How can I make an array of intengers where each array position is a digit of my number?

问题

So what I'm thinking about doing is a little program where the user inputs a binary number (i.e 1010010100111000001), and then it is stored in an array, where each array position is associated with a digit of my number. Example: bit[0] = 1, bit[1] = 0 etc. The code I have in mind is this but i don't believe it works:

int bit [31];
scanf("%i", bit);

Help please!

回答1:

The main problem is how to ensure the user enters a valid binary number, i.e. exclusively 0s and 1s. You can try to read the inputs character by character and check the input is valid by using the standard sscanf format: %[01]. You should also limit the size of the input to a single character. Hence the final format: %1[01]

#include <stdio.h>
#include <stdlib.h>

int main(int argc, char* argv[]) {
    int bits[31];
    char input[2];
    int i;
    int j;

   fprintf(stdout, "Enter at most 31 bits: ");
   for(i=0; i<31 && 1 == fscanf(stdin, "%1[01]", input); ++i)
        bits[i] = (input[0]=='1' ? 1 : 0);

   fprintf(stdout, "You entered: \n");
   for(j=0; j<i; ++j)
       fputc(bits[j]?'1':'0', stdout);
   fprintf(stdout, "\n");
   return 0;
}

回答2:

Without the error check, the code you need is

char bitstream[31];
scanf("%30s", bitstream);

回答3:

#include <stdio.h>

int main(void) {
    char line[64];
    int bit[32];
    int i, n;

    scanf("%32[01]", line);//fgets(line, sizeof(line), stdin);//input up to newline
    for(i = 0; i < 32; ++i){
        if(1!=sscanf(&line[i], "%1d", &bit[i]))//read one digit
            break;
    }
    n = i;
    for(i = 0; i < n; ++i){
        printf("%d", bit[i]);
    }
    putchar('\n');

    return 0;
}

来源：https://stackoverflow.com/questions/29432565/how-can-i-make-an-array-of-intengers-where-each-array-position-is-a-digit-of-my

标签

arrays

integer