ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0xa,%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L17
L47:
  mov    $0x41,%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L38
L42:
  mov    $0x20,%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L43
L19:
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
L17:
  lea    D10(%rip),%rdi
  call   <printf@plt>
  jmp    L47
L43:
  lea    D12(%rip),%rdi
  call   <printf@plt>
  jmp    L19
L38:
  lea    D14(%rip),%rdi
  call   <printf@plt>
  jmp    L42
```

tool_calls: name: retrieve_data
arguments: label: D10
type_infered: string
name: retrieve_data
arguments: label: D12
type_infered: string
name: retrieve_data
arguments: label: D14
type_infered: string

tool: D10: Variavel 1 tem apenas espaco em branco (white-space character)!


tool: D12: Variavel 3 tem apenas espaco em branco (white-space character)!


tool: D14: Variavel 2 tem apenas espaco em branco (white-space character)!


assistant: ```c
int main() {
  char var1 = '\n';
  char var2 = 'A';
  char var3 = ' ';
  if (isspace(var1))
    printf("Variavel 1 tem apenas espaco em branco (white-space character)!\n");
  if (isspace(var2))
    printf("Variavel 2 tem apenas espaco em branco (white-space character)!\n");
  if (isspace(var3))
    printf("Variavel 3 tem apenas espaco em branco (white-space character)!\n");
  return (0);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0xa,%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L47
L4:
  mov    $0x41,%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L46
L52:
  mov    $0x20,%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L33
L34:
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
L47:
  lea    D16(%rip),%rdi
  call   <printf@plt>
  jmp    L4
L33:
  lea    D18(%rip),%rdi
  call   <printf@plt>
  jmp    L34
L46:
  lea    D23(%rip),%rdi
  call   <printf@plt>
  jmp    L52
```

tool_calls: name: extract_symbolic_data
arguments: tag: D16
type_guessed: string
name: extract_symbolic_data
arguments: tag: D18
type_guessed: string
name: extract_symbolic_data
arguments: tag: D23
type_guessed: string

tool: D16: Variavel 1 tem apenas espaco em branco (white-space character)!


tool: D18: Variavel 3 tem apenas espaco em branco (white-space character)!


tool: D23: Variavel 2 tem apenas espaco em branco (white-space character)!


assistant: ```c
int main() {
  char var1 = '\n';
  char var2 = 'A';
  char var3 = ' ';
  if (isspace(var1))
    printf("Variavel 1 tem apenas espaco em branco (white-space character)!\n");
  if (isspace(var2))
    printf("Variavel 2 tem apenas espaco em branco (white-space character)!\n");
  if (isspace(var3))
    printf("Variavel 3 tem apenas espaco em branco (white-space character)!\n");
  return (0);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    $0x0,%eax
  call   <test@plt>
  test   %rax,%rax
  je     L24
  mov    $0x1,%eax
  jmp    L11
L24:
  cmpl   $0x1,-0x4(%rbp)
  jne    L31
  mov    $0x0,%eax
  jmp    L11
L31:
  mov    -0x10(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <test_file@plt>
L11:
  leave
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  if (test())
    return 1;
  if (argc == 1)
    return 0;
  return test_file(argv[1]);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebx
  mov    %rsi,%rbp
  mov    $0x0,%eax
  call   <test@plt>
  mov    %rax,%rdx
  mov    $0x1,%eax
  test   %rdx,%rdx
  jne    L14
  cmp    $0x1,%ebx
  je     L24
  mov    0x8(%rbp),%rdi
  call   <test_file@plt>
L14:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L24:
  mov    $0x0,%eax
  jmp    L14
```

assistant: ```c
int main(int argc, char **argv) {
  if (test())
    return 1;
  if (argc == 1)
    return 0;
  return test_file(argv[1]);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  xor    %eax,%eax
  mov    %rsi,%rbp
  push   %rbx
  mov    %edi,%ebx
  sub    $0x8,%rsp
  call   <test@plt>
  test   %rax,%rax
  jne    L24
  sub    $0x1,%ebx
  je     L17
  mov    0x8(%rbp),%rdi
  pop    %rcx
  pop    %rbx
  pop    %rbp
  jmp    1060 <test_file@plt>
L17:
  xor    %eax,%eax
L19:
  pop    %rdx
  pop    %rbx
  pop    %rbp
  ret
L24:
  mov    $0x1,%eax
  jmp    L19
```

