How to check to ensure you have an integer before

问题:

I wish to take an integer as a command line argument, but if the user passes a non-integer string, this will cause a stack overflow. What is the standard way to ensure atoi() will be successful?

回答1:

You can use:

long int strtol(const char *nptr, char **endptr, int base);

Then check if *endptr != nptr. This means that the string at least begins with the integer. You can also check that *endptr points to terminating zero, which means that the whole string was successfully parsed.

回答2:

atoi() will (should) not cause a stack overflow if the string contains characters other than digits. It will simply convert to an int any digit it finds from the beginning of the string until there is no more.

  int x = atoi("12monkeys"); // x is 12
  int y = atoi("monkeys12"); // y is 0

You may check that there is no integer overflow (number outside the range of [-2^31, 2^31-1] on a modern (current) PC architecture).

edit (comments)

While the C standards warns about an undefined behavior if the value cannot be represented, the most common C recent compilers (gcc, MS...) do not crash if the value is not acceptable (unless the char * pointer is null or wrong of course).

Anyway, you can implement your own atoi() easily (with the same limitations as in my answer)

    #include <ctype.h>

    int myatoi(char *s) {
       int res = 0, minus = *s == '-';
       if (minus) s++;

       while (isdigit(*s)) {
          res = res*10 + (*s++ - '0');
       }

       return minus ? -res : res;
    }

回答3:

This does not cause stack overflow. atoi returns 0 if it can't find a number at the start of the string. Your (non-)handling of the 0 is what causes the stack overflow.

回答4:

Cause a stack overflow? Well I suppose that's one possible result of the undefined behavior if the value in the string overflows the range of int. In practice though it usually just wraps or returns a bogus result.

If you want better error checking, use strtol instead of atoi. It has well-defined behavior on overflow (it sets errno, which you need to clear to 0 before calling strtol so you can distinguish between error returns and legitimate values being returned) and you can examine the point in the string at which it stopped conversion to see if the full string was an integer or whether there's additional content past the end.

回答5:

I don't think a standard atoi will stackoverflow, but there's no way to tell if you don't have an integer with it. Use strtol instead -- it's possible to deal with non-integers.

http://en.wikipedia.org/wiki/Strtol

回答6:

Either you can do this and enter Undefined Behavior land, you could write a simple validation function like so:

/* returns 0 on success, 1 on failure. */

int verify(char * string)
{
    int x = 0;
    int len = strlen(string);

    while(x < len) {
           if(!isdigit(*(string+x)))
           return 1;

           ++x;
    }

    return 0;
}

Please note you do have to call this function before you call atoi(), and you need string.h and stdio.h.

回答7:

atoi() is converting the string to an integer if it contains only digit chars, otherwise it will return 0.

回答8:

This might help you. Check strtol available in stdlib.h