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

```asm
<__make_fd_async>:
  endbr64
  mov    D3(%rip),%rax
  push   %rbp
  xor    %edx,%edx
  mov    %edi,%ebp
  mov    (%rax),%esi
  call   <fcntl@plt>
  cmp    $0xffffffff,%eax
  je     L8
  mov    D18(%rip),%rdx
  mov    %ebp,%edi
  or     (%rdx),%eax
  mov    %eax,%edx
  mov    D8(%rip),%rax
  mov    (%rax),%esi
  call   <fcntl@plt>
  cmp    $0xffffffff,%eax
  sete   %al
  movzbl %al,%eax
  neg    %eax
L8:
  pop    %rbp
  ret
```

assistant: ```c
int __make_fd_async(int fd) {
  int current_flags = fcntl(fd, F_GETFL, 0);
  if ((current_flags) == -1) {
    return -1;
  };
  int new_flags = current_flags | O_NONBLOCK;
  if ((fcntl(fd, F_SETFL, new_flags)) == -1) {
    return -1;
  };
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<fibo>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %edi,-0x14(%rbp)
  cmpl   $0x1,-0x14(%rbp)
  jg     L19
  mov    -0x14(%rbp),%eax
  jmp    L15
L19:
  mov    -0x14(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,%edi
  call   <fibo>
  mov    %eax,%ebx
  mov    -0x14(%rbp),%eax
  sub    $0x2,%eax
  mov    %eax,%edi
  call   <fibo>
  add    %ebx,%eax
L15:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int fibo(int n) {
  if (n <= 1)
    return n;
  else
    return fibo(n - 1) + fibo(n - 2);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<fibo>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebx
  cmp    $0x1,%edi
  jg     L13
L1:
  mov    %ebx,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L13:
  lea    -0x1(%rdi),%edi
  call   <fibo>
  mov    %eax,%ebp
  lea    -0x2(%rbx),%edi
  call   <fibo>
  lea    0x0(%rbp,%rax,1),%ebx
  jmp    L1
```

assistant: ```c
int fibo(int n) {
  if (n <= 1)
    return n;
  else
    return fibo(n - 1) + fibo(n - 2);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<oml_stralloc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  add    $0x1,%rax
  mov    %rax,%rdi
  call   <oml_malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    -0x8(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *oml_stralloc(size_t len) {
  char *ret = oml_malloc(len + 1);
  memset(ret, 0, len + 1);
  return ret;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<oml_stralloc>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  lea    0x1(%rdi),%rbp
  mov    %rbp,%rdi
  call   <oml_malloc@plt>
  mov    %rax,%rbx
  mov    %rbp,%rdx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  mov    %rbx,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
char *oml_stralloc(size_t len) {
  char *ret = oml_malloc(len + 1);
  memset(ret, 0, len + 1);
  return ret;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<oml_stralloc>:
  endbr64
  push   %r13
  lea    0x1(%rdi),%r13
  push   %r12
  mov    %r13,%rdi
  sub    $0x8,%rsp
  call   <oml_malloc@plt>
  mov    %r13,%rdx
  xor    %esi,%esi
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <memset@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *oml_stralloc(size_t len) {
  char *ret = oml_malloc(len + 1);
  memset(ret, 0, len + 1);
  return ret;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<oml_stralloc>:
  endbr64
  push   %r13
  lea    0x1(%rdi),%r13
  push   %r12
  mov    %r13,%rdi
  sub    $0x8,%rsp
  call   <oml_malloc@plt>
  mov    %r13,%rdx
  xor    %esi,%esi
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <memset@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *oml_stralloc(size_t len) {
  char *ret = oml_malloc(len + 1);
  memset(ret, 0, len + 1);
  return ret;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<strlen>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movq   $0x0,-0x8(%rbp)
  jmp    L10
L0:
  addq   $0x1,-0x8(%rbp)
L10:
  mov    -0x18(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L0
  mov    -0x8(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
size_t strlen(const char *str) {
  size_t len = 0;
  while (str[len] != 0) {
    len++;
  }
  return len;
}
```
user: What is the c source code of the assembly code below:

```asm
<strlen>:
  endbr64
  cmpb   $0x0,(%rdi)
  je     L15
  mov    $0x0,%eax
L11:
  add    $0x1,%rax
  cmpb   $0x0,(%rdi,%rax,1)
  jne    L11
  ret
L15:
  mov    $0x0,%eax
  ret
```

