ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 这段汇编代码用C语言表示是什么样的:

```asm
<judge>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  movl   $0x0,-0x4(%rbp)
  movl   $0x0,-0xc(%rbp)
  movl   $0x0,-0x8(%rbp)
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x55555556,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x55555556,%rdx,%rdx
  shr    $0x20,%rdx
  mov    %eax,%esi
  sar    $0x1f,%esi
  mov    %edx,%ecx
  sub    %esi,%ecx
  mov    %ecx,%edx
  add    %edx,%edx
  add    %ecx,%edx
  sub    %edx,%eax
  mov    %eax,%ecx
  mov    -0x4(%rbp),%eax
  add    %ecx,%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  jmp    L31
L40:
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x55555556,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  add    %eax,-0xc(%rbp)
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x55555556,%rdx,%rdx
  shr    $0x20,%rdx
  mov    %eax,%esi
  sar    $0x1f,%esi
  mov    %edx,%ecx
  sub    %esi,%ecx
  mov    %ecx,%edx
  add    %edx,%edx
  add    %ecx,%edx
  sub    %edx,%eax
  mov    %eax,%ecx
  mov    -0x4(%rbp),%eax
  add    %ecx,%eax
  mov    %eax,-0x8(%rbp)
L31:
  cmpl   $0x2,-0x8(%rbp)
  jg     L40
  cmpl   $0x2,-0x8(%rbp)
  je     L27
  cmpl   $0x2,-0x8(%rbp)
  jg     L15
  cmpl   $0x0,-0x8(%rbp)
  je     L46
  cmpl   $0x1,-0x8(%rbp)
  jmp    L15
L27:
  addl   $0x1,-0xc(%rbp)
  jmp    L15
L46:
  nop
L15:
  mov    -0xc(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int judge(int a) {
  int h = 0, sum = 0;
  int r = 0;
  h = a / 3;
  r = a % 3 + h;
  sum = h;
  while (r >= 3) {
    h = r / 3;
    sum += h;
    r = r % 3 + h;
  }
  switch (r) {
  case 0:
    break;
  case 1:
    break;
  case 2:
    sum += 1;
    break;
  }
  return sum;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<judge>:
  endbr64
  movslq %edi,%rax
  imul   $0x55555556,%rax,%rax
  shr    $0x20,%rax
  mov    %edi,%edx
  sar    $0x1f,%edx
  sub    %edx,%eax
  lea    (%rax,%rax,1),%edx
  sub    %edx,%edi
  cmp    $0x2,%edi
  jle    L19
L18:
  movslq %edi,%rdx
  imul   $0x55555556,%rdx,%rdx
  shr    $0x20,%rdx
  mov    %edi,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  add    %edx,%eax
  add    %edx,%edx
  sub    %edx,%edi
  cmp    $0x2,%edi
  jg     L18
L19:
  cmp    $0x2,%edi
  sete   %dl
  movzbl %dl,%edx
  add    %edx,%eax
  ret
```

assistant: ```c
int judge(int a) {
  int h = 0, sum = 0;
  int r = 0;
  h = a / 3;
  r = a % 3 + h;
  sum = h;
  while (r >= 3) {
    h = r / 3;
    sum += h;
    r = r % 3 + h;
  }
  switch (r) {
  case 0:
    break;
  case 1:
    break;
  case 2:
    sum += 1;
    break;
  }
  return sum;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<judge>:
  endbr64
  movslq %edi,%rax
  mov    %edi,%edx
  imul   $0x55555556,%rax,%rax
  sar    $0x1f,%edx
  shr    $0x20,%rax
  sub    %edx,%eax
  lea    (%rax,%rax,1),%edx
  sub    %edx,%edi
  cmp    $0x2,%edi
  jle    L11
  mov    $0xaaaaaaab,%ecx
  nopl   0x0(%rax,%rax,1)
