ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Demonstrate the C code that generates the assembly output below:

```asm
<nile_Sleep_doze>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  jmp    L18
L13:
  mov    $0x0,%eax
  call   <nile_pause@plt>
L18:
  mov    -0x4(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0x4(%rbp)
  test   %eax,%eax
  jne    L13
  nop
  nop
  leave
  ret
```

assistant: ```c
void nile_Sleep_doze(int npauses) {
  while (npauses--)
    nile_pause();
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<nile_Sleep_doze>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  test   %edi,%edi
  je     L4
L2:
  mov    $0x0,%eax
  call   <nile_pause@plt>
  sub    $0x1,%ebx
  jne    L2
L4:
  pop    %rbx
  ret
```

assistant: ```c
void nile_Sleep_doze(int npauses) {
  while (npauses--)
    nile_pause();
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<nile_Sleep_doze>:
  endbr64
  test   %edi,%edi
  je     L12
  push   %rbx
  mov    %edi,%ebx
  nopl   0x0(%rax,%rax,1)
L9:
  xor    %eax,%eax
  call   <nile_pause@plt>
  sub    $0x1,%ebx
  jne    L9
  pop    %rbx
  ret
  xchg   %ax,%ax
L12:
  ret
```

assistant: ```c
void nile_Sleep_doze(int npauses) {
  while (npauses--)
    nile_pause();
}
```
user: What is the c source code of the assembly code below?

```asm
<nile_Sleep_doze>:
  endbr64
  test   %edi,%edi
  je     L12
  push   %rbx
  mov    %edi,%ebx
  nopl   0x0(%rax,%rax,1)
L9:
  xor    %eax,%eax
  call   <nile_pause@plt>
  sub    $0x1,%ebx
  jne    L9
  pop    %rbx
  ret
  xchg   %ax,%ax
L12:
  ret
```

assistant: ```c
void nile_Sleep_doze(int npauses) {
  while (npauses--)
    nile_pause();
}
```
user: What is the c source code of the assembly code below:

```asm
<puts>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  jmp    L10
L5:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <putchar@plt>
  addq   $0x1,-0x8(%rbp)
L10:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L5
  nop
  nop
  leave
  ret
```

assistant: ```c
void puts(char *str) {
  while (*str != 0) {
    putchar(*str);
    str++;
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<puts>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  movzbl (%rdi),%edi
  test   %dil,%dil
  je     L13
L3:
  movsbl %dil,%edi
  call   <putchar@plt>
  add    $0x1,%rbx
  movzbl (%rbx),%edi
  test   %dil,%dil
  jne    L3
L13:
  pop    %rbx
  ret
```

assistant: ```c
void puts(char *str) {
  while (*str != 0) {
    putchar(*str);
    str++;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<puts>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  movsbl (%rdi),%edi
  test   %dil,%dil
  je     L14
L0:
  call   <putchar@plt>
  movsbl 0x1(%rbx),%edi
  add    $0x1,%rbx
  test   %dil,%dil
  jne    L0
L14:
  pop    %rbx
  ret
```

assistant: ```c
void puts(char *str) {
  while (*str != 0) {
    putchar(*str);
    str++;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<puts>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  movsbl (%rdi),%edi
  test   %dil,%dil
  je     L18
L2:
  call   <putchar@plt>
  movsbl 0x1(%rbx),%edi
  add    $0x1,%rbx
  test   %dil,%dil
  jne    L2
L18:
  pop    %rbx
  ret
```

assistant: ```c
void puts(char *str) {
  while (*str != 0) {
    putchar(*str);
    str++;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L0
  call   <__stack_chk_fail@plt>
L0:
  leave
  ret
```

assistant: ```c
int main(void) {
  int a;
  &a - 1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

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

assistant: ```c
int main(void) {
  int a;
  &a - 1;
}
```
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(void) {
  int a;
  &a - 1;
}
```
user: Converte this assembly snippet into its equivalent C code:

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

assistant: ```c
int main(void) {
  int a;
  &a - 1;
}
```
user: What is the c source code of the assembly code below?

