ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What would the C source look like for the assembly code provided here:

```asm
<editdist_dp>:
  endbr64
  mov    %ecx,%eax
  test   %edx,%edx
  je     L13
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %edx,%ebp
  mov    %ecx,%ebx
  mov    %r8,%r12
  mov    %edx,%eax
  test   %ecx,%ecx
  je     L29
  movslq %ecx,%r13
  movslq %edx,%r15
  mov    %r13,%rax
  imul   %r15,%rax
  lea    (%r8,%rax,4),%r14
  mov    (%r14,%r13,4),%eax
  test   %eax,%eax
  je     L30
L29:
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L30:
  lea    -0x1(%rdx),%eax
  mov    %eax,0x14(%rsp)
  mov    %eax,%edx
  mov    %rsi,(%rsp)
  mov    %rdi,0x8(%rsp)
  call   <editdist_dp>
  mov    %eax,0x18(%rsp)
  lea    -0x1(%rbx),%ecx
  mov    %r12,%r8
  mov    %ecx,0x1c(%rsp)
  mov    %ebp,%edx
  mov    (%rsp),%rsi
  mov    0x8(%rsp),%rdi
  call   <editdist_dp>
  mov    %eax,%ebp
  mov    %r12,%r8
  mov    0x1c(%rsp),%ecx
  mov    0x14(%rsp),%edx
  mov    (%rsp),%rsi
  mov    0x8(%rsp),%r12
  mov    %r12,%rdi
  call   <editdist_dp>
  mov    (%rsp),%rsi
  movzbl (%rsi,%r13,1),%ecx
  cmp    %cl,(%r12,%r15,1)
  setne  %dl
  movzbl %dl,%edx
  add    %eax,%edx
  lea    0x1(%rbp),%esi
  mov    0x18(%rsp),%edi
  add    $0x1,%edi
  call   <minthree@plt>
  mov    %eax,(%r14,%r13,4)
  jmp    L29
L13:
  ret
```

assistant: ```c
int editdist_dp(char str1[], char str2[], int m, int n, int table[][n]) {
  if (m == 0) {
    return n;
  }
  if (n == 0) {
    return m;
  }
  if (table[m][n] != 0) {
    return table[m][n];
  }
  int left = editdist_dp(str1, str2, m - 1, n, table) + 1;
  int right = editdist_dp(str1, str2, m, n - 1, table) + 1;
  int corner = editdist_dp(str1, str2, m - 1, n - 1, table) + (str1[m] == str2[n] ? 0 : 1);
  int res = minthree(left, right, corner);
  table[m][n] = res;
  return res;
}
```
user: Translate the below assembly code into C programming language:

