ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r15
  mov    $0xf423f,%r15d
  push   %r14
  push   %r13
  mov    $0x1,%r13d
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  lea    D35(%rip),%rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x14(%rsp),%rax
  movl   $0x0,(%rsp)
  lea    0x10(%rsp),%r14
  movl   $0x0,0x4(%rsp)
  mov    %rax,0x8(%rsp)
  jmp    L1
  nopw   0x0(%rax,%rax,1)
L76:
  test   %eax,%eax
  jle    L71
  cmp    %eax,%r15d
  setg   %cl
  test   %cl,%cl
  cmovne %r13d,%r12d
  cmovne %eax,%r15d
L71:
  cmp    %eax,(%rsp)
  jge    L59
  mov    %r13d,0x4(%rsp)
  mov    %eax,(%rsp)
L59:
  add    %eax,%ebp
  add    $0x1,%r13d
L1:
  mov    %r14,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  test   %rax,%rax
  je     L42
  mov    0x10(%rsp),%eax
  test   %eax,%eax
  je     L42
  jns    L76
  mov    0x8(%rsp),%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    0x10(%rsp),%ecx
  mov    %ecx,%eax
  neg    %eax
  cmovs  %ecx,%eax
  add    0x14(%rsp),%eax
  mov    %eax,0x10(%rsp)
  jmp    L76
  nopl   0x0(%rax,%rax,1)
L42:
  mov    %r15d,%edx
  mov    %r12d,%esi
  lea    D24(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x4(%rsp),%esi
  mov    (%rsp),%edx
  xor    %eax,%eax
  lea    D36(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  mov    %ebp,%esi
  lea    D27(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L18
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L18:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D35
type: string
name: unpack_data
arguments: data_tag: D24
type: string
name: unpack_data
arguments: data_tag: D36
type: string
name: unpack_data
arguments: data_tag: D27
type: string

tool: D35: %d

tool: D24: Melhor volta: %d - %d seg


tool: D36: Pior volta: %d - %d seg


tool: D27: Tempo total: %d seg


assistant: ```c
int main() {
  int num, maior = 0, menor = 999999, total = 0;
  int maior_count = 0, menor_count = 0, count = 1;
  while (scanf("%d", &num)) {
    if (num == 0)
      break;
    if (num < 0) {
      int next;
      scanf("%d", &next);
      num = abs(num);
      num += next;
    }
    if (num < menor && num > 0) {
      menor = num;
      menor_count = count;
    }
    if (num > maior) {
      maior = num;
      maior_count = count;
    }
    total += num;
    count++;
  }
  printf("Melhor volta: %d - %d seg\n", menor_count, menor);
  printf("Pior volta: %d - %d seg\n", maior_count, maior);
  printf("Tempo total: %d seg\n", total);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r15
  mov    $0xf423f,%r15d
  push   %r14
  push   %r13
  mov    $0x1,%r13d
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  lea    D5(%rip),%rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x14(%rsp),%rax
  movl   $0x0,(%rsp)
  lea    0x10(%rsp),%r14
  movl   $0x0,0x4(%rsp)
  mov    %rax,0x8(%rsp)
  jmp    L11
  nopw   0x0(%rax,%rax,1)
L50:
  test   %eax,%eax
  jle    L59
  cmp    %eax,%r15d
  setg   %cl
  test   %cl,%cl
  cmovne %r13d,%r12d
  cmovne %eax,%r15d
L59:
  cmp    %eax,(%rsp)
  jge    L53
  mov    %r13d,0x4(%rsp)
  mov    %eax,(%rsp)
L53:
  add    %eax,%ebp
  add    $0x1,%r13d
L11:
  mov    %r14,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  test   %rax,%rax
  je     L27
  mov    0x10(%rsp),%eax
  test   %eax,%eax
  je     L27
  jns    L50
  mov    0x8(%rsp),%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    0x10(%rsp),%ecx
  mov    %ecx,%eax
  neg    %eax
  cmovs  %ecx,%eax
  add    0x14(%rsp),%eax
  mov    %eax,0x10(%rsp)
  jmp    L50
  nopl   0x0(%rax,%rax,1)
L27:
  mov    %r15d,%edx
  mov    %r12d,%esi
  lea    D23(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x4(%rsp),%esi
  mov    (%rsp),%edx
  xor    %eax,%eax
  lea    D39(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  mov    %ebp,%esi
  lea    D32(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L21
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L21:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D5
data_type: string
name: unpack_data
arguments: data_tag: D23
data_type: string
name: unpack_data
arguments: data_tag: D39
data_type: string
name: unpack_data
arguments: data_tag: D32
data_type: string