assistant: ```c
size_t strlen(const char *str) {
  size_t len = 0;
  while (str[len] != 0) {
    len++;
  }
  return len;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<strlen>:
  endbr64
  xor    %eax,%eax
  cmpb   $0x0,(%rdi)
  je     L15
  nopl   0x0(%rax,%rax,1)
L3:
  add    $0x1,%rax
  cmpb   $0x0,(%rdi,%rax,1)
  jne    L3
  ret
  nopl   0x0(%rax,%rax,1)
L15:
  ret
```

assistant: ```c
size_t strlen(const char *str) {
  size_t len = 0;
  while (str[len] != 0) {
    len++;
  }
  return len;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<strlen>:
  endbr64
  xor    %eax,%eax
  cmpb   $0x0,(%rdi)
  je     L14
  nopl   0x0(%rax,%rax,1)
L3:
  add    $0x1,%rax
  cmpb   $0x0,(%rdi,%rax,1)
  jne    L3
  ret
  nopl   0x0(%rax,%rax,1)
L14:
  ret
```

assistant: ```c
size_t strlen(const char *str) {
  size_t len = 0;
  while (str[len] != 0) {
    len++;
  }
  return len;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<fastcos>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movss  %xmm0,-0x4(%rbp)
  movss  -0x4(%rbp),%xmm0
  movss  %xmm0,-0x8(%rbp)
  flds   -0x8(%rbp)
  fsin
  fstps  -0x4(%rbp)
  movss  -0x4(%rbp),%xmm0
  pop    %rbp
  ret
```

assistant: ```c
float fastcos(float x) {
  asm("fsin" : "+t"(x));
  return x;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<fastcos>:
  endbr64
  movss  %xmm0,-0x4(%rsp)
  flds   -0x4(%rsp)
  fsin
  fstps  -0x4(%rsp)
  movss  -0x4(%rsp),%xmm0
  ret
```

assistant: ```c
float fastcos(float x) {
  asm("fsin" : "+t"(x));
  return x;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<fastcos>:
  endbr64
  movss  %xmm0,-0x4(%rsp)
  flds   -0x4(%rsp)
  fsin
  fstps  -0x4(%rsp)
  movss  -0x4(%rsp),%xmm0
  ret
```

assistant: ```c
float fastcos(float x) {
  asm("fsin" : "+t"(x));
  return x;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<fastcos>:
  endbr64
  movss  %xmm0,-0x4(%rsp)
  flds   -0x4(%rsp)
  fsin
  fstps  -0x4(%rsp)
  movss  -0x4(%rsp),%xmm0
  ret
```

