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

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x60,%rsp
  mov    $0xa,%edi
  call   <malloc@plt>
  mov    %rax,-0x58(%rbp)
  mov    $0xa,%edi
  call   <malloc@plt>
  mov    %rax,-0x50(%rbp)
  mov    $0xa,%edi
  call   <malloc@plt>
  mov    %rax,-0x48(%rbp)
  mov    $0xa,%edi
  call   <malloc@plt>
  mov    %rax,-0x40(%rbp)
  mov    $0xa,%edi
  call   <malloc@plt>
  mov    %rax,-0x38(%rbp)
  mov    $0xa,%edi
  call   <malloc@plt>
  mov    %rax,-0x30(%rbp)
  mov    $0xa,%edi
  call   <malloc@plt>
  mov    %rax,-0x28(%rbp)
  mov    $0xa,%edi
  call   <malloc@plt>
  mov    %rax,-0x20(%rbp)
  mov    $0xa,%edi
  call   <malloc@plt>
  mov    %rax,-0x18(%rbp)
  mov    $0xa,%edi
  call   <malloc@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x58(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    -0x50(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    -0x48(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    -0x40(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    -0x38(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    -0x30(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    $0x3e8,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main(void) {
  char *big;
  char *a0 = malloc(10);
  char *a1 = malloc(10);
  char *a2 = malloc(10);
  char *a3 = malloc(10);
  char *a4 = malloc(10);
  char *a5 = malloc(10);
  char *a6 = malloc(10);
  char *a7 = malloc(10);
  char *a8 = malloc(10);
  char *a9 = malloc(10);
  free(a0);
  free(a1);
  free(a2);
  free(a3);
  free(a4);
  free(a5);
  free(a6);
  free(a7);
  free(a8);
  free(a9);
  big = malloc(1000);
  free(big);
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    $0xa,%edi
  call   <malloc@plt>
  mov    %rax,%r15
  mov    $0xa,%edi
  call   <malloc@plt>
  mov    %rax,%r14
  mov    $0xa,%edi
  call   <malloc@plt>
  mov    %rax,%r13
  mov    $0xa,%edi
  call   <malloc@plt>
  mov    %rax,%r12
  mov    $0xa,%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  mov    $0xa,%edi
  call   <malloc@plt>
  mov    %rax,%rbx
  mov    $0xa,%edi
  call   <malloc@plt>
  mov    %rax,(%rsp)
  mov    $0xa,%edi
  call   <malloc@plt>
  mov    %rax,0x8(%rsp)
  mov    $0xa,%edi
  call   <malloc@plt>
  mov    %rax,0x10(%rsp)
  mov    $0xa,%edi
  call   <malloc@plt>
  mov    %rax,0x18(%rsp)
  mov    %r15,%rdi
  call   <free@plt>
  mov    %r14,%rdi
  call   <free@plt>
  mov    %r13,%rdi
  call   <free@plt>
  mov    %r12,%rdi
  call   <free@plt>
  mov    %rbp,%rdi
  call   <free@plt>
  mov    %rbx,%rdi
  call   <free@plt>
  mov    (%rsp),%rdi
  call   <free@plt>
  mov    0x8(%rsp),%rdi
  call   <free@plt>
  mov    0x10(%rsp),%rdi
  call   <free@plt>
  mov    0x18(%rsp),%rdi
  call   <free@plt>
  mov    $0x3e8,%edi
  call   <malloc@plt>
  mov    %rax,%rdi
  call   <free@plt>
  mov    $0x0,%eax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
int main(void) {
  char *big;
  char *a0 = malloc(10);
  char *a1 = malloc(10);
  char *a2 = malloc(10);
  char *a3 = malloc(10);
  char *a4 = malloc(10);
  char *a5 = malloc(10);
  char *a6 = malloc(10);
  char *a7 = malloc(10);
  char *a8 = malloc(10);
  char *a9 = malloc(10);
  free(a0);
  free(a1);
  free(a2);
  free(a3);
  free(a4);
  free(a5);
  free(a6);
  free(a7);
  free(a8);
  free(a9);
  big = malloc(1000);
  free(big);
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %r15
  mov    $0xa,%edi
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  call   <malloc@plt>
  mov    $0xa,%edi
  mov    %rax,0x18(%rsp)
  call   <malloc@plt>
  mov    $0xa,%edi
  mov    %rax,0x10(%rsp)
  call   <malloc@plt>
  mov    $0xa,%edi
  mov    %rax,0x8(%rsp)
  call   <malloc@plt>
  mov    $0xa,%edi
  mov    %rax,(%rsp)
  call   <malloc@plt>
  mov    $0xa,%edi
  mov    %rax,%rbx
  call   <malloc@plt>
  mov    $0xa,%edi
  mov    %rax,%r15
  call   <malloc@plt>
  mov    $0xa,%edi
  mov    %rax,%r14
  call   <malloc@plt>
  mov    $0xa,%edi
  mov    %rax,%r13
  call   <malloc@plt>
  mov    $0xa,%edi
  mov    %rax,%r12
  call   <malloc@plt>
  mov    0x18(%rsp),%rdi
  mov    %rax,%rbp
  call   <free@plt>
  mov    0x10(%rsp),%rdi
  call   <free@plt>
  mov    0x8(%rsp),%rdi
  call   <free@plt>
  mov    (%rsp),%rdi
  call   <free@plt>
  mov    %rbx,%rdi
  call   <free@plt>
  mov    %r15,%rdi
  call   <free@plt>
  mov    %r14,%rdi
  call   <free@plt>
  mov    %r13,%rdi
  call   <free@plt>
  mov    %r12,%rdi
  call   <free@plt>
  mov    %rbp,%rdi
  call   <free@plt>
  mov    $0x3e8,%edi
  call   <malloc@plt>