tool: D5: %d

tool: D23: Melhor volta: %d - %d seg


tool: D39: Pior volta: %d - %d seg


tool: D32: Tempo total: %d seg


assistant: ```c
int main() {
  int num, maior = 0, menor = 999999, total = 0;
  int maior_count = 0, menor_count = 0, count = 1;
  while (scanf("%d", &num)) {
    if (num == 0)
      break;
    if (num < 0) {
      int next;
      scanf("%d", &next);
      num = abs(num);
      num += next;
    }
    if (num < menor && num > 0) {
      menor = num;
      menor_count = count;
    }
    if (num > maior) {
      maior = num;
      maior_count = count;
    }
    total += num;
    count++;
  }
  printf("Melhor volta: %d - %d seg\n", menor_count, menor);
  printf("Pior volta: %d - %d seg\n", maior_count, maior);
  printf("Tempo total: %d seg\n", total);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<vscale>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  movss  %xmm0,-0xc(%rbp)
  mov    -0x8(%rbp),%rax
  movss  (%rax),%xmm0
  mulss  -0xc(%rbp),%xmm0
  mov    -0x8(%rbp),%rax
  movss  %xmm0,(%rax)
  mov    -0x8(%rbp),%rax
  add    $0x4,%rax
  movss  (%rax),%xmm0
  mov    -0x8(%rbp),%rax
  add    $0x4,%rax
  mulss  -0xc(%rbp),%xmm0
  movss  %xmm0,(%rax)
  mov    -0x8(%rbp),%rax
  add    $0x8,%rax
  movss  (%rax),%xmm0
  mov    -0x8(%rbp),%rax
  add    $0x8,%rax
  mulss  -0xc(%rbp),%xmm0
  movss  %xmm0,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void vscale(float *v, float div) {
  v[0] *= div;
  v[1] *= div;
  v[2] *= div;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<vscale>:
  endbr64
  movaps %xmm0,%xmm1
  mulss  (%rdi),%xmm1
  movss  %xmm1,(%rdi)
  movaps %xmm0,%xmm1
  mulss  0x4(%rdi),%xmm1
  movss  %xmm1,0x4(%rdi)
  mulss  0x8(%rdi),%xmm0
  movss  %xmm0,0x8(%rdi)
  ret
```

assistant: ```c
void vscale(float *v, float div) {
  v[0] *= div;
  v[1] *= div;
  v[2] *= div;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<vscale>:
  endbr64
  movss  (%rdi),%xmm1
  mulss  %xmm0,%xmm1
  movss  %xmm1,(%rdi)
  movss  0x4(%rdi),%xmm1
  mulss  %xmm0,%xmm1
  mulss  0x8(%rdi),%xmm0
  movss  %xmm1,0x4(%rdi)
  movss  %xmm0,0x8(%rdi)
  ret
```

assistant: ```c
void vscale(float *v, float div) {
  v[0] *= div;
  v[1] *= div;
  v[2] *= div;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<vscale>:
  endbr64
  movq   (%rdi),%xmm1
  movaps %xmm0,%xmm2
  shufps $0xe0,%xmm2,%xmm2
  mulss  0x8(%rdi),%xmm0
  mulps  %xmm2,%xmm1
  movss  %xmm0,0x8(%rdi)
  movlps %xmm1,(%rdi)
  ret
```

assistant: ```c
void vscale(float *v, float div) {
  v[0] *= div;
  v[1] *= div;
  v[2] *= div;
}
```
user: Translate the below assembly code into C programming language:

```asm
<getTimeStamp>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x0,%edi
  call   <time@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int getTimeStamp() {
  time_t stime = time(NULL);
  return stime;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<getTimeStamp>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%edi
  call   <time@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
int getTimeStamp() {
  time_t stime = time(NULL);
  return stime;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<getTimeStamp>:
  endbr64
  xor    %edi,%edi
  jmp    1050 <time@plt>
```

