ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x3f8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x3e8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  jmp    L35
L21:
  movsbl (%rbx,%rbp,1),%edi
  sub    $0x7,%edi
  call   <putchar@plt>
  add    $0x1,%rbx
L8:
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jg     L21
  mov    $0xa,%edi
  call   <putchar@plt>
L35:
  mov    %rbp,%rdi
  call   <gets@plt>
  test   %rax,%rax
  je     L47
  mov    $0x0,%ebx
  jmp    L8
L47:
  mov    0x3e8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L15
  mov    $0x0,%eax
  add    $0x3f8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L15:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main() {
  int i;
  char s[1000];
  while (gets(s)) {
    for (i = 0; i < strlen(s); i++)
      printf("%c", s[i] - 7);
    printf("\n");
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x3f8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x3e8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%rdi
  call   <gets@plt>
  test   %rax,%rax
  je     L30
L43:
  xor    %ebx,%ebx
  jmp    L33
  nopl   0x0(%rax)
L38:
  movsbl 0x0(%rbp,%rbx,1),%edi
  add    $0x1,%rbx
  sub    $0x7,%edi
  call   <putchar@plt>
L33:
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jg     L38
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    %rbp,%rdi
  call   <gets@plt>
  test   %rax,%rax
  jne    L43
L30:
  mov    0x3e8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L25
  add    $0x3f8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L25:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main() {
  int i;
  char s[1000];
  while (gets(s)) {
    for (i = 0; i < strlen(s); i++)
      printf("%c", s[i] - 7);
    printf("\n");
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x3f8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x3e8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%rdi
  call   <gets@plt>
  test   %rax,%rax
  je     L21
L42:
  xor    %ebx,%ebx
  jmp    L36
  nopl   0x0(%rax)
L14:
  movsbl 0x0(%rbp,%rbx,1),%edi
  add    $0x1,%rbx
  sub    $0x7,%edi
  call   <putchar@plt>
L36:
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jg     L14
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    %rbp,%rdi
  call   <gets@plt>
  test   %rax,%rax
  jne    L42
L21:
  mov    0x3e8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x3f8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main() {
  int i;
  char s[1000];
  while (gets(s)) {
    for (i = 0; i < strlen(s); i++)
      printf("%c", s[i] - 7);
    printf("\n");
  }
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<kol_board_puts>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  mov    %rdi,-0x20(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L16
L13:
  mov    -0xc(%rbp),%edx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%edx
  mov    $0x3f,%eax
  mov    $0x1,%esi
  mov    %esi,%ebx
  mov    %edx,%ecx
  int    $0x40
  addl   $0x1,-0xc(%rbp)
L16:
  mov    -0xc(%rbp),%edx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L13
  nop
  nop
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
void kol_board_puts(char *s) {
  unsigned i;
  i = 0;
  while (*(s + i)) {
    asm("int $0x40" ::"a"(63), "b"(1), "c"(*(s + i)));
    i++;
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<kol_board_puts>:
  endbr64
  movzbl (%rdi),%ecx
  test   %cl,%cl
  je     L14
  push   %rbx
  mov    $0x0,%edx
  mov    $0x3f,%eax
  mov    $0x1,%ebx
L8:
  int    $0x40
  add    $0x1,%edx
  mov    %edx,%ecx
  movzbl (%rdi,%rcx,1),%ecx
  test   %cl,%cl
  jne    L8
  pop    %rbx
  ret
L14:
  ret
```

assistant: ```c
void kol_board_puts(char *s) {
  unsigned i;
  i = 0;
  while (*(s + i)) {
    asm("int $0x40" ::"a"(63), "b"(1), "c"(*(s + i)));
    i++;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<kol_board_puts>:
  endbr64
  movzbl (%rdi),%ecx
  test   %cl,%cl
  je     L16
  push   %rbx
  xor    %edx,%edx
  mov    $0x3f,%eax
  mov    $0x1,%ebx
  nopl   0x0(%rax,%rax,1)
L7:
  int    $0x40
  lea    0x1(%rdx),%ecx
  mov    %rcx,%rdx
  movzbl (%rdi,%rcx,1),%ecx
  test   %cl,%cl
  jne    L7
  pop    %rbx
  ret
  nopw   0x0(%rax,%rax,1)
L16:
  ret
```

