ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<strcpy>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L18
L14:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    -0x4(%rbp),%edx
  movslq %edx,%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  addl   $0x1,-0x4(%rbp)
L18:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L14
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void strcpy(char *dest, const char *src) {
  int i;
  for (i = 0; src[i] != 0; ++i) {
    dest[i] = src[i];
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<strcpy>:
  endbr64
  movzbl (%rsi),%edx
  test   %dl,%dl
  je     L11
  mov    $0x0,%eax
L5:
  mov    %dl,(%rdi,%rax,1)
  add    $0x1,%rax
  movzbl (%rsi,%rax,1),%edx
  test   %dl,%dl
  jne    L5
L11:
  ret
```

assistant: ```c
void strcpy(char *dest, const char *src) {
  int i;
  for (i = 0; src[i] != 0; ++i) {
    dest[i] = src[i];
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<strcpy>:
  endbr64
  movzbl (%rsi),%edx
  test   %dl,%dl
  je     L19
  xor    %eax,%eax
  nopl   (%rax)
L6:
  mov    %dl,(%rdi,%rax,1)
  add    $0x1,%rax
  movzbl (%rsi,%rax,1),%edx
  test   %dl,%dl
  jne    L6
L19:
  ret
```

assistant: ```c
void strcpy(char *dest, const char *src) {
  int i;
  for (i = 0; src[i] != 0; ++i) {
    dest[i] = src[i];
  }
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<strcpy>:
  endbr64
  movzbl (%rsi),%edx
  test   %dl,%dl
  je     L0
  xor    %eax,%eax
  nopl   (%rax)
L11:
  mov    %dl,(%rdi,%rax,1)
  add    $0x1,%rax
  movzbl (%rsi,%rax,1),%edx
  test   %dl,%dl
  jne    L11
L0:
  ret
```

assistant: ```c
void strcpy(char *dest, const char *src) {
  int i;
  for (i = 0; src[i] != 0; ++i) {
    dest[i] = src[i];
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<cmatrix_mrotz>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x90,%rsp
  mov    %rdi,-0x78(%rbp)
  mov    %rsi,-0x80(%rbp)
  movsd  %xmm0,-0x88(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x80(%rbp),%rax
  cmp    -0x78(%rbp),%rax
  jne    L21
  mov    -0x80(%rbp),%rcx
  lea    -0x50(%rbp),%rax
  mov    $0x48,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  lea    -0x50(%rbp),%rax
  mov    %rax,-0x80(%rbp)
L21:
  mov    -0x88(%rbp),%rax
  movq   %rax,%xmm0
  call   <cos@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x60(%rbp)
  mov    -0x88(%rbp),%rax
  movq   %rax,%xmm0
  call   <sin@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x58(%rbp)
  movl   $0x1,-0x64(%rbp)
  jmp    L24
L16:
  mov    -0x64(%rbp),%eax
  sub    $0x1,%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x80(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  mulsd  -0x60(%rbp),%xmm0
  mov    -0x64(%rbp),%eax
  add    $0x2,%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x80(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm1
  mulsd  -0x58(%rbp),%xmm1
  mov    -0x64(%rbp),%eax
  sub    $0x1,%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x78(%rbp),%rax
  add    %rdx,%rax
  subsd  %xmm1,%xmm0
  movsd  %xmm0,(%rax)
  mov    -0x64(%rbp),%eax
  sub    $0x1,%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x80(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  movapd %xmm0,%xmm1
  mulsd  -0x58(%rbp),%xmm1
  mov    -0x64(%rbp),%eax
  add    $0x2,%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x80(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  mulsd  -0x60(%rbp),%xmm0
  mov    -0x64(%rbp),%eax
  add    $0x2,%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x78(%rbp),%rax
  add    %rdx,%rax
  addsd  %xmm1,%xmm0
  movsd  %xmm0,(%rax)
  mov    -0x64(%rbp),%eax
  add    $0x5,%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x80(%rbp),%rax
  add    %rax,%rdx
  mov    -0x64(%rbp),%eax
  add    $0x5,%eax
  cltq
  lea    0x0(,%rax,8),%rcx
  mov    -0x78(%rbp),%rax
  add    %rcx,%rax
  movsd  (%rdx),%xmm0
  movsd  %xmm0,(%rax)
  addl   $0x1,-0x64(%rbp)
L24:
  cmpl   $0x3,-0x64(%rbp)
  jle    L16
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L19
  call   <__stack_chk_fail@plt>
L19:
  leave
  ret
```

