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

```asm
<calculate_variance>:
  endbr64
  movapd %xmm0,%xmm3
  test   %esi,%esi
  jle    L2
  cmp    $0x1,%esi
  je     L26
  mov    %esi,%edx
  movapd %xmm0,%xmm4
  pxor   %xmm0,%xmm0
  mov    %rdi,%rax
  shr    %edx
  unpcklpd %xmm4,%xmm4
  sub    $0x1,%edx
  shl    $0x4,%rdx
  lea    0x10(%rdi,%rdx,1),%rdx
L5:
  movupd (%rax),%xmm1
  add    $0x10,%rax
  subpd  %xmm4,%xmm1
  mulpd  %xmm1,%xmm1
  addsd  %xmm1,%xmm0
  unpckhpd %xmm1,%xmm1
  addsd  %xmm1,%xmm0
  cmp    %rax,%rdx
  jne    L5
  mov    %esi,%eax
  and    $0xfffffffe,%eax
  and    $0x1,%esi
  je     L35
L44:
  cltq
  movsd  (%rdi,%rax,8),%xmm1
  subsd  %xmm3,%xmm1
  mulsd  %xmm1,%xmm1
  addsd  %xmm1,%xmm0
  ret
  nop
L35:
  ret
  nopl   0x0(%rax)
L2:
  pxor   %xmm0,%xmm0
  ret
L26:
  pxor   %xmm0,%xmm0
  xor    %eax,%eax
  jmp    L44
```

assistant: ```c
double calculate_variance(double *sample, double mean, int len) {
  int i;
  double var;
  double x;
  var = 0.0;
  for (i = 0; i < len; i++) {
    x = sample[i] - mean;
    var += x * x;
  }
  return var;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<calloc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  cmpq   $0x0,-0x20(%rbp)
  je     L7
  mov    D28(%rip),%rax
  mov    (%rax),%rax
  mov    $0x0,%edx
  divq   -0x20(%rbp)
  cmp    %rax,-0x18(%rbp)
  jbe    L7
  mov    D5(%rip),%rax
  mov    (%rax),%edx
  mov    D9(%rip),%rax
  mov    %edx,(%rax)
  mov    $0x0,%eax
  jmp    L3
L7:
  mov    $0x0,%eax
  call   <lock@plt>
  mov    -0x18(%rbp),%rax
  imul   -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <malloc_internal@plt>
  mov    %rax,-0x8(%rbp)
  mov    $0x0,%eax
  call   <unlock@plt>
  mov    -0x8(%rbp),%rax
L3:
  leave
  ret
```

assistant: ```c
void *calloc(size_t a, size_t b) {
  if (b > 0 && a > SIZE_MAX / b) {
    errno = ENOMEM;
    return NULL;
  }
  lock();
  void *result = malloc_internal(a * b);
  unlock();
  return result;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<calloc>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    %rsi,%rbx
  test   %rsi,%rsi
  je     L4
  mov    D7(%rip),%rax
  mov    (%rax),%rax
  mov    $0x0,%edx
  div    %rsi
  cmp    %rdi,%rax
  jb     L11
L4:
  mov    $0x0,%eax
  call   <lock@plt>
  imul   %rbp,%rbx
  mov    %rbx,%rdi
  call   <malloc_internal@plt>
  mov    %rax,%rbx
  mov    $0x0,%eax
  call   <unlock@plt>
L1:
  mov    %rbx,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L11:
  mov    D14(%rip),%rax
  mov    (%rax),%edx
  mov    D11(%rip),%rax
  mov    %edx,(%rax)
  mov    $0x0,%ebx
  jmp    L1
```

