ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Translate the below assembly code into C programming language:

```asm
<expectationToProbability>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movsd  %xmm0,-0x8(%rbp)
  movsd  D14(%rip),%xmm0
  comisd -0x8(%rbp),%xmm0
  jbe    L3
  movsd  -0x8(%rbp),%xmm0
  jmp    L19
L3:
  movsd  D14(%rip),%xmm0
L19:
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  pop    %rbp
  ret
```

tool_calls: name: read_data
arguments: data_tag: D14
type: f64

tool: D14: 0.999

assistant: ```c
double expectationToProbability(double e) {
  if (e < 0.999)
    return e;
  else
    return 0.999;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<expectationToProbability>:
  endbr64
  movsd  D1(%rip),%xmm1
  comisd %xmm0,%xmm1
  ja     L4
  movapd %xmm1,%xmm0
L4:
  ret
```

tool_calls: name: read_data
arguments: tag: D1
type: f64

tool: D1: 0.999

assistant: ```c
double expectationToProbability(double e) {
  if (e < 0.999)
    return e;
  else
    return 0.999;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<expectationToProbability>:
  endbr64
  minsd  D1(%rip),%xmm0
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D1
datatype: f64

tool: D1: 0.999

assistant: ```c
double expectationToProbability(double e) {
  if (e < 0.999)
    return e;
  else
    return 0.999;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<expectationToProbability>:
  endbr64
  minsd  D7(%rip),%xmm0
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D7
datatype: f64

tool: D7: 0.999

assistant: ```c
double expectationToProbability(double e) {
  if (e < 0.999)
    return e;
  else
    return 0.999;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<muggle_str_toll>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  cmpq   $0x0,-0x18(%rbp)
  je     L59
  cmpq   $0x0,-0x20(%rbp)
  jne    L99
L59:
  mov    $0x0,%eax
  jmp    L74
L99:
  mov    D0(%rip),%rax
  movq   $0x0,(%rax)
  mov    -0x24(%rbp),%edx
  lea    -0x10(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strtoll@plt>
  mov    -0x20(%rbp),%rdx
  mov    %rax,(%rdx)
  mov    -0x10(%rbp),%rax
  cmp    %rax,-0x18(%rbp)
  jne    L80
  mov    $0x0,%eax
  jmp    L74
L80:
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L17
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <muggle_str_lstrip_idx@plt>
  cmp    $0xffffffff,%eax
  je     L17
  mov    $0x0,%eax
  jmp    L74
L17:
  mov    -0x20(%rbp),%rax
  mov    (%rax),%rdx
  mov    D11(%rip),%rax
  mov    (%rax),%rax
  cmp    %rax,%rdx
  je     L38
  mov    -0x20(%rbp),%rax
  mov    (%rax),%rdx
  mov    D24(%rip),%rax
  mov    (%rax),%rax
  cmp    %rax,%rdx
  jne    L62
L38:
  mov    $0x0,%eax
  jmp    L74
L62:
  mov    $0x1,%eax
L74:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L50
  call   <__stack_chk_fail@plt>
L50:
  leave
  ret
```

