ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请从给定的汇编指令推导出C源代码：

```asm
<minmax>:
  endbr64
  push   %r13
  xor    %eax,%eax
  mov    %esi,%r13d
  mov    $0x8,%esi
  push   %r12
  mov    %rdi,%r12
  lea    D7(%rip),%rdi
  push   %rbp
  mov    %rcx,%rbp
  push   %rbx
  mov    %rdx,%rbx
  sub    $0x8,%rsp
  call   <printf@plt>
  mov    %r12,%rsi
  lea    D12(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  movl   $0x3e8,(%r12)
  movl   $0x3e8,(%rbx)
  movl   $0x3e8,0x0(%rbp)
  cmp    $0x1,%r13d
  jle    L11
  lea    -0x2(%r13),%edx
  lea    0x4(%r12),%rax
  lea    0x8(%r12,%rdx,4),%rdx
  nopw   0x0(%rax,%rax,1)
L31:
  mov    (%rax),%esi
  cmp    (%rbx),%esi
  jle    L28
  mov    %esi,(%rbx)
L28:
  cmp    0x0(%rbp),%esi
  jge    L12
  mov    %esi,0x0(%rbp)
L12:
  add    $0x4,%rax
  cmp    %rdx,%rax
  jne    L31
L11:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D7
data_type: string
name: unpack_data
arguments: data_label: D12
data_type: string

tool: D7: minmax sizeof(b) = %lu


tool: D12: maxmin *b = %p


assistant: ```c
void minmax(int *b, int len, int *max, int *min) {
  printf("minmax sizeof(b) = %lu\n", sizeof(b));
  printf("maxmin *b = %p\n", b);
  b[0] = 1000;
  *min = *max = b[0];
  for (int i = 1; i < len; i++) {
    if (b[i] > *max) {
      *max = b[i];
    }
    if (b[i] < *min) {
      *min = b[i];
    }
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<minmax>:
  endbr64
  push   %r13
  xor    %eax,%eax
  mov    %esi,%r13d
  mov    $0x8,%esi
  push   %r12
  mov    %rdi,%r12
  lea    D15(%rip),%rdi
  push   %rbp
  mov    %rcx,%rbp
  push   %rbx
  mov    %rdx,%rbx
  sub    $0x8,%rsp
  call   <printf@plt>
  mov    %r12,%rsi
  lea    D5(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  movl   $0x3e8,(%r12)
  movl   $0x3e8,(%rbx)
  movl   $0x3e8,0x0(%rbp)
  cmp    $0x1,%r13d
  jle    L20
  lea    -0x2(%r13),%edx
  lea    0x4(%r12),%rax
  lea    0x8(%r12,%rdx,4),%rdx
  nopw   0x0(%rax,%rax,1)
L30:
  mov    (%rax),%esi
  cmp    (%rbx),%esi
  jle    L32
  mov    %esi,(%rbx)
L32:
  cmp    0x0(%rbp),%esi
  jge    L10
  mov    %esi,0x0(%rbp)
L10:
  add    $0x4,%rax
  cmp    %rdx,%rax
  jne    L30
L20:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: disassemble_data
arguments: label: D15
datatype: string
name: disassemble_data
arguments: label: D5
datatype: string

tool: D15: minmax sizeof(b) = %lu


tool: D5: maxmin *b = %p


assistant: ```c
void minmax(int *b, int len, int *max, int *min) {
  printf("minmax sizeof(b) = %lu\n", sizeof(b));
  printf("maxmin *b = %p\n", b);
  b[0] = 1000;
  *min = *max = b[0];
  for (int i = 1; i < len; i++) {
    if (b[i] > *max) {
      *max = b[i];
    }
    if (b[i] < *min) {
      *min = b[i];
    }
  }
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<Com_Chop>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cltq
  lea    -0x1(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x8(%rbp)
  jmp    L6
L0:
  mov    -0x8(%rbp),%rax
  lea    -0x1(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  movb   $0x0,(%rax)
L6:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L0
  mov    -0x18(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *Com_Chop(char *s) {
  char *right;
  right = s + strlen(s) - 1;
  while (isspace(*right))
    *right-- = 0;
  return s;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<Com_Chop>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  call   <strlen@plt>
  cltq
  lea    -0x1(%rbp,%rax,1),%rbx
  jmp    L16
L18:
  sub    $0x1,%rbx
  movb   $0x0,0x1(%rbx)
L16:
  movsbl (%rbx),%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L18
  mov    %rbp,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
char *Com_Chop(char *s) {
  char *right;
  right = s + strlen(s) - 1;
  while (isspace(*right))
    *right-- = 0;
  return s;
}
```
user: What is the c source code of the assembly code below?

```asm
<Com_Chop>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  cltq
  lea    -0x1(%r12,%rax,1),%rbx
  jmp    L16
  nopl   0x0(%rax)
L17:
  movb   $0x0,(%rbx)
  sub    $0x1,%rbx
L16:
  movsbl (%rbx),%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L17
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  ret
```