  mov    %rax,%rdi
  call   <free@plt>
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
int main(void) {
  char *big;
  char *a0 = malloc(10);
  char *a1 = malloc(10);
  char *a2 = malloc(10);
  char *a3 = malloc(10);
  char *a4 = malloc(10);
  char *a5 = malloc(10);
  char *a6 = malloc(10);
  char *a7 = malloc(10);
  char *a8 = malloc(10);
  char *a9 = malloc(10);
  free(a0);
  free(a1);
  free(a2);
  free(a3);
  free(a4);
  free(a5);
  free(a6);
  free(a7);
  free(a8);
  free(a9);
  big = malloc(1000);
  free(big);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  push   %r15
  mov    $0xa,%edi
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  call   <malloc@plt>
  mov    $0xa,%edi
  mov    %rax,0x18(%rsp)
  call   <malloc@plt>
  mov    $0xa,%edi
  mov    %rax,0x10(%rsp)
  call   <malloc@plt>
  mov    $0xa,%edi
  mov    %rax,0x8(%rsp)
  call   <malloc@plt>
  mov    $0xa,%edi
  mov    %rax,(%rsp)
  call   <malloc@plt>
  mov    $0xa,%edi
  mov    %rax,%rbx
  call   <malloc@plt>
  mov    $0xa,%edi
  mov    %rax,%r15
  call   <malloc@plt>
  mov    $0xa,%edi
  mov    %rax,%r14
  call   <malloc@plt>
  mov    $0xa,%edi
  mov    %rax,%r13
  call   <malloc@plt>
  mov    $0xa,%edi
  mov    %rax,%r12
  call   <malloc@plt>
  mov    0x18(%rsp),%rdi
  mov    %rax,%rbp
  call   <free@plt>
  mov    0x10(%rsp),%rdi
  call   <free@plt>
  mov    0x8(%rsp),%rdi
  call   <free@plt>
  mov    (%rsp),%rdi
  call   <free@plt>
  mov    %rbx,%rdi
  call   <free@plt>
  mov    %r15,%rdi
  call   <free@plt>
  mov    %r14,%rdi
  call   <free@plt>
  mov    %r13,%rdi
  call   <free@plt>
  mov    %r12,%rdi
  call   <free@plt>
  mov    %rbp,%rdi
  call   <free@plt>
  mov    $0x3e8,%edi
  call   <malloc@plt>
  mov    %rax,%rdi
  call   <free@plt>
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
int main(void) {
  char *big;
  char *a0 = malloc(10);
  char *a1 = malloc(10);
  char *a2 = malloc(10);
  char *a3 = malloc(10);
  char *a4 = malloc(10);
  char *a5 = malloc(10);
  char *a6 = malloc(10);
  char *a7 = malloc(10);
  char *a8 = malloc(10);
  char *a9 = malloc(10);
  free(a0);
  free(a1);
  free(a2);
  free(a3);
  free(a4);
  free(a5);
  free(a6);
  free(a7);
  free(a8);
  free(a9);
  big = malloc(1000);
  free(big);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  cmpl   $0x3,-0x14(%rbp)
  je     L46
  lea    D24(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x1,%eax
  jmp    L38
L46:
  mov    -0x20(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  test   %rax,%rax
  je     L16
  mov    -0x20(%rbp),%rax
  add    $0x10,%rax
  mov    (%rax),%rax
  test   %rax,%rax
  je     L16
  mov    -0x20(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  mov    $0xa,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <strtol@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x20(%rbp),%rax
  add    $0x10,%rax
  mov    (%rax),%rax
  mov    $0xa,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <strtol@plt>
  mov    %rax,-0x8(%rbp)
  nop
  mov    -0x8(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <hermite@plt>
  mov    %eax,%esi
  lea    D27(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  jmp    L38
L16:
  lea    D24(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x1,%eax
L38:
  leave
  ret
```

tool_calls: name: read_data
arguments: label: D24
type_infered: string
name: read_data
arguments: label: D27
type_infered: string

tool: D24: n and x needed

tool: D27: Hermite(n,x) = %d


assistant: ```c
int main(int argc, char **argv) {
  long n, x;
  if (argc != 3) {
    printf("n and x needed\n");
    return 1;
  }
  if (argv[1] && argv[2]) {
    n = strtol(argv[1], NULL, 10);
    x = strtol(argv[2], NULL, 10);
  } else {
    printf("n and x needed\n");
    return 1;
  }
  printf("Hermite(n,x) = %d\n", hermite(n, x));
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  cmp    $0x3,%edi
  jne    L38
  mov    %rsi,%rbx
  mov    0x8(%rsi),%rdi
  test   %rdi,%rdi
  je     L41
  cmpq   $0x0,0x10(%rsi)
  je     L41
  mov    $0xa,%edx
  mov    $0x0,%esi
  call   <strtol@plt>
  mov    %rax,%rbp
  mov    0x10(%rbx),%rdi
  mov    $0xa,%edx
  mov    $0x0,%esi
  call   <strtol@plt>
  mov    %rax,%rsi
  mov    %rbp,%rdi
  call   <hermite@plt>
  mov    %eax,%esi
  lea    D0(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
L30:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L38:
  lea    D1(%rip),%rdi
  call   <puts@plt>
  mov    $0x1,%eax
  jmp    L30
L41:
  lea    D1(%rip),%rdi
  call   <puts@plt>
  mov    $0x1,%eax
  jmp    L30
```

