ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What is the c source code of the assembly code below?

```asm
<strntoumax>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%r14
  mov    %edx,%r13d
  lea    (%rdi,%rcx,1),%r12
  cmp    %r12,%rdi
  jae    L241
L257:
  movzbl (%rbx),%edi
  call   <isspace@plt>
  test   %rax,%rax
  je     L143
  add    $0x1,%rbx
  cmp    %rbx,%r12
  jne    L257
  mov    %r12,%rbx
L241:
  mov    $0x0,%r15d
L149:
  test   %r13d,%r13d
  jne    L106
  lea    0x2(%rbx),%rax
  cmp    %r12,%rax
  jae    L195
  mov    $0xa,%r13d
  cmpb   $0x30,(%rbx)
  jne    L199
  movzbl 0x1(%rbx),%edx
  and    $0xffffffdf,%edx
  cmp    $0x58,%dl
  je     L110
  lea    0x1(%rbx),%rax
  mov    $0xa,%r13d
  cmp    %rax,%r12
  jbe    L199
  mov    %rax,%rbx
  mov    $0x8,%r13d
  jmp    L193
L143:
  cmp    %rbx,%r12
  jbe    L229
  movzbl (%rbx),%eax
  lea    -0x2b(%rax),%edx
  mov    $0x0,%r15d
  test   $0xfd,%dl
  jne    L149
  cmp    $0x2d,%al
  sete   %r15b
  movzbl %r15b,%r15d
  add    $0x1,%rbx
  jmp    L149
L229:
  mov    $0x0,%r15d
  jmp    L149
L195:
  lea    0x1(%rbx),%rax
  cmp    %r12,%rax
  jae    L133
  mov    $0xa,%r13d
  cmpb   $0x30,(%rbx)
  jne    L199
  mov    %rax,%rbx
  mov    $0x8,%r13d
  jmp    L193
L106:
  cmp    $0x10,%r13d
  je     L33
L199:
  cmp    %r12,%rbx
  jae    L103
L193:
  mov    $0x0,%ebp
L54:
  movzbl (%rbx),%edi
  call   <digitval@plt>
  cmp    %r13d,%eax
  jge    L242
  test   %eax,%eax
  js     L242
  imul   %r13d,%ebp
  add    %eax,%ebp
  add    $0x1,%rbx
  cmp    %rbx,%r12
  ja     L54
L242:
  test   %r14,%r14
  je     L172
  mov    %rbx,(%r14)
L172:
  mov    %ebp,%eax
  neg    %eax
  test   %r15d,%r15d
  cmovne %eax,%ebp
  mov    %ebp,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L33:
  lea    0x2(%rbx),%rax
  cmp    %r12,%rax
  jae    L199
  cmpb   $0x30,(%rbx)
  jne    L199
  movzbl 0x1(%rbx),%edx
  and    $0xffffffdf,%edx
  cmp    $0x58,%dl
  jne    L199
  mov    %rax,%rbx
  jmp    L193
L133:
  mov    $0xa,%r13d
  jmp    L199
L103:
  mov    $0x0,%ebp
  jmp    L242
L110:
  mov    %rax,%rbx
  mov    $0x10,%r13d
  jmp    L193
```