assistant: ```c
char *Com_Chop(char *s) {
  char *right;
  right = s + strlen(s) - 1;
  while (isspace(*right))
    *right-- = 0;
  return s;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<Com_Chop>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  cltq
  lea    -0x1(%r12,%rax,1),%rbx
  jmp    L3
  nopl   0x0(%rax)
L19:
  movb   $0x0,(%rbx)
  sub    $0x1,%rbx
L3:
  movsbl (%rbx),%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L19
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  ret
```

assistant: ```c
char *Com_Chop(char *s) {
  char *right;
  right = s + strlen(s) - 1;
  while (isspace(*right))
    *right-- = 0;
  return s;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<strings_main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    $0x0,%eax
  call   <strings_strbody@plt>
  mov    $0x0,%eax
  call   <strings_strbival@plt>
  mov    $0x0,%eax
  call   <strings_strinsert@plt>
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int strings_main(void *rtime, int argc, char **argv) {
  (void)rtime;
  (void)argc;
  (void)argv;
  strings_strbody();
  strings_strbival();
  strings_strinsert();
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<strings_main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <strings_strbody@plt>
  mov    $0x0,%eax
  call   <strings_strbival@plt>
  mov    $0x0,%eax
  call   <strings_strinsert@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int strings_main(void *rtime, int argc, char **argv) {
  (void)rtime;
  (void)argc;
  (void)argv;
  strings_strbody();
  strings_strbival();
  strings_strinsert();
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<strings_main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <strings_strbody@plt>
  xor    %eax,%eax
  call   <strings_strbival@plt>
  xor    %eax,%eax
  call   <strings_strinsert@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int strings_main(void *rtime, int argc, char **argv) {
  (void)rtime;
  (void)argc;
  (void)argv;
  strings_strbody();
  strings_strbival();
  strings_strinsert();
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<strings_main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <strings_strbody@plt>
  xor    %eax,%eax
  call   <strings_strbival@plt>
  xor    %eax,%eax
  call   <strings_strinsert@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int strings_main(void *rtime, int argc, char **argv) {
  (void)rtime;
  (void)argc;
  (void)argv;
  strings_strbody();
  strings_strbival();
  strings_strinsert();
  return 0;
}
```
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
  movl   $0x4,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int main() {
  signed int a = 4;
  return a;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  mov    $0x4,%eax
  ret
```

assistant: ```c
int main() {
  signed int a = 4;
  return a;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  mov    $0x4,%eax
  ret
```

assistant: ```c
int main() {
  signed int a = 4;
  return a;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  mov    $0x4,%eax
  ret
```

assistant: ```c
int main() {
  signed int a = 4;
  return a;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<normaliseP2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  pxor   %xmm0,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movl   $0x0,-0x10(%rbp)
  jmp    L24
L62:
  movl   $0x0,-0xc(%rbp)
  jmp    L21
L75:
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0xc(%rbp),%eax
  cltq
  shl    $0x3,%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  movsd  -0x8(%rbp),%xmm1
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  addl   $0x1,-0xc(%rbp)
L21:
  mov    -0xc(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L75
  addl   $0x1,-0x10(%rbp)
L24:
  mov    -0x10(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L62
  movl   $0x0,-0x10(%rbp)
  jmp    L25
L19:
  movl   $0x0,-0xc(%rbp)
  jmp    L61
L47:
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0xc(%rbp),%eax
  cltq
  shl    $0x3,%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0xc(%rbp),%eax
  cltq
  shl    $0x3,%rax
  add    %rdx,%rax
  divsd  -0x8(%rbp),%xmm0
  movsd  %xmm0,(%rax)
  addl   $0x1,-0xc(%rbp)
L61:
  mov    -0xc(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L47
  addl   $0x1,-0x10(%rbp)
L25:
  mov    -0x10(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L19
  nop
  pop    %rbp
  ret
```