L9:
  mov    %edi,%edx
  imul   %rcx,%rdx
  shr    $0x21,%rdx
  add    %edx,%eax
  add    %edx,%edx
  sub    %edx,%edi
  cmp    $0x2,%edi
  jg     L9
L11:
  xor    %edx,%edx
  cmp    $0x2,%edi
  sete   %dl
  add    %edx,%eax
  ret
```

assistant: ```c
int judge(int a) {
  int h = 0, sum = 0;
  int r = 0;
  h = a / 3;
  r = a % 3 + h;
  sum = h;
  while (r >= 3) {
    h = r / 3;
    sum += h;
    r = r % 3 + h;
  }
  switch (r) {
  case 0:
    break;
  case 1:
    break;
  case 2:
    sum += 1;
    break;
  }
  return sum;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<judge>:
  endbr64
  movslq %edi,%rax
  mov    %edi,%edx
  imul   $0x55555556,%rax,%rax
  sar    $0x1f,%edx
  shr    $0x20,%rax
  sub    %edx,%eax
  lea    (%rax,%rax,1),%edx
  sub    %edx,%edi
  cmp    $0x2,%edi
  jle    L19
  mov    $0xaaaaaaab,%ecx
  nopl   0x0(%rax,%rax,1)
L3:
  mov    %edi,%edx
  imul   %rcx,%rdx
  shr    $0x21,%rdx
  add    %edx,%eax
  add    %edx,%edx
  sub    %edx,%edi
  cmp    $0x2,%edi
  jg     L3
L19:
  xor    %edx,%edx
  cmp    $0x2,%edi
  sete   %dl
  add    %edx,%eax
  ret
```

assistant: ```c
int judge(int a) {
  int h = 0, sum = 0;
  int r = 0;
  h = a / 3;
  r = a % 3 + h;
  sum = h;
  while (r >= 3) {
    h = r / 3;
    sum += h;
    r = r % 3 + h;
  }
  switch (r) {
  case 0:
    break;
  case 1:
    break;
  case 2:
    sum += 1;
    break;
  }
  return sum;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<modexp>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movq   $0x1,-0x8(%rbp)
  jmp    L8
L29:
  mov    -0x20(%rbp),%rax
  and    $0x1,%eax
  test   %rax,%rax
  je     L14
  mov    -0x8(%rbp),%rax
  imul   -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rcx
  movabs $0x89705f3112a28fe5,%rdx
  mov    %rcx,%rax
  imul   %rdx
  lea    (%rdx,%rcx,1),%rax
  sar    $0x1d,%rax
  mov    %rcx,%rdx
  sar    $0x3f,%rdx
  sub    %rdx,%rax
  imul   $0x3b9aca07,%rax,%rdx
  mov    %rcx,%rax
  sub    %rdx,%rax
  mov    %rax,-0x8(%rbp)
L14:
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdx
  shr    $0x3f,%rdx
  add    %rdx,%rax
  sar    %rax
  mov    %rax,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  imul   %rax,%rax
  mov    %rax,-0x18(%rbp)
  mov    -0x18(%rbp),%rcx
  movabs $0x89705f3112a28fe5,%rdx
  mov    %rcx,%rax
  imul   %rdx
  lea    (%rdx,%rcx,1),%rax
  sar    $0x1d,%rax
  mov    %rcx,%rdx
  sar    $0x3f,%rdx
  sub    %rdx,%rax
  imul   $0x3b9aca07,%rax,%rdx
  mov    %rcx,%rax
  sub    %rdx,%rax
  mov    %rax,-0x18(%rbp)
L8:
  cmpq   $0x0,-0x20(%rbp)
  jne    L29
  mov    -0x8(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
long long int modexp(long long int base, long long int exp) {
  long long int res = 1;
  while (exp) {
    if (exp % 2) {
      res *= base;
      res %= 1000000007;
    }
    exp /= 2;
    base *= base;
    base %= 1000000007;
  }
  return res;
}
```
user: What is the c source code of the assembly code below?

