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

```asm
<foo>:
  endbr64
  mov    $0x0,%eax
L3:
  movapd %xmm0,%xmm3
  mulsd  (%rdx,%rax,1),%xmm3
  movapd %xmm3,%xmm5
  mulsd  %xmm1,%xmm5
  movsd  (%rdi,%rax,1),%xmm4
  subsd  %xmm5,%xmm4
  movsd  %xmm4,(%rdi,%rax,1)
  mulsd  %xmm2,%xmm3
  movsd  (%rsi,%rax,1),%xmm4
  subsd  %xmm3,%xmm4
  movsd  %xmm4,(%rsi,%rax,1)
  add    $0x8,%rax
  cmp    $0x20,%rax
  jne    L3
  ret
```

assistant: ```c
void foo(double (*__restrict p)[4], double (*__restrict q)[4], double *__restrict prim, double scale, double pp, double pq) {
  int md, mc, mb, ma, p_index = 0;
  for (md = 0; md < 1; md++)
    for (mc = 0; mc < 1; mc++)
      for (mb = 0; mb < 1; mb++)
        for (ma = 0; ma < 4; ma++) {
          double tmp = scale * prim[p_index++];
          p[md][ma] = p[md][ma] - tmp * pp;
          q[mc][ma] = q[mc][ma] - tmp * pq;
        }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<foo>:
  endbr64
  xor    %eax,%eax
L7:
  movsd  (%rdx,%rax,1),%xmm3
  movsd  (%rdi,%rax,1),%xmm4
  mulsd  %xmm0,%xmm3
  movapd %xmm3,%xmm5
  mulsd  %xmm1,%xmm5
  mulsd  %xmm2,%xmm3
  subsd  %xmm5,%xmm4
  movsd  %xmm4,(%rdi,%rax,1)
  movsd  (%rsi,%rax,1),%xmm4
  subsd  %xmm3,%xmm4
  movsd  %xmm4,(%rsi,%rax,1)
  add    $0x8,%rax
  cmp    $0x20,%rax
  jne    L7
  ret
```

assistant: ```c
void foo(double (*__restrict p)[4], double (*__restrict q)[4], double *__restrict prim, double scale, double pp, double pq) {
  int md, mc, mb, ma, p_index = 0;
  for (md = 0; md < 1; md++)
    for (mc = 0; mc < 1; mc++)
      for (mb = 0; mb < 1; mb++)
        for (ma = 0; ma < 4; ma++) {
          double tmp = scale * prim[p_index++];
          p[md][ma] = p[md][ma] - tmp * pp;
          q[mc][ma] = q[mc][ma] - tmp * pq;
        }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<foo>:
  endbr64
  movapd %xmm0,%xmm4
  movupd (%rdx),%xmm3
  unpcklpd %xmm1,%xmm1
  unpcklpd %xmm2,%xmm2
  movupd 0x10(%rdx),%xmm0
  unpcklpd %xmm4,%xmm4
  movapd %xmm2,%xmm6
  movupd 0x10(%rdi),%xmm5
  mulpd  %xmm4,%xmm3
  mulpd  %xmm4,%xmm0
  movapd %xmm1,%xmm4
  mulpd  %xmm3,%xmm6
  mulpd  %xmm0,%xmm4
  mulpd  %xmm0,%xmm2
  movupd 0x10(%rsi),%xmm0
  mulpd  %xmm3,%xmm1
  subpd  %xmm4,%xmm5
  movupd (%rsi),%xmm4
  subpd  %xmm2,%xmm0
  movupd (%rdi),%xmm2
  subpd  %xmm6,%xmm4
  subpd  %xmm1,%xmm2
  movups %xmm5,0x10(%rdi)
  movups %xmm0,0x10(%rsi)
  movups %xmm4,(%rsi)
  movups %xmm2,(%rdi)
  ret
```

assistant: ```c
void foo(double (*__restrict p)[4], double (*__restrict q)[4], double *__restrict prim, double scale, double pp, double pq) {
  int md, mc, mb, ma, p_index = 0;
  for (md = 0; md < 1; md++)
    for (mc = 0; mc < 1; mc++)
      for (mb = 0; mb < 1; mb++)
        for (ma = 0; ma < 4; ma++) {
          double tmp = scale * prim[p_index++];
          p[md][ma] = p[md][ma] - tmp * pp;
          q[mc][ma] = q[mc][ma] - tmp * pq;
        }
}
```
user: What is the c source code of the assembly code below:

```asm
<cio_stack_alloc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    D27(%rip),%rax
  mov    (%rax),%edx
  mov    D53(%rip),%rax
  mov    (%rax),%eax
  or     %eax,%edx
  mov    D57(%rip),%rax
  mov    (%rax),%eax
  mov    %edx,%ecx
  or     %eax,%ecx
  mov    D2(%rip),%rax
  mov    (%rax),%edx
  mov    D68(%rip),%rax
  mov    (%rax),%eax
  or     %eax,%edx
  mov    -0x18(%rbp),%rax
  mov    $0x0,%r9d
  mov    $0xffffffff,%r8d
  mov    %rax,%rsi
  mov    $0x0,%edi
  call   <mmap@plt>
  mov    %rax,-0x8(%rbp)
  mov    D8(%rip),%rax
  mov    (%rax),%rax
  cmp    %rax,-0x8(%rbp)
  jne    L0
  mov    $0x0,%eax
  jmp    L30
L0:
  mov    D54(%rip),%rax
  mov    (%rax),%edx
  mov    -0x20(%rbp),%rcx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <mprotect@plt>
  test   %rax,%rax
  jns    L4
  mov    -0x18(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <munmap@plt>
  mov    $0x0,%eax
  jmp    L30
L4:
  mov    -0x8(%rbp),%rax
L30:
  leave
  ret
```

assistant: ```c
void *cio_stack_alloc(size_t size, size_t guard_size) {
  void *ptr = mmap(NULL, size, PROT_READ | PROT_WRITE, MAP_PRIVATE | MAP_ANONYMOUS | MAP_STACK, -1, 0);
  if (ptr == MAP_FAILED)
    return NULL;
  if (mprotect(ptr, guard_size, PROT_NONE) < 0) {
    munmap(ptr, size);
    return NULL;
  }
  return ptr;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<cio_stack_alloc>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    %rsi,%r12
  mov    D4(%rip),%rax
  mov    (%rax),%ecx
  mov    D8(%rip),%rax
  or     (%rax),%ecx
  mov    D35(%rip),%rax
  or     (%rax),%ecx
  mov    D24(%rip),%rax
  mov    (%rax),%edx
  mov    D61(%rip),%rax
  or     (%rax),%edx
  mov    $0x0,%r9d
  mov    $0xffffffff,%r8d
  mov    %rdi,%rsi
  mov    $0x0,%edi
  call   <mmap@plt>
  mov    %rax,%rbx
  mov    D34(%rip),%rax
  cmp    %rbx,(%rax)
  je     L23
  mov    D7(%rip),%rax
  mov    (%rax),%edx
  mov    %r12,%rsi
  mov    %rbx,%rdi
  call   <mprotect@plt>
  test   %rax,%rax
  js     L25
L5:
  mov    %rbx,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L25:
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <munmap@plt>
  mov    $0x0,%ebx
  jmp    L5
L23:
  mov    $0x0,%ebx
  jmp    L5
```

