ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What is the c source code of the assembly code below:

```asm
<reverse_tab>:
  endbr64
  cmpq   $0x0,(%rdi)
  mov    %rdi,%rax
  je     L22
  lea    0x8(%rdi),%rdx
  nopl   0x0(%rax)
L21:
  mov    (%rdx),%rcx
  test   %rcx,%rcx
  je     L8
  mov    %rcx,-0x8(%rdx)
  add    $0x8,%rdx
  movq   $0x0,-0x8(%rdx)
  cmpq   $0x0,-0x8(%rdx)
  jne    L21
  ret
  nopl   0x0(%rax,%rax,1)
L8:
  add    $0x8,%rdx
  cmpq   $0x0,-0x8(%rdx)
  jne    L21
L22:
  ret
```

assistant: ```c
char **reverse_tab(char **tab) {
  int i;
  int j;
  j = 0;
  i = 0;
  while (tab[i]) {
    if (tab[i + 1] != NULL) {
      tab[i] = tab[i + 1];
      tab[i + 1] = NULL;
    }
    j++;
    i++;
  }
  return (tab);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<diff>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  subq   $0x1,-0x28(%rbp)
  movq   $0x0,-0x8(%rbp)
  jmp    L5
L16:
  mov    -0x8(%rbp),%rax
  add    $0x1,%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  mov    -0x8(%rbp),%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm1
  mov    -0x8(%rbp),%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  subsd  %xmm1,%xmm0
  movsd  %xmm0,(%rax)
  addq   $0x1,-0x8(%rbp)
L5:
  mov    -0x8(%rbp),%rax
  cmp    -0x28(%rbp),%rax
  jb     L16
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void diff(double *y, double *f, int32_t sz) {
  --sz;
  for (int32_t i = 0; i < sz; i++)
    f[i] = y[i + 1] - y[i];
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<diff>:
  endbr64
  sub    $0x1,%rdx
  je     L5
  mov    $0x0,%eax
L0:
  add    $0x1,%rax
  movsd  (%rdi,%rax,8),%xmm0
  subsd  -0x8(%rdi,%rax,8),%xmm0
  movsd  %xmm0,-0x8(%rsi,%rax,8)
  cmp    %rdx,%rax
  jne    L0
L5:
  ret
```

assistant: ```c
void diff(double *y, double *f, int32_t sz) {
  --sz;
  for (int32_t i = 0; i < sz; i++)
    f[i] = y[i + 1] - y[i];
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<diff>:
  endbr64
  sub    $0x1,%rdx
  je     L12
  xor    %eax,%eax
  nopl   0x0(%rax)
L4:
  add    $0x1,%rax
  movsd  (%rdi,%rax,8),%xmm0
  subsd  -0x8(%rdi,%rax,8),%xmm0
  movsd  %xmm0,-0x8(%rsi,%rax,8)
  cmp    %rdx,%rax
  jne    L4
L12:
  ret
```

assistant: ```c
void diff(double *y, double *f, int32_t sz) {
  --sz;
  for (int32_t i = 0; i < sz; i++)
    f[i] = y[i + 1] - y[i];
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<diff>:
  endbr64
  mov    %rdi,%rcx
  mov    %rdx,%rdi
  sub    $0x1,%rdi
  je     L15
  lea    0x8(%rcx),%r8
  mov    %rsi,%rax
  sub    %r8,%rax
  cmp    $0x8,%rax
  jbe    L38
  sub    $0x2,%rdx
  cmp    $0x1,%rdx
  jbe    L38
  mov    %rdi,%rdx
  xor    %eax,%eax
  shr    %rdx
  shl    $0x4,%rdx
  nopw   0x0(%rax,%rax,1)
L54:
  movupd 0x8(%rcx,%rax,1),%xmm0
  movupd (%rcx,%rax,1),%xmm1
  subpd  %xmm1,%xmm0
  movups %xmm0,(%rsi,%rax,1)
  add    $0x10,%rax
  cmp    %rax,%rdx
  jne    L54
  mov    %rdi,%rax
  and    $0xfffffffffffffffe,%rax
  and    $0x1,%edi