```asm
<getPermutation>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %esi,-0x28(%rbp)
  mov    -0x24(%rbp),%eax
  cltq
  shl    $0x2,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x10(%rbp)
  movl   $0x0,-0x1c(%rbp)
  jmp    L79
L7:
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  movl   $0x0,(%rax)
  addl   $0x1,-0x1c(%rbp)
L79:
  mov    -0x1c(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jl     L7
  mov    -0x24(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0x1c(%rbp)
  jmp    L27
L64:
  mov    -0x24(%rbp),%eax
  sub    $0x1,%eax
  sub    -0x1c(%rbp),%eax
  mov    %eax,%edi
  call   <fact@plt>
  mov    %eax,-0x14(%rbp)
  movl   $0x0,-0x18(%rbp)
  jmp    L32
L41:
  mov    -0x18(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  test   %eax,%eax
  jne    L84
  mov    -0x28(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jg     L53
  mov    -0x18(%rbp),%eax
  lea    0x31(%rax),%ecx
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    %ecx,%edx
  mov    %dl,(%rax)
  mov    -0x18(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  movl   $0x1,(%rax)
  jmp    L94
L53:
  mov    -0x14(%rbp),%eax
  sub    %eax,-0x28(%rbp)
  jmp    L80
L84:
  nop
L80:
  addl   $0x1,-0x18(%rbp)
L32:
  mov    -0x18(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jl     L41
L94:
  addl   $0x1,-0x1c(%rbp)
L27:
  mov    -0x1c(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jl     L64
  mov    -0x24(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *getPermutation(int n, int k) {
  int *used;
  used = malloc(sizeof(int) * n);
  int i;
  for (i = 0; i < n; ++i) {
    used[i] = 0;
  }
  char *ret;
  ret = malloc(sizeof(char) * (n + 1));
  for (i = 0; i < n; ++i) {
    int j, num;
    num = fact(n - 1 - i);
    for (j = 0; j < n; ++j) {
      if (used[j])
        continue;
      if (k <= num) {
        ret[i] = '1' + j;
        used[j] = 1;
        break;
      } else {
        k -= num;
      }
    }
  }
  ret[n] = 0;
  free(used);
  return ret;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<getPermutation>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %edi,%ebp
  mov    %esi,%ebx
  movslq %edi,%rax
  mov    %rax,0x8(%rsp)
  lea    0x0(,%rdi,4),%edi
  call   <malloc@plt>
  mov    %rax,%r13
  test   %ebp,%ebp
  jle    L68
  lea    -0x1(%rbp),%r14d
  mov    %ebp,%eax
  lea    0x0(%r13,%rax,4),%rdx
  mov    %r13,%rax
L55:
  movl   $0x0,(%rax)
  add    $0x4,%rax
  cmp    %rdx,%rax
  jne    L55
  lea    0x1(%rbp),%edi
  call   <malloc@plt>
  mov    %rax,%r15
  add    $0x2,%r14
  mov    $0x1,%r12d
  jmp    L58
L33:
  add    $0x31,%ecx
  mov    %cl,-0x1(%r15,%r12,1)
  movl   $0x1,(%rdx)
L26:
  add    $0x1,%r12
  cmp    %r14,%r12
  je     L64
L58:
  mov    %ebp,%edi
  sub    %r12d,%edi
  call   <fact@plt>
  mov    %r13,%rdx
  mov    $0x0,%ecx
  jmp    L54
L86:
  add    $0x1,%ecx
  add    $0x4,%rdx
  cmp    %ecx,%ebp
  je     L26
L54:
  cmpl   $0x0,(%rdx)
  jne    L86
  cmp    %ebx,%eax
  jge    L33
  sub    %eax,%ebx
  jmp    L86
L68:
  lea    0x1(%rbp),%edi
  call   <malloc@plt>
  mov    %rax,%r15
L64:
  mov    0x8(%rsp),%rax
  movb   $0x0,(%r15,%rax,1)
  mov    %r13,%rdi
  call   <free@plt>
  mov    %r15,%rax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
char *getPermutation(int n, int k) {
  int *used;
  used = malloc(sizeof(int) * n);
  int i;
  for (i = 0; i < n; ++i) {
    used[i] = 0;
  }
  char *ret;
  ret = malloc(sizeof(char) * (n + 1));
  for (i = 0; i < n; ++i) {
    int j, num;
    num = fact(n - 1 - i);
    for (j = 0; j < n; ++j) {
      if (used[j])
        continue;
      if (k <= num) {
        ret[i] = '1' + j;
        used[j] = 1;
        break;
      } else {
        k -= num;
      }
    }
  }
  ret[n] = 0;
  free(used);
  return ret;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<getPermutation>:
  endbr64
  push   %r15
  movslq %edi,%rax
  push   %r14
  lea    0x0(,%rax,4),%edi
  mov    %rax,%r15
  mov    %esi,%r14d
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    0x1(%r15),%ebx
  sub    $0x18,%rsp
  mov    %rax,0x8(%rsp)
  call   <malloc@plt>
  mov    %rax,%rbp
  test   %r15d,%r15d
  jle    L19
  mov    %r15d,%edx
  xor    %esi,%esi
  mov    %rax,%rdi
  mov    $0x1,%r13d
  shl    $0x2,%rdx
  call   <memset@plt>
  mov    %ebx,%edi
  call   <malloc@plt>
  mov    %rax,%r12
  nopl   (%rax)
L52:
  mov    %r15d,%edi
  sub    %r13d,%edi
  call   <fact@plt>
  mov    %rbp,%rcx
  xor    %edx,%edx
L9:
  mov    (%rcx),%esi
  test   %esi,%esi
  jne    L2
  cmp    %r14d,%eax
  jge    L55
  sub    %eax,%r14d
L2:
  add    $0x1,%edx
  add    $0x4,%rcx
  cmp    %edx,%r15d
  jne    L9
  add    $0x1,%r13
  cmp    %rbx,%r13
  jne    L52
L50:
  mov    0x8(%rsp),%rax
  mov    %rbp,%rdi
  movb   $0x0,(%r12,%rax,1)
  call   <free@plt>
  add    $0x18,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L55:
  add    $0x31,%edx
  mov    %dl,-0x1(%r12,%r13,1)
  add    $0x1,%r13
  movl   $0x1,(%rcx)
  cmp    %rbx,%r13
  jne    L52
  jmp    L50
L19:
  mov    %ebx,%edi
  call   <malloc@plt>
  mov    %rax,%r12
  jmp    L50
```

assistant: ```c
char *getPermutation(int n, int k) {
  int *used;
  used = malloc(sizeof(int) * n);
  int i;
  for (i = 0; i < n; ++i) {
    used[i] = 0;
  }
  char *ret;
  ret = malloc(sizeof(char) * (n + 1));
  for (i = 0; i < n; ++i) {
    int j, num;
    num = fact(n - 1 - i);
    for (j = 0; j < n; ++j) {
      if (used[j])
        continue;
      if (k <= num) {
        ret[i] = '1' + j;
        used[j] = 1;
        break;
      } else {
        k -= num;
      }
    }
  }
  ret[n] = 0;
  free(used);
  return ret;
}
```
user: Translate the below assembly code into C programming language:

```asm
<getPermutation>:
  endbr64
  push   %r15
  movslq %edi,%rax
  push   %r14
  lea    0x0(,%rax,4),%edi
  mov    %rax,%r15
  mov    %esi,%r14d
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    0x1(%r15),%ebx
  sub    $0x18,%rsp
  mov    %rax,0x8(%rsp)
  call   <malloc@plt>
  mov    %rax,%rbp
  test   %r15d,%r15d
  jle    L9
  mov    %r15d,%edx
  xor    %esi,%esi
  mov    %rax,%rdi
  mov    $0x1,%r13d
  shl    $0x2,%rdx
  call   <memset@plt>
  mov    %ebx,%edi
  call   <malloc@plt>
  mov    %rax,%r12
  nopl   (%rax)
L71:
  mov    %r15d,%edi
  sub    %r13d,%edi
  call   <fact@plt>
  mov    %rbp,%rcx
  xor    %edx,%edx
L47:
  mov    (%rcx),%esi
  test   %esi,%esi
  jne    L6
  cmp    %r14d,%eax
  jge    L68
  sub    %eax,%r14d
L6:
  add    $0x1,%edx
  add    $0x4,%rcx
  cmp    %edx,%r15d
  jne    L47
  add    $0x1,%r13
  cmp    %rbx,%r13
  jne    L71
L16:
  mov    0x8(%rsp),%rax
  mov    %rbp,%rdi
  movb   $0x0,(%r12,%rax,1)
  call   <free@plt>
  add    $0x18,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L68:
  add    $0x31,%edx
  mov    %dl,-0x1(%r12,%r13,1)
  add    $0x1,%r13
  movl   $0x1,(%rcx)
  cmp    %rbx,%r13
  jne    L71
  jmp    L16
L9:
  mov    %ebx,%edi
  call   <malloc@plt>
  mov    %rax,%r12
  jmp    L16
```

