ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Converte this assembly snippet into its equivalent C code:

```asm
<radixSort>:
  endbr64
  push   %r15
  push   %r14
  mov    %esi,%r14d
  push   %r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  lea    0x4(,%rsi,4),%ebx
  sub    $0x78,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%r13
  lea    0x60(%rsp),%rbp
  mov    %r13,0x8(%rsp)
  xchg   %ax,%ax
L46:
  mov    %ebx,%edi
  add    $0x8,%r13
  call   <malloc@plt>
  mov    %rax,-0x8(%r13)
  movl   $0x0,(%rax)
  cmp    %rbp,%r13
  jne    L46
  lea    -0x1(%r14),%eax
  mov    $0x1,%ebx
  lea    0x4(%r12,%rax,4),%r13
  nopl   0x0(%rax,%rax,1)
L4:
  mov    %r12,%r15
  test   %r14d,%r14d
  jle    L16
  nopl   0x0(%rax,%rax,1)
L23:
  mov    (%r15),%edi
  mov    %ebx,%esi
  add    $0x4,%r15
  call   <getNumOfPos@plt>
  cltq
  mov    0x10(%rsp,%rax,8),%rcx
  mov    (%rcx),%eax
  add    $0x1,%eax
  mov    %eax,(%rcx)
  mov    -0x4(%r15),%esi
  cltq
  mov    %esi,(%rcx,%rax,4)
  cmp    %r13,%r15
  jne    L23
L16:
  mov    0x8(%rsp),%r9
  xor    %esi,%esi
  nopl   0x0(%rax)
L76:
  mov    (%r9),%rdx
  mov    (%rdx),%eax
  test   %eax,%eax
  jle    L67
  movslq %esi,%rax
  lea    (%r12,%rax,4),%r8
  mov    $0x1,%eax
  nopl   (%rax)
L55:
  mov    (%rdx,%rax,4),%ecx
  lea    (%rsi,%rax,1),%edi
  mov    %ecx,-0x4(%r8,%rax,4)
  add    $0x1,%rax
  cmp    %eax,(%rdx)
  jge    L55
  mov    %edi,%esi
L67:
  add    $0x8,%r9
  movl   $0x0,(%rdx)
  cmp    %rbp,%r9
  jne    L76
  add    $0x1,%ebx
  cmp    $0xb,%ebx
  jne    L4
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L31
  add    $0x78,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L31:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void radixSort(int A[], int N) {
  int i, j, k;
  int pos, index, numOfBit;
  int *bucket[10];
  for (i = 0; i < 10; i++) {
    bucket[i] = (int *)malloc(sizeof(int) * (N + 1));
    bucket[i][0] = 0;
  }
  for (pos = 1; pos <= 10; pos++) {
    for (i = 0; i < N; i++) {
      numOfBit = getNumOfPos(A[i], pos);
      index = ++bucket[numOfBit][0];
      bucket[numOfBit][index] = A[i];
    }
    for (i = 0, j = 0; i < 10; i++) {
      for (k = 1; k <= bucket[i][0]; k++)
        A[j++] = bucket[i][k];
      bucket[i][0] = 0;
    }
  }
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<prtBin>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  movl   $0x80000000,-0x4(%rbp)
  jmp    L8
L11:
  mov    -0x14(%rbp),%eax
  and    -0x4(%rbp),%eax
  test   %eax,%eax
  setne  %al
  movzbl %al,%eax
  mov    %eax,%esi
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  shrl   -0x4(%rbp)
L8:
  cmpl   $0x0,-0x4(%rbp)
  jne    L11
  mov    $0xa,%edi
  call   <putchar@plt>
  nop
  leave
  ret
```

tool_calls: name: unpack_data
arguments: tag: D8
data_type: string

tool: D8: %d

assistant: ```c
void prtBin(int n) {
  unsigned mask = 1u << 31;
  for (; mask; mask >>= 1)
    printf("%d", n & mask ? 1 : 0);
  printf("\n");
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<prtBin>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%r12d
  mov    $0x20,%ebp
  mov    $0x80000000,%ebx
  lea    D2(%rip),%r13
L3:
  test   %ebx,%r12d
  setne  %sil
  movzbl %sil,%esi