tool_calls: name: disassemble_data
arguments: tag: D0
datatype: string
name: disassemble_data
arguments: tag: D1
datatype: string

tool: D0: Hermite(n,x) = %d


tool: D1: n and x needed

assistant: ```c
int main(int argc, char **argv) {
  long n, x;
  if (argc != 3) {
    printf("n and x needed\n");
    return 1;
  }
  if (argv[1] && argv[2]) {
    n = strtol(argv[1], NULL, 10);
    x = strtol(argv[2], NULL, 10);
  } else {
    printf("n and x needed\n");
    return 1;
  }
  printf("Hermite(n,x) = %d\n", hermite(n, x));
  return 0;
}
```
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
  push   %rbx
  sub    $0x8,%rsp
  cmp    $0x3,%edi
  jne    L35
  mov    0x8(%rsi),%rdi
  mov    %rsi,%rbx
  test   %rdi,%rdi
  je     L35
  cmpq   $0x0,0x10(%rsi)
  je     L35
  mov    $0xa,%edx
  xor    %esi,%esi
  call   <strtol@plt>
  mov    0x10(%rbx),%rdi
  mov    $0xa,%edx
  xor    %esi,%esi
  mov    %rax,%rbp
  call   <strtol@plt>
  mov    %rbp,%rdi
  mov    %rax,%rsi
  call   <hermite@plt>
  lea    D2(%rip),%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  jmp    L16
L35:
  lea    D8(%rip),%rdi
  call   <puts@plt>
  mov    $0x1,%eax
L16:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D2
datatype: string
name: retrieve_data
arguments: tag: D8
datatype: string

tool: D2: Hermite(n,x) = %d


tool: D8: n and x needed

assistant: ```c
int main(int argc, char **argv) {
  long n, x;
  if (argc != 3) {
    printf("n and x needed\n");
    return 1;
  }
  if (argv[1] && argv[2]) {
    n = strtol(argv[1], NULL, 10);
    x = strtol(argv[2], NULL, 10);
  } else {
    printf("n and x needed\n");
    return 1;
  }
  printf("Hermite(n,x) = %d\n", hermite(n, x));
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  cmp    $0x3,%edi
  jne    L17
  mov    0x8(%rsi),%rdi
  mov    %rsi,%rbx
  test   %rdi,%rdi
  je     L17
  cmpq   $0x0,0x10(%rsi)
  je     L17
  mov    $0xa,%edx
  xor    %esi,%esi
  call   <strtol@plt>
  mov    0x10(%rbx),%rdi
  mov    $0xa,%edx
  xor    %esi,%esi
  mov    %rax,%rbp
  call   <strtol@plt>
  mov    %rbp,%rdi
  mov    %rax,%rsi
  call   <hermite@plt>
  lea    D9(%rip),%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  jmp    L6
L17:
  lea    D17(%rip),%rdi
  call   <puts@plt>
  mov    $0x1,%eax
L6:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D9
type_infered: string
name: disassemble_data
arguments: data_tag: D17
type_infered: string

tool: D9: Hermite(n,x) = %d


tool: D17: n and x needed

assistant: ```c
int main(int argc, char **argv) {
  long n, x;
  if (argc != 3) {
    printf("n and x needed\n");
    return 1;
  }
  if (argv[1] && argv[2]) {
    n = strtol(argv[1], NULL, 10);
    x = strtol(argv[2], NULL, 10);
  } else {
    printf("n and x needed\n");
    return 1;
  }
  printf("Hermite(n,x) = %d\n", hermite(n, x));
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<func_if_else>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  movl   $0x0,-0x4(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  je     L1
  movl   $0x1,-0x4(%rbp)
  jmp    L11
L1:
  movl   $0x2,-0x4(%rbp)
L11:
  nop
  pop    %rbp
  ret
```