assistant: ```c
void *calloc(size_t a, size_t b) {
  if (b > 0 && a > SIZE_MAX / b) {
    errno = ENOMEM;
    return NULL;
  }
  lock();
  void *result = malloc_internal(a * b);
  unlock();
  return result;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<calloc>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rsi,%rbx
  test   %rsi,%rsi
  je     L10
  mov    D2(%rip),%rax
  xor    %edx,%edx
  mov    (%rax),%rax
  div    %rsi
  cmp    %rdi,%rax
  jb     L13
L10:
  imul   %rbp,%rbx
  xor    %eax,%eax
  call   <lock@plt>
  mov    %rbx,%rdi
  call   <malloc_internal@plt>
  mov    %rax,%r12
  xor    %eax,%eax
  call   <unlock@plt>
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L13:
  mov    D0(%rip),%rax
  xor    %r12d,%r12d
  mov    (%rax),%edx
  mov    D1(%rip),%rax
  mov    %edx,(%rax)
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void *calloc(size_t a, size_t b) {
  if (b > 0 && a > SIZE_MAX / b) {
    errno = ENOMEM;
    return NULL;
  }
  lock();
  void *result = malloc_internal(a * b);
  unlock();
  return result;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<calloc>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rsi,%rbx
  test   %rsi,%rsi
  je     L6
  mov    D18(%rip),%rax
  xor    %edx,%edx
  mov    (%rax),%rax
  div    %rsi
  cmp    %rdi,%rax
  jb     L0
L6:
  imul   %rbp,%rbx
  xor    %eax,%eax
  call   <lock@plt>
  mov    %rbx,%rdi
  call   <malloc_internal@plt>
  mov    %rax,%r12
  xor    %eax,%eax
  call   <unlock@plt>
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L0:
  mov    D11(%rip),%rax
  xor    %r12d,%r12d
  mov    (%rax),%edx
  mov    D4(%rip),%rax
  mov    %edx,(%rax)
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void *calloc(size_t a, size_t b) {
  if (b > 0 && a > SIZE_MAX / b) {
    errno = ENOMEM;
    return NULL;
  }
  lock();
  void *result = malloc_internal(a * b);
  unlock();
  return result;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  add    $0xffffffffffffff80,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x74(%rbp)
  lea    -0x70(%rbp),%rax
  mov    %rax,%rsi
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  movl   $0x0,-0x78(%rbp)
  lea    -0x70(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  sub    $0x1,%eax
  mov    %eax,-0x74(%rbp)
  jmp    L13
L16:
  mov    -0x78(%rbp),%eax
  cltq
  movzbl -0x70(%rbp,%rax,1),%eax
  mov    %al,-0x79(%rbp)
  mov    -0x74(%rbp),%eax
  cltq
  movzbl -0x70(%rbp,%rax,1),%edx
  mov    -0x78(%rbp),%eax
  cltq
  mov    %dl,-0x70(%rbp,%rax,1)
  mov    -0x74(%rbp),%eax
  cltq
  movzbl -0x79(%rbp),%edx
  mov    %dl,-0x70(%rbp,%rax,1)
  addl   $0x1,-0x78(%rbp)
  subl   $0x1,-0x74(%rbp)
L13:
  mov    -0x78(%rbp),%eax
  cmp    -0x74(%rbp),%eax
  jl     L16
  lea    -0x70(%rbp),%rax
  mov    %rax,%rsi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L25
  call   <__stack_chk_fail@plt>
L25:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: label: D4
type_infered: string
name: disassemble_data
arguments: label: D7
type_infered: string

tool: D4:  %s

tool: D7: %s

assistant: ```c
int main() {
  char str[100], temp;
  int i, j = 0;
  scanf(" %s", &str);
  i = 0;
  j = strlen(str) - 1;
  while (i < j) {
    temp = str[i];
    str[i] = str[j];
    str[j] = temp;
    i++;
    j--;
  }
  printf("%s", str);
  return (0);
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x70,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  mov    %rbx,%rsi
  lea    D15(%rip),%rdi
  call   <scanf@plt>
  mov    %rbx,%rdi
  call   <strlen@plt>
  lea    -0x1(%rax),%ecx
  test   %ecx,%ecx
  jle    L17
  mov    %rsp,%rsi
  movslq %eax,%rdx
  add    %rbx,%rdx
L2:
  movzbl (%rsi),%edi
  movzbl -0x1(%rdx),%r8d
  mov    %r8b,(%rsi)
  mov    %dil,-0x1(%rdx)
  mov    %ecx,%r8d
  sub    $0x1,%ecx
  add    $0x1,%rsi
  sub    $0x1,%rdx
  mov    %eax,%edi
  sub    %r8d,%edi
  cmp    %edi,%ecx
  jg     L2
L17:
  mov    %rsp,%rsi
  lea    D3(%rip),%rdi
  call   <printf@plt>
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L21
  mov    $0x0,%eax
  add    $0x70,%rsp
  pop    %rbx
  ret
L21:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D15
datatype: string
name: read_data
arguments: data_label: D3
datatype: string

