ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Translate the below assembly code into C programming language:

```asm
<append_strtab>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  cmpb   $0x0,(%rsi)
  je     L5
  mov    $0x1,%eax
L25:
  mov    %eax,%edx
  add    $0x1,%rax
  cmpb   $0x0,-0x1(%rsi,%rax,1)
  jne    L25
L0:
  lea    0x1(%rdx),%ebx
  mov    0x0(%rbp),%r12
  mov    %ebx,%edx
  mov    %r12,%rdi
  call   <memcpy@plt>
  movslq %ebx,%rdx
  add    %r12,%rdx
  mov    %rdx,0x0(%rbp)
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L5:
  mov    $0x0,%edx
  jmp    L0
```

assistant: ```c
char *append_strtab(char **strtab, char *str) {
  int nbytes;
  char *res;
  nbytes = 0;
  while (str[nbytes] != '\0') {
    ++nbytes;
  }
  ++nbytes;
  res = *strtab;
  memcpy(res, str, nbytes);
  *strtab = res + nbytes;
  return res;
}
```
user: What is the c source code of the assembly code below?

```asm
<append_strtab>:
  endbr64
  cmpb   $0x0,(%rsi)
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  je     L6
  mov    $0x1,%eax
  nopl   (%rax)
L11:
  mov    %eax,%edx
  add    $0x1,%rax
  cmpb   $0x0,-0x1(%rsi,%rax,1)
  jne    L11
  add    $0x1,%edx
  movslq %edx,%rbp
L2:
  mov    (%rbx),%r12
  mov    %r12,%rdi
  call   <memcpy@plt>
  lea    (%r12,%rbp,1),%rax
  mov    %rax,(%rbx)
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   (%rax)
L6:
  mov    $0x1,%ebp
  mov    $0x1,%edx
  jmp    L2
```

assistant: ```c
char *append_strtab(char **strtab, char *str) {
  int nbytes;
  char *res;
  nbytes = 0;
  while (str[nbytes] != '\0') {
    ++nbytes;
  }
  ++nbytes;
  res = *strtab;
  memcpy(res, str, nbytes);
  *strtab = res + nbytes;
  return res;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<append_strtab>:
  endbr64
  cmpb   $0x0,(%rsi)
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  je     L16
  mov    $0x1,%eax
  nopl   (%rax)
L14:
  mov    %eax,%edx
  add    $0x1,%rax
  cmpb   $0x0,-0x1(%rsi,%rax,1)
  jne    L14
  add    $0x1,%edx
  movslq %edx,%rbp
L18:
  mov    (%rbx),%r12
  mov    %r12,%rdi
  call   <memcpy@plt>
  lea    (%r12,%rbp,1),%rax
  mov    %rax,(%rbx)
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   (%rax)
L16:
  mov    $0x1,%ebp
  mov    $0x1,%edx
  jmp    L18
```

assistant: ```c
char *append_strtab(char **strtab, char *str) {
  int nbytes;
  char *res;
  nbytes = 0;
  while (str[nbytes] != '\0') {
    ++nbytes;
  }
  ++nbytes;
  res = *strtab;
  memcpy(res, str, nbytes);
  *strtab = res + nbytes;
  return res;
}
```
user: Translate the below assembly code into C programming language:

```asm
<miller_rabin_test>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %esi,-0x28(%rbp)
  mov    -0x24(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0xc(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L69
L63:
  addl   $0x1,-0x4(%rbp)
  mov    -0xc(%rbp),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  mov    %eax,-0xc(%rbp)
L69:
  mov    -0xc(%rbp),%eax
  and    $0x1,%eax
  test   %eax,%eax
  je     L63
  jmp    L17
L32:
  mov    -0x24(%rbp),%eax
  sub    $0x2,%eax
  mov    %eax,%edi
  call   <rand_num@plt>
  add    $0x2,%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x24(%rbp),%edx
  mov    -0xc(%rbp),%ecx
  mov    -0x14(%rbp),%eax
  mov    %ecx,%esi
  mov    %eax,%edi
  call   <exp_mod@plt>
  mov    %eax,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,-0x10(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L75
L30:
  mov    -0x14(%rbp),%eax
  imul   %eax,%eax
  cltd
  idivl  -0x24(%rbp)
  mov    %edx,-0x14(%rbp)
  cmpl   $0x1,-0x14(%rbp)
  jne    L1
  cmpl   $0x1,-0x10(%rbp)
  je     L1
  mov    -0x24(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0x10(%rbp)
  je     L1
  mov    $0x0,%eax
  jmp    L97
L1:
  cmpl   $0x1,-0x14(%rbp)
  je     L116
  mov    -0x14(%rbp),%eax
  mov    %eax,-0x10(%rbp)
  addl   $0x1,-0x8(%rbp)
L75:
  mov    -0x8(%rbp),%eax
  cmp    -0x4(%rbp),%eax
  jl     L30
  jmp    L47
L116:
  nop
L47:
  cmpl   $0x1,-0x14(%rbp)
  je     L17
  mov    $0x0,%eax
  jmp    L97
L17:
  mov    -0x28(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0x28(%rbp)
  test   %eax,%eax
  jne    L32
  mov    $0x1,%eax
L97:
  leave
  ret
```

assistant: ```c
int miller_rabin_test(uint32_t n, uint8_t times) {
  uint32_t x, pre, u = n - 1, i, k = 0;
  while (!(u & 1)) {
    k++;
    u /= 2;
  }
  while (times--) {
    x = rand_num(n - 2) + 2;
    pre = x = exp_mod(x, u, n);
    for (i = 0; i < k; i++) {
      x = (uint64_t)x * x % n;
      if ((1 == x) && (pre != 1) && (pre != (n - 1))) {
        return 0;
      } else if (1 == x) {
        break;
      }
      pre = x;
    }
    if (x != 1) {
      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
<miller_rabin_test>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebx
  lea    -0x1(%rdi),%r12d
  mov    %r12d,%ebp
  and    $0x1,%ebp
  jne    L170
  mov    %r12d,%r13d
L87:
  add    $0x1,%ebp
  mov    %r13d,%eax
  shr    $0x1f,%eax
  add    %r13d,%eax
  sar    %eax
  mov    %eax,%r13d
  test   $0x1,%al
  je     L87
L31:
  lea    -0x1(%rsi),%r14d
  lea    -0x2(%rbx),%r15d
  jmp    L3
L170:
  mov    %r12d,%r13d
  mov    $0x0,%ebp
  jmp    L31
L0:
  cmp    $0x1,%edx
  je     L7
  mov    %edx,%ecx
L146:
  add    $0x1,%esi
  cmp    %ebp,%esi
  je     L28
  mov    %ecx,%eax
  imul   %ecx,%eax
  cltd
  idiv   %ebx
  cmp    $0x1,%ecx
  setne  %dil
  cmp    %ecx,%r12d
  setne  %al
  test   %al,%dil
  je     L0
  cmp    $0x1,%edx
  jne    L0
  mov    $0x0,%eax
L51:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L144:
  mov    %eax,%ecx
L28:
  cmp    $0x1,%ecx
  jne    L117
L7:
  sub    $0x1,%r14d
L3:
  cmp    $0xffffffff,%r14d
  je     L14
  mov    %r15d,%edi
  call   <rand_num@plt>
  lea    0x2(%rax),%edi
  mov    %ebx,%edx
  mov    %r13d,%esi
  call   <exp_mod@plt>
  mov    %eax,%esi
  test   %ebp,%ebp
  jle    L144
  imul   %eax,%eax
  cltd
  idiv   %ebx
  mov    %edx,%ecx
  cmp    $0x1,%esi
  setne  %dil
  cmp    %esi,%r12d
  setne  %al
  test   %al,%dil
  je     L54
  cmp    $0x1,%edx
  je     L66
L54:
  cmp    $0x1,%edx
  je     L7
  mov    $0x0,%esi
  jmp    L146
L14:
  mov    $0x1,%eax
  jmp    L51
L66:
  mov    $0x0,%eax
  jmp    L51
L117:
  mov    $0x0,%eax
  jmp    L51
```