assistant: ```c
char *getPermutation(int n, int k) {
  int *used;
  used = malloc(sizeof(int) * n);
  int i;
  for (i = 0; i < n; ++i) {
    used[i] = 0;
  }
  char *ret;
  ret = malloc(sizeof(char) * (n + 1));
  for (i = 0; i < n; ++i) {
    int j, num;
    num = fact(n - 1 - i);
    for (j = 0; j < n; ++j) {
      if (used[j])
        continue;
      if (k <= num) {
        ret[i] = '1' + j;
        used[j] = 1;
        break;
      } else {
        k -= num;
      }
    }
  }
  ret[n] = 0;
  free(used);
  return ret;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<ft_str_is_lowercase>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L18
L16:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x60,%al
  jle    L33
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x7a,%al
  jle    L37
L33:
  mov    $0x0,%eax
  jmp    L48
L37:
  addl   $0x1,-0x4(%rbp)
L18:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L16
  mov    $0x1,%eax
L48:
  pop    %rbp
  ret
```

assistant: ```c
int ft_str_is_lowercase(char *str) {
  int i;
  i = 0;
  while (str[i] != '\0') {
    if (str[i] < 'a' || str[i] > 'z')
      return (0);
    i++;
  }
  return (1);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<ft_str_is_lowercase>:
  endbr64
  movzbl (%rdi),%eax
  test   %al,%al
  je     L16
  add    $0x1,%rdi
L13:
  sub    $0x61,%eax
  cmp    $0x19,%al
  ja     L20
  add    $0x1,%rdi
  movzbl -0x1(%rdi),%eax
  test   %al,%al
  jne    L13
  mov    $0x1,%eax
  ret
L16:
  mov    $0x1,%eax
  ret
L20:
  mov    $0x0,%eax
  ret
```