assistant: ```c
int main(int argc, char **argv) {
  if (test())
    return 1;
  if (argc == 1)
    return 0;
  return test_file(argv[1]);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<main>:
  endbr64
  push   %rbp
  xor    %eax,%eax
  mov    %rsi,%rbp
  push   %rbx
  mov    %edi,%ebx
  sub    $0x8,%rsp
  call   <test@plt>
  test   %rax,%rax
  jne    L0
  sub    $0x1,%ebx
  je     L3
  mov    0x8(%rbp),%rdi
  pop    %rcx
  pop    %rbx
  pop    %rbp
  jmp    1060 <test_file@plt>
L3:
  xor    %eax,%eax
L17:
  pop    %rdx
  pop    %rbx
  pop    %rbp
  ret
L0:
  mov    $0x1,%eax
  jmp    L17
```

assistant: ```c
int main(int argc, char **argv) {
  if (test())
    return 1;
  if (argc == 1)
    return 0;
  return test_file(argv[1]);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<replace>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %esi,%eax
  mov    %rdx,-0x48(%rbp)
  mov    %al,-0x3c(%rbp)
  movl   $0x0,-0x2c(%rbp)
  mov    -0x38(%rbp),%rax
  mov    %rax,-0x28(%rbp)
  jmp    L24
L18:
  mov    -0x28(%rbp),%rax
  movzbl (%rax),%eax
  cmp    %al,-0x3c(%rbp)
  sete   %al
  movzbl %al,%eax
  add    %eax,-0x2c(%rbp)
  addq   $0x1,-0x28(%rbp)
L24:
  mov    -0x28(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L18
  mov    -0x48(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x38(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %rax,%rcx
  mov    -0x10(%rbp),%rax
  lea    -0x1(%rax),%rdx
  mov    -0x2c(%rbp),%eax
  cltq
  imul   %rdx,%rax
  add    %rcx,%rax
  add    $0x1,%rax
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,-0x20(%rbp)
  mov    -0x38(%rbp),%rax
  mov    %rax,-0x18(%rbp)
  jmp    L26
L29:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  cmp    %al,-0x3c(%rbp)
  jne    L47
  mov    -0x10(%rbp),%rdx
  mov    -0x48(%rbp),%rcx
  mov    -0x20(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0x10(%rbp),%rax
  add    %rax,-0x20(%rbp)
  jmp    L10
L47:
  mov    -0x20(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x20(%rbp)
  mov    -0x18(%rbp),%rdx
  movzbl (%rdx),%edx
  mov    %dl,(%rax)
L10:
  addq   $0x1,-0x18(%rbp)
L26:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L29
  mov    -0x20(%rbp),%rax
  movb   $0x0,(%rax)
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *replace(const char *s, char ch, const char *repl) {
  int count = 0;
  for (const char *t = s; *t; t++)
    count += (*t == ch);
  size_t rlen = strlen(repl);
  char *res = malloc(strlen(s) + (rlen - 1) * count + 1);
  char *ptr = res;
  for (const char *t = s; *t; t++) {
    if (*t == ch) {
      memcpy(ptr, repl, rlen);
      ptr += rlen;
    } else {
      *ptr++ = *t;
    }
  }
  *ptr = 0;
  return res;
}
```
user: What is the c source code of the assembly code below:

```asm
<replace>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%rbp
  movzbl (%rdi),%ebx