assistant: ```c
int muggle_str_toll(const char *str, long long *pval, int base) {
  char *endptr;
  if (str == NULL || pval == NULL) {
    return 0;
  }
  errno = 0;
  *pval = strtoll(str, &endptr, base);
  if (endptr == str) {
    return 0;
  } else if (*endptr != '\0') {
    if (muggle_str_lstrip_idx(endptr) != -1) {
      return 0;
    }
  }
  if (*pval == LLONG_MAX || *pval == LLONG_MIN) {
    return 0;
  }
  return 1;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<muggle_str_toll>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  test   %rdi,%rdi
  je     L98
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  test   %rsi,%rsi
  je     L98
  mov    D2(%rip),%rax
  movq   $0x0,(%rax)
  mov    %rsp,%rsi
  call   <strtoll@plt>
  mov    %rax,0x0(%rbp)
  mov    (%rsp),%rdi
  cmp    %rbx,%rdi
  je     L78
  cmpb   $0x0,(%rdi)
  jne    L0
L16:
  mov    0x0(%rbp),%rdx
  mov    $0x0,%eax
  mov    D4(%rip),%rcx
  cmp    (%rcx),%rdx
  je     L43
  mov    D10(%rip),%rax
  cmp    %rdx,(%rax)
  setne  %al
  movzbl %al,%eax
  jmp    L43
L0:
  call   <muggle_str_lstrip_idx@plt>
  mov    %eax,%edx
  mov    $0x0,%eax
  cmp    $0xffffffff,%edx
  jne    L43
  jmp    L16
L98:
  mov    $0x0,%eax
L43:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L77
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L78:
  mov    $0x0,%eax
  jmp    L43
L77:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int muggle_str_toll(const char *str, long long *pval, int base) {
  char *endptr;
  if (str == NULL || pval == NULL) {
    return 0;
  }
  errno = 0;
  *pval = strtoll(str, &endptr, base);
  if (endptr == str) {
    return 0;
  } else if (*endptr != '\0') {
    if (muggle_str_lstrip_idx(endptr) != -1) {
      return 0;
    }
  }
  if (*pval == LLONG_MAX || *pval == LLONG_MIN) {
    return 0;
  }
  return 1;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<muggle_str_toll>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  test   %rdi,%rdi
  je     L91
  mov    %rsi,%rbp
  test   %rsi,%rsi
  je     L91
  mov    D17(%rip),%rax
  mov    %rsp,%rsi
  mov    %rdi,%rbx
  movq   $0x0,(%rax)
  call   <strtoll@plt>
  mov    (%rsp),%rdi
  mov    %rax,0x0(%rbp)
  cmp    %rbx,%rdi
  je     L91
  cmpb   $0x0,(%rdi)
  jne    L1
L49:
  mov    D22(%rip),%rdx
  cmp    %rax,(%rdx)
  je     L91
  mov    D23(%rip),%rdx
  cmp    %rax,(%rdx)
  je     L91
  mov    $0x1,%eax
  jmp    L21
  nopl   0x0(%rax)
L91:
  xor    %eax,%eax
L21:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L89
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax)
L1:
  call   <muggle_str_lstrip_idx@plt>
  cmp    $0xffffffff,%eax
  jne    L91
  mov    0x0(%rbp),%rax
  jmp    L49
L89:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int muggle_str_toll(const char *str, long long *pval, int base) {
  char *endptr;
  if (str == NULL || pval == NULL) {
    return 0;
  }
  errno = 0;
  *pval = strtoll(str, &endptr, base);
  if (endptr == str) {
    return 0;
  } else if (*endptr != '\0') {
    if (muggle_str_lstrip_idx(endptr) != -1) {
      return 0;
    }
  }
  if (*pval == LLONG_MAX || *pval == LLONG_MIN) {
    return 0;
  }
  return 1;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<muggle_str_toll>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  test   %rdi,%rdi
  je     L16
  mov    %rsi,%rbp
  test   %rsi,%rsi
  je     L16
  mov    D0(%rip),%rax
  mov    %rsp,%rsi
  mov    %rdi,%rbx
  movq   $0x0,(%rax)
  call   <strtoll@plt>
  mov    (%rsp),%rdi
  mov    %rax,0x0(%rbp)
  cmp    %rbx,%rdi
  je     L16
  cmpb   $0x0,(%rdi)
  jne    L3
L15:
  mov    D15(%rip),%rdx
  cmp    %rax,(%rdx)
  je     L16
  mov    D18(%rip),%rdx
  cmp    %rax,(%rdx)
  je     L16
  mov    $0x1,%eax
  jmp    L23
  nopl   0x0(%rax)
L16:
  xor    %eax,%eax
L23:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L7
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax)
L3:
  call   <muggle_str_lstrip_idx@plt>
  cmp    $0xffffffff,%eax
  jne    L16
  mov    0x0(%rbp),%rax
  jmp    L15
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int muggle_str_toll(const char *str, long long *pval, int base) {
  char *endptr;
  if (str == NULL || pval == NULL) {
    return 0;
  }
  errno = 0;
  *pval = strtoll(str, &endptr, base);
  if (endptr == str) {
    return 0;
  } else if (*endptr != '\0') {
    if (muggle_str_lstrip_idx(endptr) != -1) {
      return 0;
    }
  }
  if (*pval == LLONG_MAX || *pval == LLONG_MIN) {
    return 0;
  }
  return 1;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %edi,-0x34(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  cmpl   $0x3,-0x34(%rbp)
  je     L14
  mov    -0x40(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,%rsi
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0xffffffff,%edi
  call   <exit@plt>
L14:
  mov    -0x40(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rcx
  lea    -0x20(%rbp),%rax
  mov    $0x10,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strncpy@plt>
  mov    -0x40(%rbp),%rax
  add    $0x10,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <atoi@plt>
  mov    %eax,-0x24(%rbp)
  mov    -0x24(%rbp),%edx
  lea    -0x20(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <startTcpServer@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L6
  call   <__stack_chk_fail@plt>
L6:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: label: D4
datatype: string