```asm
<editdist_dp>:
  endbr64
  mov    %ecx,%eax
  test   %edx,%edx
  je     L29
  push   %r15
  mov    %edx,%eax
  push   %r14
  push   %r13
  push   %r12
  mov    %edx,%r12d
  push   %rbp
  push   %rbx
  mov    %ecx,%ebx
  sub    $0x28,%rsp
  test   %ecx,%ecx
  je     L3
  movslq %ecx,%r13
  movslq %edx,%r14
  mov    %r13,%rax
  imul   %r14,%rax
  lea    (%r8,%rax,4),%r15
  mov    (%r15,%r13,4),%eax
  test   %eax,%eax
  jne    L3
  lea    -0x1(%rdx),%r10d
  mov    %rsi,%rbp
  sub    $0x1,%ebx
  mov    %r8,0x10(%rsp)
  mov    %r10d,%edx
  mov    %r10d,0x1c(%rsp)
  mov    %rdi,0x8(%rsp)
  call   <editdist_dp>
  mov    0x10(%rsp),%r8
  mov    %r12d,%edx
  mov    %ebx,%ecx
  mov    0x8(%rsp),%rdi
  mov    %rbp,%rsi
  mov    %eax,0x18(%rsp)
  call   <editdist_dp>
  mov    0x10(%rsp),%r8
  mov    0x1c(%rsp),%edx
  mov    %ebx,%ecx
  mov    0x8(%rsp),%rdi
  mov    %rbp,%rsi
  mov    %eax,%r12d
  call   <editdist_dp>
  mov    0x8(%rsp),%rdi
  movzbl 0x0(%rbp,%r13,1),%ecx
  xor    %edx,%edx
  lea    0x1(%r12),%esi
  cmp    %cl,(%rdi,%r14,1)
  mov    0x18(%rsp),%edi
  setne  %dl
  add    %eax,%edx
  add    $0x1,%edi
  call   <minthree@plt>
  mov    %eax,(%r15,%r13,4)
L3:
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L29:
  ret
```

assistant: ```c
int editdist_dp(char str1[], char str2[], int m, int n, int table[][n]) {
  if (m == 0) {
    return n;
  }
  if (n == 0) {
    return m;
  }
  if (table[m][n] != 0) {
    return table[m][n];
  }
  int left = editdist_dp(str1, str2, m - 1, n, table) + 1;
  int right = editdist_dp(str1, str2, m, n - 1, table) + 1;
  int corner = editdist_dp(str1, str2, m - 1, n - 1, table) + (str1[m] == str2[n] ? 0 : 1);
  int res = minthree(left, right, corner);
  table[m][n] = res;
  return res;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<editdist_dp>:
  endbr64
  mov    %ecx,%eax
  test   %edx,%edx
  je     L18
  push   %r15
  mov    %edx,%eax
  push   %r14
  push   %r13
  push   %r12
  mov    %edx,%r12d
  push   %rbp
  push   %rbx
  mov    %ecx,%ebx
  sub    $0x28,%rsp
  test   %ecx,%ecx
  je     L4
  movslq %ecx,%r13
  movslq %edx,%r14
  mov    %r13,%rax
  imul   %r14,%rax
  lea    (%r8,%rax,4),%r15
  mov    (%r15,%r13,4),%eax
  test   %eax,%eax
  jne    L4
  lea    -0x1(%rdx),%r10d
  mov    %rsi,%rbp
  sub    $0x1,%ebx
  mov    %r8,0x10(%rsp)
  mov    %r10d,%edx
  mov    %r10d,0x1c(%rsp)
  mov    %rdi,0x8(%rsp)
  call   <editdist_dp>
  mov    0x10(%rsp),%r8
  mov    %r12d,%edx
  mov    %ebx,%ecx
  mov    0x8(%rsp),%rdi
  mov    %rbp,%rsi
  mov    %eax,0x18(%rsp)
  call   <editdist_dp>
  mov    0x10(%rsp),%r8
  mov    0x1c(%rsp),%edx
  mov    %ebx,%ecx
  mov    0x8(%rsp),%rdi
  mov    %rbp,%rsi
  mov    %eax,%r12d
  call   <editdist_dp>
  mov    0x8(%rsp),%rdi
  movzbl 0x0(%rbp,%r13,1),%ecx
  xor    %edx,%edx
  lea    0x1(%r12),%esi
  cmp    %cl,(%rdi,%r14,1)
  mov    0x18(%rsp),%edi
  setne  %dl
  add    %eax,%edx
  add    $0x1,%edi
  call   <minthree@plt>
  mov    %eax,(%r15,%r13,4)
L4:
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L18:
  ret
```

assistant: ```c
int editdist_dp(char str1[], char str2[], int m, int n, int table[][n]) {
  if (m == 0) {
    return n;
  }
  if (n == 0) {
    return m;
  }
  if (table[m][n] != 0) {
    return table[m][n];
  }
  int left = editdist_dp(str1, str2, m - 1, n, table) + 1;
  int right = editdist_dp(str1, str2, m, n - 1, table) + 1;
  int corner = editdist_dp(str1, str2, m - 1, n - 1, table) + (str1[m] == str2[n] ? 0 : 1);
  int res = minthree(left, right, corner);
  table[m][n] = res;
  return res;
}
```
user: Translate the below assembly code into C programming language:

```asm
<numanuma__get_time_now>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  mov    D7(%rip),%rax
  mov    (%rax),%eax
  lea    -0x30(%rbp),%rdx
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <clock_gettime@plt>
  mov    -0x30(%rbp),%ebx
  mov    $0x0,%eax
  call   <numanuma__get_time_tps@plt>
  imul   %ebx,%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    -0x18(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L2
  call   <__stack_chk_fail@plt>
L2:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
long long numanuma__get_time_now(void) {
  struct timespec ts;
  clock_gettime(CLOCK_MONOTONIC, &ts);
  return ts.tv_sec * numanuma__get_time_tps() + ts.tv_nsec;
}
```
user: What is the c source code of the assembly code below?

```asm
<numanuma__get_time_now>:
  endbr64
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  mov    D7(%rip),%rax
  mov    (%rax),%edi
  call   <clock_gettime@plt>
  mov    (%rsp),%ebx
  mov    $0x0,%eax
  call   <numanuma__get_time_tps@plt>
  imul   %eax,%ebx
  movslq %ebx,%rax
  add    0x8(%rsp),%rax
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L6
  add    $0x20,%rsp
  pop    %rbx
  ret
L6:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
long long numanuma__get_time_now(void) {
  struct timespec ts;
  clock_gettime(CLOCK_MONOTONIC, &ts);
  return ts.tv_sec * numanuma__get_time_tps() + ts.tv_nsec;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<numanuma__get_time_now>:
  endbr64
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    D5(%rip),%rax
  mov    %rsp,%rsi
  mov    (%rax),%edi
  call   <clock_gettime@plt>
  mov    (%rsp),%ebx
  xor    %eax,%eax
  call   <numanuma__get_time_tps@plt>
  imul   %eax,%ebx
  movslq %ebx,%rax
  add    0x8(%rsp),%rax
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L5
  add    $0x20,%rsp
  pop    %rbx
  ret
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
long long numanuma__get_time_now(void) {
  struct timespec ts;
  clock_gettime(CLOCK_MONOTONIC, &ts);
  return ts.tv_sec * numanuma__get_time_tps() + ts.tv_nsec;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<numanuma__get_time_now>:
  endbr64
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    D9(%rip),%rax
  mov    %rsp,%rsi
  mov    (%rax),%edi
  call   <clock_gettime@plt>
  mov    (%rsp),%ebx
  xor    %eax,%eax
  call   <numanuma__get_time_tps@plt>
  imul   %eax,%ebx
  movslq %ebx,%rax
  add    0x8(%rsp),%rax
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L2
  add    $0x20,%rsp
  pop    %rbx
  ret
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
long long numanuma__get_time_now(void) {
  struct timespec ts;
  clock_gettime(CLOCK_MONOTONIC, &ts);