```asm
<modexp>:
  endbr64
  mov    %rdi,%rcx
  mov    %rsi,%r10
  test   %rsi,%rsi
  je     L21
  mov    $0x1,%edi
  movabs $0x89705f3112a28fe5,%r9
  jmp    L15
L2:
  mov    %r10,%rsi
  shr    $0x3f,%rsi
  add    %r10,%rsi
  sar    %rsi
  mov    %rsi,%r10
  imul   %rcx,%rcx
  mov    %rcx,%r8
  mov    %rcx,%rax
  imul   %r9
  lea    (%rdx,%rcx,1),%rcx
  sar    $0x1d,%rcx
  mov    %r8,%rax
  sar    $0x3f,%rax
  sub    %rax,%rcx
  imul   $0x3b9aca07,%rcx,%rax
  mov    %r8,%rcx
  sub    %rax,%rcx
  test   %rsi,%rsi
  je     L14
L15:
  test   $0x1,%r10b
  je     L2
  imul   %rcx,%rdi
  mov    %rdi,%rsi
  mov    %rdi,%rax
  imul   %r9
  lea    (%rdx,%rdi,1),%rdi
  sar    $0x1d,%rdi
  mov    %rsi,%rax
  sar    $0x3f,%rax
  sub    %rax,%rdi
  imul   $0x3b9aca07,%rdi,%rax
  mov    %rsi,%rdi
  sub    %rax,%rdi
  jmp    L2
L21:
  mov    $0x1,%edi
L14:
  mov    %rdi,%rax
  ret
```

assistant: ```c
long long int modexp(long long int base, long long int exp) {
  long long int res = 1;
  while (exp) {
    if (exp % 2) {
      res *= base;
      res %= 1000000007;
    }
    exp /= 2;
    base *= base;
    base %= 1000000007;
  }
  return res;
}
```
user: What is the c source code of the assembly code below:

```asm
<modexp>:
  endbr64
  mov    %rdi,%rcx
  mov    $0x1,%r8d
  test   %rsi,%rsi
  je     L8
  movabs $0x89705f3112a28fe5,%r9
  nopl   0x0(%rax)
L0:
  test   $0x1,%sil
  je     L29
  mov    %r8,%rdi
  imul   %rcx,%rdi
  mov    %rdi,%rax
  imul   %r9
  mov    %rdi,%rax
  sar    $0x3f,%rax
  lea    (%rdx,%rdi,1),%r8
  sar    $0x1d,%r8
  sub    %rax,%r8
  imul   $0x3b9aca07,%r8,%rax
  sub    %rax,%rdi
  mov    %rdi,%r8
L29:
  imul   %rcx,%rcx
  mov    %rsi,%rax
  shr    $0x3f,%rax
  add    %rax,%rsi
  mov    %rcx,%rax
  mov    %rcx,%rdi
  mul    %r9
  shr    $0x1d,%rdx
  imul   $0x3b9aca07,%rdx,%rax
  sub    %rax,%rdi
  sar    %rsi
  mov    %rdi,%rcx
  jne    L0
L8:
  mov    %r8,%rax
  ret
```

assistant: ```c
long long int modexp(long long int base, long long int exp) {
  long long int res = 1;
  while (exp) {
    if (exp % 2) {
      res *= base;
      res %= 1000000007;
    }
    exp /= 2;
    base *= base;
    base %= 1000000007;
  }
  return res;
}
```
user: Translate the below assembly code into C programming language:

```asm
<modexp>:
  endbr64
  mov    %rdi,%rcx
  mov    $0x1,%r8d
  test   %rsi,%rsi
  je     L19
  movabs $0x89705f3112a28fe5,%r9
  nopl   0x0(%rax)
L28:
  test   $0x1,%sil
  je     L0
  mov    %r8,%rdi
  imul   %rcx,%rdi
  mov    %rdi,%rax
  imul   %r9
  mov    %rdi,%rax
  sar    $0x3f,%rax
  lea    (%rdx,%rdi,1),%r8
  sar    $0x1d,%r8
  sub    %rax,%r8
  imul   $0x3b9aca07,%r8,%rax
  sub    %rax,%rdi
  mov    %rdi,%r8
L0:
  imul   %rcx,%rcx
  mov    %rsi,%rax
  shr    $0x3f,%rax
  add    %rax,%rsi
  mov    %rcx,%rax
  mov    %rcx,%rdi
  mul    %r9
  shr    $0x1d,%rdx
  imul   $0x3b9aca07,%rdx,%rax
  sub    %rax,%rdi
  sar    %rsi
  mov    %rdi,%rcx
  jne    L28
L19:
  mov    %r8,%rax
  ret
```

assistant: ```c
long long int modexp(long long int base, long long int exp) {
  long long int res = 1;
  while (exp) {
    if (exp % 2) {
      res *= base;
      res %= 1000000007;
    }
    exp /= 2;
    base *= base;
    base %= 1000000007;
  }
  return res;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<linear_img>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %ecx,-0x2c(%rbp)
