CODEWITHSUNDEEP
carrays
VenatorScientiae
VenatorScientiae

Reputation: 13

C - How to make sure that an array contains 8 integers with the value 0 or 1?

That's my code so far: 

int main(void) {
    int byte[8], position = 0, match8 = 0;
    printf("Please enter 1 byte as a binary sequence: ");
    match8 = scanf("%1d%1d%1d%1d%1d%1d%1d%1d", &byte[0], &byte[1], &byte[2], &byte[3], &byte[4], &byte[5], &byte[6],&byte[7]);
    return 0;
}

Now I want to make sure that the user made a correct input and if not the program should say "Error! Please enter 1 byte as a binary sequence: " and check the input again.

I tried:

while(getchar() != '\n' || match8 != 8) {
    ...do something
}

But I can't figure out how to check if the array contains only 0 and 1.

I hope you can help me with my problem :)

Upvotes: 0

Views: 86

Answers (4)

chux
chux

Reputation: 153348

Suggest to avoid scanf() and use fgets() to read a line of user input

char buf[100];
if (fgets(buf, sizeof buf, stdin)) {

Then use strspn() to test buf[] for only '0' or '1'

size_t strspn(const char *s1, const char *s2);
The strspn function computes the length of the maximum initial segment of the string pointed to by s1 which consists entirely of characters from the string pointed to by s2. C11 7.23.5.6

  size_t offset = strspn(buf, "01");
  if (buf[offset] == '\n') {
    buf[offset--] = '\0'; // eat trailing \n
  }
  if (offset == 8 && buf[offset] == '\0') {
    sscanf(buf, "%1d%1d%1d%1d%1d%1d%1d%1d", 
        &byte[0], &byte[1], &byte[2], &byte[3], &byte[4], &byte[5], &byte[6],&byte[7]);
    // Success!
  }
}

Upvotes: 0

BLUEPIXY
BLUEPIXY

Reputation: 40145

try this

char bits[9], ch;
int byte[8];

while(1){
    printf("Please enter 1 byte as a binary sequence: ");fflush(stdout);
    bits[8] = 1, ch = 0;
    if(2 == scanf(" %8[01]%c", bits, &ch) && ch == '\n' && bits[8] == 0)
        break;
    if(ch != '\n')
        while(getchar() != '\n');//clear input
    printf("Error! ");
}
sscanf(bits, "%1d%1d%1d%1d%1d%1d%1d%1d", &byte[0], &byte[1], &byte[2], &byte[3], &byte[4], &byte[5], &byte[6],&byte[7]);

Upvotes: 3

xing
xing

Reputation: 2498

A scanset %8[01] could be used to restrict input to the characters 0 and 1. The 8 will also limit input to 8 characters. The characters can be converted to numbers with number = onezero[i] - '0';

#include <stdio.h>
#include <string.h>

int main( void) {
    char onezero[9] = "";
    int valid = 0;

    do {
        printf("Please enter 1 byte as a binary sequence: ");
        if ( ( 1 == ( valid = scanf ( " %8[01]", onezero))) && ( 8 == strlen ( onezero))) {
            printf ( "input was %s\n", onezero);
        }
        else {
            if ( valid == EOF) {
                break;
            }
            while ( getchar ( ) != '\n') {}
            valid = 0;
        }
    } while ( !valid);

    return 0;
}

Upvotes: 1

Roy Avidan
Roy Avidan

Reputation: 768

try this...

int main(void) {

int byte[8]={0};

    printf("Please enter 1 byte as a binary sequence: ");
    for(int i=0; i<8; i++) 
{
byte[i]=getchar()-'0';
if(byte[i]!=0&&byte[i]!=1) /*error*/;
}
return 0;
}

Upvotes: -1

Related Questions