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

```asm
<delete_zeros>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  xor    %r13d,%r13d
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  mov    %rdi,%rbp
  mov    $0x8,%edi
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  call   <malloc@plt>
  mov    %rax,%r15
  mov    %rax,%r14
  jmp    L74
  nop
L73:
  cmp    $0x5,%r12d
  je     L21
  movslq %r13d,%rax
  xor    %r12d,%r12d
  add    $0x1,%r13d
  movb   $0x30,(%r15,%rax,1)
L94:
  add    $0x1,%rbx
L74:
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jle    L45
  cmpb   $0x31,0x0(%rbp,%rbx,1)
  jne    L73
  cmp    $0x6,%r12d
  je     L89
  movslq %r13d,%rax
  add    $0x1,%r12d
  add    $0x1,%r13d
  movb   $0x31,(%r15,%rax,1)
  jmp    L94
  nopl   0x0(%rax)
L21:
  xor    %r12d,%r12d
  jmp    L94
  nopl   (%rax)
L45:
  mov    %r15,%rdi
  call   <strlen@plt>
  movslq %eax,%rbx
  mov    %rbx,%r12
  add    %r15,%rbx
  sub    %ebx,%r12d
  jmp    L70
  cs nopw 0x0(%rax,%rax,1)
L14:
  movb   $0x30,(%rbx)
  add    $0x1,%rbx
L70:
  mov    %rbp,%rdi
  call   <strlen@plt>
  lea    (%r12,%rbx,1),%edx
  cmp    %edx,%eax
  jg     L14
L43:
  add    $0x8,%rsp
  mov    %r14,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L89:
  lea    D6(%rip),%r14
  jmp    L43
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D6
type_infered: string

tool: D6: Error: consecutive ones can't be more than 5.

assistant: ```c
char *delete_zeros(char *input) {
  int i, current, one_count;
  char *output;
  output = (char *)malloc(sizeof(input));
  current = 0;
  one_count = 0;
  for (i = 0; i < strlen(input); i++) {
    if (input[i] == '1') {
      if (one_count > 5) {
        return "Error: consecutive ones can't be more than 5.";
      } else {
        one_count++;
        output[current] = '1';
        current++;
      }
    } else {
      if (one_count != 5) {
        output[current] = '0';
        current++;
      }
      one_count = 0;
    }
  }
  for (i = strlen(output); i < strlen(input); i++) {
    output[i] = '0';
  }
  return output;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    -0x10(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x10(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x10(%rbp),%rax
  mov    (%rax),%eax
  cltq
  mov    %rax,-0x8(%rbp)
  mov    -0x10(%rbp),%rax
  mov    %eax,%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,(%rax)
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
main() {
  int *p, *q;
  p = q;
  *p = *q;
  p = *q;
  *p = q;
}
```
user: Converte this assembly snippet into its equivalent C code:

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

assistant: ```c
main() {
  int *p, *q;
  p = q;
  *p = *q;
  p = *q;
  *p = q;
}
```
user: 下面的汇编指令对应的C源代码是什么：

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

assistant: ```c
main() {
  int *p, *q;
  p = q;
  *p = *q;
  p = *q;
  *p = q;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

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

assistant: ```c
main() {
  int *p, *q;
  p = q;
  *p = *q;
  p = *q;
  *p = q;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<bitcount>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  setne  %al
  movzbl %al,%eax
  mov    %eax,-0x4(%rbp)
  jmp    L9
L19:
  addl   $0x1,-0x4(%rbp)
L9:
  mov    -0x14(%rbp),%eax
  sub    $0x1,%eax
  and    %eax,-0x14(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  jne    L19
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned bitcount(unsigned x) {
  unsigned num = (x == 0 ? 0 : 1);
  while ((x &= (x - 1)) != 0) {
    ++num;
  }
  return num;
}
```
user: What is the c source code of the assembly code below:

```asm
<bitcount>:
  endbr64
  test   %edi,%edi
  setne  %dl
  movzbl %dl,%edx
  lea    -0x1(%rdi),%eax
  and    %eax,%edi
  je     L16
L19:
  add    $0x1,%edx
  lea    -0x1(%rdi),%eax
  and    %eax,%edi
  jne    L19
L16:
  mov    %edx,%eax
  ret
```