assistant: ```c
float fastcos(float x) {
  asm("fsin" : "+t"(x));
  return x;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %edi,-0x44(%rbp)
  mov    %rsi,-0x50(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    $0x18,%edi
  call   <malloc@plt>
  mov    %rax,-0x30(%rbp)
  mov    -0x30(%rbp),%rax
  movsd  D64(%rip),%xmm0
  movsd  %xmm0,(%rax)
  mov    -0x30(%rbp),%rax
  add    $0x8,%rax
  movsd  D6(%rip),%xmm0
  movsd  %xmm0,(%rax)
  mov    -0x30(%rbp),%rax
  add    $0x10,%rax
  movsd  D49(%rip),%xmm0
  movsd  %xmm0,(%rax)
  movsd  D40(%rip),%xmm0
  movsd  %xmm0,-0x20(%rbp)
  movsd  D69(%rip),%xmm0
  movsd  %xmm0,-0x18(%rbp)
  movsd  D45(%rip),%xmm0
  movsd  %xmm0,-0x10(%rbp)
  movl   $0x3,-0x38(%rbp)
  movl   $0x1,-0x34(%rbp)
  lea    -0x34(%rbp),%rdi
  lea    -0x20(%rbp),%rcx
  lea    -0x34(%rbp),%rdx
  mov    -0x30(%rbp),%rsi
  lea    -0x38(%rbp),%rax
  mov    %rdi,%r8
  mov    %rax,%rdi
  call   <ddot_@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x28(%rbp)
  mov    -0x28(%rbp),%rax
  movq   %rax,%xmm0
  lea    D11(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L7
  call   <__stack_chk_fail@plt>
L7:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: label: D64
datatype: f64
name: disassemble_data
arguments: label: D6
datatype: f64
name: disassemble_data
arguments: label: D49
datatype: f64
name: disassemble_data
arguments: label: D40
datatype: f64
name: disassemble_data
arguments: label: D69
datatype: f64
name: disassemble_data
arguments: label: D45
datatype: f64
name: disassemble_data
arguments: label: D11
datatype: string

tool: D64: 1

tool: D6: 2

tool: D49: 3

tool: D40: 4

tool: D69: 5

tool: D45: 6

tool: D11:  The dot product is: %f 


assistant: ```c
int main(int argc, char **argv) {
  double *a = (double *)malloc(3 * sizeof(double));
  a[0] = 1.0;
  a[1] = 2.0;
  a[2] = 3.0;
  double b[3] = {4.0, 5.0, 6.0};
  int N = 3, one = 1;
  double dot_product = ddot_(&N, a, &one, b, &one);
  printf(" The dot product is: %f \n", dot_product);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    $0x18,%edi
  call   <malloc@plt>
  mov    %rax,%rsi
  mov    D0(%rip),%rax
  mov    %rax,(%rsi)
  mov    D52(%rip),%rax
  mov    %rax,0x8(%rsi)
  mov    D12(%rip),%rax
  mov    %rax,0x10(%rsi)
  mov    D28(%rip),%rax
  mov    %rax,0x10(%rsp)
  mov    D16(%rip),%rax
  mov    %rax,0x18(%rsp)
  mov    D4(%rip),%rax
  mov    %rax,0x20(%rsp)
  movl   $0x3,0x8(%rsp)
  movl   $0x1,0xc(%rsp)
  lea    0xc(%rsp),%rdx
  lea    0x10(%rsp),%rcx
  lea    0x8(%rsp),%rdi
  mov    %rdx,%r8
  call   <ddot_@plt>
  lea    D45(%rip),%rdi
  call   <printf@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  mov    $0x0,%eax
  add    $0x38,%rsp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D0
type_infered: f64
name: retrieve_data
arguments: tag: D52
type_infered: f64
name: retrieve_data
arguments: tag: D12
type_infered: f64
name: retrieve_data
arguments: tag: D28
type_infered: f64
name: retrieve_data
arguments: tag: D16
type_infered: f64
name: retrieve_data
arguments: tag: D4
type_infered: f64
name: retrieve_data
arguments: tag: D45
type_infered: string