tool: D4: usage: %s ip port


assistant: ```c
int main(int argc, char **argv) {
  char ip[16];
  int port;
  if (argc != 3) {
    printf("usage: %s ip port\n", argv[0]);
    exit(-1);
  }
  strncpy(ip, argv[1], 16);
  port = atoi(argv[2]);
  startTcpServer(ip, port);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rsi,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  cmp    $0x3,%edi
  jne    L7
L16:
  mov    0x8(%rbx),%rsi
  mov    %rsp,%rbp
  mov    $0x10,%edx
  mov    %rbp,%rdi
  call   <strncpy@plt>
  mov    0x10(%rbx),%rdi
  call   <atoi@plt>
  mov    %eax,%esi
  mov    %rbp,%rdi
  call   <startTcpServer@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L21
  mov    $0x0,%eax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  ret
L7:
  mov    (%rsi),%rsi
  lea    D0(%rip),%rdi
  call   <printf@plt>
  mov    $0xffffffff,%edi
  call   <exit@plt>
  jmp    L16
L21:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D0
type_guessed: string

tool: D0: usage: %s ip port


assistant: ```c
int main(int argc, char **argv) {
  char ip[16];
  int port;
  if (argc != 3) {
    printf("usage: %s ip port\n", argv[0]);
    exit(-1);
  }
  strncpy(ip, argv[1], 16);
  port = atoi(argv[2]);
  startTcpServer(ip, port);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  cmp    $0x3,%edi
  je     L11
  mov    (%rsi),%rsi
  lea    D5(%rip),%rdi
  call   <printf@plt>
  or     $0xffffffff,%edi
  call   <exit@plt>
L11:
  mov    0x8(%rbx),%rsi
  mov    %rsp,%rbp
  mov    $0x10,%edx
  mov    %rbp,%rdi
  call   <strncpy@plt>
  mov    0x10(%rbx),%rdi
  call   <atoi@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  call   <startTcpServer@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L13
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L13:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D5
data_type: string

tool: D5: usage: %s ip port


assistant: ```c
int main(int argc, char **argv) {
  char ip[16];
  int port;
  if (argc != 3) {
    printf("usage: %s ip port\n", argv[0]);
    exit(-1);
  }
  strncpy(ip, argv[1], 16);
  port = atoi(argv[2]);
  startTcpServer(ip, port);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  cmp    $0x3,%edi
  je     L17
  mov    (%rsi),%rsi
  lea    D5(%rip),%rdi
  call   <printf@plt>
  or     $0xffffffff,%edi
  call   <exit@plt>
L17:
  mov    0x8(%rbx),%rsi
  mov    %rsp,%rbp
  mov    $0x10,%edx
  mov    %rbp,%rdi
  call   <strncpy@plt>
  mov    0x10(%rbx),%rdi
  call   <atoi@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  call   <startTcpServer@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L12
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L12:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D5
data_type: string