  test   %bl,%bl
  je     L56
  mov    %rdx,%r15
  mov    %esi,%r13d
  mov    %rdi,%rdx
  mov    %ebx,%eax
  mov    $0x0,%r12d
L31:
  cmp    %al,%r13b
  sete   %al
  movzbl %al,%eax
  add    %eax,%r12d
  add    $0x1,%rdx
  movzbl (%rdx),%eax
  test   %al,%al
  jne    L31
  mov    %r15,%rdi
  call   <strlen@plt>
  mov    %rax,%r14
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %rax,%rdx
  lea    -0x1(%r14),%rax
  movslq %r12d,%r12
  imul   %r12,%rax
  lea    0x1(%rax,%rdx,1),%rdi
  call   <malloc@plt>
  mov    %rax,0x8(%rsp)
  mov    %rax,%r12
  jmp    L46
L50:
  mov    %r14,%rdx
  mov    %r15,%rsi
  mov    %r12,%rdi
  call   <memcpy@plt>
  add    %r14,%r12
L23:
  add    $0x1,%rbp
  movzbl 0x0(%rbp),%ebx
  test   %bl,%bl
  je     L43
L46:
  cmp    %bl,%r13b
  je     L50
  mov    %bl,(%r12)
  lea    0x1(%r12),%r12
  jmp    L23
L56:
  call   <strlen@plt>
  lea    0x1(%rax),%rdi
  call   <malloc@plt>
  mov    %rax,%r12
  mov    %rax,0x8(%rsp)
L43:
  movb   $0x0,(%r12)
  mov    0x8(%rsp),%rax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
char *replace(const char *s, char ch, const char *repl) {
  int count = 0;
  for (const char *t = s; *t; t++)
    count += (*t == ch);
  size_t rlen = strlen(repl);
  char *res = malloc(strlen(s) + (rlen - 1) * count + 1);
  char *ptr = res;
  for (const char *t = s; *t; t++) {
    if (*t == ch) {
      memcpy(ptr, repl, rlen);
      ptr += rlen;
    } else {
      *ptr++ = *t;
    }
  }
  *ptr = 0;
  return res;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<replace>:
  endbr64
  push   %r15
  push   %r14
  mov    %esi,%r14d
  push   %r13
  mov    %rdx,%r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x18,%rsp
  movzbl (%rdi),%ebx
  call   <strlen@plt>
  mov    %r13,%rdi
  mov    %rax,%r12
  call   <strlen@plt>
  lea    0x1(%r12),%rdi
  test   %bl,%bl
  je     L7
  mov    %rax,%r15
  lea    -0x1(%rax),%rsi
  mov    %rbp,%rdx
  mov    %ebx,%eax
  xor    %ecx,%ecx
  nopl   0x0(%rax)
L55:
  cmp    %al,%r14b
  sete   %al
  add    $0x1,%rdx
  movzbl %al,%eax
  add    %eax,%ecx
  movzbl (%rdx),%eax
  test   %al,%al
  jne    L55
  movslq %ecx,%rcx
  imul   %rsi,%rcx
  add    %rcx,%rdi
  call   <malloc@plt>
  mov    %rax,%r8
  mov    %rax,%r12
  jmp    L28
  nopl   (%rax)
L65:
  add    $0x1,%rbp
  mov    %bl,(%r12)
  add    $0x1,%r12
  movzbl 0x0(%rbp),%ebx
  test   %bl,%bl
  je     L66
L28:
  cmp    %bl,%r14b
  jne    L65
  mov    %r12,%rdi
  mov    %r15,%rdx
  mov    %r13,%rsi
  add    $0x1,%rbp
  mov    %r8,0x8(%rsp)
  add    %r15,%r12
  call   <memcpy@plt>
  movzbl 0x0(%rbp),%ebx
  mov    0x8(%rsp),%r8
  test   %bl,%bl
  jne    L28
L66:
  movb   $0x0,(%r12)
  add    $0x18,%rsp
  mov    %r8,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L7:
  call   <malloc@plt>
  mov    %rax,%r12
  mov    %rax,%r8
  jmp    L66
```