tool: D0: 1

tool: D52: 2

tool: D12: 3

tool: D28: 4

tool: D16: 5

tool: D4: 6

tool: D45:  The dot product is: %f 


assistant: ```c
int main(int argc, char **argv) {
  double *a = (double *)malloc(3 * sizeof(double));
  a[0] = 1.0;
  a[1] = 2.0;
  a[2] = 3.0;
  double b[3] = {4.0, 5.0, 6.0};
  int N = 3, one = 1;
  double dot_product = ddot_(&N, a, &one, b, &one);
  printf(" The dot product is: %f \n", dot_product);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  sub    $0x38,%rsp
  mov    $0x18,%edi
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  call   <malloc@plt>
  lea    0xc(%rsp),%rdx
  lea    0x8(%rsp),%rdi
  mov    %rax,%rsi
  mov    D17(%rip),%rax
  lea    0x10(%rsp),%rcx
  mov    %rdx,%r8
  mov    %rax,(%rsi)
  mov    D39(%rip),%rax
  movl   $0x3,0x8(%rsp)
  mov    %rax,0x8(%rsi)
  mov    D54(%rip),%rax
  movl   $0x1,0xc(%rsp)
  mov    %rax,0x10(%rsi)
  mov    D24(%rip),%rax
  mov    %rax,0x10(%rsp)
  mov    D44(%rip),%rax
  mov    %rax,0x18(%rsp)
  mov    D62(%rip),%rax
  mov    %rax,0x20(%rsp)
  call   <ddot_@plt>
  lea    D8(%rip),%rdi
  call   <printf@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  xor    %eax,%eax
  add    $0x38,%rsp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D17
type_guessed: f64
name: parse_data
arguments: tag: D39
type_guessed: f64
name: parse_data
arguments: tag: D54
type_guessed: f64
name: parse_data
arguments: tag: D24
type_guessed: f64
name: parse_data
arguments: tag: D44
type_guessed: f64
name: parse_data
arguments: tag: D62
type_guessed: f64
name: parse_data
arguments: tag: D8
type_guessed: string

tool: D17: 1

tool: D39: 2

tool: D54: 3

tool: D24: 4

tool: D44: 5

tool: D62: 6

tool: D8:  The dot product is: %f 


assistant: ```c
int main(int argc, char **argv) {
  double *a = (double *)malloc(3 * sizeof(double));
  a[0] = 1.0;
  a[1] = 2.0;
  a[2] = 3.0;
  double b[3] = {4.0, 5.0, 6.0};
  int N = 3, one = 1;
  double dot_product = ddot_(&N, a, &one, b, &one);
  printf(" The dot product is: %f \n", dot_product);
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  sub    $0x38,%rsp
  mov    $0x18,%edi
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  call   <malloc@plt>
  lea    0xc(%rsp),%rdx
  lea    0x8(%rsp),%rdi
  movapd D49(%rip),%xmm0
  mov    %rax,%rsi
  lea    0x10(%rsp),%rcx
  mov    %rdx,%r8
  movups %xmm0,(%rax)
  mov    D3(%rip),%rax
  movapd D19(%rip),%xmm0
  movl   $0x3,0x8(%rsp)
  mov    %rax,0x10(%rsi)
  mov    D5(%rip),%rax
  movl   $0x1,0xc(%rsp)
  mov    %rax,0x20(%rsp)
  movaps %xmm0,0x10(%rsp)
  call   <ddot_@plt>
  lea    D42(%rip),%rdi
  call   <printf@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  xor    %eax,%eax
  add    $0x38,%rsp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D49
type_infered: f64[2]
name: retrieve_data
arguments: tag: D3
type_infered: f64
name: retrieve_data
arguments: tag: D19
type_infered: f64[2]
name: retrieve_data
arguments: tag: D5
type_infered: f64
name: retrieve_data
arguments: tag: D42
type_infered: string