assistant: ```c
void *cio_stack_alloc(size_t size, size_t guard_size) {
  void *ptr = mmap(NULL, size, PROT_READ | PROT_WRITE, MAP_PRIVATE | MAP_ANONYMOUS | MAP_STACK, -1, 0);
  if (ptr == MAP_FAILED)
    return NULL;
  if (mprotect(ptr, guard_size, PROT_NONE) < 0) {
    munmap(ptr, size);
    return NULL;
  }
  return ptr;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<cio_stack_alloc>:
  endbr64
  mov    D56(%rip),%rax
  push   %r13
  xor    %r9d,%r9d
  mov    %rsi,%r13
  push   %r12
  mov    %rdi,%rsi
  mov    $0xffffffff,%r8d
  push   %rbp
  mov    (%rax),%ecx
  mov    %rdi,%rbp
  xor    %edi,%edi
  mov    D49(%rip),%rax
  or     (%rax),%ecx
  mov    D61(%rip),%rax
  or     (%rax),%ecx
  mov    D26(%rip),%rax
  mov    (%rax),%edx
  mov    D1(%rip),%rax
  or     (%rax),%edx
  call   <mmap@plt>
  mov    %rax,%r12
  mov    D69(%rip),%rax
  cmp    %r12,(%rax)
  je     L0
  mov    D3(%rip),%rax
  mov    %r13,%rsi
  mov    %r12,%rdi
  mov    (%rax),%edx
  call   <mprotect@plt>
  test   %rax,%rax
  js     L7
L20:
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L0:
  xor    %r12d,%r12d
  pop    %rbp
  mov    %r12,%rax
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L7:
  mov    %r12,%rdi
  mov    %rbp,%rsi
  xor    %r12d,%r12d
  call   <munmap@plt>
  jmp    L20
```

assistant: ```c
void *cio_stack_alloc(size_t size, size_t guard_size) {
  void *ptr = mmap(NULL, size, PROT_READ | PROT_WRITE, MAP_PRIVATE | MAP_ANONYMOUS | MAP_STACK, -1, 0);
  if (ptr == MAP_FAILED)
    return NULL;
  if (mprotect(ptr, guard_size, PROT_NONE) < 0) {
    munmap(ptr, size);
    return NULL;
  }
  return ptr;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<cio_stack_alloc>:
  endbr64
  mov    D29(%rip),%rax
  push   %r13
  xor    %r9d,%r9d
  mov    %rsi,%r13
  push   %r12
  mov    %rdi,%rsi
  mov    $0xffffffff,%r8d
  push   %rbp
  mov    (%rax),%ecx
  mov    %rdi,%rbp
  xor    %edi,%edi
  mov    D60(%rip),%rax
  or     (%rax),%ecx
  mov    D9(%rip),%rax
  or     (%rax),%ecx
  mov    D65(%rip),%rax
  mov    (%rax),%edx
  mov    D57(%rip),%rax
  or     (%rax),%edx
  call   <mmap@plt>
  mov    %rax,%r12
  mov    D22(%rip),%rax
  cmp    %r12,(%rax)
  je     L11
  mov    D12(%rip),%rax
  mov    %r13,%rsi
  mov    %r12,%rdi
  mov    (%rax),%edx
  call   <mprotect@plt>
  test   %rax,%rax
  js     L13
L22:
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L11:
  xor    %r12d,%r12d
  pop    %rbp
  mov    %r12,%rax
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L13:
  mov    %r12,%rdi
  mov    %rbp,%rsi
  xor    %r12d,%r12d
  call   <munmap@plt>
  jmp    L22
```

assistant: ```c
void *cio_stack_alloc(size_t size, size_t guard_size) {
  void *ptr = mmap(NULL, size, PROT_READ | PROT_WRITE, MAP_PRIVATE | MAP_ANONYMOUS | MAP_STACK, -1, 0);
  if (ptr == MAP_FAILED)
    return NULL;
  if (mprotect(ptr, guard_size, PROT_NONE) < 0) {
    munmap(ptr, size);
    return NULL;
  }
  return ptr;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    $0x0,%eax
  call   <hash@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%edx
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_label: D0
type_guessed: string

tool: D0: Start -> %d, 0x%x


assistant: ```c
int main(int argc, char *argv[]) {
  int h = hash();
  printf("Start -> %d, 0x%x\n", h, h);
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <hash@plt>
  mov    %eax,%esi
  mov    %eax,%edx
  lea    D6(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: tag: D6
type: string

