ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 下面的汇编指令对应的C源代码是什么：

```asm
<contaRepeticao>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  movl   $0x0,-0xc(%rbp)
  movl   $0x0,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L23
L8:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    %eax,-0x20(%rbp)
  jne    L31
  addl   $0x1,-0xc(%rbp)
  mov    -0x8(%rbp),%edx
  mov    -0xc(%rbp),%eax
  cmp    %eax,%edx
  cmovge %edx,%eax
  mov    %eax,-0x8(%rbp)
  jmp    L5
L31:
  movl   $0x0,-0xc(%rbp)
L5:
  addl   $0x1,-0x4(%rbp)
L23:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L8
  mov    -0x8(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int contaRepeticao(int v[], int tamanhoVetor, int numero) {
  int currentStreak = 0, maxStreak = 0;
  for (int i = 0; i < tamanhoVetor; ++i) {
    if (v[i] == numero) {
      ++currentStreak;
      maxStreak = (currentStreak > maxStreak) ? currentStreak : maxStreak;
    } else {
      currentStreak = 0;
    }
  }
  return maxStreak;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<contaRepeticao>:
  endbr64
  test   %esi,%esi
  jle    L31
  mov    %rdi,%rax
  lea    -0x1(%rsi),%ecx
  lea    0x4(%rdi,%rcx,4),%rdi
  mov    $0x0,%esi
  mov    $0x0,%ecx
  jmp    L11
L25:
  add    $0x1,%ecx
  cmp    %ecx,%esi
  cmovl  %ecx,%esi
L15:
  add    $0x4,%rax
  cmp    %rdi,%rax
  je     L26
L11:
  cmp    %edx,(%rax)
  je     L25
  mov    $0x0,%ecx
  jmp    L15
L31:
  mov    $0x0,%esi
L26:
  mov    %esi,%eax
  ret
```

assistant: ```c
int contaRepeticao(int v[], int tamanhoVetor, int numero) {
  int currentStreak = 0, maxStreak = 0;
  for (int i = 0; i < tamanhoVetor; ++i) {
    if (v[i] == numero) {
      ++currentStreak;
      maxStreak = (currentStreak > maxStreak) ? currentStreak : maxStreak;
    } else {
      currentStreak = 0;
    }
  }
  return maxStreak;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<contaRepeticao>:
  endbr64
  test   %esi,%esi
  jle    L12
  lea    -0x1(%rsi),%eax
  xor    %r8d,%r8d
  lea    0x4(%rdi,%rax,4),%rcx
  xor    %eax,%eax
  jmp    L36
  nopw   0x0(%rax,%rax,1)
L25:
  add    $0x4,%rdi
  xor    %eax,%eax
  cmp    %rcx,%rdi
  je     L16
L36:
  cmp    %edx,(%rdi)
  jne    L25
  add    $0x1,%eax
  cmp    %eax,%r8d
  cmovl  %eax,%r8d
  add    $0x4,%rdi
  cmp    %rcx,%rdi
  jne    L36
L16:
  mov    %r8d,%eax
  ret
  cs nopw 0x0(%rax,%rax,1)
L12:
  xor    %r8d,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
int contaRepeticao(int v[], int tamanhoVetor, int numero) {
  int currentStreak = 0, maxStreak = 0;
  for (int i = 0; i < tamanhoVetor; ++i) {
    if (v[i] == numero) {
      ++currentStreak;
      maxStreak = (currentStreak > maxStreak) ? currentStreak : maxStreak;
    } else {
      currentStreak = 0;
    }
  }
  return maxStreak;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<contaRepeticao>:
  endbr64
  test   %esi,%esi
  jle    L27
  lea    -0x1(%rsi),%eax
  xor    %r8d,%r8d
  lea    0x4(%rdi,%rax,4),%rcx
  xor    %eax,%eax
  jmp    L0
  nopw   0x0(%rax,%rax,1)
L10:
  add    $0x4,%rdi
  xor    %eax,%eax
  cmp    %rdi,%rcx
  je     L13
L0:
  cmp    (%rdi),%edx
  jne    L10
  add    $0x1,%eax
  cmp    %eax,%r8d
  cmovl  %eax,%r8d
  add    $0x4,%rdi
  cmp    %rdi,%rcx
  jne    L0
L13:
  mov    %r8d,%eax
  ret
  cs nopw 0x0(%rax,%rax,1)
L27:
  xor    %r8d,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
int contaRepeticao(int v[], int tamanhoVetor, int numero) {
  int currentStreak = 0, maxStreak = 0;
  for (int i = 0; i < tamanhoVetor; ++i) {
    if (v[i] == numero) {
      ++currentStreak;
      maxStreak = (currentStreak > maxStreak) ? currentStreak : maxStreak;
    } else {
      currentStreak = 0;
    }
  }
  return maxStreak;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x12c,-0x4(%rbp)
  jmp    L17
L4:
  mov    -0x4(%rbp),%eax
  sub    $0x2b,%eax
  pxor   %xmm1,%xmm1
  cvtsi2sd %eax,%xmm1
  movsd  D5(%rip),%xmm0
  mulsd  %xmm0,%xmm1
  movq   %xmm1,%rdx
  mov    -0x4(%rbp),%eax
  movq   %rdx,%xmm0
  mov    %eax,%esi
  lea    D17(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  subl   $0x14,-0x4(%rbp)
L17:
  cmpl   $0x0,-0x4(%rbp)
  jg     L4
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: read_data
arguments: data_tag: D5
type_infered: byte[8]
name: read_data
arguments: data_tag: D17
type_infered: string