assistant: ```c
uintmax_t strntoumax(const char *nptr, char **endptr, int base, size_t n) {
  const unsigned char *p = (const unsigned char *)nptr;
  const unsigned char *end = p + n;
  int minus = 0;
  uintmax_t v = 0;
  int d;
  while (p < end && isspace(*p))
    p++;
  if (p < end) {
    char c = p[0];
    if (c == '-' || c == '+') {
      minus = (c == '-');
      p++;
    }
  }
  if (base == 0) {
    if (p + 2 < end && p[0] == '0' && (p[1] == 'x' || p[1] == 'X')) {
      p += 2;
      base = 16;
    } else if (p + 1 < end && p[0] == '0') {
      p += 1;
      base = 8;
    } else {
      base = 10;
    }
  } else if (base == 16) {
    if (p + 2 < end && p[0] == '0' && (p[1] == 'x' || p[1] == 'X')) {
      p += 2;
    }
  }
  while (p < end && (d = digitval(*p)) >= 0 && d < base) {
    v = v * base + d;
    p += 1;
  }
  if (endptr)
    *endptr = (char *)p;
  return minus ? -v : v;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<lz68k_out_number>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  cmpl   $0x7f,-0x14(%rbp)
  jg     L17
  mov    -0x14(%rbp),%eax
  mov    %al,-0xb(%rbp)
  mov    -0x20(%rbp),%rdx
  lea    -0xb(%rbp),%rax
  mov    %rdx,%rcx
  mov    $0x1,%edx
  mov    $0x1,%esi
  mov    %rax,%rdi
  call   <fwrite@plt>
  jmp    L19
L17:
  mov    -0x14(%rbp),%eax
  sar    $0x8,%eax
  or     $0xffffff80,%eax
  mov    %al,-0xa(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %al,-0x9(%rbp)
  mov    -0x20(%rbp),%rdx
  lea    -0xa(%rbp),%rax
  mov    %rdx,%rcx
  mov    $0x2,%edx
  mov    $0x1,%esi
  mov    %rax,%rdi
  call   <fwrite@plt>
L19:
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L15
  call   <__stack_chk_fail@plt>
L15:
  leave
  ret
```

assistant: ```c
void lz68k_out_number(uint32_t number, FILE *dst) {
  if (number < 128) {
    char n = (char)number;
    fwrite(&n, 1, 1, dst);
    return;
  }
  char dn[2];
  dn[0] = (char)(number >> 8 | 0x80);
  dn[1] = (char)(number & 0xFF);
  fwrite(dn, 1, 2, dst);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<lz68k_out_number>:
  endbr64
  sub    $0x18,%rsp
  mov    %rsi,%rcx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  cmp    $0x7f,%edi
  jle    L4
  mov    %edi,%eax
  sar    $0x8,%eax
  or     $0xffffff80,%eax
  mov    %al,0x6(%rsp)
  mov    %dil,0x7(%rsp)
  lea    0x6(%rsp),%rdi
  mov    $0x2,%edx
  mov    $0x1,%esi
  call   <fwrite@plt>
L28:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L26
  add    $0x18,%rsp
  ret
L4:
  mov    %dil,0x5(%rsp)
  lea    0x5(%rsp),%rdi
  mov    $0x1,%edx
  mov    $0x1,%esi
  call   <fwrite@plt>
  jmp    L28
L26:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void lz68k_out_number(uint32_t number, FILE *dst) {
  if (number < 128) {
    char n = (char)number;
    fwrite(&n, 1, 1, dst);
    return;
  }
  char dn[2];
  dn[0] = (char)(number >> 8 | 0x80);
  dn[1] = (char)(number & 0xFF);
  fwrite(dn, 1, 2, dst);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<lz68k_out_number>:
  endbr64
  sub    $0x18,%rsp
  mov    %rsi,%rcx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  cmp    $0x7f,%edi
  jle    L26
  mov    %edi,%eax
  mov    %dil,0x7(%rsp)
  mov    $0x2,%edx
  lea    0x6(%rsp),%rdi
  sar    $0x8,%eax
  mov    $0x1,%esi
  or     $0xffffff80,%eax
  mov    %al,0x6(%rsp)
  call   <fwrite@plt>
L14:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L10
  add    $0x18,%rsp
  ret
  nopw   0x0(%rax,%rax,1)
L26:
  mov    %dil,0x5(%rsp)
  mov    $0x1,%edx
  lea    0x5(%rsp),%rdi
  mov    $0x1,%esi
  call   <fwrite@plt>
  jmp    L14
L10:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void lz68k_out_number(uint32_t number, FILE *dst) {
  if (number < 128) {
    char n = (char)number;
    fwrite(&n, 1, 1, dst);
    return;
  }
  char dn[2];
  dn[0] = (char)(number >> 8 | 0x80);
  dn[1] = (char)(number & 0xFF);
  fwrite(dn, 1, 2, dst);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<lz68k_out_number>:
  endbr64
  sub    $0x18,%rsp
  mov    %rsi,%rcx
  mov    %fs:0x28,%rdx
  mov    %rdx,0x8(%rsp)
  xor    %edx,%edx
  cmp    $0x7f,%edi
  jle    L0
  mov    %edi,%esi
  xor    %edx,%edx
  mov    %edi,%eax
  sar    $0x8,%esi
  lea    0x6(%rsp),%rdi
  or     $0xffffff80,%esi
  mov    %sil,%dl
  mov    $0x1,%esi
  mov    %al,%dh
  mov    %dx,0x6(%rsp)
  mov    $0x2,%edx
  call   <fwrite@plt>
L22:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L26
  add    $0x18,%rsp
  ret
  nop
L0:
  mov    %dil,0x5(%rsp)
  mov    $0x1,%edx
  lea    0x5(%rsp),%rdi
  mov    $0x1,%esi
  call   <fwrite@plt>
  jmp    L22
L26:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void lz68k_out_number(uint32_t number, FILE *dst) {
  if (number < 128) {
    char n = (char)number;
    fwrite(&n, 1, 1, dst);
    return;
  }
  char dn[2];
  dn[0] = (char)(number >> 8 | 0x80);
  dn[1] = (char)(number & 0xFF);
  fwrite(dn, 1, 2, dst);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<VisualRank>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  add    $0xffffff80,%eax
  cmp    $0x5,%eax
  ja     L68
  mov    %eax,%eax
  lea    0x0(,%rax,4),%rdx
  lea    D8(%rip),%rax
  mov    (%rdx,%rax,1),%eax
  cltq
  lea    D8(%rip),%rdx
  add    %rdx,%rax
  notrack jmp *%rax
  mov    $0x5,%eax
  jmp    L49
  mov    $0x4,%eax
  jmp    L49
  mov    $0x3,%eax
  jmp    L49
  mov    $0x2,%eax
  jmp    L49
  mov    $0x1,%eax
  jmp    L49
  mov    $0x0,%eax
  jmp    L49
L68:
  mov    $0xffffffff,%eax
L49:
  pop    %rbp
  ret
```