assistant: ```c
void func_if_else(int c) {
  int a = 0;
  if (c) {
    a = 1;
  } else {
    a = 2;
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<func_if_else>:
  endbr64
  ret
```

assistant: ```c
void func_if_else(int c) {
  int a = 0;
  if (c) {
    a = 1;
  } else {
    a = 2;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<func_if_else>:
  endbr64
  ret
```

assistant: ```c
void func_if_else(int c) {
  int a = 0;
  if (c) {
    a = 1;
  } else {
    a = 2;
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x0,-0xc(%rbp)
  jmp    L0
L13:
  mov    $0xa,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  addl   $0x1,-0xc(%rbp)
L0:
  cmpl   $0x63,-0xc(%rbp)
  jle    L13
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    $0x0,%edi
  call   <exit@plt>
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
main(void) {
  char *array;
  int i;
  for (i = 0; i < 100; i++)
    array = malloc(10 * sizeof(char));
  free(array);
  exit(0);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbx
  mov    $0x64,%ebx
L0:
  mov    $0xa,%edi
  call   <malloc@plt>
  sub    $0x1,%ebx
  jne    L0
  mov    %rax,%rdi
  call   <free@plt>
  mov    $0x0,%edi
  call   <exit@plt>
  mov    $0x0,%eax
  pop    %rbx
  ret
```

assistant: ```c
main(void) {
  char *array;
  int i;
  for (i = 0; i < 100; i++)
    array = malloc(10 * sizeof(char));
  free(array);
  exit(0);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  push   %rbx
  mov    $0x64,%ebx
  nopw   0x0(%rax,%rax,1)
L3:
  mov    $0xa,%edi
  call   <malloc@plt>
  sub    $0x1,%ebx
  jne    L3
  mov    %rax,%rdi
  call   <free@plt>
  xor    %edi,%edi
  call   <exit@plt>
  xor    %eax,%eax
  pop    %rbx
  ret
```

assistant: ```c
main(void) {
  char *array;
  int i;
  for (i = 0; i < 100; i++)
    array = malloc(10 * sizeof(char));
  free(array);
  exit(0);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbx
  mov    $0x64,%ebx
  nopw   0x0(%rax,%rax,1)
L4:
  mov    $0xa,%edi
  call   <malloc@plt>
  sub    $0x1,%ebx
  jne    L4
  mov    %rax,%rdi
  call   <free@plt>
  xor    %edi,%edi
  call   <exit@plt>
  xor    %eax,%eax
  pop    %rbx
  ret
```

assistant: ```c
main(void) {
  char *array;
  int i;
  for (i = 0; i < 100; i++)
    array = malloc(10 * sizeof(char));
  free(array);
  exit(0);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<msleep>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x24(%rbp),%eax
  mov    %eax,%eax
  imul   $0x10624dd3,%rax,%rax
  shr    $0x20,%rax
  shr    $0x6,%eax
  mov    %eax,-0x18(%rbp)
  mov    -0x24(%rbp),%edx
  mov    %edx,%eax
  imul   $0x10624dd3,%rax,%rax
  shr    $0x20,%rax
  shr    $0x6,%eax
  imul   $0x3e8,%eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  imul   $0xf4240,%eax,%eax
  mov    %eax,-0x14(%rbp)
  lea    -0x10(%rbp),%rdx
  lea    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <nanosleep@plt>
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L0
  call   <__stack_chk_fail@plt>
L0:
  leave
  ret
```

