ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 下面的汇编指令对应的C源代码是什么:

```asm
<Hash>:
  endbr64
  movsbl (%rdi),%edx
  test   %dl,%dl
  je     L12
  xor    %ecx,%ecx
  nopl   (%rax)
L4:
  add    $0x1,%rdi
  add    %edx,%ecx
  movsbl (%rdi),%edx
  test   %dl,%dl
  jne    L4
  mov    %ecx,%eax
  neg    %eax
  cmovs  %ecx,%eax
  cltd
  idiv   %esi
  mov    %edx,%r8d
  mov    %r8d,%eax
  ret
  xchg   %ax,%ax
L12:
  xor    %r8d,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
int Hash(char *name, int nEntries) {
  register int val = 0;
  while (*name)
    val += *name++;
  if (val < 0)
    val = -val;
  return (val % nEntries);
}
```
user: What is the c source code of the assembly code below?

```asm
<Hash>:
  endbr64
  movsbl (%rdi),%edx
  test   %dl,%dl
  je     L9
  xor    %ecx,%ecx
  nopl   (%rax)
L16:
  add    $0x1,%rdi
  add    %edx,%ecx
  movsbl (%rdi),%edx
  test   %dl,%dl
  jne    L16
  mov    %ecx,%eax
  neg    %eax
  cmovs  %ecx,%eax
  cltd
  idiv   %esi
  mov    %edx,%r8d
  mov    %r8d,%eax
  ret
  xchg   %ax,%ax
L9:
  xor    %r8d,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
int Hash(char *name, int nEntries) {
  register int val = 0;
  while (*name)
    val += *name++;
  if (val < 0)
    val = -val;
  return (val % nEntries);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<DHparams_print_fp>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    $0x0,%eax
  call   <BIO_s_file@plt>
  mov    %eax,%edi
  call   <BIO_new@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L9
  mov    D29(%rip),%rax
  mov    (%rax),%edx
  mov    D0(%rip),%rax
  mov    (%rax),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <DHerr@plt>
  mov    $0x0,%eax
  jmp    L2
L9:
  mov    D1(%rip),%rax
  mov    (%rax),%edx
  mov    -0x18(%rbp),%rcx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <BIO_set_fp@plt>
  mov    -0x20(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <DHparams_print@plt>
  mov    %eax,-0xc(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <BIO_free@plt>
  mov    -0xc(%rbp),%eax
L2:
  leave
  ret
```

assistant: ```c
int DHparams_print_fp(FILE *fp, const DH *x) {
  BIO *b;
  int ret;
  if ((b = BIO_new(BIO_s_file())) == NULL) {
    DHerr(DH_F_DHPARAMS_PRINT_FP, ERR_R_BUF_LIB);
    return 0;
  }
  BIO_set_fp(b, fp, BIO_NOCLOSE);
  ret = DHparams_print(b, x);
  BIO_free(b);
  return ret;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<DHparams_print_fp>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    %rsi,%r12
  mov    $0x0,%eax
  call   <BIO_s_file@plt>
  mov    %eax,%edi
  call   <BIO_new@plt>
  test   %rax,%rax
  je     L17
  mov    %rax,%rbx
  mov    D10(%rip),%rax
  mov    (%rax),%edx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <BIO_set_fp@plt>
  mov    %r12,%rsi
  mov    %rbx,%rdi
  call   <DHparams_print@plt>
  mov    %eax,%ebp
  mov    %rbx,%rdi
  call   <BIO_free@plt>
L8:
  mov    %ebp,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L17:
  mov    D13(%rip),%rax
  mov    (%rax),%esi
  mov    D14(%rip),%rax
  mov    (%rax),%edi
  call   <DHerr@plt>
  mov    $0x0,%ebp
  jmp    L8
```

assistant: ```c
int DHparams_print_fp(FILE *fp, const DH *x) {
  BIO *b;
  int ret;
  if ((b = BIO_new(BIO_s_file())) == NULL) {