tool: {"D5": [0x72,0x1c,0xc7,0x71,0x1c,0xc7,0xe1,0x3f]}

tool: D17: %4d %5.1f


assistant: ```c
int main() {
  int fahr;
  for (fahr = 300; fahr > 0; fahr -= 20)
    printf("%4d %5.1f\n", fahr, (5.0 / 9.0) * (fahr - 43));
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    $0x12c,%ebx
  lea    D5(%rip),%rbp
L3:
  lea    -0x2b(%rbx),%eax
  pxor   %xmm0,%xmm0
  cvtsi2sd %eax,%xmm0
  mulsd  D4(%rip),%xmm0
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  sub    $0x14,%ebx
  jne    L3
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D5
type_infered: string
name: disassemble_data
arguments: tag: D4
type_infered: byte[8]

tool: D5: %4d %5.1f


tool: {"D4": [0x72,0x1c,0xc7,0x71,0x1c,0xc7,0xe1,0x3f]}

assistant: ```c
int main() {
  int fahr;
  for (fahr = 300; fahr > 0; fahr -= 20)
    printf("%4d %5.1f\n", fahr, (5.0 / 9.0) * (fahr - 43));
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  lea    D18(%rip),%rbp
  push   %rbx
  mov    $0x12c,%ebx
  sub    $0x8,%rsp
  cs nopw 0x0(%rax,%rax,1)
L2:
  lea    -0x2b(%rbx),%eax
  pxor   %xmm0,%xmm0
  mov    %ebx,%esi
  mov    %rbp,%rdi
  cvtsi2sd %eax,%xmm0
  mulsd  D19(%rip),%xmm0
  call   <printf@plt>
  sub    $0x14,%ebx
  jne    L2
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: label: D18
type: string
name: parse_data
arguments: label: D19
type: byte[8]