assistant: ```c
int msleep(unsigned ms) {
  struct timespec req, rem;
  req.tv_sec = ms / 1000;
  req.tv_nsec = (ms % 1000) * 1000000;
  return nanosleep(&req, &rem);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<msleep>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %edi,%eax
  imul   $0x10624dd3,%rax,%rax
  shr    $0x26,%rax
  mov    %eax,0x8(%rsp)
  imul   $0x3e8,%eax,%eax
  sub    %eax,%edi
  imul   $0xf4240,%edi,%edi
  mov    %edi,0xc(%rsp)
  lea    0x10(%rsp),%rsi
  lea    0x8(%rsp),%rdi
  call   <nanosleep@plt>
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L8
  add    $0x28,%rsp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int msleep(unsigned ms) {
  struct timespec req, rem;
  req.tv_sec = ms / 1000;
  req.tv_nsec = (ms % 1000) * 1000000;
  return nanosleep(&req, &rem);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<msleep>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %edi,%eax
  lea    0x10(%rsp),%rsi
  imul   $0x10624dd3,%rax,%rax
  shr    $0x26,%rax
  mov    %eax,0x8(%rsp)
  imul   $0x3e8,%eax,%eax
  sub    %eax,%edi
  imul   $0xf4240,%edi,%edi
  mov    %edi,0xc(%rsp)
  lea    0x8(%rsp),%rdi
  call   <nanosleep@plt>
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L5
  add    $0x28,%rsp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int msleep(unsigned ms) {
  struct timespec req, rem;
  req.tv_sec = ms / 1000;
  req.tv_nsec = (ms % 1000) * 1000000;
  return nanosleep(&req, &rem);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<msleep>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %edi,%eax
  lea    0x10(%rsp),%rsi
  imul   $0x10624dd3,%rax,%rax
  shr    $0x26,%rax
  imul   $0x3e8,%eax,%edx
  movd   %eax,%xmm0
  sub    %edx,%edi
  imul   $0xf4240,%edi,%edi
  movd   %edi,%xmm1
  lea    0x8(%rsp),%rdi
  punpckldq %xmm1,%xmm0
  movq   %xmm0,0x8(%rsp)
  call   <nanosleep@plt>
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L2
  add    $0x28,%rsp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int msleep(unsigned ms) {
  struct timespec req, rem;
  req.tv_sec = ms / 1000;
  req.tv_nsec = (ms % 1000) * 1000000;
  return nanosleep(&req, &rem);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<prv_getBase64Size>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  movabs $0xaaaaaaaaaaaaaaab,%rdx
  mul    %rdx
  mov    %rdx,%rax
  shr    %rax
  shl    $0x2,%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x18(%rbp),%rcx
  movabs $0xaaaaaaaaaaaaaaab,%rdx
  mov    %rcx,%rax
  mul    %rdx
  mov    %rdx,%rax
  shr    %rax
  mov    %rax,%rdx
  add    %rdx,%rdx
  add    %rax,%rdx
  mov    %rcx,%rax
  sub    %rdx,%rax
  test   %rax,%rax
  je     L7
  addq   $0x4,-0x8(%rbp)
L7:
  mov    -0x8(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
size_t prv_getBase64Size(size_t dataLen) {
  size_t result_len;
  result_len = 4 * (dataLen / 3);
  if (dataLen % 3)
    result_len += 4;
  return result_len;
}
```
user: Translate the below assembly code into C programming language:

```asm
<prv_getBase64Size>:
  endbr64
  movabs $0xaaaaaaaaaaaaaaab,%rdx
  mov    %rdi,%rax
  mul    %rdx
  shr    %rdx
  lea    0x0(,%rdx,4),%rax
  lea    (%rdx,%rdx,2),%rcx
  lea    0x4(%rax),%rdx
  cmp    %rcx,%rdi
  cmovne %rdx,%rax
  ret
```

assistant: ```c
size_t prv_getBase64Size(size_t dataLen) {
  size_t result_len;
  result_len = 4 * (dataLen / 3);
  if (dataLen % 3)
    result_len += 4;
  return result_len;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<prv_getBase64Size>:
  endbr64
  movabs $0xaaaaaaaaaaaaaaab,%rdx
  mov    %rdi,%rax
  mul    %rdx
  shr    %rdx
  lea    (%rdx,%rdx,2),%rcx
  lea    0x0(,%rdx,4),%rax
  lea    0x4(%rax),%rdx
  cmp    %rcx,%rdi
  cmovne %rdx,%rax
  ret
```

assistant: ```c
size_t prv_getBase64Size(size_t dataLen) {
  size_t result_len;
  result_len = 4 * (dataLen / 3);
  if (dataLen % 3)
    result_len += 4;
  return result_len;
}
```
user: What is the c source code of the assembly code below:

```asm
<prv_getBase64Size>:
  endbr64
  movabs $0xaaaaaaaaaaaaaaab,%rdx
  mov    %rdi,%rax
  mul    %rdx
  shr    %rdx
  lea    (%rdx,%rdx,2),%rcx
  lea    0x0(,%rdx,4),%rax
  lea    0x4(%rax),%rdx
  cmp    %rcx,%rdi
  cmovne %rdx,%rax
  ret
```

assistant: ```c
size_t prv_getBase64Size(size_t dataLen) {
  size_t result_len;
  result_len = 4 * (dataLen / 3);
  if (dataLen % 3)
    result_len += 4;
  return result_len;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<_cmsQuickFloor>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  movsd  %xmm0,-0x28(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movsd  D1(%rip),%xmm0
  movsd  %xmm0,-0x18(%rbp)
  movsd  -0x28(%rbp),%xmm0
  addsd  -0x18(%rbp),%xmm0
  movsd  %xmm0,-0x10(%rbp)
  mov    -0x10(%rbp),%eax
  sar    $0x10,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L1
  call   <__stack_chk_fail@plt>
L1:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D1
type_infered: byte[8]

tool: {"D1": [0x00,0x00,0x00,0x00,0x00,0x00,0x38,0x42]}

assistant: ```c
int _cmsQuickFloor(double val) {
  const double _lcms_double2fixmagic = 68719476736.0 * 1.5;
  union {
    double val;
    int halves[2];
  } temp;
  temp.val = val + _lcms_double2fixmagic;
  return temp.halves[0] >> 16;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<_cmsQuickFloor>:
  endbr64
  addsd  D2(%rip),%xmm0
  movq   %xmm0,%rax
  sar    $0x10,%eax
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D2
type_guessed: byte[8]

tool: {"D2": [0x00,0x00,0x00,0x00,0x00,0x00,0x38,0x42]}

assistant: ```c
int _cmsQuickFloor(double val) {
  const double _lcms_double2fixmagic = 68719476736.0 * 1.5;
  union {
    double val;
    int halves[2];
  } temp;
  temp.val = val + _lcms_double2fixmagic;
  return temp.halves[0] >> 16;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<_cmsQuickFloor>:
  endbr64
  addsd  D7(%rip),%xmm0
  movq   %xmm0,%rax
  sar    $0x10,%eax
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D7
type: byte[8]

tool: {"D7": [0x00,0x00,0x00,0x00,0x00,0x00,0x38,0x42]}

assistant: ```c
int _cmsQuickFloor(double val) {
  const double _lcms_double2fixmagic = 68719476736.0 * 1.5;
  union {
    double val;
    int halves[2];
  } temp;
  temp.val = val + _lcms_double2fixmagic;
  return temp.halves[0] >> 16;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<_cmsQuickFloor>:
  endbr64
  addsd  D9(%rip),%xmm0
  movq   %xmm0,%rax
  sar    $0x10,%eax
  ret
```

tool_calls: name: read_data
arguments: data_label: D9
datatype: byte[8]

tool: {"D9": [0x00,0x00,0x00,0x00,0x00,0x00,0x38,0x42]}

assistant: ```c
int _cmsQuickFloor(double val) {
  const double _lcms_double2fixmagic = 68719476736.0 * 1.5;
  union {
    double val;
    int halves[2];
  } temp;
  temp.val = val + _lcms_double2fixmagic;
  return temp.halves[0] >> 16;
}
```