    DHerr(DH_F_DHPARAMS_PRINT_FP, ERR_R_BUF_LIB);
    return 0;
  }
  BIO_set_fp(b, fp, BIO_NOCLOSE);
  ret = DHparams_print(b, x);
  BIO_free(b);
  return ret;
}
```
user: What is the c source code of the assembly code below:

```asm
<DHparams_print_fp>:
  endbr64
  push   %r13
  xor    %eax,%eax
  mov    %rsi,%r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  call   <BIO_s_file@plt>
  mov    %eax,%edi
  call   <BIO_new@plt>
  test   %rax,%rax
  je     L4
  mov    %rax,%rbp
  mov    D16(%rip),%rax
  mov    %r12,%rsi
  mov    %rbp,%rdi
  mov    (%rax),%edx
  call   <BIO_set_fp@plt>
  mov    %r13,%rsi
  mov    %rbp,%rdi
  call   <DHparams_print@plt>
  mov    %rbp,%rdi
  mov    %eax,%r12d
  call   <BIO_free@plt>
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax,%rax,1)
L4:
  mov    D11(%rip),%rax
  xor    %r12d,%r12d
  mov    (%rax),%esi
  mov    D29(%rip),%rax
  mov    (%rax),%edi
  call   <DHerr@plt>
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int DHparams_print_fp(FILE *fp, const DH *x) {
  BIO *b;
  int ret;
  if ((b = BIO_new(BIO_s_file())) == NULL) {
    DHerr(DH_F_DHPARAMS_PRINT_FP, ERR_R_BUF_LIB);
    return 0;
  }
  BIO_set_fp(b, fp, BIO_NOCLOSE);
  ret = DHparams_print(b, x);
  BIO_free(b);
  return ret;
}
```
user: What is the c source code of the assembly code below?

```asm
<DHparams_print_fp>:
  endbr64
  push   %r13
  xor    %eax,%eax
  mov    %rsi,%r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  call   <BIO_s_file@plt>
  mov    %eax,%edi
  call   <BIO_new@plt>
  test   %rax,%rax
  je     L6
  mov    %rax,%rbp
  mov    D18(%rip),%rax
  mov    %r12,%rsi
  mov    %rbp,%rdi
  mov    (%rax),%edx
  call   <BIO_set_fp@plt>
  mov    %r13,%rsi
  mov    %rbp,%rdi
  call   <DHparams_print@plt>
  mov    %rbp,%rdi
  mov    %eax,%r12d
  call   <BIO_free@plt>
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax,%rax,1)
L6:
  mov    D24(%rip),%rax
  xor    %r12d,%r12d
  mov    (%rax),%esi
  mov    D11(%rip),%rax
  mov    (%rax),%edi
  call   <DHerr@plt>
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int DHparams_print_fp(FILE *fp, const DH *x) {
  BIO *b;
  int ret;
  if ((b = BIO_new(BIO_s_file())) == NULL) {
    DHerr(DH_F_DHPARAMS_PRINT_FP, ERR_R_BUF_LIB);
    return 0;
  }
  BIO_set_fp(b, fp, BIO_NOCLOSE);
  ret = DHparams_print(b, x);
  BIO_free(b);
  return ret;
}
```
user: Translate the below assembly code into C programming language:

```asm
<cifrar>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L11
L9:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  lea    0x3(%rax),%ecx
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    %ecx,%edx
  mov    %dl,(%rax)
  addl   $0x1,-0x4(%rbp)
L11:
  mov    -0x4(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jl     L9
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void cifrar(char mesg[], char msg_c[], int size) {
  int i;
  for (i = 0; i < size; i++) {
    msg_c[i] = mesg[i] + 3;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<cifrar>:
  endbr64
  test   %edx,%edx
  jle    L3
  mov    %edx,%edx
  mov    $0x0,%eax
L11:
  movzbl (%rdi,%rax,1),%ecx
  add    $0x3,%ecx
  mov    %cl,(%rsi,%rax,1)
  add    $0x1,%rax
  cmp    %rdx,%rax
  jne    L11
L3:
  ret
```

assistant: ```c
void cifrar(char mesg[], char msg_c[], int size) {
