  mov    -0x4(%rbp),%edx
  movslq %edx,%rdx
  shl    $0x2,%rdx
  add    %rcx,%rdx
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  addl   $0x1,-0x4(%rbp)
L10:
  mov    -0x30(%rbp),%eax
  add    $0x1,%eax
  imul   -0x28(%rbp),%eax
  cmp    %eax,-0x4(%rbp)
  jl     L24
  addl   $0x1,-0x8(%rbp)
L6:
  mov    -0x2c(%rbp),%eax
  add    $0x1,%eax
  imul   -0x28(%rbp),%eax
  cmp    %eax,-0x8(%rbp)
  jl     L4
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void BringBlockBack(int **big, int **small, int size, int param, int indexi, int indexj) {
  int i, j;
  for (i = indexi * param; i < param * (indexi + 1); i++)
    for (j = indexj * param; j < param * (indexj + 1); j++)
      big[i][j] = small[i - indexi * param][j - indexj * param];
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<BringBlockBack>:
  endbr64
  imul   %ecx,%r8d
  lea    (%rcx,%r8,1),%edx
  cmp    %edx,%r8d
  jge    L19
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  mov    %ecx,%ebp
  mov    %r9d,%r11d
  imul   %ecx,%r11d
  lea    (%rcx,%r11,1),%r12d
  movslq %r8d,%r8
  lea    (%rdi,%r8,8),%rsi
  lea    -0x1(%r12),%eax
  sub    %r11d,%eax
  lea    0x4(,%rax,4),%r9
  movslq %r11d,%r8
  shl    $0x2,%r8
  mov    $0x0,%r10d
L37:
  cmp    %r12d,%r11d
  jge    L38
  movslq %r10d,%rax
  lea    (%rbx,%rax,8),%rdi
  mov    $0x0,%eax
L33:
  mov    (%rdi),%rdx
  mov    (%rdx,%rax,1),%ecx
  mov    %rax,%rdx
  add    (%rsi),%rdx
  mov    %ecx,(%rdx,%r8,1)
  add    $0x4,%rax
  cmp    %r9,%rax
  jne    L33
L38:
  add    $0x1,%r10d
  add    $0x8,%rsi
  cmp    %ebp,%r10d
  jne    L37
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L19:
  ret
```

assistant: ```c
void BringBlockBack(int **big, int **small, int size, int param, int indexi, int indexj) {
  int i, j;
  for (i = indexi * param; i < param * (indexi + 1); i++)
    for (j = indexj * param; j < param * (indexj + 1); j++)
      big[i][j] = small[i - indexi * param][j - indexj * param];
}
```
user: What is the c source code of the assembly code below?

```asm
<BringBlockBack>:
  endbr64
  imul   %ecx,%r8d
  lea    (%rcx,%r8,1),%edx
  cmp    %edx,%r8d
  jge    L44
  imul   %ecx,%r9d
  push   %r12
  movslq %r8d,%r8
  mov    %rsi,%r10
  push   %rbp
  lea    (%rdi,%r8,8),%r12
  xor    %r8d,%r8d
  push   %rbx
  lea    -0x1(%rcx),%ebx
  movslq %r9d,%rbp
  lea    (%rcx,%r9,1),%esi
  mov    %rbp,%r11
  neg    %r11
  shl    $0x2,%r11
  nopw   0x0(%rax,%rax,1)
L39:
  cmp    %esi,%r9d
  jge    L46
  movslq %r8d,%rax
  mov    (%r12,%r8,8),%rdi
  mov    (%r10,%rax,8),%rcx
  mov    %rbp,%rax
  add    %r11,%rcx
  cs nopw 0x0(%rax,%rax,1)
L45:
  mov    (%rcx,%rax,4),%edx
  mov    %edx,(%rdi,%rax,4)
  add    $0x1,%rax
  cmp    %eax,%esi
  jg     L45
L46:
  lea    0x1(%r8),%rax
  cmp    %rbx,%r8
  je     L22
  mov    %rax,%r8
  jmp    L39
  nopl   0x0(%rax)
L22:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L44:
  ret
```