assistant: ```c
unsigned bitcount(unsigned x) {
  unsigned num = (x == 0 ? 0 : 1);
  while ((x &= (x - 1)) != 0) {
    ++num;
  }
  return num;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<bitcount>:
  endbr64
  xor    %r8d,%r8d
  lea    -0x1(%rdi),%eax
  test   %edi,%edi
  setne  %r8b
  and    %eax,%edi
  je     L9
  nopl   0x0(%rax)
L0:
  lea    -0x1(%rdi),%eax
  add    $0x1,%r8d
  and    %eax,%edi
  jne    L0
L9:
  mov    %r8d,%eax
  ret
```

assistant: ```c
unsigned bitcount(unsigned x) {
  unsigned num = (x == 0 ? 0 : 1);
  while ((x &= (x - 1)) != 0) {
    ++num;
  }
  return num;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<bitcount>:
  endbr64
  xor    %r8d,%r8d
  lea    -0x1(%rdi),%eax
  test   %edi,%edi
  setne  %r8b
  and    %eax,%edi
  je     L6
  nopl   0x0(%rax)
L9:
  lea    -0x1(%rdi),%eax
  add    $0x1,%r8d
  and    %eax,%edi
  jne    L9
L6:
  mov    %r8d,%eax
  ret
```

assistant: ```c
unsigned bitcount(unsigned x) {
  unsigned num = (x == 0 ? 0 : 1);
  while ((x &= (x - 1)) != 0) {
    ++num;
  }
  return num;
}
```
user: What is the c source code of the assembly code below?

```asm
<rTrim>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  je     L32
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x8(%rbp)
  jmp    L9
L14:
  subq   $0x1,-0x8(%rbp)
L9:
  mov    -0x8(%rbp),%rax
  cmp    -0x18(%rbp),%rax
  jb     L15
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%esi
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  jne    L14
L15:
  mov    -0x8(%rbp),%rax
  add    $0x1,%rax
  movb   $0x0,(%rax)
L32:
  mov    -0x18(%rbp),%rax
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D3
data_type: string

tool: D3: 
	 

assistant: ```c
char *rTrim(char *buf) {
  if (buf) {
    char *tp = buf + strlen(buf);
    while (tp >= buf && strchr("\n\r\t ", *tp))
      tp--;
    *(tp + 1) = 0;
  }
  return buf;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<rTrim>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  test   %rdi,%rdi
  je     L2
  call   <strlen@plt>
  movslq %eax,%rbx
  lea    D5(%rip),%r12
  add    %rbp,%rbx
  jb     L29
L21:
  movsbl (%rbx),%esi
  mov    %r12,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  je     L29
  sub    $0x1,%rbx
  cmp    %rbx,%rbp
  jbe    L21
L29:
  movb   $0x0,0x1(%rbx)
L2:
  mov    %rbp,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: parse_data
arguments: label: D5
type: string

tool: D5: 
	 

assistant: ```c
char *rTrim(char *buf) {
  if (buf) {
    char *tp = buf + strlen(buf);
    while (tp >= buf && strchr("\n\r\t ", *tp))
      tp--;
    *(tp + 1) = 0;
  }
  return buf;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<rTrim>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  test   %rdi,%rdi
  je     L29
  call   <strlen@plt>
  lea    D5(%rip),%rbp
  movslq %eax,%rbx
  add    %r12,%rbx
  jae    L46
  jmp    L16
  cs nopw 0x0(%rax,%rax,1)
L43:
  sub    $0x1,%rbx
  cmp    %rbx,%r12
  ja     L16
L46:
  movsbl (%rbx),%esi
  mov    %rbp,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  jne    L43
L16:
  movb   $0x0,0x1(%rbx)
L29:
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D5
type: string

tool: D5: 
	 

assistant: ```c
char *rTrim(char *buf) {
  if (buf) {
    char *tp = buf + strlen(buf);
    while (tp >= buf && strchr("\n\r\t ", *tp))
      tp--;
    *(tp + 1) = 0;
  }
  return buf;
}
```
user: What is the c source code of the assembly code below:

```asm
<rTrim>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  test   %rdi,%rdi
  je     L4
  call   <strlen@plt>
  lea    D3(%rip),%rbp
  movslq %eax,%rbx
  add    %r12,%rbx
  jae    L22
  jmp    L0
  cs nopw 0x0(%rax,%rax,1)
L46:
  sub    $0x1,%rbx
  cmp    %rbx,%r12
  ja     L0
L22:
  movsbl (%rbx),%esi
  mov    %rbp,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  jne    L46
L0:
  movb   $0x0,0x1(%rbx)
L4:
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: parse_data
arguments: data_label: D3
type_guessed: string

tool: D3: 
	 

assistant: ```c
char *rTrim(char *buf) {
  if (buf) {
    char *tp = buf + strlen(buf);
    while (tp >= buf && strchr("\n\r\t ", *tp))
      tp--;
    *(tp + 1) = 0;
  }
  return buf;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  lea    D10(%rip),%rdi
  sub    $0x8,%rsp
  call   <puts@plt>
  lea    D2(%rip),%rdi
  call   <puts@plt>
L8:
  jmp    L8
```

tool_calls: name: disassemble_data
arguments: data_label: D10
type_infered: string
name: disassemble_data
arguments: data_label: D2
type_infered: string

tool: D10: char is signed

tool: D2: This machine is little endian

assistant: ```c
int main(void) {
  char a = 1, a1 = 1, a2 = 1;
  short b = 1, b1;
  int c = 1, c1, x = 1;
  long d = 1, d1;
  float e = 1, e1 = 1, e2 = 1, g = 1, g1;
  double f = 1, f1 = 1, f2 = 1, h = 1, h1;
  int i = 0, j = 0, k = 0, l = 0, m = 0, n = 0, o = 0, p = 0;
  char *y = (char *)&x;
  while (a1 > 0) {
    a = a1;
    a1 = a * 2;
    a2 = a1 / 2;
    i++;
  }
  if (a2 != 0)
    printf("char is signed\n");
  else
    printf("char is unsigned\n");
  if (y[0] == 0)
    printf("This machine is BIG endian\n");
  else
    printf("This machine is little endian\n");
  while (b > 0) {
    b = b * 2;
    b1 = b / 2;
    j++;
  }
  while (c > 0) {
    c = c * 2;
    c1 = c / 2;
    k++;
  }
  while (d > 0) {
    d = d * 2;
    d1 = d / 2;
    l++;
  }
  while (e == e2) {
    e = e1;
    e1 = e * 2;
    e2 = e1 / 2;
    m++;
  }
  while (g > 0) {
    g = g / 2;
    g1 = g * 2;
    o++;
  }
  while (f == f2) {
    f = f1;
    f1 = f * 2;
    f2 = f1 / 2;
    n++;
  }
  while (h > 0) {
    h = h / 2;
    h1 = h * 2;
    p++;
  }
  i = i - 1;
  j = j - 1;
  k = k - 1;
  l = l - 1;
  m = m - 1;
  n = n - 1;
  o = o - 1;
  p = p - 1;
  printf("The largest positive power of 2 represented with char  is %d\n", i);
  printf("The largest positive power of 2 represented with short is %d\n", j);
  printf("The largest positive power of 2 represented with int is %d\n", k);
  printf("The largest positive power of 2 represented with long is %d\n", l);
  printf("The largest positive and negative powers of 2 represented with float are %d and %d respectively\n", m, o);
  printf("The largest positive and negative powers of 2 represented with double are %d and %d respectively\n", n, p);
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  lea    D15(%rip),%rdi
  sub    $0x8,%rsp
  call   <puts@plt>
  lea    D1(%rip),%rdi
  call   <puts@plt>
L7:
  jmp    L7
```

tool_calls: name: read_data
arguments: data_tag: D15
datatype: string
name: read_data
arguments: data_tag: D1
datatype: string