assistant: ```c
int VisualRank(int class) {
  switch (class) {
  case PseudoColor:
    return 5;
  case TrueColor:
    return 4;
  case DirectColor:
    return 3;
  case StaticColor:
    return 2;
  case GrayScale:
    return 1;
  case StaticGray:
    return 0;
  }
  return -1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<VisualRank>:
  endbr64
  add    $0xffffff80,%edi
  cmp    $0x5,%edi
  ja     L4
  mov    %edi,%edi
  lea    D1(%rip),%rdx
  movslq (%rdx,%rdi,4),%rax
  add    %rdx,%rax
  notrack jmp *%rax
  mov    $0x4,%eax
  ret
  mov    $0x3,%eax
  ret
  mov    $0x2,%eax
  ret
  mov    $0x1,%eax
  ret
  mov    $0x0,%eax
  ret
L4:
  mov    $0xffffffff,%eax
  ret
  mov    $0x5,%eax
  ret
```

assistant: ```c
int VisualRank(int class) {
  switch (class) {
  case PseudoColor:
    return 5;
  case TrueColor:
    return 4;
  case DirectColor:
    return 3;
  case StaticColor:
    return 2;
  case GrayScale:
    return 1;
  case StaticGray:
    return 0;
  }
  return -1;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<VisualRank>:
  endbr64
  add    $0xffffff80,%edi
  mov    $0xffffffff,%eax
  cmp    $0x5,%edi
  ja     L8
  lea    D7(%rip),%rax
  mov    (%rax,%rdi,4),%eax
L8:
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D7
datatype: string

tool: D7: 

assistant: ```c
int VisualRank(int class) {
  switch (class) {
  case PseudoColor:
    return 5;
  case TrueColor:
    return 4;
  case DirectColor:
    return 3;
  case StaticColor:
    return 2;
  case GrayScale:
    return 1;
  case StaticGray:
    return 0;
  }
  return -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
<VisualRank>:
  endbr64
  add    $0xffffff80,%edi
  mov    $0xffffffff,%eax
  cmp    $0x5,%edi
  ja     L6
  lea    D0(%rip),%rax
  mov    (%rax,%rdi,4),%eax
L6:
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D0
datatype: string

tool: D0: 

assistant: ```c
int VisualRank(int class) {
  switch (class) {
  case PseudoColor:
    return 5;
  case TrueColor:
    return 4;
  case DirectColor:
    return 3;
  case StaticColor:
    return 2;
  case GrayScale:
    return 1;
  case StaticGray:
    return 0;
  }
  return -1;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,%esi
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x0,-0x4(%rbp)
  jmp    L3
L18:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x4(%rbp)
L3:
  mov    -0x4(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L18
L0:
  mov    $0x1,%edi
  call   <sleep@plt>
  jmp    L0
```

tool_calls: name: unpack_data
arguments: label: D19
type: string
name: unpack_data
arguments: label: D3
type: string