tool: {"D49": [1.0,2.0]}

tool: D3: 3

tool: {"D19": [4.0,5.0]}

tool: D5: 6

tool: D42:  The dot product is: %f 


assistant: ```c
int main(int argc, char **argv) {
  double *a = (double *)malloc(3 * sizeof(double));
  a[0] = 1.0;
  a[1] = 2.0;
  a[2] = 3.0;
  double b[3] = {4.0, 5.0, 6.0};
  int N = 3, one = 1;
  double dot_product = ddot_(&N, a, &one, b, &one);
  printf(" The dot product is: %f \n", dot_product);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<sigma>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L17
L4:
  mov    -0x14(%rbp),%eax
  add    %eax,-0x4(%rbp)
  subl   $0x1,-0x14(%rbp)
L17:
  cmpl   $0x0,-0x14(%rbp)
  jg     L4
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int sigma(int num) {
  int soma;
  for (soma = 0; num >= 1; num--)
    soma = soma + num;
  return soma;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<sigma>:
  endbr64
  test   %edi,%edi
  jle    L4
  mov    $0x0,%eax
L19:
  add    %edi,%eax
  sub    $0x1,%edi
  jne    L19
  ret
L4:
  mov    $0x0,%eax
  ret
```

assistant: ```c
int sigma(int num) {
  int soma;
  for (soma = 0; num >= 1; num--)
    soma = soma + num;
  return soma;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<sigma>:
  endbr64
  xor    %eax,%eax
  test   %edi,%edi
  jle    L7
  nopw   0x0(%rax,%rax,1)
L17:
  add    %edi,%eax
  sub    $0x1,%edi
  jne    L17
  ret
  nopl   0x0(%rax,%rax,1)
L7:
  ret
```

assistant: ```c
int sigma(int num) {
  int soma;
  for (soma = 0; num >= 1; num--)
    soma = soma + num;
  return soma;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<sigma>:
  endbr64
  test   %edi,%edi
  jle    L7
  mov    %edi,%ecx
  cmp    $0xa,%edi
  jle    L104
  movd   %edi,%xmm4
  mov    %edi,%edx
  movdqa D2(%rip),%xmm3
  xor    %eax,%eax
  pshufd $0x0,%xmm4,%xmm1
  paddd  D4(%rip),%xmm1
  shr    $0x2,%edx
  pxor   %xmm0,%xmm0
  nopl   0x0(%rax,%rax,1)
L102:
  movdqa %xmm1,%xmm2
  add    $0x1,%eax
  paddd  %xmm3,%xmm1
  paddd  %xmm2,%xmm0
  cmp    %eax,%edx
  jne    L102
  movdqa %xmm0,%xmm1
  mov    %edi,%edx
  psrldq $0x8,%xmm1
  and    $0xfffffffc,%edx
  paddd  %xmm1,%xmm0
  sub    %edx,%edi
  movdqa %xmm0,%xmm1
  psrldq $0x4,%xmm1
  paddd  %xmm1,%xmm0
  movd   %xmm0,%eax
  cmp    %edx,%ecx
  je     L17
L110:
  mov    %edi,%edx
  add    %edi,%eax
  sub    $0x1,%edx
  je     L24
  add    %edx,%eax
  mov    %edi,%edx
  sub    $0x2,%edx
  je     L24
  add    %edx,%eax
  mov    %edi,%edx
  sub    $0x3,%edx
  je     L24
  add    %edx,%eax
  mov    %edi,%edx
  sub    $0x4,%edx
  je     L24
  add    %edx,%eax
  mov    %edi,%edx
  sub    $0x5,%edx
  je     L24
  add    %edx,%eax
  mov    %edi,%edx
  sub    $0x6,%edx
  je     L24
  add    %edx,%eax
  mov    %edi,%edx
  sub    $0x7,%edx
  je     L24
  add    %edx,%eax
  mov    %edi,%edx
  sub    $0x8,%edx
  je     L24
  add    %edx,%eax
  lea    -0x9(%rax,%rdi,1),%eax
  ret
  nopl   0x0(%rax,%rax,1)
L7:
  xor    %eax,%eax
L24:
  ret
  nopl   0x0(%rax,%rax,1)
L17:
  ret
L104:
  xor    %eax,%eax
  jmp    L110
```

tool_calls: name: read_data
arguments: data_label: D2
datatype: byte[16]
name: read_data
arguments: data_label: D4
datatype: byte[16]

tool: {"D2": [0xfc,0xff,0xff,0xff,0xfc,0xff,0xff,0xff,0xfc,0xff,0xff,0xff,0xfc,0xff,0xff,0xff]}

tool: {"D4": [0x00,0x00,0x00,0x00,0xff,0xff,0xff,0xff,0xfe,0xff,0xff,0xff,0xfd,0xff,0xff,0xff]}

assistant: ```c
int sigma(int num) {
  int soma;
  for (soma = 0; num >= 1; num--)
    soma = soma + num;
  return soma;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<bytes_from_multiplier>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0x0,-0x10(%rbp)
  movq   $0x1,-0x18(%rbp)
  lea    -0x20(%rbp),%rcx
  mov    -0x28(%rbp),%rax
  mov    $0xa,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strtol@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x20(%rbp),%rax
  test   %rax,%rax
  je     L40
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  cmp    $0x4d,%eax
  je     L1
  cmp    $0x4d,%eax
  jg     L33
  cmp    $0x47,%eax
  je     L42
  cmp    $0x4b,%eax
  je     L50
  jmp    L33
L42:
  movq   $0x40000000,-0x18(%rbp)
  jmp    L40
L1:
  movq   $0x100000,-0x18(%rbp)
  jmp    L40
L50:
  movq   $0x400,-0x18(%rbp)
  jmp    L40
L33:
  nop
L40:
  mov    -0x10(%rbp),%rax
  imul   -0x18(%rbp),%rax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L10
  call   <__stack_chk_fail@plt>
L10:
  leave
  ret
```