assistant: ```c
int ft_str_is_lowercase(char *str) {
  int i;
  i = 0;
  while (str[i] != '\0') {
    if (str[i] < 'a' || str[i] > 'z')
      return (0);
    i++;
  }
  return (1);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<ft_str_is_lowercase>:
  endbr64
  movzbl (%rdi),%eax
  test   %al,%al
  je     L1
  add    $0x1,%rdi
  jmp    L12
  nopl   0x0(%rax)
L4:
  movzbl (%rdi),%eax
  add    $0x1,%rdi
  test   %al,%al
  je     L1
L12:
  sub    $0x61,%eax
  cmp    $0x19,%al
  jbe    L4
  xor    %eax,%eax
  ret
  nopl   (%rax)
L1:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int ft_str_is_lowercase(char *str) {
  int i;
  i = 0;
  while (str[i] != '\0') {
    if (str[i] < 'a' || str[i] > 'z')
      return (0);
    i++;
  }
  return (1);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<ft_str_is_lowercase>:
  endbr64
  movzbl (%rdi),%eax
  test   %al,%al
  je     L23
  add    $0x1,%rdi
  jmp    L24
  nopl   0x0(%rax)
L33:
  movzbl (%rdi),%eax
  add    $0x1,%rdi
  test   %al,%al
  je     L23
L24:
  sub    $0x61,%eax
  cmp    $0x19,%al
  jbe    L33
  xor    %eax,%eax
  ret
  nopl   (%rax)
L23:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int ft_str_is_lowercase(char *str) {
  int i;
  i = 0;
  while (str[i] != '\0') {
    if (str[i] < 'a' || str[i] > 'z')
      return (0);
    i++;
  }
  return (1);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<Delete>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x1,%eax
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D2
type_infered: string

tool: D2: Calling from Delete


assistant: ```c
int Delete(int argc, char **argv) {
  printf("Calling from Delete\n");
  return 1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<Delete>:
  endbr64
  sub    $0x8,%rsp
  lea    D9(%rip),%rdi
  call   <printf@plt>
  mov    $0x1,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: label: D9
data_type: string