  return ts.tv_sec * numanuma__get_time_tps() + ts.tv_nsec;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<individualise>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  mov    %ecx,-0x28(%rbp)
  mov    %r8,-0x30(%rbp)
  mov    %r9,-0x38(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L41
L17:
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    %eax,-0x28(%rbp)
  je     L19
  addl   $0x1,-0x8(%rbp)
L41:
  mov    -0x8(%rbp),%eax
  cmp    0x10(%rbp),%eax
  jl     L17
  jmp    L15
L19:
  nop
L15:
  mov    -0x8(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  jmp    L6
L87:
  subl   $0x1,-0x4(%rbp)
L6:
  cmpl   $0x0,-0x4(%rbp)
  jle    L18
  mov    -0x4(%rbp),%eax
  cltq
  shl    $0x2,%rax
  lea    -0x4(%rax),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    %eax,-0x24(%rbp)
  jl     L87
L18:
  mov    -0x30(%rbp),%rax
  mov    -0x4(%rbp),%edx
  mov    %edx,(%rax)
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    %eax,-0x24(%rbp)
  jge    L78
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x8(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x28(%rbp),%eax
  mov    %eax,(%rdx)
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rax,%rdx
  mov    -0x24(%rbp),%eax
  mov    %eax,(%rdx)
  mov    -0x38(%rbp),%rax
  mov    (%rax),%eax
  lea    0x1(%rax),%edx
  mov    -0x38(%rbp),%rax
  mov    %edx,(%rax)
  jmp    L16
L78:
  nop
L16:
  pop    %rbp
  ret
```

assistant: ```c
void individualise(int *lab, int *ptn, int level, int v, int *pos, int *numcells, int n) {
  int i, j;
  for (i = 0; i < n; ++i)
    if (lab[i] == v)
      break;
  for (j = i; j > 0 && ptn[j - 1] > level; --j) {
  };
  *pos = j;
  if (ptn[j] <= level)
    return;
  lab[i] = lab[j];
  lab[j] = v;
  ptn[j] = level;
  ++*numcells;
}
```
user: Translate the below assembly code into C programming language:

```asm
<individualise>:
  endbr64
  mov    %rsi,%r11
  mov    %edx,%esi
  mov    %ecx,%r10d
  mov    0x8(%rsp),%ecx
  test   %ecx,%ecx
  jle    L3
  mov    %rdi,%rax
  mov    $0x0,%edx
L14:
  cmp    %r10d,(%rax)
  je     L11
  add    $0x1,%edx
  add    $0x4,%rax
  cmp    %edx,%ecx
  jne    L14
  mov    %ecx,%edx
L11:
  test   %edx,%edx
  jle    L41
  movslq %edx,%rax
  lea    (%r11,%rax,4),%rcx
  mov    %edx,%eax
L31:
  cmp    %esi,-0x4(%rcx)
  jle    L56
  sub    $0x4,%rcx
  sub    $0x1,%eax
  jne    L31
L56:
  mov    %eax,(%r8)
  cltq
  shl    $0x2,%rax
  add    %rax,%r11
  cmp    %esi,(%r11)
  jle    L68
  add    %rdi,%rax
  mov    (%rax),%ecx
  movslq %edx,%rdx
  mov    %ecx,(%rdi,%rdx,4)
  mov    %r10d,(%rax)
  mov    %esi,(%r11)
  addl   $0x1,(%r9)
L68:
  ret
L3:
  mov    $0x0,%edx
  mov    $0x0,%eax
  jmp    L56
L41:
  mov    %edx,%eax
  jmp    L56
```