tool: D15:  %s

tool: D3: %s

assistant: ```c
int main() {
  char str[100], temp;
  int i, j = 0;
  scanf(" %s", &str);
  i = 0;
  j = strlen(str) - 1;
  while (i < j) {
    temp = str[i];
    str[i] = str[j];
    str[j] = temp;
    i++;
    j--;
  }
  printf("%s", str);
  return (0);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D14(%rip),%rdi
  sub    $0x70,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%rsi
  call   <scanf@plt>
  mov    %rbp,%rdi
  call   <strlen@plt>
  lea    -0x1(%rax),%ecx
  test   %ecx,%ecx
  jle    L9
  movslq %eax,%rdx
  mov    %rbp,%rsi
  add    %rbp,%rdx
  nopl   0x0(%rax,%rax,1)
L23:
  movzbl (%rsi),%edi
  movzbl -0x1(%rdx),%r8d
  sub    $0x1,%ecx
  add    $0x1,%rsi
  sub    $0x1,%rdx
  mov    %r8b,-0x1(%rsi)
  mov    %dil,(%rdx)
  mov    %eax,%edi
  sub    %ecx,%edi
  sub    $0x1,%edi
  cmp    %edi,%ecx
  jg     L23
L9:
  mov    %rbp,%rsi
  lea    D13(%rip),%rdi
  call   <printf@plt>
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L19
  add    $0x70,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L19:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D14
data_type: string
name: read_data
arguments: tag: D13
data_type: string

tool: D14:  %s

tool: D13: %s

assistant: ```c
int main() {
  char str[100], temp;
  int i, j = 0;
  scanf(" %s", &str);
  i = 0;
  j = strlen(str) - 1;
  while (i < j) {
    temp = str[i];
    str[i] = str[j];
    str[j] = temp;
    i++;
    j--;
  }
  printf("%s", str);
  return (0);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D2(%rip),%rdi
  sub    $0x70,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%rsi
  call   <scanf@plt>
  mov    %rbp,%rdi
  call   <strlen@plt>
  lea    -0x1(%rax),%ecx
  test   %ecx,%ecx
  jle    L13
  movslq %eax,%rdx
  mov    %rbp,%rsi
  add    %rbp,%rdx
  nopl   0x0(%rax,%rax,1)
L25:
  movzbl (%rsi),%edi
  movzbl -0x1(%rdx),%r8d
  sub    $0x1,%ecx
  add    $0x1,%rsi
  sub    $0x1,%rdx
  mov    %r8b,-0x1(%rsi)
  mov    %dil,(%rdx)
  mov    %eax,%edi
  sub    %ecx,%edi
  sub    $0x1,%edi
  cmp    %edi,%ecx
  jg     L25
L13:
  mov    %rbp,%rsi
  lea    D18(%rip),%rdi
  call   <printf@plt>
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  add    $0x70,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D2
datatype: string
name: disassemble_data
arguments: data_tag: D18
datatype: string

tool: D2:  %s

tool: D18: %s

assistant: ```c
int main() {
  char str[100], temp;
  int i, j = 0;
  scanf(" %s", &str);
  i = 0;
  j = strlen(str) - 1;
  while (i < j) {
    temp = str[i];
    str[i] = str[j];
    str[j] = temp;
    i++;
    j--;
  }
  printf("%s", str);
  return (0);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<pfblk>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x8(%rbp)
  jmp    L9
L4:
  mov    -0x28(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x20(%rbp),%rax
  mov    %edx,(%rax)
  addq   $0x4,-0x28(%rbp)
  addq   $0x4,-0x20(%rbp)
L9:
  mov    -0x28(%rbp),%rax
  cmp    -0x8(%rbp),%rax
  jbe    L4
  nop
  pop    %rbp
  ret
```

assistant: ```c
void pfblk(int nblks, int *xblk, int *list) {
  int *stop;
  stop = list + nblks;
  for (; list <= stop; list++, xblk++)
    *xblk = *list;
  return;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<pfblk>:
  endbr64
  movslq %edi,%rdi
  lea    (%rdx,%rdi,4),%rcx
  cmp    %rcx,%rdx
  ja     L18
L2:
  mov    (%rdx),%eax
  mov    %eax,(%rsi)
  add    $0x4,%rdx
  add    $0x4,%rsi
  cmp    %rdx,%rcx
  jae    L2
L18:
  ret
```

