C - 如果有超过 1 个或

Question

I've written a function to check if a string array is a name (every char can only be a->z or A->Z or space) However if I run my program it doesn't work properly. If I try to debug it, I see that it jumps to the line return 0; even if the first char in the array is 'M'. Can anyone help me?

int je_meno(char array[]) {
    int i=0;
    while(array[i]!=0){
        if(array[i]<32 || 32<array[i]<65 || 90<array[i]<97 || 122<array[i]){
            return 0;
        }
        i++;
    }
    return 1; }

Answer 1

Can it even compile the chain comparisons? Try simplifing the condition to something like:

if( !(array[i]>=65 && array[i] <=90 || array[i]>=97 && array[i]<=122 || array[i]==32) )
{
   return 0;
}

Basicaly if array[i] is NOT(capital character, small character or space) return 0

Answer 2

You could use functions isalpha() and isdigit() like this:

return !(isalpha(c) || isdigit(c) || (c == ' '));

Third expression is if you want to include spaces in your name.

So, line with if statement should be like this:

if !(isalpha(array[i]) || isdigit(array[i]) || (array[i] == ' ')) {

I think you'll need to include <ctype.h>.