assistant: ```c
void cmatrix_mrotz(double *mdest, const double *morg, double angle) {
  double temp[3 * 3];
  if (morg == mdest) {
    memcpy(temp, morg, sizeof(temp));
    morg = temp;
  }
  double cosa = cos(angle);
  double sina = sin(angle);
  for (int k = 1; k <= 3; k++) {
    mdest[((((1) - 1) * 3) + ((k)-1))] = cosa * morg[((((1) - 1) * 3) + ((k)-1))] - sina * morg[((((2) - 1) * 3) + ((k)-1))];
    mdest[((((2) - 1) * 3) + ((k)-1))] = sina * morg[((((1) - 1) * 3) + ((k)-1))] + cosa * morg[((((2) - 1) * 3) + ((k)-1))];
    mdest[((((3) - 1) * 3) + ((k)-1))] = morg[((((3) - 1) * 3) + ((k)-1))];
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<cmatrix_mrotz>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x68,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  movq   %xmm0,%r12
  mov    %fs:0x28,%rax
  mov    %rax,0x58(%rsp)
  xor    %eax,%eax
  cmp    %rdi,%rsi
  je     L4
L18:
  lea    0x8(%rsp),%rdi
  mov    %rsp,%rsi
  movq   %r12,%xmm0
  call   <sincos@plt>
  movsd  (%rsp),%xmm2
  movsd  0x8(%rsp),%xmm1
  mov    %rbp,%rax
  mov    %rbx,%rdx
  add    $0x18,%rbp
L9:
  movapd %xmm2,%xmm0
  mulsd  (%rax),%xmm0
  movapd %xmm1,%xmm3
  mulsd  0x18(%rax),%xmm3
  subsd  %xmm3,%xmm0
  movsd  %xmm0,(%rdx)
  movapd %xmm1,%xmm0
  mulsd  (%rax),%xmm0
  movapd %xmm2,%xmm3
  mulsd  0x18(%rax),%xmm3
  addsd  %xmm3,%xmm0
  movsd  %xmm0,0x18(%rdx)
  movsd  0x30(%rax),%xmm0
  movsd  %xmm0,0x30(%rdx)
  add    $0x8,%rax
  add    $0x8,%rdx
  cmp    %rbp,%rax
  jne    L9
  mov    0x58(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L31
  add    $0x68,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L4:
  lea    0x10(%rsp),%r13
  mov    $0x48,%edx
  mov    %r13,%rdi
  call   <memcpy@plt>
  mov    %r13,%rbp
  jmp    L18
L31:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void cmatrix_mrotz(double *mdest, const double *morg, double angle) {
  double temp[3 * 3];
  if (morg == mdest) {
    memcpy(temp, morg, sizeof(temp));
    morg = temp;
  }
  double cosa = cos(angle);
  double sina = sin(angle);
  for (int k = 1; k <= 3; k++) {
    mdest[((((1) - 1) * 3) + ((k)-1))] = cosa * morg[((((1) - 1) * 3) + ((k)-1))] - sina * morg[((((2) - 1) * 3) + ((k)-1))];
    mdest[((((2) - 1) * 3) + ((k)-1))] = sina * morg[((((1) - 1) * 3) + ((k)-1))] + cosa * morg[((((2) - 1) * 3) + ((k)-1))];
    mdest[((((3) - 1) * 3) + ((k)-1))] = morg[((((3) - 1) * 3) + ((k)-1))];
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<cmatrix_mrotz>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x70,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  cmp    %rdi,%rsi
  je     L11
L22:
  lea    0x10(%rsp),%rsi
  lea    0x18(%rsp),%rdi
  call   <sincos@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rax
  lea    0x18(%rbp),%rdx
  movsd  0x10(%rsp),%xmm2
  movsd  0x18(%rsp),%xmm1
L3:
  movsd  (%rsi),%xmm0
  movsd  0x18(%rsi),%xmm3
  add    $0x8,%rsi
  add    $0x8,%rax
  mulsd  %xmm1,%xmm3
  mulsd  %xmm2,%xmm0
  subsd  %xmm3,%xmm0
  movsd  %xmm0,-0x8(%rax)
  movsd  -0x8(%rsi),%xmm0
  movsd  0x10(%rsi),%xmm3
  mulsd  %xmm1,%xmm0
  mulsd  %xmm2,%xmm3
  addsd  %xmm3,%xmm0
  movsd  %xmm0,0x10(%rax)
  movsd  0x28(%rsi),%xmm0
  movsd  %xmm0,0x28(%rax)
  cmp    %rsi,%rdx
  jne    L3
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x70,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   (%rax)
L11:
  lea    0x20(%rsp),%r12
  mov    $0x48,%edx
  movsd  %xmm0,0x8(%rsp)
  mov    %r12,%rdi
  mov    %r12,%rbp
  call   <memcpy@plt>
  movsd  0x8(%rsp),%xmm0
  jmp    L22
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void cmatrix_mrotz(double *mdest, const double *morg, double angle) {
  double temp[3 * 3];
  if (morg == mdest) {
    memcpy(temp, morg, sizeof(temp));
    morg = temp;
  }
  double cosa = cos(angle);
  double sina = sin(angle);
  for (int k = 1; k <= 3; k++) {
    mdest[((((1) - 1) * 3) + ((k)-1))] = cosa * morg[((((1) - 1) * 3) + ((k)-1))] - sina * morg[((((2) - 1) * 3) + ((k)-1))];
    mdest[((((2) - 1) * 3) + ((k)-1))] = sina * morg[((((1) - 1) * 3) + ((k)-1))] + cosa * morg[((((2) - 1) * 3) + ((k)-1))];
    mdest[((((3) - 1) * 3) + ((k)-1))] = morg[((((3) - 1) * 3) + ((k)-1))];
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<indexOf>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,%eax
  mov    %edx,-0x20(%rbp)
  mov    %al,-0x1c(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <len@plt>
  cmp    %eax,-0x20(%rbp)
  jle    L35
  mov    $0xffffffff,%eax
  jmp    L14
L35:
  mov    -0x20(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  jmp    L28
L3:
  mov    -0x4(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x4(%rbp)
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    %al,-0x1c(%rbp)
  jne    L28
  mov    -0x4(%rbp),%eax
  sub    $0x1,%eax
  jmp    L14
L28:
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <len@plt>
  cmp    %eax,-0x4(%rbp)
  jl     L3
  mov    $0xffffffff,%eax
L14:
  leave
  ret
```

assistant: ```c
int indexOf(char *data, char c, int startPosition) {
  if (len(data) < startPosition)
    return -1;
  int position = startPosition;
  while (position < len(data)) {
    if (data[position++] == c) {
      return position - 1;
    }
  }
  return -1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<indexOf>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    %edx,%ebx
  mov    %esi,%r12d
  call   <len@plt>
  cmp    %ebx,%eax
  jl     L46
  movslq %ebx,%rbx
L31:
  mov    %ebx,%r13d
  mov    %rbp,%rdi
  call   <len@plt>
  cmp    %ebx,%eax
  jle    L2
  add    $0x1,%rbx
  cmp    %r12b,-0x1(%rbp,%rbx,1)
  jne    L31
  jmp    L0
L2:
  mov    $0xffffffff,%r13d
L0:
  mov    %r13d,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L46:
  mov    $0xffffffff,%r13d
  jmp    L0
```

