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

```asm
<main>:
  endbr64
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <megaInit@plt>
  cmp    $0x2,%ebp
  je     L39
  lea    D25(%rip),%rdi
  mov    $0x1,%esi
  xor    %eax,%eax
  call   <printf@plt>
  or     $0xffffffff,%edi
  call   <exit@plt>
L39:
  mov    0x8(%rbx),%rdi
  mov    $0xa,%edx
  xor    %esi,%esi
  call   <strtoul@plt>
  lea    0x6(%rsp),%rsi
  lea    0x4(%rsp),%rdi
  mov    %ax,0x4(%rsp)
  call   <RandomFunc@plt>
  movzwl 0x6(%rsp),%esi
  cmp    $0xe8a6,%si
  je     L24
L37:
  xor    %eax,%eax
  lea    D8(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L17
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L24:
  lea    D9(%rip),%rdi
  call   <puts@plt>
  movzwl 0x6(%rsp),%esi
  jmp    L37
L17:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D25
datatype: string
name: unpack_data
arguments: data_label: D8
datatype: string
name: unpack_data
arguments: data_label: D9
datatype: string

tool: D25: Call this program with %i arguments


tool: D8: %u


tool: D9: You win!

assistant: ```c
int main(int argc, char *argv[]) {
  unsigned short input[1];
  unsigned short output[1];
  int randomFuns_i5;
  unsigned short randomFuns_value6;
  int randomFuns_main_i7;
  {
    megaInit();
    if (argc != 2) {
      printf("Call this program with %i arguments\n", 1);
      exit(-1);
    } else {
    }
    randomFuns_i5 = 0;
    while (randomFuns_i5 < 1) {
      randomFuns_value6 = (unsigned short)strtoul(argv[randomFuns_i5 + 1], 0, 10);
      input[randomFuns_i5] = randomFuns_value6;
      randomFuns_i5++;
    }
    RandomFunc(input, output);
    if (output[0] == 59558) {
      printf("You win!\n");
    } else {
    }
    randomFuns_main_i7 = 0;
    while (randomFuns_main_i7 < 1) {
      printf("%u\n", output[randomFuns_main_i7]);
      randomFuns_main_i7++;
    }
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<Draw_V_Line>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    %edx,-0x1c(%rbp)
  mov    %ecx,%eax
  mov    %r8d,%ecx
  mov    %r9d,%edx
  mov    %al,-0x20(%rbp)
  mov    %ecx,%eax
  mov    %al,-0x24(%rbp)
  mov    %edx,%eax
  mov    %al,-0x28(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  jmp    L5
L7:
  movzbl -0x28(%rbp),%r8d
  movzbl -0x24(%rbp),%ecx
  movzbl -0x20(%rbp),%edx
  mov    -0x4(%rbp),%esi
  mov    -0x1c(%rbp),%eax
  sub    $0x8,%rsp
  mov    0x18(%rbp),%edi
  push   %rdi
  mov    0x10(%rbp),%r9
  mov    %eax,%edi
  call   <Mark_Pixel@plt>
  add    $0x10,%rsp
  addl   $0x1,-0x4(%rbp)
L5:
  mov    -0x4(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jle    L7
  nop
  nop
  leave
  ret
```

assistant: ```c
void Draw_V_Line(int Ymin, int Ymax, int X, unsigned char R, unsigned char G, unsigned char B, unsigned char *Frame, int Width) {
  int Y;
  for (Y = Ymin; Y <= Ymax; Y++)
    Mark_Pixel(X, Y, R, G, B, Frame, Width);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<Draw_V_Line>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %edx,0xc(%rsp)
  mov    0x58(%rsp),%r15d
  cmp    %esi,%edi
  jg     L4
  mov    %edi,%ebx
  mov    %esi,%r12d
  mov    %ecx,%r13d
