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

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %edi,-0x34(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x24(%rbp),%rax
  mov    %rax,-0x20(%rbp)
  lea    -0x20(%rbp),%rax
  mov    %rax,-0x18(%rbp)
  lea    -0x24(%rbp),%rax
  mov    %rax,%rsi
  lea    D30(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x25(%rbp),%rax
  mov    %rax,%rsi
  lea    D43(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x20(%rbp),%rax
  mov    %rax,%rsi
  lea    D37(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0xc(%rbp),%rax
  mov    %rax,%rsi
  lea    D49(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D41(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L9
  call   <__stack_chk_fail@plt>
L9:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: tag: D30
type: string
name: unpack_data
arguments: tag: D43
type: string
name: unpack_data
arguments: tag: D37
type: string
name: unpack_data
arguments: tag: D49
type: string
name: unpack_data
arguments: tag: D41
type: string

tool: D30: &a = %p


tool: D43: &c = %p


tool: D37: &p = %p


tool: D49: &d = %p


tool: D41: &q = %p


assistant: ```c
int main(int argc, char **argv) {
  int a;
  char c;
  int *p = &a;
  char d[4];
  int **q = &p;
  printf("&a = %p\n", &a);
  printf("&c = %p\n", &c);
  printf("&p = %p\n", &p);
  printf("&d = %p\n", &d);
  printf("&q = %p\n", &q);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rsi
  mov    %rsi,0x10(%rsp)
  lea    0x10(%rsp),%rbx
  mov    %rbx,0x18(%rsp)
  lea    D24(%rip),%rdi
  call   <printf@plt>
  lea    0xb(%rsp),%rsi
  lea    D27(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %rbx,%rsi
  lea    D30(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0x24(%rsp),%rsi
  lea    D33(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0x18(%rsp),%rsi
  lea    D19(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  mov    $0x0,%eax
  add    $0x30,%rsp
  pop    %rbx
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D24
type: string
name: unpack_data
arguments: label: D27
type: string
name: unpack_data
arguments: label: D30
type: string
name: unpack_data
arguments: label: D33
type: string
name: unpack_data
arguments: label: D19
type: string

tool: D24: &a = %p


tool: D27: &c = %p


tool: D30: &p = %p


tool: D33: &d = %p


tool: D19: &q = %p


assistant: ```c
int main(int argc, char **argv) {
  int a;
  char c;
  int *p = &a;
  char d[4];
  int **q = &p;
  printf("&a = %p\n", &a);
  printf("&c = %p\n", &c);
  printf("&p = %p\n", &p);
  printf("&d = %p\n", &d);
  printf("&q = %p\n", &q);
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D7(%rip),%rdi
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rsi
  lea    0x10(%rsp),%rbp
  mov    %rsi,0x10(%rsp)
  mov    %rbp,0x18(%rsp)
  call   <printf@plt>
  lea    0xb(%rsp),%rsi
  lea    D46(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rsi
  lea    D19(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x24(%rsp),%rsi
  lea    D29(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  lea    0x18(%rsp),%rsi
  lea    D33(%rip),%rdi
  call   <printf@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  add    $0x30,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D7
datatype: string
name: extract_symbolic_data
arguments: label: D46
datatype: string
name: extract_symbolic_data
arguments: label: D19
datatype: string
name: extract_symbolic_data
arguments: label: D29
datatype: string
name: extract_symbolic_data
arguments: label: D33
datatype: string

tool: D7: &a = %p


tool: D46: &c = %p


tool: D19: &p = %p


tool: D29: &d = %p


tool: D33: &q = %p


assistant: ```c
int main(int argc, char **argv) {
  int a;
  char c;
  int *p = &a;
  char d[4];
  int **q = &p;
  printf("&a = %p\n", &a);
  printf("&c = %p\n", &c);
  printf("&p = %p\n", &p);
  printf("&d = %p\n", &d);
  printf("&q = %p\n", &q);
  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
  lea    D35(%rip),%rdi
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rsi
  lea    0x10(%rsp),%rbp
  mov    %rsi,0x10(%rsp)
  mov    %rbp,0x18(%rsp)
  call   <printf@plt>
  lea    0xb(%rsp),%rsi
  lea    D34(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rsi
  lea    D44(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x24(%rsp),%rsi
  lea    D41(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  lea    0x18(%rsp),%rsi
  lea    D11(%rip),%rdi
  call   <printf@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x30,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D35
type_guessed: string
name: read_data
arguments: tag: D34
type_guessed: string
name: read_data
arguments: tag: D44
type_guessed: string
name: read_data
arguments: tag: D41
type_guessed: string
name: read_data
arguments: tag: D11
type_guessed: string

