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

```asm
<min>:
  endbr64
  minsd  %xmm0,%xmm1
  movapd %xmm1,%xmm0
  ret
```

assistant: ```c
double min(const double a, const double b) {
  if (b < a)
    return b;
  return a;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<min>:
  endbr64
  minsd  %xmm0,%xmm1
  movapd %xmm1,%xmm0
  ret
```

assistant: ```c
double min(const double a, const double b) {
  if (b < a)
    return b;
  return a;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<delete_arg>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  cmp    %eax,-0x24(%rbp)
  jl     L8
  mov    $0x0,%eax
  jmp    L5
L8:
  mov    -0x24(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  lea    -0x1(%rax),%edx
  mov    -0x18(%rbp),%rax
  mov    %edx,(%rax)
  jmp    L32
L39:
  mov    -0x24(%rbp),%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    -0x24(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,8),%rcx
  mov    -0x20(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rax),%rax
  mov    %rax,(%rdx)
  addl   $0x1,-0x24(%rbp)
L32:
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  cmp    %eax,-0x24(%rbp)
  jl     L39
  mov    -0x24(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movq   $0x0,(%rax)
  mov    -0x8(%rbp),%rax
L5:
  pop    %rbp
  ret
```

assistant: ```c
const char *delete_arg(int *argcp, char *argv[], int i) {
  const char *ret;
  if (i > *argcp - 1)
    return (NULL);
  ret = argv[i];
  (*argcp)--;
  for (; i < *argcp; i++)
    argv[i] = argv[i + 1];
  argv[i] = NULL;
  return (ret);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<delete_arg>:
  endbr64
  mov    (%rdi),%ecx
  cmp    %edx,%ecx
  jle    L16
  movslq %edx,%rax
  lea    (%rsi,%rax,8),%rax
  mov    (%rax),%r8
  sub    $0x1,%ecx
  mov    %ecx,(%rdi)
  cmp    %edx,%ecx
  jle    L5
L8:
  mov    0x8(%rax),%rcx
  mov    %rcx,(%rax)
  add    $0x1,%edx
  add    $0x8,%rax
  cmp    %edx,(%rdi)
  jg     L8
L5:
  movslq %edx,%rdx
  movq   $0x0,(%rsi,%rdx,8)
L11:
  mov    %r8,%rax
  ret
L16:
  mov    $0x0,%r8d
  jmp    L11
```

assistant: ```c
const char *delete_arg(int *argcp, char *argv[], int i) {
  const char *ret;
  if (i > *argcp - 1)
    return (NULL);
  ret = argv[i];
  (*argcp)--;
  for (; i < *argcp; i++)
    argv[i] = argv[i + 1];
  argv[i] = NULL;
  return (ret);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<delete_arg>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    (%rdi),%eax
  cmp    %edx,%eax
  jle    L15
  movslq %edx,%rcx
  lea    -0x1(%rax),%ebp
  mov    %rsi,%rbx
  shl    $0x3,%rcx
  lea    (%rsi,%rcx,1),%r8
  mov    (%r8),%r12
  mov    %ebp,(%rdi)
  cmp    %edx,%ebp
  jle    L3
  sub    %edx,%eax
  mov    %r8,%rdi
  lea    0x8(%rsi,%rcx,1),%rsi
  movslq %ebp,%rbp
  sub    $0x2,%eax
  lea    0x8(,%rax,8),%rdx
  call   <memmove@plt>
  lea    (%rbx,%rbp,8),%r8
L3:
  movq   $0x0,(%r8)
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L15:
  xor    %r12d,%r12d
  pop    %rbx
  pop    %rbp
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
const char *delete_arg(int *argcp, char *argv[], int i) {
  const char *ret;
  if (i > *argcp - 1)
    return (NULL);
  ret = argv[i];
  (*argcp)--;
  for (; i < *argcp; i++)
    argv[i] = argv[i + 1];
  argv[i] = NULL;
  return (ret);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<delete_arg>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    (%rdi),%eax
  cmp    %edx,%eax
  jle    L5
  movslq %edx,%rcx
  lea    -0x1(%rax),%ebp
  mov    %rsi,%rbx
  shl    $0x3,%rcx
  lea    (%rsi,%rcx,1),%r8
  mov    (%r8),%r12
  mov    %ebp,(%rdi)
  cmp    %edx,%ebp
  jle    L4
  sub    %edx,%eax
  mov    %r8,%rdi
  lea    0x8(%rsi,%rcx,1),%rsi
  movslq %ebp,%rbp
  sub    $0x2,%eax
  lea    0x8(,%rax,8),%rdx
  call   <memmove@plt>
  lea    (%rbx,%rbp,8),%r8
L4:
  movq   $0x0,(%r8)
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L5:
  xor    %r12d,%r12d
  pop    %rbx
  pop    %rbp
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
const char *delete_arg(int *argcp, char *argv[], int i) {
  const char *ret;
  if (i > *argcp - 1)
    return (NULL);
  ret = argv[i];
  (*argcp)--;
  for (; i < *argcp; i++)
    argv[i] = argv[i + 1];
  argv[i] = NULL;
  return (ret);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x0,-0x8(%rbp)
  mov    -0x8(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  jmp    L45
L6:
  cmpl   $0x20,-0x4(%rbp)
  jne    L28
  cmpl   $0x0,-0x8(%rbp)
  jne    L28
  movl   $0x1,-0x8(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <putchar@plt>
L28:
  cmpl   $0x20,-0x4(%rbp)
  je     L45
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <putchar@plt>
  movl   $0x0,-0x8(%rbp)
L45:
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %eax,-0x4(%rbp)
  mov    D8(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,-0x4(%rbp)
  jne    L6
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
main() {
  int c, is;
  c = is = 0;
  while ((c = getchar()) != EOF) {
    if (c == ' ') {
      if (is == 0) {
        is = 1;
        putchar(c);
      }
    }
    if (c != ' ') {
      putchar(c);
      is = 0;
    }
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    $0x1,%ebp
L66:
  mov    $0x0,%ebx
  jmp    L2
L38:
  cmp    $0x20,%eax
  jne    L57
L2:
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    D9(%rip),%rdx
  cmp    %eax,(%rdx)
  je     L8
  test   %ebx,%ebx
  jne    L38
  cmp    $0x20,%eax
  jne    L38
  mov    $0x20,%edi
  call   <putchar@plt>
  mov    %ebp,%ebx
  jmp    L2
L57:
  mov    %eax,%edi
  call   <putchar@plt>
  jmp    L66
L8:
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
main() {
  int c, is;
  c = is = 0;
  while ((c = getchar()) != EOF) {
    if (c == ' ') {
      if (is == 0) {
        is = 1;
        putchar(c);
      }
    }
    if (c != ' ') {
      putchar(c);
      is = 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    $0x8,%rsp
  mov    D6(%rip),%rbx
L50:
  xor    %eax,%eax
  xor    %ebp,%ebp
  call   <getchar@plt>
  cmp    %eax,(%rbx)
  je     L53
L30:
  cmp    $0x20,%eax
  je     L41
  mov    %eax,%edi
  call   <putchar@plt>
  jmp    L50
  nopl   0x0(%rax)
L41:
  test   %ebp,%ebp
  je     L23
L57:
  xor    %eax,%eax
  mov    $0x1,%ebp
  call   <getchar@plt>
  cmp    %eax,(%rbx)
  jne    L30
L53:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
  nopl   (%rax)
L23:
  mov    $0x20,%edi
  call   <putchar@plt>
  jmp    L57
```