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  shr    %ebx
  sub    $0x1,%ebp
  jne    L3
  mov    $0xa,%edi
  call   <putchar@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D2
type_infered: string

tool: D2: %d

assistant: ```c
void prtBin(int n) {
  unsigned mask = 1u << 31;
  for (; mask; mask >>= 1)
    printf("%d", n & mask ? 1 : 0);
  printf("\n");
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<prtBin>:
  endbr64
  push   %r13
  lea    D4(%rip),%r13
  push   %r12
  mov    %edi,%r12d
  push   %rbp
  mov    $0x20,%ebp
  push   %rbx
  mov    $0x80000000,%ebx
  sub    $0x8,%rsp
  nopw   0x0(%rax,%rax,1)
L8:
  xor    %esi,%esi
  test   %ebx,%r12d
  mov    %r13,%rdi
  setne  %sil
  xor    %eax,%eax
  shr    %ebx
  call   <printf@plt>
  sub    $0x1,%ebp
  jne    L8
  add    $0x8,%rsp
  mov    $0xa,%edi
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    1060 <putchar@plt>
```

tool_calls: name: read_data
arguments: tag: D4
type: string

tool: D4: %d

assistant: ```c
void prtBin(int n) {
  unsigned mask = 1u << 31;
  for (; mask; mask >>= 1)
    printf("%d", n & mask ? 1 : 0);
  printf("\n");
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<prtBin>:
  endbr64
  push   %r13
  lea    D4(%rip),%r13
  push   %r12
  mov    %edi,%r12d
  push   %rbp
  mov    $0x20,%ebp
  push   %rbx
  mov    $0x80000000,%ebx
  sub    $0x8,%rsp
  nopw   0x0(%rax,%rax,1)
L7:
  xor    %esi,%esi
  test   %ebx,%r12d
  mov    %r13,%rdi
  setne  %sil
  xor    %eax,%eax
  shr    %ebx
  call   <printf@plt>
  sub    $0x1,%ebp
  jne    L7
  add    $0x8,%rsp
  mov    $0xa,%edi
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    1060 <putchar@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D4
data_type: string

tool: D4: %d

assistant: ```c
void prtBin(int n) {
  unsigned mask = 1u << 31;
  for (; mask; mask >>= 1)
    printf("%d", n & mask ? 1 : 0);
  printf("\n");
}
```
user: What is the c source code of the assembly code below:

```asm
<vktrace_copy_and_append>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    %rdx,-0x48(%rbp)
  cmpq   $0x0,-0x38(%rbp)
  je     L50
  mov    -0x38(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  jmp    L31
L50:
  mov    $0x0,%eax
L31:
  mov    %rax,-0x28(%rbp)
  cmpq   $0x0,-0x48(%rbp)
  je     L56
  mov    -0x48(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L56
  cmpq   $0x0,-0x40(%rbp)
  je     L56
  mov    -0x40(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  jmp    L46
L56:
  mov    $0x0,%eax
L46:
  mov    %rax,-0x20(%rbp)
  cmpq   $0x0,-0x48(%rbp)
  je     L7
  mov    -0x48(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  jmp    L10
L7:
  mov    $0x0,%eax
L10:
  mov    %rax,-0x18(%rbp)
  mov    -0x28(%rbp),%rdx
  mov    -0x20(%rbp),%rax
  add    %rax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  add    $0x1,%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  mov    %eax,%edi
  call   <vktrace_malloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L4
  mov    -0x28(%rbp),%rdx
  mov    -0x38(%rbp),%rcx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strncpy@plt>
  mov    -0x8(%rbp),%rdx
  mov    -0x28(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0x20(%rbp),%rdx
  mov    -0x40(%rbp),%rax
  mov    %rax,%rsi