assistant: ```c
int indexOf(char *data, char c, int startPosition) {
  if (len(data) < startPosition)
    return -1;
  int position = startPosition;
  while (position < len(data)) {
    if (data[position++] == c) {
      return position - 1;
    }
  }
  return -1;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<indexOf>:
  endbr64
  push   %r13
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  movslq %edx,%rbx
  sub    $0x8,%rsp
  call   <len@plt>
  cmp    %ebx,%eax
  jge    L1
  jmp    L24
  nopw   0x0(%rax,%rax,1)
L11:
  add    $0x1,%rbx
  cmp    %r12b,-0x1(%rbp,%rbx,1)
  je     L34
L1:
  mov    %rbp,%rdi
  mov    %ebx,%r13d
  call   <len@plt>
  cmp    %ebx,%eax
  jg     L11
L24:
  mov    $0xffffffff,%r13d
L34:
  add    $0x8,%rsp
  mov    %r13d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int indexOf(char *data, char c, int startPosition) {
  if (len(data) < startPosition)
    return -1;
  int position = startPosition;
  while (position < len(data)) {
    if (data[position++] == c) {
      return position - 1;
    }
  }
  return -1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<indexOf>:
  endbr64
  push   %r13
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  movslq %edx,%rbx
  sub    $0x8,%rsp
  call   <len@plt>
  cmp    %ebx,%eax
  jge    L35
  jmp    L15
  nopw   0x0(%rax,%rax,1)
L20:
  add    $0x1,%rbx
  cmp    %r12b,-0x1(%rbp,%rbx,1)
  je     L10
L35:
  mov    %rbp,%rdi
  mov    %ebx,%r13d
  call   <len@plt>
  cmp    %ebx,%eax
  jg     L20
L15:
  mov    $0xffffffff,%r13d
L10:
  add    $0x8,%rsp
  mov    %r13d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int indexOf(char *data, char c, int startPosition) {
  if (len(data) < startPosition)
    return -1;
  int position = startPosition;
  while (position < len(data)) {
    if (data[position++] == c) {
      return position - 1;
    }
  }
  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
<inc_it>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    -0x20(%rbp),%rax
  mov    (%rax),%eax
  lea    0x1(%rax),%edx
  mov    -0x20(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x20(%rbp),%rax
  add    $0x4,%rax
  mov    (%rax),%eax
  mov    %eax,%ebx
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  add    %ebx,%eax
  lea    0x1(%rax),%edx
  mov    -0x20(%rbp),%rax
  add    $0x4,%rax
  mov    %edx,(%rax)
  nop
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
void inc_it(int *a, int *b, unsigned char *key) {
  b[0]++;
  b[1] += strlen((const char *)key) + 1;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<inc_it>:
  endbr64
  push   %rbx
  mov    %rsi,%rbx
  mov    %rdx,%rdi
  addl   $0x1,(%rsi)
  call   <strlen@plt>
  mov    0x4(%rbx),%edx
  lea    0x1(%rdx,%rax,1),%eax
  mov    %eax,0x4(%rbx)
  pop    %rbx
  ret
```

assistant: ```c
void inc_it(int *a, int *b, unsigned char *key) {
  b[0]++;
  b[1] += strlen((const char *)key) + 1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<inc_it>:
  endbr64
  addl   $0x1,(%rsi)
  mov    %rdx,%rdi
  push   %rbx
  mov    %rsi,%rbx
  call   <strlen@plt>
  mov    0x4(%rbx),%edx
  lea    0x1(%rdx,%rax,1),%eax
  mov    %eax,0x4(%rbx)
  pop    %rbx
  ret
```