  je     L15
  add    $0x1,%rax
  movsd  (%rcx,%rax,8),%xmm0
  subsd  -0x8(%rcx,%rax,8),%xmm0
  movsd  %xmm0,-0x8(%rsi,%rax,8)
  ret
  xchg   %ax,%ax
L38:
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L11:
  add    $0x1,%rax
  movsd  (%rcx,%rax,8),%xmm0
  subsd  -0x8(%rcx,%rax,8),%xmm0
  movsd  %xmm0,-0x8(%rsi,%rax,8)
  cmp    %rdi,%rax
  jne    L11
L15:
  ret
```

assistant: ```c
void diff(double *y, double *f, int32_t sz) {
  --sz;
  for (int32_t i = 0; i < sz; i++)
    f[i] = y[i + 1] - y[i];
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<CopyString>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L8
L19:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x24(%rbp),%ecx
  mov    -0x4(%rbp),%edx
  add    %ecx,%edx
  movslq %edx,%rcx
  mov    -0x20(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  mov    -0x4(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    -0x24(%rbp),%eax
  add    %edx,%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  addl   $0x1,-0x4(%rbp)
L8:
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cmp    %eax,-0x4(%rbp)
  jl     L19
  nop
  nop
  leave
  ret
```

assistant: ```c
void CopyString(char *src, char *dst, int index) {
  int i;
  for (i = 0; i < strlen(src); i++) {
    dst[index + i] = src[i];
    dst[index + (i + 1)] = 0x00;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<CopyString>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    %rsi,%r13
  mov    $0x0,%ebx
  movslq %edx,%r12
  jmp    L15
L9:
  movzbl 0x0(%rbp,%rbx,1),%edx
  lea    0x0(%r13,%r12,1),%rax
  mov    %dl,(%rax,%rbx,1)
  movb   $0x0,0x1(%rax,%rbx,1)
  add    $0x1,%rbx
L15:
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jg     L9
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
void CopyString(char *src, char *dst, int index) {
  int i;
  for (i = 0; i < strlen(src); i++) {
    dst[index + i] = src[i];
    dst[index + (i + 1)] = 0x00;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<CopyString>:
  endbr64
  push   %r13
  mov    %rsi,%r13
  push   %r12
  movslq %edx,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  jmp    L4
  nopl   0x0(%rax,%rax,1)
L16:
  movzbl 0x0(%rbp,%rbx,1),%edx
  lea    0x0(%r13,%r12,1),%rax
  movb   $0x0,0x1(%rax,%rbx,1)
  mov    %dl,(%rax,%rbx,1)
  add    $0x1,%rbx
L4:
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jg     L16
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
void CopyString(char *src, char *dst, int index) {
  int i;
  for (i = 0; i < strlen(src); i++) {
    dst[index + i] = src[i];
    dst[index + (i + 1)] = 0x00;
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<CopyString>:
  endbr64
  push   %r13
  mov    %rsi,%r13
  push   %r12
  movslq %edx,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  jmp    L10
  nopl   0x0(%rax,%rax,1)
L12:
  movzbl 0x0(%rbp,%rbx,1),%edx
  lea    0x0(%r13,%r12,1),%rax
  movb   $0x0,0x1(%rax,%rbx,1)
  mov    %dl,(%rax,%rbx,1)
