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

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rdx
  mov    %rdx,-0x38(%rbp)
  xor    %edx,%edx
  movl   $0x2710,-0x44(%rbp)
  cmp    $0x1,%edi
  jg     L138
L143:
  mov    %rsp,%rax
L41:
  cmp    %rax,%rsp
  je     L132
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L41
L138:
  mov    0x8(%rsi),%rdi
  call   <atoi@plt>
  mov    %eax,-0x44(%rbp)
  jmp    L143
L132:
  sub    $0x140,%rsp
  orq    $0x0,0x138(%rsp)
  mov    %rsp,%r13
  mov    %rsp,%rax
L91:
  cmp    %rax,%rsp
  je     L3
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L91
L3:
  sub    $0x140,%rsp
  orq    $0x0,0x138(%rsp)
  mov    %rsp,%r14
  mov    $0x1,%eax
L39:
  mov    %rax,-0x8(%r13,%rax,8)
  mov    %rax,%rdx
  add    $0x1,%rax
  mov    %rax,-0x8(%r14,%rdx,8)
  cmp    $0x29,%rax
  jne    L39
  cmpl   $0x0,-0x44(%rbp)
  jle    L146
  mov    $0x0,%r15d
  jmp    L31
L116:
  add    %rdi,%rbx
  mov    %rbx,%rax
  cqto
  idiv   %rsi
  cmp    %rcx,%rdx
  jne    L116
L74:
  call   <lcm@plt>
  mov    %rax,%rdi
  add    $0x8,%r12
  cmp    $0x140,%r12
  je     L118
L67:
  mov    (%r14,%r12,1),%rsi
  mov    0x0(%r13,%r12,1),%rcx
  mov    %rbx,%rax
  cqto
  idiv   %rsi
  cmp    %rcx,%rdx
  jne    L116
  jmp    L74
L118:
  add    $0x1,%r15d
  cmp    %r15d,-0x44(%rbp)
  je     L146
L31:
  mov    (%r14),%rdi
  mov    0x0(%r13),%rbx
  mov    $0x8,%r12d
  jmp    L67
L146:
  mov    %rbx,%rsi
  lea    D9(%rip),%rdi
  call   <printf@plt>
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L102
  mov    $0x0,%eax
  lea    -0x28(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
L102:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D9
type: string

tool: D9: %ld


assistant: ```c
int main(int argc, char *argv[]) {
  int rep = 10000;
  if (argc >= 2) {
    rep = atoi(argv[1]);
  }
  int n = 40;
  long residual[n];
  long divisor[n];
  int i;
  for (i = 0; i < n; i++) {
    residual[i] = i + 1;
    divisor[i] = i + 2;
  }
  long a, r;
  int iter;
  for (iter = 0; iter < rep; iter++) {
    a = divisor[0];
    r = residual[0];
    for (i = 1; i < n; i++) {
      while (r % divisor[i] != residual[i]) {
        r = r + a;
      }
      a = lcm(a, divisor[i]);
    }
  }
  printf("%ld\n", r);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rdx
  mov    %rdx,-0x38(%rbp)
  xor    %edx,%edx
  movl   $0x2710,-0x44(%rbp)
  cmp    $0x1,%edi
  jle    L1
  mov    0x8(%rsi),%rdi
  call   <atoi@plt>
  mov    %eax,-0x44(%rbp)
L1:
  mov    %rsp,%rax
  cmp    %rax,%rsp
  je     L5
L115:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rax,%rsp
  jne    L115
L5:
  sub    $0x140,%rsp
  orq    $0x0,0x138(%rsp)
  lea    0x7(%rsp),%rbx
  mov    %rsp,%rax
  mov    %rbx,%rcx
  and    $0xfffffffffffffff8,%rbx
  shr    $0x3,%rcx
L25:
  cmp    %rax,%rsp
  je     L99
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L25
L99:
  sub    $0x140,%rsp
  orq    $0x0,0x138(%rsp)
  mov    $0x1,%eax
  lea    0x7(%rsp),%r12
  mov    %r12,%rsi
  and    $0xfffffffffffffff8,%r12
  shr    $0x3,%rsi
  nopl   0x0(%rax,%rax,1)
L52:
  mov    %rax,%rdx
  mov    %rax,-0x8(%rbx,%rax,8)
  add    $0x1,%rax
  mov    %rax,-0x8(%r12,%rdx,8)
  cmp    $0x29,%rax
  jne    L52
  mov    -0x44(%rbp),%eax
  test   %eax,%eax
  jle    L102
  mov    0x0(,%rsi,8),%rax
  xor    %r13d,%r13d
  mov    %rax,-0x58(%rbp)
  mov    0x0(,%rcx,8),%rax
  mov    %rax,-0x50(%rbp)
L14:
  mov    -0x50(%rbp),%r15
  mov    -0x58(%rbp),%rdi
  mov    $0x8,%r14d
  nopw   0x0(%rax,%rax,1)