  int i;
  for (i = 0; i < size; i++) {
    msg_c[i] = mesg[i] + 3;
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<cifrar>:
  endbr64
  test   %edx,%edx
  jle    L18
  movslq %edx,%rdx
  xor    %eax,%eax
  nopl   (%rax)
L10:
  movzbl (%rdi,%rax,1),%ecx
  add    $0x3,%ecx
  mov    %cl,(%rsi,%rax,1)
  add    $0x1,%rax
  cmp    %rax,%rdx
  jne    L10
L18:
  ret
```

assistant: ```c
void cifrar(char mesg[], char msg_c[], int size) {
  int i;
  for (i = 0; i < size; i++) {
    msg_c[i] = mesg[i] + 3;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<cifrar>:
  endbr64
  mov    %rdi,%rcx
  test   %edx,%edx
  jle    L36
  lea    0x1(%rdi),%r9
  mov    %rsi,%rdi
  lea    -0x1(%rdx),%eax
  mov    %edx,%r8d
  sub    %r9,%rdi
  cmp    $0xe,%rdi
  jbe    L111
  cmp    $0x6,%eax
  jbe    L111
  cmp    $0xe,%eax
  jbe    L125
  mov    %edx,%edi
  movdqa D9(%rip),%xmm1
  xor    %eax,%eax
  shr    $0x4,%edi
  shl    $0x4,%rdi
  xchg   %ax,%ax
L59:
  movdqu (%rcx,%rax,1),%xmm0
  paddb  %xmm1,%xmm0
  movups %xmm0,(%rsi,%rax,1)
  add    $0x10,%rax
  cmp    %rax,%rdi
  jne    L59
  mov    %edx,%edi
  and    $0xfffffff0,%edi
  mov    %edi,%eax
  cmp    %edi,%edx
  je     L36
  mov    %edx,%r8d
  sub    %edi,%r8d
  lea    -0x1(%r8),%r9d
  cmp    $0x6,%r9d
  jbe    L61
L4:
  movq   D9(%rip),%xmm1
  movq   (%rcx,%rdi,1),%xmm0
  paddb  %xmm1,%xmm0
  movq   %xmm0,(%rsi,%rdi,1)
  mov    %r8d,%edi
  and    $0xfffffff8,%edi
  add    %edi,%eax
  cmp    %edi,%r8d
  je     L36
L61:
  movslq %eax,%r8
  movzbl (%rcx,%r8,1),%edi
  add    $0x3,%edi
  mov    %dil,(%rsi,%r8,1)
  lea    0x1(%rax),%edi
  cmp    %edi,%edx
  jle    L36
  movslq %edi,%rdi
  movzbl (%rcx,%rdi,1),%r10d
  lea    0x3(%r10),%r8d
  mov    %r8b,(%rsi,%rdi,1)
  lea    0x2(%rax),%edi
  cmp    %edi,%edx
  jle    L36
  movslq %edi,%rdi
  movzbl (%rcx,%rdi,1),%r11d
  lea    0x3(%r11),%r8d
  mov    %r8b,(%rsi,%rdi,1)
  lea    0x3(%rax),%edi
  cmp    %edi,%edx
  jle    L36
  movslq %edi,%rdi
  movzbl (%rcx,%rdi,1),%r11d
  lea    0x3(%r11),%r8d
  mov    %r8b,(%rsi,%rdi,1)
  lea    0x4(%rax),%edi
  cmp    %edi,%edx
  jle    L36
  movslq %edi,%rdi
  movzbl (%rcx,%rdi,1),%r10d
  lea    0x3(%r10),%r8d
  mov    %r8b,(%rsi,%rdi,1)
  lea    0x5(%rax),%edi
  cmp    %edi,%edx
  jle    L36
  movslq %edi,%rdi
  add    $0x6,%eax
  movzbl (%rcx,%rdi,1),%r11d
  lea    0x3(%r11),%r8d
  mov    %r8b,(%rsi,%rdi,1)
  cmp    %eax,%edx
  jle    L36
  cltq
  movzbl (%rcx,%rax,1),%edx
  add    $0x3,%edx
  mov    %dl,(%rsi,%rax,1)
  ret
  nopw   0x0(%rax,%rax,1)
L111:
  mov    %edx,%edx
  xor    %eax,%eax
  nopl   0x0(%rax)
L37:
  movzbl (%rcx,%rax,1),%edi
  add    $0x3,%edi
  mov    %dil,(%rsi,%rax,1)
  add    $0x1,%rax
  cmp    %rdx,%rax
  jne    L37
L36:
  ret
L125:
  xor    %edi,%edi
  xor    %eax,%eax
  jmp    L4
```

tool_calls: name: parse_data
arguments: data_label: D9
data_type: byte[16]

tool: {"D9": [0x03,0x03,0x03,0x03,0x03,0x03,0x03,0x03,0x03,0x03,0x03,0x03,0x03,0x03,0x03,0x03]}

assistant: ```c
void cifrar(char mesg[], char msg_c[], int size) {
  int i;
  for (i = 0; i < size; i++) {
    msg_c[i] = mesg[i] + 3;
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<sc>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  call   <strlen@plt>
  mov    %eax,%ebx