assistant: ```c
int miller_rabin_test(uint32_t n, uint8_t times) {
  uint32_t x, pre, u = n - 1, i, k = 0;
  while (!(u & 1)) {
    k++;
    u /= 2;
  }
  while (times--) {
    x = rand_num(n - 2) + 2;
    pre = x = exp_mod(x, u, n);
    for (i = 0; i < k; i++) {
      x = (uint64_t)x * x % n;
      if ((1 == x) && (pre != 1) && (pre != (n - 1))) {
        return 0;
      } else if (1 == x) {
        break;
      }
      pre = x;
    }
    if (x != 1) {
      return 0;
    }
  }
  return 1;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<miller_rabin_test>:
  endbr64
  push   %r15
  push   %r14
  lea    -0x1(%rdi),%r14d
  push   %r13
  mov    %r14d,%r13d
  mov    %r14d,%r15d
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  sub    $0x8,%rsp
  and    $0x1,%r13d
  jne    L36
  nopl   0x0(%rax,%rax,1)
L40:
  mov    %r15d,%eax
  add    $0x1,%r13d
  shr    $0x1f,%eax
  add    %r15d,%eax
  mov    %eax,%r15d
  sar    %r15d
  test   $0x2,%al
  je     L40
L133:
  lea    -0x2(%rbp),%ebx
  test   %r12d,%r12d
  jne    L96
  jmp    L110
  nopl   0x0(%rax)
L168:
  cmp    $0x1,%esi
  je     L41
  mov    %esi,%ecx
  nopl   0x0(%rax,%rax,1)
L180:
  cmp    %ecx,%r14d
  jne    L193
L164:
  sub    $0x1,%r12d
  je     L110
L96:
  mov    %ebx,%edi
  call   <rand_num@plt>
  mov    %r15d,%esi
  mov    %ebp,%edx
  lea    0x2(%rax),%edi
  call   <exp_mod@plt>
  mov    %eax,%esi
  test   %r13d,%r13d
  je     L106
  imul   %eax,%eax
  cltd
  idiv   %ebp
  mov    %edx,%ecx
  cmp    $0x1,%edx
  je     L168
L141:
  xor    %edi,%edi
  add    $0x1,%edi
  cmp    %r13d,%edi
  je     L89
  nopl   (%rax)
L80:
  mov    %ecx,%eax
  imul   %ecx,%eax
  cltd
  idiv   %ebp
  cmp    $0x1,%edx
  jne    L46
  cmp    $0x1,%ecx
  jne    L180
  cmp    $0x1,%edx
  je     L164
L46:
  add    $0x1,%edi
  mov    %edx,%ecx
  cmp    %r13d,%edi
  jne    L80
L89:
  cmp    $0x1,%ecx
  je     L164
L193:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nop
L41:
  cmp    $0x1,%edx
  jne    L141
  jmp    L164
  nopw   0x0(%rax,%rax,1)
L110:
  add    $0x8,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L106:
  mov    %eax,%ecx
  cmp    $0x1,%ecx
  je     L164
  jmp    L193
  nopl   0x0(%rax)
L36:
  xor    %r13d,%r13d
  jmp    L133
```