assistant: ```c
void kol_board_puts(char *s) {
  unsigned i;
  i = 0;
  while (*(s + i)) {
    asm("int $0x40" ::"a"(63), "b"(1), "c"(*(s + i)));
    i++;
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<kol_board_puts>:
  endbr64
  movzbl (%rdi),%ecx
  test   %cl,%cl
  je     L6
  push   %rbx
  xor    %edx,%edx
  mov    $0x3f,%eax
  mov    $0x1,%ebx
  nopl   0x0(%rax,%rax,1)
L5:
  int    $0x40
  lea    0x1(%rdx),%ecx
  mov    %rcx,%rdx
  movzbl (%rdi,%rcx,1),%ecx
  test   %cl,%cl
  jne    L5
  pop    %rbx
  ret
  nopw   0x0(%rax,%rax,1)
L6:
  ret
```

assistant: ```c
void kol_board_puts(char *s) {
  unsigned i;
  i = 0;
  while (*(s + i)) {
    asm("int $0x40" ::"a"(63), "b"(1), "c"(*(s + i)));
    i++;
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<caller2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x8,%rsp
  mov    $0xfeeb,%edi
  call   <increment@plt>
  mov    %eax,%r13d
  mov    %r13d,%edi
  call   <increment@plt>
  mov    %eax,%r12d
  mov    %r12d,%edi
  call   <callee0@plt>
  mov    %eax,%ebx
  lea    (%rbx,%r12,1),%eax
  mov    %eax,%edi
  call   <callee1@plt>
  mov    %eax,%ebx
  lea    (%rbx,%r12,1),%eax
  mov    %eax,%edi
  call   <callee2@plt>
  mov    %eax,%ebx
  lea    (%rbx,%r12,1),%eax
  mov    %eax,%edi
  call   <callee3@plt>
  mov    %eax,%ebx
  lea    (%rbx,%r12,1),%eax
  mov    %eax,%edi
  call   <callee4@plt>
  mov    %eax,%ebx
  lea    (%rbx,%r12,1),%eax
  mov    %eax,%edi
  call   <callee5@plt>
  mov    %eax,%ebx
  lea    (%rbx,%r13,1),%eax
  add    %r12d,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %rbp
  ret
```