assistant: ```c
char *replace(const char *s, char ch, const char *repl) {
  int count = 0;
  for (const char *t = s; *t; t++)
    count += (*t == ch);
  size_t rlen = strlen(repl);
  char *res = malloc(strlen(s) + (rlen - 1) * count + 1);
  char *ptr = res;
  for (const char *t = s; *t; t++) {
    if (*t == ch) {
      memcpy(ptr, repl, rlen);
      ptr += rlen;
    } else {
      *ptr++ = *t;
    }
  }
  *ptr = 0;
  return res;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<replace>:
  endbr64
  push   %r15
  push   %r14
  mov    %esi,%r14d
  push   %r13
  mov    %rdx,%r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x18,%rsp
  movzbl (%rdi),%ebx
  call   <strlen@plt>
  mov    %r13,%rdi
  mov    %rax,%r12
  call   <strlen@plt>
  lea    0x1(%r12),%rdi
  test   %bl,%bl
  je     L40
  mov    %rax,%r15
  lea    -0x1(%rax),%rsi
  mov    %rbp,%rdx
  mov    %ebx,%eax
  xor    %ecx,%ecx
  nopl   0x0(%rax)
L33:
  cmp    %al,%r14b
  sete   %al
  add    $0x1,%rdx
  movzbl %al,%eax
  add    %eax,%ecx
  movzbl (%rdx),%eax
  test   %al,%al
  jne    L33
  movslq %ecx,%rcx
  imul   %rsi,%rcx
  add    %rcx,%rdi
  call   <malloc@plt>
  mov    %rax,%r8
  mov    %rax,%r12
  jmp    L45
  nopl   (%rax)
L51:
  add    $0x1,%rbp
  mov    %bl,(%r12)
  lea    0x1(%r12),%rax
  movzbl 0x0(%rbp),%ebx
  test   %bl,%bl
  je     L5
L47:
  mov    %rax,%r12
L45:
  cmp    %bl,%r14b
  jne    L51
  add    $0x1,%rbp
  mov    %r15,%rdx
  mov    %r13,%rsi
  mov    %r12,%rdi
  mov    %r8,0x8(%rsp)
  call   <memcpy@plt>
  movzbl 0x0(%rbp),%ebx
  mov    0x8(%rsp),%r8
  lea    (%r12,%r15,1),%rax
  test   %bl,%bl
  jne    L47
L5:
  movb   $0x0,(%rax)
  add    $0x18,%rsp
  mov    %r8,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L40:
  call   <malloc@plt>
  mov    %rax,%r8
  jmp    L5
```

assistant: ```c
char *replace(const char *s, char ch, const char *repl) {
  int count = 0;
  for (const char *t = s; *t; t++)
    count += (*t == ch);
  size_t rlen = strlen(repl);
  char *res = malloc(strlen(s) + (rlen - 1) * count + 1);
  char *ptr = res;
  for (const char *t = s; *t; t++) {
    if (*t == ch) {
      memcpy(ptr, repl, rlen);
      ptr += rlen;
    } else {
      *ptr++ = *t;
    }
  }
  *ptr = 0;
  return res;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<internetParseDottedQuad>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <isIpv4Address@plt>
  test   %eax,%eax
  jne    L5
  mov    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  call   <errAbort@plt>
L5:
  movl   $0x0,-0x1c(%rbp)
  jmp    L16
L22:
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x30(%rbp),%rax
  lea    (%rdx,%rax,1),%rbx
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <atoi@plt>
  mov    %al,(%rbx)
  mov    -0x18(%rbp),%rax
  mov    $0x2e,%esi
  mov    %rax,%rdi
  call   <strchr@plt>
  add    $0x1,%rax
  mov    %rax,-0x18(%rbp)
  addl   $0x1,-0x1c(%rbp)
L16:
  cmpl   $0x3,-0x1c(%rbp)
  jle    L22
  nop
  nop
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

tool_calls: name: parse_data
arguments: label: D6
data_type: string

tool: D6: %s is not a dotted quad

assistant: ```c
void internetParseDottedQuad(char *dottedQuad, unsigned char quad[4]) {
  char *s = dottedQuad;
  int i;
  if (!isIpv4Address(s))
    errAbort("%s is not a dotted quad", s);
  for (i = 0; i < 4; ++i) {
    quad[i] = atoi(s);
    s = strchr(s, '.') + 1;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<internetParseDottedQuad>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %rsi,%r12
  call   <isIpv4Address@plt>
  test   %eax,%eax
  je     L20
L2:
  mov    %r12,%rbp
  add    $0x4,%r12
L24:
  mov    %rbx,%rdi
  call   <atoi@plt>
  mov    %al,0x0(%rbp)
  mov    $0x2e,%esi
  mov    %rbx,%rdi
  call   <strchr@plt>
  lea    0x1(%rax),%rbx
  add    $0x1,%rbp
  cmp    %rbp,%r12
  jne    L24
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L20:
  mov    %rbx,%rsi
  lea    D1(%rip),%rdi
  call   <errAbort@plt>
  jmp    L2
```