tool: D6: Start -> %d, 0x%x


assistant: ```c
int main(int argc, char *argv[]) {
  int h = hash();
  printf("Start -> %d, 0x%x\n", h, h);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <hash@plt>
  lea    D1(%rip),%rdi
  mov    %eax,%esi
  mov    %eax,%edx
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: label: D1
datatype: string

tool: D1: Start -> %d, 0x%x


assistant: ```c
int main(int argc, char *argv[]) {
  int h = hash();
  printf("Start -> %d, 0x%x\n", h, h);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <hash@plt>
  lea    D8(%rip),%rdi
  mov    %eax,%esi
  mov    %eax,%edx
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D8
datatype: string

tool: D8: Start -> %d, 0x%x


assistant: ```c
int main(int argc, char *argv[]) {
  int h = hash();
  printf("Start -> %d, 0x%x\n", h, h);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  enter  $0xffff,$0x1f
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main() {
  __asm__ __volatile__("enter $65535, $31" : : :);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  enter  $0xffff,$0x1f
  mov    $0x0,%eax
  ret
```

assistant: ```c
int main() {
  __asm__ __volatile__("enter $65535, $31" : : :);
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  enter  $0xffff,$0x1f
  xor    %eax,%eax
  ret
```

assistant: ```c
int main() {
  __asm__ __volatile__("enter $65535, $31" : : :);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  enter  $0xffff,$0x1f
  xor    %eax,%eax
  ret
```

assistant: ```c
int main() {
  __asm__ __volatile__("enter $65535, $31" : : :);
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<xt_core_string_append_char>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,%eax
  mov    %al,-0x2c(%rbp)
  movsbl -0x2c(%rbp),%eax
  mov    %eax,%edi
  call   <assert@plt>
  cmpq   $0x0,-0x28(%rbp)
  je     L46
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  add    $0x1,%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  add    $0x1,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x18(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  je     L58
  mov    -0x10(%rbp),%rdx
  mov    -0x28(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0x18(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  add    %rax,%rdx
  movzbl -0x2c(%rbp),%eax
  mov    %al,(%rdx)
  mov    -0x18(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  jmp    L26
L58:
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  call   <xt_core_trace@plt>
L26:
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  jmp    L22
L46:
  mov    $0x2,%edi
  call   <malloc@plt>
  mov    %rax,-0x18(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  je     L20
  mov    -0x18(%rbp),%rax
  movzbl -0x2c(%rbp),%edx
  mov    %dl,(%rax)
  mov    -0x18(%rbp),%rax
  add    $0x1,%rax
  movb   $0x0,(%rax)
  jmp    L22
L20:
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  call   <xt_core_trace@plt>
L22:
  mov    -0x18(%rbp),%rax
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D10
type: string

tool: D10: malloc

assistant: ```c
char *xt_core_string_append_char(char *original, char addition) {
  assert(addition);
  char *new_string;
  unsigned long original_size;
  unsigned long new_string_size;
  if (original) {
    original_size = strlen(original);
    new_string_size = original_size + 1;
    new_string = malloc(new_string_size + 1);
    if (new_string) {
      memcpy(new_string, original, original_size);
      *(new_string + original_size) = addition;
      *(new_string + new_string_size) = '\0';
    } else {
      xt_core_trace("malloc");
    }
    free(original);
  } else {
    new_string = malloc(1 + 1);
    if (new_string) {
      *(new_string) = addition;
      *(new_string + 1) = '\0';
    } else {
      xt_core_trace("malloc");
    }
  }
  return new_string;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<xt_core_string_append_char>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    %esi,%r12d
  movsbl %sil,%edi
  call   <assert@plt>
  test   %rbp,%rbp
  je     L52
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %rax,%r13
  lea    0x2(%rax),%edi
  call   <malloc@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L7
  mov    %r13,%rdx
  mov    %rbp,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    %r12b,(%rbx,%r13,1)
  movb   $0x0,0x1(%rbx,%r13,1)
L34:
  mov    %rbp,%rdi
  call   <free@plt>
L2:
  mov    %rbx,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L7:
  lea    D5(%rip),%rdi
  call   <xt_core_trace@plt>
  jmp    L34
L52:
  mov    $0x2,%edi
  call   <malloc@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L36
  mov    %r12b,(%rax)
  movb   $0x0,0x1(%rax)
  jmp    L2
L36:
  lea    D5(%rip),%rdi
  call   <xt_core_trace@plt>
  jmp    L2
```