tool: D35: &a = %p


tool: D34: &c = %p


tool: D44: &p = %p


tool: D41: &d = %p


tool: D11: &q = %p


assistant: ```c
int main(int argc, char **argv) {
  int a;
  char c;
  int *p = &a;
  char d[4];
  int **q = &p;
  printf("&a = %p\n", &a);
  printf("&c = %p\n", &c);
  printf("&p = %p\n", &p);
  printf("&d = %p\n", &d);
  printf("&q = %p\n", &q);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<multiply134>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  mov    %edi,-0x44(%rbp)
  mov    %esi,-0x48(%rbp)
  mov    %rdx,-0x50(%rbp)
  mov    %rcx,-0x58(%rbp)
  mov    %r8,-0x60(%rbp)
  mov    -0x44(%rbp),%eax
  movslq %eax,%rdx
  sub    $0x1,%rdx
  mov    %rdx,-0x38(%rbp)
  movslq %eax,%rdx
  mov    %rdx,%r14
  mov    $0x0,%r15d
  mov    -0x44(%rbp),%edx
  movslq %edx,%rcx
  sub    $0x1,%rcx
  mov    %rcx,-0x30(%rbp)
  movslq %edx,%rcx
  mov    %rcx,%r12
  mov    $0x0,%r13d
  mov    -0x44(%rbp),%ecx
  movslq %ecx,%rsi
  sub    $0x1,%rsi
  mov    %rsi,-0x28(%rbp)
  movslq %ecx,%rsi
  mov    %rsi,%r10
  mov    $0x0,%r11d
  movl   $0x0,-0x40(%rbp)
  jmp    L2
L34:
  movl   $0x0,-0x3c(%rbp)
  jmp    L9
L3:
  mov    -0x40(%rbp),%esi
  movslq %esi,%rdi
  movslq %eax,%rsi
  imul   %rdi,%rsi
  lea    0x0(,%rsi,4),%rdi
  mov    -0x50(%rbp),%rsi
  add    %rsi,%rdi
  mov    -0x3c(%rbp),%esi
  movslq %esi,%rsi
  mov    (%rdi,%rsi,4),%esi
  imul   -0x48(%rbp),%esi
  mov    %esi,%r9d
  mov    -0x40(%rbp),%esi
  movslq %esi,%rdi
  movslq %edx,%rsi
  imul   %rdi,%rsi
  lea    0x0(,%rsi,4),%rdi
  mov    -0x58(%rbp),%rsi
  add    %rsi,%rdi
  mov    -0x3c(%rbp),%esi
  movslq %esi,%rsi
  mov    (%rdi,%rsi,4),%r8d
  mov    -0x40(%rbp),%esi
  movslq %esi,%rdi
  movslq %ecx,%rsi
  imul   %rdi,%rsi
  lea    0x0(,%rsi,4),%rdi
  mov    -0x60(%rbp),%rsi
  add    %rsi,%rdi
  add    %r9d,%r8d
  mov    -0x3c(%rbp),%esi
  movslq %esi,%rsi
  mov    %r8d,(%rdi,%rsi,4)
  addl   $0x1,-0x3c(%rbp)
L9:
  mov    -0x3c(%rbp),%esi
  cmp    -0x44(%rbp),%esi
  jl     L3
  addl   $0x1,-0x40(%rbp)
L2:
  mov    -0x40(%rbp),%esi
  cmp    -0x44(%rbp),%esi
  jl     L34
  nop
  nop
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
```

assistant: ```c
void multiply134(int n, int a, int tab1[n][n], int tab2[n][n], int tab3[n][n]) {
  int i, j, k;
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) {
      tab3[i][j] = a * tab1[i][j] + tab2[i][j];
    }
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<multiply134>:
  endbr64
  test   %edi,%edi
  jle    L13
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%ebx
  mov    %esi,%r10d
  movslq %edi,%r12
  shl    $0x2,%r12
  mov    %rdx,%rbp
  mov    $0x0,%r13d
  mov    %edi,%r11d
L3:
  mov    %rbp,%r9
  mov    %rcx,%rdi
  mov    %r8,%rsi
  mov    $0x0,%eax
L27:
  mov    %r10d,%edx
  imul   (%r9,%rax,4),%edx
  add    (%rdi,%rax,4),%edx
  mov    %edx,(%rsi,%rax,4)
  add    $0x1,%rax
  cmp    %rax,%r11
  jne    L27
  add    $0x1,%r13d
  add    %r12,%rbp
  add    %r12,%rcx
  add    %r12,%r8
  cmp    %r13d,%ebx
  jne    L3
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L13:
  ret
```

