user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    D37(%rip),%rdi
  call   <printf@plt>
  lea    0x14(%rsp),%rsi
  lea    D36(%rip),%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    0x14(%rsp),%eax
  lea    0x0(,%rax,8),%edi
  call   <malloc@plt>
  mov    %rax,%r15
  mov    0x14(%rsp),%eax
  test   %eax,%eax
  jle    L42
  mov    $0x0,%ebx
L44:
  lea    0x4(,%rax,4),%edi
  call   <malloc@plt>
  mov    %rax,(%r15,%rbx,8)
  mov    0x14(%rsp),%eax
  add    $0x1,%rbx
  cmp    %ebx,%eax
  jg     L44
  test   %eax,%eax
  jle    L42
  mov    %r15,%rbp
  mov    %r15,%r12
  mov    $0x0,%r14d
  lea    D35(%rip),%r13
L23:
  add    $0x1,%r14d
  mov    %r14d,%esi
  lea    D43(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  cmpl   $0x0,0x14(%rsp)
  js     L139
  mov    $0x0,%ebx
L81:
  mov    (%r12),%rax
  lea    (%rax,%rbx,4),%rsi
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  add    $0x1,%rbx
  cmp    %ebx,0x14(%rsp)
  jge    L81
L139:
  mov    0x14(%rsp),%eax
  add    $0x8,%r12
  cmp    %eax,%r14d
  jl     L23
  test   %eax,%eax
  jle    L42
  mov    $0x0,%r12d
  jmp    L89
L106:
  mov    0x0(%rbp),%rax
  movss  (%rax,%rbx,4),%xmm0
  call   <fabs@plt>
  addss  0xc(%rsp),%xmm0
  movss  %xmm0,0xc(%rsp)
  add    $0x1,%rbx
  cmp    %ebx,0x14(%rsp)
  jg     L106
L74:
  mov    0x0(%rbp),%rax
  movss  (%rax,%r12,4),%xmm0
  call   <fabs@plt>
  movss  0xc(%rsp),%xmm2
  subss  %xmm0,%xmm2
  movd   %xmm2,%ebx
  mov    0x0(%rbp),%rax
  movss  (%rax,%r12,4),%xmm0
  call   <fabs@plt>
  movd   %ebx,%xmm3
  comiss %xmm0,%xmm3
  jae    L105
  add    $0x1,%r12
  add    $0x8,%rbp
  cmp    %r12d,0x14(%rsp)
  jle    L42
L89:
  mov    $0x0,%ebx
  pxor   %xmm4,%xmm4
  movss  %xmm4,0xc(%rsp)
  cmpl   $0x0,0x14(%rsp)
  jg     L106
  jmp    L74
L105:
  lea    D32(%rip),%rdi
  call   <puts@plt>
L80:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L38
  mov    $0x0,%eax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L42:
  mov    0x14(%rsp),%esi
  mov    %r15,%rdi
  call   <solve@plt>
  jmp    L80
L38:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D37
type: string
name: parse_data
arguments: data_label: D36
type: string
name: parse_data
arguments: data_label: D35
type: string
name: parse_data
arguments: data_label: D43
type: string
name: parse_data
arguments: data_label: D32
type: string

tool: D37: Enter no. of variables to be solved: 

tool: D36: %d

tool: D35: %f

tool: D43: Enter the co-efficient of variables and constant term of eqn. %d one by one: 


tool: D32: The method will not converge

assistant: ```c
int main(void) {
  float **arr;
  int i, j, n, flag = 0;
  float S;
  printf("Enter no. of variables to be solved: ");
  scanf("%d", &n);
  arr = (float **)malloc(n * sizeof(float *));
  for (i = 0; i < n; i++)
    arr[i] = malloc((n + 1) * sizeof(float));
  for (i = 0; i < n; i++) {
    printf("Enter the co-efficient of variables and constant term of eqn. %d one by one: \n", i + 1);
    for (j = 0; j < (n + 1); j++)
      scanf("%f", &arr[i][j]);
  }
  for (i = 0; i < n; i++) {
    S = 0;
    for (j = 0; j < n; j++)
      S += fabs(arr[i][j]);
    S -= fabs(arr[i][i]);
    if (fabs(arr[i][i]) <= S) {
      flag = 1;
      break;
    }
  }
  if (flag)
    printf("The method will not converge\n");
  else
    solve(arr, n);
  return 0;
}
```