assistant: ```c
void BringBlockBack(int **big, int **small, int size, int param, int indexi, int indexj) {
  int i, j;
  for (i = indexi * param; i < param * (indexi + 1); i++)
    for (j = indexj * param; j < param * (indexj + 1); j++)
      big[i][j] = small[i - indexi * param][j - indexj * param];
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<BringBlockBack>:
  endbr64
  imul   %ecx,%r8d
  push   %r15
  push   %r14
  push   %r13
  lea    (%rcx,%r8,1),%r14d
  push   %r12
  push   %rbp
  push   %rbx
  cmp    %r14d,%r8d
  jge    L91
  mov    %r9d,%eax
  mov    %ecx,%r10d
  imul   %ecx,%eax
  lea    (%rcx,%rax,1),%r9d
  cmp    %r9d,%eax
  jge    L91
  movslq %r8d,%rcx
  mov    %rdi,%rbx
  movslq %eax,%rdi
  mov    %r10d,-0x3c(%rsp)
  mov    %rcx,%rdx
  lea    0x0(,%rdi,4),%rbp
  mov    %rdi,-0x10(%rsp)
  neg    %rdx
  lea    (%rsi,%rdx,8),%r12
  mov    %r10d,%edx
  mov    %r10d,%esi
  and    $0xfffffffc,%esi
  shr    $0x2,%edx
  lea    -0x1(%rdx),%edi
  lea    (%rax,%rsi,1),%edx
  movslq %esi,%r15
  mov    %esi,-0x40(%rsp)
  movslq %edx,%rsi
  add    $0x1,%rdi
  shl    $0x2,%r15
  shl    $0x2,%rsi
  shl    $0x4,%rdi
  mov    %rsi,-0x38(%rsp)
  lea    0x1(%rdx),%esi
  mov    %esi,%r13d
  mov    %esi,-0x30(%rsp)
  movslq %esi,%rsi
  shl    $0x2,%rsi
  sub    %eax,%r13d
  mov    %rsi,-0x28(%rsp)
  lea    0x2(%rdx),%esi
  movslq %r13d,%r13
  mov    %esi,%edx
  mov    %esi,-0x2c(%rsp)
  shl    $0x2,%r13
  sub    %eax,%edx
  movslq %esi,%rax
  shl    $0x2,%rax
  movslq %edx,%rdx
  mov    %rax,-0x20(%rsp)
  mov    %rbp,%rax
  shl    $0x2,%rdx
  neg    %rax
  mov    %rax,-0x18(%rsp)
  lea    -0x1(%r10),%eax
  mov    %rdx,%r10
  mov    %eax,-0x44(%rsp)
  nop
L94:
  mov    (%rbx,%rcx,8),%rsi
  mov    (%r12,%rcx,8),%rax
  lea    (%rsi,%rbp,1),%rdx
  lea    0x4(%rax),%r11
  mov    %rdx,%r8
  sub    %r11,%r8
  cmp    $0x8,%r8
  jbe    L21
  cmpl   $0x2,-0x44(%rsp)
  jbe    L21
  xor    %r8d,%r8d
  nopw   0x0(%rax,%rax,1)
L86:
  movdqu (%rax,%r8,1),%xmm0
  movups %xmm0,(%rdx,%r8,1)
  add    $0x10,%r8
  cmp    %rdi,%r8
  jne    L86
  mov    -0x40(%rsp),%r11d
  cmp    %r11d,-0x3c(%rsp)
  je     L82
  mov    (%rax,%r15,1),%edx
  mov    -0x38(%rsp),%r11
  mov    %edx,(%rsi,%r11,1)
  cmp    -0x30(%rsp),%r9d
  jle    L82
  mov    (%rax,%r13,1),%edx
  mov    -0x28(%rsp),%r11
  mov    %edx,(%rsi,%r11,1)
  cmp    -0x2c(%rsp),%r9d
  jle    L82
  mov    (%rax,%r10,1),%eax
  mov    -0x20(%rsp),%rdx
  mov    %eax,(%rsi,%rdx,1)
L82:
  add    $0x1,%rcx
  cmp    %ecx,%r14d
  jg     L94
L91:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L21:
  mov    -0x10(%rsp),%rdx
  add    -0x18(%rsp),%rax
  nopw   0x0(%rax,%rax,1)
L88:
  mov    (%rax,%rdx,4),%r8d
  mov    %r8d,(%rsi,%rdx,4)
  add    $0x1,%rdx
  cmp    %edx,%r9d
  jg     L88
  jmp    L82
```

assistant: ```c
void BringBlockBack(int **big, int **small, int size, int param, int indexi, int indexj) {
  int i, j;
  for (i = indexi * param; i < param * (indexi + 1); i++)
    for (j = indexj * param; j < param * (indexj + 1); j++)
      big[i][j] = small[i - indexi * param][j - indexj * param];
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<jac_chain_nm8_free>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int jac_chain_nm8_free(void *mem) {
  (void)mem;
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<jac_chain_nm8_free>:
  endbr64
  mov    $0x0,%eax
  ret
```