assistant: ```c
int caller2(void) {
  register int l1 = increment(0xfeeb);
  register int l2 = increment(l1);
  register int n;
  n = callee0(l2);
  n = callee1(n + l2);
  n = callee2(n + l2);
  n = callee3(n + l2);
  n = callee4(n + l2);
  n = callee5(n + l2);
  return n + l1 + l2;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<caller2>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    $0xfeeb,%edi
  call   <increment@plt>
  mov    %eax,%ebp
  mov    %eax,%edi
  call   <increment@plt>
  mov    %eax,%ebx
  mov    %eax,%edi
  call   <callee0@plt>
  lea    (%rbx,%rax,1),%edi
  call   <callee1@plt>
  lea    (%rbx,%rax,1),%edi
  call   <callee2@plt>
  lea    (%rbx,%rax,1),%edi
  call   <callee3@plt>
  lea    (%rbx,%rax,1),%edi
  call   <callee4@plt>
  lea    (%rbx,%rax,1),%edi
  call   <callee5@plt>
  add    %ebp,%eax
  add    %ebx,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int caller2(void) {
  register int l1 = increment(0xfeeb);
  register int l2 = increment(l1);
  register int n;
  n = callee0(l2);
  n = callee1(n + l2);
  n = callee2(n + l2);
  n = callee3(n + l2);
  n = callee4(n + l2);
  n = callee5(n + l2);
  return n + l1 + l2;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<caller2>:
  endbr64
  push   %rbp
  mov    $0xfeeb,%edi
  push   %rbx
  sub    $0x8,%rsp
  call   <increment@plt>
  mov    %eax,%edi
  mov    %eax,%ebp
  call   <increment@plt>
  mov    %eax,%edi
  mov    %eax,%ebx
  call   <callee0@plt>
  lea    (%rbx,%rax,1),%edi
  call   <callee1@plt>
  lea    (%rbx,%rax,1),%edi
  call   <callee2@plt>
  lea    (%rbx,%rax,1),%edi
  call   <callee3@plt>
  lea    (%rbx,%rax,1),%edi
  call   <callee4@plt>
  lea    (%rbx,%rax,1),%edi
  call   <callee5@plt>
  add    $0x8,%rsp
  add    %ebp,%eax
  add    %ebx,%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int caller2(void) {
  register int l1 = increment(0xfeeb);
  register int l2 = increment(l1);
  register int n;
  n = callee0(l2);
  n = callee1(n + l2);
  n = callee2(n + l2);
  n = callee3(n + l2);
  n = callee4(n + l2);
  n = callee5(n + l2);
  return n + l1 + l2;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<caller2>:
  endbr64
  push   %rbp
  mov    $0xfeeb,%edi
  push   %rbx
  sub    $0x8,%rsp
  call   <increment@plt>
  mov    %eax,%edi
  mov    %eax,%ebp
  call   <increment@plt>
  mov    %eax,%edi
  mov    %eax,%ebx
  call   <callee0@plt>
  lea    (%rbx,%rax,1),%edi
  call   <callee1@plt>
  lea    (%rbx,%rax,1),%edi
  call   <callee2@plt>
  lea    (%rbx,%rax,1),%edi
  call   <callee3@plt>
  lea    (%rbx,%rax,1),%edi
  call   <callee4@plt>
  lea    (%rbx,%rax,1),%edi
  call   <callee5@plt>
  add    $0x8,%rsp
  add    %ebp,%eax
  add    %ebx,%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int caller2(void) {
  register int l1 = increment(0xfeeb);
  register int l2 = increment(l1);
  register int n;
  n = callee0(l2);
  n = callee1(n + l2);
  n = callee2(n + l2);
  n = callee3(n + l2);
  n = callee4(n + l2);
  n = callee5(n + l2);
  return n + l1 + l2;
}
```
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)
  cmpl   $0x3,-0x14(%rbp)
  je     L24
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0xffffffff,%eax
  jmp    L22
L24:
  mov    -0x20(%rbp),%rax
  add    $0x10,%rax
  mov    (%rax),%rax
  lea    D20(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <fopen@plt>
  mov    %rax,-0x10(%rbp)
  cmpq   $0x0,-0x10(%rbp)
  jne    L50
  mov    -0x20(%rbp),%rax
  add    $0x10,%rax
  mov    (%rax),%rax
  mov    %rax,%rsi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0xffffffff,%eax
  jmp    L22
L50:
  mov    -0x20(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  lea    D36(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <fopen@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L21
  mov    -0x20(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  mov    %rax,%rsi
  lea    D44(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0xffffffff,%eax
  jmp    L22
L21:
  mov    -0x20(%rbp),%rax
  add    $0x10,%rax
  mov    (%rax),%rsi
  mov    -0x20(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  mov    -0x8(%rbp),%rcx
  mov    -0x10(%rbp),%rdx
  mov    %rax,%rdi
  call   <transformInfile@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <fclose@plt>
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <fclose@plt>
  mov    $0x0,%eax
L22:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D9
type: string
name: extract_symbolic_data
arguments: tag: D20
type: string
name: extract_symbolic_data
arguments: tag: D7
type: string
name: extract_symbolic_data
arguments: tag: D36
type: string
name: extract_symbolic_data
arguments: tag: D44
type: string

tool: D9: Usage: a.out outputFileName inputFileName 

tool: D20: r

tool: D7: 
Unable to open file: %s!


tool: D36: w

tool: D44: 
Unable to open output file: %s!


assistant: ```c
int main(int argc, char *argv[]) {
  FILE *fin;
  FILE *fout;
  if (argc != 3) {
    printf("Usage: a.out outputFileName inputFileName \n");
    return -1;
  }
  fin = fopen(argv[2], "r");
  if (fin == NULL) {
    printf("\nUnable to open file: %s!\n", argv[2]);
    return -1;
  }
  fout = fopen(argv[1], "w");
  if (fout == NULL) {