tool: D19: argc=%d


tool: D3: argv[%d]=%s


assistant: ```c
int main(int argc, char *argv[]) {
  printf("argc=%d\n", argc);
  int i;
  for (i = 0; i < argc; i++)
    printf("argv[%d]=%s\n", i, argv[i]);
  while (1)
    sleep(1);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebp
  mov    %rsi,%r12
  mov    %edi,%esi
  lea    D19(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  test   %ebp,%ebp
  jle    L8
  mov    %ebp,%ebp
  mov    $0x0,%ebx
  lea    D11(%rip),%r13
L9:
  mov    (%r12,%rbx,8),%rdx
  mov    %ebx,%esi
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x1,%rbx
  cmp    %rbp,%rbx
  jne    L9
L8:
  mov    $0x1,%edi
  call   <sleep@plt>
  jmp    L8
```

tool_calls: name: retrieve_data
arguments: data_tag: D19
datatype: string
name: retrieve_data
arguments: data_tag: D11
datatype: string

tool: D19: argc=%d


tool: D11: argv[%d]=%s


assistant: ```c
int main(int argc, char *argv[]) {
  printf("argc=%d\n", argc);
  int i;
  for (i = 0; i < argc; i++)
    printf("argv[%d]=%s\n", i, argv[i]);
  while (1)
    sleep(1);
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %r13
  xor    %eax,%eax
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  movslq %edi,%rbp
  lea    D12(%rip),%rdi
  push   %rbx
  mov    %ebp,%esi
  sub    $0x8,%rsp
  call   <printf@plt>
  test   %ebp,%ebp
  jle    L11
  xor    %ebx,%ebx
  lea    D2(%rip),%r13
  nopl   0x0(%rax)
L4:
  mov    (%r12,%rbx,8),%rdx
  mov    %ebx,%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    %rbp,%rbx
  jne    L4
L11:
  mov    $0x1,%edi
  call   <sleep@plt>
  mov    $0x1,%edi
  call   <sleep@plt>
  jmp    L11
```

tool_calls: name: unpack_data
arguments: tag: D12
type_guessed: string
name: unpack_data
arguments: tag: D2
type_guessed: string

tool: D12: argc=%d


tool: D2: argv[%d]=%s


assistant: ```c
int main(int argc, char *argv[]) {
  printf("argc=%d\n", argc);
  int i;
  for (i = 0; i < argc; i++)
    printf("argv[%d]=%s\n", i, argv[i]);
  while (1)
    sleep(1);
  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
  push   %r13
  xor    %eax,%eax
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  movslq %edi,%rbp
  lea    D3(%rip),%rdi
  push   %rbx
  mov    %ebp,%esi
  sub    $0x8,%rsp
  call   <printf@plt>
  test   %ebp,%ebp
  jle    L0
  xor    %ebx,%ebx
  lea    D2(%rip),%r13
  nopl   0x0(%rax)
L15:
  mov    (%r12,%rbx,8),%rdx
  mov    %ebx,%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    %rbx,%rbp
  jne    L15
L0:
  mov    $0x1,%edi
  call   <sleep@plt>
  mov    $0x1,%edi
  call   <sleep@plt>
  jmp    L0
```

tool_calls: name: unpack_data
arguments: tag: D3
type: string
name: unpack_data
arguments: tag: D2
type: string

tool: D3: argc=%d


tool: D2: argv[%d]=%s


assistant: ```c
int main(int argc, char *argv[]) {
  printf("argc=%d\n", argc);
  int i;
  for (i = 0; i < argc; i++)
    printf("argv[%d]=%s\n", i, argv[i]);
  while (1)
    sleep(1);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0xc(%rbp),%rax
  mov    %rax,%rsi
  lea    D15(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0xc(%rbp),%eax
  cmp    $0x5,%eax
  jne    L19
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D15(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L19:
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L18
  call   <__stack_chk_fail@plt>
L18:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D15
type_guessed: string

tool: D15: %d

assistant: ```c
int main() {
  int a;
  scanf("%d", &a);
  if (a == 5)
    printf("%d", a);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  lea    D5(%rip),%rdi
  call   <scanf@plt>
  cmpl   $0x5,0x4(%rsp)
  je     L5
L19:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  mov    $0x0,%eax
  add    $0x18,%rsp
  ret
L5:
  mov    $0x5,%esi
  lea    D5(%rip),%rdi
  call   <printf@plt>
  jmp    L19
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D5
data_type: string

tool: D5: %d

assistant: ```c
int main() {
  int a;
  scanf("%d", &a);
  if (a == 5)
    printf("%d", a);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D4(%rip),%rbp
  mov    %rbp,%rdi
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  cmpl   $0x5,0x4(%rsp)
  je     L17
L29:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L19
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L17:
  mov    $0x5,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  jmp    L29
L19:
  call   <__stack_chk_fail@plt>
```