  mov    %r8d,%r14d
  movzbl %r9b,%ebp
L1:
  movzbl %r14b,%ecx
  movzbl %r13b,%edx
  sub    $0x8,%rsp
  push   %r15
  mov    0x60(%rsp),%r9
  mov    %ebp,%r8d
  mov    %ebx,%esi
  mov    0x1c(%rsp),%edi
  call   <Mark_Pixel@plt>
  add    $0x1,%ebx
  add    $0x10,%rsp
  cmp    %ebx,%r12d
  jge    L1
L4:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void Draw_V_Line(int Ymin, int Ymax, int X, unsigned char R, unsigned char G, unsigned char B, unsigned char *Frame, int Width) {
  int Y;
  for (Y = Ymin; Y <= Ymax; Y++)
    Mark_Pixel(X, Y, R, G, B, Frame, Width);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<Draw_V_Line>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %esi,0xc(%rsp)
  mov    0x50(%rsp),%r15
  cmp    %esi,%edi
  jg     L12
  mov    %edi,%r14d
  mov    %edx,%r13d
  movzbl %r9b,%ebx
  movzbl %r8b,%r12d
  movzbl %cl,%ebp
L4:
  sub    $0x8,%rsp
  mov    %ebp,%edx
  mov    %r14d,%esi
  mov    %r15,%r9
  mov    0x60(%rsp),%eax
  mov    %ebx,%r8d
  mov    %r12d,%ecx
  mov    %r13d,%edi
  add    $0x1,%r14d
  push   %rax
  call   <Mark_Pixel@plt>
  pop    %rax
  pop    %rdx
  cmp    %r14d,0xc(%rsp)
  jge    L4
L12:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void Draw_V_Line(int Ymin, int Ymax, int X, unsigned char R, unsigned char G, unsigned char B, unsigned char *Frame, int Width) {
  int Y;
  for (Y = Ymin; Y <= Ymax; Y++)
    Mark_Pixel(X, Y, R, G, B, Frame, Width);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<Draw_V_Line>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %esi,0xc(%rsp)
  mov    0x50(%rsp),%r15
  cmp    %esi,%edi
  jg     L11
  mov    %edi,%r14d
  mov    %edx,%r13d
  movzbl %r9b,%ebx
  movzbl %r8b,%r12d
  movzbl %cl,%ebp
L1:
  sub    $0x8,%rsp
  mov    %ebp,%edx
  mov    %r14d,%esi
  mov    %r15,%r9
  mov    0x60(%rsp),%eax
  mov    %ebx,%r8d
  mov    %r12d,%ecx
  mov    %r13d,%edi
  add    $0x1,%r14d
  push   %rax
  call   <Mark_Pixel@plt>
  pop    %rax
  pop    %rdx
  cmp    %r14d,0xc(%rsp)
  jge    L1
L11:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void Draw_V_Line(int Ymin, int Ymax, int X, unsigned char R, unsigned char G, unsigned char B, unsigned char *Frame, int Width) {
  int Y;
  for (Y = Ymin; Y <= Ymax; Y++)
    Mark_Pixel(X, Y, R, G, B, Frame, Width);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<strcasecmp>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
L53:
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  movzbl (%rax),%eax
  movsbl %al,%eax
  and    $0xff,%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x20(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x20(%rbp)
  movzbl (%rax),%eax
  movsbl %al,%eax
  and    $0xff,%eax
  mov    %eax,-0x4(%rbp)
  cmpl   $0x40,-0x8(%rbp)
  jle    L32
  cmpl   $0x5a,-0x8(%rbp)
  jg     L32
  addl   $0x20,-0x8(%rbp)
L32:
  cmpl   $0x40,-0x4(%rbp)
  jle    L27
  cmpl   $0x5a,-0x4(%rbp)
  jg     L27
  addl   $0x20,-0x4(%rbp)
L27:
  mov    -0x8(%rbp),%eax
  cmp    -0x4(%rbp),%eax
  jne    L57
  cmpl   $0x0,-0x8(%rbp)
  jne    L53
L57:
  mov    -0x8(%rbp),%eax
  sub    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int strcasecmp(const char *a, const char *b) {