    printf("\nUnable to open output file: %s!\n", argv[1]);
    return -1;
  }
  transformInfile(argv[1], argv[2], fin, fout);
  fclose(fin);
  fclose(fout);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  cmp    $0x3,%edi
  jne    L19
  mov    %rsi,%rbx
  mov    0x10(%rsi),%rdi
  lea    D38(%rip),%rsi
  call   <fopen@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  je     L26
  mov    0x8(%rbx),%rdi
  lea    D19(%rip),%rsi
  call   <fopen@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L56
  mov    0x10(%rbx),%rsi
  mov    0x8(%rbx),%rdi
  mov    %rax,%rcx
  mov    %rbp,%rdx
  call   <transformInfile@plt>
  mov    %rbp,%rdi
  call   <fclose@plt>
  mov    %r12,%rdi
  call   <fclose@plt>
  mov    $0x0,%eax
L2:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L19:
  lea    D40(%rip),%rdi
  call   <puts@plt>
  mov    $0xffffffff,%eax
  jmp    L2
L26:
  mov    0x10(%rbx),%rsi
  lea    D14(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0xffffffff,%eax
  jmp    L2
L56:
  mov    0x8(%rbx),%rsi
  lea    D15(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0xffffffff,%eax
  jmp    L2
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D38
datatype: string
name: extract_symbolic_data
arguments: data_label: D19
datatype: string
name: extract_symbolic_data
arguments: data_label: D40
datatype: string
name: extract_symbolic_data
arguments: data_label: D14
datatype: string
name: extract_symbolic_data
arguments: data_label: D15
datatype: string

tool: D38: r

tool: D19: w

tool: D40: Usage: a.out outputFileName inputFileName 

tool: D14: 
Unable to open file: %s!


tool: D15: 
Unable to open output file: %s!


assistant: ```c
int main(int argc, char *argv[]) {
  FILE *fin;
  FILE *fout;
  if (argc != 3) {
    printf("Usage: a.out outputFileName inputFileName \n");
    return -1;
  }
  fin = fopen(argv[2], "r");
  if (fin == NULL) {
    printf("\nUnable to open file: %s!\n", argv[2]);
    return -1;
  }
  fout = fopen(argv[1], "w");
  if (fout == NULL) {
    printf("\nUnable to open output file: %s!\n", argv[1]);
    return -1;
  }
  transformInfile(argv[1], argv[2], fin, fout);
  fclose(fin);
  fclose(fout);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  cmp    $0x3,%edi
  jne    L52
  mov    0x10(%rsi),%rdi
  mov    %rsi,%rbx
  lea    D46(%rip),%rsi
  call   <fopen@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  je     L46
  mov    0x8(%rbx),%rdi
  lea    D23(%rip),%rsi
  call   <fopen@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L58
  mov    0x10(%rbx),%rsi
  mov    0x8(%rbx),%rdi
  mov    %rax,%rcx
  mov    %rbp,%rdx
  call   <transformInfile@plt>
  mov    %rbp,%rdi
  call   <fclose@plt>
  mov    %r12,%rdi
  call   <fclose@plt>
  xor    %eax,%eax
L32:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L52:
  lea    D37(%rip),%rdi
  call   <puts@plt>
  or     $0xffffffff,%eax
  jmp    L32
L58:
  mov    0x8(%rbx),%rsi
  lea    D14(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  or     $0xffffffff,%eax
  jmp    L32
L46:
  mov    0x10(%rbx),%rsi
  lea    D15(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  or     $0xffffffff,%eax
  jmp    L32
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D46
type_guessed: string
name: extract_symbolic_data
arguments: data_tag: D23
type_guessed: string
name: extract_symbolic_data
arguments: data_tag: D37
type_guessed: string
name: extract_symbolic_data
arguments: data_tag: D14
type_guessed: string
name: extract_symbolic_data
arguments: data_tag: D15
type_guessed: string

tool: D46: r

tool: D23: w

tool: D37: Usage: a.out outputFileName inputFileName 

tool: D14: 
Unable to open output file: %s!


tool: D15: 
Unable to open file: %s!


assistant: ```c
int main(int argc, char *argv[]) {
  FILE *fin;
  FILE *fout;
  if (argc != 3) {
    printf("Usage: a.out outputFileName inputFileName \n");
    return -1;
  }
  fin = fopen(argv[2], "r");
  if (fin == NULL) {
    printf("\nUnable to open file: %s!\n", argv[2]);
    return -1;
  }
  fout = fopen(argv[1], "w");
  if (fout == NULL) {
    printf("\nUnable to open output file: %s!\n", argv[1]);
    return -1;
  }
  transformInfile(argv[1], argv[2], fin, fout);
  fclose(fin);
  fclose(fout);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  cmp    $0x3,%edi
  jne    L11
  mov    0x10(%rsi),%rdi
  mov    %rsi,%rbx
  lea    D28(%rip),%rsi
  call   <fopen@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  je     L32
  mov    0x8(%rbx),%rdi
  lea    D18(%rip),%rsi
  call   <fopen@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L42
  mov    0x10(%rbx),%rsi
  mov    0x8(%rbx),%rdi
  mov    %rax,%rcx
  mov    %rbp,%rdx
  call   <transformInfile@plt>
  mov    %rbp,%rdi
  call   <fclose@plt>
  mov    %r12,%rdi
  call   <fclose@plt>
  xor    %eax,%eax
L6:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L11:
  lea    D17(%rip),%rdi
  call   <puts@plt>
  or     $0xffffffff,%eax
  jmp    L6
L42:
  mov    0x8(%rbx),%rsi
  lea    D47(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  or     $0xffffffff,%eax
  jmp    L6
L32:
  mov    0x10(%rbx),%rsi
  lea    D46(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  or     $0xffffffff,%eax
  jmp    L6
```