  mov    $0x4,%edi
  call   <malloc@plt>
  mov    %rax,%r12
  mov    %ebx,%edx
  mov    %rbp,%rsi
  mov    %rax,%rdi
  call   <strncpy@plt>
  test   %ebx,%ebx
  jle    L46
  mov    %rbp,%rdi
  mov    $0x1,%r8d
  mov    $0x2,%r10d
  mov    $0x0,%r9d
  lea    -0x2(%rbp),%r13
  jmp    L109
L8:
  cmp    $0x1,%edx
  je     L36
  sub    %edx,%ebx
  lea    0x1(%rbx),%eax
  cmp    %r10d,%eax
  jl     L80
  movslq %r10d,%rax
  add    %rbp,%rax
  mov    %ebx,%ecx
  sub    %r9d,%ecx
  lea    -0x1(%rcx),%ecx
  lea    0x3(%rdi,%rcx,1),%r14
  movslq %edx,%rsi
L139:
  movzbl -0x2(%rax,%rsi,1),%ecx
  mov    %cl,(%rax)
  add    $0x1,%rax
  cmp    %r14,%rax
  jne    L139
L80:
  add    $0x2,%ebx
  movslq %ebx,%rax
  movb   $0x0,0x0(%rbp,%rax,1)
L84:
  add    $0x30,%edx
  mov    %dl,0x1(%r11)
L7:
  add    $0x2,%r9d
  add    $0x2,%r10
  add    $0x2,%rdi
  add    $0x2,%r8
  cmp    %r9d,%ebx
  jle    L18
L109:
  cmp    %r8d,%ebx
  jl     L7
  mov    %rdi,%r11
  movzbl (%rdi),%ecx
  mov    %r8,%rax
  mov    $0x1,%edx
L91:
  cmp    0x0(%rbp,%rax,1),%cl
  jne    L8
  add    $0x1,%edx
  add    $0x1,%rax
  cmp    %eax,%ebx
  jge    L91
  jmp    L7
L36:
  lea    -0x1(%rbx),%ecx
  cmp    %eax,%ecx
  jl     L127
  movslq %ecx,%rcx
  add    %rbp,%rcx
  movslq %ebx,%rsi
  add    %r13,%rsi
  lea    -0x1(%rbx),%r14d
  sub    %eax,%r14d
  sub    %r14,%rsi
L115:
  movzbl (%rcx),%eax
  mov    %al,0x1(%rcx)
  sub    $0x1,%rcx
  cmp    %rcx,%rsi
  jne    L115
L127:
  add    $0x1,%ebx
  jmp    L84
L18:
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %eax,%ebx
  mov    %r12,%rdi
  call   <strlen@plt>
  cmp    %eax,%ebx
  cmovle %rbp,%r12
L62:
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L46:
  mov    %rbp,%r12
  jmp    L62
```

assistant: ```c
char *sc(char str[]) {
  int i, j, k;
  int len;
  int count;
  char *old;
  len = strlen(str);
  old = malloc(sizeof(len + 1));
  strncpy(old, str, len);
  if (len < 1)
    return str;
  for (i = 0; i < len; i++, i++) {
    for (count = 1, j = i + 1; j < len + 1; j++)
      if (str[i] == str[j])
        count++;
      else {
        if (count == 1) {
          for (k = len - 1; k >= j; k--)
            str[k + 1] = str[k];
          len++;
        } else {
          for (k = i + 2; k < len - count + 2; k++)
            str[k] = str[k + count - 2];
          len = len - count + 2;
          str[len] = '\0';
        }
        str[i + 1] = '0' + count;
        break;
      }
  }
  if (strlen(str) > strlen(old))
    return old;
  else
    return str;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<sc>:
  endbr64
  push   %r15
  xor    %r15d,%r15d
  push   %r14
  xor    %r14d,%r14d
  push   %r13
  mov    %rdi,%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  mov    $0x4,%edi
  mov    %eax,%ebp
  call   <malloc@plt>
  mov    %ebp,%edx
  mov    %r13,%rsi
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <strncpy@plt>
  test   %ebp,%ebp
  jle    L98
  nopl   0x0(%rax)
L5:
  mov    %r15d,%eax
  add    $0x2,%r15d
  add    $0x1,%eax
  cmp    %eax,%ebp
  jl     L85
  movzbl 0x0(%r13,%r14,1),%edx
  lea    0x1(%r14),%rax
  mov    $0x1,%ebx
  jmp    L40
  nopw   0x0(%rax,%rax,1)
L67:
  add    $0x1,%rax
  add    $0x1,%ebx
  cmp    %eax,%ebp
  jl     L85
L40:
  mov    %eax,%ecx
