This C program gives a weird result:
#include <stdio.h>
#include <string.h>
int main(int argc, char *argv[])
{
char str1[5] = "abcde";
char str2[5] = " haha";
printf("%s\n", str1);
return 0;
}
when I run this code I get:
abcde haha
I only want to print the first string as can be seen from the code.
Why does it print both of them?
"abcde" is actually 6 bytes long because of the null terminating character in C strings. When you do this:
char str1[5] = "abcde";
You aren't storing the null terminating character so it is not a proper string.
When you do this:
char str1[5] = "abcde";
char str2[5] = " haha";
printf("%s\n", str1);
It just happens to be that the second string is stored right after the first, although this is not required. By calling printf on a string that isn't null terminated you have already caused undefined behavior.
Update:
As stated in the comments by clcto this can be avoided by not explicitly specifying the size of the array and letting the compiler determine it based off of the string:
char str1[] = "abcde";
or use a pointer instead if that works for your use case, although they are not the same:
const char *str1 = "abcde";
Both strings str1 and str2 are not null terminated. Therefore the statement
printf("%s\n", str1);
will invoke undefined behavior.
printf prints the characters in a string one by one until it encounters a '\0' which is not present in your string. In this case printf continues past the end of the string until it finds a null character somewhere in the memory. In your case it seems that printf past the end of string "abcde" and continues to print the characters from second string " haha" which is by chance located just after first string in the memory.
Better to change the block
char str1[5] = "abcde";
char str2[5] = " haha";
to
char str1[] = "abcde";
char str2[] = " haha";
to avoid this problem.
Technically, this behavior is not unexpected, it is undefined: your code is passing a pointer to a C string that lacks null terminator to printf, which is undefined behavior.
In your case, though, it happens that the compiler places two strings back-to-back in memory, so printf runs into null terminator after printing str2, which explains the result that you get.
If you would like to print only the first string, add space for null terminator, like this:
char str1[6] = "abcde";
Better yet, let the compiler compute the correct size for you:
char str1[] = "abcde";