ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %r15
  mov    $0x97,%ecx
  push   %r14
  lea    D6(%rip),%r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x4e8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x4d8(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rbx
  lea    0x8(%rsp),%rsi
  mov    %rbx,%rdi
  rep stos %rax,%es:(%rdi)
  movl   $0x0,(%rdi)
  mov    %r14,%rdi
  call   <scanf@plt>
  movslq 0x8(%rsp),%rax
  test   %eax,%eax
  jle    L4
  mov    $0x1,%r15d
  lea    0xc(%rsp),%r12
  lea    D5(%rip),%rbp
  mov    $0x3,%r13d
  jmp    L53
  nopl   0x0(%rax)
L44:
  movl   $0x65,0x10(%rbx)
  movl   $0x65,0x14(%rbx)
L33:
  movslq 0x8(%rsp),%rax
  add    $0x1,%r15d
  add    $0xc,%rbx
  cmp    %r15d,%eax
  jl     L4
L53:
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    (%rbx),%ecx
  mov    0x4(%rbx),%eax
  lea    -0x1(%r15),%esi
  mov    0x8(%rbx),%edi
  cmp    %eax,%ecx
  cmovle %ecx,%eax
  cmp    %eax,%edi
  mov    %eax,%edx
  cmovle %edi,%edx
  add    $0x1,%edx
  mov    %edx,0xc(%rbx)
  movslq 0xc(%rsp),%rdx
  test   %edx,%edx
  je     L44
  cmp    $0x3,%edx
  je     L25
  mov    %r13d,%edi
  movslq %esi,%rsi
  movslq %r15d,%rax
  sub    %edx,%edi
  lea    (%rsi,%rsi,2),%rsi
  lea    (%rax,%rax,2),%rax
  movslq %edi,%rdi
  add    %rax,%rdx
  add    %rdi,%rsi
  mov    0x10(%rsp,%rsi,4),%esi
  cmp    %ecx,%esi
  cmovg  %ecx,%esi
  add    %rdi,%rax
  mov    %esi,0x10(%rsp,%rdx,4)
  movl   $0x65,0x10(%rsp,%rax,4)
  jmp    L33
  nopl   0x0(%rax)
L25:
  cmp    %edi,%ecx
  mov    %eax,0x14(%rbx)
  cmovle %ecx,%edi
  mov    %edi,0x10(%rbx)
  jmp    L33
  nopl   0x0(%rax,%rax,1)
L4:
  lea    (%rax,%rax,2),%rax
  mov    %r14,%rdi
  shl    $0x2,%rax
  mov    0x10(%rsp,%rax,1),%esi
  mov    0x14(%rsp,%rax,1),%edx
  mov    0x18(%rsp,%rax,1),%eax
  cmp    %edx,%esi
  cmovg  %edx,%esi
  cmp    %eax,%esi
  cmovg  %eax,%esi
  call   <printf@plt>
  mov    0x4d8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L40
  add    $0x4e8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L40:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D6
type_infered: string
name: extract_symbolic_data
arguments: tag: D5
type_infered: string

tool: D6: %d


tool: D5: %d

assistant: ```c
int main() {
  int n, i, c, t[101][3] = {0};
  scanf("%d\n", &n);
  for (i = 1; i <= n; ++i) {
    scanf("%d", &c);
    t[i][0] = (t[i - 1][0] > (t[i - 1][1] > t[i - 1][2] ? t[i - 1][2] : t[i - 1][1]) ? (t[i - 1][1] > t[i - 1][2] ? t[i - 1][2] : t[i - 1][1]) : t[i - 1][0]) + 1;
    if (!c) {
      t[i][1] = 101;
      t[i][2] = 101;
    } else if (c == 3) {
      t[i][1] = (t[i - 1][0] > t[i - 1][2] ? t[i - 1][2] : t[i - 1][0]);
      t[i][2] = (t[i - 1][0] > t[i - 1][1] ? t[i - 1][1] : t[i - 1][0]);
    } else {
      t[i][c] = (t[i - 1][0] > t[i - 1][3 - c] ? t[i - 1][3 - c] : t[i - 1][0]);
      t[i][3 - c] = 101;
    }
  }
  printf("%d\n", (t[n][0] > (t[n][1] > t[n][2] ? t[n][2] : t[n][1]) ? (t[n][1] > t[n][2] ? t[n][2] : t[n][1]) : t[n][0]));
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<glmMaxIndex2f>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movss  %xmm0,-0x4(%rbp)
  movss  %xmm1,-0x8(%rbp)
  mov    %edi,-0xc(%rbp)
  mov    %esi,-0x10(%rbp)
  movss  -0x4(%rbp),%xmm0
  comiss -0x8(%rbp),%xmm0
  jbe    L16
  mov    -0xc(%rbp),%eax
  jmp    L4
L16:
  mov    -0x10(%rbp),%eax
L4:
  pop    %rbp
  ret
```

assistant: ```c
int glmMaxIndex2f(float a, float b, int i1, int i2) {
  if (a > b)
    return i1;
  return i2;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<glmMaxIndex2f>:
  endbr64
  ucomiss %xmm1,%xmm0
  mov    %edi,%eax
  cmovbe %esi,%eax
  ret
```