assistant: ```c
int miller_rabin_test(uint32_t n, uint8_t times) {
  uint32_t x, pre, u = n - 1, i, k = 0;
  while (!(u & 1)) {
    k++;
    u /= 2;
  }
  while (times--) {
    x = rand_num(n - 2) + 2;
    pre = x = exp_mod(x, u, n);
    for (i = 0; i < k; i++) {
      x = (uint64_t)x * x % n;
      if ((1 == x) && (pre != 1) && (pre != (n - 1))) {
        return 0;
      } else if (1 == x) {
        break;
      }
      pre = x;
    }
    if (x != 1) {
      return 0;
    }
  }
  return 1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<miller_rabin_test>:
  endbr64
  push   %r15
  push   %r14
  lea    -0x1(%rdi),%r14d
  push   %r13
  mov    %r14d,%r13d
  mov    %r14d,%r15d
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  sub    $0x8,%rsp
  and    $0x1,%r13d
  jne    L33
  nopl   0x0(%rax,%rax,1)
L95:
  mov    %r15d,%eax
  add    $0x1,%r13d
  shr    $0x1f,%eax
  add    %r15d,%eax
  mov    %eax,%r15d
  sar    %r15d
  test   $0x2,%al
  je     L95
L29:
  lea    -0x2(%rbp),%ebx
  test   %r12d,%r12d
  jne    L99
  jmp    L193
  nopl   0x0(%rax)
L173:
  cmp    $0x1,%esi
  je     L142
  mov    %esi,%ecx
  nopl   0x0(%rax,%rax,1)
L194:
  cmp    %ecx,%r14d
  jne    L120
L131:
  sub    $0x1,%r12d
  je     L193
L99:
  mov    %ebx,%edi
  call   <rand_num@plt>
  mov    %r15d,%esi
  mov    %ebp,%edx
  lea    0x2(%rax),%edi
  call   <exp_mod@plt>
  mov    %eax,%esi
  test   %r13d,%r13d
  je     L37
  imul   %eax,%eax
  cltd
  idiv   %ebp
  mov    %edx,%ecx
  cmp    $0x1,%edx
  je     L173
L61:
  xor    %edi,%edi
  add    $0x1,%edi
  cmp    %r13d,%edi
  je     L172
  nopl   (%rax)
L112:
  mov    %ecx,%eax
  imul   %ecx,%eax
  cltd
  idiv   %ebp
  cmp    $0x1,%edx
  jne    L123
  cmp    $0x1,%ecx
  jne    L194
  cmp    $0x1,%edx
  je     L131
L123:
  add    $0x1,%edi
  mov    %edx,%ecx
  cmp    %r13d,%edi
  jne    L112
L172:
  cmp    $0x1,%ecx
  je     L131
L120:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nop
L142:
  cmp    $0x1,%edx
  jne    L61
  jmp    L131
  nopw   0x0(%rax,%rax,1)
L193:
  add    $0x8,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L37:
  mov    %eax,%ecx
  cmp    $0x1,%ecx
  je     L131
  jmp    L120
  nopl   0x0(%rax)
L33:
  xor    %r13d,%r13d
  jmp    L29
```