tool: D5: usage: %s ip port


assistant: ```c
int main(int argc, char **argv) {
  char ip[16];
  int port;
  if (argc != 3) {
    printf("usage: %s ip port\n", argv[0]);
    exit(-1);
  }
  strncpy(ip, argv[1], 16);
  port = atoi(argv[2]);
  startTcpServer(ip, port);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<problem10>:
  endbr64
  push   %rbx
  mov    $0x7a1200,%edi
  call   <malloc@plt>
  mov    %rax,%rbx
  mov    $0x1e8480,%edi
  call   <sqrt@plt>
  mov    %eax,%r9d
  mov    $0x0,%eax
L107:
  mov    %eax,(%rbx,%rax,4)
  add    $0x1,%rax
  cmp    $0x1e8480,%rax
  jne    L107
  cmp    $0x1,%r9d
  jle    L17
  mov    $0x2,%edi
  mov    $0x1e847e,%r11d
  lea    0x8(%rbx),%r10
  jmp    L42
L108:
  add    $0x4,%rsi
  cmp    %r8,%rsi
  je     L15
L99:
  mov    (%rsi),%eax
  cltd
  idiv   %edi
  test   %edx,%edx
  jne    L108
  movl   $0x0,(%rsi)
  jmp    L108
L15:
  cmpl   $0x0,(%rbx,%rcx,4)
  jne    L78
  add    $0x1,%rcx
  cmp    $0x1e847f,%ecx
  jle    L15
  jmp    L38
L78:
  mov    %ecx,%edi
L38:
  cmp    %edi,%r9d
  jl     L17
L42:
  lea    0x1(%rdi),%ecx
  cmp    $0x1e847f,%ecx
  jg     L38
  movslq %ecx,%rcx
  lea    (%rbx,%rcx,4),%rsi
  mov    %r11d,%eax
  sub    %edi,%eax
  movslq %edi,%rdx
  add    %rdx,%rax
  lea    (%r10,%rax,4),%r8
  jmp    L99
L17:
  lea    0x8(%rbx),%rax
  lea    0x7a1200(%rbx),%rsi
  mov    $0x0,%edx
L30:
  movslq (%rax),%rcx
  add    %rcx,%rdx
  add    $0x4,%rax
  cmp    %rsi,%rax
  jne    L30
  mov    $0x1e8480,%esi
  lea    D2(%rip),%rdi
  call   <printf@plt>
  mov    %rbx,%rdi
  call   <free@plt>
  mov    $0x0,%eax
  pop    %rbx
  ret
```

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

tool: D2: The sum of the all primes under %d is %ld


assistant: ```c
int problem10(void) {
  int num_primes = 2000000;
  int *primes = malloc(sizeof *primes * num_primes);
  int i, j, sr = sqrt(num_primes) + 1;
  long sum;
  for (i = 0; i < num_primes; i++)
    primes[i] = i;
  for (i = 2; i < sr;) {
    for (j = i + 1; j < num_primes; j++) {
      if (primes[j] % i == 0) {
        primes[j] = 0;
      }
    }
    for (j = i + 1; j < num_primes; j++) {
      if (primes[j] != 0) {
        i = j;
        break;
      }
    }
  }
  for (sum = 0, i = 2; i < num_primes; i++) {
    if (primes[i] != 0) {
      sum += primes[i];
    }
  }
  printf("The sum of the all primes under %d is %ld\n", num_primes, sum);
  free(primes);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<problem10>:
  endbr64
  push   %rbp
  mov    $0x7a1200,%edi
  call   <malloc@plt>
  mov    $0x1e8480,%edi
  mov    %rax,%rbp
  call   <sqrt@plt>
  mov    %eax,%r10d
  xor    %eax,%eax
  nopl   0x0(%rax)
L41:
  mov    %eax,0x0(%rbp,%rax,4)
  add    $0x1,%rax
  cmp    $0x1e8480,%rax
  jne    L41
  lea    0x8(%rbp),%rcx
  cmp    $0x1,%r10d
  jle    L86
  mov    $0x2,%r8d
  lea    0x8(%rbp),%rcx
  mov    $0x1e847e,%r11d
  cs nopw 0x0(%rax,%rax,1)
L75:
  lea    0x1(%r8),%esi
  cmp    $0x1e847f,%esi
  jg     L21
  mov    %r11d,%eax
  movslq %r8d,%rdx
  movslq %esi,%rsi
  sub    %r8d,%eax
  lea    0x0(%rbp,%rsi,4),%rdi
  add    %rdx,%rax
  lea    (%rcx,%rax,4),%r9
  nopl   0x0(%rax)
L12:
  mov    (%rdi),%eax
  cltd
  idiv   %r8d
  test   %edx,%edx
  jne    L11
  movl   $0x0,(%rdi)
L11:
  add    $0x4,%rdi
  cmp    %r9,%rdi
  jne    L12
  mov    0x0(%rbp,%rsi,4),%eax
  test   %eax,%eax