assistant: ```c
int glmMaxIndex2f(float a, float b, int i1, int i2) {
  if (a > b)
    return i1;
  return i2;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<glmMaxIndex2f>:
  endbr64
  ucomiss %xmm1,%xmm0
  mov    %edi,%eax
  cmovbe %esi,%eax
  ret
```

assistant: ```c
int glmMaxIndex2f(float a, float b, int i1, int i2) {
  if (a > b)
    return i1;
  return i2;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<glmMaxIndex2f>:
  endbr64
  ucomiss %xmm1,%xmm0
  mov    %edi,%eax
  cmovbe %esi,%eax
  ret
```

assistant: ```c
int glmMaxIndex2f(float a, float b, int i1, int i2) {
  if (a > b)
    return i1;
  return i2;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x4,%edi
  call   <queen@plt>
  mov    %eax,%esi
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D3
type: string

tool: D3: %d


assistant: ```c
int main() {
  printf("%d\n", queen(4));
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x4,%edi
  call   <queen@plt>
  mov    %eax,%esi
  lea    D3(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: data_tag: D3
data_type: string

tool: D3: %d


assistant: ```c
int main() {
  printf("%d\n", queen(4));
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x4,%edi
  call   <queen@plt>
  lea    D1(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D1
type_guessed: string

tool: D1: %d


assistant: ```c
int main() {
  printf("%d\n", queen(4));
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x4,%edi
  call   <queen@plt>
  lea    D3(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