tool: D18: %4d %5.1f


tool: {"D19": [0x72,0x1c,0xc7,0x71,0x1c,0xc7,0xe1,0x3f]}

assistant: ```c
int main() {
  int fahr;
  for (fahr = 300; fahr > 0; fahr -= 20)
    printf("%4d %5.1f\n", fahr, (5.0 / 9.0) * (fahr - 43));
}
```
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
  lea    D11(%rip),%rbp
  push   %rbx
  mov    $0x12c,%ebx
  sub    $0x8,%rsp
  cs nopw 0x0(%rax,%rax,1)
L8:
  lea    -0x2b(%rbx),%eax
  pxor   %xmm0,%xmm0
  mov    %ebx,%esi
  mov    %rbp,%rdi
  cvtsi2sd %eax,%xmm0
  mulsd  D10(%rip),%xmm0
  call   <printf@plt>
  sub    $0x14,%ebx
  jne    L8
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: read_data
arguments: tag: D11
type: string
name: read_data
arguments: tag: D10
type: byte[8]

tool: D11: %4d %5.1f


tool: {"D10": [0x72,0x1c,0xc7,0x71,0x1c,0xc7,0xe1,0x3f]}

assistant: ```c
int main() {
  int fahr;
  for (fahr = 300; fahr > 0; fahr -= 20)
    printf("%4d %5.1f\n", fahr, (5.0 / 9.0) * (fahr - 43));
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<sort2>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%r12
  mov    %rsi,%r13
  mov    %edx,%r14d
  mov    %r8,%rbp
  test   %edx,%edx
  jle    L18
  mov    %rdi,%rax
  lea    -0x1(%rdx),%edx
  lea    0x4(%rdi,%rdx,4),%rsi
L24:
  movslq (%rax),%rdx
  addl   $0x1,0x0(%rbp,%rdx,4)
  add    $0x4,%rax
  cmp    %rsi,%rax
  jne    L24
  test   %ecx,%ecx
  jle    L85
L76:
  lea    0x4(%rbp),%rax
  lea    -0x1(%rcx),%edx
  lea    0x8(%rbp,%rdx,4),%rcx
L30:
  mov    -0x4(%rax),%edx
  add    %edx,(%rax)
  add    $0x4,%rax
  cmp    %rcx,%rax
  jne    L30
  mov    $0x4,%esi
  mov    %r14d,%edi
  call   <calloc@plt>
  mov    %rax,%rbx
  mov    %rax,%r15
  mov    $0x4,%esi
  mov    %r14d,%edi
  call   <calloc@plt>
  mov    %rax,0x8(%rsp)
  test   %r14d,%r14d
  jle    L53
L44:
  mov    %rbx,%rcx
  mov    %r14d,%r14d
  lea    (%rbx,%r14,4),%r8
  mov    %r12,%rdi
  sub    %rbx,%rdi
  sub    %rbx,%rax
  mov    %r13,%rsi
  sub    %rbx,%rsi
L5:
  mov    (%rdi,%rbx,1),%edx
  mov    %edx,(%rbx)
  mov    (%rsi,%rbx,1),%edx
  mov    %edx,(%rbx,%rax,1)
  add    $0x4,%rbx
  cmp    %r8,%rbx
  jne    L5
L16:
  movslq (%rdi,%rcx,1),%rdx
  lea    0x0(%rbp,%rdx,4),%rsi
  mov    (%rsi),%ebx
  lea    -0x1(%rbx),%edx
  mov    %edx,(%rsi)
  movslq %edx,%rdx
  mov    (%rcx),%esi
  mov    %esi,(%r12,%rdx,4)
  mov    (%rcx,%rax,1),%esi
  mov    %esi,0x0(%r13,%rdx,4)
  add    $0x4,%rcx
  cmp    %r8,%rcx
  jne    L16
L53:
  mov    %r15,%rdi
  call   <free@plt>
  mov    0x8(%rsp),%rdi
  call   <free@plt>
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L85:
  mov    $0x4,%esi
  mov    %r14d,%edi
  call   <calloc@plt>
  mov    %rax,%rbx
  mov    %rax,%r15
  mov    $0x4,%esi
  mov    %r14d,%edi
  call   <calloc@plt>
  mov    %rax,0x8(%rsp)
  jmp    L44
L18:
  test   %ecx,%ecx
  jg     L76
  mov    $0x4,%esi
  mov    %r14d,%edi
  call   <calloc@plt>
  mov    %rax,%r15
  mov    $0x4,%esi
  mov    %r14d,%edi
  call   <calloc@plt>
  mov    %rax,0x8(%rsp)
  jmp    L53
```