  add    $0x1,%rbx
L10:
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jg     L12
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
void CopyString(char *src, char *dst, int index) {
  int i;
  for (i = 0; i < strlen(src); i++) {
    dst[index + i] = src[i];
    dst[index + (i + 1)] = 0x00;
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<enc3>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x30(%rbp),%rax
  mov    %rax,-0x18(%rbp)
  mov    -0x28(%rbp),%rax
  mov    (%rax),%eax
  lea    -0xc(%rbp),%rdx
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <lendian2byte@plt>
  movzbl -0xc(%rbp),%eax
  shr    $0x2,%al
  mov    %eax,%edx
  mov    -0x18(%rbp),%rax
  mov    %dl,(%rax)
  movzbl -0xc(%rbp),%eax
  and    $0x3,%eax
  shl    $0x4,%eax
  mov    %eax,%ecx
  movzbl -0xb(%rbp),%eax
  shr    $0x4,%al
  mov    %eax,%edx
  mov    -0x18(%rbp),%rax
  add    $0x1,%rax
  add    %ecx,%edx
  mov    %dl,(%rax)
  movzbl -0xb(%rbp),%eax
  and    $0xf,%eax
  lea    0x0(,%rax,4),%ecx
  movzbl -0xa(%rbp),%eax
  shr    $0x6,%al
  mov    %eax,%edx
  mov    -0x18(%rbp),%rax
  add    $0x2,%rax
  add    %ecx,%edx
  mov    %dl,(%rax)
  movzbl -0xa(%rbp),%edx
  mov    -0x18(%rbp),%rax
  add    $0x3,%rax
  and    $0x3f,%edx
  mov    %dl,(%rax)
  movzbl -0x9(%rbp),%edx
  mov    -0x18(%rbp),%rax
  add    $0x4,%rax
  shr    $0x2,%dl
  mov    %dl,(%rax)
  movzbl -0x9(%rbp),%eax
  and    $0x3,%eax
  mov    %eax,%edx
  mov    -0x18(%rbp),%rax
  add    $0x5,%rax
  shl    $0x4,%edx
  mov    %dl,(%rax)
  movl   $0x0,-0x1c(%rbp)
  jmp    L19
L10:
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%edx
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  add    $0x20,%edx
  mov    %dl,(%rax)
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x3f,%al
  jbe    L13
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%edx
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  add    $0x1,%edx
  mov    %dl,(%rax)
L13:
  addl   $0x1,-0x1c(%rbp)
L19:
  cmpl   $0x5,-0x1c(%rbp)
  jle    L10
  mov    -0x30(%rbp),%rax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L4
  call   <__stack_chk_fail@plt>
L4:
  leave
  ret
```

assistant: ```c
char *enc3(int *v1, char *v2) {
  int i;
  unsigned char v5;
  unsigned char *a2 = (unsigned char *)v2;
  unsigned char a1[4];
  lendian2byte(*v1, a1);
  a2[0] = a1[0] >> 2;
  a2[1] = (a1[0] & 3) * 16 + (a1[1] >> 4);
  a2[2] = (a1[1] & 0xF) * 4 + (a1[2] >> 6);
  a2[3] = a1[2] & 0x3F;
  a2[4] = a1[3] >> 2;
  a2[5] = (a1[3] & 3) * 16;
  for (i = 0; i < 6; i++) {
    a2[i] += 32;
    if (a2[i] >= 0x40)
      a2[i] += 1;
  }
  return v2;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<enc3>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %rsi,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  mov    (%rdi),%edi
  call   <lendian2byte@plt>
  movzbl 0x4(%rsp),%eax
  mov    %eax,%edx
  shr    $0x2,%dl
  mov    %dl,(%rbx)
  movzbl 0x5(%rsp),%edx
  and    $0x3,%eax