  cmp    0x0(%r13,%rax,1),%dl
  je     L67
  cmp    $0x1,%ebx
  je     L22
  sub    %ebx,%ebp
  lea    0x1(%rbp),%eax
  cmp    %r15d,%eax
  jl     L35
  mov    %ebp,%edx
  movslq %r15d,%rax
  movslq %ebx,%rsi
  sub    %r14d,%edx
  add    %r13,%rax
  lea    -0x1(%rdx),%ecx
  lea    0x3(%r13,%r14,1),%rdx
  add    %rdx,%rcx
L117:
  movzbl -0x2(%rax,%rsi,1),%edx
  add    $0x1,%rax
  mov    %dl,-0x1(%rax)
  cmp    %rcx,%rax
  jne    L117
L35:
  add    $0x2,%ebp
  movslq %ebp,%rax
  movb   $0x0,0x0(%r13,%rax,1)
L119:
  add    $0x30,%ebx
  mov    %bl,0x1(%r13,%r14,1)
L85:
  add    $0x2,%r14
  cmp    %r15d,%ebp
  jg     L5
  mov    %r13,%rdi
  call   <strlen@plt>
  mov    %r12,%rdi
  mov    %eax,%ebx
  call   <strlen@plt>
  cmp    %eax,%ebx
  jle    L98
L53:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L98:
  mov    %r13,%r12
  jmp    L53
  nopl   (%rax)
L22:
  lea    -0x1(%rbp),%eax
  cmp    %eax,%ecx
  jg     L20
  mov    %ebp,%edx
  cltq
  movslq %ebp,%rdi
  sub    %ecx,%edx
  sub    $0x1,%edx
  sub    %rdx,%rax
  sub    %rdx,%rdi
  lea    0x1(%rdx),%r8
  lea    0x0(%r13,%rax,1),%rsi
  add    %r13,%rdi
  mov    %r8,%rdx
  call   <memmove@plt>
L20:
  add    $0x1,%ebp
  jmp    L119
```

assistant: ```c
char *sc(char str[]) {
  int i, j, k;
  int len;
  int count;
  char *old;
  len = strlen(str);
  old = malloc(sizeof(len + 1));
  strncpy(old, str, len);
  if (len < 1)
    return str;
  for (i = 0; i < len; i++, i++) {
    for (count = 1, j = i + 1; j < len + 1; j++)
      if (str[i] == str[j])
        count++;
      else {
        if (count == 1) {
          for (k = len - 1; k >= j; k--)
            str[k + 1] = str[k];
          len++;
        } else {
          for (k = i + 2; k < len - count + 2; k++)
            str[k] = str[k + count - 2];
          len = len - count + 2;
          str[len] = '\0';
        }
        str[i + 1] = '0' + count;
        break;
      }
  }
  if (strlen(str) > strlen(old))
    return old;
  else
    return str;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<gen_jump_case>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <output_with_tab@plt>
  mov    $0x0,%eax
  call   <newline@plt>
  nop
  pop    %rbp
  ret
```

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

tool: D2: jmp _case

assistant: ```c
void gen_jump_case(void) {
  output_with_tab("jmp _case");
  newline();
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<gen_jump_case>:
  endbr64
  sub    $0x8,%rsp
  lea    D5(%rip),%rdi
  call   <output_with_tab@plt>
  mov    $0x0,%eax
  call   <newline@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: tag: D5
data_type: string

tool: D5: jmp _case

assistant: ```c
void gen_jump_case(void) {
  output_with_tab("jmp _case");
  newline();
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<gen_jump_case>:
  endbr64
  sub    $0x8,%rsp
  lea    D6(%rip),%rdi
  call   <output_with_tab@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1060 <newline@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D6
datatype: string

tool: D6: jmp _case

assistant: ```c
void gen_jump_case(void) {
  output_with_tab("jmp _case");
  newline();
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<gen_jump_case>:
  endbr64
  sub    $0x8,%rsp
  lea    D2(%rip),%rdi