  mov    %rcx,%rdi
  call   <strncpy@plt>
  mov    -0x28(%rbp),%rdx
  mov    -0x20(%rbp),%rax
  add    %rax,%rdx
  mov    -0x8(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0x18(%rbp),%rdx
  mov    -0x48(%rbp),%rax
  mov    %rax,%rsi
  mov    %rcx,%rdi
  call   <strncpy@plt>
L4:
  mov    -0x10(%rbp),%rax
  lea    -0x1(%rax),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *vktrace_copy_and_append(const char *pBaseString, const char *pSeparator, const char *pAppendString) {
  size_t baseSize = (pBaseString != NULL) ? strlen(pBaseString) : 0;
  size_t separatorSize = ((pAppendString != NULL) && strlen(pAppendString) && (pSeparator != NULL)) ? strlen(pSeparator) : 0;
  size_t appendSize = (pAppendString != NULL) ? strlen(pAppendString) : 0;
  size_t bufferSize = baseSize + separatorSize + appendSize + 1;
  char *retVal = (char *)vktrace_malloc(sizeof(char) * bufferSize);
  if (retVal != NULL) {
    strncpy(retVal, pBaseString, baseSize);
    strncpy(&retVal[baseSize], pSeparator, separatorSize);
    strncpy(&retVal[baseSize + separatorSize], pAppendString, appendSize);
  }
  retVal[bufferSize - 1] = '\0';
  return retVal;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<vktrace_copy_and_append>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,%r13
  mov    %rsi,0x18(%rsp)
  mov    %rdx,%rbp
  mov    $0x0,%r12d
  test   %rdi,%rdi
  je     L50
  call   <strlen@plt>
  mov    %rax,%r12
L50:
  test   %rbp,%rbp
  je     L4
  cmpb   $0x0,0x0(%rbp)
  je     L10
  mov    0x18(%rsp),%rdi
  test   %rdi,%rdi
  je     L10
  call   <strlen@plt>
  mov    %rax,0x10(%rsp)
  jmp    L47
L4:
  movq   $0x0,0x10(%rsp)
  movq   $0x0,0x8(%rsp)
  jmp    L73
L48:
  mov    %r12,%rdx
  mov    %r13,%rsi
  mov    %rax,%rdi
  call   <strncpy@plt>
  lea    (%rbx,%r12,1),%rdi
  mov    0x10(%rsp),%rdx
  mov    0x18(%rsp),%rsi
  call   <strncpy@plt>
  lea    (%rbx,%r15,1),%rdi
  mov    0x8(%rsp),%rdx
  mov    %rbp,%rsi
  call   <strncpy@plt>
  jmp    L79
L10:
  movq   $0x0,0x10(%rsp)
L47:
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %rax,0x8(%rsp)
L73:
  mov    0x10(%rsp),%rax
  lea    (%r12,%rax,1),%r15
  mov    0x8(%rsp),%rax
  lea    (%r15,%rax,1),%r14
  lea    0x1(%r14),%edi
  call   <vktrace_malloc@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  jne    L48
L79:
  movb   $0x0,(%rbx,%r14,1)
  mov    %rbx,%rax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
char *vktrace_copy_and_append(const char *pBaseString, const char *pSeparator, const char *pAppendString) {
  size_t baseSize = (pBaseString != NULL) ? strlen(pBaseString) : 0;
  size_t separatorSize = ((pAppendString != NULL) && strlen(pAppendString) && (pSeparator != NULL)) ? strlen(pSeparator) : 0;
  size_t appendSize = (pAppendString != NULL) ? strlen(pAppendString) : 0;
  size_t bufferSize = baseSize + separatorSize + appendSize + 1;
  char *retVal = (char *)vktrace_malloc(sizeof(char) * bufferSize);
  if (retVal != NULL) {
    strncpy(retVal, pBaseString, baseSize);
    strncpy(&retVal[baseSize], pSeparator, separatorSize);
    strncpy(&retVal[baseSize + separatorSize], pAppendString, appendSize);
  }
  retVal[bufferSize - 1] = '\0';
  return retVal;
}
```
user: What is the c source code of the assembly code below?