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

tool: D3: %d


assistant: ```c
int main() {
  printf("%d\n", queen(4));
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<threadFn2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x70,%rsp
  mov    %rdi,-0x68(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x68(%rbp),%rax
  mov    %rax,-0x50(%rbp)
  lea    -0x40(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <gettimeofday@plt>
  movl   $0x0,-0x54(%rbp)
  jmp    L21
L22:
  lea    -0x30(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <gettimeofday@plt>
  mov    $0x4,%edi
  call   <malloc@plt>
  mov    %rax,-0x48(%rbp)
  mov    -0x48(%rbp),%rax
  mov    -0x54(%rbp),%edx
  mov    %edx,(%rax)
  mov    -0x48(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  lea    -0x20(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <gettimeofday@plt>
  addl   $0x1,-0x54(%rbp)
L21:
  cmpl   $0x3e7,-0x54(%rbp)
  jle    L22
  mov    -0x50(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <gettimeofday@plt>
  jmp    L5
L25:
  mov    -0x50(%rbp),%rax
  mov    (%rax),%rax
  lea    0x3e8(%rax),%rdx
  mov    -0x50(%rbp),%rax
  mov    %rdx,(%rax)
  mov    -0x50(%rbp),%rax
  mov    0x8(%rax),%eax
  lea    -0x1(%rax),%edx
  mov    -0x50(%rbp),%rax
  mov    %edx,0x8(%rax)
L5:
  mov    -0x40(%rbp),%rdx
  mov    -0x50(%rbp),%rax
  mov    (%rax),%rax
  cmp    %rax,%rdx
  jg     L25
  mov    -0x50(%rbp),%rax
  mov    0x8(%rax),%eax
  mov    -0x38(%rbp),%ecx
  sub    %ecx,%eax
  mov    %eax,%edx
  mov    -0x50(%rbp),%rax
  mov    %edx,0x8(%rax)
  mov    -0x50(%rbp),%rax
  mov    (%rax),%rax
  mov    -0x40(%rbp),%rcx
  sub    %rcx,%rax
  mov    %rax,%rdx
  mov    -0x50(%rbp),%rax
  mov    %rdx,(%rax)
  mov    $0x0,%edi
  call   <pthread_exit@plt>
  nop
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L13
  call   <__stack_chk_fail@plt>
L13:
  leave
  ret
```

assistant: ```c
void *threadFn2(void *in_arg) {
  struct timeval *end = (struct timeval *)in_arg;
  struct timeval begin;
  struct timeval ignored0, ignored1;
  gettimeofday(&begin, NULL);
  for (int i = 0; i < 1000; i++) {
    gettimeofday(&ignored0, NULL);
    int *ptr = malloc(sizeof(int));
    *ptr = i;
    free(ptr);
    gettimeofday(&ignored1, NULL);
  }
  gettimeofday(end, NULL);
  while (begin.tv_usec > end->tv_usec) {
    end->tv_usec += 1000;
    end->tv_sec -= 1;
  }
  end->tv_sec -= begin.tv_sec;
  end->tv_usec -= begin.tv_usec;
  pthread_exit(NULL);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<threadFn2>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x40,%rsp
  mov    %rdi,%r12
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  mov    $0x0,%esi
  call   <gettimeofday@plt>
  mov    $0x0,%ebx
  lea    0x10(%rsp),%rbp
L27:
  mov    $0x0,%esi
  mov    %rbp,%rdi
  call   <gettimeofday@plt>
  mov    $0x4,%edi
  call   <malloc@plt>
  mov    %rax,%rdi
  mov    %ebx,(%rax)
  call   <free@plt>
  lea    0x20(%rsp),%rdi
  mov    $0x0,%esi
  call   <gettimeofday@plt>
  add    $0x1,%ebx
  cmp    $0x3e8,%ebx
  jne    L27
  mov    $0x0,%esi
  mov    %r12,%rdi
  call   <gettimeofday@plt>
  mov    (%rsp),%rcx
  mov    (%r12),%rax
  cmp    %rax,%rcx
  jle    L1
  mov    0x8(%r12),%edx
  sub    $0x1,%edx
L22:
  add    $0x3e8,%rax
  mov    %edx,%esi
  sub    $0x1,%edx
  cmp    %rcx,%rax
  jl     L22
  mov    %esi,0x8(%r12)
L1:
  mov    0x8(%rsp),%edx
  sub    %edx,0x8(%r12)
  sub    %rcx,%rax
  mov    %rax,(%r12)
  mov    $0x0,%edi
  call   <pthread_exit@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L38
  mov    $0x0,%eax
  add    $0x40,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L38:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void *threadFn2(void *in_arg) {
  struct timeval *end = (struct timeval *)in_arg;
  struct timeval begin;
  struct timeval ignored0, ignored1;
  gettimeofday(&begin, NULL);
  for (int i = 0; i < 1000; i++) {
    gettimeofday(&ignored0, NULL);
    int *ptr = malloc(sizeof(int));
    *ptr = i;
    free(ptr);
    gettimeofday(&ignored1, NULL);
  }
  gettimeofday(end, NULL);
  while (begin.tv_usec > end->tv_usec) {
    end->tv_usec += 1000;
    end->tv_sec -= 1;
  }
  end->tv_sec -= begin.tv_sec;
  end->tv_usec -= begin.tv_usec;
  pthread_exit(NULL);
}
```
user: What is the c source code of the assembly code below?

```asm
<threadFn2>:
  endbr64
  push   %r13
  xor    %esi,%esi
  mov    %rdi,%r13
  push   %r12
  push   %rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x48,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  lea    0x10(%rsp),%r12
  lea    0x20(%rsp),%rbp
  call   <gettimeofday@plt>
  nopw   0x0(%rax,%rax,1)
L37:
  xor    %esi,%esi
  mov    %r12,%rdi
  call   <gettimeofday@plt>
  mov    $0x4,%edi
  call   <malloc@plt>
  mov    %ebx,(%rax)
  mov    %rax,%rdi
  add    $0x1,%ebx
  call   <free@plt>
  xor    %esi,%esi
  mov    %rbp,%rdi
  call   <gettimeofday@plt>
  cmp    $0x3e8,%ebx
  jne    L37
  xor    %esi,%esi
  mov    %r13,%rdi
  call   <gettimeofday@plt>
  mov    0x8(%r13),%ecx
  mov    (%rsp),%rsi
  mov    0x0(%r13),%rax
  lea    -0x1(%rcx),%edx
  cmp    %rax,%rsi
  jle    L20
  nopl   0x0(%rax)
L27:
  add    $0x3e8,%rax
  mov    %edx,%ecx
  sub    $0x1,%edx
  cmp    %rsi,%rax
  jl     L27
L20:
  sub    0x8(%rsp),%ecx
  sub    %rsi,%rax
  xor    %edi,%edi
  mov    %ecx,0x8(%r13)
  mov    %rax,0x0(%r13)
  call   <pthread_exit@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  add    $0x48,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L4:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void *threadFn2(void *in_arg) {
  struct timeval *end = (struct timeval *)in_arg;
  struct timeval begin;
  struct timeval ignored0, ignored1;
  gettimeofday(&begin, NULL);
  for (int i = 0; i < 1000; i++) {
    gettimeofday(&ignored0, NULL);
    int *ptr = malloc(sizeof(int));
    *ptr = i;
    free(ptr);
    gettimeofday(&ignored1, NULL);
  }
  gettimeofday(end, NULL);
  while (begin.tv_usec > end->tv_usec) {
    end->tv_usec += 1000;
    end->tv_sec -= 1;
  }
  end->tv_sec -= begin.tv_sec;
  end->tv_usec -= begin.tv_usec;
  pthread_exit(NULL);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<compare_int>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  mov    -0x4(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jne    L29
  mov    $0x0,%eax
  jmp    L12
L29:
  mov    -0x4(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jge    L26
  mov    $0xffffffff,%eax
  jmp    L12
L26:
  mov    $0x1,%eax
L12:
  pop    %rbp
  ret
```

