如何使用 C 检查字节数组中的字节是否包含 0？

Question

I have a byte array of strings . How can I check if a byte in that array contains 0? For example:

char byte_arr[5] = "abc00";

The last two bytes of the above array contains 0. How can I detect that in c ?

Answer 1

Simple algorithm to follow

count_zero = 0
for each element a, in byte_array
    check if a is '0'
        count_zero++  

Answer 2

Even if we make this code compile (assuming that uint8 is defined somewhere)

uint8 byte_arr[5] = "abc00";

then no bytes in this array (except the 5th one) has value zero. Last two hold value of the character '0' which depends of the character encoding used.

To check if a particular element of the array holds some value you need to compare it with that value

if(byte_arr[4] == '0') printf("Success!!!!\n");
else printf("Failure!!!!\n");

Answer 3

For starters your character array does not contain a string.

char byte_arr[5] = "abc00";

So to determine whether the character '0' is present in the array you have to traverse it elements using its size.

In this case you can use a loop as for example while loop

size_t i = 0;
char c = '0';

while ( i < 5 && byte_arr[i] != c ) i++;

if ( i != 5 ) printf( "The character %c is present at position %zu\n, c, i );

If you need to determine positions of all elements that contain the character '0' or count them then you can use for example a for loop.

#include <stdio.h>

int main(void) 
{
    enum { N = 5 };
    char byte_arr[N] = "abc00";
    char c = '0';
    
    size_t count = 0;
    
    for ( size_t i = 0; i < N; i++ )
    {
        if ( byte_arr[i] == c )
        {
            ++count;
            printf( "At the position %zu there is the character '%c'\n", i, c );
        }
    }
    
    printf( "There are %zu elements in the array that contains '%c'.\n", count, c );
    
    return 0;
}

The program output is

At the position 3 there is the character '0'
At the position 4 there is the character '0'
There are 2 elements in the array that contains '0'.

If you would declare the array for example like

char byte_arr[] = "abc00";

then the array will conmtain a string. In this case you can use the standard string function strchr that checks whether a given character is present in a string. For example

#include <string.h>

//...

char byte_arr[] = "abc00";
char c = '0';

char *p = strchr( byte_arr, c );

if ( p != NULL ) printf( "The character %c is present at position %zu\n, c, ( size_t )( p - byte_arr ) );
 

Answer 4

There's a standard function for this, in string.h , if the task is just to determine whether or not there are any:

if ( memchr(byte_arr, '0', sizeof byte_arr) )
    printf("There was a zero.\n");

Answer 5

You can loop the array until you reach the end or find the desired char:

int main(void) {
    char byte_arr[5] = "abc00";
    for (int i =0; i<5; ++i)
    {
        if(byte_arr[i]=='0'){
            printf("Found!!");
            break;
        }
    }
    return 0;
}