  mov    %r8d,-0x30(%rbp)
  mov    %r9d,-0x34(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L0
L33:
  mov    -0x18(%rbp),%rax
  movzwl (%rax),%eax
  cwtl
  imul   -0x2c(%rbp),%eax
  mov    %eax,%edx
  mov    -0x20(%rbp),%rax
  movzwl (%rax),%eax
  cwtl
  imul   -0x30(%rbp),%eax
  add    %eax,%edx
  mov    -0x34(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jns    L30
  cmpl   $0x0,0x10(%rbp)
  je     L22
  movl   $0x0,-0x4(%rbp)
  jmp    L30
L22:
  mov    -0x4(%rbp),%eax
  mov    %eax,%edx
  neg    %edx
  cmovns %edx,%eax
  mov    %eax,-0x4(%rbp)
L30:
  cmpl   $0xfff,-0x4(%rbp)
  jle    L9
  movl   $0xfff,-0x4(%rbp)
L9:
  mov    -0x4(%rbp),%eax
  mov    %eax,%edx
  mov    -0x28(%rbp),%rax
  mov    %dx,(%rax)
  addq   $0x2,-0x28(%rbp)
  addq   $0x2,-0x18(%rbp)
  addq   $0x2,-0x20(%rbp)
  addl   $0x1,-0x8(%rbp)
L0:
  mov    -0x8(%rbp),%eax
  cmp    0x18(%rbp),%eax
  jl     L33
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void linear_img(short int *src0, short int *src1, short int *dst, int c0, int c1, int c2, int ov_flag, int npix) {
  int i, res;
  for (i = 0; i < npix; ++i) {
    res = (*src0) * c0 + (*src1) * c1 + c2;
    if (res < 0) {
      if (ov_flag)
        res = 0;
      else
        res = (((res) > 0) ? ((res)) : (-(res)));
    }
    if (res > 4096 - 1)
      res = 4096 - 1;
    *dst = res;
    dst++;
    src0++;
    src1++;
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<linear_img>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rsi,%r10
  mov    0x20(%rsp),%ebp
  mov    0x28(%rsp),%esi
  test   %esi,%esi
  jle    L0
  mov    %rdx,%r11
  mov    %ecx,%ebx
  mov    %esi,%esi
  mov    $0x0,%edx
  mov    $0x0,%r12d
  jmp    L58
L9:
  mov    $0xfff,%ecx
  cmp    %ecx,%eax
  cmovle %eax,%ecx
L45:
  mov    %cx,(%r11,%rdx,2)
  add    $0x1,%rdx
  cmp    %rsi,%rdx
  je     L0
L58:
  movswl (%rdi,%rdx,2),%eax
  imul   %ebx,%eax
  movswl (%r10,%rdx,2),%ecx
  imul   %r8d,%ecx
  add    %ecx,%eax
  add    %r9d,%eax
  jns    L9
  mov    %r12d,%ecx
  test   %ebp,%ebp
  jne    L45
  mov    %eax,%ecx
  neg    %ecx
  cmovns %ecx,%eax
  jmp    L9
L0:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void linear_img(short int *src0, short int *src1, short int *dst, int c0, int c1, int c2, int ov_flag, int npix) {
  int i, res;
  for (i = 0; i < npix; ++i) {
    res = (*src0) * c0 + (*src1) * c1 + c2;
    if (res < 0) {
      if (ov_flag)
        res = 0;
      else
        res = (((res) > 0) ? ((res)) : (-(res)));
    }
    if (res > 4096 - 1)
      res = 4096 - 1;
    *dst = res;
    dst++;
    src0++;
    src1++;
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<linear_img>:
  endbr64
  push   %rbp
  push   %rbx
  movslq 0x20(%rsp),%rbx
  mov    0x18(%rsp),%ebp
  test   %ebx,%ebx
  jle    L19
  mov    %rdx,%r10
  mov    %ecx,%r11d
  xor    %edx,%edx
  jmp    L52
  nopl   (%rax)
L24:
  mov    $0xfff,%ecx
  cmp    %ecx,%eax
  cmovle %eax,%ecx
L8:
  mov    %cx,(%r10,%rdx,2)
  add    $0x1,%rdx
  cmp    %rbx,%rdx
  je     L19
L52:
  movswl (%rdi,%rdx,2),%eax
  movswl (%rsi,%rdx,2),%ecx
  imul   %r11d,%eax
  imul   %r8d,%ecx
  add    %ecx,%eax
  add    %r9d,%eax
  jns    L24
  xor    %ecx,%ecx
  test   %ebp,%ebp
  jne    L8
  neg    %eax
  jmp    L24
  nopl   0x0(%rax)
L19:
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void linear_img(short int *src0, short int *src1, short int *dst, int c0, int c1, int c2, int ov_flag, int npix) {
  int i, res;
  for (i = 0; i < npix; ++i) {
    res = (*src0) * c0 + (*src1) * c1 + c2;
    if (res < 0) {
      if (ov_flag)
        res = 0;
      else
        res = (((res) > 0) ? ((res)) : (-(res)));
    }
    if (res > 4096 - 1)
      res = 4096 - 1;
    *dst = res;
    dst++;
    src0++;
    src1++;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<yy_1_Ticks>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void yy_1_Ticks(char *yytext, int yyleng) {
  ;
  ;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<yy_1_Ticks>:
  endbr64
  ret
```

assistant: ```c
void yy_1_Ticks(char *yytext, int yyleng) {
  ;
  ;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<yy_1_Ticks>:
  endbr64
  ret
```

assistant: ```c
void yy_1_Ticks(char *yytext, int yyleng) {
  ;
  ;
}
```
user: Translate the below assembly code into C programming language:

```asm
<dec2bin>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x1d0,%rsp
  mov    %edi,-0x1c4(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x1b8(%rbp)
  jmp    L12
L6:
  mov    -0x1c4(%rbp),%eax
  cltd
  shr    $0x1f,%edx
  add    %edx,%eax
  and    $0x1,%eax
  sub    %edx,%eax
  cltq
  mov    %rax,-0x1a8(%rbp)
  mov    -0x1a8(%rbp),%rax
  mov    %eax,%edx
  mov    -0x1b8(%rbp),%eax
  cltq
  mov    %edx,-0x1a0(%rbp,%rax,4)
  mov    -0x1c4(%rbp),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  mov    %eax,-0x1c4(%rbp)
  addl   $0x1,-0x1b8(%rbp)
L12:
  cmpl   $0x0,-0x1c4(%rbp)
  jg     L6
  movl   $0x0,-0x1b4(%rbp)
  jmp    L30
L23:
  mov    -0x1a0(%rbp),%eax
  lea    0x30(%rax),%edx
  mov    -0x1b8(%rbp),%eax
  sub    -0x1b4(%rbp),%eax
  cltq
  lea    -0x1(%rax),%rcx
  mov    -0x1b0(%rbp),%rax
  add    %rcx,%rax
  mov    %dl,(%rax)
  addl   $0x1,-0x1b4(%rbp)
L30:
  mov    -0x1b8(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0x1b4(%rbp)
  jge    L23
  mov    -0x1b0(%rbp),%rax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L27
  call   <__stack_chk_fail@plt>
L27:
  leave
  ret
```

assistant: ```c
char *dec2bin(int n) {
  long r;
  int c = 0, i, b[100];
  while (n > 0) {
    r = n % 2;
    b[c] = r;
    n = n / 2;
    c++;
  }
  char *str;
  for (i = 0; i >= c - 1; i++) {
    str[c - i - 1] = b[0] + 48;
  }
  return str;
}
```
user: What is the c source code of the assembly code below:

```asm
<dec2bin>:
  endbr64
  test   %edi,%edi
  jle    L15
  mov    $0x0,%eax
L6:
  mov    %edi,%ecx
  mov    %edi,%edx
  shr    $0x1f,%edx
  add    %edx,%edi
  sar    %edi
  mov    %eax,%edx
  add    $0x1,%eax
  cmp    $0x1,%ecx
  jg     L6
  test   %edx,%edx
  jg     L36
L41:
  mov    $0x0,%eax
L21:
  add    $0x1,%eax
  cmp    %eax,%edx
  jle    L21
L36:
  mov    $0x0,%eax
  ret
L15:
  mov    $0xffffffff,%edx
  jmp    L41
```

assistant: ```c
char *dec2bin(int n) {
  long r;
  int c = 0, i, b[100];
  while (n > 0) {
    r = n % 2;
    b[c] = r;
    n = n / 2;
    c++;
  }
  char *str;
  for (i = 0; i >= c - 1; i++) {
    str[c - i - 1] = b[0] + 48;
  }
  return str;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<dec2bin>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
char *dec2bin(int n) {
  long r;
  int c = 0, i, b[100];
  while (n > 0) {
    r = n % 2;
    b[c] = r;
    n = n / 2;