L81:
  mov    (%r12,%r14,1),%rsi
  mov    (%rbx,%r14,1),%r9
  jmp    L18
  nopw   0x0(%rax,%rax,1)
L88:
  add    %rdi,%r15
L18:
  mov    %r15,%rax
  cqto
  idiv   %rsi
  cmp    %r9,%rdx
  jne    L88
  call   <lcm@plt>
  add    $0x8,%r14
  mov    %rax,%rdi
  cmp    $0x140,%r14
  jne    L81
  add    $0x1,%r13d
  cmp    %r13d,-0x44(%rbp)
  jne    L14
L102:
  mov    %r15,%rsi
  lea    D4(%rip),%rdi
  call   <printf@plt>
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L58
  lea    -0x28(%rbp),%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
L58:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D4
type_guessed: string

tool: D4: %ld


assistant: ```c
int main(int argc, char *argv[]) {
  int rep = 10000;
  if (argc >= 2) {
    rep = atoi(argv[1]);
  }
  int n = 40;
  long residual[n];
  long divisor[n];
  int i;
  for (i = 0; i < n; i++) {
    residual[i] = i + 1;
    divisor[i] = i + 2;
  }
  long a, r;
  int iter;
  for (iter = 0; iter < rep; iter++) {
    a = divisor[0];
    r = residual[0];
    for (i = 1; i < n; i++) {
      while (r % divisor[i] != residual[i]) {
        r = r + a;
      }
      a = lcm(a, divisor[i]);
    }
  }
  printf("%ld\n", r);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<addToCurTime>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x20(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <gettimeofday@plt>
  mov    -0x18(%rbp),%rsi
  mov    -0x30(%rbp),%rcx
  movabs $0x20c49ba5e353f7cf,%rdx
  mov    %rcx,%rax
  imul   %rdx
  mov    %rdx,%rax
  sar    $0x7,%rax
  sar    $0x3f,%rcx
  mov    %rcx,%rdx
  sub    %rdx,%rax
  add    %rsi,%rax
  mov    %rax,-0x18(%rbp)
  mov    -0x20(%rbp),%rsi
  mov    -0x30(%rbp),%rcx
  movabs $0x20c49ba5e353f7cf,%rdx
  mov    %rcx,%rax
  imul   %rdx
  mov    %rdx,%rax
  sar    $0x7,%rax
  mov    %rcx,%rdx
  sar    $0x3f,%rdx
  sub    %rdx,%rax
  imul   $0x3e8,%rax,%rdx
  mov    %rcx,%rax
  sub    %rdx,%rax
  imul   $0x3e8,%rax,%rax
  add    %rsi,%rax
  mov    %rax,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  cmp    $0xf423f,%rax
  jle    L7
  mov    -0x18(%rbp),%rax
  add    $0x1,%rax
  mov    %rax,-0x18(%rbp)
  mov    -0x20(%rbp),%rax
  sub    $0xf4240,%rax
  mov    %rax,-0x20(%rbp)
L7:
  mov    -0x18(%rbp),%rdx
  mov    -0x28(%rbp),%rax
  mov    %rdx,0x8(%rax)
  mov    -0x20(%rbp),%rax
  imul   $0x3e8,%rax,%rdx
  mov    -0x28(%rbp),%rax
  mov    %rdx,(%rax)
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L18
  call   <__stack_chk_fail@plt>
L18:
  leave
  ret
```

assistant: ```c
void addToCurTime(struct timespec *ts, long timediff_ms) {
  struct timeval tv;
  gettimeofday(&tv, 0);
  tv.tv_sec += timediff_ms / 1000;
  tv.tv_usec += (timediff_ms % 1000) * 1000;
  if (tv.tv_usec >= 1000000) {
    tv.tv_sec++;
    tv.tv_usec -= 1000000;
  }
  ts->tv_sec = tv.tv_sec;
  ts->tv_nsec = tv.tv_usec * 1000L;
}
```
user: Translate the below assembly code into C programming language:

```asm
<addToCurTime>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,%rbp
  mov    %rsi,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  mov    $0x0,%esi
  call   <gettimeofday@plt>
  movabs $0x20c49ba5e353f7cf,%rdx
  mov    %rbx,%rax
  imul   %rdx
  sar    $0x7,%rdx
  mov    %rbx,%rax
  sar    $0x3f,%rax
  sub    %rax,%rdx
  mov    %rdx,%rax
  add    0x8(%rsp),%rax
  mov    %rax,0x8(%rsp)
  imul   $0x3e8,%rdx,%rdx
  sub    %rdx,%rbx
  imul   $0x3e8,%rbx,%rbx
  add    (%rsp),%rbx
  mov    %rbx,(%rsp)
  cmp    $0xf423f,%rbx
  jle    L16
  add    $0x1,%rax
