在C ++中使用itoa（）函数将整数转换为字符串的时间复杂度？

Question

I want to analyze the complexity of my code and when i saw itoa() function call then i thought is it constant time or is it converted to string by doing % and / operators.(i want to convert integer to its binary string).

  int i;
  char buffer [33];
  printf("Enter a number: ");
  scanf ("%d",&i);
  itoa (i,buffer,10);  //1st
  printf ("decimal: %s\n",buffer);
  itoa (i,buffer,16);  //2nd
  printf ("hexadecimal: %s\n",buffer);
  itoa (i,buffer,2);  //3rd
  printf ("binary: %s\n",buffer);

Answer 1

The source code can be found here:

https://en.wikibooks.org/wiki/C_Programming/stdlib.h/itoa

It seems like it's doing the standard mod div approach. Also, one of the external links provides some relative performance benchmarks at the bottom of the page:

http://www.strudel.org.uk/itoa/