assistant: ```c
void individualise(int *lab, int *ptn, int level, int v, int *pos, int *numcells, int n) {
  int i, j;
  for (i = 0; i < n; ++i)
    if (lab[i] == v)
      break;
  for (j = i; j > 0 && ptn[j - 1] > level; --j) {
  };
  *pos = j;
  if (ptn[j] <= level)
    return;
  lab[i] = lab[j];
  lab[j] = v;
  ptn[j] = level;
  ++*numcells;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<individualise>:
  endbr64
  push   %rbx
  mov    %ecx,%r10d
  mov    0x10(%rsp),%ecx
  mov    %rdi,%rbx
  mov    %edx,%edi
  test   %ecx,%ecx
  jle    L93
  mov    %rbx,%rax
  xor    %edx,%edx
  jmp    L15
  nopl   0x0(%rax)
L83:
  add    $0x1,%edx
  add    $0x4,%rax
  cmp    %edx,%ecx
  je     L34
L15:
  cmp    %r10d,(%rax)
  jne    L83
  xor    %ecx,%ecx
  xor    %r11d,%r11d
  test   %edx,%edx
  je     L97
L34:
  movslq %edx,%rax
  jmp    L54
  xchg   %ax,%ax
L4:
  sub    $0x1,%rax
  mov    %eax,%ecx
  test   %eax,%eax
  je     L70
L54:
  mov    %eax,%ecx
  lea    0x0(,%rax,4),%r11
  cmp    %edi,-0x4(%rsi,%rax,4)
  jg     L4
  add    %r11,%rsi
L97:
  mov    %ecx,(%r8)
  cmp    %edi,(%rsi)
  jle    L23
  add    %rbx,%r11
  movslq %edx,%rdx
  mov    (%r11),%eax
  mov    %eax,(%rbx,%rdx,4)
  mov    %r10d,(%r11)
  mov    %edi,(%rsi)
  addl   $0x1,(%r9)
L23:
  pop    %rbx
  ret
  nopl   0x0(%rax,%rax,1)
L70:
  xor    %r11d,%r11d
  jmp    L97
L93:
  xor    %r11d,%r11d
  xor    %edx,%edx
  xor    %ecx,%ecx
  jmp    L97
```

assistant: ```c
void individualise(int *lab, int *ptn, int level, int v, int *pos, int *numcells, int n) {
  int i, j;
  for (i = 0; i < n; ++i)
    if (lab[i] == v)
      break;
  for (j = i; j > 0 && ptn[j - 1] > level; --j) {
  };
  *pos = j;
  if (ptn[j] <= level)
    return;
  lab[i] = lab[j];
  lab[j] = v;
  ptn[j] = level;
  ++*numcells;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<individualise>:
  endbr64
  push   %rbx
  movslq 0x10(%rsp),%rax
  mov    %rdi,%r10
  mov    %ecx,%r11d
  mov    %edx,%edi
  xor    %ecx,%ecx
  test   %eax,%eax
  jle    L48
  xor    %edx,%edx
  jmp    L46
  nopl   0x0(%rax)
L33:
  add    $0x1,%edx
  add    $0x4,%rcx
  cmp    %edx,%eax
  je     L61
L46:
  cmp    %r11d,(%r10,%rcx,1)
  jne    L33
  movslq %edx,%rax
  test   %edx,%edx
  jne    L81
  jmp    L44
  nopw   0x0(%rax,%rax,1)
L59:
  sub    $0x1,%rax
  mov    %eax,%edx
  test   %eax,%eax
  je     L44
L81:
  mov    %eax,%edx
  lea    0x0(,%rax,4),%rbx
  cmp    %edi,-0x4(%rsi,%rax,4)
  jg     L59
  add    %rbx,%rsi
L113:
  mov    %edx,(%r8)
  cmp    %edi,(%rsi)
  jle    L43
  add    %r10,%rbx
  mov    (%rbx),%eax
  mov    %eax,(%r10,%rcx,1)
  mov    %r11d,(%rbx)
  mov    %edi,(%rsi)
  addl   $0x1,(%r9)
L43:
  pop    %rbx
  ret
  nopl   0x0(%rax,%rax,1)
L44:
  xor    %ebx,%ebx
  jmp    L113
  nopl   0x0(%rax)
L61:
  lea    0x0(,%rax,4),%rcx
  jmp    L81
L48:
  xor    %ebx,%ebx
  xor    %edx,%edx
  jmp    L113
```