```asm
<vktrace_copy_and_append>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  mov    %rsi,%r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x28,%rsp
  test   %rdi,%rdi
  je     L11
  call   <strlen@plt>
  mov    %rax,%rbx
L11:
  test   %rbp,%rbp
  je     L49
  cmpb   $0x0,0x0(%rbp)
  je     L34
  test   %r13,%r13
  je     L34
  mov    %r13,%rdi
  call   <strlen@plt>
  mov    %rax,0x18(%rsp)
  add    %rbx,%rax
  mov    %rax,0x8(%rsp)
L61:
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    0x8(%rsp),%r14
  mov    %rax,0x10(%rsp)
  add    %rax,%r14
L22:
  lea    0x1(%r14),%edi
  call   <vktrace_malloc@plt>
  mov    %rax,%r15
  test   %rax,%rax
  jne    L29
L24:
  movb   $0x0,(%r15,%r14,1)
  add    $0x28,%rsp
  mov    %r15,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L34:
  mov    %rbx,0x8(%rsp)
  movq   $0x0,0x18(%rsp)
  jmp    L61
L29:
  mov    %rbx,%rdx
  mov    %r12,%rsi
  mov    %rax,%rdi
  call   <strncpy@plt>
  mov    0x18(%rsp),%rdx
  lea    (%r15,%rbx,1),%rdi
  mov    %r13,%rsi
  call   <strncpy@plt>
  mov    0x8(%rsp),%rdi
  mov    0x10(%rsp),%rdx
  mov    %rbp,%rsi
  add    %r15,%rdi
  call   <strncpy@plt>
  jmp    L24
  cs nopw 0x0(%rax,%rax,1)
L49:
  mov    %rbx,0x8(%rsp)
  mov    %rbx,%r14
  movq   $0x0,0x18(%rsp)
  movq   $0x0,0x10(%rsp)
  jmp    L22
```

assistant: ```c
char *vktrace_copy_and_append(const char *pBaseString, const char *pSeparator, const char *pAppendString) {
  size_t baseSize = (pBaseString != NULL) ? strlen(pBaseString) : 0;
  size_t separatorSize = ((pAppendString != NULL) && strlen(pAppendString) && (pSeparator != NULL)) ? strlen(pSeparator) : 0;
  size_t appendSize = (pAppendString != NULL) ? strlen(pAppendString) : 0;
  size_t bufferSize = baseSize + separatorSize + appendSize + 1;
  char *retVal = (char *)vktrace_malloc(sizeof(char) * bufferSize);
  if (retVal != NULL) {
    strncpy(retVal, pBaseString, baseSize);
    strncpy(&retVal[baseSize], pSeparator, separatorSize);
    strncpy(&retVal[baseSize + separatorSize], pAppendString, appendSize);
  }
  retVal[bufferSize - 1] = '\0';
  return retVal;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<vktrace_copy_and_append>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  mov    %rsi,%r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x28,%rsp
  test   %rdi,%rdi
  je     L23
  call   <strlen@plt>
  mov    %rax,%rbx
L23:
  test   %rbp,%rbp
  je     L21
  cmpb   $0x0,0x0(%rbp)
  je     L33
  test   %r13,%r13
  je     L33
  mov    %r13,%rdi
  call   <strlen@plt>
  mov    %rax,0x18(%rsp)
  add    %rbx,%rax
  mov    %rax,0x8(%rsp)
L5:
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    0x8(%rsp),%r14
  mov    %rax,0x10(%rsp)
  add    %rax,%r14
L38:
  lea    0x1(%r14),%edi
  call   <vktrace_malloc@plt>
  mov    %rax,%r15
  test   %rax,%rax
  jne    L13
L11:
  movb   $0x0,(%r15,%r14,1)
  add    $0x28,%rsp
  mov    %r15,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L33:
  mov    %rbx,0x8(%rsp)
  movq   $0x0,0x18(%rsp)
  jmp    L5
L13:
  mov    %rbx,%rdx
  mov    %r12,%rsi
  mov    %rax,%rdi
  call   <strncpy@plt>
  mov    0x18(%rsp),%rdx
  lea    (%r15,%rbx,1),%rdi
  mov    %r13,%rsi
  call   <strncpy@plt>
  mov    0x8(%rsp),%rdi
  mov    0x10(%rsp),%rdx
  mov    %rbp,%rsi
  add    %r15,%rdi
  call   <strncpy@plt>
  jmp    L11
  cs nopw 0x0(%rax,%rax,1)
L21:
  mov    %rbx,0x8(%rsp)
  mov    %rbx,%r14
  movq   $0x0,0x18(%rsp)
  movq   $0x0,0x10(%rsp)
  jmp    L38
```

assistant: ```c
char *vktrace_copy_and_append(const char *pBaseString, const char *pSeparator, const char *pAppendString) {
  size_t baseSize = (pBaseString != NULL) ? strlen(pBaseString) : 0;
  size_t separatorSize = ((pAppendString != NULL) && strlen(pAppendString) && (pSeparator != NULL)) ? strlen(pSeparator) : 0;
  size_t appendSize = (pAppendString != NULL) ? strlen(pAppendString) : 0;
  size_t bufferSize = baseSize + separatorSize + appendSize + 1;
  char *retVal = (char *)vktrace_malloc(sizeof(char) * bufferSize);
  if (retVal != NULL) {
    strncpy(retVal, pBaseString, baseSize);
    strncpy(&retVal[baseSize], pSeparator, separatorSize);
    strncpy(&retVal[baseSize + separatorSize], pAppendString, appendSize);
  }
  retVal[bufferSize - 1] = '\0';
  return retVal;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<_ZNSt6vectorIPiSaIS0_EEC1Ev>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <_ZNSt6vectorIPiSaIS0_EEC2Ev@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L8
  call   <__stack_chk_fail@plt>
L8:
  leave
  ret
```

assistant: ```c
void _ZNSt6vectorIPiSaIS0_EEC1Ev(struct l_class_OC_std_KD__KD_vector *llvm_cbe_this) {
  struct l_class_OC_std_KD__KD_vector *llvm_cbe_this_2e_addr;
  struct l_class_OC_std_KD__KD_vector *llvm_cbe_this1;
  *(&llvm_cbe_this_2e_addr) = llvm_cbe_this;
  llvm_cbe_this1 = *(&llvm_cbe_this_2e_addr);
  _ZNSt6vectorIPiSaIS0_EEC2Ev(llvm_cbe_this1);
  return;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<_ZNSt6vectorIPiSaIS0_EEC1Ev>:
  endbr64
  sub    $0x8,%rsp
  call   <_ZNSt6vectorIPiSaIS0_EEC2Ev@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void _ZNSt6vectorIPiSaIS0_EEC1Ev(struct l_class_OC_std_KD__KD_vector *llvm_cbe_this) {
  struct l_class_OC_std_KD__KD_vector *llvm_cbe_this_2e_addr;
  struct l_class_OC_std_KD__KD_vector *llvm_cbe_this1;
  *(&llvm_cbe_this_2e_addr) = llvm_cbe_this;
  llvm_cbe_this1 = *(&llvm_cbe_this_2e_addr);
  _ZNSt6vectorIPiSaIS0_EEC2Ev(llvm_cbe_this1);
  return;
}
```
user: Translate the below assembly code into C programming language:

```asm
<_ZNSt6vectorIPiSaIS0_EEC1Ev>:
  endbr64
  jmp    1050 <_ZNSt6vectorIPiSaIS0_EEC2Ev@plt>
```

assistant: ```c
void _ZNSt6vectorIPiSaIS0_EEC1Ev(struct l_class_OC_std_KD__KD_vector *llvm_cbe_this) {
  struct l_class_OC_std_KD__KD_vector *llvm_cbe_this_2e_addr;
  struct l_class_OC_std_KD__KD_vector *llvm_cbe_this1;
  *(&llvm_cbe_this_2e_addr) = llvm_cbe_this;
  llvm_cbe_this1 = *(&llvm_cbe_this_2e_addr);
  _ZNSt6vectorIPiSaIS0_EEC2Ev(llvm_cbe_this1);
  return;
}
```
user: What is the c source code of the assembly code below?