assistant: ```c
void pfblk(int nblks, int *xblk, int *list) {
  int *stop;
  stop = list + nblks;
  for (; list <= stop; list++, xblk++)
    *xblk = *list;
  return;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<pfblk>:
  endbr64
  movslq %edi,%rdi
  lea    (%rdx,%rdi,4),%rcx
  cmp    %rcx,%rdx
  ja     L19
L4:
  mov    (%rdx),%eax
  add    $0x4,%rdx
  add    $0x4,%rsi
  mov    %eax,-0x4(%rsi)
  cmp    %rdx,%rcx
  jae    L4
L19:
  ret
```

assistant: ```c
void pfblk(int nblks, int *xblk, int *list) {
  int *stop;
  stop = list + nblks;
  for (; list <= stop; list++, xblk++)
    *xblk = *list;
  return;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<pfblk>:
  endbr64
  movslq %edi,%rdi
  mov    %rdx,%r8
  mov    %rsi,%rcx
  shl    $0x2,%rdi
  add    %rdi,%r8
  jb     L44
  lea    0x4(%rdx),%rax
  sub    %rax,%rsi
  cmp    $0x8,%rsi
  jbe    L68
  cmp    $0xb,%rdi
  jbe    L68
  shr    $0x2,%rdi
  xor    %eax,%eax
  add    $0x1,%rdi
  mov    %rdi,%rsi
  shr    $0x2,%rsi
  shl    $0x4,%rsi
  nopw   0x0(%rax,%rax,1)
L27:
  movdqu (%rdx,%rax,1),%xmm0
  movups %xmm0,(%rcx,%rax,1)
  add    $0x10,%rax
  cmp    %rsi,%rax
  jne    L27
  mov    %rdi,%rax
  and    $0xfffffffffffffffc,%rax
  lea    0x0(,%rax,4),%rsi
  add    %rsi,%rcx
  add    %rsi,%rdx
  cmp    %rax,%rdi
  je     L44
  mov    (%rdx),%eax
  mov    %eax,(%rcx)
  lea    0x4(%rdx),%rax
  cmp    %rax,%r8
  jb     L44
  mov    0x4(%rdx),%eax
  mov    %eax,0x4(%rcx)
  lea    0x8(%rdx),%rax
  cmp    %rax,%r8
  jb     L44
  mov    0x8(%rdx),%eax
  mov    %eax,0x8(%rcx)
  ret
  nopw   0x0(%rax,%rax,1)
L16:
  add    $0x4,%rax
L68:
  mov    (%rdx),%edx
  add    $0x4,%rcx
  mov    %edx,-0x4(%rcx)
  mov    %rax,%rdx
  cmp    %rax,%r8
  jae    L16
L44:
  ret
```

assistant: ```c
void pfblk(int nblks, int *xblk, int *list) {
  int *stop;
  stop = list + nblks;
  for (; list <= stop; list++, xblk++)
    *xblk = *list;
  return;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<calculBarre>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  movsd  %xmm0,-0x20(%rbp)
  movsd  -0x20(%rbp),%xmm0
  movsd  D28(%rip),%xmm1
  divsd  %xmm1,%xmm0
  cvttsd2si %xmm0,%eax
  mov    %eax,%edi
  call   <ceil@plt>
  mov    %eax,%ecx
  mov    -0x14(%rbp),%eax
  cltd
  idiv   %ecx
  mov    %eax,%edi
  call   <ceil@plt>
  mov    %eax,-0x4(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L1
L19:
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x8(%rbp)
L1:
  mov    -0x8(%rbp),%eax
  cmp    -0x4(%rbp),%eax
  jl     L19
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D28
type_guessed: byte[8]
name: extract_symbolic_data
arguments: data_tag: D8
type_guessed: string
name: extract_symbolic_data
arguments: data_tag: D7
type_guessed: string

tool: {"D28": [0x00,0x00,0x00,0x00,0x00,0x00,0x49,0x40]}

tool: D8: X

tool: D7: 