assistant: ```c
void individualise(int *lab, int *ptn, int level, int v, int *pos, int *numcells, int n) {
  int i, j;
  for (i = 0; i < n; ++i)
    if (lab[i] == v)
      break;
  for (j = i; j > 0 && ptn[j - 1] > level; --j) {
  };
  *pos = j;
  if (ptn[j] <= level)
    return;
  lab[i] = lab[j];
  lab[j] = v;
  ptn[j] = level;
  ++*numcells;
}
```
user: What is the c source code of the assembly code below:

```asm
<main_main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x2,-0x8(%rbp)
  movl   $0xffffffff,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  cmp    %eax,-0x8(%rbp)
  jae    L18
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L13
L18:
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L13:
  nop
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D2
data_type: string
name: extract_symbolic_data
arguments: data_label: D7
data_type: string

tool: D2: OK.


tool: D7: Fail.


assistant: ```c
void main_main() {
  unsigned int a = 2;
  int b = -1;
  if (b > a)
    printf("OK.\n");
  else
    printf("Fail.\n");
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main_main>:
  endbr64
  sub    $0x8,%rsp
  lea    D1(%rip),%rdi
  call   <printf@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D1
type_guessed: string

tool: D1: OK.


assistant: ```c
void main_main() {
  unsigned int a = 2;
  int b = -1;
  if (b > a)
    printf("OK.\n");
  else
    printf("Fail.\n");
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main_main>:
  endbr64
  lea    D2(%rip),%rdi
  jmp    1050 <printf@plt>
```

tool_calls: name: parse_data
arguments: label: D2
type_infered: string

tool: D2: OK.


assistant: ```c
void main_main() {
  unsigned int a = 2;
  int b = -1;
  if (b > a)
    printf("OK.\n");
  else
    printf("Fail.\n");
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main_main>:
  endbr64
  lea    D3(%rip),%rdi
  jmp    1050 <printf@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D3
type_guessed: string

tool: D3: OK.


assistant: ```c
void main_main() {
  unsigned int a = 2;
  int b = -1;
  if (b > a)
    printf("OK.\n");
  else
    printf("Fail.\n");
}
```
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)
  mov    -0x20(%rbp),%rax
  mov    (%rax),%rax
  mov    $0x2f,%esi
  mov    %rax,%rdi
  call   <strrchr@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L17
  mov    -0x20(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L6
L17:
  addq   $0x1,-0x8(%rbp)
L6:
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <daemonize@plt>
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main(int arc, char *argv[]) {
  char *cmd;
  if ((cmd = strrchr(argv[0], '/')) == NULL) {
    cmd = argv[0];
  } else {
    ++cmd;
  }
  daemonize(cmd);
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %rbx
  mov    %rsi,%rbx
  mov    $0x2f,%esi
  mov    (%rbx),%rdi
  call   <strrchr@plt>
  lea    0x1(%rax),%rdi
  test   %rax,%rax
  je     L18
L11:
  call   <daemonize@plt>
  mov    $0x0,%eax
  pop    %rbx
  ret
L18:
  mov    (%rbx),%rdi
  jmp    L11
```

assistant: ```c
int main(int arc, char *argv[]) {
  char *cmd;
  if ((cmd = strrchr(argv[0], '/')) == NULL) {
    cmd = argv[0];
  } else {
    ++cmd;
  }
  daemonize(cmd);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  push   %rbx
  mov    %rsi,%rbx
  mov    $0x2f,%esi
  mov    (%rbx),%rdi
  call   <strrchr@plt>
  lea    0x1(%rax),%rdi
  test   %rax,%rax
  je     L0
L13:
  call   <daemonize@plt>
  xor    %eax,%eax
  pop    %rbx
  ret
L0:
  mov    (%rbx),%rdi
  jmp    L13
```