tool_calls: name: read_data
arguments: data_label: D5
type_infered: string

tool: D5: malloc

assistant: ```c
char *xt_core_string_append_char(char *original, char addition) {
  assert(addition);
  char *new_string;
  unsigned long original_size;
  unsigned long new_string_size;
  if (original) {
    original_size = strlen(original);
    new_string_size = original_size + 1;
    new_string = malloc(new_string_size + 1);
    if (new_string) {
      memcpy(new_string, original, original_size);
      *(new_string + original_size) = addition;
      *(new_string + new_string_size) = '\0';
    } else {
      xt_core_trace("malloc");
    }
    free(original);
  } else {
    new_string = malloc(1 + 1);
    if (new_string) {
      *(new_string) = addition;
      *(new_string + 1) = '\0';
    } else {
      xt_core_trace("malloc");
    }
  }
  return new_string;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<xt_core_string_append_char>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  movsbl %sil,%edi
  push   %rbx
  mov    %esi,%ebx
  sub    $0x8,%rsp
  call   <assert@plt>
  test   %rbp,%rbp
  je     L12
  mov    %rbp,%rdi
  call   <strlen@plt>
  lea    0x2(%rax),%edi
  mov    %rax,%r13
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L35
  mov    %r13,%rdx
  mov    %rbp,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    %bl,(%r12,%r13,1)
  movb   $0x0,0x1(%r12,%r13,1)
L13:
  mov    %rbp,%rdi
  call   <free@plt>
L24:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L35:
  lea    D10(%rip),%rdi
  call   <xt_core_trace@plt>
  jmp    L13
  xchg   %ax,%ax
L12:
  mov    $0x2,%edi
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L21
  mov    %bl,(%rax)
  movb   $0x0,0x1(%rax)
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  cs nopw 0x0(%rax,%rax,1)
L21:
  lea    D10(%rip),%rdi
  call   <xt_core_trace@plt>
  jmp    L24
```

tool_calls: name: retrieve_data
arguments: label: D10
type: string

tool: D10: malloc

assistant: ```c
char *xt_core_string_append_char(char *original, char addition) {
  assert(addition);
  char *new_string;
  unsigned long original_size;
  unsigned long new_string_size;
  if (original) {
    original_size = strlen(original);
    new_string_size = original_size + 1;
    new_string = malloc(new_string_size + 1);
    if (new_string) {
      memcpy(new_string, original, original_size);
      *(new_string + original_size) = addition;
      *(new_string + new_string_size) = '\0';
    } else {
      xt_core_trace("malloc");
    }
    free(original);
  } else {
    new_string = malloc(1 + 1);
    if (new_string) {
      *(new_string) = addition;
      *(new_string + 1) = '\0';
    } else {
      xt_core_trace("malloc");
    }
  }
  return new_string;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<xt_core_string_append_char>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  movsbl %sil,%edi
  push   %rbx
  mov    %esi,%ebx
  sub    $0x8,%rsp
  call   <assert@plt>
  test   %rbp,%rbp
  je     L41
  mov    %rbp,%rdi
  call   <strlen@plt>
  lea    0x2(%rax),%edi
  mov    %rax,%r13
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L27
  mov    %r13,%rdx
  mov    %rbp,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    %bl,(%r12,%r13,1)
  movb   $0x0,0x1(%r12,%r13,1)
L26:
  mov    %rbp,%rdi
  call   <free@plt>
L33:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L27:
  lea    D19(%rip),%rdi
  call   <xt_core_trace@plt>
  jmp    L26
  xchg   %ax,%ax
L41:
  mov    $0x2,%edi
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L10
  mov    %bl,(%rax)
  movb   $0x0,0x1(%rax)
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  cs nopw 0x0(%rax,%rax,1)
L10:
  lea    D19(%rip),%rdi
  call   <xt_core_trace@plt>
  jmp    L33
```