    c++;
  }
  char *str;
  for (i = 0; i >= c - 1; i++) {
    str[c - i - 1] = b[0] + 48;
  }
  return str;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<dec2bin>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
char *dec2bin(int n) {
  long r;
  int c = 0, i, b[100];
  while (n > 0) {
    r = n % 2;
    b[c] = r;
    n = n / 2;
    c++;
  }
  char *str;
  for (i = 0; i >= c - 1; i++) {
    str[c - i - 1] = b[0] + 48;
  }
  return str;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<vslprintf>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %ecx,-0x2c(%rbp)
  mov    -0x2c(%rbp),%ecx
  mov    -0x28(%rbp),%rdx
  mov    -0x20(%rbp),%rsi
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <vsnprintf@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x20(%rbp),%rax
  lea    -0x1(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int vslprintf(char *dst, size_t size, const char *fmt, va_list ap) {
  int ret;
  ret = vsnprintf(dst, size, fmt, ap);
  dst[size - 1] = '\0';
  return ret;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<vslprintf>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    %rsi,%rbx
  call   <vsnprintf@plt>
  movb   $0x0,-0x1(%rbp,%rbx,1)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int vslprintf(char *dst, size_t size, const char *fmt, va_list ap) {
  int ret;
  ret = vsnprintf(dst, size, fmt, ap);
  dst[size - 1] = '\0';
  return ret;
}
```
user: What is the c source code of the assembly code below:

```asm
<vslprintf>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x8,%rsp
  call   <vsnprintf@plt>
  movb   $0x0,-0x1(%rbp,%rbx,1)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int vslprintf(char *dst, size_t size, const char *fmt, va_list ap) {
  int ret;
  ret = vsnprintf(dst, size, fmt, ap);
  dst[size - 1] = '\0';
  return ret;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<vslprintf>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x8,%rsp
  call   <vsnprintf@plt>
  movb   $0x0,-0x1(%rbp,%rbx,1)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int vslprintf(char *dst, size_t size, const char *fmt, va_list ap) {
  int ret;
  ret = vsnprintf(dst, size, fmt, ap);
  dst[size - 1] = '\0';
  return ret;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  mov    %edx,-0xc(%rbp)
  mov    %ecx,-0x10(%rbp)
  lea    -0x8(%rbp),%rax
  mov    %rax,%rsi
  lea    D39(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  jmp    L41
L3:
  movl   $0x0,-0xc(%rbp)
  jmp    L58
L1:
  lea    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D39(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  cmpl   $0x0,-0xc(%rbp)
  je     L2
  mov    -0x10(%rbp),%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  sub    $0x4,%eax
  jmp    L35
L2:
  mov    -0x10(%rbp),%eax
  sub    $0x4,%eax
L35:
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%edx
  neg    %edx
  cmovns %edx,%eax
  add    %eax,-0xc(%rbp)
L58:
  mov    -0x8(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0x8(%rbp)
  test   %eax,%eax
  jne    L1
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D27(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L41:
  lea    -0x8(%rbp),%rax
  mov    %rax,%rsi
  lea    D39(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  cmp    $0xffffffff,%eax
  jne    L3
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D39