assistant: ```c
int main(int arc, char *argv[]) {
  char *cmd;
  if ((cmd = strrchr(argv[0], '/')) == NULL) {
    cmd = argv[0];
  } else {
    ++cmd;
  }
  daemonize(cmd);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbx
  mov    %rsi,%rbx
  mov    $0x2f,%esi
  mov    (%rbx),%rdi
  call   <strrchr@plt>
  lea    0x1(%rax),%rdi
  test   %rax,%rax
  je     L13
L10:
  call   <daemonize@plt>
  xor    %eax,%eax
  pop    %rbx
  ret
L13:
  mov    (%rbx),%rdi
  jmp    L10
```

assistant: ```c
int main(int arc, char *argv[]) {
  char *cmd;
  if ((cmd = strrchr(argv[0], '/')) == NULL) {
    cmd = argv[0];
  } else {
    ++cmd;
  }
  daemonize(cmd);
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<log_with_date>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x60,%rsp
  mov    %rdi,-0x58(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    $0x0,%edi
  call   <time@plt>
  mov    %eax,-0x4c(%rbp)
  lea    -0x4c(%rbp),%rax
  mov    %rax,%rdi
  call   <gmtime@plt>
  mov    %rax,-0x48(%rbp)
  mov    -0x48(%rbp),%rdx
  lea    -0x40(%rbp),%rax
  mov    %rdx,%rcx
  lea    D12(%rip),%rdx
  mov    $0x32,%esi
  mov    %rax,%rdi
  call   <strftime@plt>
  mov    -0x58(%rbp),%rdx
  lea    -0x40(%rbp),%rax
  mov    %rax,%rsi
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L5
  call   <__stack_chk_fail@plt>
L5:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D12
type: string
name: extract_symbolic_data
arguments: label: D8
type: string

tool: D12: %Y-%m-%d %H:%M:%S

tool: D8: [%s] %s


assistant: ```c
void log_with_date(char *line) {
  char date[50];
  struct tm *current_time;
  time_t now = time(0);
  current_time = gmtime(&now);
  strftime(date, sizeof(date), "%Y-%m-%d %H:%M:%S", current_time);
  printf("[%s] %s\n", date, line);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<log_with_date>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x58,%rsp
  mov    %rdi,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  mov    $0x0,%edi
  call   <time@plt>
  mov    %eax,0xc(%rsp)
  lea    0xc(%rsp),%rdi
  call   <gmtime@plt>
  mov    %rax,%rcx
  lea    0x10(%rsp),%rbp
  lea    D10(%rip),%rdx
  mov    $0x32,%esi
  mov    %rbp,%rdi
  call   <strftime@plt>
  mov    %rbx,%rdx
  mov    %rbp,%rsi
  lea    D17(%rip),%rdi
  call   <printf@plt>
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x58,%rsp
  pop    %rbx
  pop    %rbp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D10
datatype: string
name: retrieve_data
arguments: label: D17
datatype: string

tool: D10: %Y-%m-%d %H:%M:%S

tool: D17: [%s] %s


assistant: ```c
void log_with_date(char *line) {
  char date[50];
  struct tm *current_time;
  time_t now = time(0);
  current_time = gmtime(&now);
  strftime(date, sizeof(date), "%Y-%m-%d %H:%M:%S", current_time);
  printf("[%s] %s\n", date, line);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<log_with_date>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  xor    %edi,%edi
  push   %rbp
  sub    $0x58,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rbp
  call   <time@plt>
  lea    0xc(%rsp),%rdi
  mov    %eax,0xc(%rsp)
  call   <gmtime@plt>
  lea    D18(%rip),%rdx
  mov    $0x32,%esi
  mov    %rbp,%rdi
  mov    %rax,%rcx
  call   <strftime@plt>
  mov    %r12,%rdx
  mov    %rbp,%rsi
  lea    D19(%rip),%rdi
  call   <printf@plt>
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x58,%rsp
  pop    %rbp
  pop    %r12
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D18
type_guessed: string
name: retrieve_data
arguments: label: D19
type_guessed: string

tool: D18: %Y-%m-%d %H:%M:%S

tool: D19: [%s] %s


assistant: ```c
void log_with_date(char *line) {
  char date[50];
  struct tm *current_time;
  time_t now = time(0);
  current_time = gmtime(&now);
  strftime(date, sizeof(date), "%Y-%m-%d %H:%M:%S", current_time);
  printf("[%s] %s\n", date, line);
}
```
user: What is the c source code of the assembly code below:

```asm
<log_with_date>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  xor    %edi,%edi
  push   %rbp
  sub    $0x58,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rbp
  call   <time@plt>
  lea    0xc(%rsp),%rdi
  mov    %eax,0xc(%rsp)
  call   <gmtime@plt>
  lea    D6(%rip),%rdx
  mov    $0x32,%esi
  mov    %rbp,%rdi
  mov    %rax,%rcx
  call   <strftime@plt>
  mov    %r12,%rdx
  mov    %rbp,%rsi
  lea    D14(%rip),%rdi
  call   <printf@plt>
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  add    $0x58,%rsp
  pop    %rbp
  pop    %r12
  ret
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D6
type_guessed: string
name: unpack_data
arguments: data_label: D14
type_guessed: string

tool: D6: %Y-%m-%d %H:%M:%S

tool: D14: [%s] %s


assistant: ```c
void log_with_date(char *line) {
  char date[50];
  struct tm *current_time;
  time_t now = time(0);
  current_time = gmtime(&now);
  strftime(date, sizeof(date), "%Y-%m-%d %H:%M:%S", current_time);
  printf("[%s] %s\n", date, line);
}
```