assistant: ```c
void normaliseP2(double **P, int N) {
  int i, j;
  double sumP = 0.;
  for (i = 0; i < N; i++)
    for (j = 0; j < N; j++)
      sumP += P[i][j];
  for (i = 0; i < N; i++)
    for (j = 0; j < N; j++)
      P[i][j] /= sumP;
  return;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<normaliseP2>:
  endbr64
  test   %esi,%esi
  jle    L16
  mov    %rdi,%rcx
  lea    -0x1(%rsi),%r9d
  shl    $0x3,%r9
  lea    0x8(%rdi,%r9,1),%r8
  pxor   %xmm1,%xmm1
  add    $0x8,%r9
L14:
  mov    (%rdi),%rax
  lea    (%r9,%rax,1),%rdx
L37:
  addsd  (%rax),%xmm1
  add    $0x8,%rax
  cmp    %rdx,%rax
  jne    L37
  add    $0x8,%rdi
  cmp    %r8,%rdi
  jne    L14
  mov    %esi,%esi
  shl    $0x3,%rsi
L40:
  mov    $0x0,%eax
L26:
  mov    %rax,%rdx
  add    (%rcx),%rdx
  movsd  (%rdx),%xmm0
  divsd  %xmm1,%xmm0
  movsd  %xmm0,(%rdx)
  add    $0x8,%rax
  cmp    %rax,%rsi
  jne    L26
  add    $0x8,%rcx
  cmp    %r8,%rcx
  jne    L40
  ret
L16:
  ret
```

assistant: ```c
void normaliseP2(double **P, int N) {
  int i, j;
  double sumP = 0.;
  for (i = 0; i < N; i++)
    for (j = 0; j < N; j++)
      sumP += P[i][j];
  for (i = 0; i < N; i++)
    for (j = 0; j < N; j++)
      P[i][j] /= sumP;
  return;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<normaliseP2>:
  endbr64
  test   %esi,%esi
  jle    L3
  lea    -0x1(%rsi),%eax
  mov    %rdi,%rcx
  pxor   %xmm1,%xmm1
  shl    $0x3,%rax
  lea    0x8(%rdi,%rax,1),%r8
  lea    0x8(%rax),%rsi
  nop
L18:
  mov    (%rcx),%rax
  lea    (%rsi,%rax,1),%rdx
  nopw   0x0(%rax,%rax,1)
L8:
  addsd  (%rax),%xmm1
  add    $0x8,%rax
  cmp    %rdx,%rax
  jne    L8
  add    $0x8,%rcx
  cmp    %r8,%rcx
  jne    L18
  cs nopw 0x0(%rax,%rax,1)
L22:
  mov    (%rdi),%rax
  lea    (%rsi,%rax,1),%rdx
  nopw   0x0(%rax,%rax,1)
L33:
  movsd  (%rax),%xmm0
  add    $0x8,%rax
  divsd  %xmm1,%xmm0
  movsd  %xmm0,-0x8(%rax)
  cmp    %rax,%rdx
  jne    L33
  add    $0x8,%rdi
  cmp    %r8,%rdi
  jne    L22
  ret
L3:
  ret
```