assistant: ```c
main() {
  int c, is;
  c = is = 0;
  while ((c = getchar()) != EOF) {
    if (c == ' ') {
      if (is == 0) {
        is = 1;
        putchar(c);
      }
    }
    if (c != ' ') {
      putchar(c);
      is = 0;
    }
  }
}
```
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
  push   %rbx
  sub    $0x8,%rsp
  mov    D5(%rip),%rbx
L57:
  xor    %eax,%eax
  xor    %ebp,%ebp
  call   <getchar@plt>
  cmp    %eax,(%rbx)
  je     L47
L33:
  cmp    $0x20,%eax
  je     L55
  mov    %eax,%edi
  call   <putchar@plt>
  jmp    L57
  nopl   0x0(%rax)
L55:
  test   %ebp,%ebp
  je     L38
L14:
  xor    %eax,%eax
  mov    $0x1,%ebp
  call   <getchar@plt>
  cmp    %eax,(%rbx)
  jne    L33
L47:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
  nopl   (%rax)
L38:
  mov    $0x20,%edi
  call   <putchar@plt>
  jmp    L14
```

assistant: ```c
main() {
  int c, is;
  c = is = 0;
  while ((c = getchar()) != EOF) {
    if (c == ' ') {
      if (is == 0) {
        is = 1;
        putchar(c);
      }
    }
    if (c != ' ') {
      putchar(c);
      is = 0;
    }
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<boil>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x8(%rbp),%rax
  mov    (%rax),%eax
  imul   %eax,%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,(%rax)
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x8(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,%edi
  call   <display_status@plt>
  nop
  leave
  ret
```

tool_calls: name: read_data
arguments: data_tag: D9
datatype: string

tool: D9: In boil(), I got: 

assistant: ```c
void boil(int *x) {
  *x = *x * *x;
  printf("In boil(), I got: ");
  display_status(*x);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<boil>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  mov    (%rdi),%eax
  imul   %eax,%eax
  mov    %eax,(%rdi)
  lea    D7(%rip),%rdi
  call   <printf@plt>
  mov    (%rbx),%edi
  call   <display_status@plt>
  pop    %rbx
  ret
```

tool_calls: name: unpack_data
arguments: label: D7
datatype: string

tool: D7: In boil(), I got: 

assistant: ```c
void boil(int *x) {
  *x = *x * *x;
  printf("In boil(), I got: ");
  display_status(*x);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<boil>:
  endbr64
  push   %rbx
  mov    (%rdi),%eax
  mov    %rdi,%rbx
  imul   %eax,%eax
  mov    %eax,(%rdi)
  lea    D9(%rip),%rdi
  call   <printf@plt>
  mov    (%rbx),%edi
  pop    %rbx
  jmp    1060 <display_status@plt>
```

tool_calls: name: parse_data
arguments: label: D9
type: string

tool: D9: In boil(), I got: 

assistant: ```c
void boil(int *x) {
  *x = *x * *x;
  printf("In boil(), I got: ");
  display_status(*x);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<boil>:
  endbr64
  push   %rbx
  mov    (%rdi),%eax
  mov    %rdi,%rbx
  imul   %eax,%eax
  mov    %eax,(%rdi)
  lea    D4(%rip),%rdi
  call   <printf@plt>
  mov    (%rbx),%edi
  pop    %rbx
  jmp    1060 <display_status@plt>
```

tool_calls: name: unpack_data
arguments: tag: D4
data_type: string

tool: D4: In boil(), I got: 

assistant: ```c
void boil(int *x) {
  *x = *x * *x;
  printf("In boil(), I got: ");
  display_status(*x);
}
```
user: Translate the below assembly code into C programming language:

```asm
<is_palindrome>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %edi,-0x24(%rbp)
  mov    -0x24(%rbp),%eax
  mov    %eax,%edi
  call   <nr_of_digits@plt>
  mov    %eax,-0x14(%rbp)
  movl   $0x0,-0x18(%rbp)
  jmp    L21
L4:
  mov    -0x24(%rbp),%edx
  mov    -0x18(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <digit@plt>
  mov    %rax,%rbx
  mov    -0x14(%rbp),%eax
  sub    $0x1,%eax
  sub    -0x18(%rbp),%eax
  mov    -0x24(%rbp),%edx
  mov    %edx,%esi
  mov    %eax,%edi
  call   <digit@plt>
  cmp    %rax,%rbx
  je     L24
  mov    $0x0,%eax
  jmp    L35
L24:
  addl   $0x1,-0x18(%rbp)
L21:
  mov    -0x14(%rbp),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  cmp    %eax,-0x18(%rbp)
  jl     L4
  mov    $0x1,%eax
L35:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
bool is_palindrome(int n) {
  int nr = nr_of_digits(n);
  int i;
  for (i = 0; i < nr / 2; i++) {
    if (digit(i, n) != digit(nr - 1 - i, n)) {
      return false;
    }
  }
  return true;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<is_palindrome>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%r12d
  call   <nr_of_digits@plt>
  mov    %eax,%r14d
  shr    $0x1f,%r14d
  add    %eax,%r14d
  sar    %r14d
  cmp    $0x1,%eax
  jle    L34
  mov    $0x0,%ebx
  lea    -0x1(%rax),%r13d
L35:
  mov    %r12d,%esi
  mov    %ebx,%edi
  call   <digit@plt>
  mov    %rax,%rbp
  mov    %r13d,%edi
  sub    %ebx,%edi
  mov    %r12d,%esi
  call   <digit@plt>
  cmp    %rax,%rbp
  jne    L28
  add    $0x1,%ebx
  cmp    %r14d,%ebx
  jl     L35
  mov    $0x1,%eax
  jmp    L33
L34:
  mov    $0x1,%eax
  jmp    L33
L28:
  mov    $0x0,%eax
L33:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
bool is_palindrome(int n) {
  int nr = nr_of_digits(n);
  int i;
  for (i = 0; i < nr / 2; i++) {
    if (digit(i, n) != digit(nr - 1 - i, n)) {
      return false;
    }
  }
  return true;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<is_palindrome>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  mov    %edi,%r12d
  push   %rbp
  push   %rbx
  xor    %ebx,%ebx
  call   <nr_of_digits@plt>
  mov    %eax,%r13d
  lea    -0x1(%rax),%r14d
  shr    $0x1f,%r13d
  add    %eax,%r13d
  sar    %r13d
  cmp    $0x1,%eax
  jg     L17
  jmp    L38
  xchg   %ax,%ax
L21:
  add    $0x1,%ebx
  cmp    %r13d,%ebx
  jge    L38
L17:
  mov    %r12d,%esi
  mov    %ebx,%edi
  call   <digit@plt>
  mov    %r14d,%edi
  mov    %r12d,%esi
  sub    %ebx,%edi
  mov    %rax,%rbp
  call   <digit@plt>
  cmp    %rax,%rbp
  je     L21
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopw   0x0(%rax,%rax,1)
L38:
  pop    %rbx
  mov    $0x1,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
bool is_palindrome(int n) {
  int nr = nr_of_digits(n);
  int i;
  for (i = 0; i < nr / 2; i++) {
    if (digit(i, n) != digit(nr - 1 - i, n)) {
      return false;
    }
  }
  return true;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<is_palindrome>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  mov    %edi,%r12d
  push   %rbp
  push   %rbx
  xor    %ebx,%ebx
  call   <nr_of_digits@plt>
  mov    %eax,%r13d
  lea    -0x1(%rax),%r14d
  shr    $0x1f,%r13d
  add    %eax,%r13d
  sar    %r13d
  cmp    $0x1,%eax
  jg     L34
  jmp    L19
  xchg   %ax,%ax
L22:
  add    $0x1,%ebx
  cmp    %r13d,%ebx