tool: D15: char is signed

tool: D1: This machine is little endian

assistant: ```c
int main(void) {
  char a = 1, a1 = 1, a2 = 1;
  short b = 1, b1;
  int c = 1, c1, x = 1;
  long d = 1, d1;
  float e = 1, e1 = 1, e2 = 1, g = 1, g1;
  double f = 1, f1 = 1, f2 = 1, h = 1, h1;
  int i = 0, j = 0, k = 0, l = 0, m = 0, n = 0, o = 0, p = 0;
  char *y = (char *)&x;
  while (a1 > 0) {
    a = a1;
    a1 = a * 2;
    a2 = a1 / 2;
    i++;
  }
  if (a2 != 0)
    printf("char is signed\n");
  else
    printf("char is unsigned\n");
  if (y[0] == 0)
    printf("This machine is BIG endian\n");
  else
    printf("This machine is little endian\n");
  while (b > 0) {
    b = b * 2;
    b1 = b / 2;
    j++;
  }
  while (c > 0) {
    c = c * 2;
    c1 = c / 2;
    k++;
  }
  while (d > 0) {
    d = d * 2;
    d1 = d / 2;
    l++;
  }
  while (e == e2) {
    e = e1;
    e1 = e * 2;
    e2 = e1 / 2;
    m++;
  }
  while (g > 0) {
    g = g / 2;
    g1 = g * 2;
    o++;
  }
  while (f == f2) {
    f = f1;
    f1 = f * 2;
    f2 = f1 / 2;
    n++;
  }
  while (h > 0) {
    h = h / 2;
    h1 = h * 2;
    p++;
  }
  i = i - 1;
  j = j - 1;
  k = k - 1;
  l = l - 1;
  m = m - 1;
  n = n - 1;
  o = o - 1;
  p = p - 1;
  printf("The largest positive power of 2 represented with char  is %d\n", i);
  printf("The largest positive power of 2 represented with short is %d\n", j);
  printf("The largest positive power of 2 represented with int is %d\n", k);
  printf("The largest positive power of 2 represented with long is %d\n", l);
  printf("The largest positive and negative powers of 2 represented with float are %d and %d respectively\n", m, o);
  printf("The largest positive and negative powers of 2 represented with double are %d and %d respectively\n", n, p);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<fact_do>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  movl   $0x1,-0x4(%rbp)
L2:
  mov    -0x4(%rbp),%eax
  imul   -0x14(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  subl   $0x1,-0x14(%rbp)
  cmpl   $0x1,-0x14(%rbp)
  jg     L2
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int fact_do(int n) {
  int result = 1;
  do {
    result *= n;
    n = n - 1;
  } while (n > 1);
  return result;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<fact_do>:
  endbr64
  mov    $0x1,%eax
L7:
  imul   %edi,%eax
  sub    $0x1,%edi
  cmp    $0x1,%edi
  jg     L7
  ret
```

assistant: ```c
int fact_do(int n) {
  int result = 1;
  do {
    result *= n;
    n = n - 1;
  } while (n > 1);
  return result;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<fact_do>:
  endbr64
  mov    $0x1,%eax
  nopl   0x0(%rax)
L7:
  imul   %edi,%eax
  sub    $0x1,%edi
  cmp    $0x1,%edi
  jg     L7
  ret
```

assistant: ```c
int fact_do(int n) {
  int result = 1;
  do {
    result *= n;
    n = n - 1;
  } while (n > 1);
  return result;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<fact_do>:
  endbr64
  mov    $0x1,%eax
  nopl   0x0(%rax)
L2:
  imul   %edi,%eax
  sub    $0x1,%edi
  cmp    $0x1,%edi
  jg     L2
  ret
```

assistant: ```c
int fact_do(int n) {
  int result = 1;
  do {
    result *= n;
    n = n - 1;
  } while (n > 1);
  return result;
}
```
user: Translate the below assembly code into C programming language:

```asm
<eliminaDup>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    -0x1c(%rbp),%eax