tool: D9: Calling from Delete


assistant: ```c
int Delete(int argc, char **argv) {
  printf("Calling from Delete\n");
  return 1;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<Delete>:
  endbr64
  sub    $0x8,%rsp
  lea    D0(%rip),%rdi
  call   <printf@plt>
  mov    $0x1,%eax
  add    $0x8,%rsp
  ret
```

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

tool: D0: Calling from Delete


assistant: ```c
int Delete(int argc, char **argv) {
  printf("Calling from Delete\n");
  return 1;
}
```
user: What is the c source code of the assembly code below:

```asm
<Delete>:
  endbr64
  sub    $0x8,%rsp
  lea    D2(%rip),%rdi
  call   <printf@plt>
  mov    $0x1,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: label: D2
type_guessed: string

tool: D2: Calling from Delete


assistant: ```c
int Delete(int argc, char **argv) {
  printf("Calling from Delete\n");
  return 1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<binaryToFloat>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    %rdx,-0x20(%rbp)
  pxor   %xmm0,%xmm0
  movss  %xmm0,-0xc(%rbp)
  movl   $0x1,-0x8(%rbp)
  jmp    L19
L7:
  mov    -0x8(%rbp),%eax
  neg    %eax
  mov    %eax,%esi
  mov    $0x2,%edi
  call   <pow@plt>
  movd   %xmm0,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x14(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    -0x8(%rbp),%eax
  add    %edx,%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x31,%al
  jne    L3
  movss  -0xc(%rbp),%xmm0
  addss  -0x4(%rbp),%xmm0
  movss  %xmm0,-0xc(%rbp)
L3:
  addl   $0x1,-0x8(%rbp)
L19:
  mov    -0x8(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jl     L7
  movss  -0xc(%rbp),%xmm0
  leave
  ret
```

assistant: ```c
float binaryToFloat(int firstBit, int bits, char binary[bits]) {
  float sum = 0;
  for (int i = 1; i < bits; i++) {
    float divisor = (pow(2, -1 * i));
    if (binary[i + (firstBit - 1)] == '1')
      sum += divisor;
  }
  return sum;
}
```
user: What is the c source code of the assembly code below?

```asm
<binaryToFloat>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  cmp    $0x1,%esi
  jle    L25
  movslq %edi,%rdi
  lea    (%rdx,%rdi,1),%rbp
  lea    0x1(%rdx,%rdi,1),%r12
  lea    -0x2(%rsi),%eax
  add    %rax,%r12
  mov    %rbp,%rbx
  mov    D9(%rip),%r13d
  sub    $0x1,%ebp
  jmp    L8
L44:
  add    $0x1,%rbx
  cmp    %r12,%rbx
  je     L10
L8:
  mov    %ebp,%esi
  sub    %ebx,%esi
  mov    $0x2,%edi
  call   <pow@plt>
  cmpb   $0x31,(%rbx)
  jne    L44
  movd   %r13d,%xmm1
  addss  %xmm0,%xmm1
  movd   %xmm1,%r13d
  jmp    L44
L25:
  mov    D9(%rip),%r13d
L10:
  movd   %r13d,%xmm0
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: read_data
arguments: data_tag: D9
type_infered: f32