assistant: ```c
int miller_rabin_test(uint32_t n, uint8_t times) {
  uint32_t x, pre, u = n - 1, i, k = 0;
  while (!(u & 1)) {
    k++;
    u /= 2;
  }
  while (times--) {
    x = rand_num(n - 2) + 2;
    pre = x = exp_mod(x, u, n);
    for (i = 0; i < k; i++) {
      x = (uint64_t)x * x % n;
      if ((1 == x) && (pre != 1) && (pre != (n - 1))) {
        return 0;
      } else if (1 == x) {
        break;
      }
      pre = x;
    }
    if (x != 1) {
      return 0;
    }
  }
  return 1;
}
```
user: What is the c source code of the assembly code below?

```asm
<display>:
  endbr64
  sub    $0x8,%rsp
  pxor   %xmm3,%xmm3
  movapd %xmm3,%xmm2
  movapd %xmm3,%xmm1
  movapd %xmm3,%xmm0
  call   <glClearColor@plt>
  mov    D4(%rip),%rax
  mov    (%rax),%edi
  mov    D10(%rip),%rax
  or     (%rax),%edi
  call   <glClear@plt>
  mov    $0x0,%eax
  call   <drawScene@plt>
  mov    $0x0,%eax
  call   <selectObjects@plt>
  mov    $0x0,%eax
  call   <glFlush@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void display(void) {
  glClearColor(0.0, 0.0, 0.0, 0.0);
  glClear(GL_COLOR_BUFFER_BIT | GL_DEPTH_BUFFER_BIT);
  drawScene();
  selectObjects();
  glFlush();
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<display>:
  endbr64
  pxor   %xmm3,%xmm3
  sub    $0x8,%rsp
  movapd %xmm3,%xmm2
  movapd %xmm3,%xmm1
  movapd %xmm3,%xmm0
  call   <glClearColor@plt>
  mov    D14(%rip),%rax
  mov    (%rax),%edi
  mov    D16(%rip),%rax
  or     (%rax),%edi
  call   <glClear@plt>
  xor    %eax,%eax
  call   <drawScene@plt>
  xor    %eax,%eax
  call   <selectObjects@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1090 <glFlush@plt>
```

assistant: ```c
void display(void) {
  glClearColor(0.0, 0.0, 0.0, 0.0);
  glClear(GL_COLOR_BUFFER_BIT | GL_DEPTH_BUFFER_BIT);
  drawScene();
  selectObjects();
  glFlush();
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<display>:
  endbr64
  pxor   %xmm3,%xmm3
  sub    $0x8,%rsp
  movapd %xmm3,%xmm2
  movapd %xmm3,%xmm1
  movapd %xmm3,%xmm0
  call   <glClearColor@plt>
  mov    D18(%rip),%rax
  mov    (%rax),%edi
  mov    D5(%rip),%rax
  or     (%rax),%edi
  call   <glClear@plt>
  xor    %eax,%eax
  call   <drawScene@plt>
  xor    %eax,%eax
  call   <selectObjects@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1090 <glFlush@plt>
```

assistant: ```c
void display(void) {
  glClearColor(0.0, 0.0, 0.0, 0.0);
  glClear(GL_COLOR_BUFFER_BIT | GL_DEPTH_BUFFER_BIT);
  drawScene();
  selectObjects();
  glFlush();
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<bar>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movl   $0x68,D5(%rip)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void bar() {
  static int funclocal = 4;
  static int funclocal_bss;
  funclocal_bss = 104;
}
```
user: Translate the below assembly code into C programming language:

```asm
<bar>:
  endbr64
  ret
```

assistant: ```c
void bar() {
  static int funclocal = 4;
  static int funclocal_bss;
  funclocal_bss = 104;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<bar>:
  endbr64
  ret
```

assistant: ```c
void bar() {
  static int funclocal = 4;
  static int funclocal_bss;
  funclocal_bss = 104;
}
```
user: What is the c source code of the assembly code below?

```asm
<bar>:
  endbr64
  ret
```

assistant: ```c
void bar() {
  static int funclocal = 4;
  static int funclocal_bss;
  funclocal_bss = 104;
}
```
user: What is the c source code of the assembly code below:

```asm
<jx_parse_file>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  lea    D0(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <fopen@plt>
  mov    %rax,-0x10(%rbp)
  cmpq   $0x0,-0x10(%rbp)
  jne    L5
  mov    $0x0,%eax
  jmp    L11
L5:
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <jx_parse_stream@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <fclose@plt>
  mov    -0x8(%rbp),%rax
L11:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D0
type: string

tool: D0: r

assistant: ```c
struct jx *jx_parse_file(const char *name) {
  FILE *file = fopen(name, "r");
  if (!file)
    return 0;
  struct jx *j = jx_parse_stream(file);
  fclose(file);
  return j;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<jx_parse_file>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  lea    D5(%rip),%rsi
  call   <fopen@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L13
  mov    %rax,%rdi
  call   <jx_parse_stream@plt>
  mov    %rax,%rbp
  mov    %rbx,%rdi
  call   <fclose@plt>
L0:
  mov    %rbp,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L13:
  mov    %rax,%rbp
  jmp    L0
```