tool_calls: name: retrieve_data
arguments: tag: D28
data_type: string
name: retrieve_data
arguments: tag: D18
data_type: string
name: retrieve_data
arguments: tag: D17
data_type: string
name: retrieve_data
arguments: tag: D47
data_type: string
name: retrieve_data
arguments: tag: D46
data_type: string

tool: D28: r

tool: D18: w

tool: D17: Usage: a.out outputFileName inputFileName 

tool: D47: 
Unable to open output file: %s!


tool: D46: 
Unable to open file: %s!


assistant: ```c
int main(int argc, char *argv[]) {
  FILE *fin;
  FILE *fout;
  if (argc != 3) {
    printf("Usage: a.out outputFileName inputFileName \n");
    return -1;
  }
  fin = fopen(argv[2], "r");
  if (fin == NULL) {
    printf("\nUnable to open file: %s!\n", argv[2]);
    return -1;
  }
  fout = fopen(argv[1], "w");
  if (fout == NULL) {
    printf("\nUnable to open output file: %s!\n", argv[1]);
    return -1;
  }
  transformInfile(argv[1], argv[2], fin, fout);
  fclose(fin);
  fclose(fout);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<factorial>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jne    L9
  mov    $0x1,%eax
  jmp    L2
L9:
  mov    -0x4(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,%edi
  call   <factorial>
  imul   -0x4(%rbp),%eax
L2:
  leave
  ret
```

assistant: ```c
int factorial(int n) {
  if (n == 0) {
    return 1;
  } else {
    return n * factorial(n - 1);
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<factorial>:
  endbr64
  mov    $0x1,%eax
  test   %edi,%edi
  jne    L6
  ret
L6:
  push   %rbx
  mov    %edi,%ebx
  lea    -0x1(%rdi),%edi
  call   <factorial>
  imul   %ebx,%eax
  pop    %rbx
  ret
```

assistant: ```c
int factorial(int n) {
  if (n == 0) {
    return 1;
  } else {
    return n * factorial(n - 1);
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<factorial>:
  endbr64
  mov    $0x1,%eax
  test   %edi,%edi
  je     L8
  nopl   (%rax)
L18:
  imul   %edi,%eax
  sub    $0x1,%edi
  jne    L18
L8:
  ret
```

assistant: ```c
int factorial(int n) {
  if (n == 0) {
    return 1;
  } else {
    return n * factorial(n - 1);
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<factorial>:
  endbr64
  mov    $0x1,%eax
  test   %edi,%edi
  je     L11
  nopl   (%rax)
L15:
  imul   %edi,%eax
  sub    $0x1,%edi
  jne    L15
L11:
  ret
```

assistant: ```c
int factorial(int n) {
  if (n == 0) {
    return 1;
  } else {
    return n * factorial(n - 1);
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<avpriv_atomic_int_add_and_fetch>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%edx
  mov    -0xc(%rbp),%eax
  add    %eax,%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%eax
  pop    %rbp
  ret
```

assistant: ```c
int avpriv_atomic_int_add_and_fetch(volatile int *ptr, int inc) {
  *ptr += inc;
  return *ptr;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<avpriv_atomic_int_add_and_fetch>:
  endbr64
  mov    (%rdi),%eax
  add    %esi,%eax
  mov    %eax,(%rdi)
  mov    (%rdi),%eax
  ret
```

assistant: ```c
int avpriv_atomic_int_add_and_fetch(volatile int *ptr, int inc) {
  *ptr += inc;
  return *ptr;
}
```
user: What is the c source code of the assembly code below?

