#include<stdio.h>
#include<string.h>
void terminateString(char *str){
str[3] = 0;
printf("string after termination is:%s\n",str);
}
int main(){
char str[]="abababcdfef";
terminateString(str);
return 0;
}
Output:
string after termination is:aba
We are only assigning element at index '3' to 0, but why are all characters after that index are ignored? Can someone please explain this behavior?
We are only assigning element at index '3' to 0, but why do all characters after that index are ignored? Can someone please explain this behavior?
The convention with a zero-terminated string is that the 0 byte is what indicates the end of the string. So when printf() encounters the zero-byte at position 3, it stops printing.
The ISO C standard defines a string as follows (see, for example, C11 7.1.1 Definition of terms), emphasis is mine:
A string is a contiguous sequence of characters terminated by and including the first null character.
Hence, when you have the character sequence abababcdfef\0, that is indeed a string.
However, when you put a null at offset 3, the string is not aba\0abcdfef\0 but, by virtue of the fact it's only a string up to and including the first null, it is aba\0.
C-string is Null-terminated string. With null-terminated it means "a null character terminates (indicates the end of) the string".
A null character is a character with all its bits set to 0, or \0, presented in memory as 0x00.
When you set str[3] = 0 you're changing str[3] to the terminator token, so when printf reads the terminator, it thinks the string is end and only prints "aba".
What you are demonstrating is the difference in c++ between strings and char arrays. Strings are a sequence of characters that continue up to and including the first null character. A character array is a memory allocation unit. A string might not use the entire character array allocated for it (indeed it is possible that it may even exceed the bounds of the containing array). If you want to diagnostically print an array rather than a string, you would need to iterate over the array in a loop. See below:
#include<stdio.h>
#include<string.h>
void terminateString(char *str){
str[3] = 0;
printf("string after termination is:%s\n",str);
}
int main(){
char str[]="abababcdfef";
terminateString(str);
for (int i = 0; i < sizeof(str)/sizeof(str[0]); i++) {
(str[i] != 0) ? printf("%c ", str[i]) : printf("\\0 ");
}
printf("\n");
return 0;
}
// OUTPUT
// string after termination is:aba
// a b a \0 a b c d f e f \0
c/c++ doesn't really distinguish between 0, '\0', and NULL, they're all just 0 in memory. c style strings are a sequence of characters that end with '\0', so every function that works with them ends after it finds this char. When you assign str[3]=0; it's the same as str[3]='\0'; i.e. stop the string after 3 chars. If you want the letter 0, do str[3]='0';, where the single quotes let the compiler know you want the character 0, ascii 48
Edit:
Note that NULL is a macro that evaluates to 0, not the same as
apparently starting with C++11 nullptrNULL can evaluate to nullptr, in C or C++98, it is 0
http://www.cplusplus.com/reference/cstring/NULL/