  shl    $0x4,%eax
  mov    %edx,%ecx
  shr    $0x4,%cl
  add    %ecx,%eax
  mov    %al,0x1(%rbx)
  movzbl 0x6(%rsp),%eax
  and    $0xf,%edx
  mov    %eax,%ecx
  shr    $0x6,%cl
  lea    (%rcx,%rdx,4),%edx
  mov    %dl,0x2(%rbx)
  and    $0x3f,%eax
  mov    %al,0x3(%rbx)
  movzbl 0x7(%rsp),%eax
  mov    %eax,%edx
  shr    $0x2,%dl
  mov    %dl,0x4(%rbx)
  and    $0x3,%eax
  shl    $0x4,%eax
  mov    %al,0x5(%rbx)
  mov    %rbx,%rax
  lea    0x6(%rbx),%rsi
L7:
  movzbl (%rax),%ecx
  lea    0x20(%rcx),%edx
  add    $0x21,%ecx
  cmp    $0x40,%dl
  cmovae %ecx,%edx
  mov    %dl,(%rax)
  add    $0x1,%rax
  cmp    %rsi,%rax
  jne    L7
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L17
  mov    %rbx,%rax
  add    $0x10,%rsp
  pop    %rbx
  ret
L17:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char *enc3(int *v1, char *v2) {
  int i;
  unsigned char v5;
  unsigned char *a2 = (unsigned char *)v2;
  unsigned char a1[4];
  lendian2byte(*v1, a1);
  a2[0] = a1[0] >> 2;
  a2[1] = (a1[0] & 3) * 16 + (a1[1] >> 4);
  a2[2] = (a1[1] & 0xF) * 4 + (a1[2] >> 6);
  a2[3] = a1[2] & 0x3F;
  a2[4] = a1[3] >> 2;
  a2[5] = (a1[3] & 3) * 16;
  for (i = 0; i < 6; i++) {
    a2[i] += 32;
    if (a2[i] >= 0x40)
      a2[i] += 1;
  }
  return v2;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<enc3>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  sub    $0x10,%rsp
  mov    (%rdi),%edi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <lendian2byte@plt>
  movzbl 0x4(%rsp),%eax
  movzbl 0x5(%rsp),%ecx
  mov    %eax,%edx
  mov    %ecx,%esi
  and    $0x3,%eax
  and    $0xf,%ecx
  shr    $0x4,%sil
  shl    $0x4,%eax
  add    %esi,%eax
  shr    $0x2,%dl
  mov    %al,0x1(%r12)
  movzbl 0x6(%rsp),%eax
  mov    %dl,(%r12)
  mov    %eax,%esi
  and    $0x3f,%eax
  shr    $0x6,%sil
  mov    %al,0x3(%r12)
  movzbl 0x7(%rsp),%eax
  lea    (%rsi,%rcx,4),%ecx
  lea    0x6(%r12),%rsi
  mov    %cl,0x2(%r12)
  mov    %eax,%ecx
  and    $0x3,%eax
  shl    $0x4,%eax
  shr    $0x2,%cl
  mov    %al,0x5(%r12)
  mov    %r12,%rax
  mov    %cl,0x4(%r12)
  jmp    L48
  nopl   0x0(%rax)
L3:
  mov    %cl,(%rax)
  add    $0x1,%rax
  cmp    %rsi,%rax
  je     L10
L47:
  movzbl (%rax),%edx
L48:
  lea    0x20(%rdx),%ecx
  cmp    $0x3f,%cl
  jbe    L3
  add    $0x21,%edx
  add    $0x1,%rax
  mov    %dl,-0x1(%rax)
  cmp    %rsi,%rax
  jne    L47
L10:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  add    $0x10,%rsp
  mov    %r12,%rax
  pop    %r12
  ret
L4:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char *enc3(int *v1, char *v2) {
  int i;
  unsigned char v5;
  unsigned char *a2 = (unsigned char *)v2;
  unsigned char a1[4];
  lendian2byte(*v1, a1);
  a2[0] = a1[0] >> 2;
  a2[1] = (a1[0] & 3) * 16 + (a1[1] >> 4);
  a2[2] = (a1[1] & 0xF) * 4 + (a1[2] >> 6);
  a2[3] = a1[2] & 0x3F;
  a2[4] = a1[3] >> 2;
  a2[5] = (a1[3] & 3) * 16;
  for (i = 0; i < 6; i++) {
    a2[i] += 32;
    if (a2[i] >= 0x40)
      a2[i] += 1;
  }