assistant: ```c
void calculBarre(int nbLettre, double sommeEtudiants) {
  int incrementation;
  incrementation = ceil(nbLettre / ceil(sommeEtudiants / 50));
  int i;
  for (i = 0; i < incrementation; i++) {
    printf("X");
  }
  printf("\n");
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<calculBarre>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%ebx
  divsd  D9(%rip),%xmm0
  cvttsd2si %xmm0,%edi
  call   <ceil@plt>
  mov    %eax,%ecx
  mov    %ebx,%eax
  cltd
  idiv   %ecx
  mov    %eax,%edi
  call   <ceil@plt>
  test   %eax,%eax
  jle    L4
  mov    %eax,%ebp
  mov    $0x0,%ebx
  lea    D16(%rip),%r12
L6:
  mov    %r12,%rdi
  call   <printf@plt>
  add    $0x1,%ebx
  cmp    %ebx,%ebp
  jne    L6
L4:
  lea    D24(%rip),%rdi
  call   <printf@plt>
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: read_data
arguments: tag: D9
data_type: byte[8]
name: read_data
arguments: tag: D16
data_type: string
name: read_data
arguments: tag: D24
data_type: string

tool: {"D9": [0x00,0x00,0x00,0x00,0x00,0x00,0x49,0x40]}

tool: D16: X

tool: D24: 


assistant: ```c
void calculBarre(int nbLettre, double sommeEtudiants) {
  int incrementation;
  incrementation = ceil(nbLettre / ceil(sommeEtudiants / 50));
  int i;
  for (i = 0; i < incrementation; i++) {
    printf("X");
  }
  printf("\n");
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<calculBarre>:
  endbr64
  push   %r12
  divsd  D18(%rip),%xmm0
  push   %rbp
  push   %rbx
  mov    %edi,%ebx
  cvttsd2si %xmm0,%edi
  call   <ceil@plt>
  mov    %eax,%r8d
  mov    %ebx,%eax
  cltd
  idiv   %r8d
  mov    %eax,%edi
  call   <ceil@plt>
  test   %eax,%eax
  jle    L11
  mov    %eax,%ebp
  xor    %ebx,%ebx
  lea    D28(%rip),%r12
  nopw   0x0(%rax,%rax,1)
L3:
  mov    %r12,%rdi
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    %ebx,%ebp
  jne    L3
L11:
  pop    %rbx
  lea    D20(%rip),%rdi
  pop    %rbp
  pop    %r12
  jmp    1070 <printf@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D18
data_type: byte[8]
name: retrieve_data
arguments: data_tag: D28
data_type: string
name: retrieve_data
arguments: data_tag: D20
data_type: string

tool: {"D18": [0x00,0x00,0x00,0x00,0x00,0x00,0x49,0x40]}

tool: D28: X

tool: D20: 


assistant: ```c
void calculBarre(int nbLettre, double sommeEtudiants) {
  int incrementation;
  incrementation = ceil(nbLettre / ceil(sommeEtudiants / 50));
  int i;
  for (i = 0; i < incrementation; i++) {
    printf("X");
  }
  printf("\n");
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<calculBarre>:
  endbr64
  push   %r12
  divsd  D14(%rip),%xmm0
  push   %rbp
  push   %rbx
  mov    %edi,%ebx
  cvttsd2si %xmm0,%edi
  call   <ceil@plt>
  mov    %eax,%r8d
  mov    %ebx,%eax
  cltd
  idiv   %r8d
  mov    %eax,%edi
  call   <ceil@plt>
  test   %eax,%eax
  jle    L10
  mov    %eax,%ebp
  xor    %ebx,%ebx
  lea    D0(%rip),%r12
  nopw   0x0(%rax,%rax,1)
L6:
  mov    %r12,%rdi
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    %ebx,%ebp
  jne    L6
L10:
  pop    %rbx
  lea    D27(%rip),%rdi
  pop    %rbp
  pop    %r12
  jmp    1070 <printf@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D14
type_infered: byte[8]
name: unpack_data
arguments: data_tag: D0
type_infered: string
name: unpack_data
arguments: data_tag: D27
type_infered: string

tool: {"D14": [0x00,0x00,0x00,0x00,0x00,0x00,0x49,0x40]}

tool: D0: X

tool: D27: 


assistant: ```c
void calculBarre(int nbLettre, double sommeEtudiants) {
  int incrementation;
  incrementation = ceil(nbLettre / ceil(sommeEtudiants / 50));
  int i;
  for (i = 0; i < incrementation; i++) {
    printf("X");
  }
  printf("\n");
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<filter2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x2e,%al
  jne    L12
  mov    $0x0,%eax
  jmp    L1
L12:
  mov    $0x1,%eax
L1:
  pop    %rbp
  ret
```

assistant: ```c
int filter2(const struct dirent *dir) {
  char *s = dir->d_name;
  if (s[0] == '.')
    return 0;
  return 1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<filter2>:
  endbr64
  mov    (%rdi),%rax
  cmpb   $0x2e,(%rax)
  setne  %al
  movzbl %al,%eax
  ret
```