tool_calls: name: disassemble_data
arguments: data_tag: D5
type_infered: string

tool: D5: r

assistant: ```c
struct jx *jx_parse_file(const char *name) {
  FILE *file = fopen(name, "r");
  if (!file)
    return 0;
  struct jx *j = jx_parse_stream(file);
  fclose(file);
  return j;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<jx_parse_file>:
  endbr64
  push   %r12
  lea    D4(%rip),%rsi
  push   %rbp
  sub    $0x8,%rsp
  call   <fopen@plt>
  test   %rax,%rax
  je     L4
  mov    %rax,%rbp
  mov    %rax,%rdi
  call   <jx_parse_stream@plt>
  mov    %rbp,%rdi
  mov    %rax,%r12
  call   <fclose@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
  nopl   (%rax)
L4:
  xor    %r12d,%r12d
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D4
type_infered: string

tool: D4: r

assistant: ```c
struct jx *jx_parse_file(const char *name) {
  FILE *file = fopen(name, "r");
  if (!file)
    return 0;
  struct jx *j = jx_parse_stream(file);
  fclose(file);
  return j;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<jx_parse_file>:
  endbr64
  push   %r12
  lea    D2(%rip),%rsi
  push   %rbp
  sub    $0x8,%rsp
  call   <fopen@plt>
  test   %rax,%rax
  je     L1
  mov    %rax,%rbp
  mov    %rax,%rdi
  call   <jx_parse_stream@plt>
  mov    %rbp,%rdi
  mov    %rax,%r12
  call   <fclose@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
  nopl   (%rax)
L1:
  xor    %r12d,%r12d
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
```

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

tool: D2: r

assistant: ```c
struct jx *jx_parse_file(const char *name) {
  FILE *file = fopen(name, "r");
  if (!file)
    return 0;
  struct jx *j = jx_parse_stream(file);
  fclose(file);
  return j;
}
```
user: Translate the below assembly code into C programming language:

```asm
<eq>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  movq   $0x0,-0x8(%rbp)
  jmp    L14
L33:
  mov    -0x8(%rbp),%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x8(%rbp),%rax
  lea    0x0(,%rax,4),%rcx
  mov    -0x20(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%eax
  cmp    %eax,%edx
  je     L38
  mov    $0x0,%eax
  jmp    L21
L38:
  addq   $0x1,-0x8(%rbp)
L14:
  mov    -0x8(%rbp),%rax
  cmp    -0x28(%rbp),%rax
  jb     L33
  mov    $0x1,%eax
L21:
  pop    %rbp
  ret
```

assistant: ```c
int eq(int arr1[], int arr2[], size_t sz) {
  for (size_t i = 0; i < sz; i++)
    if (arr1[i] != arr2[i])
      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
<eq>:
  endbr64
  test   %rdx,%rdx
  je     L23
  mov    $0x0,%eax
L9:
  mov    (%rsi,%rax,4),%ecx
  cmp    %ecx,(%rdi,%rax,4)
  jne    L19
  add    $0x1,%rax
  cmp    %rax,%rdx
  jne    L9
  mov    $0x1,%eax
  ret
L23:
  mov    $0x1,%eax
  ret
L19:
  mov    $0x0,%eax
  ret
```

assistant: ```c
int eq(int arr1[], int arr2[], size_t sz) {
  for (size_t i = 0; i < sz; i++)
    if (arr1[i] != arr2[i])
      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
<eq>:
  endbr64
  test   %rdx,%rdx
  je     L11
  xor    %eax,%eax
  jmp    L3
  nopl   (%rax)
L9:
  add    $0x1,%rax
  cmp    %rax,%rdx
  je     L11
L3:
  mov    (%rsi,%rax,4),%ecx
  cmp    %ecx,(%rdi,%rax,4)
  je     L9
  xor    %eax,%eax
  ret
  nopl   0x0(%rax)
L11:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int eq(int arr1[], int arr2[], size_t sz) {
  for (size_t i = 0; i < sz; i++)
    if (arr1[i] != arr2[i])
      return 0;
  return 1;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<eq>:
  endbr64
  test   %rdx,%rdx
  je     L2
  xor    %eax,%eax
  jmp    L10
  nopl   (%rax)
L14:
  add    $0x1,%rax
  cmp    %rax,%rdx
  je     L2
L10:
  mov    (%rsi,%rax,4),%ecx
  cmp    %ecx,(%rdi,%rax,4)
  je     L14
  xor    %eax,%eax
  ret
  nopl   0x0(%rax)
L2:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int eq(int arr1[], int arr2[], size_t sz) {
  for (size_t i = 0; i < sz; i++)
    if (arr1[i] != arr2[i])
      return 0;
  return 1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<SubtractNumbers>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  movl   $0x0,-0x4(%rbp)
  mov    -0x14(%rbp),%eax
  sub    -0x18(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int SubtractNumbers(int number_1, int number_2) {
  int result = 0;
  result = number_1 - number_2;
  return result;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<SubtractNumbers>:
  endbr64
  mov    %edi,%eax
  sub    %esi,%eax
  ret
```