  return v2;
}
```
user: What is the c source code of the assembly code below:

```asm
<enc3>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  sub    $0x10,%rsp
  mov    (%rdi),%edi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <lendian2byte@plt>
  movzbl 0x4(%rsp),%edx
  movzbl 0x5(%rsp),%edi
  movzbl 0x6(%rsp),%esi
  mov    %edx,%r9d
  mov    %edi,%eax
  and    $0x3,%edx
  and    $0xf,%edi
  shr    $0x4,%al
  shl    $0x4,%edx
  add    %eax,%edx
  mov    %esi,%eax
  shr    $0x2,%r9b
  and    $0x3f,%esi
  shr    $0x6,%al
  lea    0x20(%r9),%r8d
  add    $0x21,%r9d
  lea    (%rax,%rdi,4),%edi
  movzbl 0x7(%rsp),%eax
  mov    %eax,%ecx
  and    $0x3,%eax
  shr    $0x2,%cl
  shl    $0x4,%eax
  cmp    $0x40,%r8b
  cmovae %r9d,%r8d
  mov    %r8b,(%r12)
  lea    0x20(%rdx),%r8d
  add    $0x21,%edx
  cmp    $0x40,%r8b
  cmovae %edx,%r8d
  lea    0x20(%rdi),%edx
  add    $0x21,%edi
  cmp    $0x40,%dl
  cmovae %edi,%edx
  mov    %r8b,0x1(%r12)
  mov    %dl,0x2(%r12)
  lea    0x20(%rsi),%edx
  add    $0x21,%esi
  cmp    $0x40,%dl
  cmovae %esi,%edx
  mov    %dl,0x3(%r12)
  lea    0x20(%rcx),%edx
  add    $0x21,%ecx
  cmp    $0x40,%dl
  cmovae %ecx,%edx
  mov    %dl,0x4(%r12)
  lea    0x20(%rax),%edx
  add    $0x21,%eax
  cmp    $0x3f,%dl
  cmovbe %edx,%eax
  mov    %al,0x5(%r12)
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  add    $0x10,%rsp
  mov    %r12,%rax
  pop    %r12
  ret
L4:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char *enc3(int *v1, char *v2) {
  int i;
  unsigned char v5;
  unsigned char *a2 = (unsigned char *)v2;
  unsigned char a1[4];
  lendian2byte(*v1, a1);
  a2[0] = a1[0] >> 2;
  a2[1] = (a1[0] & 3) * 16 + (a1[1] >> 4);
  a2[2] = (a1[1] & 0xF) * 4 + (a1[2] >> 6);
  a2[3] = a1[2] & 0x3F;
  a2[4] = a1[3] >> 2;
  a2[5] = (a1[3] & 3) * 16;
  for (i = 0; i < 6; i++) {
    a2[i] += 32;
    if (a2[i] >= 0x40)
      a2[i] += 1;
  }
  return v2;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<valid>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x14(%rbp)
  jmp    L21
L55:
  lea    -0xf(%rbp),%rax
  mov    %rax,%rsi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  movzbl -0xf(%rbp),%eax
  cmp    $0x6d,%al
  je     L10
  lea    -0xf(%rbp),%rax
  mov    %rax,%rdi
  call   <atoi@plt>
  mov    -0x14(%rbp),%edx
  add    $0x1,%edx
  cmp    %edx,%eax
  je     L10
  mov    $0x0,%eax
  jmp    L22
L10:
  addl   $0x1,-0x14(%rbp)
L21:
  mov    -0x14(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jl     L55
  mov    $0x1,%eax
L22:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L56
  call   <__stack_chk_fail@plt>
L56:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D7
datatype: string

tool: D7: %s

assistant: ```c
int valid(int n) {
  for (int i = 0; i < n; i++) {
    char next[7];
    scanf("%s", next);
    if (next[0] != 'm' && atoi(next) != i + 1)
      return 0;
  }
  return 1;
}
```
user: What is the c source code of the assembly code below?