  int ca, cb;
  do {
    ca = *a++ & 0xff;
    cb = *b++ & 0xff;
    if (ca >= 'A' && ca <= 'Z')
      ca += 'a' - 'A';
    if (cb >= 'A' && cb <= 'Z')
      cb += 'a' - 'A';
  } while (ca == cb && ca != '\0');
  return ca - cb;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<strcasecmp>:
  endbr64
  mov    $0x0,%ecx
L10:
  movzbl (%rdi,%rcx,1),%eax
  movzbl (%rsi,%rcx,1),%edx
  lea    -0x41(%rax),%r9d
  lea    0x20(%rax),%r8d
  cmp    $0x1a,%r9d
  cmovb  %r8d,%eax
  lea    -0x41(%rdx),%r9d
  lea    0x20(%rdx),%r8d
  cmp    $0x1a,%r9d
  cmovb  %r8d,%edx
  add    $0x1,%rcx
  cmp    %edx,%eax
  jne    L1
  test   %eax,%eax
  jne    L10
L1:
  sub    %edx,%eax
  ret
```

assistant: ```c
int strcasecmp(const char *a, const char *b) {
  int ca, cb;
  do {
    ca = *a++ & 0xff;
    cb = *b++ & 0xff;
    if (ca >= 'A' && ca <= 'Z')
      ca += 'a' - 'A';
    if (cb >= 'A' && cb <= 'Z')
      cb += 'a' - 'A';
  } while (ca == cb && ca != '\0');
  return ca - cb;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<strcasecmp>:
  endbr64
  xor    %ecx,%ecx
  cs nopw 0x0(%rax,%rax,1)
L8:
  movzbl (%rdi,%rcx,1),%eax
  movzbl (%rsi,%rcx,1),%edx
  lea    -0x41(%rax),%r9d
  lea    0x20(%rax),%r8d
  cmp    $0x1a,%r9d
  lea    -0x41(%rdx),%r9d
  cmovb  %r8d,%eax
  cmp    $0x1a,%r9d
  lea    0x20(%rdx),%r8d
  cmovb  %r8d,%edx
  add    $0x1,%rcx
  cmp    %edx,%eax
  jne    L7
  test   %eax,%eax
  jne    L8
L7:
  sub    %edx,%eax
  ret
```

assistant: ```c
int strcasecmp(const char *a, const char *b) {
  int ca, cb;
  do {
    ca = *a++ & 0xff;
    cb = *b++ & 0xff;
    if (ca >= 'A' && ca <= 'Z')
      ca += 'a' - 'A';
    if (cb >= 'A' && cb <= 'Z')
      cb += 'a' - 'A';
  } while (ca == cb && ca != '\0');
  return ca - cb;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<strcasecmp>:
  endbr64
  xor    %ecx,%ecx
  cs nopw 0x0(%rax,%rax,1)
L7:
  movzbl (%rdi,%rcx,1),%eax
  movzbl (%rsi,%rcx,1),%edx
  lea    -0x41(%rax),%r9d
  lea    0x20(%rax),%r8d
  cmp    $0x1a,%r9d
  lea    -0x41(%rdx),%r9d
  cmovb  %r8d,%eax
  cmp    $0x1a,%r9d
  lea    0x20(%rdx),%r8d
  cmovb  %r8d,%edx
  add    $0x1,%rcx
  test   %eax,%eax
  je     L3
  cmp    %eax,%edx
  je     L7
L3:
  sub    %edx,%eax
  ret
```

assistant: ```c
int strcasecmp(const char *a, const char *b) {
  int ca, cb;
  do {
    ca = *a++ & 0xff;
    cb = *b++ & 0xff;
    if (ca >= 'A' && ca <= 'Z')
      ca += 'a' - 'A';
    if (cb >= 'A' && cb <= 'Z')
      cb += 'a' - 'A';
  } while (ca == cb && ca != '\0');
  return ca - cb;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<freestr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  movq   $0x0,-0x8(%rbp)
  nop
  leave
  ret
```

assistant: ```c
void freestr(void *string) {
  char *str = *(char **)string;
  free(str);
  str = NULL;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<freestr>:
  endbr64
  sub    $0x8,%rsp
  mov    (%rdi),%rdi
  call   <free@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void freestr(void *string) {
  char *str = *(char **)string;
  free(str);
  str = NULL;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<freestr>:
  endbr64