assistant: ```c
void normaliseP2(double **P, int N) {
  int i, j;
  double sumP = 0.;
  for (i = 0; i < N; i++)
    for (j = 0; j < N; j++)
      sumP += P[i][j];
  for (i = 0; i < N; i++)
    for (j = 0; j < N; j++)
      P[i][j] /= sumP;
  return;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<normaliseP2>:
  endbr64
  test   %esi,%esi
  jle    L7
  mov    %esi,%r8d
  mov    %rdi,%rsi
  push   %rbx
  pxor   %xmm0,%xmm0
  lea    -0x1(%r8),%eax
  lea    0x8(%rdi,%rax,8),%r9
  mov    %r8d,%eax
  mov    %r8d,%edi
  shr    %eax
  and    $0xfffffffe,%edi
  lea    -0x1(%rax),%r10d
  mov    %r10,%r11
  mov    %rsi,%r10
  shl    $0x4,%r11
  lea    0x10(%r11),%rbx
  nopl   (%rax)
L26:
  mov    (%r10),%rdx
  cmp    $0x1,%r8d
  je     L79
  mov    %rdx,%rax
  lea    (%rbx,%rdx,1),%rcx
  nopl   0x0(%rax)
L104:
  movsd  (%rax),%xmm1
  add    $0x10,%rax
  addsd  %xmm1,%xmm0
  movsd  -0x8(%rax),%xmm1
  addsd  %xmm1,%xmm0
  cmp    %rcx,%rax
  jne    L104
  movslq %edi,%rax
  cmp    %r8d,%edi
  je     L100
L52:
  addsd  (%rdx,%rax,8),%xmm0
L100:
  add    $0x8,%r10
  cmp    %r9,%r10
  jne    L26
  movapd %xmm0,%xmm2
  lea    0x10(%r11),%r10
  unpcklpd %xmm2,%xmm2
  nopl   0x0(%rax)
L43:
  mov    (%rsi),%rdx
  cmp    $0x1,%r8d
  je     L0
  mov    %rdx,%rax
  lea    (%r10,%rdx,1),%rcx
  nopl   0x0(%rax,%rax,1)
L23:
  movupd (%rax),%xmm1
  add    $0x10,%rax
  divpd  %xmm2,%xmm1
  movups %xmm1,-0x10(%rax)
  cmp    %rax,%rcx
  jne    L23
  movslq %edi,%rax
  cmp    %r8d,%edi
  je     L37
L75:
  lea    (%rdx,%rax,8),%rax
  movsd  (%rax),%xmm1
  divsd  %xmm0,%xmm1
  movsd  %xmm1,(%rax)
L37:
  add    $0x8,%rsi
  cmp    %r9,%rsi
  jne    L43
  pop    %rbx
  ret
  nopl   0x0(%rax,%rax,1)
L0:
  xor    %eax,%eax
  jmp    L75
  nopl   0x0(%rax)
L79:
  xor    %eax,%eax
  jmp    L52
L7:
  ret
```

assistant: ```c
void normaliseP2(double **P, int N) {
  int i, j;
  double sumP = 0.;
  for (i = 0; i < N; i++)
    for (j = 0; j < N; j++)
      sumP += P[i][j];
  for (i = 0; i < N; i++)
    for (j = 0; j < N; j++)
      P[i][j] /= sumP;
  return;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<factrec>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  jne    L16
  mov    $0x1,%eax
  jmp    L9
L16:
  mov    -0x14(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,%edi
  call   <factrec>
  mov    -0x14(%rbp),%edx
  imul   %edx,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
L9:
  leave
  ret
```

assistant: ```c
int factrec(int n) {
  int f;
  if (n == 0)
    return 1;
  f = n * factrec(n - 1);
  return f;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<factrec>:
  endbr64
  mov    $0x1,%eax
  test   %edi,%edi
  jne    L1
  ret
L1:
  push   %rbx
  mov    %edi,%ebx
  lea    -0x1(%rdi),%edi
  call   <factrec>
  imul   %ebx,%eax
  pop    %rbx
  ret
```

assistant: ```c
int factrec(int n) {
  int f;
  if (n == 0)
    return 1;
  f = n * factrec(n - 1);
  return f;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<factrec>:
  endbr64
  mov    $0x1,%eax
  test   %edi,%edi
  je     L19
  nopl   (%rax)
L0:
  imul   %edi,%eax
  sub    $0x1,%edi
  jne    L0
L19:
  ret
```

assistant: ```c
int factrec(int n) {
  int f;
  if (n == 0)
    return 1;
  f = n * factrec(n - 1);
  return f;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<factrec>:
  endbr64
  mov    $0x1,%eax
  test   %edi,%edi
  je     L16
  nopl   (%rax)
L18:
  imul   %edi,%eax
  sub    $0x1,%edi
  jne    L18
L16:
  ret
```

assistant: ```c
int factrec(int n) {
  int f;
  if (n == 0)
    return 1;
  f = n * factrec(n - 1);
  return f;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<path_038_expected_1>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  addl   $0x1,-0x4(%rbp)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void path_038_expected_1(int y) {
  do {
    y++;
  } while (0);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<path_038_expected_1>:
  endbr64
  ret
```

assistant: ```c
void path_038_expected_1(int y) {
  do {
    y++;
  } while (0);
}
```
user: What is the c source code of the assembly code below:

```asm
<path_038_expected_1>:
  endbr64
  ret
```

assistant: ```c
void path_038_expected_1(int y) {
  do {
    y++;
  } while (0);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<strcspn>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movq   $0x0,-0x10(%rbp)
  jmp    L7
L10:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%edx
  mov    -0x20(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <strchr@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L12
  addq   $0x1,-0x10(%rbp)
  addq   $0x1,-0x18(%rbp)
L7:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L10
  jmp    L35
L12:
  nop
L35:
  mov    -0x10(%rbp),%rax
  leave
  ret
```

assistant: ```c
size_t strcspn(const char *s, const char *reject) {
  size_t n;
  char *c;
  for (n = 0; *s != '\0'; n++, s++) {
    if ((c = strchr(reject, *s)))
      break;
  }
  return n;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<strcspn>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rsi,%r12
  movzbl (%rdi),%esi
  test   %sil,%sil
  je     L3
  mov    %rdi,%rbp
  mov    $0x0,%ebx
L29:
  movsbl %sil,%esi
  mov    %r12,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  jne    L30
  add    $0x1,%rbx
  movzbl 0x0(%rbp,%rbx,1),%esi
  test   %sil,%sil
  jne    L29
L30:
  mov    %rbx,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L3:
  mov    $0x0,%ebx
  jmp    L30
```

assistant: ```c
size_t strcspn(const char *s, const char *reject) {
  size_t n;
  char *c;
  for (n = 0; *s != '\0'; n++, s++) {
    if ((c = strchr(reject, *s)))
      break;
  }
  return n;
}
```
user: Translate the below assembly code into C programming language:

```asm
<strcspn>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  movsbl (%rdi),%esi
  test   %sil,%sil
  je     L5
  mov    %rdi,%rbx
  xor    %r12d,%r12d
  jmp    L21
  nopl   0x0(%rax,%rax,1)
L25:
  add    $0x1,%r12
  movsbl (%rbx,%r12,1),%esi
  test   %sil,%sil
  je     L13
L21:
  mov    %rbp,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  je     L25
L13:
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L5:
  xor    %r12d,%r12d
  jmp    L13
```

assistant: ```c
size_t strcspn(const char *s, const char *reject) {
  size_t n;
  char *c;
  for (n = 0; *s != '\0'; n++, s++) {
    if ((c = strchr(reject, *s)))
      break;
  }
  return n;
}
```
user: What is the c source code of the assembly code below?