assistant: ```c
void inc_it(int *a, int *b, unsigned char *key) {
  b[0]++;
  b[1] += strlen((const char *)key) + 1;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<inc_it>:
  endbr64
  addl   $0x1,(%rsi)
  mov    %rdx,%rdi
  push   %rbx
  mov    %rsi,%rbx
  call   <strlen@plt>
  mov    0x4(%rbx),%edx
  lea    0x1(%rdx,%rax,1),%eax
  mov    %eax,0x4(%rbx)
  pop    %rbx
  ret
```

assistant: ```c
void inc_it(int *a, int *b, unsigned char *key) {
  b[0]++;
  b[1] += strlen((const char *)key) + 1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x0,%eax
  call   <__VERIFIER_nondet_int@plt>
  mov    %eax,-0xc(%rbp)
  mov    $0x0,%eax
  call   <__VERIFIER_nondet_int@plt>
  mov    %eax,-0x8(%rbp)
  jmp    L19
L6:
  mov    $0x0,%eax
  call   <__VERIFIER_nondet_int@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0xc(%rbp),%eax
  sub    $0x1,%eax
  sub    -0x4(%rbp),%eax
  mov    %eax,-0x8(%rbp)
  mov    -0xc(%rbp),%eax
  sub    $0x1,%eax
  sub    -0x4(%rbp),%eax
  mov    %eax,-0xc(%rbp)
L19:
  cmpl   $0x0,-0x8(%rbp)
  jg     L6
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main() {
  int a;
  int b;
  int r;
  a = __VERIFIER_nondet_int();
  b = __VERIFIER_nondet_int();
  while (b > 0) {
    r = __VERIFIER_nondet_int();
    b = a - 1 - r;
    a = a - 1 - r;
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbx
  mov    $0x0,%eax
  call   <__VERIFIER_nondet_int@plt>
  mov    %eax,%ebx
  mov    $0x0,%eax
  call   <__VERIFIER_nondet_int@plt>
  test   %eax,%eax
  jle    L5
L9:
  mov    $0x0,%eax
  call   <__VERIFIER_nondet_int@plt>
  sub    $0x1,%ebx
  sub    %eax,%ebx
  test   %ebx,%ebx
  jg     L9
L5:
  mov    $0x0,%eax
  pop    %rbx
  ret
```