assistant: ```c
int filter2(const struct dirent *dir) {
  char *s = dir->d_name;
  if (s[0] == '.')
    return 0;
  return 1;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<filter2>:
  endbr64
  mov    (%rdi),%rax
  cmpb   $0x2e,(%rax)
  setne  %al
  movzbl %al,%eax
  ret
```

assistant: ```c
int filter2(const struct dirent *dir) {
  char *s = dir->d_name;
  if (s[0] == '.')
    return 0;
  return 1;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<filter2>:
  endbr64
  mov    (%rdi),%rax
  cmpb   $0x2e,(%rax)
  setne  %al
  movzbl %al,%eax
  ret
```

assistant: ```c
int filter2(const struct dirent *dir) {
  char *s = dir->d_name;
  if (s[0] == '.')
    return 0;
  return 1;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x14(%rbp),%rax
  mov    %rax,%rsi
  lea    D11(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    $0x186a0,%edi
  call   <malloc@plt>
  mov    %rax,-0x10(%rbp)
  jmp    L3
L6:
  mov    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <check@plt>
  mov    -0x14(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x14(%rbp)
L3:
  mov    -0x14(%rbp),%eax
  test   %eax,%eax
  jne    L6
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L24
  call   <__stack_chk_fail@plt>
L24:
  leave
  ret
```

tool_calls: name: read_data
arguments: label: D11
type: string
name: read_data
arguments: label: D3
type: string

tool: D11: %d

tool: D3: %s

assistant: ```c
int main() {
  int t;
  scanf("%d", &t);
  char *str = (char *)malloc(100000 * sizeof(char));
  while (t) {
    scanf("%s", str);
    check(str);
    t--;
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  lea    D2(%rip),%rdi
  call   <scanf@plt>
  mov    $0x186a0,%edi
  call   <malloc@plt>
  cmpl   $0x0,0x4(%rsp)
  je     L9
  mov    %rax,%rbx
  lea    D14(%rip),%rbp
L22:
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    %rbx,%rdi
  call   <check@plt>
  mov    0x4(%rsp),%eax
  sub    $0x1,%eax
  mov    %eax,0x4(%rsp)
  jne    L22
L9:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D2
type_guessed: string
name: parse_data
arguments: tag: D14
type_guessed: string

tool: D2: %d

tool: D14: %s

assistant: ```c
int main() {
  int t;
  scanf("%d", &t);
  char *str = (char *)malloc(100000 * sizeof(char));
  while (t) {
    scanf("%s", str);
    check(str);
    t--;
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D14(%rip),%rdi
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  mov    $0x186a0,%edi
  call   <malloc@plt>
  mov    0x4(%rsp),%edx
  test   %edx,%edx
  je     L20
  mov    %rax,%rbx
  lea    D0(%rip),%rbp
  nopw   0x0(%rax,%rax,1)
L29:
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    %rbx,%rdi
  call   <check@plt>
  subl   $0x1,0x4(%rsp)
  jne    L29
L20:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L11
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L11:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D14
type: string
name: disassemble_data
arguments: data_label: D0
type: string

tool: D14: %d

tool: D0: %s

assistant: ```c
int main() {
  int t;
  scanf("%d", &t);
  char *str = (char *)malloc(100000 * sizeof(char));
  while (t) {
    scanf("%s", str);
    check(str);
    t--;
  }
  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
  push   %rbp
  lea    D11(%rip),%rdi
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  mov    $0x186a0,%edi
  call   <malloc@plt>
  mov    0x4(%rsp),%edx
  test   %edx,%edx
  je     L9
  mov    %rax,%rbx
  lea    D8(%rip),%rbp
  nopw   0x0(%rax,%rax,1)
L28:
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    %rbx,%rdi
  call   <check@plt>
  subl   $0x1,0x4(%rsp)
  jne    L28
L9:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D11
type: string
name: disassemble_data
arguments: label: D8
type: string

tool: D11: %d

tool: D8: %s

assistant: ```c
int main() {
  int t;
  scanf("%d", &t);
  char *str = (char *)malloc(100000 * sizeof(char));
  while (t) {
    scanf("%s", str);
    check(str);
    t--;
  }
  return 0;
}
```