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

tool: D4: %d

assistant: ```c
int main() {
  int a;
  scanf("%d", &a);
  if (a == 5)
    printf("%d", a);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D5(%rip),%rbp
  mov    %rbp,%rdi
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  cmpl   $0x5,0x4(%rsp)
  je     L29
L20:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L26
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L29:
  mov    $0x5,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  jmp    L20
L26:
  call   <__stack_chk_fail@plt>
```

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

tool: D5: %d

assistant: ```c
int main() {
  int a;
  scanf("%d", &a);
  if (a == 5)
    printf("%d", a);
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<EnableIO>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x3,%edi
  call   <iopl@plt>
  test   %rax,%rax
  jne    L1
  mov    $0x1,%eax
  jmp    L12
L1:
  mov    $0x0,%eax
L12:
  pop    %rbp
  ret
```

assistant: ```c
int EnableIO(void) {
  if (iopl(3) == 0) {
    return 1;
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<EnableIO>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x3,%edi
  call   <iopl@plt>
  test   %rax,%rax
  sete   %al
  movzbl %al,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int EnableIO(void) {
  if (iopl(3) == 0) {
    return 1;
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<EnableIO>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x3,%edi
  call   <iopl@plt>
  test   %rax,%rax
  sete   %al
  add    $0x8,%rsp
  movzbl %al,%eax
  ret
```

assistant: ```c
int EnableIO(void) {
  if (iopl(3) == 0) {
    return 1;
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<EnableIO>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x3,%edi
  call   <iopl@plt>
  test   %rax,%rax
  sete   %al
  add    $0x8,%rsp
  movzbl %al,%eax
  ret
```