assistant: ```c
void multiply134(int n, int a, int tab1[n][n], int tab2[n][n], int tab3[n][n]) {
  int i, j, k;
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) {
      tab3[i][j] = a * tab1[i][j] + tab2[i][j];
    }
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<multiply134>:
  endbr64
  test   %edi,%edi
  jle    L8
  movslq %edi,%r10
  push   %rbx
  mov    %esi,%r9d
  xor    %r11d,%r11d
  lea    0x0(,%r10,4),%rbx
  nopw   0x0(%rax,%rax,1)
L16:
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L6:
  mov    (%rdx,%rax,4),%esi
  imul   %r9d,%esi
  add    (%rcx,%rax,4),%esi
  mov    %esi,(%r8,%rax,4)
  add    $0x1,%rax
  cmp    %r10,%rax
  jne    L6
  add    $0x1,%r11d
  add    %rbx,%rdx
  add    %rbx,%rcx
  add    %rbx,%r8
  cmp    %r11d,%edi
  jne    L16
  pop    %rbx
  ret
L8:
  ret
```

assistant: ```c
void multiply134(int n, int a, int tab1[n][n], int tab2[n][n], int tab3[n][n]) {
  int i, j, k;
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) {
      tab3[i][j] = a * tab1[i][j] + tab2[i][j];
    }
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<multiply134>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  movslq %edi,%rbp
  push   %rbx
  test   %ebp,%ebp
  jle    L61
  mov    %rbp,%r11
  mov    %esi,%ebx
  lea    0x4(%rdx),%r10
  mov    %r8,%rax
  mov    %r11d,%edx
  mov    %r11d,%esi
  lea    0x4(%rcx),%r9
  shl    $0x2,%rbp
  and    $0xfffffffc,%esi
  shr    $0x2,%edx
  movd   %ebx,%xmm4
  xor    %r12d,%r12d
  lea    0x1(%rsi),%ecx
  lea    0x2(%rsi),%edi
  mov    %esi,-0x24(%rsp)
  movslq %esi,%r13
  lea    -0x1(%rdx),%r8d
  mov    %r11d,%esi
  lea    -0x1(%r11),%edx
  mov    %ecx,-0x20(%rsp)
  pshufd $0x0,%xmm4,%xmm2
  mov    %edi,-0x1c(%rsp)
  add    $0x1,%r8
  movslq %ecx,%r14
  mov    %edx,-0x28(%rsp)
  movdqa %xmm2,%xmm3
  shl    $0x4,%r8
  movslq %edi,%r15
  mov    %rsi,-0x10(%rsp)
  psrlq  $0x20,%xmm3
  nopl   (%rax)
L53:
  mov    %rax,%rdx
  lea    -0x4(%r10),%rcx
  lea    -0x4(%r9),%rsi
  sub    %r10,%rdx
  cmp    $0x8,%rdx
  mov    %rax,%rdx
  seta   %dil
  sub    %r9,%rdx
  cmp    $0x8,%rdx
  seta   %dl
  test   %dl,%dil
  je     L82
  xor    %edx,%edx
  cmpl   $0x2,-0x28(%rsp)
  jbe    L82
  nopl   0x0(%rax)
L8:
  movdqu (%rcx,%rdx,1),%xmm0
  movdqu (%rcx,%rdx,1),%xmm1
  psrlq  $0x20,%xmm0
  pmuludq %xmm2,%xmm1
  pmuludq %xmm3,%xmm0
  pshufd $0x8,%xmm1,%xmm1
  pshufd $0x8,%xmm0,%xmm0
  punpckldq %xmm0,%xmm1
  movdqu (%rsi,%rdx,1),%xmm0
  paddd  %xmm1,%xmm0
  movups %xmm0,(%rax,%rdx,1)
  add    $0x10,%rdx
  cmp    %r8,%rdx
  jne    L8
  cmp    -0x24(%rsp),%r11d
  je     L78
  mov    (%rcx,%r13,4),%edx
  imul   %ebx,%edx
  add    (%rsi,%r13,4),%edx
  mov    %edx,(%rax,%r13,4)
  cmp    -0x20(%rsp),%r11d
  jle    L78
  mov    (%rcx,%r14,4),%edx
  imul   %ebx,%edx
  add    (%rsi,%r14,4),%edx
  mov    %edx,(%rax,%r14,4)
  cmp    -0x1c(%rsp),%r11d
  jle    L78
  mov    (%rcx,%r15,4),%edx
  imul   %ebx,%edx
  add    (%rsi,%r15,4),%edx
  mov    %edx,(%rax,%r15,4)
L78:
  add    $0x1,%r12d
  add    %rbp,%rax
  add    %rbp,%r10
  add    %rbp,%r9
  cmp    %r12d,%r11d
  jne    L53
L61:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  cs nopw 0x0(%rax,%rax,1)
L82:
  mov    %r8,-0x18(%rsp)
  mov    -0x10(%rsp),%r8
  xor    %edx,%edx
  nopl   0x0(%rax)
L44:
  mov    (%rcx,%rdx,4),%edi
  imul   %ebx,%edi
  add    (%rsi,%rdx,4),%edi
  mov    %edi,(%rax,%rdx,4)
  add    $0x1,%rdx
  cmp    %rdx,%r8
  jne    L44
  mov    -0x18(%rsp),%r8
  jmp    L78
```