  jge    L19
L34:
  mov    %r12d,%esi
  mov    %ebx,%edi
  call   <digit@plt>
  mov    %r14d,%edi
  mov    %r12d,%esi
  sub    %ebx,%edi
  mov    %rax,%rbp
  call   <digit@plt>
  cmp    %rax,%rbp
  je     L22
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopw   0x0(%rax,%rax,1)
L19:
  pop    %rbx
  mov    $0x1,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
bool is_palindrome(int n) {
  int nr = nr_of_digits(n);
  int i;
  for (i = 0; i < nr / 2; i++) {
    if (digit(i, n) != digit(nr - 1 - i, n)) {
      return false;
    }
  }
  return true;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<pdmatrix>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    -0x14(%rbp),%eax
  cltq
  shl    $0x3,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L37
  lea    D15(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x1,%edi
  call   <exit@plt>
L37:
  mov    -0x14(%rbp),%eax
  imul   -0x18(%rbp),%eax
  cltq
  shl    $0x3,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,(%rax)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rax
  test   %rax,%rax
  jne    L21
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x1,%edi
  call   <exit@plt>
L21:
  movl   $0x1,-0xc(%rbp)
  jmp    L17
L16:
  mov    -0xc(%rbp),%eax
  cltq
  shl    $0x3,%rax
  lea    -0x8(%rax),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x18(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rcx
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rsi
  mov    -0x8(%rbp),%rax
  add    %rsi,%rax
  add    %rcx,%rdx
  mov    %rdx,(%rax)
  addl   $0x1,-0xc(%rbp)
L17:
  mov    -0xc(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L16
  mov    -0x8(%rbp),%rax
  leave
  ret
```

tool_calls: name: parse_data
arguments: label: D15
type_guessed: string
name: parse_data
arguments: label: D1
type_guessed: string

tool: D15: allocation error in pdmatrix 1.


tool: D1: allocation error in pdmatrix 2.


assistant: ```c
double **pdmatrix(int rows, int cols) {
  int i;
  double **m;
  m = (double **)malloc(rows * sizeof(double *));
  if (!m) {
    printf("allocation error in pdmatrix 1.\n");
    exit(1);
  }
  m[0] = (double *)malloc(rows * cols * sizeof(double));
  if (!m[0]) {
    printf("allocation error in pdmatrix 2.\n");
    exit(1);
  }
  for (i = 1; i < rows; i++) {
    m[i] = m[i - 1] + cols;
  }
  return m;
}
```
user: Translate the below assembly code into C programming language:

```asm
<pdmatrix>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%r12d
  mov    %esi,%ebx
  lea    0x0(,%rdi,8),%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  je     L46
L54:
  mov    %r12d,%edi
  imul   %ebx,%edi
  shl    $0x3,%edi
  call   <malloc@plt>
  mov    %rax,0x0(%rbp)
  test   %rax,%rax
  je     L47
L55:
  cmp    $0x1,%r12d
  jle    L45
  movslq %ebx,%rcx
  shl    $0x3,%rcx
  mov    %rbp,%rax
  lea    -0x1(%r12),%edx
  lea    0x0(%rbp,%rdx,8),%rsi
L0:
  mov    (%rax),%rdx
  add    $0x8,%rax
  add    %rcx,%rdx
  mov    %rdx,(%rax)
  cmp    %rsi,%rax
  jne    L0
L45:
  mov    %rbp,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L46:
  lea    D4(%rip),%rdi
  call   <printf@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  jmp    L54
L47:
  lea    D7(%rip),%rdi
  call   <printf@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  jmp    L55
```