assistant: ```c
int compare_int(int a, int b) {
  if (a == b) {
    return 0;
  } else if (a < b) {
    return -1;
  }
  return 1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<compare_int>:
  endbr64
  mov    $0x0,%eax
  cmp    %esi,%edi
  je     L9
  setge  %al
  movzbl %al,%eax
  lea    -0x1(%rax,%rax,1),%eax
L9:
  ret
```

assistant: ```c
int compare_int(int a, int b) {
  if (a == b) {
    return 0;
  } else if (a < b) {
    return -1;
  }
  return 1;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<compare_int>:
  endbr64
  xor    %eax,%eax
  cmp    %esi,%edi
  je     L2
  setge  %al
  movzbl %al,%eax
  lea    -0x1(%rax,%rax,1),%eax
L2:
  ret
```

assistant: ```c
int compare_int(int a, int b) {
  if (a == b) {
    return 0;
  } else if (a < b) {
    return -1;
  }
  return 1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<compare_int>:
  endbr64
  xor    %eax,%eax
  cmp    %esi,%edi
  je     L8
  setge  %al
  movzbl %al,%eax
  lea    -0x1(%rax,%rax,1),%eax
L8:
  ret
```

assistant: ```c
int compare_int(int a, int b) {
  if (a == b) {
    return 0;
  } else if (a < b) {
    return -1;
  }
  return 1;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movq   $0x64104004,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movl   $0xbbbbbbbb,(%rax)
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main() {
  volatile int *address = 0x64104004;
  *address = 0xbbbbbbbb;
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  movl   $0xbbbbbbbb,0x64104004
  mov    $0x0,%eax
  ret
```

assistant: ```c
int main() {
  volatile int *address = 0x64104004;
  *address = 0xbbbbbbbb;
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  movl   $0xbbbbbbbb,0x64104004
  xor    %eax,%eax
  ret
```

assistant: ```c
int main() {
  volatile int *address = 0x64104004;
  *address = 0xbbbbbbbb;
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  movl   $0xbbbbbbbb,0x64104004
  xor    %eax,%eax
  ret
```