assistant: ```c
long bytes_from_multiplier(char *inString) {
  long value = 0;
  long factor = 1;
  char *suffix;
  value = strtol(inString, &suffix, 10);
  if (suffix) {
    switch (*suffix) {
    case 'G':
      factor = 1 << 30;
      break;
    case 'M':
      factor = 1 << 20;
      break;
    case 'K':
      factor = 1 << 10;
      break;
    default:
      break;
    }
  }
  return (value * factor);
}
```
user: What is the c source code of the assembly code below:

```asm
<bytes_from_multiplier>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  mov    $0xa,%edx
  call   <strtol@plt>
  mov    (%rsp),%rcx
  mov    $0x1,%edx
  test   %rcx,%rcx
  je     L24
  movzbl (%rcx),%ecx
  mov    $0x400,%edx
  cmp    $0x4b,%cl
  je     L24
  mov    $0x100000,%edx
  cmp    $0x4d,%cl
  je     L24
  cmp    $0x47,%cl
  mov    $0x1,%edx
  mov    $0x40000000,%ecx
  cmove  %rcx,%rdx
L24:
  imul   %rdx,%rax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L13
  add    $0x18,%rsp
  ret
L13:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
long bytes_from_multiplier(char *inString) {
  long value = 0;
  long factor = 1;
  char *suffix;
  value = strtol(inString, &suffix, 10);
  if (suffix) {
    switch (*suffix) {
    case 'G':
      factor = 1 << 30;
      break;
    case 'M':
      factor = 1 << 20;
      break;
    case 'K':
      factor = 1 << 10;
      break;
    default:
      break;
    }
  }
  return (value * factor);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<bytes_from_multiplier>:
  endbr64
  sub    $0x18,%rsp
  mov    $0xa,%edx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  call   <strtol@plt>
  mov    (%rsp),%rdx
  test   %rdx,%rdx
  je     L6
  movzbl (%rdx),%edx
  sub    $0x47,%edx
  cmp    $0x6,%dl
  jbe    L20
L6:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L26
  add    $0x18,%rsp
  ret
  xchg   %ax,%ax
L20:
  movzbl %dl,%edx
  lea    D2(%rip),%rcx
  imul   (%rcx,%rdx,8),%rax
  jmp    L6
L26:
  call   <__stack_chk_fail@plt>
```

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

tool: 

assistant: ```c
long bytes_from_multiplier(char *inString) {
  long value = 0;
  long factor = 1;
  char *suffix;
  value = strtol(inString, &suffix, 10);
  if (suffix) {
    switch (*suffix) {
    case 'G':
      factor = 1 << 30;
      break;
    case 'M':
      factor = 1 << 20;
      break;
    case 'K':
      factor = 1 << 10;
      break;
    default:
      break;
    }
  }
  return (value * factor);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<bytes_from_multiplier>:
  endbr64
  sub    $0x18,%rsp
  mov    $0xa,%edx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  call   <strtol@plt>
  mov    (%rsp),%rdx
  test   %rdx,%rdx
  je     L29
  movzbl (%rdx),%edx
  sub    $0x47,%edx
  cmp    $0x6,%dl
  jbe    L32
L29:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L13
  add    $0x18,%rsp
  ret
  xchg   %ax,%ax
L32:
  movzbl %dl,%edx
  lea    D0(%rip),%rcx
  imul   (%rcx,%rdx,8),%rax
  jmp    L29
L13:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D0
type_infered: string

tool: 

assistant: ```c
long bytes_from_multiplier(char *inString) {
  long value = 0;
  long factor = 1;
  char *suffix;
  value = strtol(inString, &suffix, 10);
  if (suffix) {
    switch (*suffix) {
    case 'G':
      factor = 1 << 30;
      break;
    case 'M':
      factor = 1 << 20;
      break;
    case 'K':
      factor = 1 << 10;
      break;
    default:
      break;
    }
  }
  return (value * factor);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<gcd>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    -0x14(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jle    L11
  mov    -0x14(%rbp),%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x18(%rbp),%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x8(%rbp),%eax
  mov    %eax,-0x18(%rbp)
  jmp    L11
L28:
  mov    -0x18(%rbp),%eax
  cltd
  idivl  -0x14(%rbp)
  mov    %edx,-0x4(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,-0x18(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,-0x14(%rbp)
L11:
  cmpl   $0x0,-0x14(%rbp)
  jne    L28
  mov    -0x18(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int gcd(int n, int m) {
  if (n > m) {
    int t = n;
    n = m;
    m = t;
  }
  while (n != 0) {
    int r = m % n;
    m = n;
    n = r;
  }
  return m;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<gcd>:
  endbr64
  mov    %esi,%eax
  cmp    %esi,%edi
  jle    L39
  mov    %edi,%edx
  mov    %esi,%edi
  mov    %edx,%eax
L39:
  test   %edi,%edi
  jne    L23
  mov    %eax,%edi
  jmp    L18
L8:
  mov    %edx,%edi
L23:
  cltd
  idiv   %edi
  mov    %edi,%eax
  test   %edx,%edx
  jne    L8
L18:
  mov    %edi,%eax
  ret
```