  call   <output_with_tab@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1060 <newline@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D2
type_infered: string

tool: D2: jmp _case

assistant: ```c
void gen_jump_case(void) {
  output_with_tab("jmp _case");
  newline();
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<philospher>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
L5:
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    $0x1,%edi
  call   <sleep@plt>
  mov    -0x8(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,%edi
  call   <take_fork@plt>
  mov    $0x0,%edi
  call   <sleep@plt>
  mov    -0x8(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,%edi
  call   <put_fork@plt>
  jmp    L5
```

assistant: ```c
void *philospher(void *num) {
  while (1) {
    int *i = num;
    sleep(1);
    take_fork(*i);
    sleep(0);
    put_fork(*i);
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<philospher>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
L6:
  mov    $0x1,%edi
  call   <sleep@plt>
  mov    (%rbx),%edi
  call   <take_fork@plt>
  mov    $0x0,%edi
  call   <sleep@plt>
  mov    (%rbx),%edi
  call   <put_fork@plt>
  jmp    L6
```

assistant: ```c
void *philospher(void *num) {
  while (1) {
    int *i = num;
    sleep(1);
    take_fork(*i);
    sleep(0);
    put_fork(*i);
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<philospher>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  nopl   0x0(%rax,%rax,1)
L3:
  mov    $0x1,%edi
  call   <sleep@plt>
  mov    (%rbx),%edi
  call   <take_fork@plt>
  xor    %edi,%edi
  call   <sleep@plt>
  mov    (%rbx),%edi
  call   <put_fork@plt>
  jmp    L3
```

assistant: ```c
void *philospher(void *num) {
  while (1) {
    int *i = num;
    sleep(1);
    take_fork(*i);
    sleep(0);
    put_fork(*i);
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<philospher>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  nopl   0x0(%rax,%rax,1)
L2:
  mov    $0x1,%edi
  call   <sleep@plt>
  mov    (%rbx),%edi
  call   <take_fork@plt>
  xor    %edi,%edi
  call   <sleep@plt>
  mov    (%rbx),%edi
  call   <put_fork@plt>
  jmp    L2
```

assistant: ```c
void *philospher(void *num) {
  while (1) {
    int *i = num;
    sleep(1);
    take_fork(*i);
    sleep(0);
    put_fork(*i);
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<iotapi_handler_restart>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movw   $0x7372,-0xb(%rbp)
  movb   $0x0,-0x9(%rbp)
  lea    -0xb(%rbp),%rax
  mov    %rax,%rdi
  call   <iotapi_queue_command@plt>
  mov    %eax,-0x10(%rbp)
  cmpl   $0x0,-0x10(%rbp)
  jns    L5
  mov    $0xffffffff,%eax
  jmp    L27
L5:
  mov    $0x0,%eax
L27:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L0
  call   <__stack_chk_fail@plt>
L0:
  leave
  ret
```

assistant: ```c
int iotapi_handler_restart(void) {
  char command[3] = "rs";
  int ret = iotapi_queue_command(command);
  if (ret < 0) {
    ;
    return -1;
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<iotapi_handler_restart>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movw   $0x7372,0x5(%rsp)
  movb   $0x0,0x7(%rsp)
  lea    0x5(%rsp),%rdi
  call   <iotapi_queue_command@plt>
  sar    $0x1f,%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L5