```asm
<avpriv_atomic_int_add_and_fetch>:
  endbr64
  mov    (%rdi),%eax
  add    %esi,%eax
  mov    %eax,(%rdi)
  mov    (%rdi),%eax
  ret
```

assistant: ```c
int avpriv_atomic_int_add_and_fetch(volatile int *ptr, int inc) {
  *ptr += inc;
  return *ptr;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<avpriv_atomic_int_add_and_fetch>:
  endbr64
  mov    (%rdi),%eax
  add    %esi,%eax
  mov    %eax,(%rdi)
  mov    (%rdi),%eax
  ret
```

assistant: ```c
int avpriv_atomic_int_add_and_fetch(volatile int *ptr, int inc) {
  *ptr += inc;
  return *ptr;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<zero>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    %edx,-0x1c(%rbp)
  mov    %rcx,-0x28(%rbp)
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rdx
  sub    $0x1,%rdx
  mov    %rdx,-0x8(%rbp)
  movslq %eax,%rdx
  mov    %rdx,%r8
  mov    $0x0,%r9d
  movl   $0x0,-0x10(%rbp)
  jmp    L34
L24:
  movl   $0x0,-0xc(%rbp)
  jmp    L0
L25:
  mov    -0x10(%rbp),%edx
  movslq %edx,%rcx
  movslq %eax,%rdx
  imul   %rcx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x28(%rbp),%rdx
  add    %rdx,%rcx
  mov    -0xc(%rbp),%edx
  movslq %edx,%rdx
  movl   $0x0,(%rcx,%rdx,4)
  addl   $0x1,-0xc(%rbp)
L0:
  mov    -0xc(%rbp),%edx
  cmp    -0x18(%rbp),%edx
  jl     L25
  addl   $0x1,-0x10(%rbp)
L34:
  mov    -0x10(%rbp),%edx
  cmp    -0x14(%rbp),%edx
  jl     L24
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void zero(int M, int N, int step, int matrix[M][step]) {
  for (int i = 0; i < M; i++) {
    for (int j = 0; j < N; j++) {
      matrix[i][j] = 0;
    }
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<zero>:
  endbr64
  mov    %edi,%r9d
  movslq %edx,%rdx
  test   %edi,%edi
  jle    L11
  lea    0x0(,%rdx,4),%r8
  lea    -0x1(%rsi),%edi
  lea    0x4(%rcx,%rdi,4),%rdx
  mov    $0x0,%ecx
  not    %rdi
  shl    $0x2,%rdi
  jmp    L3
L44:
  movl   $0x0,(%rax)
  add    $0x4,%rax
  cmp    %rdx,%rax
  jne    L44
L59:
  add    $0x1,%ecx
  add    %r8,%rdx
  cmp    %ecx,%r9d
  je     L11
L3:
  lea    (%rdi,%rdx,1),%rax
  test   %esi,%esi
  jg     L44
  jmp    L59
L11:
  ret
```

assistant: ```c
void zero(int M, int N, int step, int matrix[M][step]) {
  for (int i = 0; i < M; i++) {
    for (int j = 0; j < N; j++) {
      matrix[i][j] = 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
<zero>:
  endbr64
  test   %edi,%edi
  jle    L27
  movslq %edx,%rdx
  lea    -0x1(%rsi),%eax
  push   %r14
  lea    0x4(,%rax,4),%r14
  push   %r13
  lea    0x0(,%rdx,4),%r13
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  xor    %ebx,%ebx
  nopl   (%rax)
L7:
  test   %r12d,%r12d
  jle    L2
  mov    %rcx,%rdi
  mov    %r14,%rdx
  xor    %esi,%esi
  call   <memset@plt>
  mov    %rax,%rcx
L2:
  add    $0x1,%ebx
  add    %r13,%rcx
  cmp    %ebx,%ebp
  jne    L7
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L27:
  ret
```

assistant: ```c
void zero(int M, int N, int step, int matrix[M][step]) {
  for (int i = 0; i < M; i++) {
    for (int j = 0; j < N; j++) {
      matrix[i][j] = 0;
    }
  }
}
```