assistant: ```c
int getTimeStamp() {
  time_t stime = time(NULL);
  return stime;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<getTimeStamp>:
  endbr64
  xor    %edi,%edi
  jmp    1050 <time@plt>
```

assistant: ```c
int getTimeStamp() {
  time_t stime = time(NULL);
  return stime;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<partition>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    -0x1c(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x20(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x8(%rbp)
  jmp    L22
L98:
  mov    -0x10(%rbp),%eax
  cmp    -0x20(%rbp),%eax
  je     L26
  addl   $0x1,-0x10(%rbp)
L22:
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    %eax,-0x8(%rbp)
  jl     L98
  jmp    L76
L26:
  nop
  jmp    L76
L57:
  mov    -0xc(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  je     L21
  subl   $0x1,-0xc(%rbp)
L76:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    %eax,-0x8(%rbp)
  jg     L57
  jmp    L38
L21:
  nop
L38:
  mov    -0x10(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jge    L5
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x10(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  mov    %eax,(%rdx)
  jmp    L22
L5:
  nop
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x1c(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x8(%rbp),%eax
  mov    %eax,(%rdx)
  mov    -0x10(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int partition(int a[], int lo, int hi) {
  int i = lo + 1, j = hi, v = a[lo], temp;
  while (1) {
    while (a[i] > v) {
      if (i == hi)
        break;
      i = i + 1;
    }
    while (a[j] < v) {
      if (j == lo)
        break;
      j = j - 1;
    }
    if (i >= j)
      break;
    temp = a[i];
    a[i] = a[j];
    a[j] = temp;
  }
  a[lo] = a[j];
  a[j] = v;
  return i;
}
```
user: Translate the below assembly code into C programming language:

```asm
<partition>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %esi,%r9d
  mov    %edx,%ebp
  lea    0x1(%rsi),%r10d
  movslq %esi,%rax
  lea    (%rdi,%rax,4),%r12
  mov    (%r12),%r8d
  mov    %edx,%ecx
  jmp    L74
L88:
  add    $0x1,%r10d
  jmp    L74
L38:
  cmp    %ecx,%r10d
  jge    L22
  mov    %esi,(%r11)
  mov    %edx,(%rdi)
L74:
  movslq %r10d,%rax
  lea    (%rbx,%rax,4),%r11
  mov    (%r11),%edx
  cmp    %ebp,%r10d
  je     L45
  cmp    %r8d,%edx
  jg     L88
L45:
  movslq %ecx,%rax
  lea    (%rbx,%rax,4),%rdi
  mov    (%rdi),%esi
  cmp    %r9d,%ecx
  je     L38
  cmp    %esi,%r8d
  jle    L38
  lea    -0x4(%rbx,%rax,4),%rax
L40:
  sub    $0x1,%ecx
  mov    %rax,%rdi
  mov    (%rax),%esi
  sub    $0x4,%rax
  cmp    %ecx,%r9d
  je     L38
  cmp    %r8d,%esi
  jl     L40
  jmp    L38
L22:
  mov    %esi,(%r12)
  mov    %r8d,(%rdi)
  mov    %r10d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int partition(int a[], int lo, int hi) {
  int i = lo + 1, j = hi, v = a[lo], temp;
  while (1) {
    while (a[i] > v) {
      if (i == hi)
        break;
      i = i + 1;
    }
    while (a[j] < v) {
      if (j == lo)
        break;
      j = j - 1;
    }
    if (i >= j)
      break;
    temp = a[i];
    a[i] = a[j];
    a[j] = temp;
  }
  a[lo] = a[j];
  a[j] = v;
  return i;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<partition>:
  endbr64
  movslq %esi,%rax
  push   %r13
  mov    %rdi,%r10
  mov    %edx,%ecx
  lea    (%rdi,%rax,4),%r13
  push   %r12
  lea    0x1(%rax),%r12d
  mov    %rax,%r9
  push   %rbp
  movslq %edx,%rax
  mov    %edx,%ebp
  push   %rbx
  mov    0x0(%r13),%r8d
  movslq %r12d,%rbx
  nopl   0x0(%rax)
L81:
  lea    (%r10,%rbx,4),%rdx
  lea    (%r10,%rax,4),%rdi
  mov    (%rdx),%r11d
  mov    (%rdi),%esi
  cmp    %r8d,%r11d
  jle    L78
  cmp    %ebp,%r12d
  je     L78
  add    $0x1,%r12d
  movslq %r12d,%rbx
  jmp    L81
L78:
  cmp    %r9d,%ecx
  je     L52
  cmp    %esi,%r8d
  jle    L52
  lea    -0x4(%r10,%rax,4),%rax
  jmp    L32
  nopl   0x0(%rax)
L39:
  cmp    %r8d,%esi
  jge    L52
L32:
  sub    $0x1,%ecx
  mov    (%rax),%esi
  mov    %rax,%rdi
  sub    $0x4,%rax
  cmp    %ecx,%r9d
  jne    L39
L52:
  cmp    %ecx,%r12d
  jge    L40
  mov    %esi,(%rdx)
  movslq %ecx,%rax
  mov    %r11d,(%rdi)
  jmp    L81
  nopl   (%rax)
L40:
  mov    %esi,0x0(%r13)
  mov    %r12d,%eax
  mov    %r8d,(%rdi)
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int partition(int a[], int lo, int hi) {
  int i = lo + 1, j = hi, v = a[lo], temp;
  while (1) {
    while (a[i] > v) {
      if (i == hi)
        break;
      i = i + 1;
    }
    while (a[j] < v) {
      if (j == lo)
        break;
      j = j - 1;
    }
    if (i >= j)
      break;
    temp = a[i];
    a[i] = a[j];
    a[j] = temp;
  }
  a[lo] = a[j];
  a[j] = v;
  return i;
}
```
user: Translate the below assembly code into C programming language:

```asm
<partition>:
  endbr64
  movslq %esi,%rax
  push   %r13
  mov    %rdi,%r10
  mov    %edx,%ecx
  lea    (%rdi,%rax,4),%r13
  push   %r12
  lea    0x1(%rax),%r12d
  mov    %rax,%r9
  push   %rbp
  movslq %edx,%rax
  mov    %edx,%ebp
  push   %rbx
  mov    0x0(%r13),%r8d
  movslq %r12d,%rbx
  nopl   0x0(%rax)
L31:
  lea    (%r10,%rbx,4),%rdx
  lea    (%r10,%rax,4),%rdi
  mov    (%rdx),%r11d
  mov    (%rdi),%esi
  cmp    %r8d,%r11d
  jle    L1
  cmp    %ebp,%r12d
  je     L1
  add    $0x1,%r12d
  movslq %r12d,%rbx
  jmp    L31
L1:
  cmp    %r9d,%ecx
  je     L19
  cmp    %esi,%r8d
  jle    L19
  lea    -0x4(%r10,%rax,4),%rax
  jmp    L5
  nopl   0x0(%rax)
L2:
  cmp    %r8d,%esi
  jge    L19
L5:
  sub    $0x1,%ecx
  mov    (%rax),%esi
  mov    %rax,%rdi
  sub    $0x4,%rax
  cmp    %ecx,%r9d
  jne    L2
L19:
  cmp    %ecx,%r12d
  jge    L35
  mov    %esi,(%rdx)
  movslq %ecx,%rax
  mov    %r11d,(%rdi)
  jmp    L31
  nopl   (%rax)
L35:
  mov    %esi,0x0(%r13)
  mov    %r12d,%eax
  mov    %r8d,(%rdi)
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int partition(int a[], int lo, int hi) {
  int i = lo + 1, j = hi, v = a[lo], temp;
  while (1) {
    while (a[i] > v) {
      if (i == hi)
        break;
      i = i + 1;
    }
    while (a[j] < v) {
      if (j == lo)
        break;
      j = j - 1;
    }
    if (i >= j)
      break;
    temp = a[i];
    a[i] = a[j];
    a[j] = temp;
  }
  a[lo] = a[j];
  a[j] = v;
  return i;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<exponente_modular>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  movq   $0x0,-0x10(%rbp)
  movq   $0x1,-0x8(%rbp)
  jmp    L0
L1:
  mov    -0x8(%rbp),%rax
  imul   -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  cqto
  idivq  -0x28(%rbp)
  mov    %rdx,-0x8(%rbp)
  addq   $0x1,-0x10(%rbp)
L0:
  mov    -0x10(%rbp),%rax
  cmp    -0x20(%rbp),%rax
  jl     L1
  mov    -0x8(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
uint64_t exponente_modular(uint64_t base, uint64_t exp, uint64_t mod) {
  uint64_t k = 0;
  uint64_t ret = 1;
  for (; k < exp; k++) {
    ret *= base;
    ret %= mod;
  }
  return ret;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<exponente_modular>:
  endbr64
  mov    %rdx,%r8
  test   %rsi,%rsi
  jle    L7
  mov    $0x1,%edx
  mov    $0x0,%ecx
L15:
  mov    %rdx,%rax
  imul   %rdi,%rax
  cqto
  idiv   %r8
  add    $0x1,%rcx
  cmp    %rcx,%rsi
  jne    L15
L10:
  mov    %rdx,%rax
  ret
L7:
  mov    $0x1,%edx
  jmp    L10
```