assistant: ```c
int SubtractNumbers(int number_1, int number_2) {
  int result = 0;
  result = number_1 - number_2;
  return result;
}
```
user: Translate the below assembly code into C programming language:

```asm
<SubtractNumbers>:
  endbr64
  mov    %edi,%eax
  sub    %esi,%eax
  ret
```

assistant: ```c
int SubtractNumbers(int number_1, int number_2) {
  int result = 0;
  result = number_1 - number_2;
  return result;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<SubtractNumbers>:
  endbr64
  mov    %edi,%eax
  sub    %esi,%eax
  ret
```

assistant: ```c
int SubtractNumbers(int number_1, int number_2) {
  int result = 0;
  result = number_1 - number_2;
  return result;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<execute_buffer>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L15
  lea    D24(%rip),%rax
  mov    %rax,%rdx
  mov    $0x5,%esi
  lea    D26(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <_do_log@plt>
  mov    $0x0,%eax
  jmp    L28
L15:
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <system@plt>
  test   %eax,%eax
  je     L34
  mov    -0x8(%rbp),%rax
  mov    %rax,%rcx
  lea    D18(%rip),%rax
  mov    %rax,%rdx
  mov    $0x6,%esi
  lea    D26(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <_do_log@plt>
  mov    $0x0,%eax
  jmp    L28
L34:
  mov    $0x1,%eax
L28:
  leave
  ret
```

tool_calls: name: read_data
arguments: data_tag: D24
datatype: string
name: read_data
arguments: data_tag: D26
datatype: string
name: read_data
arguments: data_tag: D18
datatype: string

tool: D24: no buffer

tool: D26: /scratch/repos/new/home/jordi_armengol_estape/c-scraper/outputs/2022-01-22/02-19-57/repos/itmm/waelzer/refs/heads/master/4.2-copy-files/execute_buffer.c

tool: D18: can't execute '%s'

assistant: ```c
bool execute_buffer(char *buffer) {
  do {
    if (!(buffer)) {
      _do_log("/scratch/repos/new/home/jordi_armengol_estape/c-scraper/outputs/2022-01-22/02-19-57/repos/itmm/waelzer/refs/heads/master/4.2-copy-files/execute_buffer.c", 5,
              ""
              "no buffer");
      return false;
    }
  } while (false);
  do {
    if (!(!system(buffer))) {
      _do_log("/scratch/repos/new/home/jordi_armengol_estape/c-scraper/outputs/2022-01-22/02-19-57/repos/itmm/waelzer/refs/heads/master/4.2-copy-files/execute_buffer.c", 6,
              ""
              "can't execute '%s'",
              buffer);
      return false;
    }
  } while (false);
  return true;
}
```