Equivalent of atoi for unsigned integers

Question

I'm doing two operations involving atoi and I'm wondering how I can do this with unsigned integers because atoi seems to convert these to signed causing a wraparound integer overflow. I want to work with 32bit unsigned integers but atoi is limiting me effectively to 31bit unsigned.

 if (multiplication_is_safe(atoi(argv[1]),atoi(argv[3])))
    {
     printf("%s * %s = %u \n", argv[1], argv[3], atoi(argv[1]) * atoi(argv[3]));
      return 0;
    }  else

Answer 1

The simple answer is to use strtoul() instead.

The longer answer is that even if all you needed was signed 32 bit integers or were happy with 31 bits for unsigned, the atoi() function is a poor fit for what you appear to be doing.

As you have already noted, the atoi() function converts a string to an integer. A normal, signed integer. However, what atoi() doesn't do is error handling. What atoi()'s specification says is "If the value cannot be represented, the behavior is undefined."

The strto*() family of functions all clearly specify how errors are handled, so you should in all cases replace atoi() with calls to strtol() (convert string to long), and in this case since you want to handle unsigned integers, you should use strtoul() (convert string to unsigned long).

Also note that if you want to handle larger numbers, there are the strtoll() and strtoull() functions, to convert your string to a long long or an unsigned long long. (And if you just want to handle the largest possible integral values without bothering with all that stuff in between, there's strtoimax() and strtoumax(), that return values of type intmax_t or uintmax_t respectively.)

POSIX Documentation:

• atoi()
• strtol()
• strtoll()
• strtoul()
• strtoull()
• strtoimax()
• strtoumax()