user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %r13
  mov    $0x8000000,%edi
  push   %r12
  push   %rbp
  call   <malloc@plt>
  mov    $0x8000000,%edi
  mov    %rax,%rbp
  call   <malloc@plt>
  mov    $0x8000000,%edi
  mov    %rax,%r12
  call   <malloc@plt>
  test   %rbp,%rbp
  mov    %rax,%r13
  sete   %al
  test   %r12,%r12
  sete   %dl
  or     %dl,%al
  jne    L79
  test   %r13,%r13
  je     L79
  mov    D22(%rip),%rax
  movdqa D22(%rip),%xmm1
  lea    0x8000000(%rbp),%rcx
  movdqa D9(%rip),%xmm2
  mov    %rax,0x0(%rbp)
  mov    %rbp,%rax
  nopw   0x0(%rax,%rax,1)
L96:
  movdqa %xmm1,%xmm0
  add    $0x10,%rax
  paddd  %xmm2,%xmm1
  pslld  $0x1,%xmm0
  movups %xmm0,-0x10(%rax)
  cmp    %rcx,%rax
  jne    L96
  movdqa D5(%rip),%xmm0
  lea    0x8000000(%r12),%rdx
  mov    %r12,%rax
  nopl   (%rax)
L68:
  movups %xmm0,(%rax)
  add    $0x10,%rax
  cmp    %rdx,%rax
  jne    L68
  lea    0x4(%r12),%rdx
  mov    %r13,%rax
  sub    %rdx,%rax
  cmp    $0x8,%rax
  jbe    L136
  lea    0x4(%rbp),%rdx
  mov    %r13,%rax
  sub    %rdx,%rax
  cmp    $0x8,%rax
  jbe    L136
  mov    %r12,%rsi
  mov    %r13,%rdx
  mov    %rbp,%rax
  sub    %rbp,%rsi
  sub    %rbp,%rdx
  nopl   0x0(%rax)
L56:
  movdqu (%rsi,%rax,1),%xmm0
  movdqu (%rax),%xmm3
  paddd  %xmm3,%xmm0
  movups %xmm0,(%rdx,%rax,1)
  add    $0x10,%rax
  cmp    %rcx,%rax
  jne    L56
L133:
  mov    %r13,%rdx
  mov    $0x1,%eax
  jmp    L90
  nopl   0x0(%rax)
L50:
  add    $0x2,%eax
  add    $0x4,%rdx
  cmp    $0x4000001,%eax
  je     L123
L90:
  cmp    %eax,(%rdx)
  je     L50
  mov    $0x1,%eax
L1:
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L79:
  lea    D32(%rip),%rdi
  call   <printf@plt>
  mov    $0x1,%eax
  jmp    L1
L136:
  mov    %r13,%rdi
  mov    %r12,%rsi
  mov    %rbp,%rax
  sub    %rbp,%rdi
  sub    %rbp,%rsi
  nop
L118:
  mov    (%rax),%edx
  add    (%rsi,%rax,1),%edx
  mov    %edx,(%rdi,%rax,1)
  add    $0x4,%rax
  cmp    %rcx,%rax
  jne    L118
  jmp    L133
L123:
  mov    %rbp,%rdi
  call   <free@plt>
  mov    %r12,%rdi
  call   <free@plt>
  mov    %r13,%rdi
  call   <free@plt>
  lea    D43(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  jmp    L1
```

tool_calls: name: extract_symbolic_data
arguments: label: D22
type_guessed: byte[16]
name: extract_symbolic_data
arguments: label: D9
type_guessed: byte[16]
name: extract_symbolic_data
arguments: label: D5
type_guessed: byte[16]
name: extract_symbolic_data
arguments: label: D32
type_guessed: string
name: extract_symbolic_data
arguments: label: D43
type_guessed: string

tool: {"D22": [0x00,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x02,0x00,0x00,0x00,0x03,0x00,0x00,0x00]}

tool: {"D9": [0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00]}

tool: {"D5": [0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00]}

tool: D32: malloc err


tool: D43: PASS


assistant: ```c
int main() {
  int i, j;
  int *a = (int *)malloc(sizeof(int) * (32 * 1024 * 1024));
  int *b = (int *)malloc(sizeof(int) * (32 * 1024 * 1024));
  int *c = (int *)malloc(sizeof(int) * (32 * 1024 * 1024));
  if (!(a && b && c)) {
    printf("malloc err\n");
    return 1;
  }
#pragma acc data copyout(a [0:N], b [0:N], c [0:N])
  {
#pragma acc kernels async(2)
    a[0] = 0;
#pragma acc kernels async(10)
    a[1] = 1;
#pragma acc wait
#pragma acc kernels loop async(2) gang
    for (j = 0; j < 1; j++) {
#pragma acc loop vector
      for (i = 0; i < (32 * 1024 * 1024); i++) {
        a[i] = i * 2;
      }
    }
#pragma acc kernels loop async(10)
    for (i = 0; i < (32 * 1024 * 1024); i++) {
      b[i] = 1;
    }
#pragma acc wait(10)
#pragma acc kernels loop async(2)
    for (i = 0; i < (32 * 1024 * 1024); i++) {
      c[i] = a[i] + b[i];
    }
#pragma acc wait
  }
  for (i = 0; i < (32 * 1024 * 1024); i++) {
    if (c[i] != i * 2 + 1) {
      return 1;
    }
  }
  free(a);
  free(b);
  free(c);
  printf("PASS\n");
  return 0;
}
```