  cltq
  shl    $0x2,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  movl   $0x0,-0x10(%rbp)
  jmp    L8
L19:
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    -0xc(%rbp),%edx
  mov    -0x8(%rbp),%rcx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <find@plt>
  test   %eax,%eax
  jne    L14
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0xc(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x8(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  addl   $0x1,-0xc(%rbp)
L14:
  addl   $0x1,-0x10(%rbp)
L8:
  mov    -0x10(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L19
  mov    -0xc(%rbp),%eax
  cltq
  shl    $0x2,%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <realloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x28(%rbp),%rax
  mov    -0xc(%rbp),%edx
  mov    %edx,(%rax)
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
int *eliminaDup(int *vect, int dim, int *dim_nodup) {
  int *newArr = malloc(sizeof(int) * dim), i, newDim;
  newDim = 0;
  for (i = 0; i < dim; i++) {
    if (!find(vect[i], newArr, newDim)) {
      newArr[newDim] = vect[i];
      newDim++;
    }
  }
  newArr = realloc(newArr, sizeof(int) * newDim);
  *dim_nodup = newDim;
  return newArr;
}
```
user: What is the c source code of the assembly code below:

```asm
<eliminaDup>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r13
  mov    %esi,%ebp
  mov    %rdx,%r14
  lea    0x0(,%rsi,4),%edi
  call   <malloc@plt>
  mov    %rax,%r12
  test   %ebp,%ebp
  jle    L2
  mov    %r13,%rbx
  lea    -0x1(%rbp),%eax
  lea    0x4(%r13,%rax,4),%r13
  mov    $0x0,%ebp
  jmp    L8
L30:
  add    $0x4,%rbx
  cmp    %r13,%rbx
  je     L28
L8:
  mov    %ebp,%edx
  mov    %r12,%rsi
  mov    (%rbx),%edi
  call   <find@plt>
  test   %eax,%eax
  jne    L30
  mov    (%rbx),%edx
  movslq %ebp,%rax
  mov    %edx,(%r12,%rax,4)
  add    $0x1,%ebp
  jmp    L30
L2:
  mov    $0x0,%ebp
L28:
  lea    0x0(,%rbp,4),%esi
  mov    %r12,%rdi
  call   <realloc@plt>
  mov    %ebp,(%r14)
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int *eliminaDup(int *vect, int dim, int *dim_nodup) {
  int *newArr = malloc(sizeof(int) * dim), i, newDim;
  newDim = 0;
  for (i = 0; i < dim; i++) {
    if (!find(vect[i], newArr, newDim)) {
      newArr[newDim] = vect[i];
      newDim++;
    }
  }
  newArr = realloc(newArr, sizeof(int) * newDim);
  *dim_nodup = newDim;
  return newArr;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<eliminaDup>:
  endbr64
  push   %r14
  push   %r13
  mov    %rdx,%r13
  push   %r12
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %rdi,%rbx
  lea    0x0(,%rsi,4),%edi
  call   <malloc@plt>
  mov    %rax,%r12
  test   %ebp,%ebp
  jle    L19
  lea    -0x1(%rbp),%eax
  xor    %ebp,%ebp
  lea    0x4(%rbx,%rax,4),%r14
  nopl   0x0(%rax)
L28:
  mov    (%rbx),%edi
  mov    %ebp,%edx
  mov    %r12,%rsi
  call   <find@plt>
  test   %eax,%eax
  jne    L16
  mov    (%rbx),%edx
  movslq %ebp,%rax
  add    $0x1,%ebp
  mov    %edx,(%r12,%rax,4)
L16:
  add    $0x4,%rbx
  cmp    %rbx,%r14
  jne    L28
  lea    0x0(,%rbp,4),%esi
L8:
  mov    %r12,%rdi
  call   <realloc@plt>
  pop    %rbx
  mov    %ebp,0x0(%r13)
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
L19:
  xor    %esi,%esi
  xor    %ebp,%ebp
  jmp    L8
```