assistant: ```c
void sort2(int *a, int *b, int n, int na, int *abegin) {
  int i, j;
  int *aaux, *baux;
  for (i = 0; i < n; i++)
    abegin[a[i]]++;
  for (i = 1; i <= na; i++)
    abegin[i] += abegin[i - 1];
  aaux = (int *)calloc(n, sizeof(int));
  baux = (int *)calloc(n, sizeof(int));
  for (i = 0; i < n; i++) {
    aaux[i] = a[i];
    baux[i] = b[i];
  };
  for (i = 0; i < n; i++) {
    j = --abegin[a[i]];
    a[j] = aaux[i];
    b[j] = baux[i];
  }
  free(aaux);
  free(baux);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<sort2>:
  endbr64
  push   %r15
  push   %r14
  mov    %edx,%r14d
  push   %r13
  mov    %r8,%r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x18,%rsp
  test   %r14d,%r14d
  jle    L0
  lea    -0x1(%r14),%edx
  mov    %rdi,%rax
  lea    0x4(%rdi,%rdx,4),%rsi
  nopl   0x0(%rax,%rax,1)
L87:
  movslq (%rax),%rdx
  add    $0x4,%rax
  addl   $0x1,0x0(%r13,%rdx,4)
  cmp    %rsi,%rax
  jne    L87
  test   %ecx,%ecx
  jle    L22
L40:
  lea    -0x1(%rcx),%edx
  lea    0x4(%r13),%rax
  lea    0x8(%r13,%rdx,4),%rcx
  xchg   %ax,%ax
L65:
  mov    -0x4(%rax),%edx
  add    %edx,(%rax)
  add    $0x4,%rax
  cmp    %rcx,%rax
  jne    L65
  mov    $0x4,%esi
  mov    %r14d,%edi
  call   <calloc@plt>
  mov    $0x4,%esi
  mov    %r14d,%edi
  mov    %rax,%rbx
  mov    %rax,%r15
  call   <calloc@plt>
  mov    %rax,%r8
  test   %r14d,%r14d
  jle    L60
L99:
  mov    %rbp,%rdi
  mov    %r12,%r10
  mov    %rbx,%rcx
  mov    %rbx,%rdx
  lea    (%rbx,%r14,4),%r9
  sub    %rbx,%rdi
  sub    %rbx,%rax
  sub    %rbx,%r10
  nop
L96:
  mov    (%rdi,%rdx,1),%esi
  mov    %esi,(%rdx)
  mov    (%r10,%rdx,1),%esi
  mov    %esi,(%rax,%rdx,1)
  add    $0x4,%rdx
  cmp    %rdx,%r9
  jne    L96
  nopl   (%rax)
L2:
  movslq (%rdi,%rcx,1),%rdx
  add    $0x4,%rcx
  lea    0x0(%r13,%rdx,4),%rsi
  mov    (%rsi),%ebx
  lea    -0x1(%rbx),%edx
  mov    %edx,(%rsi)
  mov    -0x4(%rcx),%esi
  movslq %edx,%rdx
  mov    %esi,0x0(%rbp,%rdx,4)
  mov    -0x4(%rcx,%rax,1),%esi
  mov    %esi,(%r12,%rdx,4)
  cmp    %rcx,%r9
  jne    L2
L60:
  mov    %r15,%rdi
  mov    %r8,0x8(%rsp)
  call   <free@plt>
  mov    0x8(%rsp),%rdi
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  jmp    1060 <free@plt>
  nopl   0x0(%rax,%rax,1)
L0:
  test   %ecx,%ecx
  jg     L40
  mov    $0x4,%esi
  mov    %r14d,%edi
  call   <calloc@plt>
  mov    $0x4,%esi
  mov    %r14d,%edi
  mov    %rax,%r15
  call   <calloc@plt>
  mov    %rax,%r8
  jmp    L60
  nopw   0x0(%rax,%rax,1)
L22:
  mov    $0x4,%esi
  mov    %r14d,%edi
  call   <calloc@plt>
  mov    $0x4,%esi
  mov    %r14d,%edi
  mov    %rax,%rbx
  mov    %rax,%r15
  call   <calloc@plt>
  mov    %rax,%r8
  jmp    L99
```

assistant: ```c
void sort2(int *a, int *b, int n, int na, int *abegin) {
  int i, j;
  int *aaux, *baux;
  for (i = 0; i < n; i++)
    abegin[a[i]]++;
  for (i = 1; i <= na; i++)
    abegin[i] += abegin[i - 1];
  aaux = (int *)calloc(n, sizeof(int));
  baux = (int *)calloc(n, sizeof(int));
  for (i = 0; i < n; i++) {
    aaux[i] = a[i];
    baux[i] = b[i];
  };
  for (i = 0; i < n; i++) {
    j = --abegin[a[i]];
    a[j] = aaux[i];
    b[j] = baux[i];
  }
  free(aaux);
  free(baux);
}
```
user: Translate the below assembly code into C programming language:

```asm
<memset>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L3
L12:
  mov    -0x8(%rbp),%rax
  lea    0x4(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  mov    -0x1c(%rbp),%edx
  mov    %edx,(%rax)
  addl   $0x1,-0xc(%rbp)
L3:
  mov    -0xc(%rbp),%eax
  cltq
  cmp    %rax,-0x28(%rbp)
  ja     L12
  mov    -0x18(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
void *memset(void *addr, int value, size_t len) {
  int i;
  uint8_t *ap = (uint8_t *)addr;
  for (i = 0; i < len; i++) {
    *(ap++) = value;
  }
  return addr;
}
```
user: What is the c source code of the assembly code below?

```asm
<memset>:
  endbr64
  mov    %rdi,%rax
  test   %rdx,%rdx
  je     L0
  lea    (%rdi,%rdx,4),%rcx
  mov    %rdi,%rdx
L6:
  add    $0x4,%rdx
  mov    %esi,-0x4(%rdx)
  cmp    %rdx,%rcx
  jne    L6
L0:
  ret
```