assistant: ```c
int main() {
  volatile int *address = 0x64104004;
  *address = 0xbbbbbbbb;
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<rte_ns_to_timespec>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  movq   $0x0,-0x20(%rbp)
  movq   $0x0,-0x18(%rbp)
  movq   $0x0,-0x10(%rbp)
  cmpq   $0x0,-0x30(%rbp)
  jne    L16
  mov    -0x28(%rbp),%rcx
  mov    -0x20(%rbp),%rax
  mov    -0x18(%rbp),%rdx
  mov    %rax,(%rcx)
  mov    %rdx,0x8(%rcx)
  mov    -0x10(%rbp),%rax
  mov    %rax,0x10(%rcx)
  jmp    L13
L16:
  mov    -0x30(%rbp),%rcx
  movabs $0x112e0be826d694b3,%rdx
  mov    %rcx,%rax
  imul   %rdx
  mov    %rdx,%rax
  sar    $0x1a,%rax
  sar    $0x3f,%rcx
  mov    %rcx,%rdx
  sub    %rdx,%rax
  mov    %rax,-0x20(%rbp)
  mov    -0x30(%rbp),%rcx
  movabs $0x112e0be826d694b3,%rdx
  mov    %rcx,%rax
  imul   %rdx
  mov    %rdx,%rax
  sar    $0x1a,%rax
  mov    %rcx,%rdx
  sar    $0x3f,%rdx
  sub    %rdx,%rax
  imul   $0x3b9aca00,%rax,%rdx
  mov    %rcx,%rax
  sub    %rdx,%rax
  mov    %rax,-0x18(%rbp)
  mov    -0x28(%rbp),%rcx
  mov    -0x20(%rbp),%rax
  mov    -0x18(%rbp),%rdx
  mov    %rax,(%rcx)
  mov    %rdx,0x8(%rcx)
  mov    -0x10(%rbp),%rax
  mov    %rax,0x10(%rcx)
L13:
  mov    -0x28(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
struct timespec rte_ns_to_timespec(uint64_t nsec) {
  struct timespec ts = {0, 0};
  if (nsec == 0)
    return ts;
  ts.tv_sec = nsec / 1000000000L;
  ts.tv_nsec = nsec % 1000000000L;
  return ts;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<rte_ns_to_timespec>:
  endbr64
  mov    %rsi,%rdx
  test   %rsi,%rsi
  je     L6
  movabs $0x112e0be826d694b3,%rdx
  mov    %rsi,%rax
  imul   %rdx
  sar    $0x1a,%rdx
  mov    %rsi,%rax
  sar    $0x3f,%rax
  sub    %rax,%rdx
  imul   $0x3b9aca00,%rdx,%rax
  sub    %rax,%rsi
L6:
  mov    %rdx,(%rdi)
  mov    %rsi,0x8(%rdi)
  movl   $0x0,0x10(%rdi)
  movl   $0x0,0x14(%rdi)
  mov    %rdi,%rax
  ret
```

assistant: ```c
struct timespec rte_ns_to_timespec(uint64_t nsec) {
  struct timespec ts = {0, 0};
  if (nsec == 0)
    return ts;
  ts.tv_sec = nsec / 1000000000L;
  ts.tv_nsec = nsec % 1000000000L;
  return ts;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<rte_ns_to_timespec>:
  endbr64
  mov    %rdi,%r8
  xor    %edx,%edx
  test   %rsi,%rsi
  je     L0
  movabs $0x112e0be826d694b3,%rdx
  mov    %rsi,%rax
  imul   %rdx
  mov    %rsi,%rax
  sar    $0x3f,%rax
  sar    $0x1a,%rdx
  sub    %rax,%rdx
  imul   $0x3b9aca00,%rdx,%rax
  sub    %rax,%rsi
L0:
  mov    %rdx,(%r8)
  mov    %r8,%rax
  mov    %rsi,0x8(%r8)
  movq   $0x0,0x10(%r8)
  ret
```