```asm
<valid>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  test   %edi,%edi
  jle    L21
  mov    %edi,%ebp
  mov    $0x1,%ebx
  lea    D1(%rip),%r12
  jmp    L68
L79:
  lea    0x1(%rbx),%eax
  cmp    %ebp,%ebx
  je     L35
  mov    %eax,%ebx
L68:
  lea    0x1(%rsp),%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  cmpb   $0x6d,0x1(%rsp)
  je     L79
  lea    0x1(%rsp),%rdi
  call   <atoi@plt>
  cmp    %ebx,%eax
  je     L79
  mov    $0x0,%eax
  jmp    L62
L35:
  mov    $0x1,%eax
L62:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L0
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L21:
  mov    $0x1,%eax
  jmp    L62
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D1
type: string

tool: D1: %s

assistant: ```c
int valid(int n) {
  for (int i = 0; i < n; i++) {
    char next[7];
    scanf("%s", next);
    if (next[0] != 'm' && atoi(next) != i + 1)
      return 0;
  }
  return 1;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<valid>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  test   %edi,%edi
  jle    L49
  mov    %edi,%r12d
  xor    %ebx,%ebx
  lea    0x1(%rsp),%rbp
  lea    D8(%rip),%r13
  nopl   0x0(%rax,%rax,1)
L46:
  mov    %rbp,%rsi
  mov    %r13,%rdi
  add    $0x1,%ebx
  call   <scanf@plt>
  cmpb   $0x6d,0x1(%rsp)
  je     L15
  mov    %rbp,%rdi
  call   <atoi@plt>
  cmp    %ebx,%eax
  jne    L40
L15:
  cmp    %ebx,%r12d
  jne    L46
L49:
  mov    $0x1,%eax
L23:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L12
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  xchg   %ax,%ax
L40:
  xor    %eax,%eax
  jmp    L23
L12:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D8
type_infered: string

tool: D8: %s

assistant: ```c
int valid(int n) {
  for (int i = 0; i < n; i++) {
    char next[7];
    scanf("%s", next);
    if (next[0] != 'm' && atoi(next) != i + 1)
      return 0;
  }
  return 1;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<valid>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  test   %edi,%edi
  jle    L29
  mov    %edi,%r12d
  xor    %ebx,%ebx
  lea    0x1(%rsp),%rbp
  lea    D4(%rip),%r13
  nopl   0x0(%rax,%rax,1)
L50:
  mov    %rbp,%rsi
  mov    %r13,%rdi
  add    $0x1,%ebx
  call   <scanf@plt>
  cmpb   $0x6d,0x1(%rsp)
  je     L17
  mov    %rbp,%rdi
  call   <atoi@plt>
  cmp    %ebx,%eax
  jne    L15
L17:
  cmp    %ebx,%r12d
  jne    L50
L29:
  mov    $0x1,%eax
L9:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L54
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  xchg   %ax,%ax
L15:
  xor    %eax,%eax
  jmp    L9
L54:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D4
type: string

tool: D4: %s

assistant: ```c
int valid(int n) {
  for (int i = 0; i < n; i++) {
    char next[7];
    scanf("%s", next);
    if (next[0] != 'm' && atoi(next) != i + 1)
      return 0;
  }
  return 1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<construct_groups>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    -0x28(%rbp),%rax
  movl   $0x0,(%rax)
  movl   $0x0,-0x14(%rbp)
  movl   $0x1,-0x18(%rbp)
  jmp    L5
L13:
  mov    -0x18(%rbp),%eax
  cltq
  shl    $0x2,%rax
  lea    -0x4(%rax),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%ebx
  pxor   %xmm0,%xmm0
  cvtsi2sdl -0x18(%rbp),%xmm0
  cvttsd2si %xmm0,%eax
  mov    %eax,%edi
  call   <log10@plt>
  lea    (%rbx,%rax,1),%edx
  mov    -0x18(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x28(%rbp),%rax