assistant: ```c
int EnableIO(void) {
  if (iopl(3) == 0) {
    return 1;
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<decode_basepairs>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %edx,-0x34(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movb   $0x0,-0x9(%rbp)
  movl   $0x0,-0x1c(%rbp)
  jmp    L73
L91:
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  mov    %al,-0x1e(%rbp)
  movl   $0x3,-0x14(%rbp)
  jmp    L56
L95:
  movzbl -0x1e(%rbp),%eax
  and    $0x3,%eax
  mov    %eax,-0x18(%rbp)
  cmpl   $0x0,-0x18(%rbp)
  jne    L5
  movb   $0x41,-0x1d(%rbp)
  jmp    L38
L5:
  cmpl   $0x1,-0x18(%rbp)
  jne    L104
  movb   $0x43,-0x1d(%rbp)
  jmp    L38
L104:
  cmpl   $0x2,-0x18(%rbp)
  jne    L72
  movb   $0x47,-0x1d(%rbp)
  jmp    L38
L72:
  movb   $0x54,-0x1d(%rbp)
L38:
  mov    -0x14(%rbp),%eax
  cltq
  movzbl -0x1d(%rbp),%edx
  mov    %dl,-0xd(%rbp,%rax,1)
  shrb   $0x2,-0x1e(%rbp)
  subl   $0x1,-0x14(%rbp)
L56:
  cmpl   $0x0,-0x14(%rbp)
  jns    L95
  movl   $0x0,-0x18(%rbp)
  jmp    L113
L88:
  mov    -0x30(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x30(%rbp)
  mov    -0x18(%rbp),%edx
  movslq %edx,%rdx
  movzbl -0xd(%rbp,%rdx,1),%edx
  mov    %dl,(%rax)
  addl   $0x1,-0x18(%rbp)
L113:
  cmpl   $0x3,-0x18(%rbp)
  jle    L88
  addl   $0x1,-0x1c(%rbp)
L73:
  mov    -0x1c(%rbp),%eax
  cmp    -0x34(%rbp),%eax
  jl     L91
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L86
  call   <__stack_chk_fail@plt>
L86:
  leave
  ret
```

assistant: ```c
void decode_basepairs(unsigned char *s, char *dest, int n) {
  int i, m, l;
  unsigned char j;
  char a, ss[5];
  ss[4] = '\0';
  for (i = 0; i < n; i++) {
    j = s[i];
    for (l = 3; l >= 0; l--) {
      m = j % 4;
      if (m == 0)
        a = 'A';
      else if (m == 1)
        a = 'C';
      else if (m == 2)
        a = 'G';
      else
        a = 'T';
      ss[l] = a;
      j = j >> 2;
      ;
    }
    for (m = 0; m < 4; m++)
      *dest++ = ss[m];
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<decode_basepairs>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movb   $0x0,0x7(%rsp)
  test   %edx,%edx
  jle    L57
  mov    %edx,%edx
  lea    (%rsi,%rdx,4),%r12
  lea    -0x1(%rsp),%r10
  mov    $0x41,%r9d
  mov    $0x43,%r11d
  mov    $0x47,%ebp
  mov    $0x54,%ebx
  jmp    L21
L45:
  mov    %cl,0x3(%rax)
  shr    $0x2,%dl
  sub    $0x1,%rax
  cmp    %rax,%r10
  je     L41
L79:
  mov    %r9d,%ecx
  mov    %edx,%r8d
  and    $0x3,%r8d
  je     L45
  mov    %r11d,%ecx
  cmp    $0x1,%r8d
  je     L45
  cmp    $0x2,%r8d
  mov    %ebx,%ecx
  cmove  %ebp,%ecx
  jmp    L45
L41:
  mov    $0x0,%eax
L37:
  movzbl 0x3(%rsp,%rax,1),%edx
  mov    %dl,(%rsi,%rax,1)
  add    $0x1,%rax
  cmp    $0x4,%rax
  jne    L37
  add    $0x4,%rsi
  add    $0x1,%rdi
  cmp    %r12,%rsi
  je     L57
L21:
  movzbl (%rdi),%edx
  lea    0x3(%rsp),%rax
  jmp    L79
L57:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L94
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L94:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void decode_basepairs(unsigned char *s, char *dest, int n) {
  int i, m, l;
  unsigned char j;
  char a, ss[5];
  ss[4] = '\0';
  for (i = 0; i < n; i++) {
    j = s[i];
    for (l = 3; l >= 0; l--) {
      m = j % 4;
      if (m == 0)
        a = 'A';
      else if (m == 1)
        a = 'C';
      else if (m == 2)
        a = 'G';
      else
        a = 'T';
      ss[l] = a;
      j = j >> 2;
      ;
    }
    for (m = 0; m < 4; m++)
      *dest++ = ss[m];
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<decode_basepairs>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movb   $0x0,0x7(%rsp)
  test   %edx,%edx
  jle    L49
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  movslq %edx,%r9
  xor    %edi,%edi
  lea    0x3(%rsp),%r12
  lea    -0x1(%rsp),%r8
  mov    $0x47,%r11d
  mov    $0x54,%r10d
  cs nopw 0x0(%rax,%rax,1)
L29:
  movzbl (%rbx,%rdi,1),%edx
  mov    %r12,%rax
L31:
  mov    %edx,%esi
  mov    $0x41,%ecx
  and    $0x3,%esi
  je     L46
  mov    $0x43,%ecx
  cmp    $0x1,%sil
  je     L46
  cmp    $0x2,%sil
  mov    %r10d,%ecx
  cmove  %r11d,%ecx
L46:
  mov    %cl,0x3(%rax)
  sub    $0x1,%rax
  shr    $0x2,%dl
  cmp    %rax,%r8
  jne    L31
  mov    0x3(%rsp),%eax
  mov    %eax,0x0(%rbp,%rdi,4)
  add    $0x1,%rdi
  cmp    %r9,%rdi
  jne    L29
L49:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L8:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void decode_basepairs(unsigned char *s, char *dest, int n) {
  int i, m, l;
  unsigned char j;
  char a, ss[5];
  ss[4] = '\0';
  for (i = 0; i < n; i++) {
    j = s[i];
    for (l = 3; l >= 0; l--) {
      m = j % 4;
      if (m == 0)
        a = 'A';
      else if (m == 1)
        a = 'C';
      else if (m == 2)
        a = 'G';
      else
        a = 'T';
      ss[l] = a;
      j = j >> 2;
      ;
    }
    for (m = 0; m < 4; m++)
      *dest++ = ss[m];
  }
}
```