  mov    %rax,0x8(%rsp)
  sub    $0xf4240,%rbx
  mov    %rbx,(%rsp)
L16:
  mov    0x8(%rsp),%rax
  mov    %rax,0x8(%rbp)
  imul   $0x3e8,(%rsp),%rax
  mov    %rax,0x0(%rbp)
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L19
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  ret
L19:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void addToCurTime(struct timespec *ts, long timediff_ms) {
  struct timeval tv;
  gettimeofday(&tv, 0);
  tv.tv_sec += timediff_ms / 1000;
  tv.tv_usec += (timediff_ms % 1000) * 1000;
  if (tv.tv_usec >= 1000000) {
    tv.tv_sec++;
    tv.tv_usec -= 1000000;
  }
  ts->tv_sec = tv.tv_sec;
  ts->tv_nsec = tv.tv_usec * 1000L;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<addToCurTime>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rsi,%rbx
  xor    %esi,%esi
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  call   <gettimeofday@plt>
  mov    %rbx,%rax
  movabs $0x20c49ba5e353f7cf,%rdx
  imul   %rdx
  mov    %rbx,%rax
  sar    $0x3f,%rax
  sar    $0x7,%rdx
  sub    %rax,%rdx
  mov    0x8(%rsp),%rax
  add    %rdx,%rax
  imul   $0x3e8,%rdx,%rdx
  sub    %rdx,%rbx
  imul   $0x3e8,%rbx,%rbx
  add    (%rsp),%rbx
  cmp    $0xf423f,%rbx
  jle    L17
  add    $0x1,%rax
  sub    $0xf4240,%rbx
L17:
  imul   $0x3e8,%rbx,%rbx
  mov    %rax,0x8(%rbp)
  mov    %rbx,0x0(%rbp)
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L12
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  ret
L12:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void addToCurTime(struct timespec *ts, long timediff_ms) {
  struct timeval tv;
  gettimeofday(&tv, 0);
  tv.tv_sec += timediff_ms / 1000;
  tv.tv_usec += (timediff_ms % 1000) * 1000;
  if (tv.tv_usec >= 1000000) {
    tv.tv_sec++;
    tv.tv_usec -= 1000000;
  }
  ts->tv_sec = tv.tv_sec;
  ts->tv_nsec = tv.tv_usec * 1000L;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<addToCurTime>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rsi,%rbx
  xor    %esi,%esi
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  call   <gettimeofday@plt>
  mov    %rbx,%rax
  movabs $0x20c49ba5e353f7cf,%rdx
  imul   %rdx
  mov    %rbx,%rax
  sar    $0x3f,%rax
  sar    $0x7,%rdx
  sub    %rax,%rdx
  mov    0x8(%rsp),%rax
  add    %rdx,%rax
  imul   $0x3e8,%rdx,%rdx
  sub    %rdx,%rbx
  imul   $0x3e8,%rbx,%rbx
  add    (%rsp),%rbx
  cmp    $0xf423f,%rbx
  jle    L15
  add    $0x1,%rax
  sub    $0xf4240,%rbx
L15:
  imul   $0x3e8,%rbx,%rcx
  movq   %rax,%xmm1
  movq   %rcx,%xmm0
  punpcklqdq %xmm1,%xmm0
  movups %xmm0,0x0(%rbp)
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void addToCurTime(struct timespec *ts, long timediff_ms) {
  struct timeval tv;
  gettimeofday(&tv, 0);
  tv.tv_sec += timediff_ms / 1000;
  tv.tv_usec += (timediff_ms % 1000) * 1000;
  if (tv.tv_usec >= 1000000) {
    tv.tv_sec++;
    tv.tv_usec -= 1000000;
  }
  ts->tv_sec = tv.tv_sec;
  ts->tv_nsec = tv.tv_usec * 1000L;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main(int argc, char **args) {
  (void)argc;
  (void)args;
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

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

assistant: ```c
int main(int argc, char **args) {
  (void)argc;
  (void)args;
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int main(int argc, char **args) {
  (void)argc;
  (void)args;
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int main(int argc, char **args) {
  (void)argc;
  (void)args;