assistant: ```c
int gcd(int n, int m) {
  if (n > m) {
    int t = n;
    n = m;
    m = t;
  }
  while (n != 0) {
    int r = m % n;
    m = n;
    n = r;
  }
  return m;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<gcd>:
  endbr64
  mov    %esi,%eax
  mov    %edi,%r8d
  cmp    %esi,%edi
  jle    L2
  mov    %esi,%r8d
  mov    %edi,%eax
L2:
  test   %r8d,%r8d
  jne    L6
  jmp    L15
  nopl   0x0(%rax)
L0:
  mov    %edx,%r8d
L6:
  cltd
  idiv   %r8d
  mov    %r8d,%eax
  test   %edx,%edx
  jne    L0
  mov    %r8d,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L15:
  mov    %eax,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
int gcd(int n, int m) {
  if (n > m) {
    int t = n;
    n = m;
    m = t;
  }
  while (n != 0) {
    int r = m % n;
    m = n;
    n = r;
  }
  return m;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<gcd>:
  endbr64
  mov    %esi,%eax
  mov    %edi,%r8d
  cmp    %esi,%edi
  jle    L2
  mov    %esi,%r8d
  mov    %edi,%eax
L2:
  test   %r8d,%r8d
  jne    L0
  jmp    L24
  nopl   0x0(%rax)
L21:
  mov    %edx,%r8d
L0:
  cltd
  idiv   %r8d
  mov    %r8d,%eax
  test   %edx,%edx
  jne    L21
  mov    %r8d,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L24:
  mov    %eax,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
int gcd(int n, int m) {
  if (n > m) {
    int t = n;
    n = m;
    m = t;
  }
  while (n != 0) {
    int r = m % n;
    m = n;
    n = r;
  }
  return m;
}
```