tool_calls: name: retrieve_data
arguments: data_label: D4
type_guessed: string
name: retrieve_data
arguments: data_label: D7
type_guessed: string

tool: D4: allocation error in pdmatrix 1.


tool: D7: allocation error in pdmatrix 2.


assistant: ```c
double **pdmatrix(int rows, int cols) {
  int i;
  double **m;
  m = (double **)malloc(rows * sizeof(double *));
  if (!m) {
    printf("allocation error in pdmatrix 1.\n");
    exit(1);
  }
  m[0] = (double *)malloc(rows * cols * sizeof(double));
  if (!m[0]) {
    printf("allocation error in pdmatrix 2.\n");
    exit(1);
  }
  for (i = 1; i < rows; i++) {
    m[i] = m[i - 1] + cols;
  }
  return m;
}
```
user: What is the c source code of the assembly code below?

```asm
<pdmatrix>:
  endbr64
  push   %r12
  push   %rbp
  mov    %edi,%ebp
  shl    $0x3,%edi
  push   %rbx
  mov    %esi,%ebx
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L43
L57:
  mov    %ebp,%edi
  imul   %ebx,%edi
  shl    $0x3,%edi
  call   <malloc@plt>
  mov    %rax,(%r12)
  test   %rax,%rax
  je     L53
L16:
  cmp    $0x1,%ebp
  jle    L8
  movslq %ebx,%rcx
  lea    -0x1(%rbp),%edx
  mov    %r12,%rax
  shl    $0x3,%rcx
  lea    (%r12,%rdx,8),%rsi
  nopl   0x0(%rax,%rax,1)
L37:
  mov    (%rax),%rdx
  add    $0x8,%rax
  add    %rcx,%rdx
  mov    %rdx,(%rax)
  cmp    %rsi,%rax
  jne    L37
L8:
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L43:
  lea    D3(%rip),%rdi
  call   <printf@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  jmp    L57
  nopl   0x0(%rax,%rax,1)
L53:
  lea    D11(%rip),%rdi
  call   <printf@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  jmp    L16
```

tool_calls: name: retrieve_data
arguments: tag: D3
type_infered: string
name: retrieve_data
arguments: tag: D11
type_infered: string

tool: D3: allocation error in pdmatrix 1.


tool: D11: allocation error in pdmatrix 2.


assistant: ```c
double **pdmatrix(int rows, int cols) {
  int i;
  double **m;
  m = (double **)malloc(rows * sizeof(double *));
  if (!m) {
    printf("allocation error in pdmatrix 1.\n");
    exit(1);
  }
  m[0] = (double *)malloc(rows * cols * sizeof(double));
  if (!m[0]) {
    printf("allocation error in pdmatrix 2.\n");
    exit(1);
  }
  for (i = 1; i < rows; i++) {
    m[i] = m[i - 1] + cols;
  }
  return m;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<pdmatrix>:
  endbr64
  push   %r12
  push   %rbp
  mov    %edi,%ebp
  shl    $0x3,%edi
  push   %rbx
  mov    %esi,%ebx
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L21
L43:
  mov    %ebp,%edi
  imul   %ebx,%edi
  shl    $0x3,%edi
  call   <malloc@plt>
  mov    %rax,(%r12)
  test   %rax,%rax
  je     L2
L58:
  cmp    $0x1,%ebp
  jle    L6
  movslq %ebx,%rcx
  lea    -0x2(%rbp),%esi
  mov    (%r12),%rdx
  lea    0x8(%r12),%rax
  shl    $0x3,%rcx
  lea    0x10(%r12,%rsi,8),%rsi
  nop
L35:
  add    %rcx,%rdx
  add    $0x8,%rax
  mov    %rdx,-0x8(%rax)
  cmp    %rsi,%rax
  jne    L35
L6:
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L21:
  lea    D12(%rip),%rdi
  call   <printf@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  jmp    L43
  nopl   0x0(%rax,%rax,1)
L2:
  lea    D15(%rip),%rdi
  call   <printf@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  jmp    L58
```