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<ftoi_func>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <value_as_float@plt>
  leave
  ret
```

assistant: ```c
unsigned int ftoi_func(unsigned int param1, unsigned int param2) {
  (void)param2;
  return (unsigned int)(int)value_as_float(param1);
}
```
user: Demonstrate the C code that generates the assembly output below:

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

assistant: ```c
unsigned int ftoi_func(unsigned int param1, unsigned int param2) {
  (void)param2;
  return (unsigned int)(int)value_as_float(param1);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

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

assistant: ```c
unsigned int ftoi_func(unsigned int param1, unsigned int param2) {
  (void)param2;
  return (unsigned int)(int)value_as_float(param1);
}
```
user: 请将以下的汇编代码转换成C语言代码：

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

assistant: ```c
unsigned int ftoi_func(unsigned int param1, unsigned int param2) {
  (void)param2;
  return (unsigned int)(int)value_as_float(param1);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<productSum>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x38(%rbp)
  movl   $0x0,-0x34(%rbp)
  mov    $0x50,%edi
  call   <malloc@plt>
  mov    %rax,-0x28(%rbp)
  movl   $0x1,-0x40(%rbp)
  jmp    L75
L76:
  movl   $0x1,-0x3c(%rbp)
  jmp    L80
L81:
  mov    -0x40(%rbp),%eax
  imul   -0x3c(%rbp),%eax
  mov    %eax,-0x30(%rbp)
  cmpl   $0x2694,-0x30(%rbp)
  jg     L50
  mov    -0x40(%rbp),%eax
  mov    %eax,-0x20(%rbp)
  mov    -0x3c(%rbp),%eax
  mov    %eax,-0x1c(%rbp)
  mov    -0x30(%rbp),%eax
  mov    %eax,-0x18(%rbp)
  movl   $0x0,-0x14(%rbp)
  lea    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <concat_ints@plt>
  mov    %eax,-0x2c(%rbp)
  mov    -0x2c(%rbp),%eax
  mov    %eax,%edi
  call   <is_pandigit@plt>
  test   %rax,%rax
  je     L82
  mov    -0x28(%rbp),%rdx
  mov    -0x30(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <in@plt>
  test   %eax,%eax
  jne    L82
  mov    -0x34(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x34(%rbp)
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rax,%rdx
  mov    -0x30(%rbp),%eax
  mov    %eax,(%rdx)
  mov    -0x30(%rbp),%eax
  add    %eax,-0x38(%rbp)
  jmp    L82
L50:
  nop
L82:
  addl   $0x1,-0x3c(%rbp)
L80:
  cmpl   $0x7cf,-0x3c(%rbp)
  jle    L81
  addl   $0x1,-0x40(%rbp)
L75:
  cmpl   $0x7cf,-0x40(%rbp)
  jle    L76
  mov    -0x38(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L29
  call   <__stack_chk_fail@plt>
L29:
  leave
  ret
```

assistant: ```c
uint32_t productSum(void) {
  uint32_t i, j, product, concat, productSum;
  productSum = 0;
  uint32_t *productList;
  uint32_t index = 0;
  productList = malloc(20 * sizeof(uint32_t));
  for (i = 1; i < 2000; i++) {
    for (j = 1; j < 2000; j++) {
      product = i * j;
      if (product > 9876) {
        continue;
      }
      uint32_t tmp[4] = {i, j, product, 0};
      concat = concat_ints(tmp);
      if (is_pandigit(concat)) {
        if (!in(product, productList)) {
          productList[index++] = product;
          productSum += product;
        }
      }
    }
  }
  return productSum;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<productSum>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    $0x50,%edi
  call   <malloc@plt>
  mov    %rax,(%rsp)
  mov    $0x0,%r15d
  movl   $0x0,0xc(%rsp)
  mov    $0x1,%r13d
  lea    0x10(%rsp),%r14
  jmp    L51
L64:
  add    $0x1,%ebp
  add    0x8(%rsp),%ebx
  cmp    $0x7d0,%ebp