  mov    (%rdi),%rdi
  jmp    1050 <free@plt>
```

assistant: ```c
void freestr(void *string) {
  char *str = *(char **)string;
  free(str);
  str = NULL;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<RandomFunc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  add    $0x3686696a,%eax
  xor    $0xce5d000b,%eax
  mov    %eax,-0xc(%rbp)
  movl   $0x0,-0x10(%rbp)
  nop
  mov    -0xc(%rbp),%eax
  ror    $0x9,%eax
  imul   $0x1e29b9f2,%eax,%edx
  mov    -0x20(%rbp),%rax
  mov    %edx,(%rax)
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L5
  call   <__stack_chk_fail@plt>
L5:
  leave
  ret
```

assistant: ```c
void RandomFunc(unsigned int input[1], unsigned int output[1]) {
  unsigned int state[1];
  unsigned int local2;
  unsigned int local1;
  unsigned short copy11;
  {
    state[0UL] = (input[0UL] + 914778474UL) ^ 3462201355U;
    local1 = 0UL;
    while (local1 < 0U) {
      local2 = 0UL;
      while (local2 < 0U) {
        if (state[0UL] == local2 + local1) {
          state[0UL] |= ((state[local2] + state[0UL]) & 15U) << 2UL;
        } else {
          copy11 = *((unsigned short *)(&state[local1]) + 1);
          *((unsigned short *)(&state[local1]) + 1) = *((unsigned short *)(&state[local1]) + 0);
          *((unsigned short *)(&state[local1]) + 0) = copy11;
        }
        local2 += 2UL;
      }
      local1++;
    }
    output[0UL] = ((state[0UL] >> 9U) | (state[0UL] << 23U)) * 506051058U;
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<RandomFunc>:
  endbr64
  mov    (%rdi),%eax
  add    $0x3686696a,%eax
  xor    $0xce5d000b,%eax
  ror    $0x9,%eax
  imul   $0x1e29b9f2,%eax,%eax
  mov    %eax,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned int input[1], unsigned int output[1]) {
  unsigned int state[1];
  unsigned int local2;
  unsigned int local1;
  unsigned short copy11;
  {
    state[0UL] = (input[0UL] + 914778474UL) ^ 3462201355U;
    local1 = 0UL;
    while (local1 < 0U) {
      local2 = 0UL;
      while (local2 < 0U) {
        if (state[0UL] == local2 + local1) {
          state[0UL] |= ((state[local2] + state[0UL]) & 15U) << 2UL;
        } else {
          copy11 = *((unsigned short *)(&state[local1]) + 1);
          *((unsigned short *)(&state[local1]) + 1) = *((unsigned short *)(&state[local1]) + 0);
          *((unsigned short *)(&state[local1]) + 0) = copy11;
        }
        local2 += 2UL;
      }
      local1++;
    }
    output[0UL] = ((state[0UL] >> 9U) | (state[0UL] << 23U)) * 506051058U;
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<RandomFunc>:
  endbr64
  mov    (%rdi),%eax
  add    $0x3686696a,%eax
  xor    $0xce5d000b,%eax
  ror    $0x9,%eax
  imul   $0x1e29b9f2,%eax,%eax
  mov    %eax,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned int input[1], unsigned int output[1]) {
  unsigned int state[1];
  unsigned int local2;
  unsigned int local1;
  unsigned short copy11;
  {
    state[0UL] = (input[0UL] + 914778474UL) ^ 3462201355U;
    local1 = 0UL;
    while (local1 < 0U) {
      local2 = 0UL;
      while (local2 < 0U) {
        if (state[0UL] == local2 + local1) {
          state[0UL] |= ((state[local2] + state[0UL]) & 15U) << 2UL;