datatype: string
name: retrieve_data
arguments: data_tag: D27
datatype: string

tool: D39: %d

tool: D27: %d


assistant: ```c
int main(i, n, r, x) {
  for (scanf("%d", &n); ~scanf("%d", &n); printf("%d\n", r))
    for (r = 0; n--; r += abs(i = r ? i + x - 4 : x - 4))
      scanf("%d", &x);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %edi,%ebx
  mov    %esi,0xc(%rsp)
  mov    %ecx,0x8(%rsp)
  lea    0xc(%rsp),%rsi
  lea    D0(%rip),%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  lea    0xc(%rsp),%r14
  lea    D0(%rip),%r12
  lea    0x8(%rsp),%r13
  lea    D26(%rip),%r15
  jmp    L15
L69:
  mov    0x8(%rsp),%eax
  lea    -0x4(%rax),%ebx
L6:
  mov    %ebx,%eax
  neg    %eax
  cmovs  %ebx,%eax
  add    %eax,%ebp
  mov    0xc(%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,0xc(%rsp)
  test   %eax,%eax
  je     L58
L57:
  mov    %r13,%rsi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  test   %ebp,%ebp
  je     L69
  add    0x8(%rsp),%ebx
  sub    $0x4,%ebx
  jmp    L6
L58:
  mov    %ebp,%esi
  mov    %r15,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L15:
  mov    %r14,%rsi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  cmp    $0xffffffff,%eax
  je     L9
  mov    0xc(%rsp),%ebp
  lea    -0x1(%rbp),%eax
  mov    %eax,0xc(%rsp)
  test   %ebp,%ebp
  je     L58
  mov    $0x0,%ebp
  jmp    L57
L9:
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D0
datatype: string
name: parse_data
arguments: data_tag: D26
datatype: string

tool: D0: %d

tool: D26: %d


assistant: ```c
int main(i, n, r, x) {
  for (scanf("%d", &n); ~scanf("%d", &n); printf("%d\n", r))
    for (r = 0; n--; r += abs(i = r ? i + x - 4 : x - 4))
      scanf("%d", &x);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<main>:
  endbr64
  push   %r15
  xor    %eax,%eax
  push   %r14
  lea    D0(%rip),%r14
  push   %r13
  push   %r12
  push   %rbp
  lea    D4(%rip),%rbp
  push   %rbx
  mov    %edi,%ebx
  mov    %rbp,%rdi
  sub    $0x18,%rsp
  lea    0xc(%rsp),%r13
  mov    %esi,0xc(%rsp)
  mov    %r13,%rsi
  mov    %ecx,0x8(%rsp)
  call   <scanf@plt>
  nopl   0x0(%rax)
L8:
  xor    %eax,%eax
  mov    %r13,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  cmp    $0xffffffff,%eax
  je     L42
  mov    0xc(%rsp),%r15d
  lea    -0x1(%r15),%eax
  mov    %eax,0xc(%rsp)
  test   %r15d,%r15d
  je     L69
  xor    %r15d,%r15d
  lea    0x8(%rsp),%r12
  jmp    L6
  xchg   %ax,%ax
L44:
  add    0x8(%rsp),%ebx
  sub    $0x4,%ebx
L49:
  mov    %ebx,%eax
  neg    %eax
  cmovs  %ebx,%eax
  add    %eax,%r15d
  mov    0xc(%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,0xc(%rsp)
  test   %eax,%eax
  je     L69
L6:
  xor    %eax,%eax
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  test   %r15d,%r15d
  jne    L44
  mov    0x8(%rsp),%eax
  lea    -0x4(%rax),%ebx
  jmp    L49
  nopl   0x0(%rax,%rax,1)
L69:
  mov    %r15d,%esi
  mov    %r14,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L8
  nop
L42:
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D0
data_type: string
name: extract_symbolic_data
arguments: label: D4
data_type: string

tool: D0: %d


tool: D4: %d

assistant: ```c
int main(i, n, r, x) {
  for (scanf("%d", &n); ~scanf("%d", &n); printf("%d\n", r))
    for (r = 0; n--; r += abs(i = r ? i + x - 4 : x - 4))
      scanf("%d", &x);
}
```