```asm
<_ZNSt6vectorIPiSaIS0_EEC1Ev>:
  endbr64
  jmp    1050 <_ZNSt6vectorIPiSaIS0_EEC2Ev@plt>
```

assistant: ```c
void _ZNSt6vectorIPiSaIS0_EEC1Ev(struct l_class_OC_std_KD__KD_vector *llvm_cbe_this) {
  struct l_class_OC_std_KD__KD_vector *llvm_cbe_this_2e_addr;
  struct l_class_OC_std_KD__KD_vector *llvm_cbe_this1;
  *(&llvm_cbe_this_2e_addr) = llvm_cbe_this;
  llvm_cbe_this1 = *(&llvm_cbe_this_2e_addr);
  _ZNSt6vectorIPiSaIS0_EEC2Ev(llvm_cbe_this1);
  return;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <InitApp@plt>
L2:
  mov    $0x0,%eax
  call   <Sel_mode_LEDs@plt>
  jmp    L2
```

assistant: ```c
void main(void) {
  InitApp();
  while (1) {
    Sel_mode_LEDs();
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <InitApp@plt>
L8:
  mov    $0x0,%eax
  call   <Sel_mode_LEDs@plt>
  jmp    L8
```

assistant: ```c
void main(void) {
  InitApp();
  while (1) {
    Sel_mode_LEDs();
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <InitApp@plt>
  nop
L1:
  xor    %eax,%eax
  call   <Sel_mode_LEDs@plt>
  jmp    L1
```

assistant: ```c
void main(void) {
  InitApp();
  while (1) {
    Sel_mode_LEDs();
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <InitApp@plt>
  nop
L1:
  xor    %eax,%eax
  call   <Sel_mode_LEDs@plt>
  jmp    L1
```

assistant: ```c
void main(void) {
  InitApp();
  while (1) {
    Sel_mode_LEDs();
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<itoa>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x38(%rbp)
  mov    %esi,-0x3c(%rbp)
  mov    %edx,-0x40(%rbp)
  mov    -0x38(%rbp),%rax
  mov    %rax,-0x20(%rbp)
  mov    -0x40(%rbp),%eax
  cltq
  mov    %rax,-0x8(%rbp)
  movl   $0xa,-0x28(%rbp)
  cmpl   $0x64,-0x3c(%rbp)
  jne    L79
  cmpl   $0x0,-0x40(%rbp)
  jns    L79
  mov    -0x20(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x20(%rbp)
  movb   $0x2d,(%rax)
  addq   $0x1,-0x38(%rbp)
  mov    -0x40(%rbp),%eax
  neg    %eax
  cltq
  mov    %rax,-0x8(%rbp)
  jmp    L71
L79:
  cmpl   $0x78,-0x3c(%rbp)
  jne    L71
  movl   $0x10,-0x28(%rbp)
L71:
  mov    -0x28(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x8(%rbp),%rax
  mov    $0x0,%edx
  div    %rcx
  mov    %rdx,%rax
  mov    %eax,-0x24(%rbp)
  cmpl   $0x9,-0x24(%rbp)
  jg     L20
  mov    -0x24(%rbp),%eax
  add    $0x30,%eax
  mov    %eax,%ecx
  jmp    L36
L20:
  mov    -0x24(%rbp),%eax
  add    $0x57,%eax
  mov    %eax,%ecx
L36:
  mov    -0x20(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x20(%rbp)
  mov    %cl,(%rax)
  mov    -0x28(%rbp),%eax
  movslq %eax,%rsi
  mov    -0x8(%rbp),%rax
  mov    $0x0,%edx
  div    %rsi
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L71
  mov    -0x20(%rbp),%rax
  movb   $0x0,(%rax)
  mov    -0x38(%rbp),%rax
  mov    %rax,-0x18(%rbp)
  mov    -0x20(%rbp),%rax
  sub    $0x1,%rax
  mov    %rax,-0x10(%rbp)
  jmp    L75
L39:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  mov    %al,-0x29(%rbp)
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%edx
  mov    -0x18(%rbp),%rax
  mov    %dl,(%rax)
  mov    -0x10(%rbp),%rax
  movzbl -0x29(%rbp),%edx
  mov    %dl,(%rax)
  addq   $0x1,-0x18(%rbp)
  subq   $0x1,-0x10(%rbp)
L75:
  mov    -0x18(%rbp),%rax
  cmp    -0x10(%rbp),%rax
  jb     L39
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void itoa(char *buf, int base, int d) {
  char *p = buf;
  char *p1, *p2;
  unsigned long ud = d;
  int divisor = 10;