assistant: ```c
void multiply134(int n, int a, int tab1[n][n], int tab2[n][n], int tab3[n][n]) {
  int i, j, k;
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) {
      tab3[i][j] = a * tab1[i][j] + tab2[i][j];
    }
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<mcon_strndup>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movq   $0x0,-0x10(%rbp)
  jmp    L31
L38:
  nop
  addq   $0x1,-0x10(%rbp)
L31:
  mov    -0x10(%rbp),%rax
  cmp    -0x20(%rbp),%rax
  jae    L20
  mov    -0x18(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L38
L20:
  mov    -0x10(%rbp),%rax
  add    $0x1,%rax
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L22
  mov    $0x0,%eax
  jmp    L16
L22:
  mov    -0x10(%rbp),%rdx
  mov    -0x18(%rbp),%rcx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0x8(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x8(%rbp),%rax
L16:
  leave
  ret
```

assistant: ```c
char *mcon_strndup(char *str, size_t n) {
  size_t len;
  char *copy;
  for (len = 0; len < n && str[len]; len++) {
    continue;
  }
  if ((copy = malloc(len + 1)) == NULL) {
    return NULL;
  }
  memcpy(copy, str, len);
  copy[len] = '\0';
  return copy;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<mcon_strndup>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    $0x0,%ebp
  test   %rsi,%rsi
  je     L15
L3:
  cmpb   $0x0,(%r12,%rbp,1)
  jne    L54
L5:
  lea    0x1(%rbp),%rdi
  call   <malloc@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L4
  mov    %rbp,%rdx
  mov    %r12,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  movb   $0x0,(%rbx,%rbp,1)
L4:
  mov    %rbx,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L15:
  mov    %rsi,%rbp
  jmp    L5
L54:
  add    $0x1,%rbp
  cmp    %rbp,%rsi
  jne    L3
  mov    %rsi,%rbp
  jmp    L5
```