  add    $0x18,%rsp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int iotapi_handler_restart(void) {
  char command[3] = "rs";
  int ret = iotapi_queue_command(command);
  if (ret < 0) {
    ;
    return -1;
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<iotapi_handler_restart>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x5(%rsp),%rdi
  mov    $0x7372,%eax
  movb   $0x0,0x7(%rsp)
  mov    %ax,0x5(%rsp)
  call   <iotapi_queue_command@plt>
  sar    $0x1f,%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L1
  add    $0x18,%rsp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int iotapi_handler_restart(void) {
  char command[3] = "rs";
  int ret = iotapi_queue_command(command);
  if (ret < 0) {
    ;
    return -1;
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<iotapi_handler_restart>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x5(%rsp),%rdi
  mov    $0x7372,%eax
  movb   $0x0,0x7(%rsp)
  mov    %ax,0x5(%rsp)
  call   <iotapi_queue_command@plt>
  sar    $0x1f,%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L2
  add    $0x18,%rsp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int iotapi_handler_restart(void) {
  char command[3] = "rs";
  int ret = iotapi_queue_command(command);
  if (ret < 0) {
    ;
    return -1;
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<strcmp>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  jmp    L2
L23:
  addq   $0x1,-0x8(%rbp)
  addq   $0x1,-0x10(%rbp)
L2:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%edx
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  cmp    %al,%dl
  jne    L18
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L23
L18:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movzbl %al,%edx
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  movzbl %al,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  pop    %rbp
  ret
```

assistant: ```c
int strcmp(const char *l, const char *r) {
  for (; *l == *r && *l; l++, r++) {
  }
  return *(unsigned char *)l - *(unsigned char *)r;
}
```
user: What is the c source code of the assembly code below?

```asm
<strcmp>:
  endbr64
  movzbl (%rdi),%eax
  movzbl (%rsi),%ecx
  test   %al,%al
  je     L14
  cmp    %cl,%al
  jne    L14
  mov    $0x1,%edx
L27:
  movzbl (%rdi,%rdx,1),%eax
  movzbl (%rsi,%rdx,1),%ecx
  add    $0x1,%rdx
  test   %al,%al
  je     L14
  cmp    %cl,%al
  je     L27
L14:
  movzbl %al,%eax
  movzbl %cl,%ecx
  sub    %ecx,%eax
  ret
```

assistant: ```c
int strcmp(const char *l, const char *r) {
  for (; *l == *r && *l; l++, r++) {
  }
  return *(unsigned char *)l - *(unsigned char *)r;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<strcmp>:
  endbr64
  movzbl (%rdi),%edx
  movzbl (%rsi),%ecx
  mov    $0x1,%eax
  cmp    %dl,%cl
  je     L14
  jmp    L3
  nopl   (%rax)
L17:
  movzbl (%rdi,%rax,1),%edx
  add    $0x1,%rax
  movzbl -0x1(%rsi,%rax,1),%ecx
  cmp    %cl,%dl
  jne    L3
L14:
  test   %dl,%dl
  jne    L17
  xor    %eax,%eax
  sub    %ecx,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L3:
  movzbl %dl,%eax
  sub    %ecx,%eax
  ret
```

assistant: ```c
int strcmp(const char *l, const char *r) {
  for (; *l == *r && *l; l++, r++) {
  }
  return *(unsigned char *)l - *(unsigned char *)r;
}
```