assistant: ```c
void *memset(void *addr, int value, size_t len) {
  int i;
  uint8_t *ap = (uint8_t *)addr;
  for (i = 0; i < len; i++) {
    *(ap++) = value;
  }
  return addr;
}
```
user: Translate the below assembly code into C programming language:

```asm
<memset>:
  endbr64
  mov    %rdi,%rax
  test   %rdx,%rdx
  je     L14
  lea    (%rdi,%rdx,4),%rcx
  mov    %rdi,%rdx
  nopl   0x0(%rax,%rax,1)
L5:
  add    $0x4,%rdx
  mov    %esi,-0x4(%rdx)
  cmp    %rdx,%rcx
  jne    L5
L14:
  ret
```

assistant: ```c
void *memset(void *addr, int value, size_t len) {
  int i;
  uint8_t *ap = (uint8_t *)addr;
  for (i = 0; i < len; i++) {
    *(ap++) = value;
  }
  return addr;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<memset>:
  endbr64
  mov    %rdi,%r8
  test   %rdx,%rdx
  je     L67
  lea    -0x1(%rdx),%rax
  cmp    $0x2,%rax
  jbe    L35
  mov    %rdx,%rcx
  movd   %esi,%xmm1
  mov    %rdi,%rax
  shr    $0x2,%rcx
  pshufd $0x0,%xmm1,%xmm0
  shl    $0x4,%rcx
  add    %rdi,%rcx
L6:
  movups %xmm0,(%rax)
  add    $0x10,%rax
  cmp    %rcx,%rax
  jne    L6
  mov    %rdx,%rcx
  and    $0xfffffffffffffffc,%rcx
  mov    %ecx,%eax
  lea    (%r8,%rcx,4),%rdi
  cmp    %rcx,%rdx
  je     L67
L37:
  lea    0x1(%rax),%ecx
  mov    %esi,(%rdi)
  movslq %ecx,%rcx
  cmp    %rcx,%rdx
  jbe    L67
  add    $0x2,%eax
  mov    %esi,0x4(%rdi)
  cltq
  cmp    %rax,%rdx
  jbe    L67
  mov    %esi,0x8(%rdi)
L67:
  mov    %r8,%rax
  ret
L35:
  xor    %eax,%eax
  jmp    L37
```