assistant: ```c
uint64_t exponente_modular(uint64_t base, uint64_t exp, uint64_t mod) {
  uint64_t k = 0;
  uint64_t ret = 1;
  for (; k < exp; k++) {
    ret *= base;
    ret %= mod;
  }
  return ret;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<exponente_modular>:
  endbr64
  mov    %rdx,%r8
  mov    $0x1,%edx
  test   %rsi,%rsi
  jle    L9
  xor    %ecx,%ecx
  nopl   0x0(%rax,%rax,1)
L14:
  mov    %rdx,%rax
  add    $0x1,%rcx
  imul   %rdi,%rax
  cqto
  idiv   %r8
  cmp    %rcx,%rsi
  jne    L14
L9:
  mov    %rdx,%rax
  ret
```

assistant: ```c
uint64_t exponente_modular(uint64_t base, uint64_t exp, uint64_t mod) {
  uint64_t k = 0;
  uint64_t ret = 1;
  for (; k < exp; k++) {
    ret *= base;
    ret %= mod;
  }
  return ret;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<exponente_modular>:
  endbr64
  mov    %rdx,%r8
  mov    $0x1,%edx
  test   %rsi,%rsi
  jle    L4
  xor    %ecx,%ecx
  nopl   0x0(%rax,%rax,1)
L0:
  mov    %rdx,%rax
  add    $0x1,%rcx
  imul   %rdi,%rax
  cqto
  idiv   %r8
  cmp    %rcx,%rsi
  jne    L0
L4:
  mov    %rdx,%rax
  ret
```