assistant: ```c
struct timespec rte_ns_to_timespec(uint64_t nsec) {
  struct timespec ts = {0, 0};
  if (nsec == 0)
    return ts;
  ts.tv_sec = nsec / 1000000000L;
  ts.tv_nsec = nsec % 1000000000L;
  return ts;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<rte_ns_to_timespec>:
  endbr64
  mov    %rdi,%r8
  xor    %edx,%edx
  test   %rsi,%rsi
  je     L3
  movabs $0x112e0be826d694b3,%rdx
  mov    %rsi,%rax
  imul   %rdx
  mov    %rsi,%rax
  sar    $0x3f,%rax
  sar    $0x1a,%rdx
  sub    %rax,%rdx
  imul   $0x3b9aca00,%rdx,%rax
  sub    %rax,%rsi
L3:
  mov    %rdx,(%r8)
  mov    %r8,%rax
  mov    %rsi,0x8(%r8)
  movq   $0x0,0x10(%r8)
  ret
```

assistant: ```c
struct timespec rte_ns_to_timespec(uint64_t nsec) {
  struct timespec ts = {0, 0};
  if (nsec == 0)
    return ts;
  ts.tv_sec = nsec / 1000000000L;
  ts.tv_nsec = nsec % 1000000000L;
  return ts;
}
```
user: Translate the below assembly code into C programming language:

```asm
<argv_dup>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x14(%rbp),%eax
  add    $0x1,%eax
  cltq
  shl    $0x3,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x14(%rbp),%eax
  cltq
  shl    $0x3,%eax
  mov    %eax,%edx
  mov    -0x20(%rbp),%rcx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movq   $0x0,(%rax)
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char **argv_dup(int argc, const char **argv) {
  char **new_argv = malloc((argc + 1) * sizeof(*argv));
  memcpy(new_argv, argv, argc * sizeof(*argv));
  new_argv[argc] = NULL;
  return new_argv;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<argv_dup>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebx
  mov    %rsi,%r12
  lea    0x8(,%rdi,8),%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  movslq %ebx,%r13
  lea    0x0(,%rbx,8),%edx
  mov    %r12,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  movq   $0x0,0x0(%rbp,%r13,8)
  mov    %rbp,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char **argv_dup(int argc, const char **argv) {
  char **new_argv = malloc((argc + 1) * sizeof(*argv));
  memcpy(new_argv, argv, argc * sizeof(*argv));
  new_argv[argc] = NULL;
  return new_argv;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<argv_dup>:
  endbr64
  push   %r13
  movslq %edi,%r13
  push   %r12
  lea    0x8(,%r13,8),%edi
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <malloc@plt>
  lea    0x0(,%r13,8),%edx
  mov    %rbp,%rsi
  mov    %rax,%r12
  mov    %rax,%rdi
  call   <memcpy@plt>
  movq   $0x0,(%r12,%r13,8)
  mov    %r12,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char **argv_dup(int argc, const char **argv) {
  char **new_argv = malloc((argc + 1) * sizeof(*argv));
  memcpy(new_argv, argv, argc * sizeof(*argv));
  new_argv[argc] = NULL;
  return new_argv;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<argv_dup>:
  endbr64
  push   %r13
  movslq %edi,%r13
  push   %r12
  lea    0x8(,%r13,8),%edi
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <malloc@plt>
  lea    0x0(,%r13,8),%edx
  mov    %rbp,%rsi
  mov    %rax,%r12
  mov    %rax,%rdi
  call   <memcpy@plt>
  movq   $0x0,(%r12,%r13,8)
  mov    %r12,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char **argv_dup(int argc, const char **argv) {
  char **new_argv = malloc((argc + 1) * sizeof(*argv));
  memcpy(new_argv, argv, argc * sizeof(*argv));
  new_argv[argc] = NULL;
  return new_argv;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<appender>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,%eax
  mov    %al,-0x1c(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0xff,-0x8(%rbp)
  ja     L9
  mov    -0x18(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    -0x8(%rbp),%rdx
  lea    0x1(%rdx),%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  mov    -0x18(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  add    %rax,%rdx
  movzbl -0x1c(%rbp),%eax
  mov    %al,(%rdx)
L9:
  mov    -0x18(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *appender(char *str, const char c) {
  size_t len = strlen(str);
  if (len < 256) {
    str[len + 1] = str[len];
    str[len] = c;
  }
  return str;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<appender>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %esi,%ebp
  call   <strlen@plt>
  cmp    $0xff,%rax
  ja     L9
  lea    (%rbx,%rax,1),%rdx
  movzbl (%rdx),%ecx
  mov    %cl,0x1(%rbx,%rax,1)
  mov    %bpl,(%rdx)
L9:
  mov    %rbx,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
char *appender(char *str, const char c) {
  size_t len = strlen(str);
  if (len < 256) {
    str[len + 1] = str[len];
    str[len] = c;
  }
  return str;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<appender>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbx
  mov    %esi,%ebx
  sub    $0x8,%rsp
  call   <strlen@plt>
  cmp    $0xff,%rax
  ja     L3
  movb   $0x0,0x1(%r12,%rax,1)
  mov    %bl,(%r12,%rax,1)
L3:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  ret
```

assistant: ```c
char *appender(char *str, const char c) {
  size_t len = strlen(str);
  if (len < 256) {
    str[len + 1] = str[len];
    str[len] = c;
  }
  return str;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<appender>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbx
  mov    %esi,%ebx
  sub    $0x8,%rsp
  call   <strlen@plt>
  cmp    $0xff,%rax
  ja     L5
  movb   $0x0,0x1(%r12,%rax,1)
  mov    %bl,(%r12,%rax,1)
L5:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  ret
```

assistant: ```c
char *appender(char *str, const char c) {
  size_t len = strlen(str);
  if (len < 256) {
    str[len + 1] = str[len];
    str[len] = c;
  }
  return str;
}
```