tool_calls: name: unpack_data
arguments: data_label: D19
datatype: string

tool: D19: malloc

assistant: ```c
char *xt_core_string_append_char(char *original, char addition) {
  assert(addition);
  char *new_string;
  unsigned long original_size;
  unsigned long new_string_size;
  if (original) {
    original_size = strlen(original);
    new_string_size = original_size + 1;
    new_string = malloc(new_string_size + 1);
    if (new_string) {
      memcpy(new_string, original, original_size);
      *(new_string + original_size) = addition;
      *(new_string + new_string_size) = '\0';
    } else {
      xt_core_trace("malloc");
    }
    free(original);
  } else {
    new_string = malloc(1 + 1);
    if (new_string) {
      *(new_string) = addition;
      *(new_string + 1) = '\0';
    } else {
      xt_core_trace("malloc");
    }
  }
  return new_string;
}
```
user: Translate the below assembly code into C programming language:

```asm
<ft_strdup>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <ft_strlen@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  add    $0x1,%rax
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L14
  mov    -0x10(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    -0x18(%rbp),%rcx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <ft_memcpy@plt>
  jmp    L16
L14:
  mov    $0x0,%eax
L16:
  leave
  ret
```

assistant: ```c
char *ft_strdup(const char *s) {
  char *str;
  const size_t len = ft_strlen(s);
  str = malloc(len + 1);
  return ((str) ? (char *)ft_memcpy(str, s, len + 1) : NULL);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<ft_strdup>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  call   <ft_strlen@plt>
  lea    0x1(%rax),%rbp
  mov    %rbp,%rdi
  call   <malloc@plt>
  test   %rax,%rax
  je     L9
  mov    %rbp,%rdx
  mov    %rbx,%rsi
  mov    %rax,%rdi
  call   <ft_memcpy@plt>
L9:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
char *ft_strdup(const char *s) {
  char *str;
  const size_t len = ft_strlen(s);
  str = malloc(len + 1);
  return ((str) ? (char *)ft_memcpy(str, s, len + 1) : NULL);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<ft_strdup>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <ft_strlen@plt>
  lea    0x1(%rax),%r12
  mov    %r12,%rdi
  call   <malloc@plt>
  test   %rax,%rax
  je     L2
  add    $0x8,%rsp
  mov    %r12,%rdx
  mov    %rbp,%rsi
  mov    %rax,%rdi
  pop    %rbp
  pop    %r12
  jmp    1090 <ft_memcpy@plt>
L2:
  pop    %rdx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *ft_strdup(const char *s) {
  char *str;
  const size_t len = ft_strlen(s);
  str = malloc(len + 1);
  return ((str) ? (char *)ft_memcpy(str, s, len + 1) : NULL);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<ft_strdup>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <ft_strlen@plt>
  lea    0x1(%rax),%r12
  mov    %r12,%rdi
  call   <malloc@plt>
  test   %rax,%rax
  je     L2
  add    $0x8,%rsp
  mov    %r12,%rdx
  mov    %rbp,%rsi
  mov    %rax,%rdi
  pop    %rbp
  pop    %r12
  jmp    1090 <ft_memcpy@plt>
L2:
  pop    %rdx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *ft_strdup(const char *s) {
  char *str;
  const size_t len = ft_strlen(s);
  str = malloc(len + 1);
  return ((str) ? (char *)ft_memcpy(str, s, len + 1) : NULL);
}
```