  je     L23
L8:
  cmp    $0x2694,%ebx
  jg     L64
  mov    %r13d,0x10(%rsp)
  mov    %ebp,0x14(%rsp)
  mov    %ebx,0x18(%rsp)
  movl   $0x0,0x1c(%rsp)
  mov    %r14,%rdi
  call   <concat_ints@plt>
  mov    %eax,%edi
  call   <is_pandigit@plt>
  test   %rax,%rax
  je     L64
  mov    (%rsp),%rsi
  mov    %ebx,%edi
  call   <in@plt>
  test   %eax,%eax
  jne    L64
  movslq %r15d,%rax
  mov    (%rsp),%rdx
  mov    %ebx,(%rdx,%rax,4)
  add    %ebx,0xc(%rsp)
  lea    0x1(%r15),%r15d
  jmp    L64
L23:
  add    $0x1,%r13d
  cmp    $0x7d0,%r13d
  je     L52
L51:
  mov    %r13d,0x8(%rsp)
  mov    %r13d,%ebx
  mov    $0x1,%ebp
  jmp    L8
L52:
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L87
  mov    0xc(%rsp),%eax
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L87:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
uint32_t productSum(void) {
  uint32_t i, j, product, concat, productSum;
  productSum = 0;
  uint32_t *productList;
  uint32_t index = 0;
  productList = malloc(20 * sizeof(uint32_t));
  for (i = 1; i < 2000; i++) {
    for (j = 1; j < 2000; j++) {
      product = i * j;
      if (product > 9876) {
        continue;
      }
      uint32_t tmp[4] = {i, j, product, 0};
      concat = concat_ints(tmp);
      if (is_pandigit(concat)) {
        if (!in(product, productList)) {
          productList[index++] = product;
          productSum += product;
        }
      }
    }
  }
  return productSum;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<productSum>:
  endbr64
  push   %r15
  mov    $0x50,%edi
  mov    $0x1,%r15d
  push   %r14
  xor    %r14d,%r14d
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%r12
  call   <malloc@plt>
  movl   $0x0,0xc(%rsp)
  mov    %rax,%r13
L18:
  mov    %r15d,%ebx
  mov    $0x1,%ebp
  jmp    L48
  nop
L60:
  add    $0x1,%ebp
  add    %r15d,%ebx
  cmp    $0x7d0,%ebp
  je     L10
L48:
  cmp    $0x2694,%ebx
  jg     L60
  mov    %r12,%rdi
  mov    %r15d,0x10(%rsp)
  mov    %ebp,0x14(%rsp)
  mov    %ebx,0x18(%rsp)
  movl   $0x0,0x1c(%rsp)
  call   <concat_ints@plt>
  mov    %eax,%edi
  call   <is_pandigit@plt>
  test   %rax,%rax
  je     L60
  mov    %r13,%rsi
  mov    %ebx,%edi
  call   <in@plt>
  test   %eax,%eax
  jne    L60
  movslq %r14d,%rax
  add    $0x1,%ebp
  add    %ebx,0xc(%rsp)
  add    $0x1,%r14d
  mov    %ebx,0x0(%r13,%rax,4)
  add    %r15d,%ebx
  cmp    $0x7d0,%ebp
  jne    L48
L10:
  add    $0x1,%r15d
  cmp    $0x7d0,%r15d
  jne    L18
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L54
  mov    0xc(%rsp),%eax
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L54:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
uint32_t productSum(void) {
  uint32_t i, j, product, concat, productSum;
  productSum = 0;
  uint32_t *productList;
  uint32_t index = 0;
  productList = malloc(20 * sizeof(uint32_t));
  for (i = 1; i < 2000; i++) {
    for (j = 1; j < 2000; j++) {
      product = i * j;
      if (product > 9876) {
        continue;
      }
      uint32_t tmp[4] = {i, j, product, 0};
      concat = concat_ints(tmp);
      if (is_pandigit(concat)) {
        if (!in(product, productList)) {
          productList[index++] = product;
          productSum += product;
        }
      }
    }
  }
  return productSum;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<productSum>:
  endbr64
  push   %r15
  mov    $0x50,%edi
  push   %r14
  xor    %r14d,%r14d
  push   %r13
  push   %r12
  push   %rbp
  mov    $0x1,%ebp
  push   %rbx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%r12
