Random character strange defined behavior - c

Question

I'm using an array in this code because i need a string which should be always modified, that's why i'm not using a pointer, howewer everytime i run the code i get a strange behavior at the 31th iteration.
code
int i = 0;
char name[100];
srand(getpid());
while(i<100) {
name[i] += (char)'A'+(rand()%26);
printf("%d strlen\n", i+1);
printf("%s\n", name);
printf("/////\n");
i++;
}
output
/////
30 strlen
IKXVKZOLKHLTKBFFTUZCYXHYVEBZOY
/////
31 strlen
IKXVKZOLKHLTKBFFTUZCYXHYVEBZOYJ
/////
32 strlen
IKXVKZOLKHLTKBFFTUZCYXHYVEBZOYJWttime
/////
33 strlen
IKXVKZOLKHLTKBFFTUZCYXHYVEBZOYJW�time
/////
34 strlen
IKXVKZOLKHLTKBFFTUZCYXHYVEBZOYJW��ime
/////
35 strlen
IKXVKZOLKHLTKBFFTUZCYXHYVEBZOYJW���me
/////
36 strlen
IKXVKZOLKHLTKBFFTUZCYXHYVEBZOYJW����e
/////
37 strlen
IKXVKZOLKHLTKBFFTUZCYXHYVEBZOYJW�����
In other words it prints always ttime as the 31th character and then the code overwrites each character of that word and i get question mark as a result.
Going on the things get even worse look at the final output
100 strlen
IKXVKZOLKHLTKBFFTUZCYXHYVEBZOYJW�����K��ȶ������MKRLHALEV�SNNRVWNOEXUVQNJUHAEWN�W�YPMCW�N�PXHNT��0�
/////
Why does this happen?

Answer 1

Well you are printing garbage value. What the behavior will be is not known.(Undefined behavior) By that I mean, it may be the case that those garbage values (added with your random number) may be ascii values of some characters or may be those are some non-printables. You should initialize the char array (with \0's - that will serves two purpose, Providing \0 for the running string and also you can add and be sure it will be a printable) or just assign it.
name[i] = 'A'+(rand()%26);
Also put a \0 in the end of the string. Otherwise it will try to access array index out of bound until it finds \0 and it will invoke undefined behavior.
31 is not something special - it can be anything the very next time you run it.
Code:
#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
int main(void) {
size_t i = 0;
char name[100]={0}; // = "";
srand(getpid());
while(i<99) { // till 99 making sure you won't overwrite the last NUL
name[i] ='A'+(rand()%26);
printf("%d strlen\n", i+1);
printf("%s\n", name);
printf("/////\n");
i++;
}
return 0;
}
Note that we have looped till 98th index because there is NUL terminating character in the 99th index.

Answer 2

char name[100]; is not a string by default. It is just another 100 element char array.
In C a string is a always carrying (at least) one '\0' character to mark the end of the string. printf(), mostly all str*() functions and many other functions rely on this terminating '\0'.
Also what is the idea behind adding to the array elements?
name[i] += ...
Their values are not set, they are garbage. Even worth, adding to them means reading uninitialised memory 1st, which in turn provokes undefined behaviour.
So to fix your code drop the addition add the terminator by hand:
while (i < 99) {
name[i] = (char) 'A' + (rand() % 26);
name[i + 1] = '\0';
Or go for the lazy approach any initialising name to all '\0' even before starting:
char name[100] = ""; /* or ... = {0}; */
(this would allow you to stick to doing name[i] += .... Still, as all elements are 0, adding is of no use.)
In any case do not loop until the array last element (100 here), but always one less as the last element is reserved for the terminating '\0'.

Answer 3

If the char name[100] array is a local variable inside a function, its initial value is undefined. So it will contain whatever random junk was in that chunk of memory before.
Therefore when you are doing
name[i] += (char)'A'+(rand()%26);
you are actually doing
name[i] = RANDOM JUNK + (char)'A'+(rand()%26);