How do I check if a string only contains one and zero? - c

Question

I am trying to detect whether a string contains only the characters '0' and '1'. This is what I have so far:
while (indexCheck < 32) {
if ((input[indexCheck] != '0') && (input[indexCheck] != '1')) {
printf("not binary ");
indexCheck++;
} else if ((input[indexCheck] = '0') && (input[indexCheck] = '1')) {
indexCheck++;
printf("is binary ");
}
}
I know why it returns "is binary" or "not binary" for every single character in the array, but I don't know how to fix this. I want it to return "is binary" once if the string is only made of '1' and '0', and the opposite if this is false. I'm new to C so all help is appreciated.

Answer 1

Instead of looping manually through the string, you can see if it only contains certain characters by checking to see if strspn() returns the length of the string (By seeing if the index of the value it returns is the 0 terminator at the end of the string):
_Bool is_binary(const char *s) {
if (!s || !*s) {
return 0;
}
return s[strspn(s, "01")] == '\0';
}

Answer 2

I would make a function for this:
int isBinary(const char *input)
{
for (int i = 0; input[i]; ++i)
if (input[i] != '0' && input[i] != '1')
return 0;
return 1;
}
Then you can call the function:
if (isBinary("0001110110101"))
printf("is binary\n");
else
printf("is not binary\n");
https://ideone.com/tKBCbf

Answer 3

You can stop looping through the string the moment you find a character which is neither '0' nor '1'. After the loop is terminated, you check whether or not you've reached the end of the string, i.e. the current character is a null character '\0'
while (*s == '0' || *s == '1') ++s;
if (*s)
puts("not binary");
else
puts("binary");

Answer 4

You can do:
while (indexCheck < 32)
{
if ((input[indexCheck] != '0') && (input[indexCheck] != '1'))
{
break;
}
else
{
indexCheck++;
}
}
if (indexCheck == 32)
printf("is binary ");
else
printf("is not binary ");
Only when it has processed all elements and did not encounter a non 1-or-0 ends the loop with indexCheck == 32 so you can use that to determine what to print.
Note also that your else condition is not needed.

Answer 5

there is a block of code for you with comments.
#include <stdio.h>
#include <stdlib.h>
#define STRING_SIZE 32 // Better to use #define for reusability
// Function prototype
int isBinary(char * testInput);
// Main program
int main(void)
{
// Test inputs
char testInputBinary[33] = "01010101010101010101010101010101";
char testInputNotBinary[33] = "010101010101010101010101010101ab";
// Test & collect results
if (isBinary(testInputBinary))
{
printf("Binary ! \n");
}
else
{
printf("Not binary ! \n");
}
if (isBinary(testInputNotBinary))
{
printf("Binary ! \n");
}
else
{
printf("Not binary ! \n");
}
return EXIT_SUCCESS;
}
int isBinary(char * testInput)
{
int loopIdx = 0; // Loop index
int returnVal = 0; // 0: False, 1: True
// iterate over string
for (loopIdx = 0; loopIdx < STRING_SIZE; loopIdx++)
{
if(testInput[loopIdx] != '0' && testInput[loopIdx] != '1')
{
break;
}
}
// If the loop is not broken, it means all characters are in binary form
if (loopIdx == STRING_SIZE)
{
returnVal = 1;
} // No need to writing else clause since returnVal = 0 at the beginning
return returnVal;
}

Answer 6

int isBinary = 1;
while (input[indexCheck] != '\0')
{
if (input[indexCheck] != '1' && input[indexCheck] != '0')
{
isBinary = 0;
break;
}
++indexCheck;
}
if (isBinary)
{
printf("binary");
}
else
{
printf("not binary");
}
Check each element in string input. If input[index] is not 0 or 1 the flag isBinary becomes 0 and breaks while. And you do not need length of string.