assistant: ```c
uint64_t exponente_modular(uint64_t base, uint64_t exp, uint64_t mod) {
  uint64_t k = 0;
  uint64_t ret = 1;
  for (; k < exp; k++) {
    ret *= base;
    ret %= mod;
  }
  return ret;
}
```
user: What is the c source code of the assembly code below?

```asm
<getCellIdLeft>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  and    $0x7,%eax
  test   %eax,%eax
  jne    L12
  mov    $0xffffffff,%eax
  jmp    L19
L12:
  mov    -0x4(%rbp),%eax
  sub    $0x1,%eax
L19:
  pop    %rbp
  ret
```

assistant: ```c
int getCellIdLeft(int cellId) {
  if (cellId % 8 == 0)
    return -1;
  return cellId - 1;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<getCellIdLeft>:
  endbr64
  test   $0x7,%dil
  je     L2
  lea    -0x1(%rdi),%eax
  ret
L2:
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int getCellIdLeft(int cellId) {
  if (cellId % 8 == 0)
    return -1;
  return cellId - 1;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<getCellIdLeft>:
  endbr64
  test   $0x7,%dil
  je     L7
  lea    -0x1(%rdi),%eax
  ret
  xchg   %ax,%ax
L7:
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int getCellIdLeft(int cellId) {
  if (cellId % 8 == 0)
    return -1;
  return cellId - 1;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<getCellIdLeft>:
  endbr64
  test   $0x7,%dil
  je     L1
  lea    -0x1(%rdi),%eax
  ret
  xchg   %ax,%ax
L1:
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int getCellIdLeft(int cellId) {
  if (cellId % 8 == 0)
    return -1;
  return cellId - 1;
}
```
user: What is the c source code of the assembly code below?

```asm
<alignptr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    -0x1c(%rbp),%eax
  neg    %eax
  cltq
  mov    %rax,-0x8(%rbp)
  mov    -0x1c(%rbp),%eax
  sub    $0x1,%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  and    -0x8(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
void *alignptr(void *addr, uint32_t align) {
  uintptr_t mask = ~((uintptr_t)(align - 1));
  return (void *)(((uintptr_t)addr + (align - 1)) & mask);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<alignptr>:
  endbr64
  lea    -0x1(%rsi),%eax
  cltq
  add    %rdi,%rax
  neg    %esi
  movslq %esi,%rsi
  and    %rsi,%rax
  ret
```