assistant: ```c
int jac_chain_nm8_free(void *mem) {
  (void)mem;
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<jac_chain_nm8_free>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int jac_chain_nm8_free(void *mem) {
  (void)mem;
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<jac_chain_nm8_free>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int jac_chain_nm8_free(void *mem) {
  (void)mem;
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<fn104>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x0,%eax
  call   <rand@plt>
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    %edx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  mov    %edx,%ecx
  shl    $0x2,%ecx
  add    %edx,%ecx
  sub    %ecx,%eax
  mov    %eax,%edx
  lea    0x5(%rdx),%eax
  mov    $0x4,%esi
  mov    %eax,%edi
  call   <calloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  lea    D3(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    -0x8(%rbp),%rax
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D3
type_infered: string

tool: D3: fn104

assistant: ```c
char *fn104() {
  char *result;
  result = calloc(5 + rand() % 5, 4);
  strcpy((char *)result, "fn104");
  return result;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<fn104>:
  endbr64
  push   %rbx
  mov    $0x0,%eax
  call   <rand@plt>
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  sar    $0x21,%rdx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  lea    (%rdx,%rdx,4),%edx
  sub    %edx,%eax
  lea    0x5(%rax),%edi
  mov    $0x4,%esi
  call   <calloc@plt>
  mov    %rax,%rbx
  lea    D0(%rip),%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    %rbx,%rax
  pop    %rbx
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D0
data_type: string

