For loop is running too often - c

Question

I have a for loop which should run 4 times but is running 6 times.
Could you please explain the behaviour?
This is strange because stringarr1 is not changed.
Edit: I want to remove all '!' from my first string and want to save the letters in a second string.
#include <stdio.h>
#include <math.h>
#include <string.h>
int main(){
char stringarr1[] = "a!bc";
char stringarr2[] = "";
printf("%d\n", strlen(stringarr1)); // lenght --> 4
for (size_t i = 0; i < strlen(stringarr1); i++)
{
printf("i: %d\n", i);
if (stringarr1[i] != '!') {
stringarr2[strlen(stringarr2)] = stringarr1[i];
printf("info: != '!'\n");
}
}
}

Answer 1

You are overrunning the buffer for stringarr2 (length 1), which is in this case corrupting the memory-adjacent stringarr1, causing the string length to change by overwriting its nul terminator.
Then because you are reevaluating the string length on each iteration, the loop will run for a non-deterministic number of iterations - in your case just 6, but it could be worse; the behaviour you have observed is just one of several possibilities - it is undefined.
Apart from correcting the buffer length for stringarr2, it is best practice to evaluate loop-invariants once (although in this case the string length is not invariant due to a bug). So the following:
const size_t length = strlen( stringarr1 ) ;
for( size_t i = 0; i < length; i++ )
{
...
will run for 4 iterations regardless of the buffer overrun bug because the length is not reevaluated following the corruption. Re-evaluating loop-invariants can lead to very slow code execution.

Answer 2

Your code can run any number of times. You write beyond the end of stringarr2 so you may be smashing the stack and overwriting local variables. What you meant to do is probably something like this:
#include <stdio.h>
#include <math.h>
#include <string.h>
int main(){
char stringarr1[] = "a!bc";
char stringarr2[10];
int len = strlen(stringarr1);
printf("%d\n", len); // lenght --> 4
for (size_t i = 0; i < len; i++)
{
printf("i: %d\n", i);
if (stringarr1[i] != '!') {
stringarr2[len] = stringarr1[i];
printf("info: != '!'\n");
}
}
}
Like others said, it is not really clear what you are trying to accomplish here. But in C, a declaration like char s[] = "string" only allocates enough memory to store whatever is on the right hand side of the assignment. If that is an empty string like in your case, only a single byte is allocated, to store the end of string 'null' character. You need to either explicitly specify, like I did, the number of bytes to allocate as the array size, or use dynamic memory allocation.

Answer 3

The problem is that you're writing past the end of stringarr2. This triggers undefined behaviour.
To fix this, you need to allocate sufficient memory for stringarr2.

Answer 4

First, we must allocate the string to be long enough.
char stringarr1[] = "a!bc";
//save this in a variable beforehand because strlen loops over the string every time it is called
size_t len = strlen(stringarr1);
char stringarr2[1024] = { 0 };
{ 0 } initializes all characters in the string to 0, which means the last one will always be a null terminator after we add characters. This tells C string functions where the string ends.
Now we can put stuff in there. It seems like you're trying to append, so keep a separate iterator for the 2nd string. This is more efficient than calling strlen every loop.
for(size_t i = 0, j = 0; i < len; i++){
printf("i: %d\n", i);
if (stringarr1[i] != '!') {
stringarr2[j++] = stringarr1[i];
printf("info: != '!'\n");
}
}