  call   <malloc@plt>
  movl   $0x0,0xc(%rsp)
  mov    %rax,%r13
L26:
  mov    %ebp,%r15d
  mov    $0x1,%ebx
  jmp    L20
  xchg   %ax,%ax
L69:
  add    $0x1,%ebx
  add    %ebp,%r15d
  cmp    $0x7d0,%ebx
  je     L4
L20:
  movd   %ebp,%xmm0
  movd   %ebx,%xmm1
  punpckldq %xmm1,%xmm0
  cmp    $0x2694,%r15d
  jg     L69
  mov    %r12,%rdi
  mov    %r15d,0x18(%rsp)
  movl   $0x0,0x1c(%rsp)
  movq   %xmm0,0x10(%rsp)
  call   <concat_ints@plt>
  mov    %eax,%edi
  call   <is_pandigit@plt>
  test   %rax,%rax
  je     L69
  mov    %r13,%rsi
  mov    %r15d,%edi
  call   <in@plt>
  test   %eax,%eax
  jne    L69
  movslq %r14d,%rax
  add    $0x1,%ebx
  add    %r15d,0xc(%rsp)
  add    $0x1,%r14d
  mov    %r15d,0x0(%r13,%rax,4)
  add    %ebp,%r15d
  cmp    $0x7d0,%ebx
  jne    L20
L4:
  add    $0x1,%ebp
  cmp    $0x7d0,%ebp
  jne    L26
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L50
  mov    0xc(%rsp),%eax
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L50:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
uint32_t productSum(void) {
  uint32_t i, j, product, concat, productSum;
  productSum = 0;
  uint32_t *productList;
  uint32_t index = 0;
  productList = malloc(20 * sizeof(uint32_t));
  for (i = 1; i < 2000; i++) {
    for (j = 1; j < 2000; j++) {
      product = i * j;
      if (product > 9876) {
        continue;
      }
      uint32_t tmp[4] = {i, j, product, 0};
      concat = concat_ints(tmp);
      if (is_pandigit(concat)) {
        if (!in(product, productList)) {
          productList[index++] = product;
          productSum += product;
        }
      }
    }
  }
  return productSum;
}
```
user: What is the c source code of the assembly code below?

```asm
<remove_dup>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x440,%rsp
  mov    %rdi,-0x438(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x438(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x41c(%rbp)
  mov    -0x41c(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x418(%rbp)
  lea    -0x410(%rbp),%rax
  mov    $0x400,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  movl   $0x0,-0x424(%rbp)
  movl   $0x0,-0x420(%rbp)
  jmp    L0
L1:
  mov    -0x424(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x438(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  cltq
  mov    -0x410(%rbp,%rax,4),%eax
  test   %eax,%eax
  jne    L8
  mov    -0x424(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x438(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0x420(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x420(%rbp)
  movslq %eax,%rdx
  mov    -0x418(%rbp),%rax
  add    %rax,%rdx
  movzbl (%rcx),%eax
  mov    %al,(%rdx)
  mov    -0x424(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x438(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  movslq %eax,%rdx
  mov    -0x410(%rbp,%rdx,4),%edx
  add    $0x1,%edx
  cltq
  mov    %edx,-0x410(%rbp,%rax,4)
L8:
  addl   $0x1,-0x424(%rbp)
L0:
  mov    -0x424(%rbp),%eax
  cmp    -0x41c(%rbp),%eax
  jl     L1
  mov    -0x418(%rbp),%rax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L27
  call   <__stack_chk_fail@plt>
L27:
  leave
  ret
```

assistant: ```c
char *remove_dup(char str[]) {
  int len = strlen(str);
  char *res = (char *)malloc(len + 1);
  int hit[256];
  memset(hit, 0, sizeof(hit));
  for (int i = 0, end = 0; i < len; i++) {
    if (hit[str[i]] == 0) {
      res[end++] = str[i];
      hit[str[i]]++;
    }
  }
  return res;
}
```