assistant: ```c
void *memset(void *addr, int value, size_t len) {
  int i;
  uint8_t *ap = (uint8_t *)addr;
  for (i = 0; i < len; i++) {
    *(ap++) = value;
  }
  return addr;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<_cl_printf>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x8e0,%rsp
  mov    %rdi,-0x8d8(%rbp)
  mov    %rsi,-0xa8(%rbp)
  mov    %rdx,-0xa0(%rbp)
  mov    %rcx,-0x98(%rbp)
  mov    %r8,-0x90(%rbp)
  mov    %r9,-0x88(%rbp)
  test   %al,%al
  je     L14
  movaps %xmm0,-0x80(%rbp)
  movaps %xmm1,-0x70(%rbp)
  movaps %xmm2,-0x60(%rbp)
  movaps %xmm3,-0x50(%rbp)
  movaps %xmm4,-0x40(%rbp)
  movaps %xmm5,-0x30(%rbp)
  movaps %xmm6,-0x20(%rbp)
  movaps %xmm7,-0x10(%rbp)
L14:
  mov    %fs:0x28,%rax
  mov    %rax,-0xb8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x8c8(%rbp)
  mov    -0x8d8(%rbp),%rdx
  mov    -0x8c4(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <va_start@plt>
  movl   $0x0,-0x8cc(%rbp)
  jmp    L12
L50:
  mov    -0x8cc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8d8(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%edx
  mov    -0x8cc(%rbp),%eax
  cltq
  mov    %dl,-0x8c0(%rbp,%rax,1)
  mov    -0x8cc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8d8(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L24
  addl   $0x1,-0x8cc(%rbp)
L12:
  cmpl   $0x7ff,-0x8cc(%rbp)
  jle    L50
  jmp    L27
L24:
  nop
L27:
  mov    -0x8c4(%rbp),%edx
  lea    -0x8c0(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <vprintf@plt>
  mov    %eax,-0x8c8(%rbp)
  mov    $0x0,%edi
  call   <fflush@plt>
  mov    -0x8c4(%rbp),%eax
  mov    %eax,%edi
  call   <va_end@plt>
  mov    -0x8c8(%rbp),%eax
  mov    -0xb8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L23
  call   <__stack_chk_fail@plt>
L23:
  leave
  ret
```

assistant: ```c
int _cl_printf(__attribute__((address_space(3))) char *restrict fmt, ...) {
  int retval = 0;
  va_list ap;
  va_start(ap, fmt);
  char format_private[2048];
  for (int i = 0; i < 2048; ++i) {
    format_private[i] = fmt[i];
    if (fmt[i] == '\0') {
      break;
    }
  }
  retval = vprintf(format_private, ap);
  fflush(NULL);
  va_end(ap);
  return retval;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<_cl_printf>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x818,%rsp
  mov    %rdi,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x808(%rsp)
  xor    %eax,%eax
  mov    %rdi,%rsi
  mov    $0x0,%ebp
  mov    %ebp,%edi
  call   <va_start@plt>
  mov    $0x0,%eax
L16:
  movzbl (%rbx,%rax,1),%edx
  mov    %dl,(%rsp,%rax,1)
  test   %dl,%dl
  je     L9
  add    $0x1,%rax
  cmp    $0x800,%rax
  jne    L16
L9:
  mov    %rsp,%rdi
  mov    $0x0,%ebp
  mov    %ebp,%esi
  call   <vprintf@plt>
  mov    %eax,%ebx
  mov    $0x0,%edi
  call   <fflush@plt>
  mov    %ebp,%edi
  call   <va_end@plt>
  mov    0x808(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L14
  mov    %ebx,%eax
  add    $0x818,%rsp
  pop    %rbx
  pop    %rbp
  ret
L14:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int _cl_printf(__attribute__((address_space(3))) char *restrict fmt, ...) {
  int retval = 0;
  va_list ap;
  va_start(ap, fmt);
  char format_private[2048];
  for (int i = 0; i < 2048; ++i) {
    format_private[i] = fmt[i];
    if (fmt[i] == '\0') {
      break;
    }
  }
  retval = vprintf(format_private, ap);
  fflush(NULL);
  va_end(ap);
  return retval;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<_cl_printf>:
  endbr64
  push   %r12
  mov    %rdi,%rsi
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  xor    %edi,%edi
  sub    $0x810,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x808(%rsp)
  xor    %eax,%eax
  call   <va_start@plt>
  xor    %eax,%eax
  mov    %rsp,%rdi
  jmp    L35
  cs nopw 0x0(%rax,%rax,1)
L23:
  add    $0x1,%rax
  cmp    $0x800,%rax
  je     L0
L35:
  movzbl (%rbx,%rax,1),%edx
  mov    %dl,(%rdi,%rax,1)
  test   %dl,%dl
  jne    L23
L0:
  xor    %ebp,%ebp
  mov    %ebp,%esi
  call   <vprintf@plt>
  xor    %edi,%edi
  mov    %eax,%r12d
  call   <fflush@plt>
  mov    %ebp,%edi
  call   <va_end@plt>
  mov    0x808(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L11
  add    $0x810,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L11:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int _cl_printf(__attribute__((address_space(3))) char *restrict fmt, ...) {
  int retval = 0;
  va_list ap;
  va_start(ap, fmt);
  char format_private[2048];
  for (int i = 0; i < 2048; ++i) {
    format_private[i] = fmt[i];
    if (fmt[i] == '\0') {
      break;
    }
  }
  retval = vprintf(format_private, ap);
  fflush(NULL);
  va_end(ap);
  return retval;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<_cl_printf>:
  endbr64
  push   %r12
  mov    %rdi,%rsi
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  xor    %edi,%edi
  sub    $0x810,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x808(%rsp)
  xor    %eax,%eax
  call   <va_start@plt>
  xor    %eax,%eax
  mov    %rsp,%rdi
  jmp    L15
  cs nopw 0x0(%rax,%rax,1)
L11:
  add    $0x1,%rax
  cmp    $0x800,%rax
  je     L24
L15:
  movzbl (%rbx,%rax,1),%edx
  mov    %dl,(%rdi,%rax,1)
  test   %dl,%dl
  jne    L11
L24:
  xor    %ebp,%ebp
  mov    %ebp,%esi
  call   <vprintf@plt>
  xor    %edi,%edi
  mov    %eax,%r12d
  call   <fflush@plt>
  mov    %ebp,%edi
  call   <va_end@plt>
  mov    0x808(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L33
  add    $0x810,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L33:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int _cl_printf(__attribute__((address_space(3))) char *restrict fmt, ...) {
  int retval = 0;
  va_list ap;
  va_start(ap, fmt);
  char format_private[2048];
  for (int i = 0; i < 2048; ++i) {
    format_private[i] = fmt[i];
    if (fmt[i] == '\0') {
      break;
    }
  }
  retval = vprintf(format_private, ap);
  fflush(NULL);
  va_end(ap);
  return retval;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  add    $0xffffffffffffff80,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x70(%rbp),%rax
  mov    %rax,%rsi
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  lea    -0x7c(%rbp),%rax
  mov    %rax,%rsi
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    -0x7c(%rbp),%eax
  mov    %eax,-0x74(%rbp)
  cmpl   $0x1a,-0x74(%rbp)
  jle    L24
  mov    -0x74(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x4ec4ec4f,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x3,%edx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  imul   $0x1a,%edx,%edx
  sub    %edx,%eax
  mov    %eax,-0x74(%rbp)
L24:
  movl   $0x0,-0x78(%rbp)
  jmp    L7
L18:
  mov    -0x78(%rbp),%eax
  cltq
  movzbl -0x70(%rbp,%rax,1),%eax
  mov    %eax,%edx
  mov    -0x74(%rbp),%eax
  add    %edx,%eax
  mov    %eax,%edx
  mov    -0x78(%rbp),%eax
  cltq
  mov    %dl,-0x70(%rbp,%rax,1)
  mov    -0x78(%rbp),%eax
  cltq
  movzbl -0x70(%rbp,%rax,1),%eax
  cmp    $0x7a,%al
  jle    L36
  mov    -0x78(%rbp),%eax
  cltq
  movzbl -0x70(%rbp,%rax,1),%eax
  sub    $0x1a,%eax
  mov    %eax,%edx
  mov    -0x78(%rbp),%eax
  cltq
  mov    %dl,-0x70(%rbp,%rax,1)
L36:
  addl   $0x1,-0x78(%rbp)
L7:
  mov    -0x78(%rbp),%eax
  cltq
  movzbl -0x70(%rbp,%rax,1),%eax
  test   %al,%al
  jne    L18
  lea    -0x70(%rbp),%rax
  mov    %rax,%rsi
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L41
  call   <__stack_chk_fail@plt>
L41:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D2
data_type: string
name: disassemble_data
arguments: data_label: D0
data_type: string

tool: D2: %s

tool: D0: %d

assistant: ```c
int main() {
  char str[101];
  int j, n, n1;
  scanf("%s", str);
  scanf("%d", &n);
  n1 = n;
  if (n1 > 26)
    n1 = n1 % 26;
  for (j = 0; str[j] != '\0'; j++) {
    str[j] = str[j] + n1;
    if (str[j] > 'z')
      str[j] = str[j] + 'a' - 'z' - 1;
  }
  printf("%s", str);
  return 0;
}
```