assistant: ```c
int *eliminaDup(int *vect, int dim, int *dim_nodup) {
  int *newArr = malloc(sizeof(int) * dim), i, newDim;
  newDim = 0;
  for (i = 0; i < dim; i++) {
    if (!find(vect[i], newArr, newDim)) {
      newArr[newDim] = vect[i];
      newDim++;
    }
  }
  newArr = realloc(newArr, sizeof(int) * newDim);
  *dim_nodup = newDim;
  return newArr;
}
```
user: Translate the below assembly code into C programming language:

```asm
<eliminaDup>:
  endbr64
  push   %r14
  push   %r13
  mov    %rdx,%r13
  push   %r12
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %rdi,%rbx
  lea    0x0(,%rsi,4),%edi
  call   <malloc@plt>
  mov    %rax,%r12
  test   %ebp,%ebp
  jle    L16
  lea    -0x1(%rbp),%eax
  xor    %ebp,%ebp
  lea    0x4(%rbx,%rax,4),%r14
  nopl   0x0(%rax)
L9:
  mov    (%rbx),%edi
  mov    %ebp,%edx
  mov    %r12,%rsi
  call   <find@plt>
  test   %eax,%eax
  jne    L36
  mov    (%rbx),%edx
  movslq %ebp,%rax
  add    $0x1,%ebp
  mov    %edx,(%r12,%rax,4)
L36:
  add    $0x4,%rbx
  cmp    %rbx,%r14
  jne    L9
  lea    0x0(,%rbp,4),%esi
L0:
  mov    %r12,%rdi
  call   <realloc@plt>
  pop    %rbx
  mov    %ebp,0x0(%r13)
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
L16:
  xor    %esi,%esi
  xor    %ebp,%ebp
  jmp    L0
```

assistant: ```c
int *eliminaDup(int *vect, int dim, int *dim_nodup) {
  int *newArr = malloc(sizeof(int) * dim), i, newDim;
  newDim = 0;
  for (i = 0; i < dim; i++) {
    if (!find(vect[i], newArr, newDim)) {
      newArr[newDim] = vect[i];
      newDim++;
    }
  }
  newArr = realloc(newArr, sizeof(int) * newDim);
  *dim_nodup = newDim;
  return newArr;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<radius_from_diagonal>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  movsd  %xmm0,-0x18(%rbp)
  movsd  %xmm1,-0x20(%rbp)
  movsd  %xmm2,-0x28(%rbp)
  movsd  -0x18(%rbp),%xmm0
  addsd  -0x20(%rbp),%xmm0
  movsd  %xmm0,-0x10(%rbp)
  movsd  -0x20(%rbp),%xmm0
  addsd  %xmm0,%xmm0
  movsd  -0x28(%rbp),%xmm1
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movsd  -0x10(%rbp),%xmm0
  movapd %xmm0,%xmm1
  mulsd  %xmm0,%xmm1
  movsd  -0x8(%rbp),%xmm2
  movsd  D0(%rip),%xmm0
  mulsd  %xmm2,%xmm0
  mulsd  -0x8(%rbp),%xmm0
  addsd  %xmm0,%xmm1
  movq   %xmm1,%rax
  movq   %rax,%xmm0
  call   <sqrt@plt>
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: label: D0
datatype: f64

tool: D0: 0.25

assistant: ```c
double radius_from_diagonal(double radius, double thickness, double length) {
  const double radius_outer = radius + thickness;
  const double length_outer = length + 2.0 * thickness;
  return sqrt(radius_outer * radius_outer + 0.25 * length_outer * length_outer);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<radius_from_diagonal>:
  endbr64
  movapd %xmm0,%xmm3
  addsd  %xmm1,%xmm3
  addsd  %xmm1,%xmm1
  addsd  %xmm2,%xmm1
  movapd %xmm1,%xmm0
  mulsd  D4(%rip),%xmm0
  mulsd  %xmm1,%xmm0
  mulsd  %xmm3,%xmm3
  addsd  %xmm3,%xmm0
  pxor   %xmm1,%xmm1
  ucomisd %xmm0,%xmm1
  ja     L7
  sqrtsd %xmm0,%xmm0
  ret
L7:
  sub    $0x8,%rsp
  call   <sqrt@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D4
data_type: f64

tool: D4: 0.25

assistant: ```c
double radius_from_diagonal(double radius, double thickness, double length) {
  const double radius_outer = radius + thickness;
  const double length_outer = length + 2.0 * thickness;
  return sqrt(radius_outer * radius_outer + 0.25 * length_outer * length_outer);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<radius_from_diagonal>:
  endbr64
  movapd %xmm0,%xmm3
  movsd  D3(%rip),%xmm0
  addsd  %xmm1,%xmm3
  addsd  %xmm1,%xmm1
  addsd  %xmm2,%xmm1
  mulsd  %xmm3,%xmm3
  mulsd  %xmm1,%xmm0
  mulsd  %xmm1,%xmm0
  pxor   %xmm1,%xmm1
  addsd  %xmm3,%xmm0
  ucomisd %xmm0,%xmm1
  ja     L7
  sqrtsd %xmm0,%xmm0
  ret
L7:
  jmp    1050 <sqrt@plt>
```

tool_calls: name: read_data
arguments: tag: D3
data_type: f64

tool: D3: 0.25

assistant: ```c
double radius_from_diagonal(double radius, double thickness, double length) {
  const double radius_outer = radius + thickness;
  const double length_outer = length + 2.0 * thickness;
  return sqrt(radius_outer * radius_outer + 0.25 * length_outer * length_outer);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<radius_from_diagonal>:
  endbr64
  movapd %xmm0,%xmm3
  movsd  D6(%rip),%xmm0
  addsd  %xmm1,%xmm3
  addsd  %xmm1,%xmm1
  addsd  %xmm2,%xmm1
  mulsd  %xmm3,%xmm3
  mulsd  %xmm1,%xmm0
  mulsd  %xmm1,%xmm0
  pxor   %xmm1,%xmm1
  addsd  %xmm3,%xmm0
  ucomisd %xmm0,%xmm1
  ja     L7
  sqrtsd %xmm0,%xmm0
  ret
L7:
  jmp    1050 <sqrt@plt>
```

tool_calls: name: disassemble_data
arguments: label: D6
type: f64

tool: D6: 0.25

assistant: ```c
double radius_from_diagonal(double radius, double thickness, double length) {
  const double radius_outer = radius + thickness;
  const double length_outer = length + 2.0 * thickness;
  return sqrt(radius_outer * radius_outer + 0.25 * length_outer * length_outer);
}
```