If I set an array variable a[]="abc" , and then set another array variable b[]={'d','e','f'} ,my last output code is printf("%s",b) ,it's output value is "defabc",why? My output is array b but the output value will output array b first and then output array a second. The whole code is on bellow.
#include<stdio.h>
void main(){
char a[]="abc";
char b[]={'d','e','f'};
printf("%s",b);
}
The output is "defabc". And the string length of array b is 7 why?
In C all strings should be null (i.e. \0) terminated, so your second variable should look like the following:
char b[] = {'d', 'e', 'f', '\0'};
You might be curious why "defabc" is printed with your code. The answer is, all local variables are stored in a stack-based memory layout. So your memory layout looks like this:
|'d' | <-- b
|'e' |
|'f' |
|'a' | <-- a
|'b' |
|'c' |
|'\0'|
Also note that printf("%s", ...) reads until it reach a \0, so printf("%s", a) works as expected but printf("%s", b) prints "defabc".
"abc" is stored in memory as 4 bytes 'a' 'b' 'c' '\0'.You need a null terminator at the end of both strings. your second string does not have it as its defined as an array of characters.
Correct ways to declare a string
char b[] = { 'd', 'e', 'f', '\0' }; // null terminator required
or
char b[] = "def"; // null terminator added automatically
So, this code will print def as output
#include <stdio.h>
int main() {
char a[] = "abc";
char b[] = { 'd', 'e', 'f', '\0' };
printf("%s", b);
return 0;
}
char b[]={'d','e','f', '\0'};