tool_calls: name: disassemble_data
arguments: data_tag: D12
type_infered: string
name: disassemble_data
arguments: data_tag: D15
type_infered: string

tool: D12: allocation error in pdmatrix 1.


tool: D15: allocation error in pdmatrix 2.


assistant: ```c
double **pdmatrix(int rows, int cols) {
  int i;
  double **m;
  m = (double **)malloc(rows * sizeof(double *));
  if (!m) {
    printf("allocation error in pdmatrix 1.\n");
    exit(1);
  }
  m[0] = (double *)malloc(rows * cols * sizeof(double));
  if (!m[0]) {
    printf("allocation error in pdmatrix 2.\n");
    exit(1);
  }
  for (i = 1; i < rows; i++) {
    m[i] = m[i - 1] + cols;
  }
  return m;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<smallhash_function>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L49
L54:
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x63,%al
  jne    L10
  mov    $0xffffffff,%eax
  jmp    L34
L10:
  addl   $0x1,-0x8(%rbp)
L49:
  cmpl   $0x5,-0x8(%rbp)
  jle    L54
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,-0x4(%rbp)
  movl   $0x1,-0x8(%rbp)
  jmp    L36
L13:
  mov    -0x4(%rbp),%eax
  lea    0x0(,%rax,4),%ecx
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  mov    %eax,%edx
  sar    $0x7,%dl
  shr    $0x6,%dl
  add    %edx,%eax
  and    $0x3,%eax
  sub    %edx,%eax
  movsbl %al,%eax
  add    %ecx,%eax
  movslq %eax,%rdx
  imul   $0x68cb72fb,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0xb,%edx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  mov    %edx,-0x4(%rbp)
  mov    -0x4(%rbp),%edx
  imul   $0x138b,%edx,%edx
  sub    %edx,%eax
  mov    %eax,-0x4(%rbp)
  addl   $0x1,-0x8(%rbp)
L36:
  cmpl   $0x5,-0x8(%rbp)
  jle    L13
  mov    -0x4(%rbp),%eax
L34:
  pop    %rbp
  ret
```

assistant: ```c
int smallhash_function(char *lmer) {
  int x, ans;
  for (x = 0; x < 6; x++) {
    if (lmer[x] == 99)
      return -1;
  }
  ans = lmer[0];
  for (x = 1; x < 6; x++)
    ans = (4 * ans + (lmer[x] % 4)) % 5003;
  return ans;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<smallhash_function>:
  endbr64
  mov    %rdi,%rax
  lea    0x6(%rdi),%r8
L15:
  cmpb   $0x63,(%rax)
  je     L28
  add    $0x1,%rax
  cmp    %r8,%rax
  jne    L15
  movsbl (%rdi),%eax
  lea    0x1(%rdi),%rcx
L4:
  movzbl (%rcx),%edx
  mov    %edx,%esi
  sar    $0x7,%sil
  shr    $0x6,%sil
  add    %esi,%edx
  and    $0x3,%edx
  sub    %esi,%edx
  movsbl %dl,%edx
  lea    (%rdx,%rax,4),%edx
  movslq %edx,%rax
  imul   $0x68cb72fb,%rax,%rax
  sar    $0x2b,%rax
  mov    %edx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  imul   $0x138b,%eax,%esi
  mov    %edx,%eax
  sub    %esi,%eax
  add    $0x1,%rcx
  cmp    %r8,%rcx
  jne    L4
  ret
L28:
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int smallhash_function(char *lmer) {
  int x, ans;
  for (x = 0; x < 6; x++) {
    if (lmer[x] == 99)
      return -1;
  }
  ans = lmer[0];
  for (x = 1; x < 6; x++)
    ans = (4 * ans + (lmer[x] % 4)) % 5003;
  return ans;
}
```