assistant: ```c
char *mcon_strndup(char *str, size_t n) {
  size_t len;
  char *copy;
  for (len = 0; len < n && str[len]; len++) {
    continue;
  }
  if ((copy = malloc(len + 1)) == NULL) {
    return NULL;
  }
  memcpy(copy, str, len);
  copy[len] = '\0';
  return copy;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<mcon_strndup>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  mov    $0x1,%edi
  push   %rbx
  xor    %ebx,%ebx
  test   %rsi,%rsi
  je     L58
L39:
  cmpb   $0x0,0x0(%rbp,%rbx,1)
  lea    0x1(%rbx),%rdi
  jne    L38
L58:
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L51
  mov    %rbx,%rdx
  mov    %rbp,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  movb   $0x0,(%r12,%rbx,1)
L51:
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L38:
  cmp    %rdi,%rsi
  je     L14
  mov    %rdi,%rbx
  jmp    L39
  nopw   0x0(%rax,%rax,1)
L14:
  lea    0x2(%rbx),%rax
  mov    %rdi,%rbx
  mov    %rax,%rdi
  jmp    L58
```

assistant: ```c
char *mcon_strndup(char *str, size_t n) {
  size_t len;
  char *copy;
  for (len = 0; len < n && str[len]; len++) {
    continue;
  }
  if ((copy = malloc(len + 1)) == NULL) {
    return NULL;
  }
  memcpy(copy, str, len);
  copy[len] = '\0';
  return copy;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<mcon_strndup>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  mov    $0x1,%edi
  push   %rbx
  xor    %ebx,%ebx
  test   %rsi,%rsi
  je     L52
L40:
  cmpb   $0x0,0x0(%rbp,%rbx,1)
  lea    0x1(%rbx),%rdi
  jne    L17
L52:
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L1
  mov    %rbx,%rdx
  mov    %rbp,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  movb   $0x0,(%r12,%rbx,1)
L1:
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L17:
  cmp    %rdi,%rsi
  je     L37
  mov    %rdi,%rbx
  jmp    L40
  nopw   0x0(%rax,%rax,1)
L37:
  lea    0x2(%rbx),%rax
  mov    %rdi,%rbx
  mov    %rax,%rdi
  jmp    L52
```

assistant: ```c
char *mcon_strndup(char *str, size_t n) {
  size_t len;
  char *copy;
  for (len = 0; len < n && str[len]; len++) {
    continue;
  }
  if ((copy = malloc(len + 1)) == NULL) {
    return NULL;
  }
  memcpy(copy, str, len);
  copy[len] = '\0';
  return copy;
}
```
user: What is the c source code of the assembly code below?

```asm
<transition>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <trans_rand_alloc@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int transition() {
  trans_rand_alloc();
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<transition>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <trans_rand_alloc@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int transition() {
  trans_rand_alloc();
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<transition>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <trans_rand_alloc@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int transition() {
  trans_rand_alloc();
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<transition>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <trans_rand_alloc@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int transition() {
  trans_rand_alloc();
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<cleanup>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    $0x0,%eax
  call   <pthread_self@plt>
  mov    %eax,%edi
  call   <pthread_cancel@plt>
  test   %rax,%rax
  je     L6
  lea    D15(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x1,%edi
  call   <exit@plt>
L6:
  mov    -0x8(%rbp),%rax
  mov    %rax,%rsi
  lea    D14(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: label: D15
type_guessed: string
name: disassemble_data
arguments: label: D14
type_guessed: string