        } else {
          copy11 = *((unsigned short *)(&state[local1]) + 1);
          *((unsigned short *)(&state[local1]) + 1) = *((unsigned short *)(&state[local1]) + 0);
          *((unsigned short *)(&state[local1]) + 0) = copy11;
        }
        local2 += 2UL;
      }
      local1++;
    }
    output[0UL] = ((state[0UL] >> 9U) | (state[0UL] << 23U)) * 506051058U;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<RandomFunc>:
  endbr64
  mov    (%rdi),%eax
  add    $0x3686696a,%eax
  xor    $0xce5d000b,%eax
  ror    $0x9,%eax
  imul   $0x1e29b9f2,%eax,%eax
  mov    %eax,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned int input[1], unsigned int output[1]) {
  unsigned int state[1];
  unsigned int local2;
  unsigned int local1;
  unsigned short copy11;
  {
    state[0UL] = (input[0UL] + 914778474UL) ^ 3462201355U;
    local1 = 0UL;
    while (local1 < 0U) {
      local2 = 0UL;
      while (local2 < 0U) {
        if (state[0UL] == local2 + local1) {
          state[0UL] |= ((state[local2] + state[0UL]) & 15U) << 2UL;
        } else {
          copy11 = *((unsigned short *)(&state[local1]) + 1);
          *((unsigned short *)(&state[local1]) + 1) = *((unsigned short *)(&state[local1]) + 0);
          *((unsigned short *)(&state[local1]) + 0) = copy11;
        }
        local2 += 2UL;
      }
      local1++;
    }
    output[0UL] = ((state[0UL] >> 9U) | (state[0UL] << 23U)) * 506051058U;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    $0x0,%edi
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  mov    $0x0,%eax
  call   <print_version@plt>
  movl   $0x0,-0x1c(%rbp)
  movl   $0x1,-0x18(%rbp)
  jmp    L6
L15:
  mov    -0x18(%rbp),%eax
  mov    %eax,%esi
  mov    $0xa,%edi
  call   <pow@plt>
  mov    %eax,-0x10(%rbp)
  mov    -0x18(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,%esi
  mov    $0xa,%edi
  call   <pow@plt>
  mov    %eax,%ecx
  mov    $0xf4240,%eax
  cltd
  idiv   %ecx
  mov    %eax,-0xc(%rbp)
  movl   $0x0,-0x14(%rbp)
  jmp    L35
L11:
  addl   $0x1,-0x1c(%rbp)
  mov    -0x10(%rbp),%eax
  mov    %eax,%edx
  lea    D21(%rip),%rax
  mov    %rax,%rsi
  lea    D23(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x10(%rbp),%edx
  mov    -0xc(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <benchmark@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rdx
  mov    -0xc(%rbp),%eax
  movq   %rdx,%xmm0
  mov    %eax,%edi
  call   <print_results@plt>
  mov    -0x1c(%rbp),%eax
  mov    %eax,%esi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x14(%rbp)
L35:
  cmpl   $0x2,-0x14(%rbp)
  jle    L11
  addl   $0x1,-0x18(%rbp)
L6:
  cmpl   $0x6,-0x18(%rbp)
  jle    L15
  mov    -0x1c(%rbp),%edx
  mov    -0x1c(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <print_summary@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: read_data
arguments: tag: D21
datatype: string
name: read_data
arguments: tag: D23
datatype: string
name: read_data
arguments: tag: D5
datatype: string

tool: D21: snanminabs

tool: D23: # c::%s:len=%d


tool: D5: ok %d benchmark finished


assistant: ```c
int main(void) {
  double elapsed;