tool_calls: name: retrieve_data
arguments: data_label: D1
type_infered: string

tool: D1: %s is not a dotted quad

assistant: ```c
void internetParseDottedQuad(char *dottedQuad, unsigned char quad[4]) {
  char *s = dottedQuad;
  int i;
  if (!isIpv4Address(s))
    errAbort("%s is not a dotted quad", s);
  for (i = 0; i < 4; ++i) {
    quad[i] = atoi(s);
    s = strchr(s, '.') + 1;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<internetParseDottedQuad>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rsi,%rbx
  call   <isIpv4Address@plt>
  test   %eax,%eax
  je     L24
L1:
  lea    0x4(%rbx),%r12
L26:
  mov    %rbp,%rdi
  add    $0x1,%rbx
  call   <atoi@plt>
  mov    %rbp,%rdi
  mov    $0x2e,%esi
  mov    %al,-0x1(%rbx)
  call   <strchr@plt>
  lea    0x1(%rax),%rbp
  cmp    %r12,%rbx
  jne    L26
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   (%rax)
L24:
  mov    %rbp,%rsi
  lea    D3(%rip),%rdi
  call   <errAbort@plt>
  jmp    L1
```

tool_calls: name: unpack_data
arguments: data_tag: D3
data_type: string

tool: D3: %s is not a dotted quad

assistant: ```c
void internetParseDottedQuad(char *dottedQuad, unsigned char quad[4]) {
  char *s = dottedQuad;
  int i;
  if (!isIpv4Address(s))
    errAbort("%s is not a dotted quad", s);
  for (i = 0; i < 4; ++i) {
    quad[i] = atoi(s);
    s = strchr(s, '.') + 1;
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<internetParseDottedQuad>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rsi,%rbx
  call   <isIpv4Address@plt>
  test   %eax,%eax
  je     L3
L28:
  lea    0x4(%rbx),%r12
L4:
  mov    %rbp,%rdi
  add    $0x1,%rbx
  call   <atoi@plt>
  mov    %rbp,%rdi
  mov    $0x2e,%esi
  mov    %al,-0x1(%rbx)
  call   <strchr@plt>
  lea    0x1(%rax),%rbp
  cmp    %r12,%rbx
  jne    L4
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   (%rax)
L3:
  mov    %rbp,%rsi
  lea    D5(%rip),%rdi
  call   <errAbort@plt>
  jmp    L28
```

tool_calls: name: retrieve_data
arguments: tag: D5
type_guessed: string

tool: D5: %s is not a dotted quad

assistant: ```c
void internetParseDottedQuad(char *dottedQuad, unsigned char quad[4]) {
  char *s = dottedQuad;
  int i;
  if (!isIpv4Address(s))
    errAbort("%s is not a dotted quad", s);
  for (i = 0; i < 4; ++i) {
    quad[i] = atoi(s);
    s = strchr(s, '.') + 1;
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<nbr_power>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  jmp    L0
L18:
  mov    -0x10(%rbp),%rax
  mov    (%rax),%eax
  lea    0x1(%rax),%edx
  mov    -0x10(%rbp),%rax
  mov    %edx,(%rax)
L0:
  mov    -0x10(%rbp),%rax
  mov    (%rax),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L18
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void nbr_power(char *str, int *power) {
  while (str[*power] != '\0')
    *power = *power + 1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<nbr_power>:
  endbr64
  mov    (%rsi),%eax
  movslq %eax,%rdx
  cmpb   $0x0,(%rdi,%rdx,1)
  je     L0
  add    $0x1,%eax
  cltq
L6:
  mov    %eax,(%rsi)
  add    $0x1,%rax
  cmpb   $0x0,-0x1(%rdi,%rax,1)
  jne    L6
L0:
  ret
```

assistant: ```c
void nbr_power(char *str, int *power) {
  while (str[*power] != '\0')
    *power = *power + 1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<nbr_power>:
  endbr64
  movslq (%rsi),%rdx
  cmpb   $0x0,(%rdi,%rdx,1)
  mov    %rdx,%rax
  je     L5
  add    $0x1,%eax
  cltq
  nopl   (%rax)
L9:
  mov    %eax,(%rsi)
  add    $0x1,%rax
  cmpb   $0x0,-0x1(%rdi,%rax,1)
  jne    L9
L5:
  ret
```