assistant: ```c
int main() {
  int a;
  int b;
  int r;
  a = __VERIFIER_nondet_int();
  b = __VERIFIER_nondet_int();
  while (b > 0) {
    r = __VERIFIER_nondet_int();
    b = a - 1 - r;
    a = a - 1 - r;
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %rbx
  xor    %eax,%eax
  call   <__VERIFIER_nondet_int@plt>
  mov    %eax,%ebx
  xor    %eax,%eax
  call   <__VERIFIER_nondet_int@plt>
  test   %eax,%eax
  jle    L0
  nopl   0x0(%rax)
L19:
  xor    %eax,%eax
  sub    $0x1,%ebx
  call   <__VERIFIER_nondet_int@plt>
  sub    %eax,%ebx
  test   %ebx,%ebx
  jg     L19
L0:
  xor    %eax,%eax
  pop    %rbx
  ret
```

assistant: ```c
int main() {
  int a;
  int b;
  int r;
  a = __VERIFIER_nondet_int();
  b = __VERIFIER_nondet_int();
  while (b > 0) {
    r = __VERIFIER_nondet_int();
    b = a - 1 - r;
    a = a - 1 - r;
  }
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  push   %rbx
  xor    %eax,%eax
  call   <__VERIFIER_nondet_int@plt>
  mov    %eax,%ebx
  xor    %eax,%eax
  call   <__VERIFIER_nondet_int@plt>
  test   %eax,%eax
  jle    L12
  nopl   0x0(%rax)
L18:
  xor    %eax,%eax
  sub    $0x1,%ebx
  call   <__VERIFIER_nondet_int@plt>
  sub    %eax,%ebx
  test   %ebx,%ebx
  jg     L18
L12:
  xor    %eax,%eax
  pop    %rbx
  ret
```

assistant: ```c
int main() {
  int a;
  int b;
  int r;
  a = __VERIFIER_nondet_int();
  b = __VERIFIER_nondet_int();
  while (b > 0) {
    r = __VERIFIER_nondet_int();
    b = a - 1 - r;
    a = a - 1 - r;
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x3,-0x8(%rbp)
  movl   $0x5,-0x4(%rbp)
  mov    -0x8(%rbp),%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  mov    %eax,%esi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x4(%rbp),%edx
  mov    -0x8(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <add@plt>
  mov    %eax,%esi
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D7
type_infered: string
name: retrieve_data
arguments: data_tag: D10
type_infered: string

tool: D7: i1 + i2 = %i


tool: D10: add(i1, i2) = %i


assistant: ```c
int main() {
  int i1 = 3, i2 = 5;
  printf("i1 + i2 = %i\n", i1 + i2);
  printf("add(i1, i2) = %i\n", add(i1, i2));
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x8,%esi
  lea    D17(%rip),%rdi
  call   <printf@plt>
  mov    $0x5,%esi
  mov    $0x3,%edi
  call   <add@plt>
  mov    %eax,%esi
  lea    D19(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: label: D17
data_type: string
name: unpack_data
arguments: label: D19
data_type: string