  jne    L37
  nopl   0x0(%rax)
L81:
  add    $0x1,%rsi
  cmp    $0x1e847f,%esi
  jg     L21
  mov    0x0(%rbp,%rsi,4),%eax
  test   %eax,%eax
  je     L81
L37:
  mov    %esi,%r8d
L21:
  cmp    %r8d,%r10d
  jge    L75
L86:
  mov    %rcx,%rax
  lea    0x7a1200(%rbp),%rsi
  xor    %edx,%edx
  nopl   0x0(%rax,%rax,1)
L72:
  movslq (%rax),%rcx
  add    $0x4,%rax
  add    %rcx,%rdx
  cmp    %rax,%rsi
  jne    L72
  mov    $0x1e8480,%esi
  lea    D7(%rip),%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <free@plt>
  xor    %eax,%eax
  pop    %rbp
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D7
type: string

tool: D7: The sum of the all primes under %d is %ld


assistant: ```c
int problem10(void) {
  int num_primes = 2000000;
  int *primes = malloc(sizeof *primes * num_primes);
  int i, j, sr = sqrt(num_primes) + 1;
  long sum;
  for (i = 0; i < num_primes; i++)
    primes[i] = i;
  for (i = 2; i < sr;) {
    for (j = i + 1; j < num_primes; j++) {
      if (primes[j] % i == 0) {
        primes[j] = 0;
      }
    }
    for (j = i + 1; j < num_primes; j++) {
      if (primes[j] != 0) {
        i = j;
        break;
      }
    }
  }
  for (sum = 0, i = 2; i < num_primes; i++) {
    if (primes[i] != 0) {
      sum += primes[i];
    }
  }
  printf("The sum of the all primes under %d is %ld\n", num_primes, sum);
  free(primes);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<problem10>:
  endbr64
  push   %rbp
  mov    $0x7a1200,%edi
  call   <malloc@plt>
  mov    $0x1e8480,%edi
  mov    %rax,%rbp
  call   <sqrt@plt>
  movdqa D11(%rip),%xmm0
  movdqa D9(%rip),%xmm2
  lea    0x7a1200(%rbp),%rdx
  mov    %eax,%r10d
  mov    %rbp,%rax
  nopl   0x0(%rax)
L21:
  movdqa %xmm0,%xmm1
  add    $0x10,%rax
  paddd  %xmm2,%xmm0
  movups %xmm1,-0x10(%rax)
  cmp    %rdx,%rax
  jne    L21
  lea    0x8(%rbp),%r8
  cmp    $0x1,%r10d
  jle    L19
  mov    $0x2,%edi
  lea    0x8(%rbp),%r8
  mov    $0x1e847e,%r11d
  xchg   %ax,%ax
L111:
  lea    0x1(%rdi),%ecx
  cmp    $0x1e847f,%ecx
  jg     L44
  mov    %r11d,%eax
  movslq %edi,%rdx
  movslq %ecx,%rcx
  sub    %edi,%eax
  lea    0x0(%rbp,%rcx,4),%rsi
  add    %rdx,%rax
  lea    (%r8,%rax,4),%r9
  cs nopw 0x0(%rax,%rax,1)
L102:
  mov    (%rsi),%eax
  cltd
  idiv   %edi
  test   %edx,%edx
  jne    L30
  movl   $0x0,(%rsi)
L30:
  add    $0x4,%rsi
  cmp    %r9,%rsi
  jne    L102
  mov    0x0(%rbp,%rcx,4),%eax
  test   %eax,%eax
  jne    L99
L119:
  add    $0x1,%rcx
  cmp    $0x1e847f,%ecx
  jg     L10
  mov    0x0(%rbp,%rcx,4),%eax
  test   %eax,%eax
  je     L119
L99:
  mov    %ecx,%edi
L10:
  cmp    %edi,%r10d
  jge    L111
L19:
  mov    %r8,%rax
  lea    0x7a11f8(%rbp),%rdx
  pxor   %xmm1,%xmm1
  pxor   %xmm4,%xmm4
  nopl   (%rax)
L31:
  movdqu (%rax),%xmm0
  movdqa %xmm4,%xmm2
  add    $0x10,%rax
  pcmpgtd %xmm0,%xmm2
  movdqa %xmm0,%xmm3
  punpckldq %xmm2,%xmm3
  punpckhdq %xmm2,%xmm0
  paddq  %xmm3,%xmm1
  paddq  %xmm0,%xmm1
  cmp    %rdx,%rax
  jne    L31
  movdqa %xmm1,%xmm0
  movslq 0x7a11f8(%rbp),%rdx
  mov    $0x1e8480,%esi
  lea    D5(%rip),%rdi
  psrldq $0x8,%xmm0
  paddq  %xmm0,%xmm1
  movq   %xmm1,%rax
  add    %rdx,%rax
  movslq 0x7a11fc(%rbp),%rdx
  add    %rax,%rdx
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <free@plt>
  xor    %eax,%eax
  pop    %rbp
  ret
L44:
  cmp    %edi,%r10d
  jl     L19
L0:
  jmp    L0
```