  add    %rcx,%rax
  add    $0x1,%edx
  mov    %edx,(%rax)
  mov    -0x18(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  add    %eax,-0x14(%rbp)
  addl   $0x1,-0x18(%rbp)
L5:
  cmpl   $0x0,-0x14(%rbp)
  jns    L13
  nop
  nop
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
void construct_groups(int *group) {
  group[0] = 0;
  int i;
  int acc = 0;
  for (i = 1; acc >= 0; i++) {
    group[i] = group[i - 1] + (int)log10(i * 1.0) + 1;
    acc += group[i];
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<construct_groups>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r12
  movl   $0x0,(%rdi)
  mov    $0x1,%ebx
  mov    $0x0,%ebp
L9:
  mov    -0x4(%r12,%rbx,4),%r13d
  mov    %ebx,%edi
  call   <log10@plt>
  lea    0x1(%r13,%rax,1),%eax
  mov    %eax,(%r12,%rbx,4)
  add    $0x1,%rbx
  add    %eax,%ebp
  jns    L9
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
void construct_groups(int *group) {
  group[0] = 0;
  int i;
  int acc = 0;
  for (i = 1; acc >= 0; i++) {
    group[i] = group[i - 1] + (int)log10(i * 1.0) + 1;
    acc += group[i];
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<construct_groups>:
  endbr64
  push   %r13
  mov    %rdi,%r13
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  mov    $0x1,%ebp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  movl   $0x0,(%rdi)
  jmp    L7
  nopl   0x0(%rax,%rax,1)
L3:
  add    $0x1,%rbp
L7:
  mov    %ebp,%edi
  call   <log10@plt>
  lea    0x1(%rbx,%rax,1),%ebx
  mov    %ebx,0x0(%r13,%rbp,4)
  add    %ebx,%r12d
  jns    L3
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
void construct_groups(int *group) {
  group[0] = 0;
  int i;
  int acc = 0;
  for (i = 1; acc >= 0; i++) {
    group[i] = group[i - 1] + (int)log10(i * 1.0) + 1;
    acc += group[i];
  }
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<construct_groups>:
  endbr64
  push   %r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  mov    $0x1,%ebx
  sub    $0x8,%rsp
  movl   $0x0,(%rdi)
  xchg   %ax,%ax
L9:
  mov    -0x4(%r12,%rbx,4),%r13d
  mov    %ebx,%edi
  call   <log10@plt>
  lea    0x1(%r13,%rax,1),%eax
  mov    %eax,(%r12,%rbx,4)
  add    $0x1,%rbx
  add    %eax,%ebp
  jns    L9
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
void construct_groups(int *group) {
  group[0] = 0;
  int i;
  int acc = 0;
  for (i = 1; acc >= 0; i++) {
    group[i] = group[i - 1] + (int)log10(i * 1.0) + 1;
    acc += group[i];
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<stxmWithoutMask>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  mov    %ecx,-0x28(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L15
L39:
  movl   $0x0,-0x8(%rbp)
  jmp    L31
L2:
  mov    -0x20(%rbp),%rax
  movsd  (%rax),%xmm1
  mov    -0x4(%rbp),%eax
  imul   -0x24(%rbp),%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%eax
  add    %edx,%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  addsd  %xmm1,%xmm0
  mov    -0x20(%rbp),%rax
  movsd  %xmm0,(%rax)
  addl   $0x1,-0x8(%rbp)
L31:
  mov    -0x8(%rbp),%eax
  cmp    -0x28(%rbp),%eax
  jl     L2
  addl   $0x1,-0x4(%rbp)
L15:
  mov    -0x4(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jl     L39
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void stxmWithoutMask(double *x, double *out, int m, int n) {
  int col;
  int row;
  for (row = 0; row < m; row++) {
    for (col = 0; col < n; col++) {
      *out += x[row * m + col];
    }
  }
}
```