tool: D17: i1 + i2 = %i


tool: D19: add(i1, i2) = %i


assistant: ```c
int main() {
  int i1 = 3, i2 = 5;
  printf("i1 + i2 = %i\n", i1 + i2);
  printf("add(i1, i2) = %i\n", add(i1, i2));
  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
  sub    $0x8,%rsp
  mov    $0x8,%esi
  lea    D11(%rip),%rdi
  call   <printf@plt>
  mov    $0x5,%esi
  mov    $0x3,%edi
  call   <add@plt>
  lea    D9(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: tag: D11
type: string
name: read_data
arguments: tag: D9
type: string

tool: D11: i1 + i2 = %i


tool: D9: add(i1, i2) = %i


assistant: ```c
int main() {
  int i1 = 3, i2 = 5;
  printf("i1 + i2 = %i\n", i1 + i2);
  printf("add(i1, i2) = %i\n", add(i1, i2));
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x8,%esi
  lea    D6(%rip),%rdi
  call   <printf@plt>
  mov    $0x5,%esi
  mov    $0x3,%edi
  call   <add@plt>
  lea    D5(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D6
type_infered: string
name: disassemble_data
arguments: data_label: D5
type_infered: string

tool: D6: i1 + i2 = %i


tool: D5: add(i1, i2) = %i


assistant: ```c
int main() {
  int i1 = 3, i2 = 5;
  printf("i1 + i2 = %i\n", i1 + i2);
  printf("add(i1, i2) = %i\n", add(i1, i2));
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<compute_index>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    %edx,-0x1c(%rbp)
  mov    %rcx,-0x28(%rbp)
  mov    -0x28(%rbp),%rax
  mov    (%rax),%eax
  add    $0x2,%eax
  imul   -0x1c(%rbp),%eax
  mov    -0x28(%rbp),%rdx
  add    $0x4,%rdx
  mov    (%rdx),%edx
  add    $0x2,%edx
  imul   %eax,%edx
  mov    -0x28(%rbp),%rax
  mov    (%rax),%eax
  add    $0x2,%eax
  imul   -0x18(%rbp),%eax
  add    %eax,%edx
  mov    -0x14(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int compute_index(int x, int y, int z, int *subdomain) {
  int idx = z * (subdomain[0] + 2) * (subdomain[1] + 2) + y * (subdomain[0] + 2) + x;
  return idx;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<compute_index>:
  endbr64
  mov    %esi,%r8d
  mov    (%rcx),%eax
  lea    0x2(%rax),%esi
  mov    0x4(%rcx),%eax
  add    $0x2,%eax
  imul   %esi,%edx
  imul   %edx,%eax
  imul   %r8d,%esi
  add    %esi,%eax
  add    %edi,%eax
  ret
```

assistant: ```c
int compute_index(int x, int y, int z, int *subdomain) {
  int idx = z * (subdomain[0] + 2) * (subdomain[1] + 2) + y * (subdomain[0] + 2) + x;
  return idx;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<compute_index>:
  endbr64
  mov    (%rcx),%eax
  mov    %esi,%r8d
  lea    0x2(%rax),%esi
  mov    0x4(%rcx),%eax
  imul   %esi,%edx
  add    $0x2,%eax
  imul   %r8d,%esi
  imul   %edx,%eax
  add    %esi,%eax
  add    %edi,%eax
  ret
```

assistant: ```c
int compute_index(int x, int y, int z, int *subdomain) {
  int idx = z * (subdomain[0] + 2) * (subdomain[1] + 2) + y * (subdomain[0] + 2) + x;
  return idx;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<compute_index>:
  endbr64
  mov    (%rcx),%eax
  mov    %esi,%r8d
  lea    0x2(%rax),%esi
  mov    0x4(%rcx),%eax
  imul   %esi,%edx
  add    $0x2,%eax
  imul   %r8d,%esi
  imul   %edx,%eax
  add    %esi,%eax
  add    %edi,%eax
  ret
```

assistant: ```c
int compute_index(int x, int y, int z, int *subdomain) {
  int idx = z * (subdomain[0] + 2) * (subdomain[1] + 2) + y * (subdomain[0] + 2) + x;
  return idx;
}
```