tool: D15: cleanup: cancel failed

tool: D14: cleanup for %ld


assistant: ```c
void cleanup(void *arg) {
  if (pthread_cancel(pthread_self()) != 0) {
    puts("cleanup: cancel failed");
    exit(1);
  }
  printf("cleanup for %ld\n", (long int)arg);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<cleanup>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  mov    $0x0,%eax
  call   <pthread_self@plt>
  mov    %eax,%edi
  call   <pthread_cancel@plt>
  test   %rax,%rax
  jne    L2
L8:
  mov    %rbx,%rsi
  lea    D11(%rip),%rdi
  call   <printf@plt>
  pop    %rbx
  ret
L2:
  lea    D6(%rip),%rdi
  call   <puts@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  jmp    L8
```

tool_calls: name: read_data
arguments: label: D11
type_guessed: string
name: read_data
arguments: label: D6
type_guessed: string

tool: D11: cleanup for %ld


tool: D6: cleanup: cancel failed

assistant: ```c
void cleanup(void *arg) {
  if (pthread_cancel(pthread_self()) != 0) {
    puts("cleanup: cancel failed");
    exit(1);
  }
  printf("cleanup for %ld\n", (long int)arg);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<cleanup>:
  endbr64
  push   %rbp
  xor    %eax,%eax
  mov    %rdi,%rbp
  call   <pthread_self@plt>
  mov    %eax,%edi
  call   <pthread_cancel@plt>
  test   %rax,%rax
  jne    L5
  mov    %rbp,%rsi
  lea    D26(%rip),%rdi
  pop    %rbp
  jmp    10a0 <printf@plt>
  nopl   0x0(%rax,%rax,1)
L5:
  lea    D19(%rip),%rdi
  call   <puts@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  mov    %rbp,%rsi
  lea    D26(%rip),%rdi
  pop    %rbp
  jmp    10a0 <printf@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D26
type_guessed: string
name: retrieve_data
arguments: tag: D19
type_guessed: string

tool: D26: cleanup for %ld


tool: D19: cleanup: cancel failed

assistant: ```c
void cleanup(void *arg) {
  if (pthread_cancel(pthread_self()) != 0) {
    puts("cleanup: cancel failed");
    exit(1);
  }
  printf("cleanup for %ld\n", (long int)arg);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<cleanup>:
  endbr64
  push   %rbp
  xor    %eax,%eax
  mov    %rdi,%rbp
  call   <pthread_self@plt>
  mov    %eax,%edi
  call   <pthread_cancel@plt>
  test   %rax,%rax
  jne    L8
  mov    %rbp,%rsi
  lea    D20(%rip),%rdi
  pop    %rbp
  jmp    10a0 <printf@plt>
  nopl   0x0(%rax,%rax,1)
L8:
  lea    D17(%rip),%rdi
  call   <puts@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  mov    %rbp,%rsi
  lea    D20(%rip),%rdi
  pop    %rbp
  jmp    10a0 <printf@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D20
datatype: string
name: extract_symbolic_data
arguments: tag: D17
datatype: string

tool: D20: cleanup for %ld


tool: D17: cleanup: cancel failed

assistant: ```c
void cleanup(void *arg) {
  if (pthread_cancel(pthread_self()) != 0) {
    puts("cleanup: cancel failed");
    exit(1);
  }
  printf("cleanup for %ld\n", (long int)arg);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D3
type: string
name: retrieve_data
arguments: data_tag: D0
type: string

tool: D3: There is no use of listening to explanation of feelings from others

tool: D0: Get a direct experience of feeling

assistant: ```c
int main() {
  printf("There is no use of listening to explanation of feelings from others");
  printf("Get a direct experience of feeling");
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D2(%rip),%rdi
  call   <printf@plt>
  lea    D12(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: data_label: D2
type_guessed: string
name: read_data
arguments: data_label: D12
type_guessed: string

tool: D2: There is no use of listening to explanation of feelings from others

tool: D12: Get a direct experience of feeling

assistant: ```c
int main() {
  printf("There is no use of listening to explanation of feelings from others");
  printf("Get a direct experience of feeling");
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D10(%rip),%rdi
  call   <printf@plt>
  lea    D16(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D10
type: string
name: retrieve_data
arguments: tag: D16
type: string

tool: D10: There is no use of listening to explanation of feelings from others

tool: D16: Get a direct experience of feeling

assistant: ```c
int main() {
  printf("There is no use of listening to explanation of feelings from others");
  printf("Get a direct experience of feeling");
}
```