  int count;
  int iter;
  int len;
  int i;
  int j;
  srand(time(NULL));
  print_version();
  count = 0;
  for (i = 1; i <= 6; i++) {
    len = pow(10, i);
    iter = 1000000 / pow(10, i - 1);
    for (j = 0; j < 3; j++) {
      count += 1;
      printf("# c::%s:len=%d\n", "snanminabs", len);
      elapsed = benchmark(iter, len);
      print_results(iter, elapsed);
      printf("ok %d benchmark finished\n", count);
    }
  }
  print_summary(count, count);
}
```
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    $0x8,%rsp
  mov    $0x0,%edi
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  mov    $0x0,%eax
  call   <print_version@plt>
  mov    $0x3,%r12d
  mov    $0x1,%r14d
  lea    D23(%rip),%r15
  jmp    L5
L23:
  add    $0x1,%r14d
  add    $0x3,%r12d
  cmp    $0x15,%r12d
  je     L25
L5:
  lea    -0x3(%r12),%ebx
  mov    %r14d,%esi
  mov    $0xa,%edi
  call   <pow@plt>
  mov    %eax,%ebp
  lea    -0x1(%r14),%esi
  mov    $0xa,%edi
  call   <pow@plt>
  mov    %eax,%ecx
  mov    $0xf4240,%eax
  cltd
  idiv   %ecx
  mov    %eax,%r13d
L9:
  add    $0x1,%ebx
  mov    %ebp,%edx
  mov    %r15,%rsi
  lea    D15(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %ebp,%esi
  mov    %r13d,%edi
  call   <benchmark@plt>
  mov    %r13d,%edi
  call   <print_results@plt>
  mov    %ebx,%esi
  lea    D10(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  cmp    %r12d,%ebx
  jne    L9
  jmp    L23
L25:
  mov    $0x12,%esi
  mov    $0x12,%edi
  call   <print_summary@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: parse_data
arguments: label: D23
data_type: string
name: parse_data
arguments: label: D15
data_type: string
name: parse_data
arguments: label: D10
data_type: string

tool: D23: snanminabs

tool: D15: # c::%s:len=%d


tool: D10: ok %d benchmark finished


assistant: ```c
int main(void) {
  double elapsed;
  int count;
  int iter;
  int len;
  int i;
  int j;
  srand(time(NULL));
  print_version();
  count = 0;
  for (i = 1; i <= 6; i++) {
    len = pow(10, i);
    iter = 1000000 / pow(10, i - 1);
    for (j = 0; j < 3; j++) {
      count += 1;
      printf("# c::%s:len=%d\n", "snanminabs", len);
      elapsed = benchmark(iter, len);
      print_results(iter, elapsed);
      printf("ok %d benchmark finished\n", count);
    }
  }
  print_summary(count, count);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  push   %r15
  xor    %edi,%edi
  lea    D5(%rip),%r15
  push   %r14
  lea    D9(%rip),%r14
  push   %r13
  lea    D28(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  xor    %eax,%eax
  call   <print_version@plt>
  movl   $0x3,0x8(%rsp)
  movl   $0x1,0xc(%rsp)
  nopl   0x0(%rax)
L9:
  mov    0xc(%rsp),%ebp
  mov    0x8(%rsp),%eax
  mov    $0xa,%edi
  mov    %ebp,%esi
  lea    -0x3(%rax),%r12d
  call   <pow@plt>
  lea    -0x1(%rbp),%esi
  mov    $0xa,%edi
  mov    %eax,%ebx
  call   <pow@plt>
  mov    %eax,%r8d
  mov    $0xf4240,%eax
  cltd
  idiv   %r8d
  mov    %eax,%ebp
L12:
  mov    %ebx,%edx
  mov    %r15,%rsi
  mov    %r14,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %ebx,%esi
  mov    %ebp,%edi
  add    $0x1,%r12d
  call   <benchmark@plt>
