Say if I have :
unsigned char* str = "k0kg"
And 0 is the null element. When I loop through it using a for loop, how do I check if the array has a null?
I tried:
if (str[1]==0):
I also tried:
if (str[1]=="0"):
And they didn't work. :(
The loop:
for (i=0;i<num_bytes;i++){
if (str[i]!=0){
printf("null spotted\n");
}
In C, strings, by definition, are terminated by '\0', the NUL character. So all (valid) strings have a '\0' in them, at the end.
To find the position of the '\0', simply use strlen():
const char * const end = str + strlen(str);
It's odd that you are using "unsigned char" if you are dealing with normal, printable strings. If you mean that you have a memory block with bytes it in and you want to find the first 0x00 byte, then you'll need a pointer to the start of the memory and the size of the memory area, in bytes. Then, you'd use memchr():
// Where strSize is the number of bytes that str points to.
const unsigned char * const end = memchr(str, 0, strSize);
If you are actually looking for the null element then you should do the following condition :
if(str[i]=='\0')
Say if I have : unsigned char* str = "k0kg"
And 0 is the null element. When I loop through it using a for loop, how do I check if the array has a null?
You're terminology is going to confuse any C programmer. You're confusing character representations with values. You're not looking for a null character ("null element", which would be '\0'), you're looking for the character '0'. So...
int len = strlen(str);
for(int i = 0; i < len; ++i) {
if(str[i] == '0')
printf("found it");
}
use strchr
#include <stdio.h>
#include <string.h>
int main(){
char *str = "k0kg";
char *p = strchr(str, '0');
if(p){
printf("find at %d\n", (int)(p - str));//find at 1
}
if(p=strchr(str, 0)){
printf("find at %d\n", (int)(p - str));//find at 4
}
str = "k\0kg";
if(p=strchr(str, 0)){
printf("find at %d\n", (int)(p - str));//find at 1
}
return 0;
}
"0"is character zero, aka ascii char 48 (0x30 hex).0is value zero, aka null/false