tool: D0: fn104

assistant: ```c
char *fn104() {
  char *result;
  result = calloc(5 + rand() % 5, 4);
  strcpy((char *)result, "fn104");
  return result;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<fn104>:
  endbr64
  push   %r12
  xor    %eax,%eax
  call   <rand@plt>
  mov    $0x4,%esi
  movslq %eax,%rdx
  mov    %eax,%ecx
  imul   $0x66666667,%rdx,%rdx
  sar    $0x1f,%ecx
  sar    $0x21,%rdx
  sub    %ecx,%edx
  lea    (%rdx,%rdx,4),%edx
  sub    %edx,%eax
  lea    0x5(%rax),%edi
  call   <calloc@plt>
  lea    D8(%rip),%rsi
  mov    %rax,%r12
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    %r12,%rax
  pop    %r12
  ret
```

tool_calls: name: read_data
arguments: tag: D8
datatype: string

tool: D8: fn104

assistant: ```c
char *fn104() {
  char *result;
  result = calloc(5 + rand() % 5, 4);
  strcpy((char *)result, "fn104");
  return result;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<fn104>:
  endbr64
  push   %r12
  xor    %eax,%eax
  call   <rand@plt>
  mov    $0x4,%esi
  movslq %eax,%rdx
  mov    %eax,%ecx
  imul   $0x66666667,%rdx,%rdx
  sar    $0x1f,%ecx
  sar    $0x21,%rdx
  sub    %ecx,%edx
  lea    (%rdx,%rdx,4),%edx
  sub    %edx,%eax
  lea    0x5(%rax),%edi
  call   <calloc@plt>
  lea    D5(%rip),%rsi
  mov    %rax,%r12
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    %r12,%rax
  pop    %r12
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D5
type_infered: string

tool: D5: fn104

assistant: ```c
char *fn104() {
  char *result;
  result = calloc(5 + rand() % 5, 4);
  strcpy((char *)result, "fn104");
  return result;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<my_putnbr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  jns    L10
  mov    -0x14(%rbp),%eax
  neg    %eax
  mov    %eax,-0x4(%rbp)
  mov    $0x1,%edx
  lea    D3(%rip),%rax
  mov    %rax,%rsi
  mov    $0x1,%edi
  call   <write@plt>
  jmp    L5
L10:
  mov    -0x14(%rbp),%eax
  mov    %eax,-0x4(%rbp)
L5:
  cmpl   $0x9,-0x4(%rbp)
  jbe    L23
  mov    -0x4(%rbp),%eax
  mov    %eax,%edx
  mov    $0xcccccccd,%eax
  imul   %rdx,%rax
  shr    $0x20,%rax
  shr    $0x3,%eax
  mov    %eax,%edi
  call   <my_putnbr>
L23:
  mov    -0x4(%rbp),%ecx
  mov    %ecx,%edx
  mov    $0xcccccccd,%eax
  imul   %rdx,%rax
  shr    $0x20,%rax
  mov    %eax,%edx
  shr    $0x3,%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  sub    %eax,%ecx
  mov    %ecx,%edx
  lea    0x30(%rdx),%eax
  mov    %eax,%edi
  call   <my_putchar@plt>
  nop
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D3
type_infered: string

tool: D3: -

assistant: ```c
void my_putnbr(int nbr) {
  unsigned int u_nbr;
  if (nbr < 0) {
    u_nbr = nbr * -1;
    write(1, "-", 1);
  } else
    u_nbr = nbr;
  if (u_nbr > 9)
    my_putnbr(u_nbr / 10);
  my_putchar(u_nbr % 10 + '0');
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<my_putnbr>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  test   %edi,%edi
  js     L30
L9:
  cmp    $0x9,%ebx
  ja     L10
L39:
  mov    %ebx,%eax
  mov    $0xcccccccd,%edx
  imul   %rdx,%rax
  shr    $0x23,%rax
  lea    (%rax,%rax,4),%eax
  add    %eax,%eax
  sub    %eax,%ebx
  lea    0x30(%rbx),%edi
  call   <my_putchar@plt>
  pop    %rbx
  ret
L30:
  neg    %ebx
  mov    $0x1,%edx
  lea    D1(%rip),%rsi
  mov    $0x1,%edi
  call   <write@plt>
  jmp    L9
L10:
  mov    %ebx,%edi
  mov    $0xcccccccd,%eax
  imul   %rax,%rdi
  shr    $0x23,%rdi
  call   <my_putnbr>
  jmp    L39
```

tool_calls: name: extract_symbolic_data
arguments: tag: D1
type_guessed: string

tool: D1: -

assistant: ```c
void my_putnbr(int nbr) {
  unsigned int u_nbr;
  if (nbr < 0) {
    u_nbr = nbr * -1;
    write(1, "-", 1);
  } else
    u_nbr = nbr;
  if (u_nbr > 9)
    my_putnbr(u_nbr / 10);
  my_putchar(u_nbr % 10 + '0');
}
```