assistant: ```c
void *alignptr(void *addr, uint32_t align) {
  uintptr_t mask = ~((uintptr_t)(align - 1));
  return (void *)(((uintptr_t)addr + (align - 1)) & mask);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<alignptr>:
  endbr64
  lea    -0x1(%rsi),%eax
  neg    %esi
  cltq
  movslq %esi,%rsi
  add    %rdi,%rax
  and    %rsi,%rax
  ret
```

assistant: ```c
void *alignptr(void *addr, uint32_t align) {
  uintptr_t mask = ~((uintptr_t)(align - 1));
  return (void *)(((uintptr_t)addr + (align - 1)) & mask);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<alignptr>:
  endbr64
  lea    -0x1(%rsi),%eax
  neg    %esi
  cltq
  movslq %esi,%rsi
  add    %rdi,%rax
  and    %rsi,%rax
  ret
```

assistant: ```c
void *alignptr(void *addr, uint32_t align) {
  uintptr_t mask = ~((uintptr_t)(align - 1));
  return (void *)(((uintptr_t)addr + (align - 1)) & mask);
}
```
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   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x0,-0x4(%rbp)
  mov    $0x0,%eax
  call   <SSL_library_init@plt>
  mov    $0x0,%eax
  call   <cipher_get_put_tests@plt>
  or     %eax,-0x4(%rbp)
  mov    $0x0,%eax
  call   <cipher_get_by_value_tests@plt>
  or     %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  int failed = 0;
  SSL_library_init();
  failed |= cipher_get_put_tests();
  failed |= cipher_get_by_value_tests();
  return (failed);
}
```
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   %rbx
  mov    $0x0,%eax
  call   <SSL_library_init@plt>
  mov    $0x0,%eax
  call   <cipher_get_put_tests@plt>
  mov    %eax,%ebx
  mov    $0x0,%eax
  call   <cipher_get_by_value_tests@plt>
  or     %ebx,%eax
  pop    %rbx
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  int failed = 0;
  SSL_library_init();
  failed |= cipher_get_put_tests();
  failed |= cipher_get_by_value_tests();
  return (failed);
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbx
  xor    %eax,%eax
  call   <SSL_library_init@plt>
  xor    %eax,%eax
  call   <cipher_get_put_tests@plt>
  mov    %eax,%ebx
  xor    %eax,%eax
  call   <cipher_get_by_value_tests@plt>
  or     %ebx,%eax
  pop    %rbx
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  int failed = 0;
  SSL_library_init();
  failed |= cipher_get_put_tests();
  failed |= cipher_get_by_value_tests();
  return (failed);
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %rbx
  xor    %eax,%eax
  call   <SSL_library_init@plt>
  xor    %eax,%eax
  call   <cipher_get_put_tests@plt>
  mov    %eax,%ebx
  xor    %eax,%eax
  call   <cipher_get_by_value_tests@plt>
  or     %ebx,%eax
  pop    %rbx
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  int failed = 0;
  SSL_library_init();
  failed |= cipher_get_put_tests();
  failed |= cipher_get_by_value_tests();
  return (failed);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x1,-0x4(%rbp)
  cmpl   $0x2,-0x14(%rbp)
  je     L19
  mov    -0x20(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <usage@plt>
  jmp    L14
L19:
  mov    -0x20(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <xor@plt>
  mov    %eax,-0x4(%rbp)
L14:
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
main(int argc, char **argv) {
  int retval = 1;
  if (argc != 2)
    usage(argv[0]);
  else
    retval = xor(argv[1]);
  return retval;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  cmp    $0x2,%edi
  je     L19
  mov    (%rsi),%rdi
  mov    $0x0,%eax
  call   <usage@plt>
  mov    $0x1,%eax
L4:
  add    $0x8,%rsp
  ret
L19:
  mov    0x8(%rsi),%rdi
  mov    $0x0,%eax
  call   <xor@plt>
  jmp    L4
```

assistant: ```c
main(int argc, char **argv) {
  int retval = 1;
  if (argc != 2)
    usage(argv[0]);
  else
    retval = xor(argv[1]);
  return retval;
}
```