assistant: ```c
void nbr_power(char *str, int *power) {
  while (str[*power] != '\0')
    *power = *power + 1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<nbr_power>:
  endbr64
  movslq (%rsi),%rdx
  cmpb   $0x0,(%rdi,%rdx,1)
  mov    %rdx,%rax
  je     L11
  add    $0x1,%eax
  cltq
  nopl   (%rax)
L7:
  mov    %eax,(%rsi)
  add    $0x1,%rax
  cmpb   $0x0,-0x1(%rdi,%rax,1)
  jne    L7
L11:
  ret
```

assistant: ```c
void nbr_power(char *str, int *power) {
  while (str[*power] != '\0')
    *power = *power + 1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  nop
  nop
  call   <foo@plt>
  nop
  nop
  call   <bar@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main(void) {
  asm("nop\n nop\n call "
      "foo");
  asm("nop\n nop\n call "
      "bar");
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  nop
  nop
  call   <foo@plt>
  nop
  nop
  call   <bar@plt>
  mov    $0x0,%eax
  ret
```

assistant: ```c
int main(void) {
  asm("nop\n nop\n call "
      "foo");
  asm("nop\n nop\n call "
      "bar");
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  nop
  nop
  call   <foo@plt>
  nop
  nop
  call   <bar@plt>
  xor    %eax,%eax
  ret
```

assistant: ```c
int main(void) {
  asm("nop\n nop\n call "
      "foo");
  asm("nop\n nop\n call "
      "bar");
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  nop
  nop
  call   <foo@plt>
  nop
  nop
  call   <bar@plt>
  xor    %eax,%eax
  ret
```

assistant: ```c
int main(void) {
  asm("nop\n nop\n call "
      "foo");
  asm("nop\n nop\n call "
      "bar");
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<double_factorials>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0x0,-0x10(%rbp)
  lea    -0x10(%rbp),%rcx
  mov    -0x14(%rbp),%eax
  mov    $0x2,%edx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <inefficient_factorial@plt>
  mov    -0x10(%rbp),%rax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L1
  call   <__stack_chk_fail@plt>
L1:
  leave
  ret
```

assistant: ```c
double *double_factorials(const int n) {
  double *f = NULL;
  inefficient_factorial(n, &f, 2);
  return f;
}
```
user: Translate the below assembly code into C programming language:

```asm
<double_factorials>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movq   $0x0,(%rsp)
  mov    %rsp,%rsi
  mov    $0x2,%edx
  call   <inefficient_factorial@plt>
  mov    (%rsp),%rax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L2
  add    $0x18,%rsp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
double *double_factorials(const int n) {
  double *f = NULL;
  inefficient_factorial(n, &f, 2);
  return f;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<double_factorials>:
  endbr64
  sub    $0x18,%rsp
  mov    $0x2,%edx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  movq   $0x0,(%rsp)
  call   <inefficient_factorial@plt>
  mov    (%rsp),%rax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L6
  add    $0x18,%rsp
  ret
L6:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
double *double_factorials(const int n) {
  double *f = NULL;
  inefficient_factorial(n, &f, 2);
  return f;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<double_factorials>:
  endbr64
  sub    $0x18,%rsp
  mov    $0x2,%edx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  movq   $0x0,(%rsp)
  call   <inefficient_factorial@plt>
  mov    (%rsp),%rax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L7
  add    $0x18,%rsp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
double *double_factorials(const int n) {
  double *f = NULL;
  inefficient_factorial(n, &f, 2);
  return f;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<sscanf_main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <sscanf_t01@plt>
  mov    $0x0,%eax
  call   <sscanf_t02@plt>
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D0
type_infered: string

tool: D0: sscanf ok


assistant: ```c
int sscanf_main() {
  sscanf_t01();
  sscanf_t02();
  printf("sscanf ok\n");
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<sscanf_main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <sscanf_t01@plt>
  mov    $0x0,%eax
  call   <sscanf_t02@plt>
  lea    D7(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D7
type_infered: string

tool: D7: sscanf ok


assistant: ```c
int sscanf_main() {
  sscanf_t01();
  sscanf_t02();
  printf("sscanf ok\n");
  return 0;
}
```