  mov    %ebp,%edi
  call   <print_results@plt>
  xor    %eax,%eax
  mov    %r12d,%esi
  mov    %r13,%rdi
  call   <printf@plt>
  cmp    0x8(%rsp),%r12d
  jne    L12
  lea    0x3(%r12),%eax
  addl   $0x1,0xc(%rsp)
  mov    %eax,0x8(%rsp)
  cmp    $0x12,%r12d
  jne    L9
  mov    $0x12,%esi
  mov    $0x12,%edi
  call   <print_summary@plt>
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D5
type_infered: string
name: disassemble_data
arguments: data_tag: D9
type_infered: string
name: disassemble_data
arguments: data_tag: D28
type_infered: string

tool: D5: snanminabs

tool: D9: # c::%s:len=%d


tool: D28: ok %d benchmark finished


assistant: ```c
int main(void) {
  double elapsed;
  int count;
  int iter;
  int len;
  int i;
  int j;
  srand(time(NULL));
  print_version();
  count = 0;
  for (i = 1; i <= 6; i++) {
    len = pow(10, i);
    iter = 1000000 / pow(10, i - 1);
    for (j = 0; j < 3; j++) {
      count += 1;
      printf("# c::%s:len=%d\n", "snanminabs", len);
      elapsed = benchmark(iter, len);
      print_results(iter, elapsed);
      printf("ok %d benchmark finished\n", count);
    }
  }
  print_summary(count, count);
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %r15
  xor    %edi,%edi
  lea    D17(%rip),%r15
  push   %r14
  lea    D8(%rip),%r14
  push   %r13
  lea    D1(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  xor    %eax,%eax
  call   <print_version@plt>
  movl   $0x3,0x8(%rsp)
  movl   $0x1,0xc(%rsp)
  nopl   0x0(%rax)
L7:
  mov    0xc(%rsp),%ebp
  mov    0x8(%rsp),%eax
  mov    $0xa,%edi
  mov    %ebp,%esi
  lea    -0x3(%rax),%r12d
  call   <pow@plt>
  lea    -0x1(%rbp),%esi
  mov    $0xa,%edi
  mov    %eax,%ebx
  call   <pow@plt>
  mov    %eax,%r8d
  mov    $0xf4240,%eax
  cltd
  idiv   %r8d
  mov    %eax,%ebp
L2:
  mov    %ebx,%edx
  mov    %r15,%rsi
  mov    %r14,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %ebx,%esi
  mov    %ebp,%edi
  add    $0x1,%r12d
  call   <benchmark@plt>
  mov    %ebp,%edi
  call   <print_results@plt>
  xor    %eax,%eax
  mov    %r12d,%esi
  mov    %r13,%rdi
  call   <printf@plt>
  cmp    0x8(%rsp),%r12d
  jne    L2
  lea    0x3(%r12),%eax
  addl   $0x1,0xc(%rsp)
  mov    %eax,0x8(%rsp)
  cmp    $0x12,%r12d
  jne    L7
  mov    $0x12,%esi
  mov    $0x12,%edi
  call   <print_summary@plt>
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: unpack_data
arguments: tag: D17
type_infered: string
name: unpack_data
arguments: tag: D8
type_infered: string
name: unpack_data
arguments: tag: D1
type_infered: string

tool: D17: snanminabs

tool: D8: # c::%s:len=%d


tool: D1: ok %d benchmark finished


assistant: ```c
int main(void) {
  double elapsed;
  int count;
  int iter;
  int len;
  int i;
  int j;
  srand(time(NULL));
  print_version();
  count = 0;
  for (i = 1; i <= 6; i++) {
    len = pow(10, i);
    iter = 1000000 / pow(10, i - 1);
    for (j = 0; j < 3; j++) {
      count += 1;
      printf("# c::%s:len=%d\n", "snanminabs", len);
      elapsed = benchmark(iter, len);
      print_results(iter, elapsed);
      printf("ok %d benchmark finished\n", count);
    }
  }
  print_summary(count, count);
}
```