tool_calls: name: disassemble_data
arguments: data_tag: D11
data_type: byte[16]
name: disassemble_data
arguments: data_tag: D9
data_type: byte[16]
name: disassemble_data
arguments: data_tag: D5
data_type: string

tool: {"D11": [0x00,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x02,0x00,0x00,0x00,0x03,0x00,0x00,0x00]}

tool: {"D9": [0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00]}

tool: D5: The sum of the all primes under %d is %ld


assistant: ```c
int problem10(void) {
  int num_primes = 2000000;
  int *primes = malloc(sizeof *primes * num_primes);
  int i, j, sr = sqrt(num_primes) + 1;
  long sum;
  for (i = 0; i < num_primes; i++)
    primes[i] = i;
  for (i = 2; i < sr;) {
    for (j = i + 1; j < num_primes; j++) {
      if (primes[j] % i == 0) {
        primes[j] = 0;
      }
    }
    for (j = i + 1; j < num_primes; j++) {
      if (primes[j] != 0) {
        i = j;
        break;
      }
    }
  }
  for (sum = 0, i = 2; i < num_primes; i++) {
    if (primes[i] != 0) {
      sum += primes[i];
    }
  }
  printf("The sum of the all primes under %d is %ld\n", num_primes, sum);
  free(primes);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<fastsqrt3>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  movss  %xmm0,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  movd   %eax,%xmm0
  call   <fastsqrt1@plt>
  movd   %xmm0,%eax
  mov    %eax,-0x4(%rbp)
  movss  -0x14(%rbp),%xmm0
  divss  -0x4(%rbp),%xmm0
  movss  -0x4(%rbp),%xmm1
  addss  %xmm1,%xmm0
  movss  %xmm0,-0x4(%rbp)
  movss  -0x4(%rbp),%xmm1
  movss  D3(%rip),%xmm0
  mulss  %xmm0,%xmm1
  movss  -0x14(%rbp),%xmm0
  divss  -0x4(%rbp),%xmm0
  addss  %xmm1,%xmm0
  movss  %xmm0,-0x4(%rbp)
  movss  -0x4(%rbp),%xmm0
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_label: D3
type: byte[4]