tool: D9: 0

assistant: ```c
float binaryToFloat(int firstBit, int bits, char binary[bits]) {
  float sum = 0;
  for (int i = 1; i < bits; i++) {
    float divisor = (pow(2, -1 * i));
    if (binary[i + (firstBit - 1)] == '1')
      sum += divisor;
  }
  return sum;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<binaryToFloat>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  cmp    $0x1,%esi
  jle    L22
  lea    -0x1(%rdi),%eax
  lea    -0x2(%rsi),%ecx
  pxor   %xmm2,%xmm2
  movslq %edi,%rbp
  cltq
  add    %rdx,%rbp
  movss  %xmm2,0xc(%rsp)
  lea    (%rdx,%rax,1),%rbx
  add    %rcx,%rbp
  lea    -0x1(%rbx),%r12d
  nopl   0x0(%rax)
L7:
  mov    %r12d,%esi
  mov    $0x2,%edi
  sub    %ebx,%esi
  call   <pow@plt>
  cmpb   $0x31,0x1(%rbx)
  jne    L23
  addss  0xc(%rsp),%xmm0
  movss  %xmm0,0xc(%rsp)
L23:
  add    $0x1,%rbx
  cmp    %rbp,%rbx
  jne    L7
  movss  0xc(%rsp),%xmm0
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L22:
  pxor   %xmm3,%xmm3
  movss  %xmm3,0xc(%rsp)
  movss  0xc(%rsp),%xmm0
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
float binaryToFloat(int firstBit, int bits, char binary[bits]) {
  float sum = 0;
  for (int i = 1; i < bits; i++) {
    float divisor = (pow(2, -1 * i));
    if (binary[i + (firstBit - 1)] == '1')
      sum += divisor;
  }
  return sum;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<binaryToFloat>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  cmp    $0x1,%esi
  jle    L14
  lea    -0x1(%rdi),%eax
  lea    -0x2(%rsi),%ecx
  pxor   %xmm2,%xmm2
  movslq %edi,%rbp
  cltq
  add    %rdx,%rbp
  movss  %xmm2,0xc(%rsp)
  lea    (%rdx,%rax,1),%rbx
  add    %rcx,%rbp
  lea    -0x1(%rbx),%r12d
  nopl   0x0(%rax)
L5:
  mov    %r12d,%esi
  mov    $0x2,%edi
  sub    %ebx,%esi
  call   <pow@plt>
  cmpb   $0x31,0x1(%rbx)
  jne    L9
  addss  0xc(%rsp),%xmm0
  movss  %xmm0,0xc(%rsp)
L9:
  add    $0x1,%rbx
  cmp    %rbp,%rbx
  jne    L5
  movss  0xc(%rsp),%xmm0
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L14:
  pxor   %xmm3,%xmm3
  movss  %xmm3,0xc(%rsp)
  movss  0xc(%rsp),%xmm0
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
float binaryToFloat(int firstBit, int bits, char binary[bits]) {
  float sum = 0;
  for (int i = 1; i < bits; i++) {
    float divisor = (pow(2, -1 * i));
    if (binary[i + (firstBit - 1)] == '1')
      sum += divisor;
  }
  return sum;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<FruTypeString>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,%edx
  mov    %esi,%eax
  mov    %dl,-0x14(%rbp)
  mov    %al,-0x18(%rbp)
  movzbl -0x14(%rbp),%eax
  sub    $0x7,%eax
  cmp    $0x39,%eax
  ja     L57
  mov    %eax,%eax
  lea    0x0(,%rax,4),%rdx
  lea    D23(%rip),%rax
  mov    (%rdx,%rax,1),%eax
  cltq
  lea    D23(%rip),%rdx
  add    %rdx,%rax
  notrack jmp *%rax
  cmpb   $0x0,-0x18(%rbp)
  jne    L17
  lea    D10(%rip),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L52
L17:
  lea    D86(%rip),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L52
  lea    D57(%rip),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L52
  lea    D1(%rip),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L52
  lea    D56(%rip),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L52
  lea    D3(%rip),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L52
  lea    D53(%rip),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L52
  lea    D69(%rip),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L52
L57:
  lea    D103(%rip),%rax
  mov    %rax,-0x8(%rbp)
  nop
L52:
  mov    -0x8(%rbp),%rax
  pop    %rbp
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D10
data_type: string
name: retrieve_data
arguments: tag: D86
data_type: string
name: retrieve_data
arguments: tag: D57
data_type: string
name: retrieve_data
arguments: tag: D1
data_type: string
name: retrieve_data
arguments: tag: D56
data_type: string
name: retrieve_data
arguments: tag: D3
data_type: string
name: retrieve_data
arguments: tag: D53
data_type: string
name: retrieve_data
arguments: tag: D69
data_type: string
name: retrieve_data
arguments: tag: D103
data_type: string

tool: D10: Baseboard

tool: D86: Board    

tool: D57: PowerSply

tool: D1: PowerCage

tool: D56: DIMM     

tool: D3: HotSwapCt

tool: D53: ME       

tool: D69: SysInfo  

tool: D103: Component

assistant: ```c
char *FruTypeString(unsigned char frutype, unsigned char dev) {
  char *pstr;
  switch (frutype) {
  case 0x07:
  case 0x0c:
    if (dev == 0)
      pstr = "Baseboard";
    else
      pstr = "Board    ";
    break;
  case 0x0a:
    pstr = "PowerSply";
    break;
  case 0x15:
    pstr = "PowerCage";
    break;
  case 0x20:
    pstr = "DIMM     ";
    break;
  case 0x0f:
    pstr = "HotSwapCt";
    break;
  case 0x2e:
    pstr = "ME       ";
    break;
  case 0x40:
    pstr = "SysInfo  ";
    break;
  default:
    pstr = "Component";
    break;
  }
  return (pstr);
}
```