  if (base == 'd' && d < 0) {
    *p++ = '-';
    buf++;
    ud = -d;
  } else if (base == 'x') {
    divisor = 16;
  }
  do {
    int remainder = ud % divisor;
    *p++ = (remainder < 10) ? remainder + '0' : remainder + 'a' - 10;
  } while (ud /= divisor);
  *p = 0;
  p1 = buf;
  p2 = p - 1;
  while (p1 < p2) {
    char tmp = *p1;
    *p1 = *p2;
    *p2 = tmp;
    p1++;
    p2--;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<itoa>:
  endbr64
  mov    %rdi,%r8
  movslq %edx,%rdi
  cmp    $0x64,%esi
  jne    L51
  test   %edx,%edx
  jns    L51
  movb   $0x2d,(%r8)
  neg    %edx
  movslq %edx,%rdi
  lea    0x1(%r8),%r8
  mov    $0xa,%esi
L37:
  mov    %r8,%rcx
  movslq %esi,%rsi
  jmp    L32
L51:
  cmp    $0x78,%esi
  mov    $0xa,%esi
  mov    $0x10,%eax
  cmove  %eax,%esi
  jmp    L37
L22:
  mov    %rax,%rdi
  mov    %r9,%rcx
L32:
  mov    %rdi,%rax
  mov    $0x0,%edx
  div    %rsi
  lea    0x30(%rdx),%r9d
  lea    0x57(%rdx),%eax
  cmp    $0x9,%edx
  cmovle %r9d,%eax
  lea    0x1(%rcx),%r9
  mov    %al,-0x1(%r9)
  mov    %rdi,%rax
  mov    $0x0,%edx
  div    %rsi
  cmp    %rdi,%rsi
  jbe    L22
  movb   $0x0,(%r9)
  cmp    %r8,%rcx
  jbe    L57
L29:
  movzbl (%r8),%eax
  movzbl (%rcx),%edx
  mov    %dl,(%r8)
  mov    %al,(%rcx)
  add    $0x1,%r8
  sub    $0x1,%rcx
  cmp    %rcx,%r8
  jb     L29
L57:
  ret
```

assistant: ```c
void itoa(char *buf, int base, int d) {
  char *p = buf;
  char *p1, *p2;
  unsigned long ud = d;
  int divisor = 10;
  if (base == 'd' && d < 0) {
    *p++ = '-';
    buf++;
    ud = -d;
  } else if (base == 'x') {
    divisor = 16;
  }
  do {
    int remainder = ud % divisor;
    *p++ = (remainder < 10) ? remainder + '0' : remainder + 'a' - 10;
  } while (ud /= divisor);
  *p = 0;
  p1 = buf;
  p2 = p - 1;
  while (p1 < p2) {
    char tmp = *p1;
    *p1 = *p2;
    *p2 = tmp;
    p1++;
    p2--;
  }
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<itoa>:
  endbr64
  movslq %edx,%r8
  cmp    $0x64,%esi
  jne    L31
  test   %edx,%edx
  jns    L31
  neg    %edx
  movb   $0x2d,(%rdi)
  mov    $0xa,%r9d
  add    $0x1,%rdi
  movslq %edx,%r8
L17:
  mov    %rdi,%rcx
  jmp    L22
  nopw   0x0(%rax,%rax,1)
L35:
  mov    %rax,%r8
  mov    %rsi,%rcx
L22:
  xor    %edx,%edx
  mov    %r8,%rax
  div    %r9
  lea    0x30(%rdx),%r10d
  lea    0x57(%rdx),%esi
  cmp    $0x9,%rdx
  mov    %r10d,%edx
  cmova  %esi,%edx
  lea    0x1(%rcx),%rsi
  mov    %dl,-0x1(%rsi)
  cmp    %r9,%r8
  jae    L35
  movb   $0x0,(%rsi)
  cmp    %rdi,%rcx
  ja     L47
  jmp    L48
  nopl   (%rax)
L25:
  movzbl (%rcx),%edx
L47:
  movzbl (%rdi),%eax
  sub    $0x1,%rcx
  mov    %dl,(%rdi)
  add    $0x1,%rdi
  mov    %al,0x1(%rcx)
  cmp    %rcx,%rdi
  jb     L25
L48:
  ret
  nopl   0x0(%rax)
L31:
  cmp    $0x78,%esi
  mov    $0xa,%r9d
  mov    $0x10,%eax
  cmove  %rax,%r9
  jmp    L17
```

assistant: ```c
void itoa(char *buf, int base, int d) {
  char *p = buf;
  char *p1, *p2;
  unsigned long ud = d;
  int divisor = 10;
  if (base == 'd' && d < 0) {
    *p++ = '-';
    buf++;
    ud = -d;
  } else if (base == 'x') {
    divisor = 16;
  }
  do {
    int remainder = ud % divisor;
    *p++ = (remainder < 10) ? remainder + '0' : remainder + 'a' - 10;
  } while (ud /= divisor);
  *p = 0;
  p1 = buf;
  p2 = p - 1;
  while (p1 < p2) {
    char tmp = *p1;
    *p1 = *p2;
    *p2 = tmp;
    p1++;
    p2--;
  }
}
```