tool: {"D3": [0x00,0x00,0x80,0x3e]}

assistant: ```c
float fastsqrt3(float x) {
  float v = fastsqrt1(x);
  v = v + x / v;
  v = 0.25f * v + x / v;
  return v;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<fastsqrt3>:
  endbr64
  sub    $0x18,%rsp
  movss  %xmm0,0xc(%rsp)
  call   <fastsqrt1@plt>
  movss  0xc(%rsp),%xmm3
  movaps %xmm3,%xmm2
  divss  %xmm0,%xmm2
  addss  %xmm0,%xmm2
  movaps %xmm2,%xmm0
  mulss  D6(%rip),%xmm0
  movaps %xmm3,%xmm1
  divss  %xmm2,%xmm1
  addss  %xmm1,%xmm0
  add    $0x18,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D6
type_guessed: byte[4]

tool: {"D6": [0x00,0x00,0x80,0x3e]}

assistant: ```c
float fastsqrt3(float x) {
  float v = fastsqrt1(x);
  v = v + x / v;
  v = 0.25f * v + x / v;
  return v;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<fastsqrt3>:
  endbr64
  sub    $0x18,%rsp
  movss  %xmm0,0xc(%rsp)
  call   <fastsqrt1@plt>
  movss  0xc(%rsp),%xmm1
  movaps %xmm1,%xmm2
  divss  %xmm0,%xmm2
  addss  %xmm0,%xmm2
  movss  D8(%rip),%xmm0
  add    $0x18,%rsp
  divss  %xmm2,%xmm1
  mulss  %xmm2,%xmm0
  addss  %xmm1,%xmm0
  ret
```

tool_calls: name: read_data
arguments: label: D8
type: byte[4]

tool: {"D8": [0x00,0x00,0x80,0x3e]}

assistant: ```c
float fastsqrt3(float x) {
  float v = fastsqrt1(x);
  v = v + x / v;
  v = 0.25f * v + x / v;
  return v;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<fastsqrt3>:
  endbr64
  sub    $0x18,%rsp
  movss  %xmm0,0xc(%rsp)
  call   <fastsqrt1@plt>
  movss  0xc(%rsp),%xmm1
  movaps %xmm1,%xmm2
  divss  %xmm0,%xmm2
  addss  %xmm0,%xmm2
  movss  D5(%rip),%xmm0
  add    $0x18,%rsp
  divss  %xmm2,%xmm1
  mulss  %xmm2,%xmm0
  addss  %xmm1,%xmm0
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D5
type: byte[4]

tool: {"D5": [0x00,0x00,0x80,0x3e]}

assistant: ```c
float fastsqrt3(float x) {
  float v = fastsqrt1(x);
  v = v + x / v;
  v = 0.25f * v + x / v;
  return v;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<NormalizeSignature>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    $0x4,%edx
  lea    D4(%rip),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strncmp@plt>
  test   %eax,%eax
  jne    L1
  mov    -0x8(%rbp),%rax
  add    $0x3,%rax
  movb   $0x50,(%rax)
L1:
  nop
  leave
  ret
```

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

tool: D4: RSD 

assistant: ```c
void NormalizeSignature(char *Signature) {
  if (!strncmp(Signature, "RSD ", 4)) {
    Signature[3] = 'P';
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<NormalizeSignature>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  mov    $0x4,%edx
  lea    D8(%rip),%rsi
  call   <strncmp@plt>
  test   %eax,%eax
  jne    L5
  movb   $0x50,0x3(%rbx)
L5:
  pop    %rbx
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D8
type_guessed: string

tool: D8: RSD 

assistant: ```c
void NormalizeSignature(char *Signature) {
  if (!strncmp(Signature, "RSD ", 4)) {
    Signature[3] = 'P';
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<NormalizeSignature>:
  endbr64
  push   %rbx
  mov    $0x4,%edx
  lea    D9(%rip),%rsi
  mov    %rdi,%rbx
  call   <strncmp@plt>
  test   %eax,%eax
  jne    L0
  movb   $0x50,0x3(%rbx)
L0:
  pop    %rbx
  ret
```

tool_calls: name: unpack_data
arguments: label: D9
type_infered: string

tool: D9: RSD 

assistant: ```c
void NormalizeSignature(char *Signature) {
  if (!strncmp(Signature, "RSD ", 4)) {
    Signature[3] = 'P';
  }
}
```