```asm
<strcspn>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  movsbl (%rdi),%esi
  test   %sil,%sil
  je     L20
  mov    %rdi,%rbx
  xor    %r12d,%r12d
  jmp    L37
  nopl   0x0(%rax,%rax,1)
L35:
  add    $0x1,%r12
  movsbl (%rbx,%r12,1),%esi
  test   %sil,%sil
  je     L41
L37:
  mov    %rbp,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  je     L35
L41:
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L20:
  xor    %r12d,%r12d
  jmp    L41
```

assistant: ```c
size_t strcspn(const char *s, const char *reject) {
  size_t n;
  char *c;
  for (n = 0; *s != '\0'; n++, s++) {
    if ((c = strchr(reject, *s)))
      break;
  }
  return n;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x30(%rbp)
  movl   $0x0,-0x2c(%rbp)
  movl   $0x1,-0x28(%rbp)
  movl   $0x2,-0x24(%rbp)
  movl   $0x2,-0x20(%rbp)
  movl   $0x4,-0x1c(%rbp)
  movl   $0x3,-0x18(%rbp)
  movl   $0x6,-0x14(%rbp)
  movl   $0x4,-0x10(%rbp)
  movl   $0x8,-0xc(%rbp)
  movl   $0x0,-0x38(%rbp)
  jmp    L44
L9:
  movl   $0x0,-0x34(%rbp)
  jmp    L2
L46:
  mov    -0x34(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x38(%rbp),%eax
  cltq
  add    %rax,%rax
  add    %rdx,%rax
  mov    -0x30(%rbp,%rax,4),%ecx
  mov    -0x34(%rbp),%edx
  mov    -0x38(%rbp),%eax
  mov    %eax,%esi
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x34(%rbp)
L2:
  cmpl   $0x1,-0x34(%rbp)
  jle    L46
  addl   $0x1,-0x38(%rbp)
L44:
  cmpl   $0x4,-0x38(%rbp)
  jle    L9
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L40
  call   <__stack_chk_fail@plt>
L40:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D2
type_guessed: string

tool: D2: s[%d][%d] = %d


assistant: ```c
int main() {
  int s[5][2] = {{0, 0}, {1, 2}, {2, 4}, {3, 6}, {4, 8}};
  int i, j;
  for (i = 0; i < 5; i++) {
    for (j = 0; j < 2; j++) {
      printf("s[%d][%d] = %d\n", i, j, s[i][j]);
    }
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  movl   $0x0,(%rsp)
  movl   $0x0,0x4(%rsp)
  movl   $0x1,0x8(%rsp)
  movl   $0x2,0xc(%rsp)
  movl   $0x2,0x10(%rsp)
  movl   $0x4,0x14(%rsp)
  movl   $0x3,0x18(%rsp)
  movl   $0x6,0x1c(%rsp)
  movl   $0x4,0x20(%rsp)
  movl   $0x8,0x24(%rsp)
  mov    %rsp,%rbp
  mov    $0x0,%ebx
  lea    D2(%rip),%r12
L12:
  mov    0x0(%rbp),%ecx
  mov    $0x0,%edx
  mov    %ebx,%esi
  mov    %r12,%rdi
  call   <printf@plt>
  mov    0x4(%rbp),%ecx
  mov    $0x1,%edx
  mov    %ebx,%esi
  mov    %r12,%rdi
  call   <printf@plt>
  add    $0x1,%ebx
  add    $0x8,%rbp
  cmp    $0x5,%ebx
  jne    L12
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L15
  mov    $0x0,%eax
  add    $0x30,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L15:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D2
datatype: string

tool: D2: s[%d][%d] = %d


assistant: ```c
int main() {
  int s[5][2] = {{0, 0}, {1, 2}, {2, 4}, {3, 6}, {4, 8}};
  int i, j;
  for (i = 0; i < 5; i++) {
    for (j = 0; j < 2; j++) {
      printf("s[%d][%d] = %d\n", i, j, s[i][j]);
    }
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  push   %r12
  lea    D8(%rip),%r12
  push   %rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  movq   $0x0,(%rsp)
  mov    %rsp,%rbp
  movabs $0x200000001,%rax
  mov    %rax,0x8(%rsp)
  movabs $0x400000002,%rax
  mov    %rax,0x10(%rsp)
  movabs $0x600000003,%rax
  mov    %rax,0x18(%rsp)
  movabs $0x800000004,%rax
  mov    %rax,0x20(%rsp)
L18:
  mov    0x0(%rbp),%ecx
  mov    %ebx,%esi
  xor    %edx,%edx
  mov    %r12,%rdi
  add    $0x8,%rbp
  call   <printf@plt>
  mov    -0x4(%rbp),%ecx
  mov    %ebx,%esi
  mov    $0x1,%edx
  mov    %r12,%rdi
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    $0x5,%ebx
  jne    L18
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  add    $0x30,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D8
type_guessed: string

tool: D8: s[%d][%d] = %d


assistant: ```c
int main() {
  int s[5][2] = {{0, 0}, {1, 2}, {2, 4}, {3, 6}, {4, 8}};
  int i, j;
  for (i = 0; i < 5; i++) {
    for (j = 0; j < 2; j++) {
      printf("s[%d][%d] = %d\n", i, j, s[i][j]);
    }
  }
  return 0;
}
```